Alge a:
Chapte
0
Paolo Aluffi
Graduate Studies
in Mathematics
Volume 10
American Mathematical Society
Algebra:
Chapter
0
Algebra:
Chapter
0
Paolo Aluffi
Graduate Studies
in Mathematics
Volume 104
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lyyyO*
American Mathematical
Providence,
Society
Rhode Island
Editorial Board
David Cox
(Chair)
Steven G. Krantz
Rafe Mazzeo
Martin Scharlemann
2000 Mathematics Subject Classification. Primary 00-01; Secondary 12-01, 13-01, 15-01,
18-01, 20-01.
For additional information and updates
on
this book, visit
www.ams.org/bookpages/gsm-104
Library
of Congress Cataloging-in-Publication Data
Aluffi, Paolo, 1960-
/Paolo Aluffi.
p. cm.
(Graduate studies in mathematics ;
Includes index.
ISBN 978-0-8218-4781-7 (alk. paper)
Algebra: chapter 0
1.
v.
104)
Algebra—Textbooks.I. Title.
QA154.3.A527
512—dc22
2009
2009004043
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14 13 12 11 10 09
Contents
Introduction
Chapter
I.
xv
Preliminaries: Set theory and categories
1
§1.
Naive
theory
1
1.1.
Sets
1
1.2.
Inclusion of sets
3
1.3.
Operations between sets
Disjoint unions, products
Equivalence relations, partitions, quotients
4
1.4.
1.5.
set
Functions between
6
8
Exercises
§2.
5
sets
8
2.1.
Definition
2.2.
2.7.
Examples: Multisets, indexed sets
Composition of functions
Injections, surjections, bijections
Injections, surjections, bijections: Second viewpoint
Monomorphisms and epimorphisms
Basic examples
15
2.8.
Canonical decomposition
15
2.9.
Clarification
16
2.3.
2.4.
2.5.
2.6.
8
10
10
11
12
13
Exercises
17
§3.
Categories
18
3.1.
Definition
18
3.2.
Examples
20
Exercises
§4.
Morphisms
26
27
4.1.
Isomorphisms
27
4.2.
Monomorphisms and epimorphisms
29
Contents
VI
Exercises
§55.1.
5.2.
5.3.
5.4.
30
Universal properties
Initial and final objects
Universal properties
Quotients
Products
Coproducts
Exercises
§1-
Groups, first
31
33
33
35
36
5.5.
Chapter II.
31
38
encounter
Definition of group
41
41
1.5.
Groups and groupoids
Definition
Basic properties
Cancellation
Commutative groups
45
1.6.
Order
46
1.1.
1.2.
1.3.
1.4.
§2.
Examples of groups
Symmetric groups
2.2.
Dihedral groups
2.3.
Cyclic
groups and modular arithmetic
Exercises
§3.
42
43
45
48
Exercises
2.1.
41
The category Grp
49
49
52
54
56
58
Group homomorphisms
Grp: Definition
58
3.2.
3.3.
Pause for reflection
60
3.4.
Products et al.
61
3.5.
Abelian groups
3.1.
Exercises
§4.
Group homomorphisms
59
62
63
64
4.1.
Examples
64
4.2.
Homomorphisms and order
66
4.3.
Isomorphisms
66
4.4.
Homomorphisms of abelian groups
68
Exercises
§5.
5.1.
Free groups
Motivation
69
70
70
71
5.3.
Universal property
Concrete construction
5.4.
Free abelian groups
75
5.2.
72
Exercises
78
§6.
Subgroups
79
6.1.
Definition
79
vii
Contents
6.2.
Examples: Kernel and image
6.3.
Example: Subgroup generated by
Example: Subgroups of cyclic
6.5.
Monomorphisms
Exercises
6.4.
80
a
subset
groups
81
82
84
85
88
§7.
Quotient
7.1.
Normal
88
7.2.
subgroups
Quotient group
7.3.
Cosets
90
7.4.
Quotient by normal subgroups
92
7.5.
Example
kernel <^=>
94
7.6.
groups
89
normal
95
Exercises
§8.
8.1.
95
Canonical decomposition
and Lagrange's theorem
96
8.2.
Canonical decomposition
Presentations
8.3.
Subgroups of quotients
100
8.4.
HK/Hvs. K/(Hf)K)
101
8.5.
The index and
8.6.
Epimorphisms and cokernels
Lagrange's theorem
Exercises
§9.
Group
97
99
102
104
105
actions
108
9.1.
Actions
9.2.
Actions
9.3.
Transitive actions and the category G-Set
108
109
on sets
110
Exercises
113
§10. Group objects in categories
10.1. Categorical viewpoint
Exercises
115
lapter III.
119
§1.
1.1.
Rings and modules
Definition of ring
117
119
119
Definition
examples and special classes of rings
Polynomial rings
1.4.
Monoid rings
Exercises
1.2.
115
First
1.3.
121
124
126
127
2.1.
The category Ring
Ring homomorphisms
2.2.
Universal property of
2.3.
2.4.
Monomorphisms and epimorphisms
Products
133
2.5.
EndAb(G)
134
§2.
Exercises
129
129
polynomial rings
130
132
136
§3.
3.1.
3.2.
3.3.
Ideals and quotient rings
Ideals
138
138
139
Quotients
Canonical decomposition and consequences
141
Exercises
§4.
4.1.
4.2.
4.3.
143
Ideals and quotients: Remarks and examples.
ideals
Prime and maximal
Basic operations
Quotients of polynomial rings
Prime and maximal ideals
144
144
146
150
Exercises
153
§5. Modules over a ring
5.1. Definition of (left-)R-module
5.2. The category R-Mod
5.3. Submodules and quotients
5.4. Canonical decomposition and isomorphism theorems
156
156
158
160
162
Exercises
163
in i?-Mod
Products, coproducts, etc.,
Products and coproducts
6.2. Kernels and cokernels
6.3. Free modules and free algebras
6.4. Submodule generated by a subset; Noetherian modules
6.5. Finitely generated vs. finite type
164
6.1.
164
166
167
169
Exercises
172
§6.
§7.
7.1.
7.2.
7.3.
and homology
Complexes and exact sequences
174
Split exact sequences
Homology and the snake lemma
177
Complexes
Exercises
Chapter IV.
§1.
1.1.
1.2.
1.3.
1.4.
171
174
178
183
Groups, second encounter
The conjugation action
Actions of groups on sets, reminder
Center, centralizer, conjugacy classes
The Class Formula
Conjugation of subsets and subgroups
187
187
187
189
190
191
Exercises
193
§2. The Sylow theorems
2.1. Cauchy's theorem
2.2. Sylow I
194
194
196
2.3.
2.4.
Sylow II
Sylow III
197
2.5.
Applications
200
Exercises
199
202
Contents
§3.
IX
Composition
series and
205
solvability
3.1.
The Jordan-Holder theorem
205
3.2.
Composition factors; Schreier's theorem
The commutator subgroup, derived series, and solvability
207
3.3.
Exercises
§4.
4.1.
4.2.
4.3.
4.4.
213
214
The symmetric group
Cycle notation
214
Type and conjugacy classes in on
Transpositions, parity, and the alternating group
Conjugacy in An\simplicity of An and solvability of Sn
Exercises
§5.
5.1.
210
216
219
220
224
Products of groups
The direct product
226
226
5.2.
Exact sequences of groups; extension problem
228
5.3.
Internal/semidirect products
230
Exercises
§6.
6.1.
6.2.
6.3.
233
Finite abelian groups
Classification of finite abelian groups
Invariant factors and elementary divisors
Application:
Finite
234
234
237
subgroups of multiplicative
groups of fields
Exercises
240
Chapter V.
§1.
1.1.
239
Irreducibility and factorization in integral domains
Chain conditions and existence of factorizations
243
244
244
1.2.
Noetherian rings revisited
Prime and irreducible elements
1.3.
Factorization into irreducibles; domains with factorizations
248
246
Exercises
§2.
2.1.
249
UFDs, PIDs,
Euclidean domains
251
2.2.
Irreducible factors and greatest
Characterization of UFDs
2.3.
PID => UFD
254
2.4.
Euclidean domain => PID
255
common
divisor
Exercises
§3.
3.1.
3.2.
§4.
4.2.
4.3.
253
258
Intermezzo: Zorn's lemma
Set theory, reprise
Application: Existence of maximal ideals
Exercises
4.1.
251
261
261
264
265
Unique factorization in polynomial rings
Primitivity and content; Gauss's lemma
The field of fractions of an integral domain
270
UFD
273
R UFD
Exercises
=>
R[x]
267
268
276
§5.
5.1.
5.2.
5.3.
5.4.
Irreducibility of polynomials
Roots and reducibility
Adding roots; algebraically closed
Irreducibility in C[x],
Eisenstein's criterion
280
281
fields
283
285
R[z], Q[x]
288
Exercises
§6.
6.1.
6.2.
6.3.
289
Further remarks and examples
Chinese remainder theorem
Gaussian integers
Fermat's theorem
291
291
294
on sums
of squares
298
Exercises
Chapter VI.
§1.
1.1.
300
Linear algebra
305
Free modules revisited
305
R-Mod
305
306
1.3.
Linear independence and bases
Vector spaces
1.4.
Recovering B from
1.2.
308
FR(B)
309
Exercises
§2.
312
Homomorphisms of free modules,
I
314
2.1.
Matrices
2.2.
Change of
2.3.
Elementary operations and Gaussian elimination
Gaussian elimination over Euclidean domains
2.4.
314
basis
318
320
323
Exercises
§3.
3.1.
324
Homomorphisms of free modules, II
Solving systems of linear equations
327
327
3.2.
The determinant
328
3.3.
Rank and nullity
Euler characteristic and the Grothendieck group
333
3.4.
334
Exercises
§4.
338
Presentations and resolutions
340
4.1.
Torsion
340
4.2.
Finitely presented modules and free resolutions
341
4.3.
Reading
a
344
presentation
Exercises
§5.
347
Classification of finitely generated modules
over
PIDs
349
5.1.
Submodules of free modules
350
5.2.
PIDs and resolutions
353
5.3.
The classification theorem
354
Exercises
§6.
6.1.
6.2.
Linear transformations of
357
a
free module
Endomorphisms and similarity
The characteristic and minimal polynomials of
359
359
an
endomorphism
361
Contents
6.3.
XI
365
Eigenvalues, eigenvectors, eigenspaces
Exercises
§7.
368
Canonical forms
371
7.2.
Linear transformations of free modules; actions of
/c[£]-modules and the rational canonical form
7.3.
Jordan canonical form
377
7.4.
Diagonalizability
380
7.1.
polynomial rings
Exercises
Chapter VII.
§1.
1.1.
1.2.
1.3.
§2.
373
381
Fields
385
Field extensions, I
Basic definitions
385
385
Simple extensions
Finite and algebraic extensions
387
391
Exercises
2.1.
371
397
Algebraic closure, NuUstellensatz,
Algebraic closure
and
a
little algebraic geometry
400
400
2.2.
The NuUstellensatz
404
2.3.
A little
406
amne
algebraic geometry
Exercises
414
§3.
Geometric impossibilities
Constructions by straightedge and compass
3.2.
Constructible numbers and quadratic extensions
Famous impossibilities
3.3.
Exercises
417
3.1.
417
§4.
4.1.
4.2.
4.3.
Field extensions, II
Splitting fields and normal extensions
Separable polynomials
Separable extensions and embeddings in algebraic closures
Exercises
§5.
5.1.
5.2.
5.3.
§6.
425
427
428
429
433
436
438
Field extensions, III
Finite fields
440
441
and fields
Cyclotomic polynomials
Separability and simple extensions
Exercises
6.1.
422
A little Galois theory
The Galois correspondence and Galois extensions
445
449
452
454
454
6.3.
The fundamental theorem of Galois theory, I
The fundamental theorem of Galois theory, II
461
6.4.
Further remarks and examples
464
6.2.
Exercises
§7.
7.1.
Short
march through applications of Galois theory
Fundamental theorem of algebra
459
466
468
468
Constructibility of regular n-gons
symmetric functions
7.4. Solvability of polynomial equations by radicals
7.5. Galois groups of polynomials
Abelian groups as Galois groups over Q
7.6.
Exercises
7.2.
7.3.
Fundamental theorem on
Chapter VIII.
§1.
Linear algebra, reprise
Preliminaries, reprise
469
471
474
478
479
480
483
483
1.1.
Functors
483
1.2.
1.3.
Examples of functors
When are two categories 'equivalent'?
485
487
1.4.
Limits and colimits
489
1.5.
Comparing functors
492
Exercises
§2.
2.1.
2.2.
2.3.
2.4.
496
Tensor products and the Tor functors
Bilinear maps and the definition of tensor product
Adjunction with Hom and explicit computations
Exactness properties of tensor; flatness
The Tor functors.
Exercises
§3.
3.1.
3.2.
3.3.
§4.
501
504
507
509
511
Base change
Balanced maps
Bimodules; adjunction again
Restriction and extension of scalars
Exercises
4.1.
500
515
515
517
518
520
Multilinear algebra
522
522
4.3.
Multilinear, symmetric, alternating maps
Symmetric and exterior powers
Very small detour: Graded algebra
4.4.
Tensor algebras
529
4.2.
Exercises
§5.
524
527
532
Hom and duals
535
5.1.
Adjunction again
536
5.2.
Dual modules
537
5.3.
Duals of free modules
538
5.4.
Duality and
539
5.5.
Duals and
5.6.
Duality
exactness
matrices; biduality
on vector spaces
Exercises
§6.
6.1.
6.2.
6.3.
Projective and injective modules and the Ext functors
Projectives and injectives
Projective modules
Injective modules
541
542
543
545
546
547
548
Contents
xni
6.4.
The Ext functors
551
6.5.
Ext£(G,Z)
554
Exercises
Chapter
§1.
1.1.
IX.
555
559
Homological algebra
560
(Un)necessary categorical preliminaries
560
1.2.
Undesirable features of otherwise reasonable categories
Additive categories
1.3.
Abelian categories
564
1.4.
Products, coproducts, and direct
1.5.
Images; canonical decomposition of morphisms
567
sums
570
Exercises
§2.
2.1.
2.2.
2.3.
2.4.
Working
561
574
in abelian
categories
categories
The snake lemma, again
Working with 'elements' in a small abelian category
What is missing?
Exactness in abelian
Exercises
576
576
578
581
587
589
591
3.2.
and homology, again
Reminder of basic definitions; general strategy
The category of complexes
3.3.
The long exact cohomology sequence
597
3.4.
Triangles
600
§3.
3.1.
Complexes
591
594
Exercises
§4.
4.1.
4.2.
4.3.
602
Cones
and homotopies
The mapping cone of a
605
605
morphism
Quasi-isomorphisms and derived categories
Homotopy
607
611
Exercises
§5.
5.1.
614
The homotopic category. Complexes of projectives and injectives
Homotopic maps are identified in the derived category
5.3.
Definition of the homotopic category of complexes
Complexes of projective and injective objects
5.4.
vs.
5.2.
5.5.
Homotopy equivalences
Proof of Theorem 5.9
quasi-isomorphisms
in
§6.
619
K(A)
6.2.
6.3.
and injective resolutions and the derived category
Recovering A
From objects to complexes
Poor man's derived category
Projective
Exercises
§7.
7.1.
7.2.
Derived functors
Viewpoint shift
Universal property of the derived functor
616
618
Exercises
6.1.
616
620
624
626
628
629
631
635
638
641
641
643
Contents
XIV
7.3.
7.4.
7.5.
7.6.
Taking cohomology
Long exact sequence of derived functors
Rl&
Relating J^, L^,
Example: A little group
§8.
8.2.
8.3.
8.4.
8.5.
cohomology
Exactness of the total complex
Total complexes and resolutions
resolutions again and balancing Tor and Ext
Exercises
§9.
9.1.
655
658
Double complexes
Resolution by acyclic objects
Complexes of complexes
Acyclic
647
653
Exercises
8.1.
645
Further topics
Derived categories
661
662
665
670
672
675
677
680
681
9.2.
Triangulated categories
683
9.3.
Spectral sequences
686
Exercises
Index
695
699
Introduction
This text presents
or
an
introduction to algebra suitable for upper-level undergraduate
courses. While there is a very extensive offering of textbooks
beginning graduate
at this level, in my experience teaching this material I have invariably felt the
need for a self-contained text that would start 'from zero' (in the sense of not
assuming that the reader has had substantial previous exposure to the subject) but
that would impart from the very beginning a rather modern, categorically minded
viewpoint and aim at reaching a good level of depth. Many textbooks in algebra
brilliantly satisfy some, but not all, of these requirements. This book is my attempt
at providing a working alternative.
There is a widespread perception that categories should be avoided at first
blush, that the abstract language of categories should not be introduced until a
student has toiled for a few semesters through example-driven illustrations of the
nature of a subject like algebra. According to this viewpoint, categories are only
tangentially relevant to the main topics covered in a beginning course, so they can
simply be mentioned occasionally for the general edification of the reader, who will
in time learn about them
(by osmosis?). Paraphrasing
present text, 'Discussions of categories at this level
appendices.'
It will be
clear from
otherwise. In this text,
a
cursory
categories
are
glance
are
a
reviewer of
the
reason
a
draft of the
why God created
at the table of contents that I think
introduced
on page
18, after
a scant
reminder
of the basic language of naive set theory, for the main purpose of providing a context
for universal properties. These are in turn evoked constantly as basic definitions
are
introduced. The word 'universal' appears at least 100 times in the first three
chapters.
I believe that
awareness
of the categorical language, and especially
some
appreciation of universal properties, is particularly helpful in approaching a subject
such as algebra 'from the beginning'. The reader I have in mind is someone who
has reached
a
certain level of mathematical
example,
maturity—for
who needs
no
xv
Introduction
XVI
special assistance in grasping an induction argument—but
may have only been
in
to
a
manner.
is
exposed
algebra
very cursory
My experience that many upper-level
undergraduates or beginning graduate students at Florida State University and at
comparable institutions fit this description. For these students, seeing the many
introductory concepts in algebra as instances of a few powerful ideas (encapsulated in
suitable universal properties) helps to build a comforting unifying context for these
notions. The amount of categorical language needed for this catalyzing function
is very limited; for example, functors are not really necessary in this acclimatizing
stage.
Thus, in my mind the benefit of this approach is precisely that it helps a true
beginner, if it is applied with due care. This is my experience in the classroom,
and it is the main characteristic feature of this text.
The very little categorical
language introduced at the outset informs the first part of the book, introducing
in general terms groups, rings, and modules. This is followed by a (rather
traditional)
treatment of standard
topics such
as
Sylow theorems, unique factorization,
linear algebra, and field theory. The last third of the book wades into
somewhat deeper waters, dealing with tensor products and Horn (including a first
elementary
introduction to Tor and
algebra.
Ext) and including a final chapter devoted to homological
Some familiarity with categorical language appears indispensable to me
in order to appreciate this latter
material, and this is hopefully uncontroversial.
feel for this language in the earlier parts of the book, students
find the transition into these more advanced topics particularly smooth.
Having developed
a
A first version of this book
a run
of the
was
essentially
(three-semester) algebra
a
careful transcript of my lectures in
sequence at FSU. The
chapter
on
homological
added at the instigation of Ed Dunne, as were a very substantial number
of the exercises. The main body of the text has remained very close to the original
'transcript' version: I have resisted the temptation of expanding the material when
algebra was
revising it for publication. I believe that an effective introductory textbook (this
is Chapter 0, after all...) should be realistic: it must be possible to cover in class
Otherwise, the book veers into the 'reference' category;
reference book in algebra, and it would be futile (of me)
what is covered in the book.
I
never meant to
write
a
to try to ameliorate excellent available references such as
The problem
Lang's 'Algebra'.
or any motivated reader,
give
opportunity
to get quite a bit beyond what is covered in the main text. To guide in the choice of
exercises, I have marked with a > those problems that are directly referenced from
sets will
the text, and with
to a
an
teacher,
a
those problems that are referenced from other problems. A
minimalist teacher may simply assign all and only the > problems; these do nothing
more than anchor the understanding by practice and may be all that a student
can
-¦
realistically be expected to work out
intensity as this
other courses of similar
while
juggling TA duties and
two or three
The main body of the text, together
self-contained presentation of essential material. The
one.
with these exercises, forms a
other exercises, and especially the threads traced by those marked with ->, will offer
the opportunity to cover other topics, which some may well consider just as
essential: the modular group, quaternions, nilpotent groups, Artinian rings, the Jacobson radical, localization, Lagrange's theorem on four squares, projective space and
Introduction
xvn
Grassmannians, Nakayama's lemma, associated primes, the spectral theorem for
normal operators, etc., are some examples of topics that make their appearance in
the exercises. Often a topic is presented over the course of several exercises, placed
in appropriate sections of the book. For example, 'Wedderburn's little theorem'
is mentioned in Remark III. 1.16 (that is: Remark 1.16 in Chapter III); particular
presented in Exercises III.2.11 and IV.2.17, and the reader eventually
proof in Exercise VII.5.14, following preliminaries given in Exercises VII.5.12
cases are
obtains
a
label and perusal of the index should facilitate the navigation
of such topics. To help further in this process, I have decorated every exercise with
a list (added in square brackets) of the places in the book that refer to it.
For
and VII.5.13. The
example,
in
instructor
evaluating whether
to
assign Exercise V.2.25 will be
that this exercise is quoted in Exercise VII.5.18, proving a particular
of Dirchlet's theorem on primes in arithmetic progressions, and that this will
immediately
case
an
-¦
aware
turn be
groups
quoted
Q.
in
§VII.7.6, discussing
the realization of abelian groups
as
Galois
over
I have put a high priority on the requirement that this should be a self-contained
text which essentially crosses all t's and dots all i's, and does not require that the
reader have access to other texts while working through it. I have therefore made
a conscious effort to not quote other references: I have avoided as much as possible
the exquisitely tempting escape route 'For a proof, see
why this book is as thick as it is, even if so many topics
these, commutative algebra and representation theory
me to
This is the main
reason
covered in it.
Among
are perhaps the most glaring
the standard basic definitions,
omissions. The first is
which allow
'
are not
represented to the extent of
sprinkle a little algebraic geometry here and there (for example,
§VII.2), and of a few slightly more advanced topics in the exercises, but I stopped
short of covering, e.g., primary decompositions. The second is missing altogether.
It is my hope to complement this book with a 'Chapter 1' in an undetermined
future, where I will make amends for these and other shortcomings.
see
on
By its nature, this book should be quite suitable for self-study: readers working
their own will find here a self-contained starting point which should work well
prelude to future, more intensive, explorations. Such readers may be helped
the
by
following '9-fold way' diagram of logical interdependence of the chapters:
as a
IV
ii
IX
VII
in
VIII
VI
Introduction
xvm
This may however better reflect my original intention than the final product. For
a
diagram captures the web of references from
objective
a chapter to earlier chapters, with the thickness of the lines representing (roughly)
the number of references:
gauge, this alternative
more
VI
With the
providing
an
V
self-studying reader especially
in mind, I have put extra effort into
fanfare for each and every
official 'definition'; the index should
extensive index. It is not realistic to make
introduced in a text of this size by an
lone traveler find the way back to the source of unfamiliar terminology.
new term
a
help
a
Internal references
are
Remark III. 1.16 refers to
within Chapter III, the
following
an
handled in
a
Remark 1.16 in
same
hopefully transparent way.
Chapter III; if the reference
item is called Remark 1.16.
exercise indicates other exercises
that exercise. For
example, Exercise 3.1
in
or
For
example,
is made from
The list in brackets
sections in the book
Chapter
I is followed
referring to
by [5.1, §VIII.1.1,
are references to this problem in
in
1.1
in
section
Exercise 5.1
Chapter VIII, section 1.2 in Chapter IX,
Chapter I,
IX
nowhere
in
and Exercise 1.10
Chapter
else).
(and
§IX.1.2,
IX. 1.10]: this alerts the reader that there
Acknowledgments. My debt to Lang's book, to David Dummit and Richard
Foote's 'Abstract Algebra,' or to Artin's 'Algebra' will be evident to anyone who
is familiar with these sources. The chapter on homological algebra owes much to
David Eisenbud's appendix on the topic in his 'Commutative Algebra', to Gelfandhomological algebra', and to Weibel's 'An introduction to
homological algebra'. But in most cases it would simply be impossible for me to
retrace the original source of an expository idea, of a proof, of an exercise, or of
these are all likely offsprings of ideas from any
a specific pedagogical emphasis:
Manin's 'Methods of
of these and other influential references and often of associations triggered by
following the manifold strands of the World Wide Web. This is another reason
one
why,
in
a
spirit of equanimity, I resolved
to
essentially avoid references altogether.
only
In any case, I believe all the material I have presented here is standard, and I
retain absolute ownership of every error left in the end product.
Introduction
I
xix
grateful to my students for the constant feedback that led me to
particular way and who contributed essentially to its success
in my classes. Some of the students provided me with extensive lists of typos and
outright mistakes, and I would especially like to thank Kevin Meek, Jay Stryker,
am very
write this book in this
and Yong Jae Cha for their particularly helpful comments. I had the opportunity
to try out the material on homological algebra in a course given at Caltech in the
fall of 2008, while on a sabbatical from FSU, and I would like to thank Caltech
course for their hospitality and the friendly atmosphere.
also due to MSRI for hospitality during the winter of 2009, when the
last fine-tuning of the text was performed.
and the audience of the
Thanks
are
A few people spotted big and small mistakes in preliminary versions of this
book, and I will mention Georges Elencwajg, Xia Liao, and Mirroslav Yotov for
particularly precious contributions. I also commend Arlene O'Sean and the staff at
the AMS for the excellent copyediting and production work.
Special thanks
go to Ettore Aldrovandi for expert
for her encouragement and
that had
a
indispensable help, and
advice,
to Matilde Marcolli
to Ed Dunne for
suggestions
great impact in shaping the final version of this book.
Support from the Max-Planck-Institut in Bonn, from the NSA, and from
Caltech, at different stages of the preparation of this book, is gratefully acknowledged.
Chapter
Preliminaries: Set
and
theory
categories
Set theory is
a
mathematical field in itself, and its proper treatment
famous 'Zermelo-Frankel'
axioms)
competence of this writer. We will
is little
more
than
a
(say
via the
beyond the scope of this book and the
only deal with so-called 'naive' set theory, which
goes well
system of notation and terminology enabling
us
to
express
precisely mathematical definitions, statements, and their proofs.
Familiarity with this language is essential in approaching a subject such
algebra, and indeed the reader is assumed to have been previously exposed to
In this
I
chapter
we
first review
some
order to establish the notation
we
as
it.
of the language of naive set theory, mainly in
will
use
in the rest of the book.
We will then
get a small taste of the language of categories, which plays a powerful unifying role
in algebra and many other fields. Our main objective is to convey the notion of
'universal
1.
Naive set
1.1.
A
A
property', which will be
a constant
refrain throughout this book.
theory
Sets. The notion of set formalizes the intuitive idea of 'collection of
set is determined
=
B)
by the elements
it contains: two sets
A, B
are
objects'.
equal (written
if and only if they contain precisely the same elements. 'What is an
a forbidden question in naive set theory: the buck must stop somewhere.
conveniently pretend that a 'universe' of elements is available to us, and
element?' is
We
can
draw from this universe to construct the elements and sets
we need, implicitly
will
that
all
the
we
can
be
operations
explore
performed within this
assuming
universe. (This is the tricky point!) In any case, we specify a set by giving a precise
recipe determining which elements are in it. This definition is usually put between
braces and may consist of a simple, complete, list of elements:
we
A
:=
{1,2,3}
1
2
J. Preliminaries: Set
is1 the
set
theory and categories
consisting of the integers 1,2, and 3. By convention, the order2 in which
are listed, or repetitions in the list, are immaterial to the definition.
the elements
Thus, the
same set may
{1,2,3}
be written out in many ways:
=
{1,3,2}
=
{1,2,1,3,3,2,3,1,1,2,1,3}.
This way of denoting sets may be quite cumbersome and in any
work for finite
a
sets. For infinite sets, a
list in which
example, the
of the elements
some
set of even
but such
a
=
are
will only really
problem is to write
being part of a pattern—for
understood
as
{-.., -2,0,2,4,6,---},
definition is inherently ambiguous, so this leaves
some sets are simply 'too big' to be listed,
misinterpretation. Further,
example (as
one
hopefully
case
way around this
may be written
integers
E
popular
learns in advanced
calculus)
for
principle: for
simply too many
integers.
there
real numbers to be able to 'list' them as one may 'list' the
room
even
in
are
to adopt definitions that express the elements of a set as
larger (and already known) set 5, satisfying some property P.
It is often better
elements s of some
One may then write
A
(G
means
element
=
{s
e S
|
s
satisfies
P}
and this is in general precise and
of...)
unambiguous3.
We will occasionally encounter a variation on the notion of set, called 'multiset'.
A multiset is a set in which the elements are allowed to appear 'with multiplicity':
that is, a notion for which {2,2} would be distinct from {2}. The correct way to
define a multiset is by means of functions, which we will encounter soon (see
Example 2.2).
A few famous sets
are
0: the empty set, containing
N: the
no
set of natural numbers
elements;
(that is, nonnegative integers);
Z: the set of integers;
Q: the set of rational numbers;
R: the set of real numbers;
C: the set of complex numbers.
Also, the
term
element. Thus
Here
3
:=
{1}, {2},
are a
means
is used to refer to any set consisting of precisely
{3} are different sets, but they are all singletons.
singleton
few useful symbols
there exists...
(the
one
(called quantifiers):
existential
quantifier);
is a notation often used to mean that the symbol on the left-hand side is defined by
whatever is on the right-hand side.
Logically, this is just expressing the equality of the two
sides and could just as well be written '='; the extra : is a psychologically convenient decoration
inherited from computer science.
Ordered lists are denoted with round parentheses:
(1,2,3)
is not the same as
But note that there exist pathologies such as Russell's paradox, showing that
of definitions can lead to nonsense. All is well so long as S is indeed known to be
with.
(1,3,2).
even
this style
a set to
begin
1. Naive set
theory
V
for
Also,
means
3
all...
3! is used to
(the
there exists
mean
For example, the
set of even
E
in
universal
words, "all integers
a
unique...
a
may be written as
integers
{a
=
quantifier).
G Z
| (3n
Z)
G
such that there exists
a
=
an
2n}
:
integer
n
for which
a
2n". In
=
could replace 3 by 3! without changing the set—but
that has to do
with properties of Z, not with mathematical syntax. Also, it is common to adopt
this
case we
the shorthand
E
in which the existential
quantifier
{2n\neZ},
=
is understood.
Being able to parse such strings of symbols effortlessly, and being able to write
fluently, is extremely important. The reader of this book is assumed to
have already acquired this skill.
them out
Note that the order
in which
things
written may make
are
a
big difference.
For
example, the statement
(Va
G
Z) (3b
is true: it says that the result of
b
Z)
G
doubling
an
=
2a
arbitrary integer yields
integer;
an
but
(3b
is
false:
much
as
G
it says that there exists
is
every integer—there
Z) (Va
b
Z)
G
=
2a
fixed integer b which is 'simultaneously' twice
no such thing.
a
as
Note also that writing simply
b
=
2a
by itself does not convey enough information, unless the context makes it completely
clear what quantifiers are attached to a and b: indeed, as we have just seen, different
quantifiers may make this into
a true or a
false statement.
1.2. Inclusion of sets. As mentioned above, two sets are equal if and only if
they contain the same elements. We say that a set S is a subset of a set T if every
element of S is
an
element of T, in symbols,
scr.
By convention, S C T
exclude the possibility
consistently use
contained in T:
We
can
C
means
the
same
that S and
in this book.
thing: that
is
(unlike
< vs.
that is, S C T and S
^
T.
think of 'inclusion of sets' in terms of logic: S CT
s g
(the quantifier Vs
is
understood);
of T'; that is, all elements of S
Note that for all
S =>
s
means
that is, 'if s is an element of 5, then
elements of T; that is, S C T as
S and S C S.
=
T.
that
eT
are
5, 0 C
5, then S
sets
If S C T and T C
it does not
<),
T may be equal. To avoid any confusion, we will
One adopts S C T to mean that S is 'properly'
s
is
an
element
promised.
4
J. Preliminaries: Set
theory and categories
The symbol |5| denotes the number of elements of 5, if this number
oo. If S and T are finite, then
otherwise, one writes |5|
is finite;
=
SCT
The subsets of
of S.
a set
S form
a
=>
<
|5|
\T\.
set, called the power set,
or
the set
{0}.
|^(5)|
=
2'5I if S is finite (cf. Exercise 2.11).
Operations between sets. Once we have a few sets to play with,
more by applying certain standard operations. Here are a few:
1.3.
of parts
For example, the power set of the empty set 0 consists of one element:
The power set of S is denoted ^{S)\ a popular alternative is 25, and indeed
we can
obtain
U: the union;
fl: the intersection;
\:
the
difference;
II: the
disjoint union;
x:
the
(Cartesian) product;
and the important notion of 'quotient by
Most of these operations should be familiar
{1,2,4}
U
{3,4,5}
an
equivalence relation'.
to the reader: for
=
example,
{1,2,3,4,5}
while
{1,2,4}
In terms of Venn
(the
{3,4,5}
diagrams of infamous
=
{1,2}.
'new math' memory:
solid black contour indicates the set included in the
Several of these operations may be written out in
a
operation).
transparent way in terms
of logic: thus, for example,
s e
Two sets S and T
are
S fl T <^>
disjoint if S
(s
e S and s G
fl T
=
0,
T).
that is, if
no
element is
'simultaneously' in both of them.
The complement of a subset T in a
of all elements of S which are not in T.
set
S is the difference set S
T consisting
Thus, for example, the complement of the
set of even integers in Z is the set of odd integers.
The operations II, x, and quotients by equivalence relations are slightly more
mysterious, and it is very instructive to contemplate them carefully. We will look
1. Naive set
at them in
when
we
5
theory
a
naive way first and
particularly
have acquired
more
language and
come
back to them in
view them from
can
a more
a
short while
sophisticated
viewpoint.
Disjoint unions, products. One problem with these operations is that their
output may not be defined as a set, but rather as a set up to isomorphisms of sets,
that is, up to bijections. To make sense out of this, we have to talk about functions,
1.4.
and
we
will do that in
a moment.
Roughly speaking, the disjoint union of two sets S and T is a set SU.T obtained
first
by
producing 'copies' S' and T' of the sets S and T, with the property that
S' fl T'
0, and then taking the (ordinary) union of S' and T'. The careful reader
=
will feel uneasy, since this 'recipe' does not
produce
a
'copy' of
a
set, surely there
define
whatever it
one set:
are many ways to
do
so.
means to
This ambiguity will
be clarified below.
Nevertheless, note that we can say something about SII T even on these very
shaky grounds: for example, if S consists of 3 elements and T consists of 4 elements,
the reader should expect
a
that SII T consists of 7 elements.
(correctly)
Products are marred by the same kind of ambiguity, but fortunately there is
convenient convention that allows us to write down 'one' set representing the
product of
elements
the ordered
S
Thus,
if S
given S and T,
two sets S and T:
are
=
{1,2,3}
x
let S
x
pairs4 (s,t)
T
{(s, t)
such that
{3,4},
then
:=
and T
S xT
we
of elements of S and T:
=
=
s G
5,
t G
T be the set whose
T}.
{(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}.
sophisticated example, RxRis the set of pairs of real numbers, which
(as
calculus) is a good way to represent a plane. The set Z x Z could
be represented by considering the points in this plane that happen to have integer
coordinates. Incidentally, it is common to denote these sets R2, Z2; and similarly,
the product A x A of a set by itself is often denoted A2.
For
a more
we
learn in
If S and T
are
finite sets,
clearly \SxT\ \S\
\T\.
=
we can use products to obtain explicit 'copies' of sets as needed
disjoint union: for example, we could let S"
{0} x 5, V
{1} x T,
guaranteeing that S" and V are disjoint (why?); and there is an evident way to
'identify' S and S", T and V. Again, making this precise requires a little more
vocabulary.
The operations U, fl, II, x extend to operations on whole 'families' of sets: for
Also note that
for the
=
example, if S\,...,Sn
are
sets,
we
=
write
n
fl Si
Si
=
n
S2
n
n
Sn
2=1
One
can
define the ordered pair
information of the elements s, t,
of the pair).
by setting (s, t)
{s, {s, t}}: this carries the
conveying the fact that s is special (= the first element
(s, t)
as well as
as
a set
=
J. Preliminaries: Set
6
theory and categories
for the set whose elements are those elements which are simultaneously elements of
all sets Si,..., Sn; and similarly for the other operations. But note that while it is
clear from the definitions that, for example,
Si
it is not
U
S2
clear in what
so
x
Si
should be 'identified'
S2
U
S3
sense
x
(Si
=
U
S3
=
Si
U
(S2
U
S3),
the sets
S3,
(where
S2)
U
(Si
S2)
x
we can
x
Si
S3,
x
(S2
define the leftmost set
x
S3)
as
the set of 'ordered
triples' of elements of Si, S2, S3, by analogy with the definition for two sets). In
fact, again, we can really make sense of such statements only after we acquire the
language of functions. However, all such
probably expects; by virtue of
the reader
somewhat cavalier and gloss
More generally, if 5? is
over
a set
|Js,
sey
statements do turn out to be true, as
this fortunate circumstance,
such subtleties.
of sets,
we may
f]s,
]Js,
sey
sey
union, intersection, disjoint union, product of all sets in y. There
important subtleties concerning these definitions: for example, if all S G ^
for the
nonempty, does it follow that
so, but
choice.
(if
y is
infinite)
&
be
consider sets
]Js,
sey
we can
are
are
The reader probably thinks
rise^
this is a rather thorny issue, amounting to the axiom of
1S
nonempty?
By and large, such subtleties do not affect the material in this course; we will
come to terms with them in due time5, when they become more relevant to
partly
the issues at hand
(cf. §V.3).
Equivalence relations, partitions, quotients. Intuitively, a relation on
a set S is some special affinity among selections of elements of S. For
example, the relation < on the set Z is a way to compare the size of two integers:
since 2 < 5, 2 'is related to' 5 in this sense, while 5 is not related to 2 in the same
1.5.
elements of
sense.
For all practical purposes, what a relation 'means' is completely captured by
which elements are related to which elements in the set. We would really know all
complete list of all pairs (a, b) of integers
a pair, while (5,2) is not.
(2,5)
completely straightforward definition of the notion of relation:
there is to know about <
such that
This leads to a
a
relation
say
that
a
on
Z if
b. For example,
a <
on a set
and b
we
had
a
is such
S is simply a subset R of the product S
'related by i?' and write
x
S. If
(a, b)
G
i?,
we
are
aRb.
Often
we use
fancier symbols for relations, such
The reader will have
even
and
before
so it
we come
should.
to
as
employ the axiom of choice in
<, <, =, ~,
some exercises every now and
back to these issues, but this will probably happen below the
awareness
then,
level,
1. Naive set
7
theory
The prototype of
a
well-behaved relation is '=', which corresponds to the
'diagonal'
{(a, b)
Three
if
~
G S x S
properties of this
denotes for
very
G
S)
(Va
G
5) (V6
transitivity:
That is, every
a
Definition 1.1.
(Va
is
An
a
~
=
to
{(a, a) |
=
a G
S}
C 5 x 5.
turn out to be
of equality, then
particularly important:
~
satisfies
a;
S)
G
5) (V6
equal
b}
=
special relation
reflexivity: (Va
G
a
the relation
a moment
symmetry:
|
a
6 => 6
~
S) (Vc
G
itself; if
a
e
is
~
S), (a
equal
equivalence relation
on
a;
~
b and 6
6, then 6
to
~
is
c)
=>
equal
~
c.
to a; etc.
S is any relation
a set
a
~
satisfying
these three properties.
j
In terms of the corresponding subset R of S x 5, 'reflexivity' says that the
diagonal is contained in R\ 'symmetry' says that R is unchanged if flipped about
the diagonal (that is, if every (a, b) is interchanged with (6, a)); while unfortunately
'transitivity' does not have a similarly nice pictorial translation.
The datum of an equivalence relation on S turns out to be equivalent to a type
of information which looks a little different at first, that is, a partition of S. A
partition of S is a family of disjoint nonempty subsets of 5, whose union is S: for
example,
^
is
a
partition of the
=
{{1,4,7}, {2,5,8}, {3,6}, {9}}
set
{1,2,3,4,5,6,7,8,9}.
a partition of S from
the
class
of a (w.r.t. ~) is
S,
equivalence
Here is how to get
a e
a
relation
~
on
S: for every element
the subset of S defined
[a]^:={beS\b
-
by
a};
partition g?^ of S (Exercise 1.2).
Conversely (Exercise 1.3) every partition & is the partition corresponding in
this fashion to an equivalence relation. Therefore, the notions of 'equivalence
relation on 5' and 'partition of 5' are really equivalent.
then the equivalence classes form
Now
we can
respect to
~).
view g?^
as a set
a
(whose
elements
are
the equivalence classes with
This is the quotient operation mentioned in
§1.3.
Definition 1.2. The quotient of the set S with respect to the equivalence relation
is the set
S/~
:=
^
of equivalence classes of elements of S with respect to
Example
1.3. Take S
=
Z, and let
a
Then
Z/~
consists of two
~
~
b <^=>
b is
even.
equivalence classes:
Z/~
=
~.
be the relation defined by
a
~
{[0]„,[!]„}.
j
J. Preliminaries: Set
8
Indeed,
integer b is either
every
odd
hence b
hence 6 0 is even, so b
0, and b G [0]^)
and
This
is
of
course
the starting
be
1,
[1]~).
(and
even
1 is even, so b
(and
of
modular
arithmetic, which
point
or
theory and categories
~
will
we
~
cover
in due detail later
on
(§11.2.3).
j
One way to think about this operation is that the equivalence relation 'becomes
equality in the quotient': that is, two elements of the quotient 5/~ are equal if and
only if the corresponding elements in S are related by ~. In other words, taking a
quotient is
equivalence relation into an equality. This observation
'categorical terms' in a short while (§5.3).
a way to turn any
will be further formalized in
Exercises
Exercises marked with
-<
a > are
referred to from the text; exercises marked with a
These referring exercises and sections are
referred to from other exercises.
are
listed in brackets
the current exercise;
following
see
the introduction for further
clarifications, if necessary.
1.1. Locate
1.2.
>
discussion of Russell's paradox, and understand it.
a
Prove that if
g?^ defined
in
§1.5
is
~
a set 5, then the corresponding family
of
S: that is, its elements are nonempty,
partition
relation
a
is indeed
a
on
disjoint, and their union is S. [§1.5]
1.3.
>
Given
that £? is the
partition 2? on a set 5, show how
corresponding partition. [§1.5]
a
1.4. How many different
equivalence relations
to define a relation ~onS such
may be defined on the set
{1,2,3}?
1.5. Give an example of a relation that is reflexive and symmetric but not transitive.
What happens if you attempt to use this relation to define a partition on the set?
(Hint: Thinking about the second question will help you answer the first one.)
1.6. > Define
Z.
a
relation
Prove that this is
for R/~. Do the
(oi,o2)
(bub2)
~
2.
~
an
same
on
the set R of real numbers by setting
equivalence relation, and find
for the relation
& on
b1-a1eZ and b2
^=^
-
a
the plane R
a2
Z.
a
~
b <^=> b—ae
'compelling' description
R defined by declaring
x
[§11.8.1, II.8.10]
Functions between sets
2.1. Definition. A
will follow for just about every structure
introduced in this book will be to try to understand both the type of structures
and the ways in which different instances of a given structure may interact.
common
thread
we
through functions. It is tempting to think of a
B in 'dynamic' terms, as a way to 'go from A
business with relations, it is straightforward to formalize
do not need to invoke any deep 'meaning' of any given /:
Sets interact with each other
function / from
to B\
Similarly
a set
A to
to the
this notion in ways that
everything that
can
a set
be known about
a
function / is captured by the information of
2. Functions between sets
9
which element b
of B
is nothing but
subset of A
a
is the image
x B:
rf
This set
is the
Tf
{(a, b)
:=
of
any given element a
G A x B
graph of /; officially,
a
b
f(a)}
=
of A.
This information
CAxB.
graph6.
function really 'is' its
Not all subsets T C Ax B correspond to ('are') functions: we need to put
requirement on the graphs o£ functions, which can be expressed as follows:
(Va
('in
or
functional
G
a
depending
in,
G
A) (3\b G B)
f(a)
function must send each element
study of Riemann surfaces)
that / is a function from a
following picture ('diagram'):
set
A^-^B
The action of
a
a
function /
'decorated' arrow,
as
A
:
B
on an
The collection of all functions from
graph,
b.
±\fx(which
one
functions in this
A to
a set
B,
element of B,
are very
important
sense.
one
writes
/
:
A
.
element
a G
A is sometime indicated by
/(a).
A to
a set
a set
B is itself
a
set7,
denoted
take seriously the notion that a function is really the same thing
then we can view BA as a (special) subset of the power set of A x B.
we
Every set A comes
in A x A: the
diagonal
B
in
a^
BA. If
as
are not
announce
draws the
=
of A to exactly
a
'Multivalued functions' such
on a.
e.g., the
To
or
(a,6)er/,
notation')
(Va
That is,
A) (3\b G B)
one
equipped with a
identity function
very
on
special function, whose graph
as
its
is the
A
ldA: A^ A
defined by (Va G
set A determines
A) id^(a)
a
=
a.
function S
More generally, the inclusion of any subset S of a
A, simply sending every element s of S to 'itself
in A.
If S is
a
subset of A,
we
denote by
f(S)
the subset of B defined by
f(S):={beB\(3aeA)b f(a)}.
=
/(5) is the subset of B consisting of all elements that are images of elements
of S by the function /. The largest such subset, that is, f(A), is called the image
of f, denoted 'im/'.
That is,
S
Also, f\s denotes the 'restriction' of / to the subset
B defined by
(VseS):
f\s(s) f(s).
S:
this is the function
-?
=
To be precise, it is the graph
Tf together
with the information of the source A and the
part of the data of the function.
This is another 'operation among sets', not listed in §1.3. Can you
target B of /. These
are
this set? (Cf. Exercise 2.10.)
see
why
we use
BA for
theory and categories
J. Preliminaries: Set
10
That is, f\sis
where i
:
S
the composition
(in
the
sense
A is the inclusion. Note that
explained
f(S)
=
in the next
subsection) /oi,
im(/|s).
Examples: Multisets, indexed sets. The 'multisets' mentioned briefly in
simple example of a notion easily formalized by means of functions. A
2.2.
§1.1
are a
multiset may be defined by giving
positive8 integers; if m : A
function from
a
a
(regular)
set A to the set
function, the corresponding
multiset consists of the elements a e A, each taken m(a) times. Thus, the multiset
N* for which m(a)
3,
{a,a,a,6,6,6,6,6,c} is really the function m : {a,b,c}
1. As with ordinary sets, the order in which the elements are
5, m(c)
ra(6)
N* is such
N* of
a
=
=
=
listed is not part of the information carried by
notions such as inclusion, union, etc., extend in a
For another
viewpoint
on
multisets,
by the
see
Another example is given
a
multiset.
Simple set-theoretic
straightforward
way to multisets.
Exercise 3.9.
use
of 'indices'.
If
we
write let ai,... ,an
Z..., with
be integers..., we really mean consider a function a : {1,... ,n}
the understanding that a^ is shorthand for the value a(i) (for i
=
1,... n). It is
,
tempting to think of an indexed set {a^}^/ simply as a set whose elements happen
to be denoted a^, for i ranging over some 'set of indices' /; but such an indexed set
is more properly a function I —>
A, where A is some set from which we draw the
elements a^. For example, this allows
of {a^l^N, even if by coincidence ao
us to
=
consider ao and a\ as distinct elements
a\ as elements of the
It is easy to miss such subtleties, and
some
abuse of notation is
usually harmless. These distinctions play a role in
linear independence of sets of vectors; cf. §VI. 1.2.
2.3.
g
:
B
target set A.
(for example)
common
Composition of functions. Functions may be composed: if /
C are functions, then so is the operation go f defined by
(*)
(VaeA)
that is, we use / to go from A to
draw pictures such as
A
:
A
B and
(gof)(a):=g(f(a)):
B, then apply
C
^-^ —?-£
B
and
discussions of
or
g to reach C.
A
Graphically
we may
—f—>
B
gof
9°f
\g
J,
c
Such graphical representations of collections of (for example) sets connected
are called diagrams. We will draw many diagrams, and in contexts
substantially more general than the one at hand right now.
by
functions
We say that the diagrams drawn above 'commute', or 'are commutative',
meaning that if we start from A and travel to C in either of the two possible ways
prescribed by the diagram, the result of applying the functions one encounters is the
same. This is precisely the content of the statement (*).
'Some references allow 0
as a
possible multiplicity.
2. Functions between sets
11
and h : C
Composition is associative: that is, if/:A—>£?,g:£?—>C,
o
o
o
then
h
the
functions,
diagram
(g /)
(hog) f. Graphically,
are
D
=
commutes.
This important observation should be completely evident from the
definition of composition.
The identity function is very special with respect to compositions: if / : A
is any function, then idj? o /
/ and / o id^
/. Graphically, the diagrams
=
B
=
commute.
2.4. Injections, surjections, bijections. Special kinds of functions deserve
highlighting:
A function
/
:
A
B is injective
a'
that is, if
/
A function
Injections
If
/
:
injection
/(a') ^ /(a")
=*?
one-to-one)
or
if
:
/
:
A
B is surjective
G
B) (3a
(or
G
'covers the whole of £?';
are
a
surjection
6
A)
more
often drawn ^->; surjections
f(a)
=
or
onto)
:
precisely, if im/
often drawn
are
if
=
B.
-».
is both injective and surjective,
one-to-one
/
/
an
sends different elements to different elements9.
(V6
that is, if
^ a"
(or
A ^ B,
correspondence
or an
we say it is bijective (or a bisection or a
isomorphism of sets.) In this case we often write
or
A^B,
and
that A and B
'isomorphic' sets.
A is a bijection.
Of course the identity function id^ : A
If A
£?, that is, if there is a bijection / : A
£?, then the sets A and £? may
be 'identified' through /, in the sense that we can match precisely the elements a
we say
are
=
of A with the corresponding elements
and A
£?, then B is necessarily also
=
f(a)
a
of B. For example, if A is
finite set and
a
finite set
\A\ \B\.
=
This terminology allows us to make better sense of the considerations on
in §1.3: the 'copies' A', B' of the given sets A, B should simply
'disjoint union' given
'Often
one
checks this definition in the contrapositive
(Va'
A) (Va"
A)
f{a')
=
(hence equivalent) formulation,
f{a")
=?
a'
=
a".
that is,
12
J. Preliminaries: Set
theory and categories
be isomorphic sets to A, £?, respectively. The proposal given at the end
produce such disjoint 'copies' works, because (for example) the function
/ : A
-
{0}
x
of
§1.4
to
A
defined by
(VaeA)
is
manifestly
/(a)
=
(0,a)
bijection.
a
Injections, surjections, bijections: Second viewpoint. There
2.5.
is
an
alternative and instructive way to think about these notions.
If
/
:
A
B is
a
bijection, then
we can
'flip
its
graph' and define
a
function
g:B^A:
that is, we can let a
g(b) precisely when b
f(a). (The fact that / is both
ive
and
that
the
of
flip
inject
surjective guarantees
Tf is the graph of a function
=
according
=
to the definition
This function g has
a very
§2.1. Check this!)
in
given
interesting property: graphically,
/
id a and / o g
idj?. The first identity tells us that g is
is, g o /
'left-inverse'10 of /; the second tells us that g is a 'right-inverse' of /. We simply
commute; that
a
=
say that it is the inverse
=
of f,
denoted
What about the converse? If
is true, but in fact
Proposition
function has
a
be much
2.1. Assume
A^ty,
more
and let
has
a
left-inverse if
(2) f
has
a
right-inverse if and only if it
(
o
f
=>
=
)
If
prove
/
:
B has
assume
g(f(a'))
that is, g sends
to be
if it
inverse, is it
a
bijection?
This
f
A—>
B be
:
a
function.
Then
is infective.
is
surjective.
(1).
A
[dA. Now
and only
an
have inverses'.
precise.
(1) f
Proof. Let's
g
we can
f~l. Thus, 'bijections
f(a')
=
and
a
left-inverse, then there exists a g : B
^ a" are arbitrary different elements
that a!
icU(a')
f{a")
different, showing that /
=
aV a"
=
idA(a")
=
A such that
in
A\then
g(f(a"));
to different elements. This forces
f(a')
and
f{a")
is injective.
B is injective. In order to construct a function
) Now assume / : A
A, we have to assign a unique value g(b) e A for each element b G B. For
this, choose any fixed element s G A (which we can do because A ^ 0); then set
(
g
:
<^=
B
(a
9ib):=\s
if b
=
f(a)
for
some a £
A,
if^im/.
Never mind that g is drawn to the right of / in the diagram—we
say that g is a left-inverse
of / because it is written to the left of /: g o f = id^.
2. Functions between sets
13
words, if b is the image of
In
it to your fixed element
element
an
a
of A, send it back to a; otherwise, send
s.
The given assignment defines a function, precisely because / is injective: indeed,
this guarantees that every b that is the image of some a G A by / is the image of a
unique a (two distinct elements of A cannot be simultaneously sent to b by /, since
Thus every b G B is sent to
/
is
is
required of functions.
Finally, the function
b
injective).
f(a)
=
for all
a
g
is of the first type,
G A, as needed.
The proof of
Corollary 2.2.
sided) inverse.
(2)
A
is left
:
B
so
A is
unique well-defined element of A,
left-inverse of /. Indeed, if
it is sent back to
as an
function f
a
a
:
exercise
A
a
by
g\that
is, gof(a)
a G
=
a
as
A, then
=
id a(cl)
(Exercise 2.2).
B is
a
bijection if and only if
it has a
(two-
injective but not surjective, then it will not have a right-inverse,
necessarily have more than one left-inverse (this should be clear from
the argument given in the proof of Proposition 2.1). Similarly, a surjective function
will in general have many right-inverses; they are often called sections.
If a function is
and
it will
Proposition 2.1 hints that something deep is going on here. The definition
of injective and surjective maps given in §2.4 relied crucially on working directly
with the elements of
our
detected by the way
are
sets; Proposition 2.1 shows that in fact these properties
functions
are
'organized'
among sets. Even if we did not
know what 'elements' means, still we could make sense of the notions of
injectivity
and surjectivity (and hence of isomorphisms of sets) by exclusively referring to
properties of functions.
This is a more 'mature' point of view and one that will be championed when we
talk about categories. To some extent, it should cure the reader from the discomfort
of talking about 'elements', as we did in our informal introduction to sets, without
defining what these mysterious entities
are
supposed
to be.
bijection / is f~l. This symbol
bijections, but in a slightly different context:
The standard notation for the inverse of a
also used for functions that
are not
B is any function and T C B is a subset of £?, then
/
subset of A of 'all elements that map to T'; that is,
:
A
f-1(T)
If T
the
=
{q}
consists of
fiber of /
fibers
over
over q.
a
=
f~l(T)
is
if
denotes the
{aeA\f(a)eT}.
single element of £?, f~l(T) (abbreviated f~1(q)) is called
B is a bijection if it has nonempty
a function / : A
Thus
(that is, / is surjective), and these fibers are in fact
injective). In this case, this notation f~l matches nicely
all elements of B
singletons (that is, /
is
with the notation of 'inverse' mentioned above.
2.6. Monomorphisms and epimorphisms. There is yet another way to express
inject ivity and surjectivity, which appears at first more complicated than what we
have
seen so
far but which is in fact
even more
basic.
14
J. Preliminaries: Set
A function /
A
:
B is
for all
a
monomorphism (or monic) if the following holds:
sets Z and all
foa'
functions
foa"
=
a'
=>
,a"
a
:
Z
A
a".
=
Proposition 2.3. .4 function is injective if and only if it is
Proof. (
has
a
=>
) By Proposition 2.1,
left-inverse g
another
B
:
A. Now
^
theory and categories
if
a
function / : A
that a' ,a" are
assume
a
monomorphism.
£? is injective, then it
arbitrary functions from
set Z to A and that
/oa'
compose on the left
(g
since g is
a
°
by
/)
=
g
o
(/
/oa";
associativity of composition:
g, and use
° ol
=
o
a')
=
(/
o
g
o
a")
=
(g
o
f)
o
a";
left-inverse of /, this says
id^
o
a'
=
id a
a
=
a
°
ol",
and therefore
as
needed to conclude that
/
is
a
,
monomorphism.
a monomorphism.
This says something about
/
(
)
Z
Z
sets
and
functions
we
are
A;
arbitrary
arbitrary
going to use a microscopic
Z
of
this
to
be
information, choosing
portion
any singleton {p}. Then assigning
A amounts to choosing to which elements a!
functions a',a" : Z
a'(p),
<^=
Now
assume
that
is
=
a"
a"(p) we should send the single element p of Z. For this particular choice of
Z, the property defining monomorphisms,
=
foa'
=
foa"
a'
=>
=
ol'
becomes
foa'(p)
=
foa"(p)
=?
a'
=
a";
that is
a'
Q".
f(a') f(a") =?
Z
{p} to A are equal if and only if they send
=
Now
same
two functions from
element,
so
=
other
p to the
this says
f(a')
This has
=
to be
The reader should
f(a")
=>
a'
=
a".
that is, for all choices of distinct
injective, as was to be shown.
to be true for all
words, / has
=
now
a7, a",
expect that there be
a
a;, a"
in A. In
definition in the style of the
one
given for monomorphisms and which will turn out to be equivalent to 'surjective'.
This is the case: such a notion is called epimorphism. Finding it, and proving
the equivalence with the ordinary definition of 'surjective', is left to the reader11
(Exercise 2.5).
This is a particularly important exercise, and we recommend that the reader write out all
the gory details carefully.
2. Functions between sets
2.7. Basic
15
examples. The basic operations
provided
on sets
us
with several
important examples of injective and surjective functions.
Example 2.4. Let A, B be sets. Then there
are
natural projections tta, Kb'-
AxB
defined by
7rA((a,b))
for all
(a, b)
7rj3((o,6))
G A x B. Both of these maps are
2.5.
Example
:=o,
there
Similarly,
are
:=
b
(clearly) surjective.
j
natural injections from A and B to the disjoint
union:
AUB
obtained by sending
Example
surjective)
a
G
copy A' of A
isomorphic
2.6. If
~
is
A (resp., b G B) to the corresponding element
(resp., B' of B) in AJ1B.
an
equivalence relation
on a set
A, there
is
in the
j
(clearly
a
canonical projection
A
obtained by sending every
a G
»A/~
A to its equivalence class
[a]^.
j
2.8. Canonical decomposition. The reason why we focus our attention on
injective and surjective maps is that they provide the basic 'bricks' out of which any
function may be constructed.
To see
relation
~
this,
on
we
A
as
observe that every function
follows: for all a'\a" G A,
a'
(The
~
a"
^
f{a!)
reader should check that this is indeed
Theorem
2.7. Let
decomposes
as
f
follows:
:
A
B be any
/
A
:
for
all
a G
A.
equivalence
equivalence relation.)
function,
and
first function is the canonical projection A
the third function is the inclusion imfCB, and the
defined by
f([aU):=f(a)
where the
an
f{a").
=
an
B determines
define
~
as
A/~ (as
bijection
f
above.
in
Then
f
Example 2.6),
in the middle is
J. Preliminaries: Set
16
we
theory and categories
The formula defining / shows immediately that the diagram commutes;
verify in order to prove this theorem is that
so
all
have to
that formula does define
a
that function
bijection.
The first
item is
is in fact
an
a
function;
instance of
a
class of verifications of the utmost importance.
/ has a colossal built-in ambiguity: the same element in A/~
may be the equivalence class of many elements of A; applying the formula for /
requires choosing one of these elements and applying / to it. We have to prove
The formula given for
that the result of this operation is independent from this choice: that is, that all
possible choices of representatives for that equivalence class lead to the same result.
We encode this type of situation by saying that we have to verify that / is
We will often have to check that the operations we consider are well-
well-defined.
defined, in contexts very similar to the
Proof. Spelling
a'', a" in A,
\o!\~ \a"\^
means
=
so
precisely
well-defined.
To
mean
/(«')
a",
=
f(a')
f{a")
all
/(«")•
and the definition of
=
verify that, for
have to
we
as
~
has been engineered
is indeed
required here. So /
im/
/ : A/~
is
a
bijection,
we
check
is injective and surjective.
/
If
Injective:
/([a']~)
~
~
above,
=*?
the second item, that is, that
explicitly that
a'
[«"]-
=
that a'
that this would
verify
epitomized here.
out the first item discussed
M~
Now
one
a" by definition of
=
/([a"]~),
~, and then
f(a')
f{a") by definition
Therefore
[a!]~
\a"}^.
then
/(M~) =/([«"]-)
proving injectivity.
Surjective: Given
=
of /; hence
=
=*
any b G im/, there is
an
WU
=
[a"U
element
a G
A such that
f(a)
b.
=
Then
/([*]-) =/(a)
by definition of
needed.
/. Since
b
was
arbitrary
in
=6
im/,
this shows that
/
is surjective,
as
Theorem 2.7 shows that every function is the composition of
a surjection,
isomorphism, followed by an injection. While its proof is trivial, this is
result of some importance, since it is the prototype of a situation that will occur
followed by
a
an
several times in this book. It will resurface every
as 'the first isomorphism theorem'.
now
2.9. Clarification. Finally, we can begin to clarify
unions, products, and quotients, made in §1.4. Our
and then, with
one comment
definition of
names
such
about disjoint
was the
AJ1B
disjoint sets A', B' isomorphic to A, £?, respectively.
It is easy to provide a way to effectively produce such isomorphic copies (as we did
in §1.4); but it is in fact a little too easy—many
other choices are possible, and one
(conventional)
union of two
does not look any better than any other. It is in fact
more
sensible not to make
a
Exercises
17
once and for all and simply accept the fact that all of them produce
candidates
for AIIB. From this egalitarian standpoint, the result of the
acceptable
AIIB
is
not
'well-defined' as a set in the sense specified above. However,
operation
fixed choice
it is easy to see (Exercise 2.9) that AUB is well-defined up to isomorphism: that
is, that any two choices for the copies A', B' lead to isomorphic candidates for
AUB. The
same
considerations apply to products and quotients.
The main feature of sets obtained by taking disjoint unions, products,
or
really 'what elements they contain' but rather 'their relationship with
all other sets'. This will be (even) clearer when we revisit these operations and
quotients is not
others12 in the context of categories.
Exercises
2.1. > How many different
and itself?
2.2. > Prove statement
of disjoint
the
bijections
subsets of
a
(2)
in
there between
2.1. You may
Proposition
set, there is
a set
S with
n
elements
choose
a way to
one
assume that given a family
element in each member of
family13. [§2.5, V.3.3]
2.3. Prove that the inverse of
of
are
[§11.2.1]
two
2.4.
bijections
is
> Prove that
a
a
bijection
is
a
bijection and that the composition
bijection.
'isomorphism'
is
an
equivalence relation (on
any set of
sets).
[§4-1]
2.5. > Formulate a notion of
phism
in
seen
§2.6,
epimorphism,
and prove
a
phisms and surjections.
[§2.6, §4.2]
2.6.
in
With notation
determines
a
2.7. Let
:
/
as
in the
style of the
notion of
monomor-
result analogous to Proposition 2.3, for epimor-
Example 2.4, explain how
any function
/
:
A
B
section of tta
A
B be any function. Prove that the graph
of / is isomorphic
Tf
to A.
2.8. Describe
(cf. §2.8)
as
explicitly
as
of the function R
you
can
all terms in the canonical decomposition
C defined by
e2?m\ (This exercise matches
r i—?
one
assigned previously. Which one?)
2.9. > Show that if A' ^ A" and B' ^
B",
and further A'C\B'
then A! U B' ^ A!' U B". Conclude that the operation AllB
is well-defined up to isomorphism (cf. §2.9). [§2.9, 5.7]
2.10. > Show that if A and B
are
finite sets, then
=
0 and A"C\B"
(as
described in
\BA\ \B\M.[§2.1,
=
0,
§1.4)
=
2.11, §11.4.1]
The reader should also be aware that there are important variations on the operations we
have seen so far—particularly
important are the fibered flavors of products and disjoint unions.
This
(reasonable)
statement is the axiom
of choice; cf. §V.3.
J. Preliminaries: Set
18
2.11. > In view of Exercise
2.10, it is
3.
a
2A
not unreasonable to use
to denote the set
(say {0,1}). Prove
(cf. §1.2). [§1.2, III.2.3]
set A to a set with 2 elements
arbitrary
bijection between 2A and the
of functions from an
that there is
theory and categories
power set of A
Categories
The language of categories is affectionately known as abstract nonsense, so named
by Norman Steenrod. This term is essentially accurate and not necessarily
derogatory: categories refer to nonsense in the sense that they are all about the 'structure',
and not about the 'meaning', of what they represent. The emphasis is less on how
you run into a specific set you are looking at and more on how that set may sit
relationship with all other sets. Worse (or better) still, the emphasis is less on
studying sets, and functions between sets, than on studying 'things, and things
that go from things to things' without necessarily being explicit about what these
things are: they may be sets, or groups, or rings, or vector spaces, or modules, or
other objects that are so exotic that the reader has no right whatsoever to know
in
(yet).
'Categories' will intuitively look like
about them
Categories
first, and in multiple ways.
they are 'collections of objects', and
sets at
may make you think of sets, in that
further there will be notions of 'functions from categories to categories' (called
functors1^). At the same time, every category may make you think of the collection of
all sets, since there will be analogs of 'functions' among the things it contains.
3.1. Definition. The definition of a category looks complicated at first, but the
gist of it may be summarized quickly: a category consists of a collection of 'objects',
and of 'morphisms' between these objects,
satisfying
a
list of natural conditions.
The reader will note that I refrained from writing
a set
of objects, opting for
generic 'collection'. This is an annoying, but unavoidable, difficulty: for
example, we want to have a 'category of sets', in which the 'objects' are sets and
the
more
the 'morphisms' are functions between sets, and the problem is that there simply
is not a set of all sets15. In a sense, the collection of all sets is 'too big' to be a set.
are however ways to deal with such 'collections', and the technical name for
them is class. There is a 'class' of all sets (and there will be classes taking care of
There
groups, rings,
etc.).
An alternative would be to define
a
large enough
set
(called
a
universe)
and
then agree that all objects of all categories will be chosen from this gigantic entity.
In any case, all the reader needs to know about this is that there is a way to
make it work. We will use the term 'class' in the definition, but this will not affect
any proof or any other definition in this book. Further, in some of the examples
considered below the class in question is a set (we say that the category is small
in this
case),
so
the reader will feel
perfectly
at home when
contemplating these
examples.
However,
we will not consider
functors until later chapters:
functors will be in Chapter VIII.
That is
one
thing
we learn
from Russell's paradox.
our
first formal encounter with
3.
19
Categories
Definition 3.1. A category C consists of
a
class
Obj(C)
of objects of the category; and
for every two objects A, B of C,
properties listed below.
As
as
a
a set
Homc(A, B)
j
prototype to keep in mind, think of the objects
'functions'. This
look natural and
one
as
'sets' and of morphisms
example should make the defining properties of morphisms
easy to remember:
object A of C, there exists (at least)
For every
of morphisms, with the
the 'identity'
on
one
morphism
I a G Home (A,
A),
A.
morphisms: two morphisms / G Home (A, B) and g G
a morphism gf G Home (A, C).
That is, for every
Homc(i5,C)
of
of
there
is
a
function
C
C
triple
objects A, B,
(of sets)
One
can
compose
determine
Homc(i4,B)
and the image of the pair
x
Homc(£,C)
is denoted
(f,g)
This 'composition law' is associative: if
and h
G
Homc(A,C),
-?
gf.
/
G
Homc(A,£?),
g G
Home (.B,C),
Home (C,D), then
(hg)f
=
h(gf).
The identity morphisms are identities with respect to composition:
for all / G Home (A, B) we have
flA
This is really
of
a
/,
=
W
that is,
/•
=
mouthful, but again, to remember
requirement is that the sets
all
this, just think of functions
sets. One further
Home (A, B),
Homc(C,D)
be disjoint unless A
C', B
D; this is something you do not usually think about,
but again it holds for ordinary set-functions16. That is, if two functions are one and
the same, then necessarily they have the same source and the same target: source
=
and target
=
part of the datum of
are
A morphism of an object A of
Home (A, A) is denoted Endc(^).
this is
'pointed' set,
defines an 'operation'
composition gf.
a
Writing '/
understood,
set-functions:
:
a
set-function.
category C to itself is called an endomorphism;
a category tells us that
One of the axioms of
Endc(^). The reader should note that composition
Endc(^4): if f,g are elements of Endc(^4), so is their
as I a G
on
Homc(A, £?)' gets
G
tiresome in the
safely drop the index C,
A —>
B. This also allows us
one may
f
a
long
run.
If the category is
do with
or even use arrows as we
draw diagrams of morphisms in
to be a 'commutative' diagram) if
to
any category; a diagram is said to 'commute' (or
all ways to traverse it lead to the same results of composing morphisms along the
way, just
as
explained for diagrams of functions of
sets in
§2.3.
We will often use the term 'set-function' to emphasize that we are dealing with a function
in the context of sets.
J. Preliminaries: Set
20
In fact,
will
we
feel free to
now
The official definition of
a
use
diagrams
as
possible objects of categories.
in this context would be
diagram
theory and categories
a set
of objects of
a
category C, along with prescribed morphisms between these objects; the diagram
commutes if it does in the sense specified above. The specifics of the visual
representation of
diagram
a
are
of
irrelevant.
course
3.2. Examples. The reader should note that 90% of the definition of the notion
of category goes into explaining the properties of its morphisms; it is fair to say
that the morphisms are the important constituents of a category. Nevertheless,
it is psychologically irresistible to think of a category in terms of its objects: for
example, one talks about the 'category of sets'. The point is that usually the kind
of 'morphisms' one may consider are
objects: if one is talking about sets,
other than
a
(psychologically
what
function of sets? In other situations
at
least)
possibly
can one
determined by the
for 'morphism'
mean
(cf. Example
3.5 below
or
little less clear what the morphisms should be, and looking for the
3.9)
notion
'right'
may be an interesting project.
Exercise
it is
Example
a
3.2. It is
with set-functions
here and go
hopefully crystal
(as morphisms),
clear by
form
a
now
that sets
further until this assertion sheds any residual
no
(as objects), together
category; if not, the reader must stop
mystery17.
universally accepted, official notation for this important category.
or 'Sets', with some fancy decoration for
one may encounter Set, Sets, 6et, (Sets),
amusing variations on these themes. We will use 'sans-serif fonts to
There is
no
It is customary to write the word 'Set'
emphasis. For example, in the literature
and
many
denote categories; thus, Set will denote the category of sets. Thus
Obj(Set)
=
for A, B in
the class of all sets;
Obj(Set) (that is,
for A, B
sets) RomSet(A,B)
=
BA.
operations recalled in §§1.3-1.5 is not part of the
operations highlight interesting features of Set, which
may or may not be shared by other categories. We will soon come back to some of
these operations and understand more precisely what they say about Set.
j
Note that the
presence of the
definition of category: these
Example
3.3. Here is
Suppose S
is
properties. Then
a set
a
completely different example.
is a relation on S satisfying the reflexive and transitive
encode this data into a category:
and
we can
~
objects: the elements of 5;
morphisms: if a, b
are
objects (that is, if
set consisting of the element
(a, b)
a, b G
5),
G S x S if a
~
then let
Hom(a,6)
be the
6, and let Hom(a, b)
=
0
otherwise.
We will give the reader such prompts every now and then: at key times, it is more useful
to take stock of what one knows than blindly march forward hoping for the best. A difficulty at
this time signals the need to reread the previous material carefully. If the mystery persists, that's
what office hours are there for.
But typically you should be able to find your way out on your
own, based on the information we have given you, and you will most likely learn more this way.
You should give it your best try before seeking professional help.
3.
21
Categories
Note that
(unlike
in
Set)
there
few morphisms:
at most one for any
pair
objects.
We have to define 'composition of morphisms' and verify that the conditions
specified in §3.1 are satisfied. First of all, do we have 'identities'? If a is an object
(that is, if a G 5), we need to find an element
of objects, and
no
morphisms
are very
at all between 'unrelated'
la G
is reflexive: this tells us that Va,
assuming that
is, Hom(a, a) consists of the single element (a, a). So we have no choice:
This is precisely why
a
a; that
~
we must
we
are
~
let
la
As for composition, let a, 6,
/
we
have to define
a
=
c
(a, a)
us
that
Hom(a, 6),
G
Hom(a, b)
g G
since we are
(that is,
G
elements of
S)
and
Hom(6, c);
g G
G
Hom(a, c). Now,
Hom(a,6)
is nonempty, and according to the definition of morphisms
that a
6, and / is in fact the element (a, b) of A x B.
means
~
tells
Hom(6, c)
us
a
the
Hom(a,a).
corresponding morphism gf
in this category that
Similarly,
G
be objects
/
tells
Hom(a,a).
assuming that
b
~
c
and g
b and b
~
~
c
Now
(6, c).
=
=>
a
~
c
is transitive. This tells us that
~
single element (a,c). Thus
we
again have
no
choice:
Hom(a, c)
we must
consists of
let
gf:=(a,c)eRom(A,C).
Is this operation associative? If
then necessarily
f
=
/
G
(a,b),
Hom(a,6),
g
=
(b,c),
g G
Hom(6, c),
and h G
Hom(c, d),
h=(c,d)
and
#/
=
(a,c),
hg
=
(b,d)
and hence
h(gf)
=
(a,d)
=
(hg)f,
proving associativity.
The reader will have
to this
composition,
as
no
difficulties checking that la is
needed
an
identity with respect
(Exercise 3.3).
The most trivial instance of this construction is the category obtained from a
S taken with the equivalence relation '='; that is, the only morphisms are the
identity morphisms. These categories are called discrete.
set
As another example, consider the category corresponding to endowing Z with
the relation <. For example,
22
J. Preliminaries: Set
theory and categories
(randomly chosen) commutative diagram in this category. It would still
(commutative) diagram in this category if we reversed the vertical arrow 3
be
if
from
is
a
we
added
4 to 3, since
from 3 to 4, while
an arrow
we are not
allowed to draw
an arrow
3
a
or
4^3.
special—for
example, every diagram drawn in them
and
this
is
necessarily commutative,
very far from being the case in, e.g., Set.
Also note that these categories are all small.
j
These categories
are very
is
Example
example of
3.4. Here is another18
Let S again be
Obj(S)
=
a set.
^(5),
Define
a
a
small category.
category S by setting
the power set S
(cf. §1.2
and Exercise
for A, B objects of S (that is, A C S and B C S) let
(A, B) if AC B, and let Hom§(A, B) 0 otherwise.
2.11);
Hom§(A, B)
be the pair
=
The identity 1a consists of the pair (A, A) (which is one, and in fact the only
one, morphism from A to A, since A C A). Composition is obtained by stringing
inclusions: if there
are
morphisms
B^C
A^B,
S, then A
Checking the
in
case!).
Examples
the
family of
C.
C; hence A C C and there is a morphism A
specified in §3.1 should be routine (make sure this is the
C B and B C
axioms
style (but employing more sophisticated structures, such as
topological space) are hugely important in wellsuch as algebraic geometry.
j
in this
open subsets of a
established fields
example is very abstract, but thinking about it will make
everything we have seen so far; and it is a very common
will abound in the course.
variations
of
which
construction,
Let C be a category, and let A be an object of C. We are going to define a
category Qa whose objects are certain morphisms in C and whose morphisms are
certain diagrams of C (surprise!).
Example
3.5. The next
you rather comfortable with
all morphisms from any object of
Odj(Ca)
morphism / G Home(Z, A) for some object
=
a
Qa
is
an arrow
Z
A in C; these
are
C to A\thus, an object of Qa is
Z of C. Pictorially, an object of
often drawn 'top-down',
as
in
Z
f\
A
What
are morphisms in C^ going to be? There really is only one sensible way to
assign morphisms to a category with objects as above. The brave reader will want
to stop reading here and continue only after having come up with the definition
independently. There will be many similar examples lurking behind constructions
Actually, this is again
why? (Exercise 3.5.)
an instance
of the categories considered in Example 3.3. Do you
see
3.
we
23
Categories
will encounter in this book, and ideally speaking, they should appear completely
time comes. A bit of effort devoted now to understanding this
natural when the
prototype
situation will have
these notes away
now
ample reward
in the future.
Spoiler follows,
so
put
and jot down the definition of morphism in Ca-
Welcome back.
Let
/i, J2
be objects of Ca, that is, two
h
h
A
in C. Morphisms
j\ fa
are
arrows
A
defined to be commutative diagrams
<T-^Z2
Zx
A
in the 'ambient' category C.
That is,
morphisms /
in C such that f\ f^o.
g
correspond precisely
to those
morphisms
a
:
Z\
Z<i
=
Once you understand what morphisms have to be, checking that they satisfy
the axioms spelled out in §3.1 is straightforward. The identities are inherited from
A in Ca, the identity 1/ corresponds to the diagram
the identities in C: for / : Z
lz
->Z
'W
A
which commutes by virtue of the fact that C is
a
category. Composition is also
subproduct of composition in C. Two morphisms j\ fa
to putting two commutative diagrams side-by-side:
and then
it follows
(again
because C is
a
category!)
removing the central arrow, i.e.,
Zi-
-tZ*
fa
in C^
a
correspond
that the diagram obtained by
24
J. Preliminaries: Set
also
theory and categories
Check all this(!), and verify that composition in Qa is associative
this follows immediately from the fact that composition is associative in C).
commutes.
(again,
Categories constructed in this
they
are
particular
of
cases
fashion
comma
called slice categories in the literature;
are
categories.
j
apply the construction given in
Z and
the
Example 3.3, say for S
relation <. Call C this category, and choose an object A of C—that
is, an integer,
for example, A
3. Then the objects of Qa are morphisms in C with target 3, that
is, pairs (n, 3) G Z x Z with n < 3. There is a morphism
Example 3.6.
Example 3.5 to
For the sake of concreteness, let's
the category constructed in
=
~
=
(ra,3)
if and only if
<
m
case Qa may be harmlessly identified with the
j
3, with 'the same' morphisms as in C.
In this
n.
'subcategory' of integers
Example
(n,3)
-?
<
entirely similar example to the one explored in Example 3.5 may
by considering morphisms in a category C from a fixed object A to all
3.7. An
be obtained
objects in C, again with morphisms defined by suitable commutative diagrams. This
leads to coslice categories. The reader should provide details of this construction
(Exercise 3.7).
Example
and A
=
a
j
3.8. As
'concrete' instance of
a
fixed singleton
{*}.
a
category
as
in
Example 3.7, let C
=
Set
Call the resulting category Set*.
An object in Set* is then a morphism / : {*}
S in Set, where S is any set.
The information of an object in Set* consists therefore of the choice of a nonempty
set S and of
an
element
and is determined by,
Thus,
we may
s G
S—that
is, the element
/(*):
this element determines,
/.
denote objects of Set*
as
pairs (5,5), where S is
any set and
s G S is any element of S.
A
this!)
morphism between
two such
to a set-function a : S
Objects of Set*
in this book will be
are
(T, t), corresponds then (check
objects, (5, s)
T such that
o~(s)
=
t.
called 'pointed sets'. Many of the structures we will
sets. For example (as we will see) a 'group' is a
pointed
study
set
G
with, among other requirements, a distinguished element cq (its 'identity'); 'group
homomorphisms' will be functions which, among other properties, send identities to
identities; thus,
they
are
morphisms of pointed
sets in the sense considered above.
j
3.9. It is useful to contemplate a few more 'abstract' examples in the
style of Examples 3.5 and 3.7. These will be essential ingredients in the promised
revisitation of some of the operations mentioned in §1.3. Their definition will appear
disappointedly simple-minded to the reader who has mastered Examples 3.5 and 3.7.
Example
This time
can
define
we start
a new
from
a
given category C and
category Qa,b by essentially the
order to define C^:
objects A, B of C. We
procedure that we used in
two
same
3.
25
Categories
Ob](Ca,d)
=
diagrams
B
in
C; and
morphisms
z1
z2
B
are commutative
We
B
diagrams
will leave to the reader the task of
this rough description. This
example is really nothing more than a mixture of Ca and Cj3, where the two
structures interact because of the stringent requirement that the same a must make both
formalizing
sides of the diagram commute:
A
=
h°
and
pi
=
g2cr
'simultaneously'.
Flipping most of the arrows gives an analogous variation of Example 3.7,
producing a category which we may19 denote QAyD', details are left to the reader.
j
Example 3.10. As a final variation on these examples, we conclude by considering
the fibered version of Ca,d (and CA'D). Take this as a test to see if you have really
understood Ca,d—experts
would tell you that this looks fairly sophisticated for
students just learning categories, so don't get disheartened if it does not flow too
well at first (but pat yourself on the shoulder if it does!). Start with a given
category C, and this time choose
with the
same
target C. We
Obj(Ca>/3)
=
can
commutative
two fixed
morphisms
then consider
a
a :
A
category Ca^
C in C,
C', (3 : B
as follows:
diagrams
C
B
in C, and
There does not
seem to
be
an
established notation for these commonplace categories.
J. Preliminaries: Set
26
morphisms correspond
to commutative
A solid understanding of Example
this point the reader should have
theory and categories
diagrams
3.9 will make this example look just as tame; at
difficulties formalizing it (that is, explaining
no
how composition works, what identities are,
etc.).
Also left to the reader is the construction of the 'mirror' example C*
B with common source.
two morphisms a : C
A, (3 : C
,
starting
from
j
Exercises
3.1.
>
Let C be
a
category. Consider
a structure
Cop with
Obj(Cop):=Obj(C);
for A, B objects of Cop
(hence objects
of
C), Romc<>p(A,B)
:=
Homc(.B, A).
Show how to make this into a category (that is, define composition of morphisms
in Cop and verify the properties listed in §3.1).
Intuitively, the 'opposite' category Cop is simply obtained by 'reversing all the
arrows' in C. [5.1, §VIII.1.1, §IX.1.2, IX.1.10]
3.2. If A is
finite set, how
a
large
is
Endset(^4)?
precisely what it means to say that la is an identity with respect
composition in Example 3.3, and prove this assertion. [§3.2]
3.3. > Formulate
to
3.4. Can
we
define
a
category in the style of Example 3.3 using the relation
< on
the set Z?
Explain in what
Example 3.3. [§3.2]
3.5. >
in
3.6. >
(Assuming
sense
some
Example 3.4
familiarity
is
an
instance of the categories considered
with linear
algebra.)
Define
a
category V by
N and letting Homv (n,m)
the set ofmxn matrices with real
taking Obj(V)
will
for
all
m
N.
leave
the
reader
the task of making sense of a
G
entries,
n,
(We
matrix with 0 rows or columns.) Use product of matrices to define composition.
=
=
Does this category 'feel' familiar?
3.7. > Define
carefully objects
[§VI.2.1, §VIII.1.3]
and morphisms in Example 3.7, and draw the
diagram corresponding to composition.
[§3.2]
subcategory C of a category C consists of a collection of objects of C, with
morphisms Rome (A, B) C Homc(A,B) for all objects A, B in Obj(C;), such that
identities and compositions in C make C into a category. A subcategory C is full
3.8. > A
4.
27
Morphisms
if Homc(A,£?)
=
for all A, B in
H.omc(A,B)
Obj(C').
infinite sets and explain how it may be viewed
as a
Construct
a
category of
full subcategory of Set.
[4.4,
§VI. 1.1, §VIII. 1.3]
3.9. > An alternative to the notion of multiset introduced in
§2.2 is obtained by
equivalence relations; equivalent elements are taken
to be multiple instances of elements 'of the same kind'. Define a notion of morphism
between such enhanced sets, obtaining a category MSet containing (a 'copy' of) Set
considering
as
sets endowed with
full subcategory.
a
(There
may be more than one reasonable way to do this!
intentionally an open-ended exercise.) Which objects in MSet determine
ordinary multisets as defined in §2.2 and how? Spell out what a morphism of
multisets would be from this point of view. (There are several natural notions
This is
of morphisms of multisets. Try to define morphisms in MSet so that the notion
you obtain for ordinary multisets captures your intuitive understanding of these
objects.) [§2.2, §3.2, 4.5]
objects of
3.10. Since the
how to make
make
sense
sense to
of
a
a
category C
notion of
(necessarily) sets,
are not
in
'subobject'
general.
In
some
it is not clear
situations it does
talk about subobjects, and the subobjects of any given object A in
fl for a fixed, special
correspondence with the morphisms A
object fl of C, called a subobject classifier. Show that Set has a subobject classifier.
C
are
in one-to-one
3.11. > Draw the relevant
category C
diagrams and define composition and identities for the
mentioned in Example 3.9.
mentioned in Example 3.10. [§5.5, 5.12]
4.
as
in Set
highlight
we
for the category
Ca^
should
an
certain types of functions
(injective, surjective, bijective),
arbitrary category. The reader
of
morphisms by their actions on 'elements' is
defining qualities
in
the
because
option
general setting,
objects of an arbitrary category do
it is useful to try to do the
not
same
Morphisms
Just
not
Do the
'
for morphisms in
same
an
note that
(in general)
have 'elements'.
This is why
we
spent
some
time
analyzing injectivity, etc., from different
viewpoints
§§2.4-2.6. It turns out that the other viewpoints
transfer nicely into the categorical setting.
in
4.1. Isomorphisms. Let C be
a
Definition 4.1. A morphism /
G
sided)
these notions do
category.
H.omc(A, B)
inverse under composition: that is, if
gf
on
=
1a,
3g
fg
=
is
G
an
isomorphism if
Homc(i5, A)
it has
a
(two-
such that
lfl.
-J
Recall that in §2.5 the inverse of a bijection of sets / was defined 'elementwise';
in particular, there was no ambiguity in its definition, and we introduced the
notation f~l for this function. By contrast, the 'inverse' g produced in Definition 4.1
does not appear to have this uniqueness explicitly built into its definition. Luckily,
its defining property does guarantee its uniqueness, but this requires
a
verification:
J. Preliminaries: Set
28
Proposition
4.2.
The inverse
Proof. We have to
verify
given isomorphism /
:
A
of
is unique.
isomorphism
an
theory and categories
A act as inverses of a
that if both g\ and gi : B
£?, then g\
#2- The standard trick for this kind of
=
/ on the left by one of the morphisms, and on the right
by the other one; then apply associativity. The whole argument can be compressed
into one line:
verification is to compose
9i
=
0ilj3
=
9i(f92)
=
(9if)92
=
1a#2
=
92
as needed.
/ is a morphism with a leftright-inverse
necessarily / is an isomorphism, g\ g2,
and this morphism is the (unique) inverse of /.
Since the inverse of / is uniquely determined by /, there is no ambiguity in
Note that the argument
inverse g\ and
denoting
it
really
proves that if
#2, then
a
=
by f~l.
Proposition
4.3.
With notation
Each identity 1a is
an
as
above:
isomorphism and
isomorphism and further (f~l)~l
f
are
then
the
G
G
isomorphisms,
composition
If f Homc(A, B), g Homc(i5,C)
gf is an isomorphism and (gf)-1
f~lg~l•
If f
is an
isomorphism, then f~l
is its own inverse.
is an
=
-
=
Proof. These all 'prove themselves'. For example, it is immediate to
f~lg~l is a left-inverse of gf: indeed20,
The verification that
f~lg~l
Note that taking the
is also
inverse
a
right-inverse of gf
reverses
is
verify
that
analogous.
the order of composition:
(gf)~l
=
rlg-1.
Two objects A, B of a category are isomorphic if there is an isomorphism
B. An immediate corollary of Proposition 4.3 is that 'isomorphism' is an
/ : A
B.
equivalence relation21. If two objects A, B are isomorphic, one writes A
=
Example 4.4. Of course, the isomorphisms in the category Set
bijections; this was observed at the beginning of §2.5.
are
precisely the
j
Example 4.5. As noted in Proposition 4.3, identities are isomorphisms. They may
be the only isomorphisms in a category: for example, this is the case in the category
C obtained from the relation < on Z, as in Example 3.3. Indeed, for a, b objects of
b and a morphism g : b
a only
C (that is, a, b G Z), there is a morphism / : a
if a < b and b < a, that is, if a
b. So an isomorphism in C necessarily acts from
an object a to itself; but in C there is only one such morphism, that is, la.
j
=
Associativity of composition implies that parentheses
expressions, as done here (cf. Exercise 4.1).
may be shuffled at will in
longer
The reader should have checked this in Exercise 2.4, for Set; the same proof will work in
any category.
4.
29
Morphisms
Example 4.6. On the other hand, there are categories in which every morphism is
an isomorphism; such categories are called groupoids. The reader 'already knows'
many examples of groupoids; cf. Exercise 4.2.
j
an object A of a category C is an isomorphism from A to
automorphisms of A is denoted Autc(A); it is a subset of Endc(^).
An automorphism of
itself. The set of
By Proposition 4.3, composition
confers
the composition of two elements
composition
Autc(^4)
is, flA
f,g
G
In other words,
lA/
=
an
element
/
an
Autc(A);
G
identity for composition (that
Autc(A)
is
has
a group,
devote all
our
an
inverse
f~l
Autc(A).
G
for all objects A of all categories C.
attention to
groups!
Monomorphisms and epimorphisms. As pointed
4.2.
gf
/);
G
Autc(A)
soon
remarkable structure:
is
Autc(A)
contains the element 1^, which is
=
a
is associative;
every element
We will
Autc(^4)
on
out
above,
we
do not
have the option of defining for morphisms of an arbitrary category a notion such
as 'injective' in the same way as we do for set-functions in §2.4:
that definition
requires a notion of 'element', and in general no such notion is available for objects
of a category. But nothing prevents us from defining monomorphisms as we did in
§2.6,
in
an
arbitrary category:
Definition 4.7. Let C be a category. A morphism /
phism if the following holds:
G
Home (A, B) is
for all objects Z of C and all morphisms a',a"
foa'
Similarly, epimorphisms
Definition 4.8. Let C be
a
are
=
foa"
defined
as
=>
a'
=
G
a monomor-
Homc(Z, A),
a".
j
follows:
category. A morphism /
G
Home (A, B) is
an
epimor-
phism if the following holds:
for all objects Z of C and all morphisms ft', ft"
P'of
Example
=
p"of
=»
G
H.omc(B,Z),
/3'=/3".
4.9. As proven in
Proposition 2.3, in the category Set the
the
monomorphisms
precisely
injective functions. The reader should have by now checked
in
Set the epimorphisms are precisely the surjective functions (cf.
that, likewise,
Exercise 2.5). Thus, while the definitions given in §2.6 may have looked
counterintuitive at first, they work as natural 'categorical counterparts' of the ordinary notions
are
of
injective/surjective
functions.
j
Example 4.10. In the categories of Example 3.3, every morphism is both a
monomorphism and an epimorphism. Indeed, recall that there is at most one
morphism between any two objects in these categories; hence the conditions defining
j
monomorphisms and epimorphisms are vacuous.
J. Preliminaries: Set
30
a few unexpected twists in these
set-theorists. For instance, in Set, a function is
it is both injective and surjective, hence if and only
4.10 reveals
Contemplating Example
definitions, which defy
our
theory and categories
intuition
as
isomorphism if and only if
a monomorphism and an epimorphism.
But in the category defined
<
on
is
both
a
Z, every morphism
monomorphism and an epimorphism, while
by
the only isomorphisms are the identities (Example 4.5). Thus this property is a
special feature of Set, and we should not expect it to hold automatically in every
an
if
it is both
category; it will not hold in the category Ring of rings (cf. §111.2.3). It will hold in
every abelian category (of which Set is not an example!), but that is a story for a
very
distant future
(Lemma
IX. 1.9).
epimorphism, that is, surjective, if and only if
right-inverse (Proposition 2.1); this may fail in general, even in respectable
categories such as the category Grp of groups (cf. Exercise II.8.24).
Similarly,
it has
in Set
a
function is
an
a
Exercises
4.1. >
Composition
given,
e.g.,
is defined for two
A
then
one may compose
morphisms. If
more
than two morphisms
are
—f-^
B —?->
D —*->
C —^
E,
them in several ways, for example:
(ih)(gf),
(i(hg))f,
i((hg)f),
etc.
only composing two morphisms. Prove that the result
of any such nested composition is independent of the placement of the parentheses.
fi equals
(Hint: Use induction on n to show that any such choice for fnfn-i
so that at every step one is
((•••((/n/n-l)/n-2)
•••)/!)•
Carefully working
out the case n
=
5 is
helpful.) [§4.1, §11.1.3]
Example 3.3 we have seen how to construct a category from a set endowed
relation, provided this latter is reflexive and transitive. For what types of
relations is the corresponding category a groupoid (cf. Example 4.6)? [§4.1]
4.2. > In
with
a
4.3. Let
A, B be objects of
a
category C, and let /
G
Home (A, B) be
a
morphism.
Prove that if / has a right-inverse, then / is an epimorphism.
Show that the converse does not hold, by giving an explicit example of a category
and an epimorphism without a right-inverse.
4.4. Prove that the
Deduce that
one can
objects
in C and
as
composition of two monomorphisms is a monomorphism.
a subcategory Cmono of a category C by taking the same
defining Homcmono(-A, B) to be the subset of Homc(A,£?)
define
of monomorphisms, for all objects A, B. (Cf. Exercise 3.8; of course, in
general Cmono is not full in C.) Do the same for epimorphisms. Can you define
consisting
subcategory Cnonmono of C by restricting
monomorphisms?
a
to
morphisms that
are not
5. Universal properties
4.5. Give
a concrete
31
description of monomorphisms and epimorphisms
category MSet you constructed in Exercise 3.9.
of morphism you defined in that
5.
Universal
(Your
answer
will depend
in the
on
the notion
exercise!)
properties
§3 may have left the reader with the impression that
produce
large number of minute variations of the same basic ideas,
without really breaking any new ground. This may be fun in itself, but why do we
really want to explore this territory?
Categories offer a rich unifying language, giving us a bird's eye view of many
The 'abstract' examples in
at will a
one can
constructions in algebra (and other fields). In this course, this will be most apparent
in the steady appearance of constructions satisfying suitable universal properties.
For instance, we will see in a moment that products and disjoint unions (as reviewed
in §1.3 and following) are characterized by certain universal properties having to
do with the categories Ca,b and C
considered in Example 3.9.
'
Many of the concepts introduced in this course will have an explicit description
(such as the definition of product of sets given in §1.4) and an accompanying
description in terms of a universal property (such as the one we will see in §5.4). The
'explicit' description may be very useful in concrete computations or arguments,
as a rule it is the universal property that clarifies the true nature of the
but
construction. In
some cases (such as for the disjoint union) the explicit description may
turn out to depend on a seemingly arbitrary choice, while the universal property
will have no element of arbitrariness. In fact, viewing the construction in terms
of its corresponding universal property clarifies why
defined 'up to isomorphism'.
one can
Also, deeper relationships become apparent when the
only expect
constructions
it to be
are
viewed
in terms of their universal properties. For example, we will see that products of
sets and disjoint unions of sets are really 'mirror' constructions (in the sense that
reversing arrows transforms the universal property for one into that for the other).
This is not so clear (to this writer, anyway) from the explicit descriptions in §1.4.
5.1. Initial and final objects.
Definition 5.1. Let C be
if for every
VA e
We
one
category. We say that
\/Ae
exists
Obj(C)
object F
morphism A
say that an
exactly
a
object A of C there
Uomc(IjA)
:
of C is
exactly
one
final
object J
morphism /
is
an
a
of C is initial in C
A
in C:
singleton.
in C if for every
object A of C there
exists
F in C:
Obj(C)
:
Homc(A,F)
is
a
singleton.
j
One may use terminal to denote either possibility, but in general we would
advise the reader to be explicit about which 'end' of C one is considering.
A category need
not have initial or final
objects,
as
the
following example shows.
J. Preliminaries: Set
32
Example
5.2.
The category obtained
Example 3.3)
has
would be
integer
an
no
initial
an
Z with the relation <
(see
initial object in this category
i such that i < a for all
final object would be
Similarly,
is no such thing.
a
by endowing
final object. Indeed,
or
theory and categories
integers a; there is no such integer.
integer / larger than every integer, and there
an
By contrast, the category considered in Example 3.6 does have
namely the pair (3,3); it still has no initial object.
Also, initial and final objects, when they exist,
a
final object,
j
may or may not be
unique:
Example 5.3. In Set, the empty set 0 is initial (the 'empty graph' defines the
unique function from 0 to any given object!), and clearly it is the unique set that
fits this requirement (Exercise 5.2).
Set also has final objects: for every set A, there is a unique function from A
singleton {p} (that is, the 'constant' function). Every singleton is final in Set;
j
thus, final objects are not unique in this category.
to a
However, we claim that if initial/final objects exist, then they are unique up
isomorphism. We will invoke this fact frequently, so here is its official
to a unique
statement and its
Proposition
If I\,I2
If Fi, F<i
(immediate) proof:
5.4. Let C be a category.
(ire
are
both initial objects in C, then I\
=
both
final objects
Further, these isomorphisms
are
in
C, then F\
=
is
a
one
element in Home (A,
unique morphism I
Now
F2.
uniquely determined.
Proof. Recall that (by definition of category!) for
least
h-
A), namely
every
object A of C there
is at
the identity 1^. If / is initial, then there
I, which therefore
—>
must be the
identity 1/.
I\and I2 are both initial in C. Since I\is initial, there is a unique
I2 in C; we have to show that / is an isomorphism. Since I2 is
h in C. Consider gf : h —>
initial, there is a unique morphism g : I2
h; as
observed, necessarily
assume
morphism /
:
Ii
gf
since I\is initial.
By the
same
=
i/x
=
i/2
token
19
since I2 is initial. This proves that
/ : I\
I2 is
an
The proof for final objects is entirely analogous
isomorphism,
as
needed.
(Exercise 5.3).
Proposition 5.4 "explains" why, while not unique, the final objects in Set are
all isomorphic: no singleton is more 'special' than any other singleton; this is the
typical situation. There may be psychological reasons why one initial or final object
20 may look
looks more compelling than others (for example, the singleton {0}
=
to some like the most 'natural' choice among all
in how these objects sit in their category.
singletons),
but this plays
no
role
5. Universal properties
33
properties. The most natural context in which to introduce
properties requires a good familiarity with the language of functors, which we
will only introduce at a later stage (cf. §VIII. 1.1). For the purpose of the examples
we will run across in (most of) this book, the following 'working definition' should
5.2. Universal
universal
suffice.
We say that a construction satisfies a universal property (or 'is the solution to
universal problem') when it may be viewed as a terminal object of a category.
The category depends on the context and is usually explained 'in words' (and often
a
without
even
mentioning the word category).
In particularly simple
cases
this may take the form of
universal with respect to the property of mapping to
the assertion that 0 is initial in the category of Set.
a statement
such
as
0 is
sets; this is synonymous with
More often, the situation is more complex. Since being initial/final amounts to
uniqueness of certain morphisms, the 'explanation' of a universal
the existence and
property may follow the pattern, "object X is universal with respect to the
property: for
any Y such
that..., there exists
a
unique morphism Y
following
X such
that...."
The not-so-naive reader will recognize that this explanation hides the definition
of
category and the statement that X is terminal (probably final in
this case) in this new category. It is useful to learn how to translate such wordy
explanations into what they really mean. Also, the reader should keep in mind that
an accessory
it is not
uncommon to sweep
the solution to
a
is
presumably implicit
that follow.
5.3.
under the rug part of the essential information about
some key morphism): this information
will be apparent from the examples
This
given set-up.
universal problem
Quotients. Let
in any
~
be
an
(usually
equivalence relation defined
on a set
A. Let's parse
the assertion:
"The quotient A/~ is universal with respect to the property of
such a way that equivalent elements have the same image."
mapping
A to
a
set in
What
can
this possibly mean, and is it true?
The assertion is talking about functions
with Z any set,
satisfying
the property
a'
~
a"
=>
y>(a')
=
y>(a").
are objects of a category (very similar to the category defined in
Example 3.7); for convenience, let's denote such an object by (ip, Z). The only
These morphisms
reasonable
way to define
morphisms (ipi,Zi)
Z1
(</?2> ^2)
a-^Z2
/
A
is
as
commutative
diagrams
J. Preliminaries: Set
34
This
is the
same
definition considered in Example 3.7.
Does this category have initial
Claim 5.5. Denoting by
This
this
tt
is an initial
pair (it, A/~)
is what
theory and categories
our
objects?
the 'canonical projection'
defined
in
Example 2.6, the
object of this category.
writer meant
by the mysterious
assertion
copied above. Once
is understood, it is very easy to prove that the assertion is indeed correct.
Proof. Consider any
(<p, Z)
as
above. We have to prove that there exists
(<p,Z), that is,
morphism (tt,A/~)
a
a
unique
unique commutative diagram
>Z
A/~—
that is,
a
Let
making this diagram commute.
arbitrary element of A/~. If the diagram
unique function
[a]~
be
commute, then
an
cp
is indeed going to
necessarily
Tp([a]„) =ip{a)\
this tells
that Tp is indeed unique, if it exists at all—that
is,
Z.
does define a function A/~
us
if
this prescription
Hence, all we have to check is that Tp is well-defined, that is, that if
[a2]~, then <p{a\) ^(a2); and indeed
[a\]^
=
=
[ai]„
=
[a2]~
=>
oi
~
a2
=>
This is precisely the condition that morphisms in
<p(ai)
our
=
<p(a2).
category satisfy.
Note the several levels of sloppiness in the assertion considered above: it does
not tell us very explicitly what category to consider; it does not tell us that we
especially pay attention to initial objects in this category. Worst of all,
the solution to the universal problem is not really A/~, but rather the morphism
should
7r :
A
A/~.
-?
The reader should practice the skill of translating loose assertions such
given above into precise statements; it is not at all uncommon
examples at the same level of 'abuse of language' as this one.
one
as
the
to run into
reason why we get away with writing such assertions is that the context
allows
the experienced reader to parse them effectively, and they are
really
The
spelled-out version. After all, there is in general no
conceivable choice for a morphism A
A/~ other than the canonical projection;
hence, neglecting to mention it is forgivable. Also, the final object in the category
considered above is supremely uninteresting (what is it? cf. Exercise 5.5), so surely
substantial y
more
we must
concise than their
have meant the initial
one.
learn by viewing quotients in terms of their universal property?
For example, suppose
is the equivalence relation defined starting from a function
£?, as in §2.8. Then the reader will realize easily that im/ also satisfies
/ : A
What do
we
~
5. Universal properties
35
the universal property given above for A/~; therefore (by Proposition
and A/~ must be isomorphic. This is precisely the content of Theorem
the universal property sheds
in
some
light
on
5.4) im/
2.7; thus,
the 'canonical decomposition' studied
§2.8.
5.4. Products. It is also a very
good
exercise to stare at
a
familiar construction
and try to see the universal property which may be behind it. We will now encourage
you, dear reader, to contemplate the notion of the product of two sets given in §1.4
and to see if the universal property it satisfies jumps out at you. Spoiler follows,
so this is a good time to stop reading these notes and to try on your own.
Here is the universal property.
A
x
£?, with the
two natural
A, B be sets, and consider the product
Let
projections:
,A
7TA
Ax B
*^B
(see Example 2.4).
Then
for
every set Z and
morphisms
z
there
exists a unique
morphism
a :
Z
Ax B such that the diagram
—>
commutes.
In this situation,
Proof. Define
is
a
usually denoted /U
manifestly
that ttact
Further,
unique,
as
=
fA
=
(fA(z),fB(z)).
makes the diagram commute: \/zG Z
7TA<j(z)
showing
/#.
Vz G Z
a(z)
This function22
x
=
-KaUa{z)Jb{z))
and similarly ttbct
=
=
fA(z),
fs-
the definition is forced by the commutativity of the diagram;
claimed.
'Note that there is
no
'well-definedness' issue this time.
so a
is
J. Preliminaries: Set
36
theory and categories
In other words, products of sets (or, more precisely, products of sets together
with the information of their natural projections to the factors) are final objects in
the category
considered in Example 3.9, for C
Ca,b
=
Set.
What is the advantage of viewing products this way? The main advantage
is that the universal property may be stated in any category, while the definition
of products given in §1.4 only makes sense in Set (and possibly in other categories
where
'elements'). We say that a category C has (finite) products,
'with
category
(finite) products', if for all objects A, B in C the category
in
considered
Example 3.9 has final objects. Such a final object consists of the
Ca,J3
or
is
one
has
notion of
a
a
data of
object of C, usually denoted A
an
x
£?, and of
two
morphisms A
x
B
A,
AxB^B.
'product' from this perspective does not need to 'look like a
recurring example of the category obtained from < on Z, as
in Example 3.3. Does this category have products? Objects of this category are
simply integers a,b e Z; call a x b for a moment the 'categorical' product of a and
b. The universal property written out above becomes, in this case, for all z G Z
such that z < a and z <b, we have z < a x b.
Note that
product'.
a
Consider our
This universal problem does have
a
solution Va, b:
it is conventionally not
6, but rather min(a,6). It is immediate to see that min(a,6) satisfies
property. Thus this category has products, and in fact we see that the product
called
the
a x
in this category amounts to the familiar operation of taking the minimum of two
integers.
product of two
both
are
of
examples
products, taken
integers':
both
'the
same'
universal
categories; they
satisfy
property, in different
Thus there is
an
unexpected
connection between 'the Cartesian
sets' and 'the minimum of two
in different
contexts.
5.5. Coproducts. The prefix co- usually indicates that one is 'reversing all
arrows'. Just as products are final objects in the categories Ca,b obtained by
considering morphisms in C with common source, whose targets are A and £?, coproducts
will be initial objects in the categories23 C
of morphisms with common
'
target, whose
sources
are
property out before
Here it is.
and B will be
is
:
B
we
A and B.
Dear reader, look away and spell this universal
do.
Let A, B be objects of
a
category C.
A coproduct A II B of A
object of C, endowed with two morphisms %a A
and satisfying the following universal property: for all
an
A II B
and morphisms
B
'These categories
were also considered in
Sb
Example 3.9; cf. Exercise 3.11.
A II
£?,
objects Z
5. Universal properties
there exists
a
37
unique morphism
a :
All B
Z such that the diagram
commutes.
The symmetry with the universal property of products is hopefully completely
We say that a category C has coproducts if this universal problem has a
apparent.
solution for all pairs of objects A and B.
Is the reader familiar with any coproduct? Yes!
Proposition 5.6.
The disjoint union is
coproduct
a
in Set.
that the disjoint union A II B is defined as the union of two
disjoint isomorphic copies A', B' of A, B, respectively; for example, we may let
A!
{0} x A, B' {1} x B. The functions i^, iD are defined by
Proof. Recall
=
(§1.4)
=
iA(a)
where
we
view these elements
Now let
Ja
A
Z,
Jb
=
as
B
(0,o),
iB(b)
elements of
Z be
=
({0}
(1,6),
x
A)
U
({1}
arbitrary morphisms
x
B).
to a common target.
Define
a : A II B
=
({0}
x
A)
U
({1}
x
B)
Z
-?
by
f/A(a)
a(C)"\M&)
ifc
=
ifc
=
(0,a)e{0}xA,
(l,6)€{l}xB.
This definition makes the relevant diagram commute and is in fact forced upon
by this commutativity, proving that a exists and is unique.
us
This observation tells us that the category Set has coproducts and further sheds
considerable light on the mysteries of disjoint unions. For example, there was an
element of arbitrariness in our choice of 'a' disjoint union, although different choices
led to isomorphic notions. Now we see why: terminal objects of a category are not
unique in general, although they are unique up to isomorphism (Proposition 5.4);
there is not a 'most beautiful' disjoint union of two sets just as there is not a 'most
beautiful'
singleton
in Set
(cf. §5.1).
Also, an unexpected 'symmetry' between products and disjoint unions becomes
suddenly apparent from the point of view of universal properties.
The reader is invited to contemplate the notion of coproduct in the other
categories
point)
two
we have encountered. For example (and probably not surprisingly at this
the category obtained from < on Z does have coproducts: the coproduct of
objects (i.e., integers)
a, b is
simply the
maximum of a and b.
J. Preliminaries: Set
38
theory and categories
Exercises
5.1. Prove that a final
object
in
category C is initial in the opposite category Cop
a
(cf. Exercise 3.1).
5.2.
>
Prove that 0 is the unique initial object in Set.
5.3. > Prove that final
5.4.
objects
are
What are initial and final
unique
up to
[§5.1]
isomorphism. [§5.1]
in the category of
objects
'pointed sets'
(Example 3.8)? Are they unique?
5.5. > What are the final
objects
in the category considered in
§5.3? [§5.3]
corresponding to endowing (as in Example 3.3) the
positive integers with the divisibility relation. Thus there is exactly one
m in this category if and only if d divides m without remainder;
morphism d
there is no morphism between d and m otherwise. Show that this category has
5.6. > Consider the category
set Z+ of
products and coproducts. What
are
their 'conventional' names?
5.7. Redo Exercise 2.9, this time using
Proposition 5.4.
5.8. Show that in every category C the
products Ax B and B
if they
[§VII.5.1]
A
x
are
isomorphic,
Observe that they both satisfy the universal property for the
product of A and B\then use Proposition 5.4.)
5.9.
exist.
(Hint:
Let C be
category with products.
a
Find
a
reasonable candidate for the
universal property that the product A x B x C of three objects of C ought to
satisfy, and prove that both (A x B) x C and A x (B x C) satisfy this universal
property. Deduce that
(A
B)
x
x
C and A
x
(B
x
C)
are
envelope a little further still, and define
families (i.e., indexed sets) of objects of a category.
5.10. Push the
for
necessarily isomorphic.
products and coproducts
Do these exist in Set?
It is
common to
denote the product A
x
n
5.11. Let
A,
resp.
Define a relation
£?, be
a
A
x
~
on
(01,61)
(This is immediately
~
set, endowed with
B
times
an
equivalence relation
be
an
^=^
01 ~a o2 and
(A
B)/~
x
Prove that
~a, resp. ~£.
A/~a, (A
b±
~B
b2.
equivalence relation.)
Use the universal property for quotients (§5.3)
functions
A by An.
by setting
(02,62)
seen to
x
x
B)/~
to establish that there are
B/~b-
(A B)/~, with these two functions, satisfies the universal property
for the product of Aj~A and B/~D.
Conclude
x
(without
further
work)
that
(A
x
B)/~
=
(A/~A)
x
(B/~D).
Exercises
5.12.
-i
Define the notions
39
oifibered products
and
fibered coproducts,
considered in Example 3.10
objects of the categories Ca>/3, Ca
the
by stating carefully
corresponding universal properties.
(cf.
as
terminal
also Exercise
3.11),
As it happens, Set has both fibered products and coproducts. Define these
objects 'concretely', in terms of naive set theory. [II.3.9, III.6.10, III.6.11]
Chapter
first encounter
Groups,
In this chapter
we
II
introduce groups,
we
observe they form
a
category (called Grp),
study 'general' features of this category: what are the monomorphisms and
the epimorphisms in this category? what is the appropriate notion of 'equivalence
relation' and 'quotients' for a group? does a 'decomposition theorem' hold in Grp?
and
we
and other analogous questions.
In Chapter III we will acquire a similar degree of familiarity with rings and
modules. A more object-oriented analysis of Grp (for example, a treatment of the
famous Sylow theorems, 'composition series',
is deferred to Chapter IV.
or
the classification of finite abelian
groups)
1. Definition of group
Groups and groupoids.
1.1.
Joke
1.1.
Definition:
This is actually
a
A group is
a
groupoid with
a
single object.
j
viable definition, since groupoids have been defined
but most mathematicians would find it ludicrous to
perfectly
already (in Example 1.4.6);
introduce groups in this fashion,
or they will at the very least politely express
effectiveness
of doing so. In order to redeem himself, the
pedagogical
author will parse this definition right away to show what it really says. If * is the
lone object of such a groupoid G,
doubts on the
Homc(*,*)
(because
G is
this set G.
with
an
in G is
=
Autc(*)
groupoid!), and this set carries all the information about G. Call
Then (by definition of category) there is an associative operation on G,
a
identity 1*, and (by definition of groupoid, which
an
isomorphism)
every g G G has an inverse
g~l
says that every
morphism
G G.
41
42
Groups, first
II.
is1:
That is what a group
G with
a set
a
encounter
composition law satifsying
a
few key
axioms, i.e., associativity, existence of identity, and existence of inverses.
1.2. Definition. Now for the official definition. Let G be
with
a
binary operation, that is,
a
'multiplication'
a
nonempty set, endowed
map
:GxG^G.
Our notation will be
(g,h)
=:
gmh
simply gh if the
name of the operation can be understood. The careful reader
have
expected that we should write h g, in the style of what we have done
may
for categories, but this is what common conventions dictate2.
or
G, endowed with the binary operation
if
the
G
operation can be understood) is a group if
simply
Definition 1.2. The set
or
(i)
the operation
is associative, that is,
(V#, h,k
(ii)
there exists
an
G
G)
(gmh)mk
:
identity element
(3eG
(iii)
(briefly, (G,«),
G) (Vg
e
e
gm(hmk)\
=
is,
that
eG for •,
G)
:
g
eG
=
g
=
eG
g\
that is,
every element in G has an inverse with respect to •,
(\/geG)(3heG): g.h
Example
1.3. Since
we
explicitly require G to
by letting G
{e}
way to concoct a group is
function G
x
=
G in this case,
G
so
=
eG
hmg.
=
j
be nonempty, the most economical
be
there is only
a
one
singleton. There is only one
possible binary operation on
G, defined by
e
e := e.
The three axioms trivially hold for this example,
so
is
{e}
equipped with
a
unique
group structure.
This is
usually called the trivial group; purists should call
singleton gives rise to one.
any such group a
trivial group, since every
Example
1.4.
The reader should check
carefully (cf.
Exercise
j
1.2)
that
(Z,+),
(Q>+)> 0&>+) (C,+), and several variations using (for example, the subset
{+1,-1} of Z, with ordinary multiplication) all give example of groups. While
very interesting in themselves, these examples do not really capture at any
>
intuitive level 'what'
these examples
a group
really is, because they
are commutative
are too
special. For example, all
(see §1.5).
j
Prom this perspective, Joke 1.1 is a little imprecise: the group is not the groupoid G, but
rather the set of isomorphisms in G, endowed with the operation of composition of morphisms.
Not without exceptions;
see
for example permutation groups, discussed in §2.
1. Definition of group
Example
43
My readers are likely familiar with an extremely important nonexample, namely the group of invertible, nxn matrices with (say) real
2. We will generally shy away from this class of examples in these early
1.5.
commutative
entries,
n >
chapters, since we will have ample opportunities to think
approach linear algebra (starting in Chapter VI). But we
a
matrix
or two
before then. The reader should
a
b
c
d
bc^ 0, form
with real entries, and such that ad
now
about matrices when
may
we
occasionally borrow
check that 2x2 matrices
a group
under the ordinary matrix
multiplication:
&i
b\\(a<i
C\
d\) \C2 d<i)
(The condition ad
inverse?) Since, for
be
b<i\
(a\a<i + b\c<i a\b<i+b\d<i\
_
'
\C1a2-\-d1C2 C1&2
+
d\d2J
0 guarantees that the matrix is invertible.
^
What is its
example,
Ji)G?KlKiK9Gl
this group is indeed not commutative.
The group of invertible
We will encounter
nxn matrices
more
with real entries is denoted GLn (R).
j
representative examples in §2.
1.3. Basic properties. From the 'groupoid' point of view, the identity
cq would
the group
denote this element:
be denoted 1*. It is not uncommon to omit G from the notation
(when
different symbols rather than e to
on the context. In any case, for any
given group G this element is unique. That is, no other element of G can work as
is
understood)
1 and 0
an
are
and to
use
popular alternatives, depending
identity:
Proposition
1.6.
If h
G G is
an
identity of G, then h
Incidentally, this makes groups pointed sets in
group has a well-defined distinguished element.
Proof. Using first that
cq is an
this argument only
=
uses
eoh
=
cq-
the sense of
identity and then that h
h
(Amusingly,
=
is
Example 1.3.8:
an
identity,
one
every
gets3
eo-
that eo is
a
'left' identity and h is
a
'right'
identity.)
Proposition
then h\ /i2-
1.7.
The inverse is also unique:
ifh\,h2 are
both inverses
of g
in
G,
=
As previously announced,
being
we are
only considering
we
one
may omit
the symbol
for the operation, since for the time
operation. This will be done without warning in the future.
44
II.
Groups, first
encounter
actually follows from Proposition 1.4.2 (by viewing G as the set of
isomorphisms of a groupoid with a single object). The reader should construct
a stand-alone proof, using the same trick, but carefully hiding any reference to
morphims.
Proof. This
Proposition 1.7 authorizes
denoted
One
us to
give
the inverse of g: this is
a name to
usually4
g~l.
more
notational item is in order. The definition of
contemplates the 'product' of
principle have a choice
two
elements;
as to
in
multiplying
a
a group only
of
elements, one
string
the order in which products
are
may
in
executed. For example,
(gi •92)99s
stands for: apply the operation
of this operation and g^\ while
to g\ and g^, and then apply
9i
stands for: apply
this operation.
Associativity
again to the result
•(9299s)
to gi and #3, and then apply it again to g\ and to the result of
tells
us
precisely that the result of the operation on three elements
we perform it.
With this in mind, we are
does not depend on the way in which
authorized to write
9\•92993',
this expression is not ambiguous, by associativity. What about four or more
elements? The reader should have checked in Exercise 1.4.1 that all ways to associate
any number of elements leads to the same result. So we are also authorized to write
things like
9i
92
g3
•9i7''»
this is also unambiguous. The reader should keep in mind, however, that of
the order in which the elements are listed is important: in general,
9i
92
9s
9i7
7^
92
9i
Of course no such care is necessary if all gi
notation
can
then be used5:
g°
=
cq, and for a
n times
It is easy to check that then \/g G G and Vra,
=
n G
gmgn.
:But note the 'abelian' case, discussed in §1.5.
'In the abelian
case one uses
'multiples'; cf. §1.5.
9i7-
coincide; the conventional 'power'
positive integer n
n times
gm+n
9s
course
Z
1. Definition of group
1.4.
45
Cancellation. 'Cancellation' holds in groups. That is,
Proposition
1.8. Let G be
ga
ha => g
=
Proof. Both statements
are proven
and applying associativity.
ga
=
(ga)a~1
ha =>
=>
=
g
Then Va,g,/i G G
a group.
h,
=
ag
ah => g
=
=
h.
by multiplying (on the appropriate side) by a-1
For example,
=
(ha)a~1
=>
g(aa~1)
h(aa~l)
=
=>
gee
=
hec
h.
The proof for the other implication follows the
pattern.
same
Examples of operations which do not satisfy cancellation, and hence do not
define groups, abound. For instance, the operation of ordinary multiplication does
not make the set R of real numbers into a group: indeed, 0 'cannot be cancelled'
problem here is that 0 does not
multiplication. As it happens, this is the only
problem with this example: ordinary multiplication does make
since 1-0
have
an
=
2-0
even
if 1
^
2.
Of course, the
inverse in R, with respect to
R*
into
a group, as
:=
R
{0}
the reader should check immediately.
1.5. Commutative groups. One axiom not appearing in the definition of group
we say that the operation
is 'commutative' if
is commutativity:
gmh
(iv) (Vg,heG):
We
hmg.
=
say that two elements g, h 'commute' if
gh
=
hg. Thus,
in
a
commutative
group any two elements commute.
'Commutative
groups'
contexts, especially
are
important objects: they arise naturally in several
'modules
as
say in Chapter III and
the ring Z' (about which we will have
When they do so, they are usually called
over
beyond).
a
lot to
abelian
groups.
The notation used in
treating abelian groups differs somewhat from the
emphasize the 'Z-module structure', and
standard notation for groups. This is to
helpful when
we
it is
abelian group coexists with other operations—a
situation which
will encounter frequently.
an
Thus, the operation in an abelian group A is, as a rule, denoted by + and
'addition'; the identity is then called 0a; and the inverse of an element a G
called
is denoted —a
(and
is of
course
maybe
should be called the
replaced by 'multiple': 0a
na
=
a +
+ a,
'
'
v
n
=
'opposite'?).
0, and for
a
The 'power' notation
positive integer
+
(—n)a(—a)
=
s
very well not be
a
an
binary operation
n
(—a).
/
v
times
n
times
The reader should keep in mind that at this stage 'na' is
result of applying
is
A
element of A in any reasonable
sense.
a
notation, not the
Indeed, n G Z
Moreover, it may
to two elements n, a of A.
may
very
II. Groups, first encounter
46
well be that
in
na
=
if
ma even
^
n
m, in
spite of the fact that 'cancellation' works
groups.
The
qualifier 'abelian' and the
notation Oa, —cl,
etc.,
mostly used for
are
commutative groups arising in certain standard situations: for example, the notions
of rings, modules, vector spaces are defined by suitably enriching a commutative
group, which is then promoted to 'abelian' for notational convenience.
There are other situations in which commutative groups arise naturally, without
triggering the 'abelian' notation. For example, the group (R*,-) mentioned at the
end of §1.4 is commutative, but its operation is indicated by
(if at all), and its
identity element is written 1.
determining notation.
History, rather than logic, is often the main factor
1.6. Order.
Definition 1.9. An element g of a group G has finite order if6 gn
positive integer n. In this case, the order \g\is the smallest positive
=
gn
e.
=
One writes
\g\
By definition, if gn
precise:
for
e
=
some
such that
if g does not have finite order.
oo
=
for
e
n
some
positive integer
j
\g\<
n, then
n.
One can be
more
Lemma 1.10. If gn
=
Proof. As observed,
n >
exist7
then
a
r <
for
=
\g\by
\g\.Note
gr
is smaller than
>
and
0
\g\(m
n
+
\g\is
a divisor
> 0.
n—
\g\
1)
<
of n.
There must
0,
that
=
gn-\g\.m gn ^M)-m
=
By definition of order, \g\is
r
n, then
definition of order, that is,
such that
m
\g\m
n
positive integer
some
positive integer
r
that is,
e
\g\and gr
.
=
g
.
g-m
=
e
the smallest positive integer such that g^
e. Since
0 necessarily. This
e, r cannot be positive; hence r
=
=
=
says
0
that
is,
n
=
-m,
\g\
proving that
This lemma has the
encourage
n-
is indeed
following
\g\-m,
an
integer multiple of |g|
as
claimed.
immediate and useful consequence, which
we
the reader to keep firmly in mind:
Corollary 1.11. Let g be
gN
Of
n
=
an
=
e
element
<^=>
of finite order,
N is
a
and let N e Z. Then
multiple of \g\.
following prescription as ng = 0.
surreptitiously using fairly sophisticated information
about Z, namely the 'division algorithm', hence essentially the fact that Z is a Euclidean domain!
course in an abelian group we would write the
Purists may object that here
we
are
justice. We may as well
have already acquired a
thorough familiarity with the operations of addition and multiplication among integers. Shame
This is material that will have to wait until Chapter V to be given some
be open about it and admit that yes, we are assuming that the readers
on
us!
1. Definition of group
47
Definition 1.12. If G is finite as a set, its order
we
write
|G|
Cancellation implies that
if
|G|
=
\G\is
the number of its elements;
if G is infinite.
oo
=
j
\g\< \G\for
finite, consider the |G|
oo; if G is
all g G G. Indeed, this is vacuously true
+ 1 powers
g°=e,g,g*,g\...,gM
of g. These cannot all be distinct; hence
0<i<j<\G\ such
(3iJ)
By
cancellation
(that is, multiplying
on
the right by
9J-{
showing \g\< (j
We will
-
soon
i)
<
that
=
g{
gj.
=
g~l)
c,
\G\.
be able to formulate
the relation between the order of
a
much
more
precise
statement
concerning
and the order of its elements: if g G G
and |G| is finite, then the order of g divides the order of G. This will be an
immediate consequence of Lagrange's theorem; cf. Example 8.15.
a group
Another general remark concerning orders is that their behavior with respect
to the operation of the group is not always predictable: it may very well happen
that g, h have finite order in a group G, and yet \gh\ oo, or \gh\ your favorite
=
positive integer: work
On the other
the extreme
case
Proposition
hand, the situation is more constrained if g and h commute. In
h, it is easy to obtain a very precise statement:
in which g
=
1.13. Let g G G be
order Vra > 0, and in
an
element
of finite
order.
Then g™
has finite
facft
lcm(ra, |#|)
\gr>
elementary properties of gcd and 1cm:
to prove that
|gm|
cm
=
\g\
gcd(ra,|#|)'
cm^\9\)an(j
Proof. The equality of the two numbers
only need
=
out Exercise 1.12 and Exercise 2.6 if you don't believe it.
lcm(a,6)
=
ab/ gcd(a,b)
1^1
follows from
for all
a
and b. So
we
v^
The order of g™
is the least positive d for which
9md
that is
(by Corollary 1.11)
is the smallest
multiple of
=
e,
for which md is
a
which is also
a
m
multiple of \g\.In other words, ra|gm|
multiple of \g\:
ra|#m| =lcm(ra,|#|).
The stated formula follows immediately from this.
In general, for commuting elements,
Proposition
1.14.
If gh
=
hg, then \gh\divides lcm(|g|, |ft|).
The notation 1cm stands for 'least common multiple'. We are also assuming that the reader
is familiar with simple properties of gcd and 1cm.
II. Groups, first encounter
48
Proof. Let
gN
hN
=
=
\g\
=
\h\
=
m,
by Corollary
e
(gh)«
=
n.
If N is any
(gh)(gh)
(gh\
=
multiple of
common
multiple N oi
(gh)
h
gyhh_
gg
m
and n, then
m
lcm(m,n)
=
g"h"
=e
N times
N times
N times
As this holds for every
common
1.11. Since g and h commute,
and n, in particular
e.
The statement then follows from Lemma 1.10.
One cannot say more about \gh\in general, even if g and h commute
see Exercise 1.14 for an important special case.
(Exercise 1.13).
But
Exercises
1.1.
>
groupoid.
in
some
1.2.
a careful proof that every group is the group of isomorphisms of a
In particular, every group is the group of automorphisms of some object
Write
category. [§2.1]
Consider the 'sets of numbers' listed in §1.1, and decide which are made into
groups by conventional operations such as + and •. Even if the answer is negative
>
(for example, (R, •)is not a group), see if variations on the definition
(for example, (R*, •)is a group; cf. §1.4). [§1.2]
of these sets
lead to groups
1.3. Prove that
(gh)~l
1.4. Suppose that
g2
=
=
e
h~lg~l
for all elements g, h of
for all elements g of
a group
G;
a group
G.
prove that G is
commutative.
'multiplication table' of
multiplications g h:
1.5. The
a group
is
an array
e
h
e
e
h
9
9
gmh
identity element. Of
compiling the results of all
course the table depends on the order in which
listed in the top row and leftmost column.) Prove that every row
and every column of the multiplication table of a group contains all elements of the
group exactly once (like Sudoku diagrams!).
(Here
e
is the
the elements
are
2.
Examples of groups
49
Prove that there is only one possible multiplication table for G if G has
exactly 1, 2, or 3 elements. Analyze the possible multiplication tables for groups
with exactly 4 elements, and show that there are two distinct tables, up to
1.6.
-i
Use these tables to prove that all groups with < 4
reordering the elements of G.
elements are commutative.
(You
but
are
welcome to analyze groups with 5 elements using the
enough about
you will soon know
same
technique,
groups to be able to avoid such brute-force
approaches.) [2.19]
1.7. Prove
1.8.
Let G be
-i
U9eG9
=
1.11.
Corollary
a
finite group, with
exactly
element
one
/ of
order 2. Prove that
f- [4-16]
a finite group, of order n, and let m be the number of elements
of
order
mis odd. Deduce that if n is even, then
g e G
exactly 2. Prove that n
2.
G necessarily contains elements of order
1.9.
Let G be
Suppose the order of
1.10.
g is odd. What can you say about the order of
1.11. Prove that for all g, ft in a group
\g\for all a, g in
g2l
G, \gh\ \hg\.(Hint: Prove that |aga_1|
=
=
G.)
1.12. > In the group of invertible 2x2
matrices, consider
-G "J). »-(-! -!)•
Verify
that
\g\ 4, |ft|
=
1.13. > Give
even
an
3, and \gh\
=
=
gcd(\g\,|ft|)
can you say
=
[§1.6]
example showing that \gh\is
if g and ft commute.
1.14. > As a
oo.
not
necessarily equal
lcm(|g|, |ft|),
[§1.6, 1.14]
counterpoint to Exercise 1.13, prove that if g
then
1, then \gh\ \g\
|ft|. (Hint: Let N \gh\\
=
about this
to
=
and ft commute and
gN
=
(h~l)N.
What
element?) [§1.6, 1.15, §IV.2.5]
a commutative group, and let g G G be an element of maximal
that is, such that if ft G G has finite order, then |ft| < \g\.Prove that
in fact if ft has finite order in G, then |ft| divides \g\.(Hint: Argue by contradiction.
If |ft| is finite but does not divide \g\,
then there is a prime integer p such that \g\
1.15.
-i
Let G be
finite order,
=
pmr, |ft|
=
pns, with
r
compute the order of
and
s
relatively prime
gpmh8.) [§2.1,
to p and m < n. Use Exercise 1.14 to
4.11, IV.6.15]
2. Examples of groups
2.1. Symmetric groups. In §1.4.1 we have already observed that every object A
of every category C determines a group, called Autc(-A), namely the group of
automorphisms of A. In a somewhat artificial sense it is clear that every group arises in
this fashion
(cf.
Exercise
1.1);
this fact is true in
more
become apparent when we discuss group actions
and Exercise 9.17.
'meaningful'
(§9):
ways, which will
cf. especially Theorem 9.5
II. Groups, first encounter
50
In any case, this observation provides the reader with
an
infinite class of very
examples:
important
Definition 2.1. Let A be a set. The symmetric group, or group of permutations
of A, denoted Sa, is the group Autset(A). The group of permutations of the set
{1,..., n}
is denoted
by Sn.
j
The terminology is easily justified: the automorphisms of a set A are the setisomorphisms, that is, the bijections, from A to itself; applying such a bijection
amounts
precisely
to
permuting ('scrambling') the elements of A. This operation
change it (as a set), hence
may be viewed as a transformation of A which does not
a
'symmetry'.
The groups
Sa are famously large: as the reader checked in Exercise 1.2.1,
n\. For example, |SVo| > 10100, which is substantially larger than the
estimated number of elementary particles in the observable universe.
\Sn\
=
Potentially confusing point: The various conventions clash in the way the
operation in Sa should be written. From the 'automorphism' point of view, elements of
Sa are functions and should be composed as such; thus, if f,g G Sa
Autset(^4),
=
then the 'product' of
/
and g should be written g
(VpeA):
as
o
and should act
/
as
follows:
gof(p)=g(f(p)).
But the prevailing style of notation in group theory would write this element
fg, apparently reversing the order in which the operation is performed.
Everything would fall back into agreement if we adopted the convention of
functions
writing
after the elements on which they act rather than before: (p)f rather
than f(p). But one cannot change century-old habits, so we have no alternative
but to live with both conventions and to state
any
carefully
which
one we are
using
at
given time.
Contemplating the groups Sn for small values of n is an exercise of inestimable
course S\ is a trivial group; 52 consists of the two possible permutations:
value. Of
r 11-» l
i
(2^2
~
which
we
could call
e
ee
In practice
permutation group,
=
^
(2^1
and
(identity)
r 11-» 2
<
and
~
ff
/ (flip),
=
ef
e,
with operation
=
fe
=
f.
give a new name to every different element of every
have to develop a more flexible notation. There are in fact several
for this; for the time being, we will indicate an element a G Sn by
we cannot
so we
possible choices
listing the effect of applying a underneath the list 1,..., n,
elements e, / in #2 may be denoted by
-o n. t-n
V2
1
This
2
'
J
as a
matrix9. Thus the
2
1
is only a notational device—these
matrices should not be confused with the matrices
appearing in linear algebra.
2.
Examples of groups
51
style, S3
In the same notational
f/1
2
3Wl
\\1 3y
2
'
^2
consists of
2
3\ A
2
3\ A
2
3\ A
2
3\ A
2
1
'
2
'
3
'
1
'
3
3J ^3
1/
VI
2y V3
2y
For the multiplication, we will adopt the sensible (but not very
convention mentioned above and have permutations act 'on the right':
3\\
*
\2
iy J
standard)
thus, for example,
and similarly
2
That
(2
3) (3
1
=
2j
1
3
3'
(2
3) (3
1
=
1
2j
2
is,
A
2
3\/l 2
V2
1
[3
3;
3\_/l
2
3\
3
2)
~
1
vi
27
both sides of the equal sign act in the
3.
The
reader
should
now
check that
1, 2,
since the permutations
on
A
2
V3
1
3\ A
2
2y ^2
1
3\
A
2
3\
\3
2
l)
same
way
on
_
~
3)
'
That is, letting
A
2
3\
\2
1
3)
_
_
X~
y
'
~
(I
2
3\
\3
1
2;
'
then
yx
showing that the operation
S3 is
^
xy,
in S3 does not
satisfy the commutative axiom. Thus,
immediately realize that in fact Sn is
noncommutative group; the reader will
noncommutative for all n > 3.
a
While the commutation relation does not hold, other interesting relations do
hold in S3. For example,
x2
showing that S3
and 3
(the
y3
e,
=
e,
(the identity e), 2 (the element x),
(Incidentally, this shows that the result
commutativity hypothesis.) Also,
contains elements of order 1
element
y);
cf. Exercise 2.2.
of Exercise 1.15 does require the
as
=
A
2
3\
yX={3
2
l)=Xy
2
the reader may check. Using these relations, we see that every product of any
x and y, xllyl2xl3yu
•,may be reduced to a product xlyj with
assortment of
0<i<l,0<<7<2, that is,
e,
to one of the six elements
y,
y2,
x,
xy2
xy,
:
for example,
y VV
=
(y3)2y(x2fxy3y2
=
(yx)y2
=
(xy2)y2
=
xy3y
=
xy.
II. Groups, first encounter
52
On the other hand, these six elements are all distinct—this
may be checked by
cancellation and order considerations10. For example, if we had xy2
y, then we
would get x
and
this
cannot
be
since
the
relations
tell us
y~x by cancellation,
=
=
that
x
y~l
has order 2 and
has order 3.
The conclusion is that the six products displayed above must be all six elements
ofS3:
53
=
{e,x,y,xy,y2,xy2}.
In the process we have verified that S3 may also be described
'generated' by two elements x and y, with the 'relations' x2
e, y3
=
More generally,
may be written as a
a
subset A of
a group
as
=
the group
e, yx
=
xy2.
G 'generates' G if every element of G
product of elements of A and of inverses of elements of A. We
will deal with this notion more formally in §6.3 and with descriptions of groups in
terms of generators and relations in §8.2.
is a transformation which preserves a
loose
to
talk about automorphisms, when we may
just
way
be too lazy to define rigorously the relevant category. As automorphisms of objects
of a category, symmetries will naturally form groups.
2.2.
Dihedral groups. A
structure.
This is of
course
'symmetry'
a
One context in which this notion may be visualized vividly is that of 'geometric
figures' such as polygons in the plane or polyhedra in space. The relevant category
as follows: let the objects be subsets of an ordinary plane R2 and
let morphisms between two subsets A, B consist of the 'rigid motions' of the plane
(such as translations, rotations, or reflections about a line) which map A to a subset
could be defined
of B. A rigorous treatment of these notions would be too distracting at this point,
so we will appeal to the intuition of the reader, as we do every now and then.
From this perspective, the 'symmetries' of a subset of the plane
motions which map it onto itself; they clearly form a group.
are
the rigid
The dihedral groups may be defined as these groups of symmetries for the
regular polygons. Placing the polygon so that it is centered at the origin (thereby
excluding translations as possible symmetries), we see that the dihedral group for
regular n-sided polygon consists of the n rotations by 27r/n radians about the
origin and the n reflections about lines through the origin and one of the vertices.
Thus, the dihedral group for a regular n-sided polygon consists of 2n elements;
we will denote11 this group by the symbol D2n'
a
Again, contemplating these groups, at least for small values of n, is a
simple way to relate the dihedral groups to the symmetric
groups of §2.1, capturing the fact that a symmetry of a regular polygon P is
determined by the fate of the vertices of P. For example, label the vertices of an
equilateral triangle clockwise by 1, 2, 3; then a counterclockwise rotation by an
angle of 27r/3 sends vertex 1 to vertex 3, 3 to 2, and 2 to 1, and no other symmetry
wonderful exercise. There is a
of the triangle does the
same.
by explicit computation of the corresponding permutations,
trying to illustrate the fact that the relations are 'all we need to know'.
Unfortunately there does not seem to be universal agreement on this notation: some
references use the symbol Dn for what we call D2n here.
It may of course also be checked
but
we are
2.
Examples of groups
53
Visually, this looks like
In other words,
we can
associate with the counterclockwise rotation of
an
equilateral
triangle by 27r/3 the permutation
(3
Such
a
labeling defines
a
2je53-
1
function
As
S3;
-
further, this function is injective (since a symmetry is determined by the
permutation of vertices it induces). In fancier language, which we will master in due time,
we say
that Dq acts
(faithfully)
It is clear that this
can
on
the set
{1,2,3}.
(for example,
be done in several ways
we
could label the
ways). However, any such assignment will have the property
that composition of symmetries in D^n corresponds to composition of permutations
in 5n; the reader should carefully work this out for several examples involving
54.
Dq -?
S3 and D8 -?
vertices in different
A concise way
to describe the situation is that these functions are
(group)
homomorphisms (cf. §4).
Since both Dq and S3 have 6 elements and the function
Dq
53 given above is injective, it must also be surjective. Thus there are bijective
homomorphisms between Dq and 53; we say that these groups are isomorphic
(cf. §4.3). We will study these concepts
As
by
an
alternative
(and
more
very
abstract)
carefully
in the next several sections.
conclusion, denote
by x the reflection about
equilateral triangle:
way to draw the same
y the counterclockwise rotation considered above and
the line through the center and vertex 3 of
Reflecting
twice gives the
identity,
x2
as
=
does rotating three times; thus
e,
y3
yx (rotating counterclockwise by
is the same symmetry as xy2 (flipping
Further,
line)
our
=
27r/3, then nipping about the slanted
first, then rotating clockwise by 2tt/3).
That is,
yx
=
e.
xy2.
II. Groups, first encounter
54
In other words, the group Dq is also
x2
generated by
subject
two elements x,y,
to the
found for 53. Since D6 and S3
e, y3
e, yx
xy2—precisely
admit matching descriptions in terms of generators and relations (that is, matching
presentations; cf. §8.2), 'of course' they are isomorphic.
relations
=
=
as we
=
2.3. Cyclic groups and modular arithmetic.
Consider the equivalence relation on Z defined by12
(Va, b
G
Z)
:
=
a
This is called congruence modulo
b
mod
n
Let
^=>
be
n
a
(b
n
positive integer.
a).
We have encountered this relation already, for
n.
2, in Example 1.1.3. It is very easy to check that it is an equivalence relation,
for all n; the set of equivalence classes is often denoted by Zn, Z/(n), Z/nZ, or Fn.
We will opt for Z/nZ, which is not preempted by other notions13. We will denote
by [a]n the equivalence class of the integer a modulo n, or simply [a] if no ambiguity
n
=
arises.
The reader should check
that
carefully
Z/nZ
consists of
exactly
n
elements,
namely
[0]n,
We
Z/nZ.
[l]n
can use the group structure on Z to induce an (abelian) group structure on
In order to do this, we define an operation + on Z/nZ, by setting Va, b G Z
[a]
Of
[n].
,...,
[6]:=[a
+
+
6].
have to check that this prescription is well-defined; luckily, this is
the
very easy:
following small lemma does the job, as it shows that the result of
the operation does not depend on the representatives chosen for the classes.
course
we
Lemma 2.2.
If a
a' modn and b
=
(o
Proof. By hypothesis
6)
+
-
b' modn, then
{a'
=
and
n\(a' a)
(a'
=
a)
+
n
mod
(br
6); therefore 3k,£
n.
G Z such that
(b' -b)= in.
kn,
=
b')
Then
(a'
proving
that
+
n
b')
-
(a
divides
+
6)
(a!
+
(a'
=
b')
-
(a
a)
+
+
6),
(bf -b)
as
=
kn + £n
=
(k
+
£)n,
needed.
Therefore, we have a binary operation + on Z/nZ. It is immediately checked
that the resulting structure is a group. Associativity is inherited from Z:
([a]
+
and
as
[6])
+
so are
[c]
=
[a+ 6]
+
the identity
[c]
=
[0]
[(a+ 6)
+
c]
and 'inverse'
=
[a+ (6 + c)]
=
[a]
The notation n
3When n =
different concept;
p is
+
[6]
=
[a
+
6]
[b + c]
=
=
[6 + a]
=
[6]
+
([6]
+
[c]);
Z/nZ
are
commutative,
[a].
m stands for n is a divisor of m; that is, m
VIII. 1.19.
+
=
=
nk for
some
prime, Zp is the official notation for 'p-adic integers', which
see Exercise
[a]
—[a] [—a].
It is also immediately checked that the resulting groups
the abelian-style notation suggests:
[a]
+
integer k.
completely
are a
2.
Examples of groups
55
reader, who should
We trust that this material is not new to the
check all these assertions
in any
case
carefully.
The abelian groups thus obtained, together with Z, are called cyclic groups;
a popular alternative notation for the group (Z/nZ, +) is Cn.
This is adopted
when
one
wants
to
use
the
rather
than
'additive'
notation;
'multiplicative'
especially
thus
we can say
that Cn is generated by
groups are
Cyclic
element x, with the relation xn
tremendously important, and
later sections. For the time being
the group, in the
sense
we
will
come
=
e.
back to them in
record the fact that the element
we
[l]n
generates
one
Z/nZ
e
that every other element may be obtained
m > 0 is an integer, then
as a
multiple of this element. For example, if
[m]n
=
[1
+
...
+
l]n
[l]n
=
m
+
+
v
v
[l]n
=
[l]n.
m
/
v
times
m
times
phrase this fact by observing that the order of [l]n in Z/nZ
n multiples 0
(n 1) [l]n must all be
[l]n, 1 [l]n,
distinct, and hence they must fill up Z/nZ.
Equivalently,
is
we may
this implies that the
n:
Proposition
2.3.
...,
The order
IHn|
Proof. If
[m]n
=
m
|
[l]n
n
m, then
=
and more
generally
—77 v
If
n
a consequence,
does not divide m, observe again that
the order of the group.
the order of every element of Z/nZ divides n
We will see later (Example 8.15) that this is
general features of the order of elements
Corollary
if n\m,
gcd(ra,n)
[0]n.
=
is 1
and apply Proposition 1.13.
Remark 2.4. As
|Z/nZ|,
[m]n
Z/nZ
in
of [m]n
2.5.
The class
in any finite group.
[m]n generates Z/nZ if and
=
a
j
i/gcd(ra,n)
only
=
1.
This simple result is quite important. For example, if n
p is a prime integer,
it shows that every nonzero class in the group Z/pZ generates it. In any case,
it allows us to construct more examples of interesting groups. The reader should
=
(or recall;
Z/nZ, given by
check
cf. Exercise
2.14)
that there also is
[o]n [b]n
'
:=
a
well-defined multiplication
on
[ab]n.
This operation does not define a group structure on Z/nZ: indeed, the class [0]n
does not have a multiplicative inverse. On the other hand, for any positive n denote
1:
by (Z/nZ)* the subset of Z/nZ consisting of classes [m]n such that gcd(ra,n)
=
(Z/nZ)*
:=
{[m]n
This subset is clearly well-defined:
gcd(m/,n)
=
1
(Exercise 2.17),
so
G
if
Z/nZ | gcd(ra,n)
m
=
=
1}.
m' mod n, then
gcd(m, n)
the defining property of classes in
independent of representatives.
Proposition
2.6.
Multiplication makes (Z/nZ)*
into a group.
=
1
<^=>
(Z/nZ)*
is
II. Groups, first encounter
56
1, then
Simple properties of gcd's show that if gcd(rai,n)
gcd(ra2,n)
1.
if
a
divided
both
n
and
mi?7i2, then
prime integer
gcd(raira2, n)
(For example,
Proof.
=
=
=
it would necessarily divide rai or m2, and one of the two gcd's would not be 1.)
Therefore, the product of two elements in (Z/nZ)* is an element of (Z/nZ)*, and
does define
a
binary operation
(Z/nZ)*
(Z/nZ)*
x
It is clear that this operation is associative
in Z); [l]n is an element of (Z/nZ)* and is an
so
all
we
have to check is that elements of
(Z/nZ)*.
-?
(because multiplication
identity with respect
to
is associative
multiplication;
have multiplicative inverses in
(Z/nZ)*
(Z/nZ)*.
This follows from
Corollary
additive group Z/nZ, and hence
(3a
G
If
2.5.
1, then [m]n generates the
=
multiple of [m]n
some
Z)
gcd(ra,n)
:
[m]n
a
must
equal [l]n:
[l]n;
=
this implies
NnMn
Therefore
[m]n
does have
will
that
gcd(a,n)
verify
a
=
For instance, [8] 15 has
multiplicative
=
[l]n.
inverse in
Z/nZ, namely [a]n.
The reader
1, completing the proof.
a
multiplicative
inverse in
(Z/15Z)*. Tracing
the
argument given in the proof,
2-8 +
and hence [2] 15
For
We will
(-l)-15
=
l,
the multiplicative inverse of
[8] 15 is [2] 15.
n
p a positive prime integer, the group ((Z/pZ)*, •) has order (p
1).
have more to say about these groups in later sections (cf. Example 4.6).
[8] 15
=
[l]i5:
=
Exercises
2.1.
-1
One
the entry
can
associate annxn matrix Ma with
a
permutation
a G
Sn by letting
be 1 and letting all other entries be 0. For example, the matrix
corresponding to the permutation
at
(i, cr(i))
<T=(s
1
2)e53
would be
Ma
=
/o
0
i\
0
1
0
\i
0
.
0/
Prove that, with this notation,
Mar
for all
a, r G
Sn, where the product
on
=
MaMr
the right is the ordinary product of matrices.
[IV.4.13]
2.2. > Prove that if d < n, then
Sn contains elements of order d. [§2.1]
Exercises
57
2.3. For every positive integer
2.4. Define
for
a
triangle
a
n
find
an
element of order
n
in S®.
S4 by labeling vertices of a square, as we did
homomorphism D8
§2.2. List the 8 permutations in the image of this homomorphism.
in
2.5. > Describe generators and relations for all dihedral groups D<in. (Hint: Let x
be the reflection about a line through the center of a regular n-gon and a vertex, and
by 27r/n. The
let y be the counterclockwise rotation
group D^n will be
generated
and y, subject to three relations14. To see that these relations really determine
D271, use them to show that any product xHy7,2x7,3yu
equals x%yi for some i, j
by
x
with 0
1, 0
< i <
<
j
<
n.) [8.4, §IV.2.5]
2.6. > For every positive
integer
n construct a
\g\ 2, \h\ 2, and \gh\
such that
=
=
2.7. Find all elements of D^n that
of
n
plays
Verify carefully
2.10. Prove that
2.11.
>
For
containing two elements
1, D2n will do.) [§1.6]
g, h
n >
commute with every other element.
(The parity
role.)
a
2.8. Find the orders of the groups of
2.9.
group
(Hint:
n.
=
symmetries of the five 'platonic solids'.
that 'congruence mod rC is
Z/nZ
consists of
precisely
an
equivalence relation.
elements.
n
Prove that the square of every odd integer is congruent to 1 modulo 8.
[§VII.5.1]
2.12. Prove that there are no
studying the equation
to be
even.
Letting
a
[a]\
=
+
2&, b
integers a,b,c such that a2
[6]|
=
3[c]|
=
2£,
c
=
in
2.13. > Prove that if
gcd(m,n)
=
1, then there exist integers
bn
(Use Corollary 2.5.) Conversely, prove that
1. [2.15, §V.2.1, V.2.4]
gcd(m,n)
6, then
-1
would all have
=
3m2. What's
=
a
and b such that
1.
if am + bn
=
1 for some
integers
a
and
=
2.14. > State and prove an
2.15.
c
2m, you would have k2 -\-£2
am +
Z/nZ
Z/4Z,
3c2. (Hint: By
=
that?)
wrong with
on
+ b2
show that a, b,
is
a
Let
analog of Lemma 2.2, showing that the multiplication
well-defined operation.
n >
0 be
an
[§2.3, §111.1.2]
odd integer.
Prove that if
gcd(m, n)
Prove that if
gcd(r,2n)
=
1, then gcd(2m +
=
1, then
Conclude that the function
[m]n
n,
gcd(^,n)
[2m
+
2n)
=
n]2n
1.
is
=
1.
(Use
Exercise
2.13.)
(Ditto.)
a
bijection between (Z/nZ)*
and (Z/2nZ)*.
Two relations are evident. To 'see' the third one, hold your right hand in front of and
away from you, pointing your fingers at the vertices of an imaginary regular pentagon. Flip the
pentagon by turning the hand toward you; rotate it counterclockwise w.r.t. the line of sight by
72°; flip it again by pointing it away from you; and rotate it counterclockwise a second time. This
returns the hand to the initial position. What does this tell you?
II. Groups, first encounter
58
The number
</>(n)
proved that if
given later on
of elements of
(cf.
Exercise
2.16. Find the last
=
digit of 123823718238456. (Work
2.17. > Show that if
1.
(Z/nZ)* is Euler's (j)-function. The reader has just
<£(2n)
<£(n). Much more general formulas will be
V.6.8). [VII.5.11]
is odd, then
n
m
=
m' mod n, then
gcd(m, n)
in
=
Z/10Z.)
1 if and
only if gcd(ra', n)
=
[§2.3]
an injective function Z/dZ
Sn preserving the operation,
that is, such that the sum of equivalence classes in Z/nZ corresponds to the product
of the corresponding permutations.
2.18. For d <n, define
2.19. > Both
(Z/5Z)*
and
multiplication tables, and prove that
(Cf. Exercise 1.6.) [§4.3]
3.
The category
Groups
will be the
(Z/12Z)*
no
consist of 4 elements. Write their
re-ordering of the elements
will make them match.
Grp
objects of the category Grp.
In this section
we
define the
morphisms in the category and deal with simple properties of these morphisms.
3.1. Group homomorphisms. As we know,
types of information: a set G and an operation15
mG : G x G
satisfying certain properties. For two groups
a
consists of two distinct
group
G
-?
(G,mc)
and
(i7, ra#),
a group
homo-
morphism
(G,mG)
(usually given the
(f
is first of all
a
function
(H,mH)
-?
:
same name, cp
in this
case)
between the
underlying sets; but this function must 'know about' the operations mo
on H. What is the most natural
Note that the set-function
cp
:
G
H determines
(tpx<p):GxG^H
we
on
G,
ra#
requirement of this sort?
xH
a
function
:
could invoke the universal property of products to obtain this function (cf.
3.1), but since we are dealing with sets, there is no need for fancy language
Exercise
define
here—just
the function by
(V(a,b)
There is
an
e G x
G)
:
(y>
diagram combining all these
x
<p)(a,b)
=
(<p(a),<p(6)).
maps:
GxG-t—^HxH
G
5;
>H
In §1, me was denoted •;
here we need to keep track of operations on different groups, so
for a moment we will use a symbol recording the group (and evoking 'multiplication').
3. The category
59
Grp
What requirement could be
more
natural than asking that this diagram commute?
Definition 3.1. The set-function (p
diagram displayed above commutes.
This is
a
:
H defines
G
a group
homomorphism if the
j
seemingly complicated way of saying something simple: since cp and
commutativity means the following. For all a,b e G,
to travel through the diagram give
mc rr^H are functions of sets,
the two ways
(a, &)¦
(a, b)
xp(a- b)
b\
a-
>(<p(a),<p(b))
I
(p(a) (p(b)
for both operations: in G on the left, in H on the right.
the diagram means that we must get the same result in both
cases; therefore, Definition 3.1 can be rephrased as
where
we
now
write
Commutativity of
the
set-function
:
(p
G
H is
a group
(p(a b)
-
=
homomorphism if Va, 6
G G
(p(a) (p(b).
-
In other words, (p is a homomorphism if it 'preserves the structure'. This may
more familiar to our reader. As usual, the reason to bring in diagrams (as
sound
in Definition
3.1)
is that this would make it easy to transfer part of the discussion
to other categories.
If the context is clear,
one
may
simply
write
'homomorphism', omitting the
qualifier 'group'.
3.2.
Grp: Definition.
For G, H
groups16
we
define
HomGrp(G,tf)
to be the set of group
homomorphisms G
H.
If G, i7, K are groups and (p : G
K are two group
i7, ip : H
o
K is & group
it
is
to
check
that
the
homomorphisms,
composition ip
easy
(p : G
from
the
of
this
amounts
to
view,
homomorphism:
diagram point
observing that
the 'outer rectangle' in
^>
GxG—+HxH—¥KxK
TpOcp
We
operation.
are
yielding
to the usual abuse of
language that lets
us omit
explicit mention of the
II. Groups, first encounter
60
must commute if the two 'inner
rectangles'
commute.
For
arbitrary
a, b G
G, this
means,
(1>
where
(1)
o
<p)(a b)
=
1>{ip(a b)) (=} 1>{ip(a) <p(b)) (=} ^(a)) ^(b))
holds because cp is
a
homomorphism and (2) holds because ^
is
a
homo-
morphism.
Further, it is clear that composition is associative (because it is for set-functions)
and that the identity function idc : G —>
G is a group homomorphism. Therefore,
Grp
is indeed
a
category.
3.3. Pause for reflection. The careful reader might raise
prescribe the
specific function
axioms
is,
a
existence of
an
identity element
an
objection: the group
'inverse', that
eG and of an
t(g):=g-\
ta-.G^G,
Shouldn't the definition of morphism in Grp keep track of this type of data? The
we have given only keeps track of the multiplication map mG.
definition
The
reason
why
we can
get away with this is that preserving
m
automatically
preserves e and i\
Proposition
(p(eG)
3.2. Let cp
:
H be
G
a group
homomorphism. Then
=eH;
In terms of
diagrams, the second
assertion amounts to saying that
must commute.
Proof. The first item follows from the definition of
cancellation: since en
=
e-H
eH
which implies en
=
homomorphism and
e#,
<p(€g) <p(eG)
=
=
vi^G eG)
=
<p{eG) ^(ec),
<p(eG) by 'cancelling (p(eG)\
The proof of the second assertion is similar: \/g G G,
wig'1) <p(g)
implying (p(g~x)
=
=
wig'1 g)
<^(g)_1 by
=
<p(eG)
cancellation.
=
eH
=
^(g)'1 <p(g),
3. The category
61
Grp
3.4. Products et al. The categories Grp and Set look rather alike at first:
a group, we can 'forget' the information of multiplication and we are left with
given
and
is
a
set;
homomorphism,
forget that it preserves the multiplication
left with a set-function. A concise way to express this fact is that there
a group
we are
given
we can
'functor' Grp -^ Set
functors more extensively
a
in fact
(called
much later
on,
'forgetful' functor); we
starting in Chapter VIII.
a
will deal with
However, there are important differences between these two categories. For
example, recall that Set has a unique initial object (that is, 0) and this is not the
same as the final objects (that is, the singletons). Also recall that a trivial group
is a group consisting of a single element (Example 1.3).
Proposition
3.3.
Trivial groups
both initial and final in Grp.
are
This makes trivial groups 'zero-objects' of the category Grp.
Proof. It should be clear that trivial groups are final: there is only one function
from a set to a singleton, that is, the constant function; this is vacuously a group
homomorphism.
To see that trivial groups are initial, let T
{e} be a trivial group; for any
G by T(e)
cq- This is clearly a group homomorphism,
group G, define (p : T
and it is the only possible one since every group homomorphism must send the
identity to the identity (Proposition 3.2).
=
=
in fact, the product of two groups G, H
underlying sets.
To see this, we need to define a multiplication on G x H\ the catchword here
is componentwise: define the operation in G x H by performing the operation on
Here
is
is
similarity: Grp has products;
a
supported
on
the product G
x
H of the
each component separately. Explicitly, define Vgi,#2
(9uhi) (g2,h2)
:=
G, Vfti,/i2
H
(0102,^2).
This operation defines a group structure on G x H: the operation is associative, the
identity is (e<3,e#), and the inverse of (g,h) is (g_1,/i_1). All needed verifications
are
left to the reader. The group G
x
H is called the direct product of the groups G
and H.
Also note that the natural projections
GxH
H
G
(defined
as
set-functions
as
in
§1.2.4)
are group
homomorphisms: again, this follows
immediately from the definitions.
Proposition
in Grp.
3.4.
With operation
Proof. Recall (§1.5.4) that this
defined componentwise, G
means
that GxH satisfies the
x
H is
following
property: for any group A and any choice of group homomorphisms (fo
a
product
universal
A
G,
II. Groups, first encounter
62
:
(fH
A
^>
H, there exists
a
unique
group
homomorphism
cpo x cpn
making the
diagram
commute.
Now, a unique set-function cpo x ¥h exists making the diagram commute,
because the set G x H is a product of G and H in Set. So we only need to check
that (pa x (pn is a group homomorphism, and this is immediate (if cumbersome):
Va,6
e
A,
(pG x
(pH{ab)
=
(<PG(a>b),(pH(a>b))
=
=
(<PG(a>)<PG(b),(pH(a)(pH(b))
(<PG(a>),<PH(a))(<PG(b),<PH(b))
=
(<pg
x
<ph(o))(<pg
x
<ph(6)).
D
What about coproducts? They do exist in Grp, but their construction requires
handling presentations more proficiently than we do right now, and general
coproducts of groups will not be used in the rest of the book; so the reader will have to
deal with them on his or her own. For an important example, see Exercise 3.8;
will show up in Exercises 5.6 and 5.7, since
free groups are themselves
coproducts. The reader will finally produce the coproduct of any
two groups explicitly in Exercise 8.7. For now, just realize that the disjoint union,
which works as a coproduct in Set (Proposition 1.5.6), is not an option in Grp: there
is no reasonable group structure on the disjoint union. The coproduct of G and H
in Grp is denoted G * H and is called the free product of G and H.
more
particular cases of
3.5. Abelian groups. The category Ab whose objects are abelian groups, and
whose morphisms are group homomorphisms, will in a sense be more important for
us than the category Grp. In many ways, as we will see, Ab is a 'nicer' category17
than Grp.
products
Again the trivial groups are both initial and final (that is, 'zero') objects;
exist and coincide with products in Grp. But here is a difference: unlike
in Grp, coproducts in Ab coincide with products. That is, if G and H are abelian
GxH,
groups, then the product GxH (with the two natural homomorphisms G
H
GxH) satisfies the universal property for coproducts in Ab (cf. Exercise 3.3).
When working as
called their direct
There is
even
a
a
coproduct, the product G
sum
x
H of two abelian groups is often
and is denoted G 0 H.
pretty subtlety here, which may highlight the power of the language:
are commutative, the product GxH does not (necessarily) satisfy
if G and H
As
we
will see
in
due time
(Proposition III.5.3), Ab is one instance of a general class of
Z). Unlike Grp, these categories are
(for R
categories of 'modules over a commutative ring FC
=
abelian, which makes them very well-behaved. We will learn two
categories in Chapter IX.
or
three things about abelian
Exercises
63
the universal property for coproducts in
example, see Exercise 3.6.
Grp,
even
if it does in Ab. For
an
explicit
Exercises
3.1. > Let cp
there is
a
:
H be
G
in
morphism
a
a
category C with products. Explain why
unique morphism
(<p
(This morphism
3.2. Let (p
:
G
is defined
^>
H, ip
:
x
(p)
:
G
G
x
explicitly for C
H
=
H
-?
#.
x
Set in
§3.1.) [§3.1, 3.2]
K be morphisms in
^>
consider morphisms between the products G xG, H
a
category with products, and
x
H, KxK
as
in Exercise 3.1.
Prove that
(ilxp)
(This
x
(i/xp)
(il>xil>)((px(p).
=
is part of the commutativity of the
3.3. > Show that if
G, H
G, H be
3.4. Let
is trivial?
(Hint:
3.5. Prove that
abelian groups, then G
are
property for coproducts in Ab
x
§3.2.)
H satisfies the universal
(cf. §1.5.5). [§3.5, 3.6, §111.6.1]
groups, and assume that G
No. Can you construct
Q
in
diagram displayed
is not the direct
a
=
H
x
G. Can you conclude that H
counterexample?)
product of
two nontrivial groups.
product of the cyclic groups C2, C3 (cf. §2.3): C2
Exercise 3.3, this group is a coproduct of C2 and C3 in Ab. Show that
coproduct of C2 and C3 in Grp, as follows:
3.6. > Consider the
find injective homomorphisms C2
arguing by contradiction,
assume
deduce that there would be
S3, C3
C3. By
a
S3;
C3 is a coproduct of C2, C3, and
S3 with certain
homomorphism C2 x C3
that C2
a group
x
it is not
x
properties;
show that there is
no
such homomorphism.
[§3.5]
3.7. Show that there is
a
surjective homomorphism Z
coproduct in Grp; cf. §3.4.)
One can think of Z * Z
relations whatsoever.
Exercise
=
e<3,
a group
study
a
Z
C2
*
C3. (* denotes
with two generators x, y, subject to
general version of such groups in §5;
no
see
5.6.)
3.8. > Define
x2
(We
as
will
*
y3
will obtain
=
an
a group
G with two generators x, y, subject
(only)
to the relations
coproduct of C2 and C3 in Grp. (The reader
description for C2 * C3 in Exercise 9.14; it is
e<3. Prove that G is a
even more concrete
called the modular
3.9. Show that
group.) [§3.4, 9.14]
fiber products
and coproducts exist in Ab.
(Cf.
Exercise
1.5.12.)
II. Groups, first encounter
64
4.
Group homomorphisms
Examples. For any two groups G, H, the set HomGrp(G, H) is certainly
H by sending
nonempty: at the very least we can define a homomorphism G
every element of G to the identity en of H. Thus, HomGrp(G, i7) is a 'pointed set'
(cf. Example 1.3.8).
There is a purely categorical way to think about this: since Grp has zero-objects
{*} (recall that trivial groups are both initial and final in Grp; cf. Proposition 3.3),
there are unique morphisms
4.1.
G-{*},
{*}^H
in Grp; their composition is the distinguished element of HomGrp(G, H) mentioned
above; we will call this element the trivial morphism.
'meaningful' examples can be constructed by considering the groups
S3 defined in §2.2 is a
§2: for instance, the function Dq
will
all
be
instances
of the notion of group action. In
Such
homomorphism.
examples
likely
A
an
'action'
of
a
on
an
of
a category C is a homomorphism
general,
group G
object
More
encountered in
G^ Autc(A);
that is, a group G 'acts'
of that object to itself,
object if the elements of G determine isomorphisms
If C
Set, this
way compatible with compositions.
on an
in
a
=
means that elements of G determine specific permutations of the set A. For
example, the symmetries of an equilateral triangle (that is, elements of Dq) determine
permutations of the vertices of that triangle (that is, permutations of a set with
three
us
elements),
so compatibly with composition; this is what gives
53. We say that Dq 'acts on the set of vertices' of the
and they do
homomorphisms Dq
triangle.
The reader
will have much
can
(and should)
more to say
construct many more
examples of this kind. We
about actions of groups and other algebraic entities in
later sections.
example with a different flavor: the exponential function is a
homomorphism from (R, +) to the group (R>0, •)of positive real numbers, with ordinary
eaeb. A similar (and very important)
multiplication as operation. Indeed, ea+h
Here is
an
=
class of examples may be obtained as follows: let G be any group and g G G any
element of G; define an 'exponential map' eg : Z
G by
(VaeZ)
Then eg is (clearly)
if eg is surjective.
a group
:
€9(a):=ga.
homomorphism. The element
g generates G if and
only
One concrete instance of this homomorphism (in the abelian environment, thus
using multiples rather than powers) is the 'quotient' function 7rn : Z
Z/nZ,
a \-+ a
[l]n
=
[a]n
:
with the notation introduced above, this is e[i]n. This function is surjective; hence
[l]n generates Z/nZ. In fact, as observed in §2.3 (Corollary 2.5), [m]n generates
Z/nZ
if and only if
gcd(m, n)
=
1.
4.
65
Group homomorphisms
If
|
m
homomorphism
n, there is a
tt^
making
:
Z/nZ
Z/mZ
-?
the diagram
commute: that is,
the reader should check
If m\ and rri2
Z/nZ
to both
since 6
=
carefully
that this function is well-defined
Z/miZ
2-3, there is
a
homomorphism
Z/6Z
(or,
in
Z/2Z
-?
C^
'multiplicative notation', Cq
x
x
Z/3Z
C3). Explicitly,
[0]e
-
([0]2, [0]3),
[1]6
"
([1]2, [1]3),
[2]6
~
([0]2, [2]3),
[3]e
-
([1]2, [0]3),
[4]6
"
([0]2, [1]3),
[5]6
~
([1]2, [2]3).
Note that this homomorphism is a bijection; as we will see in
makes it an isomorphism; in particular, Cq is also a product
One
can concoct a
the function
(Exercise 4.1).
both divisors of n, we have homomorphisms 7rJ^ , 7r^2 from
and Z/m2Z and hence to their direct product. For instance,
are
Z/2Z
Z/4Z
(§4.3),
of C2 and C3
Z/raZ also if
homomorphism Z/nZ
-?
a moment
n
|
ra:
in
this
Grp.
for example,
defined by
[0]2
-
[0]4,
[1]2
-
[2]4
clearly a group homomorphism. Unlike 7r^, this homomorphism is not nicely
compatible18 with the homomorphisms 7rn.
On the other hand, is there a nontrivial group homomorphism (for example)
C4
C7? Note that there are 74
2,401 set-functions from C4 to C7 (cf.
Exercise 1.2.10); the question is whether any of these functions (besides the trivial
homomorphism sending everything to e) preserves the operation. We already know
that a homomorphism must send the identity to the identity (Proposition 3.2), and
that already rules out all but 216 functions (why?); still, it is unrealistic to write
is
=
all of them out
explicitly to see if any is a homomorphism.
The reader should think about this before we spill the beans
in the next
subsection.
Also, note that while 7rJ^ preserves multiplication
does not; that is, it is not a 'ring homomorphism'.
example: [1]2 [1)2
[1]2, but [2]4 [2]4
[0]4.
=
=
as well as sum, this new homomorphism
This is immediately visible in the given
II. Groups, first encounter
66
Group homomorphisms are set-functions
they must preserve many features of the
of this principle: group homomorphisms must
4.2. Homomorphisms and order.
preserving the group structure; as such,
theory.
Proposition 3.2
is
instance
an
preserve identities and inverses. It is also clear that if cp : G
H is a group
homomorphism and
must be an element
g is an element
of finite
cq
indeed, if gn
of finite order in H:
=
<p(g)n
In
order in G, then
for some n >
<P(9n)
=
fact, this observation establishes
=
a more
<p(ea)
precise
=
(p(g)
0, then
eH.
statement:
H be a group homomorphism, and let
Proposition 4.1. Let cp : G
element of finite order. Then \<p(g)\ divides \g\.
Proof. As observed,
(p(g)^
=
e#;
applying Lemma
1.10 gives the statement.
Example 4.2. There are no nontrivial homomorphisms Z/nZ
image of every element of Z/nZ must have finite order, and the
finite order
in
(Z, +)
g G G be an
Z: indeed, the
only element with
is 0.
There are no nontrivial homomorphisms (p : C±
Cj. Indeed, the orders of
elements in C4 divide 4 (cf. Proposition 2.3), and the orders of elements in Cj
divides 7. Thus, the order of each (p(g) must divide both 4 and 7; this forces
e for all g G C4.
j
\<p(g)\ 1 for all g, that is, (p(g)
=
Of
=
the order itself is not preserved: for example, 1 G Z has infinite
[l]n 7rn(l) G Z/nZ has order n (with notation as in §4.1). Order is
course
order, while
=
preserved through isomorphisms,
as we
will
see
in
a moment.
H is (of course)
Isomorphisms. An isomorphism of groups cp : G
isomorphism in Grp, that is, a group homomorphism admitting an inverse
4.3.
ip~l
an
:H^G
homomorphism. Taking our cue from Set, if a homomorphism
isomorphism, then it must in particular be a bijection between the
sets.
underlying
Luckily, the converse also holds:
which is also
of
a group
groups is an
Proposition 4.3.
Let cp
isomorphism of groups if
Proof. One implication
implication,
assume (p :
G
:
H be
G
and only
if
is immediate,
H is
a
a
it is a
homomorphism.
bijection.
group
Then
cp is an
pointed out above. For the other
bijective group homomorphism. As a bijection,
as
(p
has an inverse in Set:
ip~l
simply need to
of i7, and let g\
we
=
^_1(/ii h2)
as
needed.
=
:H^ G;
check that this is a group
<^_1(/ii), g<i
=
(p~1{h2)
^-1(^(#i) <P(92))
'
=
homomorphism. Let h\,/12 be elements
be the corresponding elements of G. Then
^-1(^(#i 92))
'
=9i'92
=
^_1(^i) ^_1(^2)
4.
67
Group homomorphisms
Example
Dq
S3 defined in §2.2 is an isomorphism of groups,
homomorphism. So is the exponential function (R, +)
4.4. The function
since it is a
bijective group
(R>0, •)mentioned in §4.1. If the exponential function eg : Z
an element g G G (as in §4.1) is an isomorphism, we say that G is
G determined
by
'infinite cyclic'
an
group.
The function
7rf
7rf
x
:
C2
Cq
x
C3 studied ('additively')
in
§4.1
is
isomorphism.
an
j
Definition 4.5. Two groups G, H are isomorphic if they are
in the sense of §1.4.1, that is (by Proposition 4.3), if there is
homomorphism G
isomorphic in Grp
a bijective group
H.
j
once and for all in §1.4.1 that 'isomorphic' is automatically
H if G and H are isomorphic.
relation.
We write G
equivalence
G; these form a group
Automorphisms of a group G are isomorphisms G
We have observed
an
=
AutGrp(G) (cf. §1.4.1), usually
Example
denoted
4.6. We have introduced
notion of isomorphism allows
us to
Aut(G).
our
give
a
template of cyclic
groups in
§2.3.
The
formal definition:
Definition 4.7. A group G is cyclic if it is isomorphic to Z
some19 n.
or to
Thus, Ci x Cs is cyclic, of order 6, since C<i
1.
(Exercise 4.9) Cm x Cn is cyclic if gcd(ra,n)
The reader will check easily (Exercise 4.3) that
and only if it contains an element of order n.
Cq.
x
C3
=
Cn
=
Z/nZ
for
j
More generally
=
a group
of order
n
is
cyclic if
somewhat surprising source of cyclic groups: if p is prime, the
group ((Z/pZ)*, •) is cyclic. We will prove a more general statement when we have
There is
a
machinery (Theorem IV.6.10), but the adventurous reader can
already enjoy a proof by working out Exercise 4.11. This is a relatively deep fact;
note that, for example, (Z/12Z)* is not cyclic (cf. Exercise 2.19 and Exercise 4.10).
The fact that (Z/pZ)* is cyclic for p prime means that there must be integers a
accumulated more
such that every nonmultiple of p is congruent to a power of a; the usual proofs of
this fact are not constructive, that is, they do not explicitly produce an integer with
this property. There is
the cyclic group
have to wait for
a very
pretty connection between the order of
an
element of
(Z/pZ)* and the so-called 'cyclotomic polynomials'; but that will
a little field theory (cf. Exercise VII.5.15).
As we have seen, the groups Dq and S3 are isomorphic. Are Cq and S3
isomorphic? There are 1^6,656 functions between the sets Cq and S3, of which 720 are
bijections and 120 are bijections preserving the identity. The reader is welcome to
list all 120 and attempt to verify by hand if any of them is a homomorphism. But
j
maybe there is a better strategy to answer such questions
Isomorphic objects of a category are essentially indistinguishable in that
category. Thus, isomorphic groups share every group-theoretic structure. In particular,
This includes the possibility that n
=
1, that is, trivial groups
are
cyclic.
II. Groups, first encounter
68
Proposition 4.8. Let cp
:
H be
G
an
isomorphism.
(VjeG): \<p(g)\ \g\;
.
=
G is commutative if and only if H is commutative.
Proof. The first assertion follows from Proposition 4.1: the order of (p(g) divides
the order of g, and on the other hand the order of g
(p~1((p(g)) must divide the
=
<p(g)\ thus the two orders must be equal.
order of
The proof of the second assertion is left to the reader.
Further instances of this principle will be assumed without explicit mention.
Example 4.9. Cq
another reason: in
^ S3, since one is commutative and the other is not. Here is
Cq there is 1 element of order one, 1 of order two, 2 of order
2
and
of
order
three,
six; in S3 the situation is different: 1 element of order one, 3
of order two, 2 of order three. Thus, none of the 120 bijections Cq
S3 preserving
the identity is a group homomorphism.
Note: Two finite commutative groups are isomorphic if and only if they have
the same number of elements of any given order, but we are not yet in a position
to prove this; the reader will verify this fact in due time (Exercise IV.6.13). The
commutativity hypothesis is necessary: there do exist pairs of nonisomorphic finite
j
groups with the same number of elements of any given order (same exercise).
Homomorphisms of abelian
4.4.
some ways
any two groups
an
abelian
G, H. In Ab,
group) for
The operation
G
H
in
{ip
+
a group
ip)(a
+
much
homomorphisms, let
e
G)
{ip
:
we are
=
=
(p(a
+
(<p(a)
b)
+
+
a group
(in fact,
G, H.
i))(a)
+
ip(a
iKa))
+
+
Note that the equality marked by !
b)
=
(<p(b)
:=
<p(a)
uses
((p(a)
+
:
ip be the function defined by
(p +
homomorphism? Yes, because Va, b
b)
HoniAb is
more:
is 'inherited' from the operation in H: if ip, ip
HoniAb(G, H)
(Vo
Is (p + ip
we can say
any two abelian groups
are two group
already mentioned that Ab
ready to highlight another
HomGrp(G, H) is a pointed set for
groups. We have
'better behaved' than Grp, and
instance of this observation. As we have seen,
is in
+
1>(b))
+
ip(a).
G G
<p(b))
=
(¥>
+
+
(^(o)
VOW
crucially the fact that
+
+
^(6))
(¥>
+
1>)(b).
H is commutative.
With this operation, HoniAb(G, H) is clearly a group: the associativity of + is
inherited from that of the operation in H\the trivial homomorphism is the identity
H is defined (not surprisingly) by
element, and the inverse20 of (p : G
(VaeG)
:
(-(p)(a)
=
-(p(a).
In fact, note that these conclusions may be drawn as soon as H is commutative:
HomGrP(G,H) is a group if H is commutative (even if G is not). In fact, if H is
Unfortunate clash of terminology! We
G.
possible function H
mean
the 'inverse'
as
in
'group inverse',
not as a
Exercises
a
69
commutative group, then HA
back to this group in §5.4.
Homset(^4,i^)
=
is
a group
for all sets A;
we
will
come
Exercises
4.1. > Check that the function
diagram commute. Verify
m | n necessary?
[§4.1]
4.3.
an
§4.1 is well-defined and makes the
homomorphism. Why is the hypothesis
C2 x C2 is
C2 x C2?
is there any nontrivial
Prove that
>
defined in
a group
homomorphism -k\ x -k\ : C4
isomorphism C4
4.2. Show that the
In fact,
7r^
that it is
a group
element of order
n.
of order
n
is
isomorphic
to
Z/nZ
not an
isomorphism.
if and only if it contains
[§4.3]
(Z, +), (Q, +), (R, +) are isomorphic to one
(R,+), (C,+) are isomorphic to one another?
4.4. Prove that no two of the groups
another.
(Cf.
Can you decide whether
Exercise
VI.1.1.)
4.5. Prove that the groups
4.6. We have
seen
(Q,+)
and
groups
4.7. Let G be
(R
-1
a group.
Let G be
{0}, •)are
not
isomorphic.
that (R, +) and (R>0, •)are isomorphic (Example 4.4). Are the
(Q>0,-) isomorphic?
a group,
G defined by g i-> g~l is a
1—?
g2 is a homomorphism
Prove that the function G
homomorphism if and only if G
if and only if G is abelian.
4.8.
{0}, •)and (C
is abelian. Prove that g
and let g G G. Prove that the function 7^
:
G
G defined
by (Va G G) : jg(a)
gag~l is an automorphism of G. (The automorphisms jg are
called 'inner' automorphisms of G.) Prove that the function G
Aut(G) defined
is
a
Prove
that
this
is
trivial if and
homomorphism.
homomorphism
by g
7g
=
only if G
is abelian.
[6.7, 7.11, IV.1.5]
positive integers such that gcd(ra,n)
Cmn ^CmX Cn. [§4.3, 4.10, §IV.6.1, V.6.8]
> Prove that if m,
4.9.
n
are
=
1, then
Use
> Let p 7^ q be odd prime integers; show that (Z/pgZ)* is not cyclic. (Hint:
Exercise 4.9 to compute the order N of (Z/pgZ)*, and show that no element
can
have order
4.10.
N.) [§4.3]
4.11. > In due time we will prove the easy fact that if p is a
the equation xd
=
1 can have at most d solutions in
Z/pZ.
prime integer, then
Assume this fact, and
prove that the multiplicative group G
(Z/pZ)* is cyclic. (Hint: Let g G G be an
1 for all h G G.
element of maximal order; use Exercise 1.15 to show that ft'pl
=
=
Therefore....) [§4.3, 4.15, 4.16, §IV.6.3]
4.12.
-.
Compute the order of [9]31
in the group
(Z/31Z)*.
Does the equation x3
9
0 have solutions in Z/31Z? (Hint: Plugging in
Z/31Z is too laborious and will not teach you much. Instead,
=
all 31 elements of
II. Groups, first encounter
70
use
the result of the first part: if
about
4.13.
is
c
a
solution of the equation, what
can you say
|c|?) [VII.5.15]
-.
Prove that
AutGrp(Z/2Z
Z/2Z)
x
^
4.14. > Prove that the order of the group of
the number of
Euler's
positive integers
^-function; cf.
that
r < n
Ss. [IV.5.14]
automorphisms of a cyclic group Cn is
relatively prime to n. (This is called
are
6.14.) [§IV.1.4, IV.1.22, §IV.2.5]
Exercise
Compute the group of automorphisms of (Z, +). Prove that if p
then AutGrp(Cp) ^ Cp_i. (Use Exercise 4.11.) [IV.5.12]
4.15.
-i
4.16.
-i
Prove Wilson's theorem:
(p
(For
one
direction,
use
is
prime,
positive integer p is prime if and only if
a
1)!
=
—1
mod p.
Exercises 1.8 and 4.11. For the other, assume d is
(p 1)!; therefore
) [IV.4.11]
a proper
divisor of p, and note that d divides
4.17. For
a
few small
(but
not too
4.18. Prove the second part of
small) primes
p, find a generator of
(Z/pZ)*.
Proposition 4.8.
5. Free groups
contemplate
new
no
Having become
more familiar with homomorphisms, we can
fancier example of a group. The motivation underlying this
construction may be summarized as follows: given a set A, whose elements have
5.1. Motivation.
now
one
special 'group-theoretic' property,
we want to construct a group
F(A) containing
A 'in the most efficient way'.
For example, if A
0, then a trivial group will do. If A
{a} is a singleton,
a trivial group will not do: because although a trivial group {a} would itself
be a singleton, that one element a in it would have to be the identity, and that is
=
=
then
certainly a very special group-theoretic property. Instead, we construct an infinite
cyclic group (a) whose elements are 'formal powers' an, n G Z, and we identify a
with the power a1:
a-2, a~l,a?
(a) :={•••
,
=
e,
a1
=
a,
a2, a3,
};
take all these powers to be distinct and define multiplication in the evident
that the exponential map
way—so
we
ea
:
Z
(a),
-?
ea(n)
:=
an
an isomorphism.
The fact that 'all powers are distinct' is the formal way to
implement the fact that there is nothing special about a: in the group F({a})
(a),
a obeys no condition other than the inevitable a0
e.
Summarizing: if A is a singleton, then we may take F(A) to be an infinite
is
=
=
cyclic
group.
The task is to formalize the heuristic motivation
given above and
construct a
group F(A) for every set A. As we often do, we will now ask the reader to put
away this book and to try to figure out on his or her own what this may mean and
how it may be accomplished.
5. Free groups
71
5.2. Universal property. Hoping that the reader has now acquired an individual
viewpoint on the issue, here is the standard answer: the heuristic motivation is
formalized by means of a suitable universal property. Given a set A, our group
F(A) will have to 'contain' A; therefore it is natural to consider the category 3?A
whose objects are pairs (j, G), where G is a group and
is
a
set-function21 from A
morphisms
to G and
(ji,Gi)-02,G2)
are commutative
diagrams of set-functions
Gl—^G2
h
A
required to be a group homomorphism.
The reader will be reminded of the categories we considered in Example 1.3.7:
the only difference here is that we are mixing objects and morphisms of one
in which ip is
(that is, Grp) with objects and morphisms of another (related) category (that
G is a way to
is, Set). The fact that we are considering all possible functions A
category
implement the fact that
we
have
we
no a
do not want to put any restriction
once
they
A
are
free
mapped
group
to a group
F(A)
on
priori group-theoretic information about A:
on
what may happen to the elements of A
we consider all possibilities at once.
G; hence
A will be
(the
group component
of) an initial object
F(A) in the 'most
in 3?A. This choice implements the fact that A should map to
way': any other way to map A to a group can be reconstructed from
by composing with a group homomorphism. In the language of universal
properties, we can state this as follows: F(A) is a free group on the set A if there is a
set-function j : A
G,
F(A) such that, for all groups G and set-functions / : A
efficient
this one,
there exists
a
unique group homomorphism (p
:
F(A)
G such that the diagram
—>
G
F(A) -^—?
commutes. By general
F(A)
up to
a
=
Z,
as
assume that
a
F(A),
let's check that if A
in
a
j is injective, identifying A with
stronger,
more
{a}
is
§5.1.
a
subset of G; the construction
would be completely analogous, and the resulting group would be the
arbitrary functions leads to
=
Z will send
The function j : A
set-function / : A
G amounts to choosing
proposed
Z. For any group G, giving
zlWe could
this universal property defines
F(A) exist?
group exists. But does
'concrete' construction of
singleton, then F(A)
a to 1 G
(Proposition 1.5.4),
isomorphism, if this
Before giving
a
nonsense
same.
useful, universal property
However, considering
II. Groups, first encounter
72
element g
f(a) G G. Now,
G making the diagram
one
(p
:
=
if
G, then there
a G
is a unique
homomorphism
Z
w
commute:
(p(l)
because this forces
=
o
(f
j(a)
=
/(a)
=
g, and then the
homomorphism condition forces (p(n)
gn. That is, cp is necessarily the exponential
map eg considered in §4.1. Therefore, infinite cyclic groups do satisfy the universal
property for free groups over a singleton.
=
Concrete construction. As
5.3.
not exist. So
we
we
know, terminal objects of
have to convince the reader that free groups
category need
a
F(A) exist,
for every
set A.
Given any set A,
we
are
going
to think of A as an
'alphabet' and
construct
'words' whose letters are elements of A or 'inverses' of elements of A. To formalize
this, consider
a set
A! isomorphic to A and disjoint from it; call a~l the element
a G A. A word on the set A is an ordered list
in A' corresponding to
(ai,a2,which
we
,an),
denote by the juxtaposition
w
=
ai<i2
-
-
-
an,
where each 'letter' ai is either an element a G A or an element
denote the set of words on A
we
include in
W(A)
hy W{A)\ the number
the 'empty word'
For example, if A
=
{a}
is
a
w
=
(),
a~l
G A'.
We will
of letters is the 'length' of w\
consisting of no letters.
singleton, then
an
n
element
ofW(A)
may look like
a~la~laaaa~laa~l.
An element of W({x,y})
may look like
xxx~lyy~lxxy~lx~lyy~lxy~lx.
Now the notation
redundant: for
we
have chosen hints that elements in
xyy~lx
are
W(A)
may be
example,
and
xx
distinct words, but they ought to end up being the same element of a group
to do with words. Therefore, we want to have a process of 'reduction' which
having
takes
a word and cleans it up by performing all cancellations. Note that we have
to do this 'by hand', since we have not come close yet to defining an operation or
making formal sense of considering a~l to be the 'inverse' of a.
is a completely
Describing the reduction process is invariably awkward—it
evident procedure, but writing it down precisely and elegantly is a challenge. We
settle for the following.
will
5. Free groups
Define
73
'elementary' reduction
an
for the first
W(A)
W(A): given w G W(A), search
of
a pair aa~l or a~la, and let
right)
r :
left to
(from
occurrence
be the word obtained by removing such
r(w)
a
pair. In the
two
examples
given above,
r(a~la~laaaa~laa~l)
=
a-1aaa~laa~l,
r(xxx~1yy~1xxy~1x~1yy~1xy~1x)
=
xyy~1xxy~1x~1yy~1xy~1x.
Note that
it;
is
a
r(w)
w
=
precisely when
'reduced word' in this
Lemma 5.1.
If w
G
W(A)
Indeed, either r(w)
Proof.
has length n, then22
w
=
w, but one cannot decrease the
identity application of
r
or
n
is the
For example,
is a reduced word.
the length of r(w) is less than the length of
w more than n/2 times, since each non-
decreases the length by two.
length of
w.
:
=
W(A) by setting R(w)
W(A)
By the lemma, R(w)
R(a~1a~1aaaa~1aa~1)
rA(a~1a~1aaaa~1aa~1)
is
always
a
=
rLf-l(w),
reduced word.
is the empty word, since
r3'(a-1aaa~laa~l)
R(xxx~1yy~1xxy~1x~1yy~1xy~1x)
and
rLf-l(w)
length of
Now define the 'reduction' R
where
'no cancellation is possible'. We say that
case.
r2(aa~1aa~1)
=
=
xxxy~1y~1x,
=
as
r{aa~l)
=
();
the reader may
check.
Let
F(A)
be the set of reduced words
R we have
We
operation
as
on
A, that is, the image of the reduction
map
just defined.
ready to (finally) define free groups 'concretely'. Define a binary
F(A) by juxtaposition & reduction: for reduced words w, w', define w
are
on
the reduction of the juxtaposition of
w
It is essentially evident that
F(A)
-
w'
is
w
:=
and
w'
w',
R(ww').
a group
under this operation:
The operation is associative.
The empty word e
() is the
reduction is necessary).
=
If
it;
is
a
identity
in
F(A),
since
ew
=
we
=
w
(no
reduced word, the inverse of w is obtained by reversing the order
w and replacing each a G A by a-1 G A! and each a-1 by a.
of
the letters of
The most cumbersome of these statements to prove
follows easily from (for example) Exercise 5.4.
formally
is associativity; it
There is a function j : A
F(A), defined by sending the element
word consisting of the single 'letter' a.
Proposition
5.2.
The pair
(j, F(A)) satisfies
on A.
\_q\denotes the largest integer < q.
a G
A to the
the universal property for free groups
II. Groups, first encounter
74
Proof. This is also
Any function /
essentially evident,
A
,-i
G to
once
has absorbed all the notation.
one
extends uniquely to a map (p : F(A)
determined
the
condition
and by the requirement that the
G,
homomorphism
by
which
fixes
its
value
on
one-letter
words a G A (as well as on
diagram commutes,
:
a group
eAf).
To check
follows. If /
more
:
A
formally that cp exists as a homomorphism, one can proceed
G is any function, we can extend / to a set-function
(p
by insisting that
on
:
one-letter words
<p(a)
=
W(A)
a or
f(a),
as
G
-?
a-1 (for
<p(a-1)
a e
A),
f(a)-\
=
and that <p is compatible with juxtaposition:
<p(ww')
<p(w)<p(w')
=
for any two words w, w'. The key point
now
is that reduction is invisible for (p\
<p(R(w)) =(p(w),
since this is
clearly the
G agrees with (p
(p(w w')
-
that is, cp is
a
=
on
for elementary reductions; therefore, since cp
reduced words, we have for w,w' G F(A)
case
<p(w w')
-
=
<p(R(ww'))
homomorphism,
as
(p(wwf)
F({a})
(p(w)(p(wf)
=
(p(w)(p(wf)
:
Z; but it is already somewhat
F({x,y}).
The best
we can
do is
graph23
to the elements of the group and whose
of generators.
=
two generators,
This is an example of the Cayley graph of a group
correspond
=
F(A)
needed.
Example 5.3. It is easy to 'visualize'
challenging for the free group on
the following: behold the infinite
=
:
edges
(cf.
Exercise
8.6):
connect vertices
a graph whose vertices
according to the action
5. Free groups
75
point (the center of the picture), then branching out in
length of 1, then branching out similarly by a length of 1/2,
then by 1/4, then by 1/8, then... (and we stopped there, to avoid cluttering the
picture too much). Then every element of F({x, y}) corresponds in a rather natural
way to exactly one dot in this diagram. Indeed, we can place the empty word at
the center; and we can agree that every x in a word takes us one step to the right,
every x~l to the left, every y up, and every y~l down. For example, the word
obtained
by starting
four directions by
yx~xyx
takes
us
at a
a
here:
\yx xyx
:
;-•
--
-
---(
*
;--
----•
*
---•
*
*
--•
--•
:
*
*
«
"
The reader will surely encounter this group elsewhere: it is the
group of the
fundamental
'figure 8'.
j
5.4. Free abelian groups. We can pose in Ab the same question answered above
for Grp: that is, ask for the abelian group Fab(A) which most efficiently contains
the set A, provided that we do not have any additional information on the elements
of A.
Of course we do know something about the elements of A this time: they
will have to commute with each other in Fab(A). This plays no role if A
{a} is
=
singleton, and therefore Fab({a})
for larger sets, so we should expect
a
Z; but the requirement is different
in general.
The formalization of the heuristic requirement is precisely the same universal
=
a
F({a})
different
=
answer
property that gave us free groups, but (of course) stated in Ab: Fab(A) is a free
abelian group on the set A if there is a set-function j : A
Fab(A) such that,
for all abelian groups G and set-functions f : A —>
G, there exists a unique group
G such that the following diagram commutes:
homomorphism ip : Fab(A)
II. Groups, first encounter
76
is unique up to isomorphism,
Again, Proposition 1.5.4 guarantees that
in
it
but
we
have
to
it
exists!
This
is
some way simpler than for
exists;
prove
if
in
the
sense
that
is
easier
to
at
least for finite sets A.
understand,
Grp,
Fab(A)
Fab(A)
To fix ideas, we will first describe the
We will denote by Z0n the direct sum
Z0
-
for
answer
-
0
-
a
finite set, say A
=
{l,---,n}.
Z;
n-times
recall
(§3.5)
coproduct).
that this group 'is the same as' the product24 Zn
There is a function j : A
Z0n, defined by
jM:=(0,-",0, 2-th 1
place
Claim 5.4. For A
{1,
n}, Z0n is
abelian group
free
Proof. Note that every element of Z0n
can
,
Yn=imiJ{i>Y indeed,
(mi,--- ,mn) (mi,0,--- ,0)
=
mij(l)
=
(mi,
,
Now let
/
mn)
:
A
We define (p : Z0n
(0,
=
as a
on
A.
(0,ra2,0,-• ,0)
+
-
-
-
+
+
(0,-• ,0,ran)
+
rnn(0,
,0,1)
\-mnj(n),
H
,0)
view it
be written uniquely in the form
mi(l,0,--- ,0)+m2(0,l,0,--- ,0)
=
and
+
we
,0,..-,0)eZ^.
a
=
(but
if and only if all m2-
G be any function from A
G by
=
are
{1,
0.
,
n}
to an abelian group G.
-?
n
(n
/
2=1
indeed,
we
have
no
2=1
choice—this
definition is forced by the needed commutativity
of the diagram
/
A
and by the homomorphism condition. Thus ip is certainly uniquely determined, and
just have to check that it is a homomorphism. This is where the commutativity
we
of G enters:
\/n
(n^rn'ijii) J
i=l
+(p
\n
n
j(i) J
^ m-/(i) ^ m-;/W ^(m-+ m-O/W
[y^m"
\i=l
i=l
/
/
n
+
=
=
i=l
i=l
because G is commutative,
=
<P
(E(m^
+
= 1
\2
as
j{i)] =<p(J2^i J(i) Em"JW)
+
/
= 1
\2
2=1
/
needed.
Indeed, it is
but
m'l)
we will
common to
denote this group by Zn, omitting the ©. No confusion is likely,
try to distinguish the two to emphasize that they play different categorical roles.
5. Free groups
77
Remark 5.5. A less hands-on, more high-brow argument can be given by
contemplating the universal property defining free abelian groups vis-a-vis the universal
property for coproducts; cf. Exercise 5.7.
has
Now for the general case: let A be any set. As we have seen, HA
a natural abelian group structure if H is an abelian group
of HA
as
j
are
arbitrary set-functions
a :
A
H. We
can
define
a
Homset(^4, H)
(§4.4); elements
=
subset H®A of HA
follows:
H®A
:=
{a
:
A
H
-?
| a(a) ^
The operation in HA induces
eH for
only finitely
many elements a G
A}.
operation in i7eA, which makes H®A into
an
a
group25.
The reader should note that H®A is the whole of HA if A is
that Z0A ^ Z0n if A
with the function
For H
=
=
{1,
Z there is
to the function j;a
:
A
{1,
n}
,
a
,
n}: indeed, (mi,
Z
sending
:
A
^
f 1
Note that for A
same
function
=
{1,
,n}
and
ran)
Z0n
a
finite set; and
may be identified
i to m^.
natural function j
Z defined by
(V*€A):
,
,a(,):={0
Z0A,
if
x
obtained by sending
=
.f ^
identifying Z0A
=
a G
A
a,
fl>
Z0n,
this function j is the
denoted j earlier.
Proposition 5.6. For every set A,
Fab(A)
^
Z0A.
Proof. The key point is again that every element of Z0A may be written uniquely
as a
finite
sum
y] maj(a),
ma
^
0 for
only finitely
many a;
a€A
once this is
understood, the argument is precisely the
'Thus H®A is
a
subgroup of HA\ cf. §6.
same as
for Claim 5.4.
II. Groups, first encounter
78
Exercises
5.1. Does the category &A defined in
§5.2
5.2. Since trivial groups T are initial in
should be initial in
of A to the
have final
Grp,
objects?
one may
If so, what
are
they?
be led to think that
(e,T)
<jPA,
(only)
for every A: e would be defined by sending every element
element in T; and for any other group G, there is a unique
homomorphism T
G. Explain why (e,T) is
not initial in
&A (unless A
=
0).
5.3. > Use the universal property of free groups to prove that the map j : A
F(A)
is injective, for all sets A. (Hint: It suffices to show that for every two elements
G such that f(a) ^ f(b).
a, b of A there is a group G and a set-function / : A
Why?
How do you construct
/
and
G?) [§III.6.3]
5.4. > In the 'concrete' construction of free groups, one can try to reduce words
by performing cancellations in any order; the process of 'elementary reductions'
used in the text (that is, from left to right) is only one possibility. Prove that the
result of iterating cancellations on a word is independent of the order in which the
cancellations are performed. Deduce the associativity of the product in F(A) from
this.
[§5.3]
5.5.
Verify explicitly
that H®A is
5.6. > Prove that the group
Z*Z of Z by itself in
a group.
F({x,y}) (visualized
the category
in Example 5.3) is a coproduct
Grp. (Hint: With due care, the universal property
for one turns into the universal property for the
other.) [§3.4, 3.7, 5.7]
5.7. > Extend the result of Exercise 5.6 to free groups
abelian groups
F({xi,..., xn})
and to free
Fab({xi,... ,xn}). [§3.4, §5.4]
5.8. Still
Fab(A)
more
generally,
Fab(B)
0
prove that
F(AUB) F(A)*F(B) and that Fab(AUB)
(That is, the constructions F, Fab 'preserve
=
for all sets A, B.
=
coproducts'.)
5.9. Let G
5.10.
-
ZeN. Prove that GxG^G.
Let F
Define
for
=
an
=
equivalence relation
some g G
that
case
Assume
as sets.
F. Prove that
~
on
F/~
is
F
a
/ if and only
and only if A is
by setting /'
finite set if
~
\i
f
f
=
2g
finite, and in
|F/~| =2^.
^
Fab(A). If A is finite, prove that B is also, and that A ^ B
result holds for free groups as well, and without any finiteness
See Exercises 7.13 and VI.1.20.)
Fab(B)
(This
hypothesis.
[7.4, 7.13]
Fab(A).
6.
79
Subgroups
6.
Subgroups
(G, •) be
6.1. Definition. Let
underlying set H is
Definition 6.1.
group
a
a group,
and let
(H, •)be
another group, whose
subset of G.
(i7, •)is
subgroup of G if the inclusion function
a
i
:
H
^->
homomorphism.
G is
a
j
For example, the trivial group consisting of the
single element
cq is a
subgroup
ofG.
If (H, •)is
a
subgroup of (G, •),then Vhi, h2
(*)
G H:
i(h1mh2)=i(h1)-i(h2).
We say that the operation
on H is 'induced' from the operation
on G; in practice
one omits explicit mention of i and of the operations, and (*) guarantees that no
ambiguity
will arise from this.
subgroup condition may be streamlined. A subset H of a group G
subgroup if the operation in G induces (by (*)) a binary operation in H
H is closed with respect to the operation in G), satisfying the group
that
(we say
axioms. Since the identity and inverses are preserved through homomorphisms
(Proposition 3.2), the identity en of H will have to coincide with the identity cq
The
determines a
of G and the inverse of an element h G H has to be the same as the inverse of that
element in G. The most economical way to say all this is
Proposition
6.2. A nonempty subset H
of
ab~l
(\/a,beH) :
Proof.
It is clear that if H is a
if b G i7, then the inverse of b
ofG.
a group
G is
a
subgroup if and only if
G H.
subgroup, then the stated condition holds: indeed,
operation
must also be in H and H is closed under the
Conversely, assume the stated condition holds; we have to check that H is
closed under the operation of G, the induced operation on H is associative, and
it admits an identity element and inverses (that is, it contains ec and is closed
under
taking
Choosing a
=
inverses in
b
=
h,
G).
we see
Since H is nonempty,
that
eG
=
hh~l
thus H contains the identity. Given
=
ab~l
any h G
G
we can
find
an
element h G H.
H\
H, choosing
a
=
eo and b
=
h shows
that
h~l
=
eGh~l
=
ab~l
G
thus H contains the inverse of any of its elements.
a
hi, b
/i^1; the stated condition says that
=
H\
Given any
fti,/i2
G
i7, choose
=
hih2
=
hi(h2)_1
=ab~l eH,
proving that H is closed under the operation.
Finally, the fact that the operation is associative in G implies immediately that
the induced operation is associative in H, concluding the proof that H, with the
induced operation, is a group.
II. Groups, first encounter
80
This criterion makes it
particularly straightforward
concerning subgroups. For example,
Lemma 6.3.
is any
If {Ha}aGA
family of subgroups of
to check
a group
simple facts
G, then
H=f]Hi
is a
subgroup of G.
Proof. This follows right away
e G Ha for all a, so e G H; and
a, b G H
(Va
=>
proving that 17 is
A)
G
:
a,
from
b G Ha
Proposition 6.2: H
=>
(Va
G
A)
:
a&-1
is nonempty, because
G #a
=>
a&_1
G
if,
subgroup of G.
a
Similarly,
Lemma 6.4. Le£ <p
G' 6e
G
:
of G'. Then (^_1(i7/)
a ^rot/p
homomorphism, and let H' be
Proof. Recall (end of §1.2.5) that (^_1(i7/) consists
17'. Since ^(e^)
ec G HT, this set is nonempty.
and <p(6) are in H', and hence
=
y?(a6-1)
thus, a6_1
G
<p>~l(H).
a
subgroup
subgroup of G.
is a
=
^(a)y?(6)-1
This implies that
(f~1(H/)
G
of all g G G such that <^(g) G
If a,6 G <p>~l(H'), then <p(a)
#':
is
a
subgroup of G, by
Proposition 6.2,
6.2.
Examples: Kernel and image. Every
group
homomorphism
ip
:
G
G'
determines two interesting subgroups:
the kernel of <p, kery? C G; and
the image of ^,
im^CG'.
Definition 6.5. The kernel of <p
mapping to the identity in G'\
ker<^
Since
subgroup
{eG'}
of G.
is
For
a
an
:=
{g
G
G G
G' is the subset of G consisting of elements
| (f(g)
=
eG<}
=
<p>~l{eG>).
j
subgroup of G', Lemma 6.4 shows that ker<p is indeed a
(even) more explicit argument, note that ker<p is nonempty,
since e^ G ker (p; and if a, b
are
ip(ab~l)
proving that ab~x
:
G ker (p.
in ker <p, then
=
ip(a)ip(b)~l
This shows that
=
eG>e~^
kenp
is
a
=
e&,
subgroup of G, by
Proposition 6.2.
The verification that im(p is a subgroup is left to the reader. In fact, the reader
should check that the image of any subgroup of G is a subgroup of G'.
We will soon (§7.1) see that kernels are 'special' subgroups. As with most
constructions of importance in algebra, they satisfy a universal property, which
may be expressed
as
follows.
6.
81
Subgroups
G' be a homomorphism. Then the
Proposition 6.6. Let (p : G
ker ip <-^ G is final in the category2^ of group homomorphisms a : K
(f
inclusion
i
:
G such that
is the trivial map.
o a
G such that (p o a is the
words, every group homomorphism a : K
homomorphism (denoted '0' in the diagram) factors uniquely through kenp:
In other
trivial
Proof. If
a :
K
G is such that <p
(poa(k)
is, a(k) G kenp. We
with restricted target.
that
can
is the trivial map, then Vk e K
o a
=
(f(a(k))
(and must)
=
then let
eG,
a :
K
ker
a
simply be
a
itself,
Proposition 6.6 indicates how one might define a notion analogous to 'kernel' in
general settings. This viewpoint will be championed much later in this book,
especially in Chapter IX.
very
Remark 6.7. The argument shows that in fact kernels of group homomorphisms
G
satisfy a somewhat stronger universal property: any set-function a : K
such that the image of (p
through ker (p.
6.3.
have
o a
is the
identity
in H must factor
(as
a
set-function)
j
Example: Subgroup generated by
unique group homomorphism
a
subset. If A C G is any subset,
we
a
ipA
:
F(A)
-
G,
by the universal property of free groups. The image of this homomorphism is a
subgroup of G, the subgroup generated by A in G, often denoted27 (A).
Of course, if G is abelian, then (pA factors through Fab(A), so we may replace
F(A) by
Fab(A)
in this
case.
The 'concrete' description of free groups (§5.3) leads to the
it consists of all products in G of the form
following description
of (A):
a\a2as
where each ai is either an element of
-
-
an
A, the
inverse of
an
element of A,
identity. This is clearly the most 'economical' way to manufacture
G, given the elements of A.
The reader should specify what the morphisms are in this category.
If A
=
{gi,... ,gr}
is a finite set, one writes
(gi,... ,gr).
a
or
the
subgroup of
II. Groups, first encounter
82
The reader who has not
(yet) developed
the following alternative description:
a taste for free groups may prefer
A is the intersection of all subgroups of G
containing A,
n
(A)
h-
H subgroup of G, H D A
Indeed, the intersection on the right-hand side is a subgroup of G by Lemma 6.3,
it contains A, and it is clearly the smallest subgroup satisfying this condition.
If A
Z and cpA
Z
G is
{g} consists of a single element, then F(A)
nothing but the 'exponential map' eg (cf. §4.1); (A)
(g) is then the image of this
=
=
=
map:
(g) =im(ep)
=
{... ,g~2,g~l,e,g,g2,... }.
The subgroup (g) is the 'cyclic subgroup generated by g': indeed, (g) is cyclic in the
sense of Definition 4.7; the reader can easily check this fact already (Exercise 6.4); it
will also be recovered
as an
immediate consequence of the construction of quotients
(cf. §7.5).
Definition 6.8. A group G is
such that G
(A).
finitely generated
if there exists
a
finite
subset ACG
=
For examples,
cyclic
j
finitely generated (in fact, they are generated
group is finitely generated if and only if there is a
groups are
by a singleton). By definition,
surjective homomorphism
a
F({l,...,n})-»G
for
a
One of the most memorable results proven in this book will give
classification of finitely generated abelian groups: we will be able to prove that
some
n.
every such group is a direct sum of cyclic groups (Theorem IV.6.6, Exercise VI.2.19,
and the generalization given in Theorem VI.5.6). The situation for general groups
more complex. The classification of finite (simple) groups is one
achievements
of twentieth-century mathematics, and it is spread over
major
is considerably
of
the
at
least 10,000 pages of research articles. To appreciate the difference in complexity,
note that there are 4% abelian groups of order 1024 up to isomorphism (as the
reader will be able to establish in due time:
49,487,365,402 if we count noncommutative
Exercise
IV.6.6); allegedly,
there
are
well28.
ones as
6.4. Example: Subgroups of cyclic groups. We are ready to determine all
subgroups of all cyclic groups, that is, all subgroups of Z and of Z/nZ, for all n > 0
(because every cyclic group is isomorphic to one of these; cf. Definition 4.7). The
result is easy to remember: subgroups of cyclic groups are themselves cyclic groups.
It is convenient to start from Z. For d G Z we let
dZ
:=
(d)
=
{m
e Z
| 3q
e
Z, m
=
that is, dZ denotes the set of integer multiples of d. Of
the 'cyclic subgroup of Z generated by d\
Proposition 6.9. Let G C Z be
a
subgroup. Then G
This comparison is a little unfair, however, since it
groups of order < 2000 have order 1024.
so
=
dq};
course
this is nothing but
dZ
some
for
happens that
more
d>0.
than 99% of all
6.
83
Subgroups
proof will actually show that if G C Z is nontrivial, then d is the smallest
element
of G, and the reader is invited to remember this useful fact.
positive
The
By Proposition 6.9, every nontrivial subgroup of Z is in fact
Putting this a little strangely, it says that every subgroup of the
on
one
free group
generator is free. It is in fact true that every subgroup of a
(finitely generated) free group is free; we will not prove this fact, although the
diligent reader will get a taste of the argument in Exercise 9.16. In any case, beware
Remark 6.10.
isomorphic to Z.
that free groups
groups
Exercise
on
two
arbitrarily
on
generators already contain subgroups isomorphic to free
is
7.12) [F, F]
F({x, y})
(unfortunately, we will not
=
generators
Proof of Proposition 6.9. If G
contain positive integers: indeed, if
can
prove this beautiful statement
{0},
=
We
Indeed, the commutator subgroup (cf.
isomorphic to a free group on infinitely many
many generators.
for F
a G
then G
G and
apply 'division with remainder'
verify
<
in
G, and
claim G
the inclusion G C dZ, let
=
dq
=
m G
dZ.
G, and
+ r,
with 0 < r < d. Since n G G and dZ C G and since G is a
r
But d is the smallest
we
to write
m
=
m
dq
subgroup,
=
that
we see
G G.
positive integer in G, and r G G is smaller than d; so r
r
0, that is, m
qd G dZ; G C dZ follows, and
be positive. This shows
done.
j
0Z. If not, note that G must
and a > 0.
0, then —aeG
=
integer29
then let d be the smallest positive
The inclusion dZ C G is clear. To
a
either).
cannot
=
The 'quotient' homomorphism 7rn : Z
the analogous result for finite cyclic groups:
Z/nZ (cf. §4.1)
allows
us to
we are
establish
Proposition 6.11. Let n > 0 be an integer and let G C Z/nZ be a subgroup. Then
G is the cyclic subgroup ofZ/nZ generated by [d]n, for some divisor dofn.
Proof. Let 7rn
:
Lemma 6.4, G' is
generated by
a
Z
a
be the quotient map, and consider G' := 7r~1(G). By
subgroup of Z; by Proposition 6.9, G' is a cyclic subgroup of Z,
Z/nZ
nonnegative integer d. It follows that
G
thus G is indeed
G'
since
n G
n,
claimed.
as
As
a
=
([d]n);
=
=
of Proposition 6.11, there is
subgroups of Z/nZ and the
are
7rn(G,)=7rn((d))
cyclic subgroup of Z/nZ, generated by a class [d]n. Further,
[n]n [0]n G G) and G' dZ, we see that d divides
(because 7rn(n)
a consequence
We
=
secretly appealing
set of
to the
=
a
positive divisors of
bijection between the
n.
For example,
'well-ordering principle'. That
should have a smallest element is one of those fact about Z—like
the
we
remainder—that
are
assuming the reader is already familiar with.
set of
Z/12Z
has
every set of positive integers
availability of division-with-
II. Groups, first encounter
84
exactly 6 subgroups, because 12 has 6 positive divisors: 1, 2, 3, 4, 6, and 12. Here
is the corresponding list of subgroups:
<[l]l2>
=
<[2]l2>
=
<[3]i2>
=
<[4]l2>
=
<[6]l2>
=
<[12]l2>
=
{[0]l2, [l]l2, [2]l2, [3]l2, [4]l2, [5]l2, [6]l2, [7]l2, [8] 12, [9]l2, [10]l2, [ll]l2},
{[0]l2, [2]i2, [4]i2, [6] 12, [8] 12, [10]l2},
{[0]i2,[3]i2,[6]i2,[9]i2},
{[0]l2,[4]i2,[8]l2},
{[0]l2,[6]l2},
{[0]l2>.
Also note that if di, &i are both divisors of n, and d\ d>2, then ([di]n) 5 ([^n)That is, the correspondence between subgroups of Z/nZ and divisors of n preserves
the natural lattice structure carried by these sets. We can draw these lattices for
Z/6Z as follows:
{[0],[1],[2],[3],[4],[5]}
{[0],[2],[4]}
{[0],[3]}
picture and subsets in the other. The reader
will draw the lattice of subgroups of S3, noting that it looks completely different
from the one for Z/6Z.
where lines connect multiples in
one
Contemplating subgroups of cyclic groups has pretty
theoretic' consequences; cf. Exercise 6.14.
6.5. Monomorphisms. We end this section with
If H
in
Grp
is
G is an
subgroup of G, the inclusion H
'categorical' sense of §1.4.2. In fact, it is easy to characterize all
a
Proposition
The
(p is a
(b) kei(p
(p
Proof,
categorical considerations.
example of a monomorphism
in the
HomGrp(G, G') (where G,
(c)
'number-
some
^->
monomorphisms (p G
(a)
(and useful)
:
(a)
following
are
are any
groups):
equivalent:
monomorphism;
=
G
6.12.
G'
{eG};
G' is infective
=^
(b):
Assume
(as
a
set-function).
(a) holds,
keripz
and consider the two parallel compositions
1G-Z->G',
Exercises
85
is the trivial map. Both cpoi and cpoe are the trivial
e. But i
e implies that ker(p
monomorphism, this implies i
where i is the inclusion and
map; since cp is a
is trivial, that is,
(b)
<p(9i)
(c):
==>
=
e
=
(b)
Assume kenp
<P(92)
(pigig^1) eG>
9\92l e keV(P => 0102-1 eG
=>
(c)
phisms
(a):
=>
injective,
Then
{ec}.
=
V(gi)v(g2)~l
=>
This shows that <p is
=
holds.
as
=
=>
eG>
=
=> 9i=92-
=
needed.
If (p is injective, then it satisfies the defining property for monomora',a" : Z
G,
in Set: that is, for any set Z and any two set-functions
(poa'
=
(poa"
This must hold in particular if Z has
homomorphisms,
so ip
is
a
^=>
a
monomorphism
a'= a".
structure
group
in
and
a7, a"
are
group
Grp.
The equivalence (a) <^=> (c) may lead the reader to think that from the point
of view of monomorphisms, Grp and Set are pretty much alike. This is not quite so:
while it is true that homomorphisms with
monomorphisms, as in Set (cf. Exercise 6.15), the
Exercise
a
left-inverse
converse
are
necessarily
Grp (cf.
is not true in
6.16).
Exercises
6.1.
(If you know about matrices.) The group of invertible n x n matrices with
entries in R is denoted GLn(R) (Example 1.5). Similarly, GLn(C) denotes the group
-i
of
n x n
invertible matrices with complex entries.
Consider the
following
sets of
matrices:
{M e GLn(R) | det(M) 1};
SLn(C) {M e GLn(C) | det(M) 1};
On(R) {M e GLn(R) | MM1 MlM
SOn(R) {Me On(R) | det M Jn};
Un(C) {Me GLn(C) | MAft AftM
SUn(C) {M e Un(C) | det(M) 1}.
SLn(R)
=
=
=
=
=
=
=
=
Jn};
=
=
=
=
=
Jn};
=
n x n identity matrix, Ml is the transpose of M, Aft is the
of
Af, and det(Af) denotes the determinant30 of Af. Find all
conjugate transpose
inclusions
possible
among these sets, and prove that in every case the smaller set
Here In stands for the
is
a
subgroup of the larger
one.
These sets of matrices have compelling geometric interpretations: for example,
S03(R) is the group of 'rotations' in R3. [8.8, 9.1, III.1.4, VI.6.16]
ones
If you are not familiar with some of these notions, that's ok: leave this exercise and similar
if that is the case. We will come back to linear algebra and matrices in Chapter VI
alone
and following.
II. Groups, first encounter
86
6.2.
-i
Prove that the set of 2
with a, 6, d in C and ad ^ 0 is
set ofnxn complex matrices
is
x 2
a
a
b
0
d
subgroup of GL2(C). More generally,
{o>ij)\<i,j<n with
subgroup of GLn(C). (These
a
matrices
matrices
are
prove that the
0 for i > j and an
ann ^ 0
a^
called 'upper triangular', for evident
=
reasons.) [IV.1.20]
6.3.
-i
Prove that every matrix in
SU2(C)
may be written in the form
a + bi
c +
di
a
bi
—c
+ di
where a, 6, c, a7 G R and a2 + 62 + c2 + a72
a three-dimensional
sphere embedded
in
=
1.
(Thus, SU2(C)
in
particular,
R4;
may be realized as
it is
simply connected.)
[8.9, III.2.5]
6.4. > Let G be
map eg
6.5.
:
Z
Let G be
{gn |
g G
G}
is
a
a
and let g G G.
cyclic group (in the
a
commutative
group,
that the image of the exponential
of Definition 4.7). [§6.3, §7.5]
Verify
a group,
G is
sense
and let
n
subgroup of G. Prove that this
>
integer. Prove that
necessarily the case if G is
0 be
is not
an
not commutative.
6.6. Prove that the union of
a
subgroup of G.
a
family of subgroups of
subgroups of
tfHCH'
or H' C H.
only
Let H, H' be
a group
{Ji>0Hi
G is not necessarily
G. Prove that H U H' is
On the other hand, let Hq C Hi C #2 C
that
a group
In fact:
is a
be subgroups of
a
subgroup of G
a group
G. Prove
subgroup of G.
4.8) form a subgroup of
Aut(G); this subgroup is denoted Inn(G). Prove that Inn(G) is cyclic if and only
if Inn(G) is trivial if and only if G is abelian. (Hint: Assume that Inn(G) is cyclic;
Show that inner automorphisms
6.7.
-1
with
notation
as
in Exercise 4.8, this
(cf.
means
Exercise
that there exists
an
element
In particular, gag~l
such that \/g G G 3n G Z jg
anaa~n
7™.
commutes with every g in G. Therefore
) Deduce that if Aut(G) is
=
G
is abelian.
=
=
a.
a
G
Thus
G
a
cyclic, then
[7.10, IV.1.5]
6.8. Prove that
an
abelian group G is finitely generated if and only if there is
a
surjective homomorphism
^^
'
v
n
for
some n.
6.9. Prove that every
not
times
finitely generated.
finitely generated subgroup of Q
is
cyclic. Prove that Q
is
Exercises
6.10.
-i
87
The set of 2
matrices with integer entries and determinant 1 is denoted
x 2
SL2(Z):
SL2(Z)
=
is
SL2(Z)
Prove that
f
<
,
such that a, 6, c, d G Z, ad
J
generated by the
s=(? ~j)
This is
(Hint:
matrix
a
m
by multiplying
,
I
are
I)
-q
1 in
,
matrices
*=(j
and
1
it suffices to show that you
SL2(Z),
1
=
little tricky. Let H be the subgroup generated by
I
m=
be
by suitably chosen elements of
H.
can
s
and t. Given
a
obtain the identity
Prove that I
J
and
in H, and note that
a
b\
c
d)[0
-q\
(l
_
"
1
J
(a
b
\c
d
-
-
qa\
qc)
and
b\
(a
\c
d)
(
1
0\
_
l)
\-q
"
(a
-
\c
-
qb
qd
Note that if c and d are both nonzero, one of these two operations may be used
to decrease the absolute value of one of them. Argue that suitable applications of
these operations reduce to the case in which c
0 or d
0. Prove directly that
=
m G
H in that
=
case.) [7.5]
Ab, the classification theorem for abelian
finitely generated abelian group is a
in
of
Ab.
The
reader
coproduct
cyclic groups
may be tempted to conjecture that
is
a
in
coproduct
Grp. Show that this is not the case,
every finitely generated group
that
is
not
a
of
S3
coproduct
by proving
cyclic groups.
6.11. Since direct
sums are
coproducts
in
groups mentioned in the text says that every
6.12. Let m,
n
be positive integers, and consider the subgroup
(m,n)
of Z they
generate. By Proposition 6.9,
(m, n)
for
some
6.13.
-1
6.14.
>
dZ
positive integer d. What is d, in relation
to m, n?
Draw and compare the lattices of subgroups of C2
lattice of subgroups of S3, and
r < m
=
If
m
that
is
are
compare it with the one for
C2 and C4. Draw the
Cq. [7.1]
x
positive integer, denote by <j>{m) the number of positive integers
relatively prime to m (that is, for which the gcd of r and m is 1);
a
this is called Euler's </>-
(or 'totient') function. For example, 0(12)
group (Z/raZ)*; cf. Proposition 2.6.
=
4. In other
words, (f)(m) is the order of the
Put together the following observations:
(j>{m)
=
the number of generators of Cm,
every element of Cn generates
a
subgroup of Cn,
the discussion following Proposition
isomorphic to Cm, for some m n),
6.11
(in particular,
every
subgroup of Cn
is
II. Groups, first encounter
88
to obtain
a
proof of the formula
^2
(f)(m)
=
n.
m>0,m|n
(For example, (f)(1) + (f)(2) + (f)(3)
[4.14, §6.4, 8.15, V.6.8, §VII.5.2]
6.15. > Prove that if
a group
is,
group homomorphism ip
monomorphism. [§6.5, 6.16]
Counterpoint
(f)(4)
(f)(6) +0(12)
+
homomorphism
a
6.16. >
+
G'
:
(p
:
=1+1+2+2+2+4=
G
G such that ip
—>
(p
homomorphism
to Exercise 6.15: the
left-inverse, that
idc, then (p is a
G' has
o
a
=
:
cp
12.)
Z/3Z
S3 given
by
w(oi)=(; 1 j),
is
a
wui)
monomorphism; show that
it has
(5; 5). *m>-(5 5 ?)
-
left-inverse in Grp.
no
subgroups will make this problem particularly
7.
Quotient
(Knowing
about normal
easy.) [§6.5]
groups
7.1. Normal subgroups. Before tackling 'quotient groups',
sense kernels are special subgroups, as claimed in §6.2.
we
should
clarify
in
what
Definition 7.1. A subgroup N of
a group
gng~l
Note that every subgroup of
a
G is normal if \/g e G,Vn e N,
G N.
j
commutative group is normal
(because
then Mg G
gng~l n G N). However,
general not all subgroups are normal: examples
may be found already in S3 (cf. Exercise 7.1). There exist noncommutative groups
in which every subgroup is normal (one example is the 'quaternionic group' Qg;
cf. Exercise III. 1.12 (iv)), but they are very rare.
G,
in
=
Lemma 7.2.
subgroup
// cp
:
G' is any group homomorphism, then kertp is
G
^(gng'1)
proving that gng~l
G
=
kenp
(Can
a
There is
a
a
subgroup of G;
=
to
verify
^(9)eG^(g)~1
=
it is normal note
eG>,
kenp.
little while; for
the reader guess?)
in
is
^(#MnM#_1)
Loosely speaking, therefore,
see
normal
ofG.
Proof. We already know that
that \/geG,\/nekenp
will
a
now
kernel
=^
normal. In fact
more is true, as we
I don't want to spoil the surprise for the reader.
convenient shorthand to express conditions such as normality: if
subset, we denote by gA, Ag, respectively, the following
g G G and A C G is any
subsets of G:
gA:={heG\(3aeA):h ga},
=
7. Quotient groups
89
Ag
:=
Then the normality condition
{heG\(3aeA)
or
in
h
=
ag}.
be expressed by
can
(ty
:
G)
G
gNg-1
:
C
TV,
number of other ways:
a
gNg~l
=
N
C
gN
or
Ng
gN
or
Ng
=
for all g e G. The reader should check that these are indeed equivalent conditions
Ng} does not mean that g commutes
(Exercise 7.3) and keep in mind that 'gN
with every element of iV; it means that if n G N, then there are elements n', n" G N,
in general different from n, such that gn
n'g (so that gN C Ng) and ng
gn"
=
=
(so
that Ng C
=
#iV).
7.2.
an
Quotient group. Recall that we have
equivalence relation (§1.1.5) and that this
(clumsily
stated in
§1.5.3).
seek
a group
quotient of
a set
by
universal property
It is natural to investigate this notion in Grp.
notion satisfies
a
on (the set underlying) a group G;
equivalence relation
G/~ and a group homomorphism tt : G
G/~ satisfying the
We consider then
we
the notion of a
an
~
appropriate universal property, that is, initial with respect to group homomorphisms
b => cp(a)
G' such that a
(f : G
ip(b).
=
~
set
It is natural to try to construct the group G/~ by defining an operation on the
The situation is tightly constrained by the requirement that the quotient
G/~.
map
[b]
=
G
:
7r
7r(6)
in
G/~ (as
are
elements of
§1.2.6) be a group homomorphism: for if [a]
G/~ (that is, equivalence classes with respect
=
then the homomorphism condition
[a]
[b]
=
it
forces
(a)
7r(6)
7r(ab)
=
[ab].
=
But is this operation well-defined? This amounts to conditions
relation, which we proceed to unearth.
For the
[a]
=
[a'],
operation
then
[ab]
=
Similarly,
G
G)
:
G
G)
a
With notation
a'
~
=>
ag
a'g.
~
a'
~
=^
ga
we
need
go!.
~
above, the operation
as
[a]
a group structure on
a
In this
it is necessary that if
of what b is; that is,
a
:
the equivalence
this is all that there is to it:
Proposition 7.3.
defines
on
for the operation to be well-defined in the second factor
(Vg
Luckily,
factor',
to be well-defined 'in the first
[a'b] regardless
(Vg
7r(a),
~),
to
case
~
a
G/~ if
=>
the quotient function
with respect to
homomorphisms
:
:=
[ab]
and only
ga
tt :
(p
[6]
G
G
~
ga
if Va, a',g
and ag
G/~
is a
~
G G
a g.
homomorphism and
G; s?/c/i £/ia£
a
~
a!
=^
(p(a)
is universal
=
<p(a').
II. Groups, first encounter
90
already noted that the condition is necessary.
with
the
stated consequences, assume Wa,a',g G G
sufficient,
Proof. We have
a
~
a
=>
~
ga
and ag
ga
To prove it is
a g.
~
Then the operation
[a]
is
well-defined, and
associativity of
([a]
.
The class
[c]
is
=
an
[g_1]
is
[ab]
G/~
=
[9}
[geo]
G/~
[g-'g]
=
is indeed
To prove that
{(ab)c)
=
is the inverse of
homomorphism: this
a
[c]
[eo]
[g-1]
This shows
[ab]
:=
[a(bc)\
=
identity with respect
[d]
The class
verify
[6]
that it defines
a group structure on
G/~.
The
is inherited from the associativity of G: Va,b,ce G
[&])
[cq]
have to
we
to this
\g],
=
[eo]
[a]
=
[be]
.
=
[a]
operation: \/g G
[9]
=
[eog]
=
.
([b]
.
[c]).
G
\g].
[g]:
M
=
and
a group,
is what led
[g]
[s"1]
\gg~1}
=
=
fa}.
have already observed that
to the definition of •.
we
us
satisfies the universal property,
tp :
tt :
G
G/~
assume
G'
G
-?
a' =^ (p(a)
homomorphism such that a
<p(a'). Since (cf. §1.5.3)
the set G/~ satisfies the corresponding universal property in Set, we know that
there exists a unique set-function
is
a group
<p
defined31 by <£([a])
a group
:=
So
<p(a)-
we
homomorphism, and this
<p([a]
for all
[a], [b]
=
~
G
[b\)
G/~,
as
=
<p([ab})
:
G/~
G\
-?
only need
to check that this function <p is in fact
is immediate:
=
<p(ab)
=
<p{a)<p{b)
=
<p([a])<p([b])
needed.
We will say that
is compatible with the group structure of G if the condition
on the quotient G/~ is
given in Proposition 7.3 holds. Since the operation
uniquely determined by the operation on G, we yield to the usual abuse of language
and omit it. If
is compatible, we will call G/~ the quotient group of G by ~.
~
~
7.3.
Cosets. The conditions obtained in
(t)
(V#
(tt)
(V5GG):
lead
G)
:
Proposition 7.3,
a~b => ga
~
6 => ag
~
a
-
gb,
bg,
complete description of all compatible relations on a group G. In fact,
a description of the relations satisfying it, and
them
the
reader should keep in mind that we have a
analyze
separately;
to a
each of these two conditions leads to
we
will
group structure on
G/~ only
Let's begin with
{]).
if both
are
satisfied.
Here is the description:
The point of §1.5.3 is precisely that this function is well-defined.
7. Quotient groups
91
Proposition 7.4. Let
be
~
equivalence relation
an
on
G, satisfying (f).
a group
Then
the equivalence class of eo
H
subgroup
is a
a~b «=> a~lb eH «=> aH
.
and
of G;
bH.
=
Proof. Let H C G be the equivalence class of the identity; H ^ 0 as cq G H. For
b and hence 6_1
e^ (applying (f), multiplying on the left
a,b E H, we have eo
~
by
6_1);
hence ab~l
~
a
~
(by (f) again, multiplying
ab~
~
a
~
the left by
on
a);
and hence
eo
by the transitivity of and since a G H. This shows
proving that H is a subgroup (by Proposition 6.2).
~
that a6_1 G i7 for all a,b E H,
b. Multiplying on the left by a-1, (f) implies
a,b e G and a
i7.
i7 is closed under the operation, this implies
that
a_16
Since
G
ec
a_16,
is,
a~lbH C H, hence bH C ai7; as
is symmetric, the same reasoning gives aH C
bH. Thus, we have proved
bH; and hence aH
Next,
assume
~
~
~
=
a
Finally,
assume
of i7, this
a
~
ai7
=
means ec
~
~
b => a_16 G # => aH
bH. Then
a
=
aeo G
a_16. Multiplying
=
bH.
6i7, and hence a~lb
on
the left by
a
e H.
shows
By definition
(by (f) again)
that
6, completing the proof.
Proposition 7.4 shows that the equivalence classes of
satisfying (f) are in fact all in the form
an
equivalence relation
aH
for
a
a
fixed subgroup H,
H deserve a
subgroup
Definition 7.5. The
a G
as a ranges
left-cosets of
G. The right-cosets of H
Now,
a
are
a
H in
subgroup
the sets Ha,
// H
is any
subgroup of
(Va,be G):
a G
a group
G
the sets aH, for
are
G.
j
a group
a~Lb
«=>
G, the relation
a~lb
~l
defined by
G H
equivalence relation satisfying (f).
Proof. This
see
by
'converse' to Proposition 7.4 holds:
Proposition 7.6.
is an
in G. These important subsets determined
name.
is
straightforward
that the relation satisfies
a~Lb =>
for all g
a~xbeH
=>
and is mostly left to the reader
(f),
(Exercise 7.8).
To
note that
a~1(g~1g)beH
=>
(ga)~1(gb)eH
=>
ga~Lgb
G G.
Taken
together, Propositions
7.4 and 7.6 show
There is a one-to-one correspondence between subgroups of G
and equivalence relations on G satisfying (f); for the relation ~l corresponding to
a subgroup H, G/~l may be described as the set of left-cosets aH of H.
Proposition 7.7.
II. Groups, first encounter
92
no difficulty producing the mirror statements (and
exhaustive
description of all equivalence relations
proofs) giving similarly
will
The
end
result
be
satisfying (ff).
The reader should have
a
There is a one-to-one correspondence between subgroups of G
and equivalence relations on G satisfying (ff); for the relation ~r corresponding to
a subgroup H, G/~r may be described as the set of right-cosets Ha of H.
Proposition 7.8.
The relation
corresponding
to H in this second way is defined
b ^^ ab~l e H ^^ Ha
a ~R
=
by
Hb.
What may be surprising at first is that the relations ~l and ~r corresponding
same subgroup H may very well not be the same relation. That is, left-cosets
to the
and right-cosets of
more
subgroup need
a
not coincide.
Of course eH
=
He
H, and
=
generally
hH
(V/i eH):
=
Hh
=
H.
Further
(Va
hence, if aH
automatically
=
true if G
Example
and the 1
G
G)
:
aeaHD Ha;
Ha. This is of course
Hb, then in fact necessarily aH
is commutative, but it is simply not the case in general.
=
7.9. Let G
=
S3, and let H be the subgroup consisting of the identity
2 switch:
<-?
Then
H
1
^3
=
2J
{\3
1
2j\s
2
1) J
1
2)'(1
3
2)}-
'
while
^(3
2) {(3
=
1
This state of affairs simply reflects the fact that the two conditions
are
different: there is
no reason to
expect that if
one
holds, the other
(f)
and
one
(ff)
should
(unless G is commutative, of course). Once more, keep in mind that both
have to hold for the quotient G/~ to be a group, compatibly with the operation in
also hold
G; cf. Proposition 7.3.
7.4.
Quotient by normal subgroups. To
stress the main
arbitrary subgroup H of G leads to two partitions of G, which
G/~L
=
{aH\aeG},
G/~R
=
point again,
we
an
have denoted
{Ha\aeG}.
The relation ~L satisfies property (f) listed at the beginning of §7.3; ~r satisfies
(ff). A priori, these relations, and hence the corresponding partitions, are different.
and ~r coincide (as is necessarily the case, for example,
translates into a condition on H: for such 'special' subgroups,
The condition that
if G
is
commutative)
~l
(f) and (ff) get back together. The good
identify and is not new to the reader.
news
is that this condition is easy to
7. Quotient groups
93
Proposition 7.10. The relations ~l, ~r
if and only if H is normal.
Proof. Two relations coincide if the
left- and
~l=~r <=>-
But this is
(cf §7.1),
corresponding
to a
corresponding partitions
right-cosets of H coincide
<=>*
H coincide
subgroup
agree. Therefore
(\/g G G) : gH
=
Hg.
of the equivalent conditions defining the notion of normal subgroup
proving the statement.
one
The innocent-looking Proposition 7.10 is of fundamental importance. If H is
normal, then the one equivalence relation ~=~L=~R corresponding to H satisfies
both (f) and (ft) (by Proposition 7.4 and its mirror statement), and hence (by
Proposition 7.3)
the quotient
G/~
has
a
=
{aH\aeG}
=
{Ha\aeG}
natural group structure.
a normal subgroup of a group G. The quotient group
of G modulo H, denoted32 G/H, is the group G/~ obtained from the relation
defined above. In terms of (left-) cosets, the product in G/H is defined by
Definition 7.11. Let H be
~
(aH)(bH)
:=
(ab)H.
The identity element cq/h °f the quotient group
H.
eGH
G/H
is the coset of the
identity,
=
j
By Proposition 7.3, the quotient function
7T:
G
G/H
-?
g G G to gH
Hg is a group homomorphism and is universal with respect
to group homomorphisms (p : G
bH ==> (p(a)
G' such that aH
<p(b). This
universal property is extremely useful, so we will grace it with theorem status:
sending
=
=
Theorem 7.12. Let H be
homomorphism cp
homomorphism (p
G
:
:
normal subgroup of a group G. Then for every group
G' such that H C ker ip there exists a unique group
a
G'
G/H
=
so
that the diagram
G
>G'
commutes.
Proof. We only need to match the stated universal property with the
proved
in
Proposition 7.3, and indeed,
H
is
equivalent
C
ker^
«=>
(V/i
a
H) : <p{h)
=
eG>
to
(Va,be G) : ab~l eH
In
G
large display
we sometime use
=>
ip(ab~l)
=
the full 'fraction' notation
eG>
¦§-.
one
we
II. Groups, first encounter
94
that is, to
(ya.be G) : ab~l eH
and finally, keeping in mind how the relation
(Vo,be G)
~
a~b =>
:
<p(a)
=>
=
ip(b)
corresponding
y?(o)
=
to i7 is
defined,
<p(6),
the condition giving the universal property in Proposition 7.3.
7.5. Example. The reader is already very familiar with an
groups LjriL. Indeed, in §2.3 we defined
examples: the cyclic
important class of
LjriL
as
the set of
equivalence classes in Z with respect to the congruence equivalence relation
(Va, b
Now
we
recognize that
e
Z)
:
(b
n
a
=
is
a)
b
mod
n
equivalent
b
^=>
n
(b
a).
to
nL,
a e
which is the relation ~l corresponding (in 'abelian' notation) to the subgroup riL
of Z. This subgroup is of course normal, since Z is abelian. The 'congruence classes
mod ri are nothing but the cosets of the subgroup riL in Z; using abelian notation
for cosets, we could write
[a]n
a +
=
(riL).
Of course the operation defined on LjriL in §2.3 matches precisely the one defined
above for quotient groups. This justifies the notation Z/nZ introduced in §2.3.
The reader
can
already appreciate
in this
simple
context the usefulness of
Theorem 7.12. Let g e G be an element of order n and consider the
exponential
map
N^gN.
eg:Z^G,
By Corollary 1.11,
ker eg
=
{N
e Z
|
N is
a
multiple of \g\} riL.
=
Theorem 7.12 then implies right away that eg factors through the quotient:
:
Z
That is, there is
an
>G
induced map
L/nL^(g).
In fact, the 'canonical decomposition' of §1.2.8 implies that this is an isomorphism
(verifying that (g) is cyclic in the sense of Definition 4.7, as the reader should have
checked 'by hand' already in Exercise
general in the next section.
Also note that
\g\
=
n=
\(g)\in
6.4).
this
We will formalize this observation in
case.
Exercises
95
7.6. kernel
^=>
in gory detail
a group
normal. If H is
G/H
and
a
a
normal subgroup,
G
now
constructed
G/H.
-?
What is the kernel of 7r? The identity of
Therefore
=
have
surjective homomorphism
7T :
kev7r
we
G/H
is the coset cqH, that is, H itself.
{geG\gH H}
=
=
H.
This observation completes the circle of ideas begun in §7.1: there we had noticed
that every kernel (of a group homomorphism) is a normal subgroup; and now
we
have verified that every normal subgroup is in fact
We encapsulate this in the slogan
a
kernel
(of
some
group
homomorphism).
kernel ^=> normal
in group
theory33,
'kernel' and 'normal subgroup'
:
equivalent concepts.
For example, every subgroup in an abelian group is the kernel of some
homomorphism: yet another indication that life is simpler in Ab than in Grp.
are
Exercises
7.1. > List all
are
subgroups of S3 (cf.
normal and which
7.2. Is the image of
are not
a group
Exercise
normal.
6.13)
and determine which subgroups
[§7.1]
homomorphism necessarily
a
normal subgroup of the
target?
7.3. >
Verify
equivalent.
that the equivalent conditions for normality given in
is
are
indeed
[§7.1]
7.4. Prove that the relation defined in Exercise 5.10
Fab(A)
§7.1
compatible with the
on a
free abelian group F
group structure. Determine the
quotient F/~
=
as a
better known group.
A' ^=> A'
on SL2(Z) by letting A
equivalence relation
±A. Prove that
is compatible with the group structure. The quotient SL2(Z)/~
is denoted PSL2(Z) and is called the modular group; it would be a serious contender
in a contest for 'the most important group in mathematics', due to its role in
algebraic geometry and number theory. Prove that PSL2(Z) is generated by the
(cosets of the) matrices
7.5.
-1
Define
an
~
~
=
~
(1 ~o)
and
(1 ~o)-
(You will not need to work very hard, if you use the result of Exercise 6.10.) Note
that the first has order 2 in PSL2(Z), the second has order 3, and their product has
infinite order.
[9.14]
We will run into analogous observations in ring theory, where we will verify that kernels
and ideals coincide, and for modules, as kernels and submodules again coincide.
II. Groups, first encounter
96
7.6. Let G be
and let
a group,
a~6
Show that in general
~
n
be
a
(3geG)ab~1 =gn.
«=>
is not
an
positive integer. Consider the relation
equivalence relation.
equivalence relation if G
corresponding subgroup of G.
Prove that
is
~
an
is commutative, and determine the
a group, n a positive integer, and let H
all
elements of order n in G. Prove that H is
generated by
7.7.
Let G be
7.8. > Prove
7.10.
G be the
subgroup
Proposition 7.6. [§7.3]
7.9. State and prove the 'mirror' statements of
to the
C
normal.
Propositions 7.4 and 7.6, leading
description of relations satisfying (ft).
-i
Let G be
a group,
a subgroup. With notation as in Exercise 6.7,
only if V7 G Inn(G), j(H) C H.
in G, then there is an interesting homomorphism
and H C G
show that H is normal in G if and
Conclude that if H is normal
Inn(G)->Aut(lT). [8.25]
[G, G] be the subgroup of G generated by all
(This is the commutator subgroup of G; we will
return to it in §IV.3.3.) Prove that [G, G] is normal in G. (Hint: With notation
as in Exercise 4.8, g
aba~1b~1 g~l
^g(aba~1b~1).) Prove that G/[G,G] is
commutative. [7.12, §IV.3.3]
7.11. > Let G be
a group,
and let
elements of the form aba~1b~1.
=
7.12. > Let F
be
a
a commutative group
homomorphism F/[F,F]
in Exercise 7.11.
free group, and let / : A
G be a set-function
G. Prove that / induces a unique
G, where [F,F] is the commutator subgroup of F defined
Theorem 7.12.) Conclude that F/[F,F] ^ Fab(A). (Use
F(A)
=
from the set A to
(Use
Proposition 1.5.4.) [§6.4, 7.13, VI.1.20]
7.13.
F(A)
to
8.
-1
^
Let A, B be sets and F(A), F(B) the corresponding free groups. Assume
F(B). If A is finite, prove that B is also and A ^ B. (Use Exercise 7.12
upgrade
Exercise
Canonical
5.10.) [5.10, VI.1.20]
decomposition
and
Lagrange's
theorem
We will collect in this section
a number of observations on the structure of quotient
All
these
results
are
groups.
straightforward, given the background work done so
far. Some of them are often given fancy names such as first isomorphism theorem
in the literature; we are not too fond of such terminology: the universal property
proven in Theorem 7.12 is really the only thing we need to take along, and it serves
us
wonderfully
well. The 'isomorphism theorems'
this universal property.
are
all immediate applications of
8. Canonical decomposition and
Lagrange's theorem
97
8.1. Canonical decomposition. The first observation comes from the
canonical decomposition for set-functions, obtained in §1.2.8: every set-functions may be
viewed as the composition of a surjective map, followed by a bijective map, followed
by an injective
results in Grp:
map. We now know
enough
to state the
Theorem 8.1. Every group homomorphism (p
:
corresponding (very useful)
G' may be decomposed
G
as
follows:
where the isomorphism <p in the middle is the homomorphism induced by (p
Theorem
(as
in
7.12).
It is important that the reader agree that we have already proved anything
that deserves to be proven here. We know that the projection on the left and the
inclusion on the right are homomorphisms and (p comes from Theorem 7.12. The
decomposition is the
same one
obtained at the level of set-functions in
G
G', the reader should instantaneously view
§1.2.8; in
particular, the function in the middle is a bijection. Since bijective homomorphisms
are isomorphisms (Proposition 4.3), it is an isomorphism.
Theorem 8.1 should induce the following Pavlovian reaction: exposed to any
group
as
homomorphism
(canonically
(p
identified
isomorphism theorem'
homomorphisms:
Corollary
:
8.2.
is the
Suppose
cp
a
G' is
G
:
G/kenp
subgroup of G'. What is usually called the first
particular case corresponding to surjective
with)
a
surjective
group
homomorphism. Then
ker<p'
Proof, irrup
=
G' in Theorem 8.1.
This result is very useful—it
comes in extremely handy when proving that two
groups are
this
isomorphic, both
section)
in theoretical contexts
(as
we
will
see
in the rest of
and in concrete instances.
Example 8.3. If H\ C G\ and Hi C C?2, then the product H\ x Hi may be
viewed as a subset of Gi x G2. It is clear that if Gi, G^ are groups and Hi, H2
are subgroups, then Hi x Hi is a subgroup of Gi
prototype application of Corollary 8.2:
Claim 8.4. If Hi
normal subgroup
of
Gi and H^
x
G2. The following claim is
a
H2 is
a
the group Gi
x
G2 are normal subgroups, then Hi
Gi and
Gi
x
G2
Hi
x
H<i
C
C
^
Gi_
G2
Hi
H<i
Indeed, composing the projections
7Ti :
Gi
x
G2
Gi,
7T2 :
Gi
x
G2
G2
x
II. Groups, first encounter
98
with the morphisms to the quotients gives surjective homomorphisms
s~i
s~i
7Ti : Gi x G2
and hence a
Gl
~
-fj-,
tti
G2
~
7T2 : Gi x G2
—>
—>
-77tl2
homomorphism
Gi
tt :
G2
x
^
G\
x
G2
——-
ill
tii
by the universal property of products. Explicitly,
*(9i,92)
in
particular,
tt
(9iHug2H2)
=
:
is surjective and
kerTr
{(gug2)
=
=
G
G1
x
G2 (9iHug2H2)
=
(HUH2)}
{(91,92) GGi xG2\gieHug2eH2}
=
ffi
x
1T2.
The claim then follows immediately from Corollary 8.2.
The result (of course) extends
should become second
Example 8.5.
H2
G2 C G2:
As
a
to more factors in the
nature and is
particular
usually left
product. Any such check
to the reader.
of Claim 8.4, take Hi
case
j
=
{e^}
C
Gi and
=
Gi
G2
x
Gi
^
where
on
the left
we
identify G2
G2
^
{eGl}X G2-Uu
~
G2
with the subgroup
{ecj
x
G2. For instance34
(cf. §4.1)
Cq_
^
C2
Cs
C3
x
^c
Cs
Example 8.6. The cyclic group Cs may be viewed as a subgroup of the dihedral
group D6: the rotations of a triangle give a copy of C3 inside D6. Then C3 is
normal in Dq, and
This
can of course be checked 'by hand'. But note that there is an evident surjective
C2, whose kernel is C3: map an element a of Dq to the
homomorphism Dq
identity in C2 if it does not flip the triangle (that is, precisely when a G C3),
and map it to the other element if it does. Corollary 8.2 implies the stated facts
immediately.
Example
8.7. One
j
can
give
its points with rotations of
a
a
circle
(denoted S1)
plane about
a
a group structure by identifying
point and adding them accordingly.
The function
p
Abuses of language such
as
:
R1
S1
-?
which
the formula which follows—in
one is not
specifying how to realize C3 as a subgroup of Ce, because there is really only
unfortunately commonplace.
explicitly
one way to do
it—are
8. Canonical decomposition and
mapping
number
a
r to
Lagrange's theorem
the result of
group homomorphism; this
multiple of 2n. Hence
99
by 2?rr radians is then a surjective
identity precisely when we rotate by an integer
is the
a
kerp
rotation
=
ZCR.
By Corollary 8.2, therefore,
this amounts to 'wrapping' R infinitely many
times around the circle, realizing R as the 'universal cover' of S1; here, Z plays the
Exercise
(Cf.
1.1.6.) Geometrically,
role of 'fundamental group' of S1.
j
group is a quotient of a free group, and every abelian
free abelian group. Indeed, every group G can be surjected
upon by a free group, and in many ways (at the very least, F(G) will do!). Abelian
groups may be likewise surjected upon by free abelian groups. Then Corollary 8.2
8.2. Presentations.
group is
a
quotient of
Every
a
produces the needed isomorphism of G with
A presentation of
a group
where A is a set and R is
is an explicit surjection
a
G is
an
a
quotient of
:
free group.
explicit isomorphism
subgroup of 'relations'.
p
a
F(A)
-»
In other words,
a
presentation
G
This is especially useful if A is small, and R may be
described very explicitly; usually this is done by listing 'enough' relations, that is,
a set 8? of words ra E R
ker p generating it in the sense that R is the smallest
of which R is the kernel.
=
normal
subgroup35
is
of
F(A) containing
presentation of
Thus,
and 8% C
a
a set
is
F(A)
&.
G is usually encoded as a pair (A\&), where A
of words, such that G
A/R with R as above.
a group
a set
=
A group is finitely presented if it admits a presentation (A, 8?) in which both
are finite.
Finitely presented groups are not (necessarily) 'small': for
A and 8%
example, the free group
on
finitely
many generators is
(trivially) finitely presented.
examples of presentations. For instance,
the free group F(A) is presented by (A|0). More interestingly, the description of
53 given in §2.1 'presents' S3 as a quotient of the free group F({x,y}) (cf.
We have already
run
into several
Example 5.3) by the smallest normal subgroup containing x2, y3, and yx
(x,y\x2,y3,xyxy) in shorthand. From this point of view, it is clear that
admitting the same presentation (example: S3 and Dq) are isomorphic.
The situation is less idyllic than it may seem at first, though: even if
presentation of
a group
no role.
xy2:
groups
a
G is known, it may be very hard to establish whether two explicit
Note that this is
plays
=
a
different requirement than the
one
adopted in §6.3, in which normality
II. Groups, first encounter
100
combinations of the generators coincide in G. This is known
and it has been shown to be undecidable in general36.
as
the word problem,
In any case, now that we know about presentations of groups, finding coproducts in Grp should be straightforward: see Exercise 8.7.
There is
of free
'mirror' statement analogous to the fact that all groups
a
groups:
are quotients
subgroup of a symmetric group.
name of Cayley's theorem; its natural
every group may be realized as a
elementary observation
This
place
goes under the
is within the discussion of group actions
8.3.
Subgroups of quotients. The lattice of subgroups (cf. §6.4) of
(cf.
Theorem
9.5).
a
be described very explicitly in terms of the lattice of subgroups
G/H
of the group G: simply keep the part of the lattice of subgroups of G corresponding
to subgroups which contain H.
quotient
can
Example 8.8. Here is the effect of this operation on the lattice of subgroups of
C12
Z/12Z (labeled by generators; cf. §6.4), after quotienting by H ([6]) ^ C2:
=
=
f
The result matches the lattice of subgroups of Cq
=
C12/C2.
Here is why this works. First note that if H C K
and H is normal in G, then H is normal in K.
are
j
subgroups of
a group
G
Proposition 8.9. Let H be a normal subgroup of a group G. Then for every
subgroup K of G containing H, K/H may be identified with a subgroup of G/H.
The function
u :
{subgroups
defined by u(K)
=
K/H
K
of G containing H}
is a
{subgroups of G/H}
bisection preserving inclusions.
Proof. The
this
sense
group K/H consists of the cosets aH G G/H with a E K, and in
it is a subset (and clearly a subgroup) of G/H. It is also clear that if
H C K C L, then
u(K)
That is, there is
no
=
K/H
C
L/H
=
u{L)\ that
general algorithm that, given
words in the generators, will establish
same element of G.
(in
a
finite
time)
a
is,
u preserves
presentation of
a
inclusions.
group
G and two
whether those two words represent the
Lagrange's theorem
8. Canonical decomposition and
Thus
produce
we
an
simply have
to
of
{subgroups
Then let K' be
a
G/H}
:
is
a
bijection, and for this
G
G/H
it suffices to
check that
v are
and
H}.
to be the subset of G:
{aeG\aHe K'},
=
is the canonical projection. Then K is a subgroup of G (by
(because H
7r_1(e) and e G K'). The reader will
contains H
u
K of G containing
{subgroups
-k~1(K')
Lemma 6.4) and
In
u
subgroup of G/H; define v(K')
K :=
tt
that
inverse function
v :
where
verify
101
=
inverses of each other.
fact, the correspondence is even nicer, in the sense that it preserves
following statement is often called the third isomorphism theorem:
normality. The
Proposition 8.10. Let H be
subgroup of G containing H.
normal in G, and in this
a
normal subgroup
Then
N/H
of
a
is normal in
group
G, and let N be
and only if N
G/H if
a
is
case
G/H
N/H
^G
N'
Proof. If N is normal, then consider the projection
G^Nthe subgroup H is contained in N, which is the kernel of this homomorphism,
we get (by the universal property of quotients, Theorem 7.12) an induced
so
homomorphism
H^ N'
The subgroup N/H of G/H is the kernel of this homomorphism; therefore it is
normal.
Conversely, if N/H is normal in G/H, consider the composition
G
G
—»
G/H
N/H'
H
The kernel of this homomorphism is N; therefore N is normal. Further, this
homomorphism is surjective; hence the stated isomorphism (G/H)/(N/H)
G/N
=
follows immediately from Corollary 8.2.
8.4.
of
HK/H
a group
vs.
K/(HDK).
G. What if H, K
Section 8.3 deals with two 'nested' subgroups H C K
nested?
are not
The notation introduced in §7.1 extends to subsets of G: if A C G, B C G,
then AB denotes the subset
AB
It would be nice if HK
are
guaranteed
simply not the
were
subgroups, but this
is
if
:={ab\aeA,b
to be a
G
B}.
subgroup of G
general, if G
case in
as soon as
H and K
is not commutative.
of the subgroups is normal. The following is often
called the second isomorphism theorem.
It is, however, the
case
one
II. Groups, first encounter
102
Proposition 8.11. Let H, K be
normal in G. Then
HK is
a
subgroups of
subgroup of G, and
a group
G, and
assume
that H is
H is normal in HK;
H n K is normal in K, and
Proof. To
verify
that HK is
a
HK
K
H
HDK
subgroup of G when
H is normal, note that HK is
the union of all cosets Hk, with k G K; that is,
HK
where
G
tt :
HK is
7r-1(7r(K)),
=
is the canonical projection. Since ir(K) is a subgroup
6.4. It is clear that H is normal in HK.
G/H
of G/H,
subgroup by Lemma
a
For the second part, consider the homomorphism
:
<p
sending
HK/H
-?
HK followed by the
(that is, the inclusion K
This
is
indeed,
surjective:
every element of
quotient).
k G K to the coset Hk
canonical projection
HK/H
K
to the
^->
may be written as a coset
Hhk,
but Hhk
=
Hk,
so
Hhk
heH,keK;
is in the image of (p.
cp(k)
=
By the omnipresent
Corollary 8.2,
HK
K
H
What is ker
ker ip
cp'
ker
(pi
=
{k
G K
tp(k)
e}
=
=
{k
G K
\Hk
=
H}
=
{k
G K
\kG H}
=
H n K,
with the stated result.
8.5.
The index and
the set of
in
left-cosets37
general, and
it is
Lagrange's
of H,
a group
even
theorem. The notation
G/H
is used to denote
when H is not normal in G. Thus
G/H
is
a set
when H is in fact normal in G.
Definition 8.12. The index of H in G, denoted [G : H], is the number of elements
j
\G/H\of G/H, when this is finite, and oo otherwise.
Thus, [G
:
H] (if finite)
denotes the number of left-cosets of H in G, regardless
of whether H is normal in G.
Lemma 8.13. Let H be
a
subgroup of
H
H
are
a group
gH,
h
Hg,
h
G. Then \/g G G the functions
gh,
i—?
hg
i—?
bisections.
This may seem an arbitrary choice (why not right-cosets?). It is. Writing from left to right
gives us a bias towards /e/ifc-actions, and G acts nicely on the left on the set of left-cosets; this will
make better sense when we get to Example 9.4. In any case, there is a bijection between the set
of left-cosets and the set of
right-coset: Exercise 9.10.
Lagrange's theorem
8. Canonical decomposition and
Proof. Both functions
that they
are
103
surjective by definition of
are
coset.
Cancellation
Corollary 8.14 (Lagrange's theorem). If G is a finite
subgroup, then \G\ [G : H] \H\.In particular, \H\is a
=
Proof. Indeed, G is the disjoint union of
by Lemma 8.13.
Lagrange's theorem is
more
\G/H\distinct
group and H C G is
divisor
Therefore, g'G'
cosets gH, and
Example
of
8.16. If
|G|
is
a
\gH\ \H\
=
useful than it may appear at first.
ec for all finite groups
=
a
of \G\.
Example 8.15. The order \g\of any element g of a finite group G is
|G|: indeed, \g\equals the order of the subgroup (g) generated by g.
Note:
implies
injective.
prime integer
p, then
G, all
a
divisor of
g G G.
necessarily G
j
=
Z/pZ.
Indeed, let g G G be any element other than the identity; then (g) is a subgroup
G, of order > 1. By Lagrange's theorem, \(g)\ p
|G|; that is, G
(g) is
=
cyclic of order
Example
=
=
p, as claimed.
8.17
little
(Fermat's
any integer. Then ap
=
j
theorem).
Let p be
a
prime integer, and let
a
be
mod p.
a
Indeed, this is immediate if a is a multiple of p; if a is not a multiple of p, then
[a]p modulo p is nonzero, so it is an element of the group (Z/pZ)*, which
the class
1. Thus
has order p
[<-1
(Example 8.15);
hence
[a]£
Warning: However, do
it does not say that if d is
[a]p
=
as
=
[i]P
claimed.
j
not ask too much of
a
divisor of
order d (the smallest counterexample
|G|,
Lagrange's theorem. For example,
a subgroup of G of
then there exists
A*, a group of order 12, which does not
contain subgroups of order 6; the reader will verify this in Exercise IV.4.17); it does
not even say that if p is a prime divisor of |G|, then there is an element of order
p in G. This latter statement happens to be true, but for 'deeper' reasons. The
abelian
and
we
is
is easy (cf. Exercise 8.17). The general case is called
will deal with it later on (cf. Theorem IV.2.1).
case
The index is
that if H C K
a
are
well-behaved invariant. It is clearly multiplicative, in the
is normal
that these numbers
(so
sense
subgroups of G, then
[G:H}
provided
Cauchy's theorem,
that HK is
(again, provided this has
a
are
=
[G:K}-[K:H],
are subgroups of G and
cf.
well;
Proposition 8.11), then
finite. Also, if H and K
subgroup
as
H
chance of making sense, that is, if the orders are finite):
this follows immediately from the isomorphism in Proposition 8.11 and index
considerations. In fact, the formula holds even without assuming that one of the
a
subgroups is normal in G. Do you
see
why? (Exercise 8.21.)
II. Groups, first encounter
104
Further, if H and G
finite, then Lemma 8.13 implies immediately that the
are
the number of left-cosets of H in G, also equals the
H.
of
It
is in fact easy to show that there always is a bijection
right-cosets
between the set G/H of left-cosets and the set of right-cosets (cf. Exercise 9.10),
index of H in G, defined
as
number of
regardless of finiteness hypotheses.
(reasonably) denoted H\G.
The set of right-cosets of H in G is often
Epimorphisms and cokernels. The reader may expect that a mirror
Proposition 6.12 should hold for group epimorphisms. This is almost true:
H is an epimorphism (in the category Grp) if and only
homomorphism cp : G
8.6.
statement to
a
if
it is surjective. However, while one implication is easy, the
epimorphism => surjective in Grp are somewhat cumbersome.
The situation is leaner
(as usual)
this is part of what makes Ab
in Ab: there is in Ab
a
proofs
good
we
know for
notion of cokernel;
'abelian category'.
an
As is often the case, the reader may now want to pause a moment and try
to guess the right definition. Keeping in mind the universal property for kernels
(Proposition 6.6), can the reader come up with the universal property defining
'cokernels'? Can the reader prove that these exist in Ab and detect epimorphisms?
Don't look ahead!
Here is how the story goes. The universal property is (of course) obtained by
G'
reversing the arrows in the property for kernels: given a homomorphism (p : G
of abelian groups, we want an abelian group coker (p equipped with a homomorphism
7r :
which is initial with respect
a :
homomorphism
coker (p:
to all
morphisms
L such that
G
coker (p
G'
a o
a
such that ao(p
=
0. That is, every
(p is the trivial map must factor
(uniquely)
through
o
Cokernels exist in Ab: because the image of (p is a subgroup of G', hence a normal
subgroup of G' since G' is abelian; the condition that a o (p is trivial says that
inap
C
kera, and hence
=
coker cp
im(p
satisfies the universal property,
The 'problem' in
the situation is more
Also
note
that,
universal property:
is constant
We
by Theorem
Grp is that
complex.
inap is not
in the abelian case,
as
on cosets
7.12.
guaranteed
to be normal in
G'/imip automatically
stated, but with respect
to any
of map.
can now state a true
mirror of
satisfies
set-function G'
Proposition 6.12,
in Ab:
a
G';
thus
stronger
L which
Exercises
105
Proposition 8.18. Let (p
following
(a)
are
(p is an
(b) coker <^
(c)
(p
Proof,
:
:
(a)
homomorphism of abelian
a
groups.
The
equivalent:
epimorphism;
is
trivial;
G' is surjective
G
G' be
G
=>
(b):
Assume
(as
set-function).
a
(a) holds,
and consider the two parallel compositions
coker (p,
G—^->
G' —^—%
e
where
7r is the canonical projection and e is the trivial map. Both no(p and eo(p are
the trivial map; since cp is an epimorphism, this implies tt
e. But tt
e implies
=
that coker (p
(b)
=>
is
=
trivial, that is, (b) holds.
(c):
If coker cp
G'/im(p
=
is trivial, then
im(p
=
G'\
hence cp is
surjective.
(a): If (p is surjective, then it satisfies the universal property for epiin
Set: for any set Z and any two set-functions a! and a" : G'
Z',
morphisms
(c)
=>
a'
o
(p
=
a"
o
(p ^=> a7
This must hold in particular if Z is endowed with
group
homomorphisms,
A cokernel
(p
:
G
so cp
is
an
may be defined in
epimorphism
=
a77.
a group structure
in
and
a', a"
are
Grp.
Grp: the universal property for the cokernel of
G' is satisfied by
G'/N, where N is the smallest38 normal subgroup
(Exercise 8.22). But Proposition 8.18 fails, because the
of G' containing map
implication
(b)
=^
(c)
does not hold: in Grp it is no longer true that
Exercise 8.23).
only surjective
homomorphisms have trivial cokernel (cf.
Exercises
8.1. If
a group
H may be realized
as a
subgroup of
H
does it follows that G\
8.2.
-i
=
G<p. Give
Extend Example 8.6
as
a
or a
8.3. Prove that every finite group is
counterexample.
Suppose G is
subgroup of index 2, that is, such that there
of H in G. Prove that H is normal in G.
G\ and Gi and if
H'
proof
follows.
two groups
are
[9.11,
a group
precisely
two
and H C G is
(say, left-)
a
cosets
IV. 1.16]
finitely presented.
The intersection of any family of normal subgroups is normal, as the reader may readily
check, so this subgroup exists.
II. Groups, first encounter
106
8.4.
is
a
presentation of the dihedral
With respect to the generators defined in Exercise 2.5, set
(Hint:
b
(a,6|a2,62, (ab)n)
Prove that
xy; prove you can get the relations
=
in Exercise 2.5, and
given here from the
ones
group D^na
you
=
x
and
obtained
conversely.)
a group G, and assume that ab has
subgroup generated by a and b in G is isomorphic
8.5. Let a, b be distinct elements of order 2 in
finite order
n >
3. Prove that the
to the dihedral group D^n8.6.
Let G be
-i
(Use
the previous
and let A be
a group,
exercise.)
of generators for G; assume A is
directed graph whose set of vertices
a set
finite. The corresponding Cayley graph39 is a
is in one-to-one correspondence with G, and two vertices #i, #2
an
edge if
are
connected by
A] this edge may be labeled a and oriented from g\
example, the graph drawn in Example 5.3 for the free group F({x,y})
g^
to #2- For
=
9\cl for an a G
generators x, y is the corresponding Cayley graph (with the convention that
horizontal edges are labeled x and point to the right and vertical edges are labeled y
on two
and point
up).
Prove that if a Cayley graph of a group is a tree, then the group is free.
Conversely, prove that free groups admit Cayley graphs that are trees. [§5.3, 9.15]
8.7. > Let
(A\3l),resp., (A'\3lf),be a presentation for
we may assume
that A, A!
are
a group
disjoint. Prove that the
G,
resp., G'
group G *G'
(cf. §8.2);
presented by
(AUA'I^U*')
satisfies the universal property for the
universal properties of both free
G
morphisms G^G*G',G'
^
8.8.
a
(If
-i
group.
8.9.
*
G'.) [§3.4, §8.2, 9.14]
you know about matrices
normal subgroup of
GLn(R),
coproduct of G and G' in Grp. (Use the
quotients to construct natural homo-
groups and
(cf.
6.1).) Prove that SLn(R) is
GLn(R)/SLn(R) as a well-known
Exercise
and 'compute'
[VI.3.3]
-.
matrix.
(Ditto.) Prove that S03(R)
SU2(C)/{±/2}, where
(Hint: It so happens that every matrix in SOa(R) can
=
J2 is the identity
be written in the
form
a2 + b2
-
c2
-
2(ad + be)
2(bd ac)
-
d2
2(bc
ad)
a2-b2 + c2- d2
2(ab + cd)
-
2(ac + bd)
2(cd ab)
a2-b2-c2 +
-
d2)
1. Proving this fact is not hard, but at this
where a, 6, c, d G R and a2+b2+c2+d2
will
find
it
stage you
probably
computationally demanding. Feel free to assume this,
=
and
Exercise 6.3 to construct
a surjective homomorphism SU2(C)
SOs(R);
the
kernel
of
this
compute
homomorphism.)
If you know a little topology, you can now conclude that the fundamental
use
group40
of
S03(R)
is G2.
Warning: This is
one
[9.1, VI.1.3]
of several alternative conventions.
If you really want to believe this fact, remember that SOs(IR) parametrizes rotations in M3.
Hold a tray with a glass of water on top of your extended right hand. You should be able to rotate
the tray clockwise by a full 360° without spilling the water, and your muscles will tell you that
the corresponding loop in
50s(]R)
is not trivial. But then you will be able to rotate the tray again
Exercises
107
8.10. View Z
x
Z
as a
subgroup of
R
x
R:
Describe the quotient
R xR
Zx Z
in terms
to those used in
analogous
group? Cf. Exercise 1.1.6.)
Example 8.7. (Can
you 'draw a
picture' of this
(Notation as in Proposition 8.10.) Prove 'by hand' (that is, without invoking
universal properties) that N is normal in G if and only if N/H is normal in G/H.
8.11.
(Notation as in Proposition 8.11.) Prove 'by hand' (that is, by using
6.2) that HK is a subgroup of G if H is normal.
8.12.
Proposition
8.13.
-i
Let G be
a
finite commutative group, and
every element of G is
a square.
8.15. Let a,
n
|G|
is odd. Prove that
[8.14]
8.14. Generalize the result of Exercise 8.13: if G is
integer relatively prime
assume
to n, then the function G
a group
G,
g
of order
gk
n
and k is
an
is surjective.
i—?
be positive integers. Prove that n divides
see Exercise 6.14. (Hint: Example 8.15.)
</>(an
1),
where </> is
Euler's </>-function;
8.16. Generalize Fermat's little theorem to congruences modulo
possibly nonprime) integers. Note that
it is not true that an
for example, 24 is not congruent to 2 modulo 4.
generalization is known as Euler's theorem.)
a
and
n:
8.17. > Assume G is
a
Prove that there exists
otherwise,
amodn for all
What is true?
(This
finite abelian group, and let p be a prime divisor of |G|.
element in G of order p. (Hint: Let g be any element
an
of G, and consider the subgroup
show that there is an element h G
use
arbitrary (that is,
=
the quotient
(g); use the fact that this subgroup is cyclic to
(g) in G of prime order q. If q p, you are done;
G/(h) and induction.) [§8.5, 8.18, 8.20, §IV.2.1]
=
an abelian group of order 2n, where n is odd.
Prove that G has
2.
element
of
order
has
at
least
for
one,
example by Exercise 8.17.
exactly
(It
Use Lagrange's theorem to establish that it cannot have more than one.) Does the
8.18. Let G be
one
same
a
conclusion hold if G is not necessarily commutative?
full 360° clockwise without spilling any water, taking it back to the original position. Thus, the
(homotopically) trivial, as it should be if the fundamental group is cyclic of
square of the loop is
order 2.
II. Groups, first encounter
108
8.19. Let G be
a
finite group, and let d be a proper divisor of |G|. Is it necessarily
an element of G of order d? Give a proof or a counterexample.
true that there exists
8.20. > Assume G is
that there exists
a
a
finite abelian group, and let d be a divisor of |G|. Prove
H C G of order d. (Hint: induction; use Exercise 8.17.)
subgroup
[§IV.2.2]
8.21. > Let
i7, K be subgroups of
a group
G. Construct
a
bijection between the
set of cosets hK with h G H and the set of left-cosets of H fi K in H. If H and K
are
finite,
prove that
[§8.5]
: G
G' be a group homomorphism, and let N be the smallest
normal subgroup containing im<^. Prove that G'/N satisfies the universal property
of coker<£ in Grp. [§8.6]
8.22. > Let cp
8.23. > Consider the
subgroup
^={(l
3)'(2
2
of 53. Show that the cokernel of the inclusion H
is
not
3)}
1
^->
S3 is trivial, although H
^>
S3
surjective. [§8.6]
8.24. > Show that
epimorphisms
Grp do
in
not
necessarily have right-inverses.
[§I.4.2]
8.25. Let H be
a
commutative normal
homomorphism from G/H
9.
Group
to
subgroup of G. Construct
Aut(H). (Cf.
Exercise
an
interesting
7.10.)
actions
9.1. Actions. As mentioned in §4.1,
category C is simply
a
an action
of
a group
G
on an
object A of
a
homomorphism
a :
G
Autc(A).
-?
The way to interpret this is that every element g G G determines a 'transformation
of A into itself, i.e., an isomorphism of A in C, and this happens compatibly with
the operation of G and composition
In
a
rather strong sense,
we
in C.
really only
care
about groups because they act
on
things: knowing that G
something about A; group actions are
one key tool in the study of geometric and algebraic entities.
In fact, group actions are one key tool in the study of groups themselves: one
of the best ways to 'understand' a group is to let it act on an object A, hoping
that the corresponding homomorphism a is an isomorphism, or at least an injective
monomorphism. For example, we were lucky with Dq in §2.2: we let Dq act on a
set with three elements (the vertices of an equilateral triangle) and observed that
the resulting a is an isomorphism. Thus Dq
S3. We would be almost as lucky
acts on A tells us
=
9. Group actions
109
by letting Dg act on the vertices of a square: then a would
an explicit subgroup of S4, which simplifies its analysis.
at least realize
Definition
category C is faithful
9.1. An action of
The
case
C
9.2. Actions
that
Autc(A)
G
a group
(or effective) if the corresponding
a :
Spelling
on sets.
on
j
it in this
chapter.
out our definition of action in case A is a set, so
is the symmetric group
Definition 9.2. An action of
focus
we
a
as
is injective.
Autc(A)
Set is already very rich, and
=
object A of
on an
G
Dg
Sa,
G
a group
we
get the following:
A is
on a set
a
set-function
p:GxA^A
such that
p(eG, a)
=
for all
a
(Vg,h
G
A and
a G
G),(Va
A)
G
p(gh,a)
:
=
p(g,p(h,a)).
j
function p satisfying these conditions, we can define a
a(g)(a) p(g,a). (This defines a(g) as a set-function A
Indeed, given
a
Homset(^4, A) by
needed.) This function
cr(gh)(a)
=
=
Thus the image of
p(gh, a)
p(g, p(h, a))
=
we
o
a
cr(g)(a)
cr(g~1)
a(g)(a);
=
acts as the inverse of
consists of invertible
Conversely, given
p(g,a)
cr(g)(p(h, a))
=
cr(g~1g)(a)
=
have verified that this is
=
A,
as
a(g)(a(h)(a))
cr(g)ocr(h)(a).
a :
and
G
preserves the operation, because
In particular, this verifies that
^{g~X)
:
=
the
a
a
G
cr(eG)(a)
=
set-functions;
p(eG, a)
a
A)
G
a.
=
is acting
(Va
as a
function
Sa,
—>
homomorphism,
homomorphism
same
=
because
cr(g):
a
:
needed.
as
Sa, define p
G
:
G
x
A
A by
argument (read backwards) shows that p satisfies the
needed properties.
It is unpleasant to carry p along. In practice,
one
requirements in Definition 9.2 amount then to eGa
=
(V(/,ft€G),(VaGA): (gh)a
'as if p defined
If G
acts on
an associative
A, then
eGa
just writes
a
=
for all
ga for
a G
p(g, a);
the
A and
g(ha),
operation.
=
a
for all
a G
A; the
action of
a group
G
on a set
A
is faithful if and only if the identity eG is the only element g of G such that ga
a
for all a G A, that is, 'fixing' every element of A. An action is free if the identity
eG is the only element fixing any element of A.
=
Example 9.3. Every group G acts in a natural way on the underlying set G. The
function p : G x G
G is simply the operation in the group:
(V#, a
G
G)
:
p(#, a)
=
ga.
II. Groups, first encounter
110
In this
case the defining property really is associativity. This is referred to as the
action by left-multiplication41 of G on itself. There is (at least) another very natural
G by
way to act with G on itself, by conjugation: define p : G x G
p(9,h) =ghg~1.
This is indeed
an
action:
p(g, p(h, k))
=
\/g,h,k G G,
gp(h, k)g~l
Example
9.4. More
left-cosets
to (ga)H.
(cf. §8.5)
=
generally, G
g(hkh~l)g~l
acts
=
(gh)k(gh)~l
by left-multiplication
of any subgroup H: act by g G G
on
=
p(gh, k).
j
the set G/H of
G/H by sending it
on
aH G
j
in
These examples of actions are extremely useful in studying groups, as we will see
Chapter IV. For instance, an immediate consequence is the following counterpart
to
§8.2:
Theorem 9.5
(Cayley's theorem). Every
is, every group may be realized
as a
group acts
subgroup of
a
faithfully
on some set.
That
permutation group.
Proof. Indeed, simply observe that the left-multiplication action of G
on
itself is
manifestly faithful.
The notion defined in Definition 9.2 is, for the sake of precision, called a
right-action would associate to each pair (g, a) with g G G and a G A
/enaction. A
element ag G A',
our
make-believe associativity would
a(gh)
for all
a G
=
(ag)h
A and g, h G G. This is a different requirement than the one
multiplication on the right in a group G gives a prototypical
Definition 9.2;
of
a
an
given in
example
right-action of G (on itself).
Every right-action
may be turned into a left-action with due care
(cf.
Therefore it is not restrictive to just consider left-actions; from
'action' will be understood to be a left-action, unless stated otherwise.
Exercise
an
now say
9.3).
now on,
9.3. Transitive actions and the category G-Set.
Definition 9.6. An action of
such that b
=
a group
G
on a set
A is transitive if Va, b G
A3g
ga.
G G
j
For example, the left-multiplication action of a group on itself is transitive.
Transitive actions are the basic ingredients making up every action; this is seen by
means of the following important concepts.
Definition 9.7. The orbit of
a G
A under
an
action of
a group
G is the set
0G(a):={ga\geG}.
This is /e/t-multiplication in the sense that the 'acting' element g of G is placed to the left
of the element a 'acted upon'.
111
9. Group actions
Definition 9.8. Let G act
a
A, and let
on a set
consists of the elements of G which fix
Stabc(a)
Orbits of
have
an
action of
an
:=
{g
G
on
a group
a G
A. The stabilizer subgroup of
a:
e G
a set
a}.
=
ga
A form
j
partition of A; and
a
induced,
sense, 'understand' all actions if we understand transitive actions.
accomplished
a
in
we
Therefore we can, in a
transitive action of G on each orbit.
This will be
moment, by studying actions related to stabilizers.
a
For any group G, sets endowed with a (left) G-action form in a natural way
A is an action (as
category G-Set: objects are pairs (p,A), where p : G x A
in Definition
and morphisms between two objects
actions. That is, a morphism
9.2)
are
set-functions which
are
compatible with the
(p,A)^(p',A')
in G-Set amounts to
set-function cp
a
G
:
A
^-^
A
x
A' such that the diagram
GxA'
->A'
commutes.
In the usual shorthand notation omitting the p's, this
means
that
\/geG,\/aeA,
g(p(a)
=
p(ga);
that is, the action 'commutes' with (p. Such functions
are
called
(G-) equivariant.
isomorphism of G-sets (defined as in §1.4.1); the
reader should expect (and should verify) that these are nothing but the equivariant
bisections.
Among G-sets we single out the sets G/H of left-cosets of subgroups H of G;
We therefore have
as
a
notion of
noted in Example 9.4, G acts
on
G/H by left-multiplication.
Every
left-action of G on a set A is isomorphic to the
the stabilizer of any a G A.
left-multiplication of G on G/H, for H
Proposition
9.9.
transitive
=
Proof. Let G act
H
=
Stabc(a).
transitively
on
a
set
We claim that there is
(p
:
an
A, let a G A be any element, and let
equivariant bijection
A
G/H
-?
defined by
(p(gH)
:=
ga
for all g eG.
Indeed, first of all
{9\192)o>
cp is well-defined:
a, and it follows that g±a
=
if
giH
=
g<2,H, then gf lg2 G H, hence
verify that cp is bijective,
g^a as needed. To
a function ip : A
G/H by sending an element ga of A to gH\ ip is welldefined because if g±a
&, so gf lg2 G H and g\H g2H. It
g^a, then
is clear that (p and ip are inverses of each other; hence (p is a bijection.
define
=
Equivariance
is immediate:
g^^gio)
(p(gf(gH))
=
=
g'ga
=
=
gf(p(gH).
112
II.
Corollary 9.10. IfO
O is a finite set and
is an orbit
of the
action
of a finite
\0\ divides
Proof. By Proposition 9.9 there is
a
encounter
group G on a set
A, then
\G\.
bijection between O and Gj Stabc(a) for
any
O; thus
element a G
|0|.|StabG(a)|
by Corollary
=
|G|
8.14.
Corollary
upgrades Lagrange's theorem to orbits of any action; it is
provides a very strong constraint on group actions.
9.10
extremely useful,
as
it
9.11. There
Example
Groups, first
are no
transitive actions of S3
on a set
with 5 elements.
Indeed, 5 does not divide 6.
j
Ultimately, almost everything we will prove in Chapter IV on the structure of
finite groups will be a consequence of 'counting arguments' stemming from
applications of Corollary 9.10 to actions of a group by conjugation or left-multiplication.
There
may seem to be an element of arbitrariness in the statement of
change the element
change, but it does so
Proposition 9.9: what if we
The stabilizer
may
Proposition 9.12. Suppose
b
=
ga.
a group
of which
G acts
we are
taking the stabilizer?
controlled way:
a
on
a set
A, and let
a
G
A,
g G
G,
Then
StebG(b)
Proof. Indeed,
assume
h G
ghg~l
G
Stabc(&). This
a
g~lb.
argument, noting that
=
StabG(a);
(ghg~l)(b)
thus
a
in
=
gStebG(a)g~1.
then
gh(g~lg)a
=
gha
=
ga
=
b
:
proves the D inclusion; C follows by the
same
=
to be normal, then it is really independent
In
of a (in any given orbit).
any case, there is an isomorphism of G-sets beween
G/H and G/(gHg~l), as follows from these considerations (and as the reader will
For example, if
Stabc(a) happens
independently check in Exercise
9.13).
Exercises
113
Exercises
9.1.
(Once
already familiar with
more, if you are
a
little linear
algebra...)
The
matrix groups listed in Exercise 6.1 all come with evident actions on a vector
space: if M is an n x n matrix with (say) real entries, multiplication to the right by
a column n-vector v returns a column n-vector Mv, and this defines a left-action
on
Rn viewed
as
the space of column n-vectors.
Prove that, through this action, matrices M G
inRn.
Find
interesting action of SL2(C)
an
on
On(R)
preserve
lengths and angles
R3. (Hint: Exercise 8.9.)
9.2. The effect of the matrices
(o _i). (_i o)
the plane is to respectively flip the plane about the y-axis and to rotate it 90°
clockwise about the origin. With this in mind, construct an action of D$ on R2.
on
9.3. If G
=
supported
on
(G, •) is
the
a
group,
(V#,/i
Verify that G°
we
define
can
'opposite'
an
group G°
=
(G, •)
G, by prescribing
same set
is indeed
e
G)
gmh:=h-g.
:
a group.
Show that the 'identity': G°
G, g
g is an
isomorphism if and only if G
i—?
is
commutative.
Show that G°
=
G
(even
if G is not
commutative!).
Show that giving a right-action of G on a set A is the
Sa, that is, a left-action of G° on A.
morphism G°
same as
giving
a
homo-
Show that the notions of left- and right-actions coincide 'on the nose' for
commutative groups. (That is, if (g, a) i—?
ag defines a right-action of a commutative
group G
on a set
A, then setting
ga
=
ag defines a
left-action).
For any group G, explain how to turn a right-action of G into a left-action
of G. (Note that the simple 'flip' ga
ag does not work in general if G is not
=
commutative.)
9.4. As mentioned in the text,
on
right-multiplication defines
itself. Find another natural right-action of
a group on
9.5. Prove that the action by left-multiplication of
9.6. Let O be
an
action of G
O is transitive.
on
orbit of
an
9.7. Prove that stabilizers
9.8. For G
a group,
action of
are
a group
G
a
right-action of a
a group on
on a set.
group
itself.
itself is free.
Prove that the induced
indeed subgroups.
verify that G-Set is indeed a category, and verify that the
are precisely the equivariant bijections.
isomorphisms in G-Set
114
II.
Groups, first
encounter
products and coproducts and that every finite object of
of
coproduct
objects of the type G/H
{left-cosets of H}, where H is
of
and
acts
on
G
G
subgroup
G/H by left-multiplication.
9.9. Prove that G-Set has
G-Set
a
is
a
=
9.10. Let H be any
the
subgroup of a
G. Prove that there is
group
bijection between
a
G/H of left-cosets of H and the set H\Gof right-cosets of H in G. (Hint:
G acts on the right on the set of right-cosets; use Exercise 9.3 and Proposition 9.9.)
set
9.11.
-i
Let G be
finite group, and let H be a subgroup of
H is normal in G,
a
dividing \G\.Prove that
the smallest prime
Interpret the action of G
o~ : G
Sp.
Then
is
G/kera
the index of ker
Show that ker
(isomorphic to)
as
p, where p is
follows:
a
homomorphism
subgroup of Sp. What does this
a
say about
in G?
a
a C
Conclude that H
G/H by left-multiplication
on
index
as
H.
ker a, by index considerations.
=
a kernel, proving that it is normal. (This exercise generalizes the result
of Exercise 8.2.) [9.12]
Thus H is
9.12.
-i
Generalize the result of Exercise 9.11, as follows. Let G be a group, and
a subgroup of index n. Prove that H contains a subgroup K that is
let H C G be
normal in G and such that
[G
divides the gcd of
K]
:
\G\and
n\.
(In particular,
[G:K}< n!.) [IV.2.23]
9.13. > Prove
'by hand' that for all subgroups
G/{gHg~l) (endowed
in G-Set. [§9.3]
H of
with the action of G by
a group
G and Vg G G,
left-multiplication)
are
G/H
and
isomorphic
Prove that the modular group
PSL2(Z) is isomorphic to the coproduct G2*
that
the
modular
G3. (Recall
group PSL2(Z) is generated by x
(? ~J) and
y
(1 ~q)' satisfying the relations x2 y3 e in PSL2(Z) (Exercise 7.5). The
task is to prove that x and y satisfy no other relation: this will show that PSL2(Z)
is presented by (x,y | x2,?/3), and we have agreed that this is a presentation for
G2 * G3 (Exercise 3.8 or 8.7). Reduce this to verifying that no products
9.14.
-i
=
=
=
(y±1x)(y±1x)---(y±1x)
=
{y±lx){y±lx)
or
{y±lx)y±l
equal the identity. This latter verification is
carried
out
traditionally
by cleverly exploiting an action42. Let the modular group
on the set of irrational real numbers by
with
one or more
factors
can
b\
(r)
a
a
N
J
Kc
Check that this does define
y(r)
=
l--j
r
an
y~1(r)
action of
=
The modular group acts on C U
suffices to act
on R
Q for the
ar +
,
-
1
,
r
{00} by
act
b
=
cr + a
PSL2(Z),
yx(r)
=
and note that
l+r,
y~lx{r)
=
1 +
r
Mobius transformations. The observation that it
purpose of this verification is due to
Roger Alperin.
10. Group
objects
in
115
categories
Now complete the verification with a case-by-case analysis. For example, a product
(y±1x)(y±1x) (y±1x)y cannot equal the identity in PSL2(Z) because if it did, it
would act as the identity on R
Q, while if r < 0, then y(r) > 0, and both yx and
send
irrationals
to
positive
positive irrationals.) [3.8]
y~lx
9.15.
-i
Prove that every
(finitely generated)
group G acts
freely
on
any
directed graph are defined
if the vertices vi, V2 are
as actions on the set of vertices preserving incidence:
connected by an edge, then so must be gvi, gv<i for every g G G.) In particular,
conclude that every free group acts freely on a tree. [9.16]
corresponding Cayley graph. (Cf.
9.16. > The
groups
group
(on
a
freely
finite
AutG_set(G)
*
on a tree.
is free.
set)
9.17. > Consider G
G.
as
a
a group
G
10.
[§6.4]
G-set, by acting with left-multiplication.
Prove that
[§2.1]
9.18. Show how to construct
will be the
on a
of the last statement in Exercise 9.15 is also true: only free
Assuming this, prove that every subgroup of a free
converse
can act
Exercise 8.6. Actions
A.
(Hint:
morphisms?)
on a set
Group objects
in
a
groupoid carrying the information of the
action of
A will be the set of objects of the groupoid. What
categories
Categorical viewpoint. The definition of group (Definition 1.2) is firmly
grounded on the category Set: A group is a set G endowed with a binary
operation
However, we have noticed along the way (for example, in §3) that what is
behind
it is a pair of functions:
really
10.1.
m:GxG^G,
l.G^G
satisfying certain properties (which translate into associativity, existence of inverses,
etc.). Much of what we have seen could be expressed exclusively in terms of these
functions, systematically replacing considerations on 'elements' by suitable
commutative diagrams and enforcing universal properties as a means to define key notions
such as the quotient of a group by a subgroup. For example, homomorphisms may
diagram: cf. Definition 3.1.
This point of view may be transferred easily to categories other than Set, and
the corresponding notions are very important in modern mathematics.
be defined
purely
in terms of the
Definition 10.1. Let C be
a
commutativity of
a
category with (finite) products and with
object 1.
A group object in C consists of
an
m:GxG^G,
object G of C and of morphisms
e :
1
G,
-?
t.G^G
a
final
II. Groups, first encounter
116
in C such that the
diagrams
(GxG)xG
mxidc)GxG
idGXm
Gx(GxG)
)G
lxGJ^GxG
GxlJ^GxG
G^^GxG^^GxG
G^GxG}^GxG
+
G
-+G
commute.
j
The morphism A
idG x Hg is the 'diagonal' morphism G
G induced by the universal property for products and the identity map(s)
G —>
G. Likewise, the other unnamed morphisms in these diagrams are all uniquely
Comments.
G
=
x
determined by suitable universal properties. For example, there is a unique
1 because 1 is final. The composition with the projection
morphism e : G —>
GJ^lxG.
is the identity;
so
G
is
lxG-
(why?);
+
therefore the projection 1
->G
x
G
^lxG
G is indeed
an
isomorphism,
as
indicated.
The reader will
definition of
hopefully realize immediately (Exercise 10.2) that our original
groups given in §1 is precisely equivalent to the definition of group
object in Set: the commutativity of the given diagrams codifies associativity and
the existence of two-sided identity and inverses.
Most interesting categories the reader will
such
encounter
(not necessarily
in this
the category of topological spaces, differentiable manifolds, algebraic
book),
varieties, schemes, etc., will carry 'their own' notion of group object. For example,
a
as
topological
object in the category of topological spaces; a Lie
in the category of differentiable manifolds, etc.
group is a group
group is a group
object
Exercises
117
Exercises
10.1. Define all the unnamed maps appearing in the diagrams in the definition of
group object, and prove they are indeed isomorphisms when so indicated. (For the
projection 1
x
G, what is left to prove is that the composition
G
1xG^G^lxG
is the identity,
as
mentioned in the
10.2. > Show that groups,
sets'.
defined in §1.2,
are
'group objects
in the category of
[§10.1]
10.3. Let
(w.r.t.
as
text.)
(G, •)be
•)such that
prove that the
a group,
(G, o)
and suppose o:GxG->Gisa group homomorphism
a group. Prove that o and
coincide. (Hint: First
is also
identity with respect
to the two
10.4. Prove that every abelian group has
operations
must be the
exactly
one structure
object
Grp?
in Ab is
same.)
of group object in
the category Ab.
10.5. By the previous exercise, a group
abelian group. What is a group object in
nothing other than
an
Chapter
Rings
1.
III
and modules
Definition of
ring
In this chapter we will do for rings and modules what we have done in Chapter II for
groups: describe them in general terms, with particular attention to distinguished
subobjects and quotients. More detailed information on these structures will be
in Chapter V we will look more carefully at several
classes
of
and
modules (over commutative rings) will take center
interesting
rings,
in
our
overview
of
linear
rapid
stage
algebra in Chapter VI and following. In this
will
we
also
include
a
brief
chapter
jaunt into homological algebra, a topic that will
entertain us greatly in Chapter IX.
deferred to later chapters:
1.1. Definition. Rings (and modules) are defined by 'decorating' abelian groups
with additional data. As motivation for the introduction of such structures, note
that all number-based examples of groups that we have encountered, such as Z or R,
operation of multiplication as well as the 'addition' making
them into (abelian) groups. The 'ring axioms' will reflect closely the properties and
compatibilities of these two operations in such examples.
are
endowed with
an
These examples
are,
however,
for the introduction of rings arises
phisms of abelian groups. Recall
special. A more sophisticated motivation
by analyzing further the structure of homomorvery
that if G, H are abelian groups, then
HomAb(G, H) is also an abelian group. In particular, if G is an abelian group, then
so is the set of endomorphisms EndAb(G)
HoniAb(G, G). More is true: mor-
(§11.4.4)
=
an object of a category to itself may be composed with each other (by
definition of category!). Thus, two operations coexist in EndAb(G): addition
(inherited from G, making EndAb(G) an abelian group), and composition. These two
phisms from
operations
are
compatible with each other
Definition 1.1. A ring
(i?, +, •)is
binary operation •,satisfying on its
having a two-sided identity, i.e.,
an
in
a sense
captured by the ring
axioms:
abelian group (i?, +) endowed with a second
the requirements of being associative and
own
119
Rings and modules
III.
120
(Vr,s,te R)
(31*
(which
:
i?) (Vr
G
make
G
(r s)
t
R)
r
:
(s *),
r
=
1*
(i?, •) a monoid),
r
=
lR
=
r
and further interacting with + via the
following
distributive properties:
(Vr, s,t
G
R)
(r
:
name
5)
£
=
r
5 + r
t
and £
(r
is often omitted in formulas, and
The notation
the
+
+
5)
=
£
r + £
usually refer
we
s.
to
j
rings by
of the underlying set.
we are calling a 'ring', others may call a 'ring with identity'
1':
or a 'ring with
it is not uncommon to exclude the axiom of existence of a
'multiplicative identity' from the list of axioms defining a ring. Rings without
identity are sometimes called rngs, but we are not sure this should be encouraged1.
The reader should check conventions carefully when approaching the literature.
Examples of structures without a multiplicative identity abound: for example, the
set 2Z of even integers, with the usual addition and multiplication, satisfies all the
ring axioms given above with the exception of the existence of 1 (and is therefore
Warning: What
a
But in these notes all rings will have 1.
rng).
Of course the multiplicative identity is necessarily unique: the argument given
for Proposition II. 1.6 works verbatim.
The identity element of the abelian group underlying a ring is denoted 0* (or
simply 0, in context) and is called the 'additive' identity. This is a special element
with respect to multiplication:
Lemma 1.2. In
a
ring R,
0.r
for
all
r G
=
0 + 0; hence,
r.0
from which
It
=
r-0
R.
Proof. Indeed, 0
proven
0
=
r
0
0
=
=
applying distributivity,
r-(0
+
0)=r.0
by cancellation (in the
group
+
r-0,
The equality 0
(i?, +)).
r
=
0 is
similarly.
is
equally
easy to check that
multiplication behaves
as
expected
on
r
'subtraction'. In fact, if —1
denotes the additive inverse of 1, then the additive inverse
r
r:
of any
G R is the result of the multiplication (—1) indeed, using distributivity,
r +
(by
Lemma
The
1.2)
term
time Mac Lane
(-1)
.
r
=
from which
'rng'
was
1
r +
(-1)
(—l)r r
=
.r
=
(l-l)-r
follows by
=
0-r
(additive)
=
0
cancellation.
introduced with this meaning by Jacobson; but essentially at the same
as the name for the category of rings with identity. Hoping to steer
introduced Rng
clear of this clash of terminology
we have
opted
to call this category
'Ring'.
1. Definition of
1.2.
*
.
examples and special classes of rings.
First
Example
in this
1.3. We
well
(as
* = *
121
ring
define
can
as * + *
=
a
structure on a trivial group
ring
{*} by letting
this is often called the zero-ring. Note that 0
*);
=
ring (cf. Exercise 1.1).
1
j
Example 1.4. More interesting examples are the number-based groups such as Z
or R, with the usual operations. These are very well known to our reader, who will
realize immediately that they satisfy the requirements given in Definition 1.1; but
they
special. Why?
are very
To
begin with,
note that
j
multiplication
is not among the requirements
we
is commutative in these
have posed
rings
on
examples; this
given
in the official definition
above.
Definition
(Vr, s
ring R is
1.5. A
G
R)
:
s = s
r
commutative if
r.
j
extremely important class of rings;
algebra
algebra studying them. We will focus on
commutative rings in later chapters; in this chapter we will develop some of the
basic theory for the more general case of arbitrary rings (with 1).
Commutative rings
(with identity)
form
an
is the subfield of
commutative
1.6. An example of a noncommutative ring that is (likely) familiar to
the readers is the ring of 2 x 2 matrices with, say, real entries: matrices can be
added 'componentwise', and they can be multiplied as recalled in Example II. 1.5;
the two operations satisfy the requirements in Definition 1.1.
Example
Square matrices of any size, and with entries in any ring, form
a
ring
(Exercise 1.4).
j
Example 1.7. The reader is already familiar with a large class of (commutative)
rings: the groups Z/nZ, endowed with the multiplication defined in §11.2.3 (that is:
[o]n [b]n
'
[^6]n;
this is well-defined
:—
Exercise
(cf.
II.2.14)) satisfy
the ring axioms
listed above.
The rings
j
Z/nZ prompt
why rings such
nonzero
in
Z, Q, R,
us to
...
highlight an important point. Another reason
special is that multiplicative cancellation by
are
elements holds in these rings. Of
since rings
fails
as
are
general
in
particular (abelian)
since
one cannot
groups; and
'cancel 0'
(Va GiJ),a^0:
a
(by
-b
=
a
holds, for example, in Z, does not follow
Indeed, this cancellation property does not
rings Z/nZ. For example,
[2]6
though [4]6 ^ [1]6.
The problem here is
[4]6
=
[8]6
=
[2]6
multiplicative cancellation clearly
Lemma
c
1.2).
==>
b
from the
which
the
additive cancellation is automatic,
course
=
But
the fact that
c,
ring axioms.
rings; it may
hold in all
=
even
well fail in
[2]6 [1]6
even
for
some
6^0 (take
a
=
that in
[2]6,
b
Z/6Z there
[S]q).
=
are
elements
a
^
0 such that
a
b
=
0
122
III.
Definition
elements b
1.8.
^
An element
in
a
0 in R for which ab
=
ring R is
a
Rings and modules
(left-)zero-divisor
a
if there exist
0.
j
The reader will have no difficulty figuring out what a n#/i£-zero-divisor should
be. The element 0 is a zero-divisor in all nonzero rings R\ the zero ring is the only
ring without zero-divisors(l).
Proposition
1.9. In
a
ring R,
R is not a left- (resp., right-) zero-divisor if and
R.
multiplication by a is an injective function R
a G
In other words,
multiplicative left-
a
is not
(resp., right-)
Proof. Let's
a
left-
only
if left (resp., right)
zero-divisor if and only if
a holds in R.
(resp., right-)
cancellation by the element
verify the 'left'
analogous). Assume a is not
(the 'right'
statement
a
statement is of course
left-zero-divisor and ab
=
ac
entirely
for b,c e R. Then, by
distributivity,
a(b
and this implies b
c
=
0 since
c)
a
=
ab
is not
a
0,
left-zero-divisor; that is, b
ac
=
proves that left-multiplication is injective in this
=
c.
This
case.
a
0
left-zero-divisor, then 3b ^ 0 such that ab
0; this
shows that left-multiplication is not injective in this case, concluding the proof.
Conversely,
Rings such
Such rings
if
as
a
is
a
=
commutative rings without
Z, Q, etc., are
special, but
are very
very
(nonzero)
=
zero-divisors.
important, and they deserve their
own
terminology:
Definition
1.10. An
integral domain
is
a nonzero
commutative ring R
(with 1)
such that
(Vo, be R)
:
ab
=
0 ^
a
=
0orb
=
0.
j
Chapter V will be entirely devoted to integral domains.
An element which is not a zero-divisor is called a non-zero-divisor.
Thus,
integral domains are those nonzero commutative rings in which every nonzero element
is a non-zero-divisor. By Proposition 1.9, multiplicative cancellation by nonzero
elements holds in integral domains. The rings Z, Q, R, C
As we have seen, some Z/nZ are not integral domains.
Here
pausing
all integral domains.
of those places where the reader can do him/herself a great favor by
moment and figuring something out: answer the question, which Z/nZ
is
a
are
one
integral domains? This is entirely within reach, given what the reader knows
question will be answered
already. Don't read ahead before figuring this out—this
are
within
a
few short paragraphs, spoiling all the fun.
There
special ring: we will see in due
time that it is a 'UFD' (unique factorization domain); in fact, it is a 'PID' (principal
ideal domain); in fact, it is more special still!, as it is a 'Euclidean domain'. All of
this will be discussed in Chapter V, particularly §V.2.
are even
subtler
reasons
why
Z is
a very
1. Definition of
123
ring
However, Q, R, C
are
are more
some, since
they
fields.
Definition 1.11. An element
it is
special than all of that and then
right-unit if 3v
a
u
is a
by
u
a
ring R is
=
1.
a
left-unit
if 3v G R such that
uv
=
Units are two-sided units.
1;
j
ring R:
and only
left- (resp., right-) unit if
surjective functions R
is a
is a
if u
of
G R such that vu
1.12. In a
Proposition
u
if left- (resp., right-) multiplication
R;
left- (resp., right-) unit, then right- (resp., left-) multiplication by
u is not a right- (resp., left-) zero-divisor;
u
is infective; that is,
the
inverse
of
two-sided unit is unique;
a
two-sided units
form
Proof. These assertions
a group
are
all
under multiplication.
straightforward.
R ng/i£-multiplication by u, so that
such that vu
1; then Vr G R
pu(r)
=
ru.
For example, denote by pu : R
If u is a right-unit, let v G R be
=
°
pu
That is, pv is
Pv(r)
=
pu(rv)
=
(rv)u
=
r(vu)
=
rlR
=
r.
right-inverse to pu, and therefore pu is surjective (Proposition 1.2.1).
Conversely, if pu is surjective, then there exists a v such that Ir
p(u)(v) vu->
a
=
so that u is a
right-unit.
This checks the first statement, for right-units.
For the second statement, denote by Xu : R
R /^-multiplication by u:
ur.
Assume
u
is
a
and
let
v
be
such
that vu
1#; then Vr G R
right-unit,
Xu(r)
=
=
^v
That is, Xv is
The
a
°
Au(r)
=
Xv(ur)
=
v(ur)
=
(vu)r
=
l#r
=
r.
left-inverse to \u,so Xu is injective (Proposition 1.2.1 again).
rest of the
proof
is left to the reader
(Exercise 1.9).
a two-sided unit u is unique, we can give it a name; of
denote it by u~l. The reader should keep in mind that inverses of left- or
right-units are not unique in general, so the 'inverse notation' is not appropriate
Since the inverse of
course we
for them.
Definition 1.13. A division ring is
two-sided unit.
a
ring
in which every
nonzero
element is
a
j
We will mostly be concerned with the commutative case, which has its
own
name:
Definition 1.14. A
nonzero
element is
a
field
is
a nonzero
commutative ring R
(with 1)
in which every
unit.
j
The whole of Chapter VII will be devoted to studying fields.
By Proposition 1.12 (second part), every field is an
conversely: indeed, Z is an integral domain, but it is not
field => integral domain,
integral domain, but
a
field. Remember:
not
124
III.
integral domain
There is a
=£> field.
situation, however, in which the
Proposition 1.15. Assume R is
domain if and only if it is a field.
Rings and modules
two notions coincide:
a finite commutative ring; then R is an
integral
Proof. One implication holds for all rings, as pointed out above; thus we only have
to verify that if R is a finite integral domain, then it is a field. This amounts to
verifying that if
a unit in R.
a
is
a
non-zero-divisor in
finite (commutative) ring i?,
a
then it is
Now, if a is a non-zero-divisor, then multiplication by a in R is injective
(Proposition 1.9); hence it is surjective, as the ring is finite, by the pigeon-hole principle;
hence a is a unit, by Proposition 1.12.
Remark 1.16. A little surprisingly, the hypothesis of commutativity in
Proposition 1.15 is actually superfluous: a theorem known as Wedderburn's little theorem
shows that
finite
division rings
this fact in a distant future
Example
are
necessarily
commutative. The reader will prove
(Exercise VII.5.14).
j
of units in the ring Z/nZ is precisely the group (Z/nZ)*
§11.2.3: indeed, a class [m]n is a unit if and only if (right-)
1.17. The group
introduced in
by [m]n is surjective (by Proposition 1.12), if and only if the map a i—?
a[m]n
1
is surjective, if and only if [m]n generates Z/nZ, if and only if gcd(ra,n)
(Corollary II.2.5), if and only if [m]n G (Z/nZ)*.
multiplication
=
In particular, those n for which all nonzero elements of Z/nZ are units (that is,
1 for all
for which Z/nZ is a field) are precisely those n G Z for which gcd(m, n)
=
that are not multiples of n; this is the case if and only if n is prime. Putting this
together with Proposition 1.15, we get the pretty classification (for integers p^O)
m
Z/pZ integral domain
^=>
Z/pZ
field ^=> p prime,
which the reader is well advised to remember firmly.
Example
1.18. The
rings Z/pZ, with
p
prime,
fact, for every prime p and every integer r
sense) multiplication on the product group
Z/pZ
x
...
x
V
>
are not
0 there is
j
the only finite fields. In
a (unique, in a suitable
Z/pZ
'
v
r
making
it into
a
times
field. A discussion of these fields will have to wait until
accumulated much
more
material
to construct small examples 'by
(cf. §VII.5.1), but the reader
hand' (cf. Exercise 1.11).
we
have
could already try
j
Polynomial rings. We will study polynomial rings in some depth, especially
over fields; they are another class of examples that is to some extent already familiar
to our reader. I will capitalize on this familiarity and avoid a truly formal (and
truly tedious) definition.
1.3.
1. Definition of
125
ring
ring. A polynomial f(x) in the indeterminate x and
finite linear combination of nonnegative 'powers' of x
Definition 1.19. Let R be
with
coefficients
in R is
a
a
with coefficients in R:
f(x)
^^ aiX%
=
=
a2X2
a0 + alx +
H
,
where all a* are elements of R (the coefficients) and we require a*
0 for i ^> 0.
if
Two polynomials are taken to be equal
all the coefficients are equal:
=
^ aiX1
=
^^ ^iX%
(^
^^
>
0)
:
°>i
=
&*•
J
The set of polynomials in x over i? is denoted by R[x], Since all but finitely
many ai are assumed to be 0, one usually employs the notation
f(x)
for
X^i>o a*x%
if
fli
0 for i >
=
a0 + aix H
=
anxn
h
n.
At this point the reader should just view all of this as a notation:
an element of an infinite direct sum of the group
really stands for
'polynomial' notation is
to impose on R[x]: if
more
f(x)=
suggestive
^2aiX%
it hints at what operations
as
and
g^
=
i>0
then
we
a
polynomial
(i?,+).
The
we are
going
^2biX^
i>0
define
f(x)
+
g(x)
:=
^(oi + 6i)xf
i>0
and
f(x)-g(x):=J2 E aA^'/c>0 i+J=/c
To
clarify
this latter definition,
how it works for small k:
see
+
aobo + (aobi + a\bo)x+ (ao&2 + a\b\
that is, business
as
a2bo)x2
+
(ao&3
f(x) g(x) equals
+ ai&2 + ^261 +
a^bo)xz
H
,
usual.
It is essentially straightforward (Exercise 1.13) to check that R[x], with these
operations, is a ring; the identity 1 of R is the identity of R[x], when viewed as a
polynomial (that is, Ir[x]
=
Ir + Ox + Ox2 H
).
The degree of a nonzero polynomial f(x)
X^i>o aix%-> denoted deg/(x), is the
d
for
which
This
notion
is very useful, but really behaves
0.
ad ^
largest integer
=
well (Exercise 1.14) only if R
R
=
is
an
integral domain: for example,
note that over
Z/6Z
deg([l]
+
deg(([l] + [2}x)
Polynomials of degree
of R in R[x], since the
0
\2]x)
=
deg([l] + [3]x)
1,
(1 + [3]*))
(together
with
operations +,
=
0)
deg([l]
are
+
=
[5]x)
1,
=
but
1^1
+ 1.
called constants; they form
on constant
polynomials
original operations in i?, up to this identification.
assign to the polynomial 0 the degree —00.
are
'copy'
nothing but the
a
It is sometimes convenient to
III.
126
in
Polynomial rings
Rings and modules
indeterminates may be obtained by iterating this
more
construction:
R[x,y,z]
:=
R[x][y][z];
elements of this ring may be written as 'ordinary' polynomials in three
indeterminates and are manipulated as usual. It can be checked easily that the construction
does not really depend on the order in which the indeterminates are listed, in the
sense that different orderings lead to isomorphic rings (in the sense soon to be
officially). Different indeterminates commute with each other; constructions
analogous to polynomial rings, but with noncommuting indeterminates, are also
very important, but we will not develop them in this book (we will glance at one
defined
such notion in
Example VIII.4.17).
We will occasionally consider a polynomial ring in infinitely many
indeterminates: for example, we will denote by R[xi,X2,...] the case of countably many
indeterminates. Keep in mind, however, that polynomials are finite linear
combinations of finite products of the indeterminates; in particular, every given element
of
] only involves finitely many indeterminates. An honest definition of
ring involves direct limits, which await us in §VIII. 1.4.
Rings of power series may be defined and are very useful; the ring of series
R[x\,X2,
this
X^o aiX%
is denoted
Q>o+Q>iZ+Q>2%2 +
-R[[#]]. Regrettably,
=
in
'..
we
x
with coefficients in i?, and evident operations,
will
only occasionally
encounter these
rings
in
this book.
The ring R[x] is (clearly) commutative if R is commutative; it is an integral
domain if R is an integral domain (Exercise 1.15); but it has no chances of being
a field even if R is a field, since x has no inverse in
R[x]. The question of which
properties of R
a
'inherited' by
are
R[x]
is subtle and important, and
we
will give it
great deal of attention in later sections.
1.4. Monoid rings. The polynomial ring is an instance of a rather general
construction, which is occasionally very useful. A semigroup is a set endowed with an
associative operation; a monoid is a semigroup with an identity element. Thus a
group
is
a
monoid in which every element has
ordinary addition form
nonnegative integers2)
Given
a
Elements of
monoid
R[M]
a
is
semigroup, while the
a
(that is,
monoid under addition.
(M, •)and
are
inverse; positive integers with
an
set N of natural numbers
a
ring i?,
we can
obtain
a new
ring R[M]
as
follows.
formal linear combinations
^2
am-rn
meM
where the 'coefficients' rm are elements of R and am
summands
sum
(hence,
as
in
i?eM). Operations
(
5Z
meM
§1.3,
in
R[M]
am'm)
abelian group
are defined by
as an
+
(^2
meM
bm-m)
=
^2
meM
'Some disagree, and insist that N should not include 0.
^
0 for at most
R[M]
(am
is
+
finitely many
nothing but the direct
bm)
-
m,
Exercises
127
(^2 CLrn'rn)'(^2
meM
brn-m)
The reader will hopefully
in
see
in fact
^2
^2
(a™>it>m2)m.
'
meM mim2=m
The identity in R[M] is Ir-Im, viewed
have 0 as coefficient.
polynomial ring
=
meM
as a
formal
sum
in which all other summands
the similarity with the construction of the
(Exercise 1.17) the polynomial ring R[x] may be
§1.3;
R[x]
R[N].
Group rings are the result of this construction when M is in fact a group. The
group ring R[Z] is a ring of 'Laurent polynomials' R[x,x~l], allowing for negative
interpreted
as
well
as
as
positive exponents.
Exercises
1.1. > Prove that if 0
1.2.
-i
Let S be
=
1 in a
ring i?, then R is
set, and define operations
a
on
a
zero-ring. [§1.2]
the power set
^(S)
of S by setting
\/A,Be&>(S)
A +
B:=(AuB)\(AnB),
A-B
=
A + B
AnB:
A-B
(where the solid black contour indicates the set included
(<^(5),+,-) is a commutative ring. [2.3, 3.15]
in the
operation).
Prove
that
1.3.
Let R be a ring, and let S be any set. Explain how to endow the set Rs of
set-functions S
R of two operations +, so as to make Rs into a ring, such that
Rs is just a copy of R if S is a sigleton. [2.3]
-i
1.4. > The set ofnxn matrices with entries in a ring R is denoted Mn(R). Prove
that componentwise addition and matrix multiplication make M.n(R) into a ring,
R
for any ring R. The notation $ln(R) is also commonly used, especially for R
or C (although this indicates one is considering them as Lie algebras) in parallel with
the analogous notation for the corresponding groups of units; cf. Exercise II.6.1. In
fact, the parallel continues with the definition of the following sets of matrices:
=
sln(R)
sln(C)
son(R)
sun(C)
=
=
=
=
{Me g[n(R) | tr(Af)
{Me Qln(C) | tr(Af)
{M
e
sin(R)
=
=
\M+ Mt
{Me sln(C) |
M + Aft
0};
0};
=
=
0};
0}.
Here tr M is the trace of M, that is, the sum of its diagonal entries. The other
notation matches the notation used in Exercise II.6.1. Can we make rings of these sets
III.
128
Rings and modules
by endowing them with ordinary addition and multiplication of matrices? (These
sets are all Lie algebras; cf. Exercise VI.1.4.) [§1.2, 2.4, 5.9, VI.1.2, VI. 1.4]
ring. If a,
1.5. Let R be a
1.6.
-i
An element
Prove that if
of
a
and b
a
Is the hypothesis ab
to hold?
a
ring R is nilpotent if an
are
=
i?, is a-\-bnecessarily
b are zero-divisors in
nilpotent
in R and ab
=
0 for
zero-divisor?
some n.
ba, then
=
a
a +
b is also nilpotent.
ba in the previous statement necessary for its conclusion
[3.12]
1.7. Prove that
factors of
is
[m]
nilpotent
in
Z/nZ
if and only if
m
is divisible
by all prime
n.
1.8. Prove that
domain. Find
1.9. > Prove
a
x
in which the equation x2
ring
Proposition
1.10. Let R be
only solutions
±1 are the
=
1.12.
is not
a
a
=
1 in an
integral
1 has more than 2 solutions.
=
[§1.2]
ring. Prove that if
a
left-inverses, then
equation x2
to the
a
R is
G
a
right-unit and has two
a right-zero-divisor.
or more
left-zero-divisor and is
1.11. > Construct a field with 4 elements: as mentioned in the text, the underlying
abelian group will have to be Z/2Z x Z/2Z; (0,0) will be the zero element, and (1,1)
will be the multiplicative identity. The question is what (0,1) (0,1), (0,1) (1,0),
(1,0) (1,0) must be, in order to get a, field. [§1.2, §V.5.1]
complex numbers
1.12. > Just as
may be viewed as combinations a +
i2
a, 6 G R and i satisfies the relation
construct
a
ring3
EI
a, b, c, d G R and i, j,
i2
=
j2
=
k2
=
—1
(and commutes with
bi, where
R),
we
by considering linear combinations a + bi + cj + dk
k commute with R and satisfy the following relations:
=
—1,
ij
=
/c, jk
—ji
=
=
i,
—kj
=
ki
=
may
where
—ik
j.
=
Addition in EI is defined componentwise, while multiplication is defined by imposing
distributivity and applying the relations. For example,
(l + i+j)-(2 + k)
(i) Verify
=
2 + 2i + 2j +
that this prescription does indeed define
(ii) Compute (a
(iii)
l'2 + i'2+j'2 + l-k + i'k+j'k
=
+ bi +
Prove that EI is
Elements of EI
subgroup of the
are
a
cj
+
dk)(a
bi
cj
dk),
a
k-j + i
=
ring.
where a,b,c,de R.
division ring.
called quaternions.
group of units of
Note that Q$ := {±1, ±i, ±j, ±/c} forms a
a noncommutative group of order 8, called
EI; it is
the quaternionic group.
(iv)
(v)
2 + 3i+j + k.
List all subgroups of Qs, and prove that they
Prove that Qs, D$
are
all normal.
isomorphic.
Prove
that
admits
the
Q%
presentation
(vi)
(x,y\x2y~2,y4:,xyx~1y).
are not
[§11.7.1, 2.4, IV.1.12, IV.5.16, IV.5.17, V.6.19]
The letter HI is chosen in honor of William Rowan Hamilton.
2. The category
1.13. >
Verify
129
Ring
that the multiplication defined in
1.14. > Let R be a
ring, and let f(x),g(x)
G
is associative.
R[x]
R[x]
be
nonzero
[§1.3]
polynomials. Prove
that
<
deg(/(x) +g(x))
Assuming
that R is
max(deg(/(x)),deg(^f(x))).
integral domain,
an
deg(/(x) g(x))
=
prove that
deg(f(x))
deg(g(x)).
+
[§1.3]
1.15. > Prove that
R[x]
is
an
integral domain if and only if R
is
an
integral domain.
[§1-3]
1.16. Let R be
(i)
Prove that
if ao is
(ii)
a
ring, and consider the ring of
a power
a^x1
series ao + a\X+
Explain
R[[x]]
is
in what
R[[x\] (cf. §1.3).
power series
is a unit in
H
unit in R. What is the inverse of 1
Prove that
1.17. >
a
x
integral domain if and only if R
an
sense
R[x]
R[[x]]
if and only
in -R [[#]]?
agrees with the monoid
is.
ring R[N], [§1.4]
2. The category Ring
Ring homomorphisms. Ring homomorphisms are defined in the natural
are rings, a function cp : R
S is a ring homomorphism if it preserves
both operations and the identity element. That is, cp must be a homomorphism of
the underlying abelian groups,
2.1.
way: if i?, S
(Vo, beR):
it must preserve the operation of
(p(a
b)
=
ip(a)
+
<p(6),
multiplication,
(p(ab)
(\/a,beR) :
and
+
=
(p(a)(p(b),
finally
<pO-r)
It is evident that rings form
a
=
is-
category, with ring homomorphisms
as
mor-
phisms. We will denote this category by Ring.
The zero-ring is clearly final in Ring. However, note that it is not initial:
because of the requirement that ring homomorphisms send 1 to 1, the only rings
to which zero-rings map homomorphically are the zero-rings (Exercise 2.1).
The category Ring does have initial objects: the ring of integers Z (with the
usual operations +, •)is initial in Ring. Indeed, for every ring R we can define a
R by
group homomorphism cp : Z
(Vn
that is,
in fact
as
a
G
Z)
<p(n)
:
=
n
1#,
the 'exponential map' e\R corresponding to 1# G R; cf.
ring homomorphism, since ip(l)
(p(mn)
=
(mn)lR
=
m(nlR)
=
=
§11.4.1.
1#, and
(mlR) (nlR)
=
(p(m) <p(n),
But (p is
Rings and modules
III.
130
where the equality
holds by the distributivity axiom4. This ring homomorphism
is unique, since it is determined by the requirement that (p(l)
Ir and by the fact
that (p preserves addition. Thus for every ring R there exists one and only one ring
=
=
homomorphism Z
i?, showing that Z is initial in Ring.
This should already convince the reader that Z is a very special ring. There is
in fact nothing arbitrary about Z: knowledge of the group Z determines the ring
structure of Z, as will be underscored below. This is one of the main reasons why
included the 'identity' axiom in the list in Definition 1.1: there are many nonisostructures of 'ring without identity' on the group (Z, +) (cf. Exercise 2.15),
we
morphic
but only
one structure
of ring with identity
Ring homomorphisms
S is
in R and (p : R
unit. Indeed, if
v
is
a
a
preserve units: that is, if u is a (left-, resp., right-) unit
ring homomorphism, then (p(u) is a (left-, resp., right-)
that
(p(v)
is
a
inverse of u, then
(right-, say)
(p(u)(p(v)
so
(right-)
(p(uv)
=
inverse of
may well be a zero-divisor:
a
for
ring homomorphism, and
=
<p(1r)
=
Is,
ip(u).
On the other hand, the image of
is
(Exercise 2.16).
non-zero-divisor by
a ring homomorphism
the
canonical
example,
projection tt : Z
Z/6Z
2 is
a
non-zero-divisor in Z, yet
a
7r(2)
=
[2]6
is a
zero-divisor.
2.2. Universal property of
universal property not unlike the
(and possibly
respect
most
useful)
to commutative
polynomial rings. Polynomial rings satisfy
case
rings;
a
for free groups explored in §11.5.2. The simplest
is for the polynomial rings Z[#i,
,xn], and with
one
we
will leave the reader the pleasure of stating and
proving fancier notions.
Let A
objects
are
=
{ai,... ,an}
be
of order
a set
pairs (j,R), where R is
a
:
j
is
a
set-function
n.
Consider the category Sia whose
ring5 and
commutative
A
R
-
(cf. §11.5.2!); morphisms
0"i,fli)->0'2,i*2)
are
commutative
diagrams
Rl
>
i?2
h
A
in which (p is a
ring homomorphism.
For example, (i, Z[#i,
,
xn\) is
an
object of &A, where
i
:
A
Z[#i,
,
xn]
sends a^ to Xk4The reader should parse this display carefully, as there is a potentially confusing mix of
multiples (such as ml#) and multiplication in R (explicitly denoted here by ¦).
two
operations:
For the reader interested in generalizations:
commute with every element of R is needed here.
only the requirement that
j(ai),... j(an)
,
2. The category
131
Ring
Proposition 2.1.
(i,Z[xi,-—
,xn])
is initial in&A-
arbitrary object of 3%a\ we
unique morphism (i,Z[#i,(j,R), that is,
,xn])
R such that
homomorphism ip : Z[#i,
xn]
Proof. Let
(j,R)
be
have to show that there is
an
there exists exactly
—>
one
a
ring
—>
,
Z[xi,--- ,xn]
i
->R
s'
commutes.
As usual in these verifications, the key point is that the requirements posed on
The postulated commutativity of the diagram means that
(p force its definition.
n.
1,
,
Then, since (p must be a ring homomorphism,
(p(xk)
j(a>k) for k
=
=
necessarily
=
where
^ :
Z
Thus,
^L(mil...iri)j(al)il---j(a7n)i™,
i? is the unique ring homomorphism
if (p exists, then it is unique.
(as
Z is initial in
obtained clearly6 preserves the operations and sends
homomorphism, concluding the proof.
=
s G
ring
Z
S. In this
case
element of
a
just
ring
ring 5,
and 'extending'
the commutativity of S is
to s
j
S be a fixed ring homomorphism, and
generally, let a : R
element commuting with a(r) for all r e R. Then there is a unique
5 extending a and sending x to s (Exercise 2.6).
homomorphism a : R[x]
Example
let
t :
we
1 to 1, so it does define a
1, Proposition 2.1 says that if s is any
Example 2.2. For n
then there is a unique ring homomorphism Z[x]
S sending x
the unique ring homomorphism
immaterial (why?).
Ring).
On the other hand, the formula
2.3. More
S be
an
In particular, we get an 'evaluation map' for polynomials over commutative
Given a polynomial f(x)
rings as follows.
X^>o a^x% ^ ^\x\ievery r G R
=
determines
an
element
/(r)
=
£>r':
i>0
this may be viewed
r.
and s
as
a(f(x)),
where
a
is obtained
as
above with
a
=
id#
:
R
R
=
i?,
polynomial f(x) determines a polynomial function f : R
good idea to keep the two concepts of 'polynomial' and
j
'polynomial function' well distinct (cf. Exercise 2.7).
Thus,
defined by
every
r i—?
It is a
/(r).
Well, it is really clear that it preserves addition; multiplication requires a bit of work, which
the reader would be well advised to perform. This is where the commutativity of R is used.
Rings and modules
III.
132
2.3.
Monomorphisms and epimorphisms. The kernel of a homomorphism
R
S of rings is
kerip
:=
{r
R\(f(r)
e
0}
=
cp
:
:
that is, it is nothing but the kernel of cp when the latter is viewed as a
homomorphism of groups. As such, ker (p is a subgroup of (i?, +); it satisfies an even stronger
requirement7, which we will explore at great length in §3.
this
The reader may hope that
is indeed the case:
Proposition
2.4. For a
an
analog
to
Proposition II.6.12 holds
ring homomorphism
:
cp
in
S, the following
R
Ring, and
are
equivalent:
(a)
(p is a
(b) ker^
(c)
(p is
monomorphism;
=
{0};
infective (as
Proof. We prove
a
set-function).
the rest to the reader. Assume (p : R
S is
the
extension
of
Example 2.2,
Applying
property
r and
we obtain unique ring homomorpshisms evr : Z[x] —>
R such that evr(x)
:
such
that
the
R
ev0 Z[x]
parallel ring homomorphisms:
ev(x) 0. Consider
a
(a)
monomorphism and
==>
r G
(b), leaving
ker (p.
=
=
evr
Z[x]
since
(p(r)
=
0
=
agree on Z and
evo
u>
:
]R—^S
the two compositions (p
agree on x)\ hence evr
<p(0),
they
o
evr, (p
o
evo agree
evo since (p is a
=
(because they
monomorphism.
Therefore
r
This proves
r G
kenp
==>
r
=
=
evr(x)
=
ev0(x)
=
0.
0, that is, (b).
R is a monomorphism, then S may
By Proposition 2.4, if S
i?; the following definition formalizes this situation.
be identified
with a subset of
Definition
2.5. A
subring S of
a
ring R is
of R and such that the inclusion function S
a
ring whose underlying set is a
R is a ring homomorphism.
^->
subset
j
Equivalently, a subring of R is a subset SCR which contains 1# and satisfies
the ring axioms with respect to the operations +, induced from R. It is
immediately checked that S C R is a subring if it is a subgroup of (i?, +), it is closed with
respect to •,and it contains 1#. As
a nonzero
ring R (because it does
Proposition
a
nonexample, the zero-ring
not contain
is not
a
subring of
Ir).
Ring and Grp are rather
phenomenon occurs
is
rings
certainly an epimorphism
2.4 may induce the reader to believe that
similar from this general categorical standpoint.
But
a new
concerning epimorphisms. A surjective map of
in Ring, since it is already an epimorphism in Set—we
have
earlier (e.g., '(c) => (a)' in Proposition II.8.18); but unlike
run
as
into this argument
in
Set, Grp, and Ab,
Note that ker (p cannot be a subring in any reasonable sense, according to our definition of
ring, since it does not contain a multiplicative identity in all but very pathological situations. It
is a subring if the identity requirement is omitted.
2. The category
133
Ring
in
epimorphisms need not be surjective
Ring. Indeed, consider the inclusion ho-
momorphism of rings
t:
Z-^Q:
is not surjective, and hence it is not an epimorphism in Set or Ab (can the reader
'describe' coker^? Exercise 2.12); but it is an epimorphism of rings. Indeed, if ai,
t
parallel ring homomorphisms
a2 are
and ai, a2 agree9 on Z,
for p, q e Z, q ^ 0,
a{
is the
same
(?)
say
=
for both10.
(cf. §1.2.6).
Warning: Thus,
a
=
ot\\i a2|z,
=
a^a^1)
Thus,
i
=
then they must agree
a(p)a(q)-1
(i
=
on
Q: because
1,2)
satisfies the categorical requirement for
epimorphisms
and
an
in
Ring, a homomorphism may well
without
epimorphism
being an isomorphism!
be both a
monomorphism
2.4. Products. Products exist in Ring: if i?i, R2 are rings, then Ri x R2 may
be defined by endowing the direct product of groups R\ x R2 (cf. §11.3.4) with
componentwise multiplication. Thus, both operations on Ri x R2 are defined
R2,
componentwise: V(ai,&i) G i?i, V(a2,b2)
(01,61)+ (o2,62)
(01,61) (o2,62)
The identity in Ri
R2 is indeed
x
:=
:=
(01 +o2,6i +62),
(01 o2,6i 62).
R2 is (l^^l^). The reader will check (Exercise 2.13) that
categorical product in Ring.
The reader should keep in mind that in general this is not the only ring
structure one can define on the direct product of underlying groups. For example,
Ri
x
mentioned in
§1.2,
a
one can
while the product ring
define
Z/pZ
x
field whose underlying group is Z/pZ x
Z/pZ is very far from being a field (why?).
a
as
Z/pZ,
The situation with coproducts is more unfortunate—dealing
with this in any
for
the
case
of
commutative
tensor
products, and
generality (even
rings) requires
by the
time
we
this question.
develop
tensor
products (§VIII.2),
will have almost
we
§2.2 suffices
template case
But the universal property reviewed in
simple examples, and the reader should work
(Exercise 2.14).
the
Incidentally—with
be tempted to consider
a
case
'direct
out one
forgotten
to deal with
now,
for fun
of abelian groups in mind (§11.3.5), the reader may
of rings', agreeing with the product in the finite
sum
Unfortunately, some references define ring epimorphisms as 'surjective ring
homomorphisms'; this should be discouraged.
As they must, since Z is initial.
Note that a(q) must have a double-sided inverse in R since q has one in Q, namely 1/q.
In particular, ai(q~1)
ot2{q~1) must agree, since they must equal this unique inverse of a(q);
cf. Proposition 1.12.
=
III.
134
Rings and modules
and maybe giving something interesting in general. This is less promising than
it looks: it does not satisfy the universal property of coproducts in Ring (why?),
case
and, further,
the 'infinite version' is not
EndAb(G).
2.5.
This is
examples of rings, and
good
as
place
a
ring because it is missing the identity.
a
expand a little on one of the main
special as the examples reviewed in
as any to
that is not quite
one
as
§1.2.
For every abelian group G, the group EndAb(G) := HoniAb(G, G) of endomorphisms of G is a ring, under the operations of addition and composition (as
discussed in the paragraph preceding Definition 1.1). Indeed, associativity follows
from the category axioms, and distributivity is immediately checked.
It is instructive
group
the endomorphism ring of the abelian
carefully
to examine
(Z,+):
Proposition
2.6.
EndAb(Z)
=
Z
as
(p
:
rings.
Proof. Consider the function
defined
Z
EndAb(Z)
-?
by
(p(a) =a(l)
Z. Then
for all group homomorphisms a : Z
addition in EndAb(Z) is defined so that Vn G Z
(a
(cf. §11.4.4);
in
0)(n)
is a
p)
+
(a
=
+
p)(l)
n G
Z;
homomorphism: the
a(n)+0(n)
in
=
a(l)
+
/J(l)
ring homomorphism. Indeed, for
a(n)
for all
=
a group
particular
<p{a
Further, ip
by a; then
+
cp is
na(l)
=
=
na
=
=
a,
y>(a)
(3
in
+
¥>(/*)•
EndAb(Z)
denote
a(l)
an
particular,
a((3(l))
=
a(3(l)
=
a(l)(3(l).
Therefore,
ip(a oP)
as
needed. Also,
Finally,
by
=
cp(idz)
(ao p)(l)
=
idz(l)
=
=
a(p(l))
=
a(l)p(l)
=
(p(a)(p(/3)
1-
Z, let ip(a) be the homomorphism
cp has an inverse: for a G
a :
Z
defined
(Vn
the reader will
G
easily check that ip
Therefore cp is a
Z)
:
a(n)
=
an;
ring homomorphism and inverse
ring isomorphism, verifying the statement.
is
a
to (p.
Z
2. The category
135
Ring
Proposition 2.6 gives
'natural': this
structure arises
'arbitrary' about
in which the
ring structure on Z is truly
from
naturally
categorical considerations, there is nothing
a sense
it.
generally, the rings EndAb(G) and other rings arising as endomorphism
of
other
structures (e.g., vector spaces) are arguably the most important class
rings
of examples of rings. The entire theory of modules is based on the observation that
EndAb(G) is a ring for every abelian group G.
More
§1.2 are expressed very concretely in these
for
the
endomorphism rings;
example,
group of units in EndAb(G) is nothing but
AutAb(G).
Every ring R interacts with the ring of endomorphisms of the underlying abelian
group (i?, +). For r e R, define left- and right-multiplication by r by An //n
respectively. That is, Va G R
Some of the notions reviewed in
Ar(a)
The
following
useful; it is
a
=
ra,
observation is nothing
/xr(^)
more
ar.
=
than easy notation juggling, but it is
'ring analog' of Cayley's theorem (Theorem II.9.5):
Proposition 2.7.
homomorphism
Let R be
a
ring. Then the function
r \->
\ris
an
injective ring
(i?,+),
that is, Ar G
X:R^ EndAb(R).
Proof. For
any
R and for all a, b G i?, distributivity gives
r G
Xr(a
+
6)
=
this shows that Ar is indeed
r(a -\-b)
an
=
ra +
rb
=
Xr(a)
endomorphism of the
+
Xr(b)
group
:
EndAb(#).
The function A
injective, since if
r
R
:
^
EndAb(-R)
—>
defined by the assignment
s, then
Ar(l)=r^5
so
that Ar
Ar is clearly
r i—?
=
Afl(l),
7^ As.
verify that A is in fact a homomorphism of rings. Recall that the
addition in EndAb(-R) is inherited from R (cf. §11.4.4): for all r, s G i?, Ar + As is
defined by
We have to
(Va
G
R)
(Ar
:
+
Xs)(a)
Thus,
the fact that A preserves addition is
distributivity: for all r, 5, a G i?,
Ar+S(a)
=
(r
+
s)a
=
an
=
Xr(a)
+
As(a).
immediate consequence of
ra-\-sa
=
Xr(a)
+
As(a).
As for multiplication, it is associativity's turn to do its job:
Xrs(a)
Of
course
=
(rs)a
=
r(sa)
=
rXs(a)
=
Xr(Xs(a))
Ai is the identity, completing the verification.
=
(Xr
o
Xs)(a).
III.
136
Rings and modules
The function \x : R
\xr is 'almost'
EndAb(-R) defined by r i—?
will
the
reader
check
that
homomorphism:
(Exercise 2.18)
—>
=
fJ>r+s
and //i
=
id#. That is,
course
ring
f^r + f^s •>
// would in fact be a
multiplication11 in R. Of
commutative (since then A
/x).
a
ring homomorphism if
we
'reversed'
this issue disappears if R happens to be
=
Exercises
2.1. > Prove that if there is a
is
a
2.2.
homomorphism from
Let R and S be
operations +,
to a
ring i?, then R
rings, and let
:
cp
S be
R
Prove that if (p ^ 0 and 5 is
(Therefore, in both
Let S be
the ring
a
function preserving both
•.
Prove that if (p is surjective, then necessarily
2.3.
zero-ring
a
zero-ring [§2.1]
a
cases (p
an
<p(lj?)
=
I5.
integral domain, then <p(1r)
is in fact
a
=
I5.
ring homomorphism).
set, and consider the power set ring £?(S) (Exercise 1.2) and
(Z/2Z)5
you constructed in Exercise 1.3. Prove that these two
rings
are
isomorphic. (Cf. Exercise 1.2.11.)
2.4. Define functions H
a +
gU(M)
-?
and H
bi + cj + dk
gb(C) (cf.
-?
(a
b
-b
a
c
a
—c
b
\—d
a +
bi + cj + dk
Thus, quaternions
2.5.
-1
The
real number
norm
N(w)
d\
—b
a)
a +
bi
c +
di^
-c +
di
a
bij
for all a, 6, c, d G R. Prove that both functions
1.12) by
—dc
d
-c
Exercises 1.4 and
may be viewed as real or
injective ring homomorphisms.
complex matrices.
of
bi + cj + dk, with a, b,c,d e R, is the
=
a
a2
quaternion w
+ b2 + c2 + d2.
=
a +
are
Prove that the function from the multiplicative group M* of nonzero quaternions
to the multiplicative group R+ of positive real numbers, defined by assigning to
each nonzero quaternion its norm, is a homomorphism. Prove that the kernel of
this homomorphism is isomorphic to SU2(C) (cf. Exercise II.6.3). [4.10, IV.5.17,
V.6.19]
This issue is entirely analogous to the business of right-actions vs. left-actions; cf. §11.9.3,
especially Exercise II.9.3.
Exercises
2.6. >
137
the 'extension property' of polynomial rings, stated in Example 2.3.
Verify
[§2-2]
2.7. > Let R
and let
Z/2Z,
=
f(x)
=
x2
x; note
f(x) ^
0. What is the
polynomial
function R^R determined by f(x)l [§2.2, §V.4.2, §V.5.1]
2.8. Prove that every
2.9.
r e
subring of
a
field is
an
integral domain.
ra for all
ring R consists of the elements a such that ar
R. Prove that the center is a subring of R.
Prove that the center of a division ring is a field. [2.11, IV.2.17, VII.5.14,
The center of
-i
a
=
VII.5.16]
2.10.
The centralizer of
-i
such that
ar
=
ra.
an
element
a
of
a
ring R consists of the elements r G R
a is a subring of i?, for every a G R.
Prove that the centralizer of
Prove that the center of R is the intersection of all its centralizers.
Prove that every centralizer in a division
ring is
a
division ring.
[2.11, IV.2.17,
VII.5.16]
2.11.
Let R be
-i
a
division ring consisting of
as follows:
p2 elements,
where p is
a
prime.
Prove that R is commutative,
If R is not commutative, then its center C (Exercise
R. Prove that \C\would then consist of p elements.
Let
r
i?,
r G
0 C.
r
Prove that the centralizer of
r
2.9)
is
a proper
(Exercise 2.10)
subring of
contains both
and C.
Deduce that the centralizer of
Derive
r
is the whole of R.
contradiction, and conclude that R had
a
(hence,
a
of Wedderburn's theorem: every finite division ring is
a
to be commutative
field).
This is
field.
a
particular
case
[IV.2.17, VII.5.16]
Q. Describe the cokernel of i in Ab
Ring (as defined by the appropriate universal property in the
given in §11.8.6). [§2.3, §5]
2.12. > Consider the inclusion map i
:
Z
^->
and its cokernel in
style of the
one
2.13. > Verify that the 'componentwise' product R\ x R2 of two rings satisfies the
universal property for products in a category, given in §1.5.4. [§2.4]
2.14.
>
Verify
that
Z[#i,#2] (along
morphisms) satisfies
Z[x] in the category of
with the evident
universal property for the coproduct of two copies of
commutative rings. Explain why it does not satisfy it in Ring.
2.15.
> For
the
[§2.4]
1, the abelian groups (Z, +) and (raZ,+)
manifestly
isomorphism. Use this
to transfer the structure of 'ring without identity' (raZ, +, •) back
an explicit formula for the 'multiplication'
this defines on Z (that
is, such that (p(a b)
(f(a) <p(b)). Explain why structures induced by different
positive integers m are nonisomorphic as 'rings without 1'.
m
>
the function ip
isomorphic:
isomorphism
onto Z: give
=
:
Z
raZ,
n
i—?
mn is a group
are
III.
138
(This shows that there
are
without identity to the group
Rings and modules
different ways to give a structure of ring
Compare this observation with Exercise 2.16.)
many
(Z, +).
[§2-1]
2.16. > Prove that there is
identity
(up to isomorphism) only one structure of ring with
(Z,+). (Hint: Let R be a ring whose underlying
the abelian group
on
By Proposition 2.7, there
group is Z.
is
an
injective ring homomorphism A
and the latter is isomorphic to Z by Proposition 2.6.
EndAb(-R),
surjective.) [§2.1, 2.15]
2.17.
-i
Let R be
ring, and let E
a
the underlying abelian group (i?,
the center of R. (Prove that if a
elements of i?, then
+).
G
R
EndAb(-R) be the ring of endomorphisms of
Prove that the center of E is isomorphic to
E commutes with all
right-multiplications by
element of i?; then use
In particular, this shows that if R is commutative, then the center of
Proposition
a
2.7.)
End#(i?) is isomorphic
2.18. >
Verify
Proposition 2.7.
is
left-multiplication by
to R.
an
[VIII.3.15]
the statements made about right-multiplication /x,
following
[§2.5]
2.19. Prove that for
as a
=
:
Prove that A is
n G
Z
a
positive integer, EndAb(Z/nZ) is isomorphic
to
Z/nZ
ring.
3. Ideals and quotient rings
3.1. Ideals. In both Set and Grp we have been able to 'classify' surjective morboth cases, a surjective morphism is, up to natural identifications, a
phisms: in
quotient by
a (suitable) equivalence relation. In Grp, we have seen that such
equivalence relations arise in fact from certain substructures: normal subgroups.
The situation in Ring is analogous. We will establish a canonical decomposition
for rings, modeled after Theorems 1.2.7 and II.8.1; the corresponding version of the
'first isomorphism theorem' (Corollary II.8.2) will identify every surjective ring
homomorphism with
a
quotient by
a
suitable substructure.
The role of normal
subgroups will be played by ideals.
Definition 3.1. Let R be
rl C I for all
r G
a
ring. A subgroup I of (-R,+) is
(Vr
it is
a
right-ideal if
A two-sided ideal is
G
It C I for all
(Vr
Of
a
left-ideal of R
if
i?; that is,
a
G
subgroup
R)(\/a G I)
r G
:
ra G
J;
are
I.
i?; that is,
R)(\/a G I)
:
/ which is both
a
left- and
a
right-ideal.
j
in a commutative ring there is no distinction between left- and rightin
ideals. Even
the general setting, we will almost exclusively be concerned with
two-sided ideals; thus I will omit qualifiers, and an ideal of a ring will implicitly be
course
two-sided.
3. Ideals and quotient rings
Remark 3.2. As
seen
in
139
§2.3,
subring of R
a
is
a
subset S C R which contains 1r
induced from R.
and satisfies the ring axioms with respect to the operations +,
Ideals are close to being subrings: they are subgroups, and they are closed
with respect to multiplication. But the only ideal of a ring R containing 1^ is
R itself: this is an immediate consequence of the 'absorption properties' stated in
Definition 3.1. Thus ideals
are
in
general
not
subrings; they
'rngs'.
are
j
Ideals are considerably more important than subrings in the development of
the theory of rings12. Of course the image of a ring homomorphism is necessarily a
subring of the target; but a lesson learned from §11.8.1 and following is that kernels
really capture the structure of a homomorphism, and kernels are ideals:
Example 3.3. Let (p
of it
:
S be any ring homomorphism. Then kenp is
R
Indeed, we know already that ker (p is a subgroup; we have to verify the
absorption properties. These are an immediate consequence of Lemma 1.2: for all
all a G kenp, we have
(p(ra)
(p(ar)
More
generally,
and
(Exercise 3.2)
soon see
(p(a)(p(r)
=
=
0
a
=
=
a
0
=
r e
R,
0,
0.
it is easy to verify that the inverse image of
is clearly an ideal of S.
an
ideal is
ideal
an
{O5}
in the context of groups, we
that 'kernels of ring homomorphisms' and 'ideals' are in fact equivalent
Similarly
will
(p(r)(p(a)
=
ideal
an
subgroups
to the situation with normal
concepts.
j
Quotients. Let / be a subgroup of the abelian group (R, +) of a ring R.
Subgroups of abelian groups are automatically normal, so we have a quotient group
3.2.
R/I,
whose elements
are
the cosets of /:
r +
(written,
of course, in additive
I
notation).
Further,
we
have
a
surjective
group
homomorphism
7r :
As we have
explored
o
it
R
+ i.
—>
r 1-4 r
—,
in great detail for groups, this construction satisfies
suitable universal property with respect to group homomorphisms
a
(Theorem II.7.12).
Of course
we are now going to ask under what circumstances this construction can
performed in Ring, satisfying the analogous universal property with respect to
ring homomorphisms.
That is, what should we ask of /, in order to have a ring structure on R/I,
so that 7r becomes a ring homomorphism?
Go figure this out for yourself, before
reading ahead!
be
'Arguably,
the
reason is
that ideals
are
precisely the submodules of
a
ring R.
III.
140
As
Since
Rings and modules
often is the case, the requirement is precisely what tells us the answer.
will have to be a ring homomorphism, the multiplication in R/I must be
so
tt
as follows: for all
(a
+
/), (b
+
I)
(o
+
/) (b
+
I)
in
R/I,
7r(o) 7r(6)
=
=
7r(a6)
=
06 + /,
1?
where the equality
This
is forced if
=
says that there is
only
tt
is to be
one
a
ring homomorphism.
sensible ring structure
(Va, beR):
(a
I)(b + I)
+
:=
on
R/I, given by
ab + I.
The reader should realize right away that if this operation is well-defined, then it
does make R/I into a ring: associativity will be inherited from the associativity in
i?, and the identity will simply be the coset 1 + / of 1.
So all is well, if the proposed operation is well-defined. But is it going
to be
well-defined?
Example
take Z
be, if / is
subgroup of Q; then
3.4. It need not
as a
arbitrary subgroup of R.
an
For example,
0 + Z =1+Z
(= Z)
as
elements of the group
0--+Z
7r :
Z^-+Z=l--+Z.
=
j
Assume then that the operation is well-defined, so that R/I is
R —>
R/I is a ring homomorphism. What does this say about II
Answer: / is the kernel of R—>
R/I,
in
+ Z is another coset, yet
and
Q/Z,
so
necessarily
/ must be
an
a
ring and
ideal,
as seen
Example 3.3!
Conversely, let
us
/ is
assume
prescription for the operation in
a' + I
recall that this
a"b"
-
means
a'b'
=
=
-
R/I
ideal of i?, and
verify
that the proposed
is well-defined. For this, suppose
a" + I
that a"
a"b"
an
b' + I
and
a' E I, b"
a"b' + a"b'
-
=
b" + J;
b' G /; then
a'b'
=
a"(b"
-
b')
+
{a"
-
a')b'
e
/,
using both the left-absorption and right-absorption properties of Definition 3.1.
This says precisely that
a'b' + 1
a"b" + /,
=
proving that the operation is well-defined.
Summarizing,
we
canonical projection
ideal of R.
tt
have verified that
:
R
Definition 3.5. This ring
Example 3.6. We
nonnegative integer
R/I
—>
R/I
a
ring, in such a way that the
ring homomorphism, if and only if / is an
R/I
is
a
is called the quotient ring of R modulo I.
know that all
n
is
subgroups of (Z, +)
(Proposition II.6.9).
are
of the form nL for
are
in fact ideals of the
a
It is immediately verified that all
are of course
The quotients Z/nZ
ring (Z, +, •)•
nothing but the rings so-denoted in §1.2 (and earlier).
subgroups of Z
j
141
3. Ideals and quotient rings
The fact that Z is initial in
let /
ker /
:
Ring
now
prompts
a
natural definition. For
Z
=
a
R be the unique ring homomorphism, defined by a *—>
nZ for a well-defined nonnegative integer n determined by R.
Definition 3.7. The characteristic of R is this nonnegative integer
is
a
Thus, the characteristic of R is n > 0 if the order
positive integer n, while the characteristic is 0 if
We have
fulfilled
now
our
promise
to
identify
of 1r
as an
-
a
ring i?,
1r. Then
n.
element
j
of (i?, +)
the order of 1r is oo.
j
the notions of ideal and kernel of
ring homomorphisms: every kernel is an ideal (cf. Example 3.3); and on the other
hand every ideal I is the kernel of the ring homomorphisms R
R/I. We have
the slogan:
kernel <^=> ideal
in the context of ring theory.
The key universal property holds in this context as it does for groups; cf.
II.7.12. We reproduce the statement here for reference, but the reader should
Theorem
realize that very little needs to be proven at this point: the needed (group)
homomorphism exists and is unique by Theorem II.7.12, and verifying it is a ring
homomorphism is immediate.
Theorem 3.8. Let I be
a two-sided ideal of a ring R.
Then for every ring
S such that I C ker cp there exists a unique ring homomorphism
homomorphism (f : R
S so that the diagram
(p : R/I
commutes.
As
a
reminder to the lazy reader, (p is defined by
<p(r
(part of) the
kenp), and it
3.3.
+
I) :=<p(r);
content of the theorem is that this function is well-defined
is
a
Canonical
(if
/ C
ring homomorphism.
decomposition and consequences. As the reader should now
key element in a standard decomposition of every ring
expect, Theorem 3.8 is the
homomorphism. This is entirely analogous to the decomposition of set-functions
studied in §1.2.8 and the decomposition of group homomorphisms obtained in §11.8.1.
Here is the statement:
Theorem 3.9.
follows:
Every ring homomorphism cp
:
R
S may be decomposed
as
142
III.
where the isomorphism
Theorem
Rings and modules
homomorphism induced by
<p in the middle is the
(p
(as
in
3.8).
The reader will realize that this statement requires no proof at this point:
at the level of groups (by Theorem II.8.1) and the maps
the
decomposition holds
all ring homomorphisms
as
are
observed earlier in this section.
The 'first isomorphism theorem' for rings is
Corollary 3.10. Suppose (p
:
S is
R
a
immediate corollary:
an
surjective ring homomorphism. Then
kev(p
develop the healthy instinctive reaction
of rings as a quotient (by an ideal), up
As in the group case, the reader should
of viewing every surjective homomorphism
to a natural identification.
Equally instinctive should be the realization that ideals of a quotient R/I are in
correspondence with ideals of R containing I. Once more, not a whole
is new here: we already know (cf. §11.8.3) that the function
one-to-one
lot
{subgroups
u :
defined by u(J)
check that J/I is
=
J/I
is
J of R containing
/}
{subgroups
bijection preserving inclusions;
a
of
R/I}
it takes
a moment to
ideal of R/I if and only if J is an ideal of R. The main content of
this observation is best packaged in the corresponding 'third isomorphism theorem':
an
R, and let J be
Proposition 3.11. Let I be an ideal of a ring
containing I. Then J/I is an ideal of R/I, and
=
ker(i?
R/J),
ip
by
Theorem 3.8. Explicitly,
ker (p
we see
from
=
that
{r
+ /
J/I
Corollary
is
| (p(r
an
+
I)
ideal
cp(r
=
+
J}
(since
I)
=
+ /
a
of R
an
induced ring homomorphism
R/J
-?
r +
=
{r
it is
have
we
R/I
:
ideal
J'
j/i
Proof. Since I C J
an
|
cp is
J;
r +
kernel),
J
manifestly surjective. Since
=
J}
=
{r
+ /
|
r e
J}
=
J/I,
and the stated isomorphism follows
3.10.
What about the 'second' isomorphism theorem?
This would be
a
relation
between the ideals
7 +J
I
J
'
in J
of the rings R/I, R/(I H J), respectively, assuming I and J are ideals of R
where it is immediately checked that I + J and I D J are indeed ideals13).
(and
The reader may want to go back to §11.8.4 for the version of this story for
Our feeling is that Ring is not the best place to play this game, since
groups.
The notation I + J should not surprise the reader:
group R,
so we
know what it
means to 'add'
them; cf. §11.7.1.
/, J
are
subgroups of the abelian
Exercises
143
(/ + «/)//, J/(I DJ)
rings according
are not even
to my conventions. Modules will
be a more natural context for this result.
In any case, the reader will benefit the most from
his/her
exploring the
matter on
own; cf. Exercise 3.17.
Exercises
image of
3.1. Prove that the
What
if its kernel is
3.2. > Let (p
that /
3.3.
=
-i
:
Show that
S be
R
(p~l(J)
Let (p
ring homomorphism
S is
R
:
cp
ideal of 5? What
an
a
subring of S.
can you say
about (p
subring of R?
a
:
a
about (p if its image is
can you say
is
an
S be
R
a
ring homomorphism, and let J be
ideal of R.
a
ring homomorphism, and let J be
need not be
(p(J)
an
ideal of S. Prove
an
ideal of R.
[§3.1]
an
ideal of S.
Assume that (p is surjective; then prove that
<p(J)
is
an
ideal of S.
Assume that (p is surjective, and let /
kenp; thus we may identify
Let J
an
ideal
of
the
previous
point. Prove that
<p(J),
R/I by
=
S
with
R/I.
=
this is just
(Of course
3.4. Let R be
Prove that R
3.5.
-i
a
=
Let J be
a
where
j
i + J'
n
a
by placing
any entry of
matrix A G
.
Let J be
ring i?, and let I
ideal of R.
over a ring R.
if
the
matrices
obtained
only
Mn(R) belongs
A in any position, and 0 elsewhere, belong to J. (Hint:
Carefully contemplate the operation
-i
an
Mn(R)
ofnxn matrices
to J if and
___
3.6.
3.11.) [4.11]
every subgroup of (i?, +) is in fact
is the characteristic of R.
two-sided ideal of the ring
a
Prove that
_.
R
rehash of Proposition
ring such that
Z/nZ,
R/i
/ooo\
/ooo\/abc\/ooo\
(
f
(oool I d
e
(lool
=
,
fn/)1
oooj.) [3.6]
two-sided ideal of the ring Mn(R) ofnxn matrices over a
(1,1) entries of matrices in J. Prove that I is
a
C R be the set of
two-sided ideal of R and J consists precisely of those matrices whose entries all
belong to /. (Hint: Exercise 3.5.) [3.9]
a
a ring, and let
of
R. Prove that
right-ideal
R. Prove that Ra is
left-ideal of R and aR is
3.7. Let R be
a G
a
a
is
a
division ring if and only if its only left-ideals and
resp. R
3.8.
>
=
left-,
resp.
right-,
a
unit if and
only if R
=
aR,
Ra.
Prove that
right-ideals
are
a
are
a
{0}
ring R is
and R.
In particular, a commutative ring R is
{0} and R. [3.9, §4.3]
a
field if and only if the only ideals of R
144
III.
3.9.
-i
Counterpoint
to Exercise 3.8:
It is not true that a
ring if and only if its only two-sided ideals
property is said
to be
are
{0}
simple; by Exercise 3.8, fields
and R. A
Rings and modules
ring R is a division
ring with this
nonzero
the only simple commutative
are
rings.
Prove that
3.10. > Let (p
nonzero
Mn(R)
:
k
simple. (Use Exercise 3.6.) [4.20]
is
R be
ring. Prove that
a
(p is
ring homomorphism, where k is
infective. [§V.4.2, §V.5.2]
ring containing C
R.
homomorphisms R
3.11. Let R be
a
3.12. > Let R be
R
is
an
a
ideal of R.
Find
commutative
(Cf.
subring. Prove that there
ring. Prove that the
noncommutative ring in which the set of
a
[3.13, 4.18, V.3.13, §VII.2.3]
3.13.
-i
Let R be
a
set of
R/N contains
reduced.) [4.6, VII.2.8]
nilpotent elements
nilpotent elements. (Such
no nonzero
an integral domain
1? [V.4.17]
of
characteristic
ring
Prove that the characteristic of
integer. Do
are
you know any
no
a
ring
nilpotent elements of
commutative ring, and let N be its nilradical
Prove that
-i
a
field and R is
Exercise 1.6. This ideal is called the nilradical of
ideal.
3.14.
as
a
an
Exercise
3.12).
ring is said
to be
(cf.
a
R.)
is not
is either 0
or a
prime
A ring R is14 Boolean if a2
a for all a G R. Prove that ^(S) is Boolean,
for every set S (cf. Exercise 1.2). Prove that every Boolean ring is commutative, and
has characteristic 2. Prove that if an integral domain R is Boolean, then R
Z/2Z.
3.15.
=
-i
=
[4.23, V.6.3]
3.16.
-i
Let S be
in T form
a set
and T C S
subset. Prove that the subsets of S contained
a
ideal of the power set ring &(S). Prove that if S is finite, then every
ideal of £?(S) is of this form. For S infinite, find an ideal of £?(S) that is not of
an
this form.
[V.1.5]
3.17. > Let /, J be ideals of a ring R. State and prove a precise
ideals (I + J)/I of R/I and J/{I n J) of R/(I n J). [§3.3]
4.
Ideals and
quotients:
Remarks and
result relating the
examples.
Prime and maximal ideals
4.1. Basic operations. It is often convenient to define ideals in terms of
a set
of
generators.
Let
R be any element of a ring.
of R. Indeed, for all r G R we have
a G
rl
as
needed.
Similarly, aR
14After George
Boole.
is
=
right-ideal.
Then the subset I
rRa C Ra
=
Ra of R is
a
left-ideal
4. Ideals and quotients: Remarks and
145
examples
In the commutative case, these two subsets coincide and are denoted (a). This
is the principal ideal generated by a. For example, the zero-ideal {0}
(0) and the
whole ring R
are
both
ideals.
principal
(1)
=
=
In general
sum
^2a
(Exercise 4.1),
Ia is
an
if
is
{Ia}aeA
family of
a
ideals of
a
ring i?, then the
ideal of R. If aa is any collection of elements of
a
commutative
ring R, then
(a>ot)oteA
:=
,(flq)
/
a£A
is the ideal
generated by the elements
(oi,...,on)
particular,
aa. In
=
(oi)
H
h
(on)
is the smallest ideal of R containing ai,..., an; this ideal consists of the elements
of R that may be written
as
\-rnan
rioi H
for r*i,..., rn G R.
/
(ai,..., an) for
=
An ideal / of
some a\,...,an G
Example 4.1. It is
quotients in
a
a
commutative ring R is
finitely generated
if
R.
good idea to get used to a bit of 'calculus' of ideals and
judicious use of the isomorphism theorems yields
terms of generators;
convenient statements. For example, let R be
denote by b the class of b in R/(a). Then
a
commutative ring, and let a, 6 G R\
(R/(a))/(b)^R/(a,b).
Indeed,
this is
a
particular
case
of Proposition 3.11, since
(a)
as
ideals of
R/(a).
j
Note that principal ideals
are
(very special) finitely generated
important that
we
give special
notions
are
so
names
to
rings
ideals.
in which
These
they
are
satisfied by every ideal.
Definition 4.2. A commutative ring R is Noetherian if every ideal of R is
finitely
generated.
j
Definition 4.3. An integral domain R is
ideal of R is principal.
a
PID
('Principal
Ideal
Domain')
j
Thus, PIDs are (very special) Noetherian rings. In due time
length with these classes of rings (cf. Chapter V); Noetherian rings
important in number theory and algebraic geometry.
The reader is already familiar with an important PID:
Proposition
if every
we
will deal at
are very
4.4. Z is a PID.
Proof. Let / C Z be an ideal. Since / is a
Proposition II.6.9. Since nL
=
(n),
subgroup,
/
=
nL for
this shows that / is principal.
some n G
Z, by
III.
146
The fact that Z is
as they do in Z:
and hence
behave
a
PID captures
if ra,
n are
precisely 'why' greatest common divisors
integers, then the ideal (ra, n) must be principal,
(ra,n)
for
some
ra G
(d)
Rings and modules
=
(d)
(positive) integer d. This integer is manifestly the gcd of
and n G (d), then d | ra and d | n, etc.
ra
and
since
n:
field, the ring of polynomials k[x] is also a PID; proving this is easy,
the
'division
with remainder' that we will run into very soon (§4.2); the reader
using
If /c is a
should work this out
on
his/her
in the general theory when
we
(Exercise 4.4)
own
review the
general
now.
This fact will be absorbed
notion of 'Euclidean domain', in
§V.2.4.
By contrast, the ring Z[x] is not a PID: indeed, the reader should be able to
verify that the ideal (2, x) cannot be generated by a single element. As we will see
in due time, greatest common divisors make good sense in a ring such as Z[x], but
the matter is a little more delicate, since this ring is not a PID15.
There
are
several
more
basic operations involving ideals; for now, the
following
two will suffice.
Again assume that {Ia}aeA is a collection of ideals of a ring R. Then the
intersection f]aeA I<* 1S (clearly) an ideal of R; it is the largest ideal contained
in all of the ideals Ia.
If J, J are ideals of i?, then IJ denotes the ideal generated by all products
ij with i G /, j G J. More generally, if Ji,... ,/n are ideals in i?, then the
'product' 11
ik
-
It
/ is
a
-
In denotes the ideal generated by all products i\ in with
h>
The reader should
§11.8.4)
-
IJ would
note the clash of notation: in the context of groups
mean
(especially
something else. Watch out!
is clear that IJ C If) J: every element ij with i e I and j G J is in / (because
right-ideal) and in J (because J is a left-ideal); therefore I D J contains all
products ij, and hence it must contain the ideal IJ they generate. Sometime the
product agrees with the intersection:
(4)
n
(3)
=
(12)
=
in Z;
(4) (3)
and sometime it does not:
(4)n(6)
The matter of whether IJ
=
=
(12)^(24)
=
(4)-(6).
I D J is often subtle;
a
prototype situation in which
this equality holds is given in Exercise 4.5.
4.2. Quotients of polynomial rings. We have already observed that the
quotient Z/nZ is our familiar ring of congruence classes modulo n. Quotients of
polynomial rings by principal ideals
are
a
good
source
of 'concrete', but maybe less
familiar, examples.
It is, however, a 'UFD', that is,
notion of gcd; cf. §V.2.1.
a
'unique factorization domain'. This suffices for
a
good
4. Ideals and quotients: Remarks and
Let R be
a
(nonzero) ring,
f(x)
=
xd
and let
ad-ixd~l
+
147
examples
+ aix + a0 G
+
R[x]
polynomial; for convenience, we are assuming that f(x) is monic, that is, its
leading coefficient (the coefficient of the highest power of x appearing in f(x)) is 1.
be
a
In terms of ideals, this is not
a serious requirement if the coefficient ring R is a
but
it
may be substantial otherwise: for example, (2x) C Z[x]
(Exercise 4.7),
cannot be generated by a monic polynomial. Also note that a monic polynomial
field
is
necessarily
deg(f(x)q(x))
a
=
(Exercise 4.8) and that if f(x)
degf(x) -\-degq(x) for all polynomials q(x).
non-zero-divisor
It is convenient to
f(x),
exist
that
assume
f(x) is monic because we can then divide by
g(x) G R[x] is another polynomial, then there
That is, if
with remainder.
polynomials q(x), r(x)
G
is monic, then
such that
R[x]
g(x)
=
f(x)q(x)+r(x)
and16 degr(x)
< deg/(x). This is simply the process of 'long division' of
polynomials, which is surely familiar to the reader, and can be performed over any
when dividing by
monic
ring
polynomials17.
The situation appears then to be similar to the situation in Z, where we also
have division with remainder. Quotients and remainders are uniquely18 determined
by g(x) and f(x):
Lemma 4.5. Let
f(x)
be
a
monic
f(x)qi(x)
+
polynomial, and
ri(x)
=
f(x)q2(x)
ri(x) and r2(x) polynomials of degree
ri(x) r2(x).
with both
and
assume
<
+
r2(x)
degf(x).
Then
qi(x)
=
q2(x)
=
Proof. Indeed,
we
have
f(x)(qi(x)
-
q2(x))
=
r2(x)
-
n(x)\
if
r2(x) 7^ n(x), then r2(x)—
r\{x)has degree < deg/(x), while f (x)(qi(x)
has degree > deg/(x), giving a contradiction. Therefore r\{x) r2(x), and
q2(x) follows right away since monic polynomials are non-zero-divisors.
=
q2(x))
q\{x)
=
The preceding considerations may be summarized in a rather efficient way in
the language of ideals and cosets. We will now restrict ourselves to the commutative
case, mostly for notational convenience, but also because this will guarantee that
ideals
two-sided ideals,
are
Note:
degr(cc)
<
so
that quotients
are
defined
as
rings (cf. §3.2).
With the convention that the degree of the polynomial 0 is
is satisfied by r(x)
0.
deg/(cc)
oo, the condition
=
The key point is that if n > d, then for all a G R we have axn
=
axn~df(x) + h(x)
for
some
polynomial h(x) of degree < n. Arguing inductively, this shows that we may perform division
by f(x) with remainder for all 'monomials' axn, and hence (by linearity) for all polynomials
g\x) e R[x].
This assertion has to be taken with a grain of salt in the noncommutative case, as different
quotients and remainders may arise if we divide 'on the left' rather than 'on the right'.
III.
148
Assume then that R
is
commutative ring.
a
f(x) is monic, then for every g(x) G
degree < degf(x) and such that
g(x)
as cosets
of the principal ideal
R[x]
(f(x))
+
=
in
(f(x))
What
there exists
r(x)
+
a
we
Rings and modules
have shown is that, if
unique polynomial r(x) of
(f(x))
R[x\.
useful group-theoretic statement. Note that
Refining
of
d
be
seen
as elements of a direct sum
<
polynomials
degree
may
this observation leads to
R®d
a
=
R0
...
0 R
:
d times
indeed, the function ip
:
R®d
R[x]
—>
defined by
ip((r0,rir- ,rd_i))
r0 + nx +
=
+
rd-ixd~l
clearly an injective homomorphism of abelian groups, hence an isomorphism
image, and this consists precisely of the polynomials of degree < d. We will
glibly identify R®d with this set of polynomials for the purpose of the discussion
is
onto its
that follows.
The next result may be seen as a way to concoct many different and
structures on the direct sum
ring
Proposition 4.6. Let R be a commutative
polynomial of degree d. Then the function
(f
defined by sending g(x) G R[x] to
an isomorphism of abelian
induces
interesting
R®d:
R[x]
:
ring, and let
f(x)
G
R[x]
be
a
monic
R®d
-?
the remainder
of
the division
of g(x) by f(x)
groups
'
(/(*))
Proof. The given function (p is well-defined by Lemma 4.5, and it is surjective
since it has a right inverse (that is, the function ip : R®d
R[x] defined above).
We claim that
9i(x)
with
degri(x)
<
=
cp is a
homomorphism of abelian
d, degr2(x)
gi(x)+92(x)
and
deg(ri(x)
+
r2(x))
<p(9i(x)
By
and
f(x)qi(x) +ri(x)
+
f(x)(qi(x)
< d: this
92{x))
=
Indeed, if
f(x)q2(x) +r2(x)
d, then
<
=
g2(x)
groups.
=
+
q2(x))
+
(ri(x)
+
r2(x))
implies (again by Lemma 4.5)
ri(x)
+
r2(x)
=
<p(gi(x))
+
<p(g2(x)).
the first isomorphism theorem for abelian groups, then, cp induces
an
isomorphism
R[x] ^ ^ed
kenp
if and only if g(x)
f(x)q(x) for some q(x) G R[x],
the principal ideal generated by f(x). This shows
On the other hand, cp(g(x))
0
that is, if and only if g(x) is in
kenp
(f(x)), concluding the proof.
=
=
=
4. Ideals and quotients: Remarks and
Example
149
examples
x
a for some a G R.
f(x) is monic of degree 1: f(x)
after
division
is
the
'evaluation' g(a)
by f(x)
simply
g(x)
4.7. Assume
Then the remainder of
=
(cf. Example 2.3): indeed,
g(x)
for some r G R
evaluating
at a
(the
a)q(x)
remainder must have degree < 1; hence it is
claimed. In particular,
(a
=
g(a)
a)q(a)
+ r
R[x]
R,
0
=
case
g(x)
-?
q(a)
+ r
only if g(x)
0 if and
=
The content of Proposition 4.6 in this
induces
+ r
a
constant);
gives
g(a)
as
(x
=
G
=
r
(x
a).
is that the evaluation map
i->
g(a)
isomorphism
an
(x
of abelian groups;
Corollary 3.10) that this
a)
verify (either by hand
isomorphism of rings.
the reader will
is in fact
an
or
by invoking
j
It is fun to analyze higher-degree examples.
For every monic
of
a
4.6
degree d, Proposition
gives
different ring (that is, R[x]/(f(x)))
f(x) R[x]
isomorphic to R®d as a group; one can then use this isomorphism to define a new
4.8.
Example
G
structure onto the group
ring
R®d.
isomorphic (as seen in Example 4.7); but
interesting
already
degree 2. For a concrete example, apply this
1:
with
x2
+
procedure
Proposition 4.6 gives an isomorphism of groups
f(x)
For d
=
1 all these structures are
arise in
structures
=
R 0 R
[X2
+
;
1)
what multiplication does this isomorphism induce on R 0 Rl Take two elements
(ao, ai), (&o> fri) of R 0 R- With the notation used in Proposition 4.6, we have
(oo, oi)
Now
a
=
ip(ao
+
oix),
(bo,bi)
=
ip(bo
+
b\x).
bit of high-school algebra gives
(a0
+
aix)(bo
+
b\x)
=
=
a0b0
(x2
(a0bi
+
aib0)x
l)ai&i
+
{{a0b0
+
+
+
-
aibix2
0161)
+
(oo&i
+
aib0)x)
which shows
(f((a0
+
aix)(b0
+
bix))
Therefore, the multiplication induced
(ao,ai) (bojbi)
This recipe may
seem
=
=
on
(a0b0
-
Oi6i,o06i +ai60)-
R 0 R by this procedure is defined by
(ao&o
ai6i,ao&i
+
ai&o)R, the
complex numbers
somewhat arbitrary, but note that upon taking R
ring of real numbers, and identifying pairs (x,y)
G R 0 R with
=
III.
150
iy, the multiplication we obtained
multiplication in C. Therefore,
x +
on
(x2 + 1)
In
other
this
words,
procedure
rings.
starting from R[x].
as
Rings and modules
R 0 R matches
precisely the ordinary
constructs the
ring C 'from scratch',
j
The point is that the polynomial equation x2 + 1
0 has no solutions in R;
the quotient R[x]/(x2 + 1) produces a ring containing a copy of R and in which the
polynomial does have roots (that is, ± the class of x in the quotient). The fact that
=
this turns out to be isomorphic to C may not be too surprising, considering that C
is precisely a ring containing a copy of R and in which x2 + 1 does have roots (that
is, ±i).
Such constructions
back to them in
§V.5.2,
the "algebraist's way" to solve equations. We will come
a sense the whole of Chapter VII will be devoted to
are
and in
this topic.
4.3. Prime and maximal ideals. Other 'qualities' of ideals are best expressed
in terms of quotients. We are still assuming that our rings are commutative—both
due to
use
unforgivable
laziness and because that is the only context in which
we
will
these notions.
Definition 4.9. Let I
^ (1)
be
an
ideal of
an
integral domain.
J is
a
prime ideal if R/I is
/ is
a
maximal ideal if
Example 4.10.
integral domain;
as we
have
seen
is
R/I
a
commutative ring R.
a
field.
j
R, the ideal (x a) is prime in R[x] if and only if R is an
R,
only if R is a field. Indeed, R[x]/(x a)
4.7.
Example
For all
a e
it is maximal if and
in
The ideal (2,x)
is maximal in
=
since
Z[x],
Z
Z[x] I Z[x]/(x)
_
M-^T-(2)"Z/2Z
is
a
field
(for
the isomorphism
=,
cf. Example
4.1).
j
Of course these notions may be translated into terms not involving quotients
at all, and it is largely a matter of aesthetic preference whether prime and
maximal ideals should be defined as in Definition 4.9 or by the following equivalent
conditions:
Proposition
4.11. Let I
I is prime
if
^ (1)
and only
be
if for
an
if
and only
if for
ICJ
of
a
commutative ring R.
all a,b e R
ab e I =>
I is maximal
ideal
=>
(a
el
or
all ideals J
(I
=
J
or
be
I);
of R
J
=
R).
Then
4. Ideals and quotients: Remarks and
Proof. The ring
R/I
is
151
examples
integral domain if and only if Va, b
an
a-b
=
0 =>
(a
=
0
6
or
G
R/I
0).
=
This condition translates immediately to the given condition in i?, with
b
b + I, since the 0 in R/I is /.
a
=
a
+1,
=
maximality, the given condition follows from the correspondence between
ideals of R/I and ideals of R containing / (§3.3) and the observation that a
As for
commutative ring is
a
field if and only if its only ideals
(0)
are
From the formulation in terms of quotients, it is
maximal
indeed, fields
are
==>
and
(1) (Exercise 3.8).
completely clear that
prime;
integral domains. This fact
is of
course easy to
check in terms of
the other description, but the argument is a little more cumbersome
Prime ideals are not necessarily maximal, but note the following:
Proposition 4.12. Let I be an ideal of a commutative
then I is prime if and only if it is maximal.
immediately from Proposition
Proof. This follows
For
example, let (n) be
(n) prime
Indeed, for
Example 1.17.
nonzero n
In general, the
<^=>
an
(n)
the ring
ideal of Z, with
maximal <^=>
Z/nZ
is
finite,
n >
n
finite,
0; then
as an
Proposition
integer.
4.12
applies; cf.
prime ideals of a commutative ring R
Speci?. We can 'draw' SpecZ as follows:
set of
spectrum of i?, denoted
is
If R/I
1.15.
is prime
so
ring R.
(Exercise 4.14).
is called19 the
Actually, attributing the fact that nonzero prime ideals of Z are maximal to
the finiteness of the quotients (as we have just done) is slightly misleading; a better
'explanation' is that Z is a PID, and this phenomenon is common to all PIDs:
Proposition 4.13. Let R be a PID,
prime if and only if it is maximal.
Proof. Maximal ideals
nonzero
prime ideals
are
are
prime in
maximal in
and let I be
ring,
PID; we
a nonzero
every
so
a
will
we
ideal in R.
only need
use
to
Then I is
verify
that
the characterization of
prime and maximal ideals obtained in Proposition 4.11. Let I
ideal in i?, with a ^ 0, and assume I C J for an ideal of R. As R is
=
Believe it or not, the term is borrowed from functional analysis.
(a)
a
be
a
PID, J
prime
=
(b)
for
Rings and modules
III.
152
some
then b G
b G R. Since I
(a)
If b G
or c G
(a),
then
(a),
(6)
(a)
=
Q
since /
C
(a);
=
(b)
(a)
J,
=
we
have that
a
=
be for
some c G
R. But
is prime.
and I
J follows. If
=
then
(a),
c G
c
=
da for
some
d G R. But then
a
from which frd
=
=
1 since cancellation
integral domain). This implies that b
be
bda,
by the nonzero
is
=
a
a
holds in R
unit, and hence J
have shown that if I C J, then either I
maximal, by Proposition 4.11.
That is,
we
=
=
J
(b)
or
(since
=
J
R is
an
R.
=
R: thus J is
Example 4.14. Let k be a field. Then nonzero prime ideals in k[x] are maximal,
since k[x] is a PID (as the reader has hopefully checked by now; cf. Exercise 4.4).
a picture of Spec/c[x] would look pretty much like the picture of SpecZ
shown above: with the maximal ideals at one level, all containing the prime (and
nonmaximal) ideal (0).
Therefore,
This picture is particularly appealing for fields such as k
C, which are
in
r G k such that
that
which
for
there
exists
G
is,
algebraically closed,
every f(x)
k[x]
=
f(r)
=
0. It would take
us too
but the reader should be
to this20.
ideals in
aware
far afield to discuss this notion at any length now;
that C is algebraically closed. We will come back
this fact, it is easy to
all and only the ideals
Assuming
C[x]
are
verify (Exercise 4.21)
that the maximal
(x-z)
where
z
little,
we
ranges over all
could
come up
complex numbers. That is, stretching our imagination
with the following picture for SpecC[x]:
(x)
(x -i)
(x
-
a
1)
-/-
There is
a
'complex line21 worth' of maximal ideals: for each
maximal
are no
(x z); the prime
other prime ideals.
(0)
z
G
C
we
have the
is contained in all the maximal ideals; and there
The picture for SpecC[x] may serve as justification for the fact that, in algebraic
geometry, C[x] is the ring corresponding to the 'affine line' C; it is the ring of an
algebraic curve. It turns out that the fact that there is exactly 'one level' of maximal
A particularly pleasant proof may be given using elementary complex analysis,
consequence
21
of Liouville's theorem,
We know: it looks like
or
a
the maximum modulus principle; cf. §V.5.3.
plane. But it is
a
line,
as a
complex entity.
as a
Exercises
ideals
153
in
(0)
over
C[x]
reflects precisely the fact that the corresponding geometric
object has dimension 1.
(Krull) dimension of a commutative ring R is the length of the
chain
of
prime ideals in R. Thus, Proposition 4.13 tells us that PIDs, such
longest
as Z, have 'dimension 1'. In the lingo of algebraic geometry, they all correspond to
In general, the
curves.
j
Example
4.15. For
where k
a
examples of rings of higher dimension, consider
/c[#i,..., xn\,
is
field. Note that there
C
(0)
in this
ring.
(Why
is
C
chains of prime ideals of length
(xi,x2)
C
...
C
n
(xi,...,xn)
these ideals prime? Cf. Exercise 4.13.) This says that
n. One can show that n is in fact the longest length of
has dimension >
k[x\,...,xn]
a
are
(xi)
are
chain of prime ideals in k[xi,..., xn]; that is, the Krull dimension of k[xi,..., xn]
precisely n. In algebraic geometry, the ring C[#i,... ,xn] corresponds to the
complex space Cn.
Dealing precisely with these notions
n-dimensional
is not so easy, however. Even the seemingly
statement
that
the
maximal
ideals
of C[#i,... , 3?77,j are all and only the ideals
simple
for
e
(xi
),
(zi,..., zn) Cn, requires a rather deep result, known
as
Hubert's Nullstellensatz.
in
§VII.2, after
We will
come
j
back to all of this and get
we
develop (much)
more
a very
small taste of algebraic geometry
machinery.
Exercises
ring, and let {Ia}aeA be
4.1. > Let R be a
^ Ia
:=
ot£A
<
^ ra
such that ra G
Ia and
family of
a
ra
=
ideals of R. We let
0 for all but
finitely
many
Prove that
a >
.
)
\ot£A
^2a Ia is
[§4.1]
an
ideal of R and that it is the smallest ideal containing all of
the ideals Ia.
homomorphic image of a Noetherian ring is Noetherian. That
S is a surjective ring homomorphism and R is Noetherian,
then S is Noetherian. [§6.4]
4.2. > Prove that the
is, prove that if cp
:
R
4.3. Prove that the ideal
(2,x)
of
Z[x]
is not
principal.
field, then k[x] is a PID. (Hint: Let I C k[x] be any ideal.
^ (0), let f(x) be a monic polynomial in / of
minimal degree. Use division with remainder to construct a proof that /
(/(#)),
arguing as in the proof of Proposition II.6.9.) [§4.1, §4.3, §V.2.4, §V.4.1, §VI.7.2,
4.4. > Prove that if k is a
If I
=
(0),
then / is principal. If /
=
§VII.1.2]
III.
154
/, J be ideals
4.5. > Let
in
a
Rings and modules
ring i?, such that I + J= (1). Prove that IJ
=
I(~)J.
[§4-1]
/, J be ideals in a ring R. Assume that R/(IJ) is reduced (that is, it has
I D J.
nilpotent elements; cf. Exercise 3.13). Prove that 13
4.6. Let
no nonzero
=
4.7. > Let R
generated by
4.8.
not
>
a
k be
=
a
or
is
a
ring and f(x)
R[x] a monic polynomial. Prove that f(x)
[§4.2, 4.9]
is
a
Let
left-zero-divisor in
f(x)
=
ddXd
dd-ig(x)
=
R[x\.
e
g(x)
f(x)g(x)
such that
0 for all i,
Q(Vd)
Define
N(zw)
a
=
Q(Vd)
N(z)N(w)
(Cf.
Prove that
Q and
a
function iV
The function iV is
subrings.
is
a
Q(y/d)
:
and that
a
{a
:=
+
Z
-?
bVd |
a, b G
by N(a
N(z) ^
'norm'; it is
is
h
bo be
Deduce that
0.
=
-\
ring, and let f(x)
a
such that
f(x)b
=
cidg(x)
=
say about
0.
(Hint:
polynomial
a nonzero
0, and then
6e?)
integer, and consider the
Q}.
subring of C.
also Exercise
Q(Vd)
bexe
=
^ 0,
not the square of an
-i
Prove that
follows. Let R be
by induction. What does this
Let d be an integer that is
subset of C defined by22
4.10.
as
Prove that 3b G i?, b
h ao, and let
-\
of minimal degree
prove
(principal)
G
zero-divisor.
right-)
4.9. Generalize the result of Exercise 4.8,
be
ideal in
k[x]
nonzero
unique monic polynomial. [§4.2, §VI.7.2]
Let R be
(left-
field. Prove that every
a
0 if
bVd)
Q(>/d),
+
z G
very useful in the
:=
z
a2
^
b2d.
Prove that
Q(Vd)
and of its
-
0.
study of
2.5.)
field and in fact the smallest subfield of C containing both
Vd. (Use N.)
Prove that
Q(Vd)
^
[V.1.17, V.2.18, V.6.13,
Q[£]/(£2
-
d). (Cf. Example 4.8.)
VII. 1.12]
4.11. Let R be a commutative
ring,
a e
R, and /i(x),..., fr{x)
G
R[x\.
Prove the equality of ideals
(ZiW,
fr(x)jX- a)
=
•.
(/i(o),..., fr{a),x- a).
Prove the useful substitution trick
(/i(x),..., fr(x),x
(Hint:
Exercise
-
a)
(/i(o),..., fr(a))'
3.3.)
4.12. > Let R be a commutative
ring and
ai,
,
R[xU...,Xn]
(xi
-
ai,...,xn
-
an elements of R. Prove that
^R
on)
[§VII.2.2]
Of
course there are
on which one is used.
two
'square
roots of cf; but the definition of
Q(Vd)
does not depend
Exercises
155
4.13. > Let R be
integral domain.
an
For all k
1,...,n prove that (#i,...,Xk)
=
is prime in R[xi,... ,xn], [§4.3]
4.14.
>
Prove 'by hand' that maximal ideals
are
prime, without using quotient
rings. [§4.3]
4.15. Let (p
an
:
S be
R
homomorphism of commutative rings, and let I C 5 be
a prime ideal in 5, then (p~l(I) is a prime ideal in R.
necessarily maximal if I is maximal.
a
ideal. Prove that if I is
Show that (p~l(I) is
not
a commutative ring, and let P be a prime ideal of R. Suppose 0 is
the only zero-divisor of R contained in P. Prove that R is an integral domain.
4.16. Let R be
(If you know a little topology...) Let K be a compact topological space,
and let R be the ring of continuous real-valued functions on K, with addition and
4.17.
-i
multiplication defined pointwise.
(i) For
p G
K, let Mp
=
{/
G
R\f(p)
=
0}.
Prove that
Mp
is
a
maximal ideal in
R.
Prove that if
(ii)
(Hint:
(iii)
/i,..., fr
Consider
/2
+
...
G R have no common zeros, then
+
Conclude that p
(The
Mp
defines
i—?
a
bijection from
(1).
in the
Mp
for
some p G
if.
if to the set of maximal ideals
set of maximal ideals of a commutative
spectrum of R; it is contained
=
/2.)
Prove that every maximal ideal M in i? is of the form
(Hint: You will use the compactness of K and (ii).)
of R.
(/i,..., fr)
ring R is called the maximal
(prime) spectrum Speci?
Relating commutative rings and 'geometric' entities such
as
defined in §4.3.
topological
spaces is
the business of
algebraic geometry.)
The compactness hypothesis is necessary:
cf. Exercise V.3.10.
[V.3.10]
a commutative ring, and let N be its nilradical (Exercise 3.12).
Prove that N is contained in every prime ideal of R. (Later on the reader will
check that the nilradical is in fact the intersection of all prime ideals of R:
4.18. Let R be
Exercise
V.3.13.)
4.19. Let R be
a
commutative ring, let P be
a
prime ideal in i?, and let Ij be
ideals of R.
(i) Assume that I\ Ir C P; prove that Ij C P for some j.
(ii) By (i), if P D f]rj=1 Ij, then P contains one of the ideals Ij. Prove
if P D H?li Iji tnen P contains one of the ideals Ij.
4.20. Let M be
a
two-sided ideal in
that M is maximal if and only if
4.21. > Let k be an
=
(x
c)
for
§VII.2.2]
4.22. Prove that
(x2
+
1)
disprove:
(not necessarily commutative) ring R.
R/M is a simple ring (cf. Exercise 3.9).
a
algebraically closed field, and let
that / is maximal if and only if /
or
is maximal in
R[x].
I C
some c G
k.
Prove
k[x] be an ideal. Prove
[§4.3, §V.5.2, §VII.2.1,
III.
156
Rings and modules
ring R has Krull dimension 0 if every prime ideal in R is maximal. Prove
that fields and Boolean rings (Exercise 3.15) have Krull dimension 0.
4.23. A
4.24. Prove that the
thus
5.
it
corresponds
Modules
ring Z[x] has Krull dimension > 2. (It is in fact exactly 2;
surface from the point of view of algebraic geometry.)
to a
over a
ring
We have emphasized the parallels between the basic theory of groups and the basic
theory of rings; there are also important differences. In Grp, one takes the quotient
of a group, by a (normal sub)group, and the results is a group: throughout the
process one never leaves
Grp. The
situation is in
a sense even
better in Ab, where
of an abelian
the normality condition is automatic; one simply takes the quotient
group by an abelian (sub)group, obtaining an abelian group.
The situation in Ring is not nearly as neat. One of the three characters in the
story is an ideal: which is not a ring according to the axioms listed in Definition 1.1,
in all but the most pathological cases. In other words, the kernel of a homomorphism
of rings is (usually) not a ring. Also, cokernels do not behave as one would hope
(cf. Exercise 2.12). Even if one relaxes the ring definition, giving up the identity
and making ideals subrings, there is no reasonably large class of examples in which
the 'ideal' condition is automatically satisfied by substructures. In short, Ring is
not a particularly pleasant category.
Modules will fix all these problems. If R is a ring and / C R is a two-sided
ideal, then all three structures i?, /, and R/I are modules over R. The category
of i?-modules is the prime example of a well-behaved category: in this category
kernels and cokernels exist and do precisely what they ought to. The category Ab
is a particular case of this construction, since it is the category of modules over
disguise (Example 5.4). The category of modules
many of the excellent properties of Ab; and we will get a
Z in
category for each ring R.
abelian category.
These
are
all
examples23
over
a
ring R will share
brand new, well-behaved
of the important notion of
5.1. Definition of
(left-)R-module. In short, i?-modules are abelian groups
action of R. To flesh out this idea, recall that 'actions' in general
homomorphisms into some kind of endomorphism structure: for example,
endowed with
denote
an
defined group actions in §11.9.1 as group homomorphisms
the groups of automorphisms of objects of a category.
we
We
can
give
an
analogous definition of the
action of
Indeed, recall that if M is an abelian group, then
a ring in a natural way (cf. §2.5). A left-action of
a
from
ring
EndAb(A^)
a
ring R
on
a
fixed group to
abelian group.
HomAb(M, M) is
on an
:=
M is then
simply
a
homomorphism of rings
a
:i?^EndAb(M);
In fact, it can be shown ('Freyd-Mitchell's embedding theorem') that every small abelian
category is equivalent to a subcategory of the category of left-modules over a ring. So to some
extent we can understand abelian
We will
come
categories by understanding categories of modules well enough.
back to this in Chapter IX.
5. Modules
we
over a
will say that
Similarly
a
157
ring
makes M into
a
left-R-module.
to the situation with group
actions, it is convenient
to
spell
out this
definition.
Claim 5.1.
datum
of
a
The datum
of a homomorphism
p: Rx M
satisfying
the
p(r, m
p(r
a as
above is precisely the
same as
the
function
following requirements: (Vr, s
+
=
ra)
+ 5,
p(rs,m)
p(l,ra)
n)
+
p(r, ri);
p(r, ra)
+
p(s, ra);
G
R) (Vra, n
G
M)
p(r,p(s,m));
=
=
=
p(r, ra)
M
-?
ra.
The proof of this claim follows precisely the pattern of the discussion in the
beginning of §11.9.2, so it is left to the reader (Exercise 5.2). Of course the relation
between p and
a
is given
by
p(r,m)
=
o~(r)(m).
As always, carrying p around is inconvenient, so it is common to omit mention of
it: p(r,m) is often just denoted rra; this makes the requirements listed above a
little more readable. Summarizing and adopting this shorthand,
Definition 5.2. A left-i?-module structure
map
Rx M
M, (r,m)
—>
r(ra
+
(r
s)m
+
(rs)m
Ira
=
n)
=
=
=
on an
abelian group M consists of
a
rra, such that
i—?
rm + m\
rm + sm;
r(sm);
ra.
j
Right-R-modules are defined analogously. The reader can glance back at §11.9,
especially Exercise II.9.3, for a reminder on right- vs. left-actions; the issues here are
analogous. Thus, for example, a right-i?-module structure may be identified with
a left-i?°-module structure, where R° is the 'opposite' ring obtained by reversing
the order of multiplication (Exercise 5.1). However, note that R and R° have no
good reasons to be isomorphic in general (while every group is isomorphic to its
opposite).
is
These issues become immaterial if R is commutative: then the identity R
isomorphism, and left-modules/right-modules are identical concepts.
an
R°
The
reader will not miss much by adopting the blanket assumption that all rings
mentioned in this section are commutative. It is occasionally important to make this
hypothesis explicit (for example in dealing with algebras, cf. Example 5.6), but most
going to review works verbatim for, say, left-modules over an
arbitrary ring as for modules over a commutative ring. We will write 'module' for
'left-module', for convenience; it will be the reader's responsibility to take care of
appropriate changes, if necessary, to adapt the various concepts to right-modules.
of the material we are
III.
158
Rings and modules
5.2. The category i?-Mod. The reader should spend some time getting familiar
with the notion of module, by proving simple properties such as
(Vra
e
M)
:
0
(Vra
e
M)
:
(-1)
m
=
0,
m
=
-ra,
ring (Exercise 5.3). The following fancier-sounding
is
also useful in developing a feel for modules:
equally
property
(but
trivial)
where M is a module over some
Proposition 5.3. Every abelian group is
Proof. Let G be
an
a
Z-module,
in
exactly
A Z-module structure
abelian group.
on
one way.
G is
a
ring homo-
morphism
Z^EndAb(G).
Since Z is initial
in
Ring (§2.1), there
exists
exactly
one
such homomorphism,
proving the statement.
Thus, 'abelian group' and 'Z-module' are one and the same notion. Quite
action of n G Z on an element a of an abelian group simply yields
the ordinary 'multiple' na\ this operation is trivially compatible with the operations
concretely, the
inZ.
A homomorphism of R-modules is a homomorphism of (abelian) groups which
compatible with the module structure. That is, if M, N are i?-modules and
N is a function, then cp is a homomorphism of R-modules if and only if
(p : M
is
M)(\/rri2 G M)
(Vrai
G
(Vr
i?)(Vra
G
G
M)
(p(mi
:
<p(rm)
:
=
+
7712)
=
(p(mi)
+
<^(m2);
r<p(m).
It is
hopefully clear that the composition of two i?-module homomorphisms is an
i?-module homomorphism and that the identity is an i?-module homomorphism:
i?-modules form
which
we
a
category
will denote24 'i?-Mod'.
Example 5.4. The category Z-Mod of Z-modules is 'the same as' the category Ab:
indeed, every abelian group is a Z-module in exactly one way (Proposition 5.3),
and Z-module homomorphisms
are
simply homomorphisms of abelian
groups.
j
k is a field, i?-modules are called k-vector spaces. We will
call the category of vector spaces over a field k '/c-Vect'; this is just another name
for /c-Mod. Morphisms in /c-Vect are often called25 linear maps. 'Linear algebra' is
the study of /c-Vect (extended to R-M06 when possible); Chapters VI and VIII will
Example
5.5. If R
=
be devoted to this subject.
If
R
left-modules
is
or
not
commutative,
j
we
should agree on whether J?-Mod denotes the category of
mean 'left-modules'.
of right-modules. We will
This term is also used for homomorphisms of R modules for
as
frequently.
more
general rings R, but
not
5. Modules
over a
159
ring
Example 5.6. Any homomorphism of rings a
S by
interesting iJ-module: define p : R x S
p(r,s)
:=
:
R
S may be used to define
an
a(r)s
r G i? and s E S. The operation on the right is simply multiplication in 5,
and the axioms of Definition 5.2 are immediate consequence of the ring axioms and
of the fact that a is a homomorphism. For instance, taking S
R and a
id#
for all
=
makes R
It is
a
(left-)
module
common to
over
write
rs
=
itself.
rather than
a(r)s.
The ring operation in S and the i?-module structure induced by
even
of S:
a are
linked
more tightly if we require R to be commutative and a to map R to the center
that is, if we require a(r), s to commute for every r G i?, s G 5. Indeed, in this
the left-module structure defined above and the right-module structure defined
analogously would coincide; further, with these requirements the ring operation in S
case
(51,52)
is
compatible with the i?-module
(ri5i)(r252)
Vfi,r2
G
i?, V5i,52
=
that
si52
structure in the sense
a(ri)5ia(r2)52
G 5:
»-?
is,
=
we can
that26
a(ri)a(r2)5i52
=
(rir2)(sis2)
'move' the action of R at will through
products in S.
Due to their importance, these examples deserve their
Definition 5.7. Let R be
homomorphism
a :
S such that
i?
own
official
commutative ring. An R-algebra is
a(R) is contained in the center of S.
a
a
name:
ring
j
i?-algebra by the name of
'is' an i?-module S with a
ring S with a compatible i?-module
The usual abuse of language leads us to refer to an
the target S of the homomorphism. Thus, an i?-algebra
compatible ring structure,
structure. An R-algebra S
There is
an
or, if you
is
a
prefer,
a
division algebra if S is
a
division ring.
evident notion of 'i?-algebra homomorphism' (preserving both the
structure), and we thus get a category i?-Alg. The situation
ring and module
simplifies substantially if S is itself commutative, in which
case
the condition
on
'Commutative R-algebras' form a category, which the
attentive reader will recognize as a coslice category (Example 1.3.7) in the category
the center is unnecessary.
of commutative
Also,
rings.
note that
Z-Alg
is just another
name
for Ring
(why?).
The polynomial rings R[xi,... ,xn], as well as all their quotients, are
commutative i?-algebras. This is a particularly important class of examples; for
an
algebraically closed field, these
geometry
are
R
=
k
the rings used in 'classical' affine algebraic
(cf. §VII.2.3).
j
a unique module structure over any ring R and is a zerothat
it
is both initial and final. As in the other main categories
i?-Mod,
is,
object
we have encountered, a bijective homomorphism of R-modules is automatically an
The trivial group 0 has
in
'More
generally, this choice makes the multiplication in S 'R-bilinear'.
III.
160
isomorphism
in i?-Mod
category i?-Mod
(for
Rings and modules
In these and many other respects, the
ring R) and Ab are similar.
(Exercise 5.12).
any commutative
just as in the category Ab, each
object of the category (cf. §11.4.4)27.
Hom#_Mod(A^ N)
M
N
let
and
be
i?-modules.
Since
Indeed,
homomorphisms of i?-modules are in
of
abelian
particular homomorphisms
groups,
IfR
is commutative, the
similarity
goes further:
may itself be seen as an
set
C
Hom^.Mod(M,N)
HomAb(M,N)
(up to natural identifications). The operation making Hom.Ab(Af> N) into an
abelian group, as in §11.4.4, clearly preserves HomR.Mod(M, N); it follows that the
latter is an abelian group. For r G R and cp G Homj?_Mod(-W, N), the prescription
as sets
(Vra
defines
a
function28
rep
:
M
R is commutative, because
(rep) (am)
Thus,
we
have
an
M)
verify
(rcp)(m)
:
:=
N. This function is
—>
(Va
G
rep (am)
i?), (Vra
=
G
(ra)cp(m)
natural action of R
a
is immediate to
=
G
an
R-module homomorphism if
M)
=
(ar)cp(m)
the abelian group
on
that this makes
rcp(m)
Kom.R.^{0(\(M,N)
=
a(rcp(m)).
Hom#_Mod(A^> N),
into
an
and it
i?-module.
Watch out: if R is not commutative, then in general Homj?_Mod(-W> N) is 'just'
abelian group. More structure is available if M, N are bimodules; cf. §VIII.3.2.
5.3.
quotients. Since i?-modules are an 'enriched' version of
through the usual constructions very quickly, by
the analogous constructions in Ab are preserved by the R-
Submodules and
abelian groups,
we can progress
just pointing out that
module structure.
A submodule N of an i?-module M is a
subgroup preserved by the
action of R.
That is, for all r G R and n G N, the element rn (defined by the R-module structure
of M) is in fact in N. Put otherwise, and perhaps more transparently, N is itself
an
i?-module, and the inclusion N
Example
5.8. We
submodules of R
Example
are
5.10.
submodule of M.
submodule
If A7" is
view R itself
or
a
abelian group
submodules
If
r
G
homomorphism.
(left-)R-module (cf. Example 5.6);
(left-)ideate of R.
as a
then precisely the
5.9. Both the kernel and the
of R-modules
Example
can
are
C M is an i?-module
image of
a
homomorphism
cp
:
M
the
j
M'
(of M, M', respectively).
R and M is
an
j
i?-module, rM
If I is any ideal of i?, then IM
=
{rm m
{rm \rG J, ra
=
G
G
M}
M}
is
a
is
a
R.
submodule of M, then it is in particular a (normal) subgroup of the
(M, +); thus we may define the quotient M/N as an abelian group.
Notational convention: One often writes Hom#(M, N) for Hom^.Mod(^)^)- We will not
adopt this convention here but we will use it freely in later chapters.
Parse the notation carefully: rip is the name of a function on the left, while rip(rn) on the
right is the action of the element r G R on the element (f(m) G N, defined by the .R-module
structure on N.
5. Modules
Of
161
ring
it would be desirable to
course
one
over a
reasonable way to do
so:
see
M
7T :
to be an i?-module
m G
M. That is,
+
N)
Claim 5.11.
on
module, and
as
usual there is only
M/N
-?
we are
r7r(ra)
=
=
7r(rra)
=
rm +
N
led to define the action of R
r(m
R-module
as a
homomorphism, and this forces
r(m
for all
this
will want the canonical projection
we
+
N)
:= rm
on
M/N by
+ N.
N, this prescription does define
For all submodules
a
structure
of
M/N.
The proof of this claim is immediate and is left to the reader. The R-module
M/N is (of course) called the quotient of M by N.
Example 5.12. If R is a ring and / is a two-sided ideal of i?, then all three of /,
i?, and the quotient ring R/I are i?-modules: / is a submodule of i?, and the rings
R and R/I are in fact i?-algebras if R is commutative (cf. Example 5.6).
j
5.13.
Example
If R is not commutative and / is just
a
(say) left-ideal,
then
the quotient R/I is not defined as a ring, but it is defined as a left- module (the
quotient of the module R by the submodule I). The action of R on R/I is given
by left-multiplication: r(a
The reader should
I)
+
now
Theorem 5.14. Let N be
ra +
=
expect
a
homomorphism of R-modules cp
a
I.
universal property for quotients, and here it is:
of an R-module M. Then for every
P such that N C ker cp there exists a unique
P so that the diagram
M/N
submodule
:
homomorphism of R-modules (p
M
:
M
>P
commutes.
As in previous appearances of such statements, this is an immediate
consequence of the set-theoretic version (§1.5.3) and of easy notation matching and
compatibility checks. For an even faster proof, one can just apply Theorem II.7.12
verify that
Since
M/N,
<p is an i?-module
every submodule N is then the kernel of the canonical
our
recurring slogan becomes,
:
Grp or Ring, being a kernel poses no restriction
substructures. Put otherwise, 'every monomorphism in i?-Mod is
of the distinguishing features of an abelian category.
as
in
projection M
in the context of i?-Mod
kernel <^=> submodule
unlike
and
homomorphism.
on
a
the relevant
kernel'; this is
one
III.
162
Rings and modules
decomposition and isomorphism theorems. The discussion
proceeds along the same lines as for (abelian) groups; the statements of the key
facts and a few comments should suffice, as the proofs are nothing but a rehashing
of the proofs of analogous statements we have encountered previously. Of course
the reader should take the following statements as assignments and provide all the
needed details.
5.4.
Canonical
now
In the
form:
context of
Theorem
5.15.
as
R-modules, the canonical decomposition takes the following
Every R-module homomorphism
:
(p
M
M' may be decomposed
follows:
where the isomorphism <p in the middle is the homomorphism induced by (p
Theorem
(as
in
5.14)-
The 'first isomorphism theorem' is the
Corollary
Then
5.16.
Suppose
cp
:
M
following
M' is
W
a
consequence:
surjective R-module homomorphism.
M
2*
.
kercp
If M is
an
R-module and A7" is
a
submodule of M, then there is
a
bijection
(cf. §11.8.3)
u :
P of M containing
{submodules
preserving inclusions,
N}
M/N
P/N
We also have
II.8.11),
in the
a
M/N}
and the 'third isomorphism theorem' holds:
Proposition 5.17. Let N be a submodule
submodule of R containing N. Then P/N is
Proposition
of
{submodules
of
an
R-module M, and let P be
of M/N, and
a
submodule
a
M
~
~P'
version of the 'second
isomorphism theorem' (cf.
with simplifications due to the fact that normality is not
an
issue
theory of modules:
Proposition
5.18. Let
N, P be submodules of
N + P is
a
submodule
of M;
N DP is
a
submodule
of P,
an
R-module M. Then
and
N + P
_
P
"
N
NOP'
More generally, it is hopefully clear that the
f]a N&
as
of any family
{Na}a
subgroups of the abelian
of M.
of submodules of
group
an
sum
^2a Na
R-module M
and intersection
(which
M; cf. for example Lemma II.6.3)
are
are
defined
submodules
Exercises
163
Exercises
a ring. The opposite ring R° is obtained from R by reversing the
that
is, the product a b in R° is defined to be ba G R. Prove that
multiplication:
the identity map R
R° is an isomorphism if and only if R is commutative. Prove
that .A/fnW is isomorphic to its opposite (not via the identity map!). Explain
how to turn right-R-modules into left-i?-modules and conversely, if R
R°. [§5.1,
5.1. > Let R be
=
VIII.5.19]
5.2. > Prove Claim 5.1.
5.3. > Let M be
module
a
-m, for all ra G M.
over a
ring R. Prove that 0
m
=
0 and that
m
(—1)
=
[§5.2]
ring. A
nonzero i?-module M is simple (or irreducible) if its only
M.
N be
and
Let M, N be simple modules, and let cp : M
{0}
of
Prove
that
either
or
is
an
i?-modules.
0
homomorphism
isomorphism.
cp
cp
5.4.
-i
Let R be
submodules
a
[§5.1]
a
are
=
(This rather
innocent statement is known
5.5. Let R be a
Prove that
ring, viewed
as an
Hom#_Mod(-R>-^0
M
iJ-module
as
Schur's
as
over
lemma.) [5.10, 6.16,
itself, and let M be
an
VI. 1.16]
iJ-module.
i?-modules.
abelian group. Prove that if G has a structure of Q-vector space,
then it has only one such structure. (Hint: First prove that every element of G has
5.6. Let G be
an
necessarily infinite order.)
5.7. Let K be
field, and let k
a
in fact
space over k
(and
that K is
extension of k.
an
5.8. What is the initial
5.9.
-i
Let R be
a
a
C K be a subfield of K.
/c-algebra)
in
a
Show that K is a vector
natural way. In this situation,
we say
object of the category i?-Alg?
commutative ring, and let M be an R-module. Prove that
on the i?-module End^_Mod(-^) makes the latter an
the operation of composition
i?-algebra in a natural way.
Prove that
Mn(R) (cf.
Exercise
1.4)
is
an
i?-algebra,
in
a
natural way.
[VI. 1.12,
VI.2.3]
5.10. Let R be
Exercise
5.4).
5.11.
is
a
>
a
commutative ring, and let M be a simple
EndjR_Mod(-^) is a division iJ-algebra.
R-module (cf.
Prove that
Let R be
a
commutative ring, and let M be
bijection between the
set of
i?[x]-module
an
R-module. Prove that there
structures on M and
EndjR_Mod(^)-
[§VI.7.1]
N be a
5.12. > Let R be a ring. Let M, N be iJ-modules, and let (p : M
homomorphism of R-modules. Assume cp is a bijection, so that it has an inverse <p~l
as a set-function.
Prove that <p~l is a homomorphism of R-modules. Conclude
that a bijective i?-module homomorphism is an isomorphism of i?-modules. [§5.2,
—>
§VI.2.1, §IX.1.3]
Rings and modules
III.
164
integral domain, and let / be
isomorphic to R as an i?-module.
5.13. Let R be
Prove that /
is
5.14. > Prove
5.15.
-i
element,
(This
is
^
(I
+
3)13
as
ring, and let /, 3 be ideals of R.
a
a
particular
p+fc5
case
a
define
of Nakayama's lemma, Exercise
commutative ring, and let / be
ring
a
Prove that
i?-modules.
commutative ring, M an i?-module, and let
determining a submodule aM of M. Prove that M
Let R be
5.17. > Let R be
jjjk q
principal ideal of R.
a nonzero
Proposition 5.18. [§5.4]
Let R be a commutative
I (R/J)
5.16.
an
an
a G
=
0
R be
<^=>
a
nilpotent
aM
=
M.
VI.3.8.) [VI.3.8]
ideal of R.
Noting that
structure on the direct sum
ReesR(I)
:=
0/J= i?0 I 0 I2 0 I3 0
.
j>0
homomorphism sending R identically to the first term in this direct sum makes
Reesn(I) into an iJ-algebra, called the Rees algebra of /. Prove that if a G R is
a non-zero-divisor, then the Rees algebra of (a) is isomorphic to the polynomial
ring R[x] (as an iJ-algebra). [5.18]
The
as in Exercise 5.17 let a G R be a non-zero-divisor, let / be
i?, and let 3 be the ideal al. Prove that Rees^(J)
Reesj?(J).
5.18. With notation
any ideal of
6.
=
in i?-Mod
Products, coproducts, etc.,
We have stated several times that categories such
as
R-Mod
are
'well-behaved'. We
will explore in this section the sense in which this can be formalized at this stage.
The bottom line is that these categories enjoy the same nice properties that we
have noted along the way for the category Ab.
We will also include here
some
general considerations
on
finitely generated
modules and algebras.
As in the previous section, we will write 'module' for 'left-module'; the reader
should make appropriate adaptations to the case of right-modules. Little will be
lost by assuming that all rings appearing here are commutative (thereby removing
the distinction between left- and
right-modules).
6.1. Products and coproducts. As in Ab, products and coproducts exist, and
finite products and coproducts coincide, in i?-Mod. Indeed, recall the construction
of the direct sum of two abelian groups (§11.3.5): if M and N are abelian groups,
then M(BN denotes their product, with componentwise operation. If M and N are
R-modules, we can give an R-module structure to M 0 N by prescribing Vr G R
r(m,n)
This defines the direct
sum
of M, N,
:=
(rm,rn).
as an
i?-module.
Note that M 0 N
together with several homomorphisms of R-modules:
7TM : M
0iV
M,
-?
ttjv : M 0 iV
iV
-?
comes
6. Products, coproducts, etc., in P-Mod
sending (ra, n)
respectively, and
to ra, n,
iM
sending
ra to
Proposition
(m,0)
6.1.
165
and
:
M
M 0 N,
N
:
^
M ® N
(0,n).
n to
Tfte direct
the product and the coproduct
Proof. Product: Let P be
iN
-?
an
si/ra
of
M (& N
satisfies
the universal properties
of both
M and N.
R-module, and let
two P-module homomorphisms. The definition of
(pM
an
P
M,
(pw
:
P
A" be
—>
P-module homomorphism
(^MX(pjv:P^M©iV
is forced by the needed commutativity of the diagram
N
That is,
(Vp
G
P)
(cpM
:
x
<Pn)(p)
:=
(<Pm(p),<Pn(p))'
an R-module homomorphism, and it is the unique one making the diagram
commute; therefore M 0 N works as a product of M and N.
Coproduct: View the preceding argument through a mirror! Let P be an P-
This is
M
P be two P-module homomorphisms.
module, and let ipM
P, ipw : A7"
The definition of an P-module homomorphism
^m © ^iv
is forced
:
M © N
P
-?
by the needed commutativity of the diagram
That is29, (Vra
G
M)(Vra
G
(ipM®ipN)(m,n)
N)
=
=
=
+ (^m
(iI>m®iI>n)(™>j0)
(^M
©
^iv)
°
*Af
(m)
+
©
(V>M
^)(0,n)
©
^iv)
°
«Jv(w)
ipM(m)+ipN(n).
P-module homomorphism: it is a homomorphism of abelian groups by
virtue of the commutativity of addition in P, and it clearly preserves the action of R.
This is
an
This should look familiar; cf. Exercise II.3.3.
III.
166
Since it is the unique i?-module
verifies that M 0 N works as a
Rings and modules
homomorphism making the diagram commute, this
coproduct of M and N.
It may seem like a good idea to write M x N rather than M 0 iV when viewing
the latter as the product of M and iV; but in due time (§VIII.2) we will encounter
M
x
N again, in the context of 'bilinear maps', and in that context M
as an R-module.
x
N is not
viewed
The reader should also work out the fibered versions of these constructions;
cf. Exercises 6.10 and 6.11.
The fact that finite products and coproducts agree in R-Mod does not extend
to the infinite
case
(Exercise 6.7).
6.2. Kernels and cokernels. The facts that i?-Mod has
a
zero-object (the
0-
abelian groups (§5.2), and it has (finite) products and
module),
coproducts make i?-Mod an additive category. The fact that i?-Mod has wellbehaved kernels and cokernels, which we review next, upgrades it to the status of
its Horn sets
abelian category.
(We
are
will
come
back to these general definitions in
§IX.l.)
general, monomorphisms and epimorphisms do not automatically satisfy
good properties, even when objects of a given category are realized by adding
In
structure to sets. For example, we have seen that the precise relationship between
'surjective morphism' and 'epimorphism' may be rather subtle: epimorphisms are
surjective in Grp, but for complicated reasons (§11.8.6); and there are epimorphisms
that are not surjective in Ring (§2.3).
The situation in i?-Mod is as simple as it can be. Recall that we have identified
universal properties for kernels and cokernels
as follows: if
(cf. §11.8.6);
in the category i?-Mod
these would go
<p:M ^N
is
of
homomorphism of i?-modules, then kenp
factoring R-module homomorphisms a : P
a
is final with respect to the property
M such that (p o a = 0:
while coker cp is initial with respect to the property of
P such that (3 o (p
0:
homomorphisms (3 : N
=
Proposition 6.2.
The
following
kernels and cokernels exist;
hold in i?-Mod:
factoring i?-module
6. Products, coproducts, etc., in i?-Mod
cp is
monomorphism
function;
<^=>
a
(p is an epimorphism
function.
<^=>
167
keicp
is trivial
<^=>
cp is injective
as
a set-
coker^
is trivial
<<=>*
(p is surjective
as
a set-
its
with the kernel
Further, every monomorphism identifies
phism, and every epimorphism identifies
phism.
source
its target with the cokernel
of some
of some
mormor-
This proposition of course simply generalizes to i?-Mod facts we know already
from our study of Ab, and a quick review should suffice for the careful reader.
Kernels exist: indeed, the 'standard' definition of kernel satisfies the universal
spelled out above (same argument as in Proposition II.6.6). Cokernels exist:
properties
indeed, let
coker (p
=
;
map
P is such that (3 o (p
(3 : N
0, then inap C ker/3; so (3 must factor uniquely
the
universal
through N/inap by
property of quotients, Theorem 5.14. That is,
does
the
universal
satisfy
property for cokernels.
N/im(p
if
=
The proofs of all the implications in the second and third points in
Proposition 6.2 follow familiar patterns from (for example) Proposition II.6.12 and
Proposition II.8.18. The last sentence of Proposition 6.2 simply reiterates the slogan
submodule <^=> kernel and its mirror statement (which is just as true). Further
details
are
left to the reader.
By the way, whatever happened to the conditions characterizing monomorphisms and epimorphisms in Set (Proposition 1.2.1)? In Set, a function is a
monomorphism if and only if it has a left-inverse, and it is an epimorphism if and only if it
has a right-inverse. We have learned not to expect any of this to happen in more
Z defined by
general categories. Modules are not an exception: the function Z
2'
is
a
without
a
and
the
left-inverse,
'multiplication by
monomorphism
projection
Z
Z/2Z is an epimorphism without a right-inverse. We will come back to this
point in §7.2.
algebras. The universal property oifree R-modules
is modeled after the properties defining the other free objects we have encountered:
the goal is to define the R-module containing a given set A 'in the most efficient
way'. Again, the situation will match the case of abelian groups closely, so the
6.3. Free modules and free
reader may want to refer back to
§11.5.4.
The universal property formalizing the heuristic requirement goes
given a set A, we seek an iJ-module FR(A), called a free R-module
A, together with
and set-functions
as
follows:
on
the set
set-function j : A
FR(A), such that for all .R-modules M
M there exists a unique i?-module homomorphism
/ : A
a
III.
168
(f
:
FR(A)
M such that the
—>
Rings and modules
diagram
M
FR(A) -^—>
nonsense guarantees that such an i?-module is unique up to
isomorphism if it exists at all (Proposition 1.5.4) and that the function j : A
FR(A) is necessarily injective (cf. Exercise II.5.3). The question is, does it exist?
commutes. Abstract
answer will not appear to be very exciting, since it generalizes directly the
of abelian groups (that is, Z-modules). Given any set A, define the (possibly
infinite) direct sum N®A of an R-module N as follows:
The
case
:={a:A^N\a(a) ^
N®A
Of
has
0 for
only finitely
many elements a G
A}.
this agrees with the definition given for abelian groups in §11.5.4;
evident i?-module structure, obtained by defining, for all r G R and a
course
an
(ra)(a)
For iV
R
function ja
we may
define
a
:=
N®A
e
A,
r(a(a)).
function j
A
:
R®A by sending
a
G
A to the
A^R:
if
(1
|0
(V*€A): *„(*):=
Claim 6.3.
FR(A)
^
x
=
a,
.f ^
&
R®A.
The proof of this claim matches precisely the proof of Proposition II.5.6 and
is left to the reader (Exercise 6.1). The key is that every element of R®A may be
written uniquely
(shorthand
module
on
for
AJ
as a
finite
sum
SaeA V7(a))' incidentally, this
are
In particular, for A
{1,..., n}
i?en defined by
i?en, with j : A -?
=
a,
finite set, Claim
jM:=(0,-",0,.2-th 1
place
6.3 states that the R-module
,0,...,0)Gi?en,
FR({1,... ,n}).
satisfies the universal property for
a
is how elements of 'the free R-
often written—this
is legal, by virtue of Claim 6.3.
This is all entirely analogous to the story for Z-modules. The situation becomes
little more interesting if we switch from R-modules to commutative R-algebras;
the category is different,
essentially the only
finite set. In this case,
have a set-function j : A
a
Proposition
6.4.
R[A]
should expect a different
will need in these notes, so
so we
one we
we
a
R[A] for the polynomial ring
defined by j(i)
Xi.
write
R[A],
is
answer.
assume
free
The finite
A
=
commutative
R-algebra
on
case
is
{1,... ,n} is
R[xi,... ,xn]; we
=
the set A.
6. Products, coproducts, etc., in i?-Mod
169
Proof. The statement translates into the
algebra S and
every set-function
homomorphism cp
commutes.
R
S
:
R[A]
Since S is
/(l),
as to map X\ to
every commutative R-
5, there exists
a
unique i?-algebra
S such that the diagram
an
R-algebra,
Then
then to
we
have
fixed homomorphism of rings
a
we may construct (p :
times the 'extension
n
following: for
A
—>
(cf. Example 5.6).
by applying
/
:
R[A]
property' of Example
i?[xi,x2]
=
i?[xi][x2]
a
:
S
R[xi,... ,xn]
a to
R[x\] so
to
map x2
/(2), etc.
—>
=
2.3: extend
so as to
uniquely determined by its requirements.
ring homomorphism and shows that it is unique. The reader
Note that each extension is
This gives (p
as a
will verify that cp is also (automatically) an i?-module homomorphism, and hence
a homomorphism of R-algebras, concluding the proof.
After the fact, the reader
and recognize that the
xn\given there was really a version of
may want to revisit
§2.2
'universal property of polynomial rings Z[#i,...,
their role as free objects in the category of commutative rings, a.k.a. commutative
Z-algebras.
It is not difficult to identify free objects in the larger category iJ-Alg: they
consist of 'noncommutative polynomial rings' R(A), with variables from the set A
but without any condition relating ab and ba for a ^ b in A. More precisely, R(A)
isomorphic to the monoid ring (cf. §1.4) over the free monoid on A, consisting of
all finite strings of elements in A, with operation defined by concatenation30. We
will encounter this ring again, but only in the distant future (Example VIII.4.17).
is
Submodule generated by a subset; Noetherian modules. Let M be
iJ-module, and let A C M be a subset of M. By the universal property of free
modules, there is a unique homomorphism of R-modules
6.4.
an
<pA
:
R®A
M.
-?
The image of this homomorphism is a submodule of M, the submodule generated by
A in M, usually denoted (A) (or (ai,..., an) if A
{ai,..., an} is finite). Thus,
=
(A)
=
{^ ra a | ra ^ 0 for only finitely many elements a G A}.
a€A
It is
hopefully
clear that
(A)
is the smallest submodule of M containing A.
finitely generated if M
(A) for a finite
of
R-modules
surjective homomorphism
The module M is
and only if there
is
a
=
#en
for
some n.
^
set
A, that is, if
M
One of the highlights of Chapter VI will be the classification of finitely
over PIDs (Theorem VI.5.6). We have already briefly mentioned
generated modules
This construction is similar to the free group
presence of inverses and of possible cancellations.
on
A, but without the complication of the
III.
170
the case for Z (recall
back in §11.6.3.
that Z is
PID and Z-modules
a
are
Rings and modules
nothing but abelian groups!)
Finitely generated R-modules are tremendously important, but they are not as
as one might hope at first. For example, it may be that a module M
finitely generated, but some submodule of M is not finitely generated!
well-behaved
is
Example
6.5. Let R
terminates.
=
Then R is
Z[xi,X2,...],
a
finitely generated
polynomial ring
as an
on
infinitely
many inde-
i?-module: indeed, 1 generates it.
However, the ideal
(xi,x2,...)
of R generated by all indeterminates is not finitely generated
as
an
i?-module
(Exercise 6.14).
j
Definition 6.6. An i?-module M is Noetherian if every submodule of M is
generated as an .R-module.
Thus,
a
finitely
j
ring R is Noetherian
Noetherian 'as
a
module
in the sense of Definition 4.2 if and only if it is
itself. The ring in Example 6.5 is not Noetherian.
over
We will study the Noetherian condition more carefully later
already see one reason why this is a good, 'solid' notion.
(§V.1.1);
on
but
we can
Proposition
6.7. Let M be
M is Noetherian
if
and only
an
if
R-module, and let N be
both N and
M/N
are
a
submodule
of M.
Then
Noetherian.
Proof. If M is Noetherian, then so is M/N (same proof as for Exercise 4.2), and so
is N (because every submodule of A7" is a submodule of M, so it is finitely generated
because M is
This proves the 'only if part of the statement.
Noetherian).
For the converse, assume N and M/N are Noetherian, and let P be a submodule
of M; we have to prove that P is finitely generated. Since P D N is a submodule
of N and N is Noetherian, Pfi N is finitely generated. By the 'second isomorphism
theorem', Proposition 5.18,
and hence
P/(P H N)
is
Noetherian, this shows that
P
P + N
PnN
N
isomorphic
P/(P
It follows that P itself is
H
N)
'
to a submodule of
is
M/N.
Since
M/N
is
finitely generated.
finitely generated, by
Exercise 6.18.
Corollary 6.8. Let R be a Noetherian ring, and let M be
Then M is Noetherian (as an R-module).
a
finitely generated
R-module.
Proof. Indeed, by hypothesis there is an onto homomorphism i?0n -» M of Rmodules; hence (by the first isomorphism theorem, Corollary 5.16) M is isomorphic
to
a
quotient of i?0n. By Proposition 6.7, it suffices
to prove that
i?0n is Noetherian.
1 by hypothesis.
This may be done by induction. The statement is true for n
n > 1, assume we know that i?®(n_1) is Noetherian; since i?®(n_1) may be
=
For
viewed
as a
submodule of
i?0n,
in such
a way
that
6. Products, coproducts, etc., in i?-Mod
171
and R is Noetherian, it follows that i?0n is Noetherian, again by
applying Proposition 6.7.
(Exercise 6.4),
6.5. Finitely generated
keep these
important to
language used
The
"S is
finite type. If S is an i?-algebra, it may be 'finitely
as an i?-module and as an i?-algebra. It is
distinct, although unfortunately the
two concepts well
to express them is very similar.
following definitions differ
finitely generated
R if there is
over
vs.
in two very different ways:
generated'
as
an onto
a
in three small details...
US is
module
homomor-
over
The mathematical difference is
seen
in
§6.3,
over a
i?0n;
Thus, a commutative31 ring S is finitely
onto
over
homomorphism of i?-modules
i?0n
for
=
A is isomorphic to R[xi,... ,xn].
generated as an R-module if there is an
the free commutative R-algebra
to
algebra
homomor-
substantial than it may appear. As we
finite set A
{1,... ,n} is isomorphic
more
the free R-module
as an
an onto
phism of i?-algebras from the free Ralgebra on a finite set to S."
phism of iJ-modules from the free Rmodule on a finite set to S."
have
finitely generated
R if there is
some n;
it is
finitely generated
as an
S
-»
R-algebra if there
is
an onto
homomorphism
of R-algebras
R[xi,...,xn]
for
5
-»
S
In other words, S is finitely generated as an R module if and only if
R®n/M for some n and a submodule M of i?0n; it is a finite-type iJ-algebra
some n.
=
if and only if S
=
R[x\,...,xn]/I
We say that S is
is clear that 'finite'
finite
=>
for
some n
and
an
in the first case32 and
'finite
type';
ideal / of
R[x\,...,xn].
of finite type
it should be just
as
in the second.
clear that the
It
converse
does not hold.
Example 6.9. The polynomial ring
finite as an i?-module.
important
a
is
a
finite-type i?-algebra,
but it is not
j
The distinction, while macroscopic in general, may evaporate in special,
cases. For example, one can prove that if k and K are fields and k C K, then
K is of finite type
6
R[x]
over
k if and only
k-vector
if
it is in fact finite
This is
finite-dimensional
space).
deep result we already mentioned in Example
important class of examples) in §VII.2.2.
David Hilbert's name is associated
finite-type i?-algebras:
one
as a
/c-module
(that is,
it is
version of EilberVs Nullstellensatz,
4.15 and that we will prove (in an
to another important result concerning
if R is Noetherian
(as
a
ring, that is,
as
an
i?-module)
We are mostly interested in the commutative case, so we will make this hypothesis here;
the only change in the general case is typographical: (¦ ¦) rather than [¦¦¦].
Also, note that a
ring is finitely generated as an algebra if and only if it is finitely generated
algebra; cf. Exercise 6.15.
This is particularly unfortunate, since S may very well be an infinite set.
commutative
commutative
as
a
III.
172
and S
is
a
then S is also Noetherian
finite-type i?-algebra,
This is
S-module).
an
an
Rings and modules
(as
a
ring, that is,
as
immediate consequence of the so-called Hubert's basis
theorem.
The proof of Hilbert's basis theorem is completely elementary: it could be given
as an exercise, with a few key hints; we will see it in §V.1.1.
here
Exercises
6.1. > Prove Claim 6.3.
[§6.3]
or disprove that if R
isomorphic to M 0 M.
6.2. Prove
not
6.3.
ring, M
Let R be a
p2
morphism such that
M
kerpeimp.
=
is ring and M is
i?-module, and
an
(Such
p.
a
p
a nonzero
:
M
M
is called
map
i?-module, then M is
an
i?-module homo-
projection.)
a
Prove that
=
ring, and let n > 1. View i?e(n_1) as
homomorphism i?®(n_1) <^-> R®n defined by
6.4. > Let R be
the injective
a
(ri,...,rn_i)
Give
a
i->
a
i?0n,
submodule of
via
(ri,...,rn_i,0).
one-line proof that
-
£t'
fle(n-i)
[§6.4]
6.5. >
(Notation
(#0^1)0^2
6.6.
-i
as
in
§6.3.)
For any ring R and any two sets Ai, A<i, prove that
R®(A1xA2)t [§VIII.2.2]
Let R be
Prove that
and
^
a
ring, and let F
Hom^.Mod(^-R)
=
i?0n be
i?-module M such that
a nonzero
a
finitely generated free i?-module.
an example of a ring R
F. On the other hand, find
HomjR_Mod(^-R)
=
0.
[6.8]
6.7. > Let A be any set.
For any
family {Ma}aGA of
and coproduct
0aGA
modules
Ma. If Ma
=
over a
ring i?, define the product YlaeA Ma
a e A, these are denoted RA, i?eA,
R for all
respectively.
^ ZeN. (Hint: Cardinality.)
Prove that ZN
[§6.1, 6.8]
6.8.
Let R be
a
ring.
If A is
any set, prove that
HomjR_Mod(-ReA5-R)
satisfies
the universal property for the product of the family {Ra}aeA, where Ra
R for
all a; thus, HomjR_Mod(-ReA,-R)
RA- Conclude that HomjR_Mod(-ReA,-R) is not
=
isomorphic
to
R®A
in
general (cf.
Exercises 6.6 and
6.7.)
N be a
6.9.
Let R be a ring, F a nonzero free i?-module, and let cp : M
homomorphism of i?-modules. Prove that cp is onto if and only if for all i?-module
-i
homomorphisms
a :
F
N there exists
an
i?-module homomorphism (3
:
F
M
Exercises
173
such that
a
=
P
o
(Free
(p.
modules
are
projective,
as we
will
see
in
Chapter VIII.)
[7.8, VI.5.5]
6.10. >
i/:
iV
^
Let M, N, and Z be P-modules, and let //
Z be homomorphisms of P-modules.
(Cf.
Exercise
1.5.12.)
:
M
Z,
-?
Prove that P-Mod has 'fibered products': there exists an P-module M xz N
M Xz N —>
M, ttn : M Xz N ^> N, such that
with P-module homomorhisms ttm
/jLottm
requirement. That is, for
N such
M, (p^ : P
homomorphisms (pM : P
a unique R-module homomorphism P
MxzN
vottn, and which is universal with respect to this
every P-module P and R-module
that fio(pM
=
^>
voipN, there exists
making the diagram
commute.
The module M Xz N may be called the
A7" along //, since the construction is
are
pull-back of M along
symmetric).
MxzN-
->N
M
->Z
commutative, but 'even better' than commutative; they
a square, as
fibered coproduct of
notion of
a
of
(or
are
often decorated by
[§6.1, 6.11, §IX.1.4]
shown here.
6.11. > Define
v
'Fiber diagrams'
P-module A, in the style of Exercise 6.10
(and
two R-modules
cf. Exercise
A
"-^N
M
-*N®AM
M, N, along
an
1.5.12)
Prove that fibered coproducts exist in P-Mod. The fibered coproduct M 0^ N is
called the push-out of M along v (or of N along //). [§6.1]
6.12. Prove
Proposition 6.2.
6.13. Prove that every
homomorphic image of a finitely generated module is finitely
generated.
(#i, #2, •)of the ring R
ideal, i.e.,
P-module). [§6.4]
6.14. > Prove that the ideal
generated (as
an
6.15. > Let R be
=
Z[#i, #2,... ]
is not
finitely
as an
a
finitely generated as
commutative algebra
commutative ring.
an
algebra
over
P.
over
Prove that
a
commutative
P-algebra S
R if and only if it is finitely generated
(Cf. §6.5.) [§6.5]
as
is
a
III.
174
6.16. > Let R be
a
Rings and modules
A (left-)R-module M is cyclic if M
ring.
=
(m)
for
some
Prove that simple modules (cf. Exercise 5.4) are cyclic. Prove that an
i?-module M is cyclic if and only if M
R/I for some (left-)ideal /. Prove that
of
a
module
is
every quotient
cyclic
cyclic. [6.17, §VI.4.1]
m
M.
G
=
6.17.
(Exercise
-i
Let M be
6.16),
a
cyclic i?-module,
so
that M
=
R/I
for
a
(left-)ideal
/
and let N be another R-module.
Prove that
Hom^.Mod(M, N)
For a, 6 G Z, prove that
^
{n
G N
| (Vo
J), an
G
Hom#_Mod(Z/aZ, Z/6Z)
=
=
0}.
Z/gcd(a,6)Z.
[7.7]
6.18. > Let M be
and
M/N
are
R-module, and let A7" be a submodule of M. Prove that if N
finitely generated, then M is finitely generated. [§6.4]
an
both
7. Complexes and homology
In many contexts, modules arise not 'one at a time' but in whole series: for example,
a real manifold of dimension d has one 'homology' group for each dimension from 0
language capable of dealing with whole sequences
of modules at once. This is the language of home-logical algebra, of which we will
get a tiny taste in this section, and a slightly heartier course in Chapter IX.
to d. It is necessary to
develop
a
7.1. Complexes and exact sequences. A chain complex of R-modules (or, for
simplicity, a complex) is a sequence of R-modules and R-module homomorphisms
di+2
such that
(Vi) : di
The notation
simplicity (but do
carried by
a
o
di+i
=
r
Afi+i
di
Mi
>
_
di-i
_
Afi_i
>
>
0.
(M#,d#)
not
di+i
may be used to denote a
forget
complex,
that the homomorphisms di
are
or
simply M# for
part of the information
complex).
A complex may be infinite in both directions; 'tails' of 0's are (usually) omitted.
Several possible alternative conventions may be used: for example, indices may be
increasing rather than decreasing, giving a cochain complex (whose homology is
cohomology; this will be our choice in Chapter IX). Such choices are clearly
mathematically immaterial, at least for the simple considerations which follow.
The homomorphisms di are called boundary, or differentials, due to important
called
examples from geometry. Note that the
di
o
defining
di+i
=
condition
0
is equivalent to the requirement
imdi+i
We carry in
our
minds
an
image such
as
C
kerdi.
7. Complexes and
175
homology
we think of a complex. The ovals are the modules Mi\ the fat black dots are
the 0 elements; the gray ovals, getting squashed to zero at each step, are the kernels;
and we thus visualize the fact that the image of 'the preceding homomorphism' falls
when
inside the kernel of 'the next homomorphism'.
of
The picture is inaccurate in that it hints that the 'difference' between the image
di+i and the kernel of di (that is, the areas colored in a lighter shade of gray)
should be the
same for all r, this is of course not the case in general. In fact, almost
the whole point about complexes is to 'measure' this difference, which is called the
homology of the complex (cf. §7.3). We say that a complex is exact 'at Mi if it has
no
homology there; that is,
im^+i
=
kerdf.
Visually,
This complex
For
appears to be exact at the oval in the middle.
example, if Mi
=
a
trivial module
(usually
complex is necessarily exact at Mi, since then
A complex is exact and is often called
modules.
denoted simply by
ker di
0.
imd^+i
=
an exact sequence
0),
then the
=
if it is exact at all its
Example 7.1. A complex
>L—^M
>0
is exact at L if and
only if
Indeed,
L, that is,
homomorphism 0
a
exactness at L is
is
a
monomorphism.
equivalent
to ker a
=
image of the trivial
to
ker a
This is equivalent to the injectivity of
a
=
0.
(Proposition 6.2).
j
Rings and modules
III.
176
Example
7.2. A
complex
>M^-^N
is exact at N if and
only if (3
Indeed, the complex
homomorphism N
is
an
epimorphism.
is exact at N if and
0, that is, im/3
=
only if im/3
kernel of the trivial
j
an exact
complex of the form
>L—^M^-^N
0
=
N.
Definition 7.3. A short exact sequence is
As
>0
>0.
j
in the previous two
examples, exactness at L and N is equivalent to a
and
(3 being surjective. The extra piece of data carried by a short
being injective
seen
exact sequence is the exactness at M, that is,
ima
=
ker/3;
by the first isomorphism theorem (Corollary 5.16),
N ^
M
we
then have
M
=
.
im a
ker (3
All in all, we have good material to work on some more Pavlovian conditioning: at
the sight of a short exact sequence as above, the reader should instinctively identify
L with a submodule of M (via the injective map a) and N with the quotient M/L
(via the isomorphism induced by the surjective map (3, under the auspices of the
first isomorphism theorem).
(p
:
Short exact sequences abound in nature. For example, a single
M' gives rise immediately to a short exact sequence
homomorphism
M
0
> ker (p
> im (p
> M
> 0
.
In fact, one important reason to focus on short exact sequences is that this
observation allows us to break up every exact complex into a large number of short exact
sequences:
contemplate the impressive diagram
0
.0
imdi+i
im
di+2
=
ker
d^+i
=
ker di
im di
=
ker di-\
7. Complexes and
The
177
homology
sequences are short exact sequences, and
diagonal
complex.
This observation simplifies many
they interlock nicely by the
exactness of the horizontal
7.2.
exact sequences. A
Split
arguments; cf. for
particular
considering the second projection from
a
of short exact sequence arises by
case
direct
example Exercise 7.5.
sum:
M2\ there is then
Mi ©M2
an exact sequence
> Mi
0
identifying Mi
obtained by
sequences
are
said to 'split';
> N
Mi
'splits' if it is isomorphic to
> 0
,
0
M2
of these sequences in the
one
sense
that there is
a
diagram
0
N
Mi
>
0
M[
>
in which the vertical maps are all
Example
> M2
with the kernel of the projection. These short exact
more generally, a short exact sequence
0
commutative
> Mi 0 M2
M[
0
M'2
>
M2
0
M'2
> 0
isomorphisms33.
7.4. The exact sequence of Z-modules
>z—^z >£
0
>o
is not split.
j
sequences give us the opportunity to go back to a question we left
at
the
end of §6.2: what should we make of the condition of 'having a
dangling
left- (resp., right-) inverse' for a homomorphism? We realized that this condition is
Splitting
stronger than the requirement of
can we
give
a more
Proposition
(p has
a
being a monomorphisms (resp.,
explicit description of such morphisms?
7.5. Let cp
:
M
left-inverse if
N be
and only
> M
0
an
if
an
R-module homomorphism. Then
the sequence
) N
> coker (p
> 0
splits.
(p has
a
right-inverse if and only if the
0
> ker (p
epimorphism);
> M
sequence
*
> iV
> 0
splits.
In fact, this last requirement is somewhat redundant; cf. Exercise 7.11.
III.
178
Proof. We
will prove the first part and leave the other
as an
Rings and modules
exercise to the reader
(Exercise 7.6).
If the sequence splits, then (p may be identified with the embedding of M into
M gives a left-inverse of (p.
sum M 0 M', and the projection M 0 M'
assume
that
has
a
left-inverse
ip:
Conversely,
cp
a
direct
>M^^N
0
M
we claim that N is isomorphic to M 0 ker ip and that cp corresponds to the
M 0 ker ip
N. The isomorphism
identification of M with the first factor: M
Then
=
M 0 ker ip
N is given
by
(m, k)
its inverse A7"
<p(ra)
/c;
M 0 ker -0 is
—>•
(i/>(ri),n
(pip(n)).
n i—>•
The element
+
i—?
n
(pip(ri)
ip{n
is in
ker-0
(pip(n))
All necessary verifications
=
are
as
it should be, since
ip{ri)
ip(pip(n)
immediate and
ip{ri)
=
ip{ri)
=
0.
left to the reader.
are
Because of Proposition 7.5, R-module homomorphisms with a left-inverse are
called split monomorphisms, and homomorphisms with a right-inverse are called
split epimorphisms.
We will
of
Grp)
in
generally,
back to split exact sequences (in the more demanding context
§IV.5.2 and then later again when we return to modules and, more
come
to abelian
categories (Chapters VIII and IX).
7.3. Homology and the snake lemma.
Definition 7.6. The i-th homology of
M.:
>
Afi+i
a
complex
>
Mi
>
Mi-i
>
of i?-modules is the i?-module
im di+1
That is, Hi(M9) is a module capturing the 'light gray annulus' in the heuristic
picture of a complex. Of course
Hi(M9)
=
0 <^=>
imdi+i
that is, the homology modules
exact'.
=
kerdi
<^=>
are a measure
the
complex M#
of the 'failure of
a
is exact at Mi
:
complex from being
7. Complexes and
179
homology
Example 7.7. In fact, homology should be thought of as a (vast) generalization
of the notions of kernel and cokernel. Indeed, consider the (very) particular case in
which M# is the
complex
->Mi
0
->o.
->Mr)
Then
Hx(M.)
2* ker
<p,
2* coker
(p.
H0(M.)
by indicating
diagram involving two short exact sequences generates a 'long
exact sequence' in homology. This is actually a particular case of a more general
to which a suitable commutative diagram involving three
construction—according
complexes yields a really long 'long exact homology sequence'. We will come back to
this general construction when we deal more extensively with homological algebra
in Chapter IX. The reader is also likely to learn about it in a course on algebraic
topology, where this fact is put to impressive use in studying invariants of manifolds.
We will end this very brief excursion into more abstract territories
how
a
commutative
In the
lemma.
form
a
simple form
we
will
analyze, this
is
affectionately
known
as
the snake
Consider two short exact sequences linked by homomorphisms,
commutative diagram34:
0
->£i
->Mi
0-
->£o
->M0
->JVi
->0
>N0
->0
so
as to
(So
Lemma 7.8
0-
(The snake lemma).
ker A
ker \x
With notation
ker
-
v
as
above, there
completely straightforward
one 'surprising' homomorphism
an exact sequence
coker \x—>
coker A —>
coker v
Remark 7.9. Most of the homomorphisms in this sequence
way from the
is
are
—>
0
induced in
corresponding homomorphisms A,
is the
one
denoted 5;
we
.
//,
a
v.
The
will discuss its definition
below.
j
Remark 7.10. In view of Example 7.7,
statement as
0
we
could have written the sequence in this
?#!(£.)¦ ->#i(M.)-
^H^N.)^
5
H0(L.)
where L9
generalizes
H0(M.)
H0(N.)
0
> Lq
-¥ 0
etc.
The snake lemma
->£i
complex 0
arbitrary complexes L#, M#, iV#, producing a 'long exact homology
is the
to
_>
,
In fact, it is better to view this diagram as three (very short) complexes linked by Rmodule homomorphisms a$, Pi so that 'the rows are exact'. In fact, one can define a category
of complexes, and this diagram is nothing but a 'short exact sequence of complexes'; this is the
approach
we will
take in Chapter IX.
III.
180
sequence' of which this is just the tail end. As mentioned
this rather straightforward generalization later (§IX.3.3).
Rings and modules
above,
we
will discuss
j
Remark 7.11. A popular version of the snake lemma does not assume that a\ is
injective and (3q is surjective: that is, we could consider a commutative diagram of
exact sequences
Li
->L0
0-
The lemma will then
ker A
Mi
>
->M0
state that there is
-¥ ker /j,
-> ker
>
->
> 0
-+JVo
'only'
an exact sequence
5
v
Ni
coker A
> coker //
> coker
v
.
Proving the snake lemma is something that should not be done in public, and
it is notoriously useless to write down the details of the verification for others to
read: the details are all essentially obvious, but they lead quickly to a notational
quagmire. Such proofs are collectively known as the sport of diagram chase, best
executed by pointing several
fingers
while enunciating the elements one
diagram on a blackboard,
and
manipulating
stating their fate35.
at different parts of a
is
Nevertheless, we should explain where the 'connecting' homomorphism 5 comes
from, since this is the heart of the statement of the snake lemma and of its proof.
Here is the whole diagram, including kernels and cokernels; thus, columns are exact
(as
well
as
the two original sequences, placed
0
> ker A
0
>
By the
horizontally):
0
ker \x
coker A
> coker \x
0
0
way, we trust that the reader now sees
0
> ker
v
> coker
v
>
0
0
why this lemma
is called the snake
lemma.
Real purists chase diagrams in arbitrary categories, thus without the benefit of talking
about 'elements', and we will practice this skill later on (Chapter IX). For example, the snake
lemma can be proven by appealing to universal property after universal property of kernels and
cokernels, without ever choosing elements anywhere. But the performing technique of pointing
fingers at a board while monologuing through the argument remains essentially the same.
7. Complexes and
181
homology
the snaking homomorphism 5. Let a G keri/. We claim that a
be mapped through the diagram all the way to coker A, along the solid arrows
marked here:
Definition of
can
0
0
0
0
^b
I0
(So
1
0
Indeed,
ker i/CiVij
so
view
Pi is surjective,
Let d
=
fi(c)
What is the
diagram,
a G
ker i/,
so
3c G Mi, mapping to b.
be the image of
image of d
u(b)
=
c
same
G
Finally, let /
G coker \x be the
Is this
5(a)
legal? At
Lo, mapping
:=
as
By the commutativity of the
v(b). However,
0. Thus, d G ker/?o- Since
therefore, 3e
We want to set
in Mo.
in the spot marked *?
it must be the
so
element b of Ni.
a as an
b
was
rows are
the image in N\ of
exact, ker/?o
=
imao;
to d.
image of
e.
f.
two steps in the chase we have taken
3c
G
Mi such that fii(c)
3e
G
I/o such that ao(e)
=
=
preimages:
6,
d.
The second step does not involve a choice: because a$ is injective by assumption,
so the element e mapping to d is uniquely determined by d. But there was a choice
involved in the first step: in order to verify that 5 is well-defined, we have to show
some other c would not affect the proposed value / for 5(a).
that choosing
This is proved by another chase. Here is the relevant part of the diagram:
0
c
->d-
>b
0
III.
182
Suppose
we
choose
a
different c' mapping to the
0
Then
/?i(c'
c)
=
0; by exactness, 3g
0
Now the
>
c
>
9
same
0.
b
(c'
-
c)
Oil
,
>
,
(cf
A(^)
0
01
c)
-
:
c)
> 0
a\(g):
=
0.
point is that, since columns form complexes,
0
b:
L\such that (c'
e
Rings and modules
g dies in coker A:
_
0
> 0
fl
OiQ
0
and it follows
changing
c to
the commutativity of the diagram and the injectivity of ao) that
cf modifies e to e + X(g) and / to / + 0
/. That is, / is indeed
(by
=
independent of the choice.
Thus 5 is well-defined!
tiny part of the proof of the snake lemma, but it probably
why reading a written-out version of a diagram chase may
uninformative.
supremely
The rest of the proof (left to the reader (!) but we are not listing this as an
This is
suffices
be
a
to demonstrate
official exercise for fear that
someone
might actually
amounts to many, many similar arguments.
turn a solution in for
grading)
The definition of the maps induced
substantially less challenging than the definition of the
connecting morphism 5 described above. Exactness at most spots in the sequence
on kernels and cokernels is
0
is also
ker \x—>
coker \x—>
ker v —>
coker A —>
coker v
ker A
reasonably straightforward;
most of the work will go into
0
proving
exactness
at ker v and coker A.
shy away from trying this, for it is excellent,
indispensable practice. Miss this opportunity and you will forever feel unsure about
manipulations.
Dear reader:
don't
such
The snake lemma streamlines several facts, which would not be hard to prove
individually, but become really straightforward once the lemma is settled. For
example,
Corollary 7.12. In the same situation presented in the snake lemma (notation as
in $7.3), assume that \x is surjective and v is infective. Then A is surjective and v
is
an
isomorphism.
Proof. Indeed,
(Proposition
\x surjective
6.2). Feeding
=>
coker//
=
0;
v
injective
=>
kerv
=
0
this information into the sequence of the snake lemma gives
an
Exercises
183
exact sequence
> ker A
0
> ker \x
Exactness implies coker A
with the stated
=
> 0
coker
v
> coker A
=
0
(Exercise 7.1);
> coker
hence A and
>
v
v are
0
surjective,
consequences.
Several more such statements may be
experiment
> 0
proved just
as
easily; the reader should
to his or her heart's content.
Exercises
7.1.
>
Assume that the complex
>0
is exact. Prove that M ^ 0.
7.2. Assume that the
>M
[§7.3]
complex
0
is exact.
>0
Prove that M
=
M
M'
> 0
>
M'.
7.3. Assume that the complex
> 0
is exact.
> L
> M
—^
> N
M'
Show that, up to natural identifications, L
=
> 0
kenp
and N
>
=
coker (p.
7.4. Construct short exact sequences of Z-modules
o
> zeN
zeN
> z
> o
and
> zeN
o
(Hint:
David Hilbert's Grand
7.5. > Assume that the
> zeN
> o.
Hotel.)
complex
>L
> M
is exact and that L and N are Noetherian.
7.6. > Prove the
> zeN
>•••
A^
Prove that M is Noetherian.
[§7.1]
'split epimorphism' part of Proposition 7.5. [§7.2]
7.7. > Let
>M
0
be
a
(i)
>iV
>P
short exact sequence of i?-modules, and let L be
Prove that there is
0
an exact
Homje_Mod(.P, L)
>0
an
i?-module.
sequence36
>
RomR_Mo6(N, L)
>
Hom^.Mod(M, L).
In general, this will be a sequence of abelian groups; if R is commutative, so that each
Hom#_Mod is an -R-module (§5.2), then it will be an exact sequence of .R-modules.
III.
184
(ii) Redo
Exercise 6.17.
Construct
(iii)
(Use
/
the exact sequence 0
R
Rings and modules
R/I
0.)
—>
example showing that the rightmost homomorphism
an
in
need
(i)
not be onto.
Show that if the original sequence splits, then the rightmost homomorphism
(iv)
in
(i)
is onto.
[7.9, VIII.3.14, §VIII.5.1]
7.8. > Prove that every exact sequence
>M
0
of R modules, with F
>N
>F
Exercise
free, splits. (Hint:
>0
6.9.) [§VIII.5.4]
7.9. Let
0
>M
>N
>F
>0
a short exact sequence of i?-modules, with F free, and let L be
Prove that there is an exact sequence
be
0 —>
YLomR_Uod(F,
(Cf.
Exercise
L)
Homi?.Mod(Ar, L)
—>
as
—>
0
.
that A and v are isomorphisms.
isomorphism. This is called the 'short
immediately from the five-lemma (cf. Exercise 7.14), as
7.10. > In the situation of the snake lemma,
well
RomR_Mo6(M, L)
—>
iJ-module.
7.7.)
Use the snake lemma and
five-lemma,'
an
as
it follows
assume
prove that /j, is an
from the snake lemma.
[VIII.6.21, IX.2.4]
7.11. > Let
(*)
be
> Mi
0
an
exact
sequence
of i?-modules.
by M2.) Suppose there
the diagram
0
> N
(This
is any R-module
II
> 0
may be called an 'extension' of M2
homomorphism N
> M2
N
'
I
Mi
II
0
> M2
-i
making
> 0
II
> M2
Mi 0 M2
0
commute, where the bottom sequence is the standard sequence of
Prove that (*) splits. [§7.2]
7.12.
0 M2
II
^
> Mi
Mi
a
direct
sum.
Practice your diagram chasing skills by proving the 'four-lemma': if
Ai
Bi
>
B0
Di
7
(3
a
^o
Ci
>
Co
>
D0
is a commutative diagram of R-modules with exact rows, a is an epimorphism, and
/?, 5 are monomorphisms, then 7 is an isomorphism. [7.13, IX.2.3]
Exercises
185
7.13. Prove another37 version of the 'four-lemma' of Exercise 7.12: if
is
a
and
Z?i
> Ci
> £>i
> Ei
B0
-*Co
-*D0
-*E0
diagram of R-modules with exact rows, ft and 5
monomorphism, then 7 is an isomorphism.
commutative
e
is
7.14.
-1
a
are
epimorphisms,
Prove the 'five-lemma': if
Ai
Ci
> Di
> Ex
> Co
> D0
> E0
Bi
r
> B0
^o
diagram of R-modules with exact rows, ft and 5 are isomorphisms,
epimorphism, and e is a monomorphism, then 7 is an isomorphism. (You can
is
a
commutative
a
is
an
avoid the needed diagram chase by pasting together results from previous
exercises.)
[7.10]
7.15.
-1
Consider the following commutative diagram of R-modules:
0
0
->Lo
->Mo
^No
->0
Li
Mi
Ni
->0
-*L0
->M0
^N0
->0
->
0
Assume that the three
rows
0
and the two rightmost columns are exact.
Second version: assume that the three rows
and the two leftmost columns
exact. This is the 'nine-lemma'. (You
snake lemma; for this,
7.16. In the
exact
same
0
are exact
Prove that the left column is exact.
are exact
0
are
can
exact; prove that the right column is
avoid
you will have to turn the
situation
as
a diagram chase by applying the
diagram by 90°.) [7.16]
in Exercise 7.15,
assume
and that the leftmost and rightmost columns
monomorphism and ft
Is the central column necessarily exact?
Prove that
a
is
a
is
an
that the three
epimorphism.
It is in fact unnecessary to prove both versions, but to realize this
matter from the
more
general
context of abelian
rows are
are exact.
one
categories; cf. Exercise IX.2.3.
has
to
view
the
III.
Rings and modules
artfully
with six copies of Z
186
(Hint: No. Place
and two
Z 0 Z in the middle, and surround it
O's.)
Assume further that the central column is
that it is then
7.17.
-i
infinite)
necessarily
complex (that is, (3
a
0);
prove
exact.
Generalize the previous two exercises
commutative
o a =
as
follows.
Consider
a
(possibly
diagram of i?-modules:
0
>
L*+i
>
0
> Li
0
U-\
in which the central column is
right columns
Afi+i
Mi
a
> Mi-i
complex and
>
Ni+1
> Ni
> Ni-i
0
> 0
0
every row is exact.
Prove that the
also complexes. Prove that if any two of the columns
are exact, so is the third.
(The first part is straightforward. The second part
will take you a couple of minutes now due to the needed diagram chases, and a
couple of seconds later, once you learn about the long exact (co)homology sequence
left and
in
§IX.3.3.) [IX.3.12]
are
Chapter
Groups,
IV
second encounter
In this chapter we return to Grp and study several topics of a less 'general' nature
than those considered in Chapter II. Most of what we do here will apply exclusively
finite groups; this is an important example in its own right, as it has spectacular
applications (for example, in Galois theory; cf. §VII.7), and it is a good subject
from the expository point of view, since it gives us the opportunity to see several
to
general concepts at work in a context that is complex enough to carry substance,
but simple enough (in this tiny selection of elementary topics) to be appreciated
easily.
1. The conjugation action
1.1. Actions of groups on sets, reminder. Groups really shine when you let
something. This section will make this point very effectively, since
them act on
surprisingly precise results
finite groups by extremely simple-minded
applications of the elementary facts concerning group actions that we established
back in §11.9.
we
will get
Recall
we
proved (Proposition II.9.9) that
every transitive
(left-)
action of
S is, up to a natural notion of isomorphism, 'left-multiplication
the set of left-cosets G/H\ Here, H may be taken to be the stabilizer Stabc(a)
a group
on
that
on
G
on a set
of any element a e S, that is (Definition II.9.8) the subgroup of G fixing a. This
fact applies to the orbits of every left-action of G on a set; in particular, the number
of elements in a finite orbit O equals the index of the stabilizer of any a G O; in
of
particular (Corollary II.9.10) the number of elements \0\
the order
we
|G|
an
orbit must divide
of G, if G is finite.
These considerations may be packaged into a useful 'counting' formula, which
formula for that action; this name is usually reserved to the
could call the class
particular case of the action of G onto itself by conjugation, which
more carefully below.
we
will explore
187
IV. Groups, second encounter
188
In order to state the formula,
G acts
assume
denote the stabilizer Stabc(a). Also, let Z be the
Z
Note that
a
a G
Z <^=> Ga
is 'trivial', in the
sense
=
Proof. The orbits form
{aeS\(\/geG):ga
G; we could say that a
that it consists of a alone.
=
Proposition 1.1. Let S be
notation as above,
where ACS has exactly
on
one
a
a
finite set,
element
for
=
5; for a G 5, let Ga
fixed points of the action:
a set
set of
a}.
G
and let G be
Z if and
a group
each nontrivial orbit
only if the orbit of
acting
of
With
S.
on
the action.
partition of 5, and Z collects the trivial orbits; hence
+
\s\ \z\
=
^2\oa\,
a£A
Oa denotes the orbit of a. By Proposition II.9.9, the order \Oa\equals the
index of the stabilizer of a, yielding the statement.
where
The main
summand [G
strength of Proposition
:
Ga]
constraint, when
some
this says when G is
(and
information is known about
is >
|G|.
1).
This
can
be
a
For example, let's
strong
see
what
a p-group:
Definition 1.2. A p-group is
integer
that, if G is finite, each
1.1 rests in the fact
divides the order of G
a
finite group whose order is
a power
of
a
p.
j
Corollary 1.3. Let G be a p-group acting
point set of the action. Then
\Z\ \S\
=
Proof. Indeed, each summand
than 1; hence it is 0 mod p.
prime
[G
:
Ga]
on a
finite
set
S, and let Z be the fixed
mod p.
in
Proposition
1.1 is a power of p
larger
For instance, in certain situations this can be used to establish1 that Z ^ 0: see
Exercise 1.1. Such immediate consequences of Proposition 1.1 will assist us below,
in the proof of Sylow's theorems.
In this sense, Proposition 1.1 is an instance of a class of results known as 'fixed point
The reader will likely encounter a few such theorems in topology courses, where the
theorems'.
role of the 'size' of a set may be played by
space.
(for example)
the Euler characteristic of
a
topological
1. The
conjugation action
189
Center, centralizer, conjugacy classes. Recall (Example II.9.3) that
every group G acts on itself in at least two interesting ways: by (left-) multiplication
and by conjugation. The latter action is defined by the following p : G x G
G:
1.2.
p(g,a) =gag~l.
As
we
know
this datum is equivalent to the datum of
(§11.9.2),
certain group
a
homomorphism:
a :
from G
to the
permutation
This action
Definition
SG
-?
group on G.
highlights several interesting objects:
1.4. The center of
Concretely,
G
G, denoted Z{G\ is
the
subgroup kera of G.
j
the center2 of G is
Z(G)
=
{geG\(\/aeG):ga
=
ag}.
Indeed, a(g)
identity
only if a(g) acts as the identity on G; that
if
if
and
a
for
all
a
G
is,
G; that is, if and only if g commutes with all
only gag~l
elements of G. In other words, the center is the set of fixed points in G under the
conjugation action.
in Sq if and
is the
=
Note that the center of a group G is automatically normal in G:
immediate to check 'by hand', but there is no need to do so since it
definition and kernels
this is
is
a
nearly
kernel by
normal.
are
A group G is commutative if and only if
conjugation action is trivial on G.
Z(G)
=
G, that is, if and only if the
In general, feel happy when you discover that the center of
a group
is not trivial:
this will often allow you to set up proofs by induction on the number of elements
of the group, by mod-ing out by the center (this is, roughly, how we will prove the
Sylow theorem). Or note the following useful fact,
trying to prove that a group is commutative:
first
Lemma 1.5.
commutative
Proof.
Let G be a
hence
(and
finite
group, and assume
is in
G/Z(G)
1.5.) As G/Z(G) is cyclic, there exists
gZ(G) generates G/Z(G). Then Va G G
aZ(G)
a
=
some
r
grz.
If now
G
a,
Z; that is, there is
b
are
in
G,
use
some s G
Z and
w G
ab
2Why
'Z'? 'Center'
=
an
is
=
an
=
Z{G)\ but
{grz){gsw)
handy when
cyclic. Then G
is
(gZ(G)Y
element
grz,
z
G
Z(G)
of the center such that
b
=
gsw
then
=
in
element g G G such
this fact to write
a
for
G/Z(G)
comes
fact trivial).
Exercise
(Cf.
that the class
for
which
gr+szw
ift gentrum auf ©eutfo.
=
{gsw){grz)
=
ba,
IV. Groups, second encounter
190
where
and b
we
have used the fact that
arbitrary, this
were
z
and
with every element of G. As
w commute
a
proves that G is commutative.
Next, the stabilizer of
a G
G under conjugation has
Definition 1.6. The centralizer
a
(or normalizer) ZG(a)
special
of
a
G
name:
G is its stabilizer
under conjugation.
j
Thus,
{g eG\gag'1
{g
G
consists of those elements in G which commute with
a.
ZG(a)
for all
G;
a G
If there is
be
in
=
fact, Z(G)
no
=
a}
=
=
f]aeG ZG(a). Clearly
ambiguity concerning the
G\ga
ag}
In particular,
a G
group G
=
Z(G)
<^>
Z(G) C ZG(a)
ZG(a) G.
=
a, the index G may
containing
dropped.
Definition
1.7.
The conjugacy class of
conjugation action.
same
G
G is the orbit
[a]
of
a
under the
conjugate if they belong
are
to the
conjugacy class.
The notation
j
fond of it. Using
Note that [a]
=
frequently, but we are not
are nothing but the equivalence classes
interesting equivalence relation.
[a] is not standard; C(a)
[a] reminds us that these
of elements of G under
if ga
a
Two elements a, b of G
=
a
{a}
ag for all g G
certain
if and only if
is used
gag-1
=
G; that is, if and only if
a
a G
more
for all g G G; that is, if and only
Z(G).
1.3. The Class Formula. The 'official' Class Formula for
particular
case
a
finite group G is the
of Proposition 1.1 for the conjugation action.
Proposition 1.8
(Class formula).
Let G be
\G\ \Z(G)\+
=
where A C G is
a set
containing
one
a
finite
group.
Then
^[G:Z(a)},
representative for each nontrivial conjugacy
class in G.
Proof. The
set of fixed
apply Proposition
points is Z(G), and the stabilizer of a
is the centralizer
The class formula is
surprisingly useful.
In
applying it, keep
in mind that every
the right (that is, both |Z(G)| and each [G : Z(a)]) is
this fact alone often suffices to draw striking conclusions about G.
summand
on
Possibly the most famous such application is to p-groups, via
Corollary 1.9. Let G be
a
nontrivial p-group.
Then G has
a
a
divisor of
Corollary
|G|;
1.3:
nontrivial center.
|Z(G)|
\G\modp and \G\> 1 is a power of p, necessarily |Z(G)|
multiple of p. As Z(G) ^ 0 (since eG G Z(G)), this implies \Z(G)\>p.
Proof. Since
a
Z(a);
1.1.
=
is
D
1. The
For
Exercise
example,
1.6)
In
action
conjugation
191
it follows
immediately (from Corollary
that if p is prime, then every group of order
general, the class formula
1.9 and Lemma 1.5; cf.
p2 is commutative.
poses a strong constraint on what can go on in a
group.
Example
1.10. Consider
a group
G of order 6; what
are
the possibilities for its
class formula?
If G is commutative, then the class formula will tell
6
=
little:
us very
6.
If G is not commutative, then its center must be trivial (as
Lagrange's theorem and Lemma 1.5); so the class formula is 6
a consequence
=
1 +
of
•,where
conjugacy classes. But each of these summands
than
smaller
than
1,
6, and must divide 6; that is, there are no
larger
collects the sizes of the nontrivial
must be
choices:
6=1+2+3
is the only possibility. The reader should check that this is indeed the class formula
for 53; in fact, S3 is the only noncommutative group of order 6 up to isomorphism
(Exercise 1.13).
j
Another useful observation is that normal subgroups must be unions of
a normal subgroup, a G H, and b
gag~l is conjugate
conjugacy classes: because if H is
to a, then
=
b G
To stick with the
contain the
|G|
=
identity and
gHg-1
=
H.
6 example, note that every subgroup of a group must
its size must divide the order of the group; it follows that
normal subgroup of a noncommutative group of order 6 cannot have order 2, since
2 cannot be written as sums of orders of conjugacy classes (including the class of
a
the
1.4.
identity).
Conjugation of
subsets and
subgroups. We
may also act
by conjugation
the power set of G: if A C G is a subset and g e G, the conjugate of A is
the subset gAg~l. By cancellation, the conjugation map a 1—?
gag~l is a bijection
on
between A and gAg~l.
This leads to terminology analogous
to the one introduced in
§1.2.
Definition 1.11. The normalizer No (A) of A is its stabilizer under conjugation.
The centralizer of A is the subgroup Zq(A) C Nq(A) fixing each element of A. j
Thus,
Va G A,
g G
gag~l
For A
ZG(A)
C
NG(A)
=
{a}
NG(A).
=
if and only if3
gAg~l
=
A, and
g G
ZG(A)
a
singleton,
we
have
No({a})
=
Zo({a})
=
Zq(o).
subgroup of G, every conjugate gHg~l of H is also
conjugate subgroups have the same order.
If H is
if and only if
a.
a
If A is finite (but
not in
general),
this condition is equivalent to gAg
a
In
general,
subgroup of G;
1 C A.
IV. Groups, second encounter
192
Remark 1.12. The definition implies immediately that H C Nq(H) and that H
is normal in G if and only if Nq(H)
G. More generally, the normalizer Nq(H)
=
of H in G is
(clearly)
of G
the largest subgroup
in which H is normal.
j
One could apply Proposition 1.1 to the conjugation action on subsets or
however, there are too many subsets, and one has little control over the
subgroup;
number of subgroups. Other numerical considerations involving the number of
conjugates of a given subset or subgroups may be very useful.
subgroup. Then (if finite) the number of subgroups
equals the index [G : Nq(H)] of the normalizer of H in G.
Lemma 1.13. Let H C G be a
conjugate
to H
Proof. This is again
an
Corollary 1.14. If[G
is finite and divides [G
:
:
immediate consequence of Proposition II.9.9.
H] is finite,
H].
then the number
of subgroups conjugate
to H
Proof.
[G:H]
=
[G: NG(H)} [NG(H) : H]
(cf. §11.8.5).
One of the celebrated Sylow theorems will strengthen this statement
substantially in the case in which H is a maximal p-group contained in a finite group G.
For
a statement
a group, see
concerning the size of the normalizer of
an
arbitrary p-subgroup of
Lemma 2.9.
Another useful numerical tool is the observation that if H and K
of
a group
conjugation
by
G and H C
g e H
conjugation is
that
Nq(K)—so
are
subgroups
gKg~l K for all g G H—then
K. Indeed, we have already observed that
=
gives an automorphism of
a bijection, and it is immediate
to
see
that it is
a
homomorphism:
Vkuk2eK
(ghg~l)(gk2g~l)
Thus, conjugation gives
a
=
gki(g~lg)k2g~l
=
g(kik2)g~l.
set-function
7
The reader will check that this is
:
H
AutGrp(i^).
-?
a group
homomorphism and will determine ker 7
(Exercise 1.21).
This is especially useful if H is finite and some information is available
concerning Autcrp(^) (for an example, see Exercise 4.14). A classic application is
presented in Exercise 1.22.
Exercises
193
Exercises
1.1. > Let p be a prime integer, let G be a p-group, and let S be a
\S\^ Omodp. If G acts on 5, prove that the action must have fixed
such that
set
points. [§1.1,
§2.3]
1.2.
Find the center of D<in.
(The
answer
depends
on
the parity of
n.
Use
a
presentation.)
1.3. Prove that the center of Sn is trivial for
a to
b
b and
^
a, and let c
Then compare the action of
r
1.4. > Let G be a group, and let N be a
in G.
3.
(Suppose
that
Sn sends
a G
be the permutation that acts solely by swapping
or and to on a.)
7^
c.
a, 6. Let
n >
subgroup of Z(G). Prove that
N is normal
[§2.2]
1.5. > Let G be a group. Prove that
G/Z(G) is isomorphic to the group Inn(G) of
II.4.8.) Then prove Lemma 1.5 again by
inner automorphisms of G. (Cf. Exercise
using the result of Exercise II.6.7. [§1.2]
integers, and let G be
1.6. > Let p, q be prime
either G is commutative
or
1.7. Prove
or
p2,
for
a
prime
disprove that if p
is
prime, then
that every group of order
a group
of order pq.
Prove that
(using Corollary 1.9)
[§1.3]
the center of G is trivial. Conclude
p, is commutative.
every group of order
p3
is
commutative.
1.8. > Let p be
contains
a
a
prime number, and let G be
normal subgroup of order
1.9.
Let p be a prime number, G
of G. Prove that H n Z(G) + {e}.
-i
1.10. Prove that if G is
a group
pk
a p-group:
\G\ pr. Prove that G
r.
[§2.2]
=
for every nonnegative k <
a p-group,
(Hint:
and H
nontrivial normal subgroup
a
Use the class
formula.) [3.11]
of odd order and g G G is conjugate to
g~l,
then
1.11. Let G be
a finite group, and suppose there exist representatives #i,... ,gr of
distinct conjugacy classes in G, such that Vi,j, giQj
gjgi. Prove that G is
commutative. (Hint: What can you say about the sizes of the conjugacy classes?)
the
r
1.12.
8
=
=
Verify
that the class formula for both Dg and Qg
2 + 2 + 2 + 2.
(Also
note that
(cf.
Exercise
III.1.12)
is
Dg ¥ QsO
1.13. > Let G be a noncommutative group of order 6. As observed in Example 1.10,
G must have trivial center and exactly two conjugacy classes, of order 2 and 3.
Prove that if
every element of a group has order < 2, then the group is
commutative. Conclude that G has an element y of order 3.
Prove that
(y)
Prove that
[y]
is normal in G.
is the conjugacy class of order 2 and
Prove that there is
an x G
G such that yx
=
xy2.
[y]
=
{y,y2}.
IV. Groups, second encounter
194
Prove that
x
has order 2.
Prove that
x
and y generate G.
Prove that G ^ 53.
[§1.3, §2.5]
1.14. Let G be a group, and assume
[G
:
Z(G)]
=
n
is finite. Let ACGbe any
subset. Prove that the number of conjugates of A is at most
1.15.
Suppose that the class formula for a group G is 60
only normal subgroups of G are {e} and G.
n.
1 + 15 + 20 + 12 + 12.
=
Prove that the
1.16.
Let G be
>
finite group, and let H C G be a subgroup of index 2. For
resp., [a]c, the conjugacy class of a in i7, resp., G. Prove
[cl]g or [°]h is half the size of [cl]g, according to whether the
a
H, denote by [a]//,
a G
that either
centralizer
[cl]h
Zq[o)
=
is not
or
is contained in H.
by Exercise II.8.2; apply Proposition
(Hint:
II.8.11.) [§4.4]
Note that H is normal in G,
Let H be a proper subgroup of a finite group G. Prove that G is not the
union of the conjugates of H. (Hint: You know the number of conjugates of H\
keep in mind that any two subgroups overlap, at least at the identity.) [1.18, 1.20]
1.17.
-i
1.18. Let S be
assume
|5|
is, such that
S
=
G/H,
a set
> 2.
a
transitive action of
a
finite group G, and
points in 5, that
(Hint: By Proposition II.9.9,
Use Exercise 1.17.)
you may assume
^
gs
endowed with
Prove that there exists a g G G without fixed
s
for all
s
G
with H proper in G.
S.
a proper subgroup of a finite group G. Prove that there exists
G whose conjugacy class is disjoint from H.
1.19. Let H be
g G
1.20. Let G
a
GL2(C), and let H be the subgroup consisting of upper triangular
(Exercise II.6.2). Prove that G is the union of the conjugates of H. Thus,
finiteness hypothesis in Exercise 1.17 is necessary. (Hint: Equivalently, prove
=
matrices
the
that every 2x2 matrix is
C is
algebraically closed;
1.21. > Let
function 7
and that
:
conjugate to a matrix
Example III.4.14.)
H, K be subgroups of
H
kerj
in H. You will use the fact that
see
G, with H
Ng(K). Verify that the
Autcrp(^) defined by conjugation is a homomorphism of groups
Hf) ZG(K), where ZG(K) is the centralizer of K. [§1.4, 1.22]
a group
C
—>
=
cyclic subgroup of G of order p.
the
order
of G and that H is normal
prime dividing
in G. Prove that H is contained in the center of G.
1.22. > Let G be a finite group, and let H be a
Assume that p is the smallest
(Hint: By
Exercise 1.21 there is
Exercise II.4.14,
2.
The
Sylow
Autcrp(^)
a
has order p
homomorphism
1. What
7
:
G
can you say
Autcrp(^); by
7?) [§1.4]
about
theorems
2.1. Cauchy's theorem. The 'Sylow theorems' consist of three statements
(cf. Definition 1.2) of a given finite group G. The form we will
concerning p-subgroups
2. The
Sylow theorems
195
give for the first of these statements will tell us that G contains p-groups of all sizes
allowed by Lagrange's theorem: if p is a prime and pk divides |G|, then G contains
a subgroup of order pk. The proof of this statement is an easy induction, provided
1 is known: that is, provided that one has established
the statement for k
=
Theorem 2.1
divisor
Let G be
(Cauchy's theorem).
of \G\.Then
G contains
element
an
group, and let p be a prime
finite
a
order p.
of
As it happens, only the abelian version of this statement is needed for the proof
of the first Sylow theorem; then the full statement of Cauchy's theorem follows
from the first
Sylow
theorem itself. Since the
Cauchy's theorem for abelian
Sylow theorems.
groups
(diligent) reader has already proved
(in Exercise II.8.17), we could directly move
on to
quick proof4 of the full statement of Cauchy's theorem
Sylow and is a good illustration of the power of the general
'class formula for arbitrary actions' (Proposition 1.1). We will present this proof,
while also encouraging the reader to go back and (re)do Exercise II.8.17 now.
However, there is
which does not rely on
a
Proof of Theorem 2.1. Consider the set S of p-tuples of elements of G:
{oi,...,op}
such that di
e. We claim that |5|
ap
chosen (arbitrarily), then ap is determined
=
-
Therefore,
as
once ai,...,
it is the inverse of a\
=
ap
ap-\.
e, then
a2
(even
if G is not
also
right-inverse
commutative):
ava\
e
=
because if a\ is
a
left-inverse to a<i
ap, then it is
to it.
S: given
[m]
where Z is the set of fixed points of this action. Fixed points
are
0 <
av-\ are
p divides the order of S as it divides the order of G.
Also note that if a\
a
|G|P_1: indeed,
=
Therefore,
we
m < p, act
by [m]
may act
with the group
Z/pZ
on
in
Z/pZ,
with
on
{oi,...,op}
by sending
as we
it to
just observed, this is still
Now
Corollary
an
element of S.
1.3 implies
|Z|
=
|S|=0
modp,
p-tuples of the
form
(*)
and
{a,..., a};
note that Z
conclude that
with
^ 0,
since
{e, ...,e}
G
Z.
\Z\ 1; therefore there exists
>
Since p > 2 and p divides
some
fl^e.
This argument is apparently due to James McKay.
\Z\,we
(*),
element in Z of the form
IV. Groups, second encounter
196
This says that there exists
an
a
G
G,
a
^
e, such that ap
=
e,
proving the
statement.
We should remark that the
the
raw statement
subgroup of G of order
such subgroups.
Claim
proof given here
p, and we are able to say
2.2. Let G be a
the number
precise result than
a cyclic
something about the number of
proves a more
of Theorem 2.1: every element of order p in G generates
finite group, let p be a
of cyclic subgroups of G of order p.
prime divisor
Then N
of \G\,and
let N be
1 mod p.
=
The proof of this fact is left to the reader (as an incentive to really understand
the proof of Theorem 2.1).
Claim 2.2, coupled with the simple observation that if there is only 1 cyclic
H of order p, then that
for interesting applications.
subgroup
subgroup
must be normal
(Exercise 2.2),
Definition 2.3. A group G is simple if its only normal subgroups
itself.
groups occupy a
Simple
up'
special place
in the
are
{e}
suffices
and G
j
theory of groups: one can 'break
simple groups; we will see
any finite group into basic constituents which are
how this is done in
simple
or
Example
§3.1. Thus,
it is important to be able to tell whether
a group
is
not5.
2.4. Let p be a
positive prime integer. If |G|
=
rap, with 1 < ra < p,
then G is not
simple.
consider
the subgroups of G with p elements. By Claim 2.2, the number
Indeed,
of such subgroups is
lmodp. Thus, if there is more than one such subgroup,
then there must be at least p + 1. Any two distinct subgroups of prime order can
only meet at the identity (why?); therefore this would account for at least
=
l +
elements in G.
Since
|G|
=
cyclic subgroup of order
proving that G is not simple.
one
(p+l)(p-l)=p2
rap <
p in
p2,
this is impossible.
G, which
Therefore there is only
must be normal as mentioned
above,
j
Sylow I. Let p be a prime integer. A p-Sylow subgroup of a finite group G
1. That is, P C G is a
prm and (p, ra)
subgroup of order pr, where |G|
p-Sylow subgroup if it is a p-group and p does not divide [G : P].
2.2.
is
a
=
=
If p does not divide the order of |G|, then G contains a p-Sylow subgroup:
namely, {e}. This is not very interesting; what is interesting is that G contains a
p-Sylow subgroup even when p does divide the order of G:
Theorem 2.5
subgroup,
for
(First Sylow theorem). Every finite
group contains a
p-Sylow
all primes p.
In fact,
mentioned at
a
complete list of all finite simple groups is known: this is the classification result
§11.6.3, arguably one of the deepest and hardest results in mathematics.
the end of
2. The
Sylow theorems
The first
Sylow
theorem follows from the seemingly stronger statement:
Ifpk
Proposition 2.6.
197
divides the order
ofG,
then G has
a
subgroup of order pk.
The statements are actually easily seen to be equivalent, by Exercise 1.8; in
any case, the standard argument proving Theorem 2.5 proves Proposition 2.6, and
we see no reason to hide this fact. Here is the argument:
Proof of Proposition 2.6. If k
k > 1 and in particular that |G| is
=
a
0, there is nothing
multiple of p.
to prove, so we may assume
again there is nothing to prove; if |G| > p
[G : H] is relatively prime to p,
then pk divides the order of H, and hence H contains a subgroup of order pk by
the induction hypothesis, and thus so does G.
Argue by
induction
on
and G contains a proper
|G|:
if
|G|
subgroup
=
p,
H such that
Therefore, we may assume that all proper subgroups of G have index divisible
by p. By the class formula (Proposition 1.8), p divides the order of the center Z(G).
By Cauchy's theorem6, 3a G Z(G) such that a has order p. The cyclic subgroup
N
(a) is contained in Z(G), and hence it is normal in G (Exercise 1.4). Therefore
=
we can
consider the quotient
Since
\G/N\
G/N.
\G\/pand pk divides |G| by hypothesis,
of G/N. By the induction hypothesis, we
=
divides the order
we
have that
may
pk~l
conclude that
contains a subgroup of order pk~l. By the structure of the subgroups of
quotient (§11.8.3, especially Proposition II.8.9), this subgroup must be of the
form P/N, for P a subgroup of G.
G/N
a
But then \P\ \P/N\ \N\ pk~l
=
=
-p
=
pk,
as
needed.
are slicker ways to prove Theorem 2.5.
We will see a pretty (and
in
alternative
but
the
above
is easy to remember and is
proof given
§2.3;
insightful)
There
a
good template for similar arguments.
Remark
diligent reader worked out in Exercise II.8.20 a stronger
Proposition 2.6, for abelian groups. The arguments are similar; the
advantage in the abelian case is that any cyclic subgroup produced by Cauchy's
theorem is automatically normal, while ensuring normality requires a few twists
and turns in the general case (and, as a result, yields a weaker statement).
j
2.7. The
statement than
2.3. Sylow II. Theorem 2.5 tells us that some maximal p-group in G attains
the largest size allowed by Lagrange's theorem, that is, the maximal power of the
prime p dividing
One
can
p-group in
be
|G|
|G|.
more
precise: the second Sylow theorem tells us that every maximal
ap-Sylow subgroup. It is as large as is allowed by Lagrange's
is in fact
theorem.
The situation is in fact even better: all p-Sylow subgroups are conjugates of
each other7. Moreover, even better than this, every p-group inside G must be
contained in a conjugate of any fixed p-Sylow subgroup.
Note that,
Of
course
as
we only need the abelian case of this theorem.
p-Sylow subgroup of G, then so are all conjugates gPg-1 of P.
mentioned in §2.1,
if P is
a
IV. Groups, second encounter
198
The
proof of this
Theorem 2.8
subgroup,
(Second Sylow theorem).
[G
P]
easy!
very
Let G be
a
finite
group, let P be
Then H is contained in
a p-group.
g G G such that H C
Proof. Act with H
:
precise result is
and let H C G be
there exists
are
very
a
ap-Sylow
conjugate of P:
gPg~l.
the set of left-cosets of P, by left-multiplication. Since there
[G : P], we know this action must have
on
cosets and p does not divide
fixed points
let gP be
(Exercise 1.1):
one
hgP
that is, g~lhgP
=
of them. This
=
means
that Vft G H:
gP;
P for all h in H; that is,
g~lHg
C
P; that is, H
C
gPg~l,
as
needed.
We
can
obtain
have constructed
a
an
even
more
complete picture of the
situation.
Suppose
we
chain
H0
{e}CH1C...CHk
=
of p-subgroups of a group G, where \Hi\ p1. By Theorem 2.8 we know that Hk
is contained in some p-Sylow subgroup, of order pr
the maximum power of p
=
=
dividing the order of G. But we claim that the chain
step at a time all the way up to the Sylow subgroup:
H0
=
{e}
C
Hx
C
C
Hk
C
Hk+1
can
C
in fact be continued
C
one
Hr;
and, further, Hk may be assumed to be normal in .fffc+i- The following lemma will
simplify the proof of this fact considerably and will also help us prove the third
Sylow theorem.
Lemma 2.9. Let H be
a p-group
contained in
[NG(H) :H]
Proof. If H is trivial, then
Nq(H)
=
=
a
finite
[G:H]
group G.
Then
mod p.
G and the two numbers
are
equal.
Assume then that H is nontrivial, and act with H on the set of left-cosets of
H in G, by left-multiplication. The fixed points of this action are the cosets gH
such that V/i G H
hgH
that is, such that g~xhg
G H for all h G
(by order considerations) gHg~l
Therefore, the set of fixed points of
NG(H).
=
H.
=
H;
gH,
words, H C gHg~x, and hence
means precisely that g G Nq(H).
in other
This
the action consists of the set of cosets of H in
The statement then follows immediately from
As
a consequence,
'still' divides
[G Hk],
:
Corollary
1.3.
if Hk is not a p-Sylow subgroup 'already', in the sense that p
then p must also divide [No(Hk) : Hk]. Another application
of Cauchy's theorem tells us how to obtain the next subgroup
More precisely, we have the following result.
Hk+i
in the chain.
2. The
Sylow theorems
199
finite group G, and assume that H
p-subgroup H' of G containing H,
Proposition 2.10. Let H be a p-subgroup of a
is not a p-Sylow subgroup. Then there exists a
such that [Hf
:
H]
=
p and H is normal in Hf.
Proof. Since H is not a
p-Sylow subgroup of G,
Since H is normal in
Lemma 2.9.
p divides
[Ng(H)
:
H], by
may consider the quotient group
Nq(H),
and
divides
the
order
of
this
p
group. By Theorem 2.1, Nq{H)/H has
Ng(H)/H,
an element of order p; this generates a subgroup of order p of Ng(H)/H, which
must be
It
(cf. §11.8.3)
is
in the form
straightforward
to
we
H'/H
verify
for
a
subgroup H' of NG(H).
that H' satisfies the stated requirements.
The statement about 'chains of p-subgroups' follows immediately from this
result.
Note that
Cauchy's
theorem and Proposition 2.10 provide
Sylow theorem.
a new
proof of
Proposition 2.6 and hence of the first
2.4. Sylow III. The third (and last) Sylow theorem gives a good handle on the
number of p-Sylow subgroups of a given finite group G. This is especially useful in
establishing the existence of normal subgroups of G: since all p-Sylow subgroups
of a group are conjugates of each other (by the second Sylow theorem), if there is
only one p-Sylow subgroup, then that subgroup must be normal8.
Theorem 2.11. Letp be
prm. Assume that
G
divides
Np
m
=
[G:P]
=
Lemma 2.9
it divides the index
multiplying by
Np,
we
we
=
^ Omodp
of P. In fact,
=
Np \NG(P) : P].
have
=
[NG(P) : P]
mod p;
get
mNp
m
m
[G: NG(P)} [NG(P) : P]
m=[G:P]
Since
finite group of order \G\
ofp-Sylow subgroups of
p-Sylow subgroups of G.
the p-Sylow subgroups of G are the conjugates of any given
By Lemma 1.13, Np is the index of the normalizer Ng(P)
(Corollary 1.14)
Now, by
a
Then the number
denote the number of
By Theorem 2.8,
p-Sylow subgroup P.
of P; thus
prime integer, and let G be
and is congruent to 1 modulo p.
m
Proof. Let
a
p does not divide m.
=
m
mod p.
and p is prime, this implies
Np
=
1
mod p,
as needed.
Of course there are other ways to prove Theorem 2.11: see for
Exercise 2.11.
'For
an
alternative viewpoint,
see Exercise 2.2.
example
IV. Groups, second encounter
200
Consequences stemming from the group actions we have
and
encountered,
especially the Sylow theorems, may be applied to establish exquisitely
facts
about
individual groups as well as whole classes of groups; this is often
precise
based on some simple but clever numerology.
2.5. Applications.
The following examples
are
exceedingly simple-minded but
will
hopefully
convey the flavor of what can be done with the tools we have built in the
two sections. More
examples
previous
may be found among the exercises at the end of this
section.
2.5.1. More
nonsimple
Claim 2.12.
1
< m < p.
groups.
Let G be a group
Then G is not
order mpr, where p is
of
a
prime integer and
simple.
(Cf. Example 2.4.)
Proof. By the third Sylow theorem, the number Np of p-Sylow subgroups divides
1. Therefore G
m and is in the form 1 + kp. Since m < p, this forces k
0, Np
has a normal subgroup of order pr\ hence it is not simple.
=
Of course the
mpr, where
Example
(m,p)
same
=
argument gives the
1 and the
2.13. There
are no
same
conclusion for every group of order
only divisor doim such that d
simple
=
=
1
mod p is d
=
1.
groups of order 2002.
Indeed9,
2002
the divisors of 2
7
13
=
2
11
7
13;
are
1,2, 7, 13, 14,26,91, 182:
of these, only 1 is congruent
order
to 1 mod 11.
subgroup of
j
The reader should not expect the third
so
Thus there is a normal
11 in every group of order 2002.
Sylow
theorem to always yield its fruits
readily, however.
Example
2.14. There are no
Note that 3
=
simple
1 mod 2 and 4
=
groups of order 12.
1 mod 3: thus the argument used above does not
guarantee the existence of either a normal
2-Sylow subgroup
or a
normal
3-Sylow
subgroup.
However,
suppose that there is more than one
3-Sylow subgroup.
Then there
must be 4, by the third Sylow theorem. Since any two such subgroups must intersect
in the identity, this accounts for exactly 8 elements of order 3. Excluding these
leaves
to
fit
us
one
with the identity and 3 elements of order 2 or 4; that is just enough
2-Sylow subgroup. This subgroup will then have to be normal.
Thus, either there is a 3-Sylow normal subgroup
subgroup—either
way, the group is not simple.
It is safe to guess that this statement has been
the world in the year 2002.
assigned
or
on
there is
a
2-Sylow
room
normal
j
hundreds of algebra tests
across
2. The
Sylow theorems
Even this
more
201
refined counting will often fail, and
Example 2.15. There
are no
simple
one
has to dig deeper.
groups of order 24.
Indeed, let G be a group of order 24, and consider its 2-Sylow subgroups; by
the third Sylow theorem, there are either 1 or 3 such subgroups. If there is 1, the
2-Sylow subgroup is normal and G is not simple. Otherwise, G acts (nontrivially)
by conjugation
on
this set of three
a proper,
The reader should practice by selecting
as
much
as
he/she
are a common
2.5.2.
can, in
general, about
feature of qualifying
Groups of order pq,
j
a
random number
groups of order
n.
n and trying to say
Beware: such problems
exams.
p < q prime.
Claim 2.16. Assume p < q are prime
of
this action gives a nontrivial
nontrivial normal subgroup of
2-Sylow subgroups;
S3, whose kernel is
homomorphism G
G—thus
again G is not simple.
integers and q^l modp. Let G be
a group
order pq. Then G is cyclic.
Sylow theorem, G has a unique (hence normal) subgroup H
Indeed, the number Np of p-Sylow subgroups must divide g, and q is
1 or q. Necessarily Np
prime, so Np
lmodp, and q ^ lmodp by hypothesis;
1.
therefore Np
Since H is normal, conjugation gives an action of G on H, hence (by
Exercise 1.21) a homomorphism 7 : G
Aut(H). Now H is cyclic of order p, so
Proof. By the third
of order
p.
=
=
=
I Aut(i7)|
=
p
1
(Exercise 4.14);
the order of
7(G)
must divide both pq and p
1,
and it follows that 7 is the trivial map.
Therefore, conjugation is trivial
on
H: that is, H C
Z(G).
Lemma 1.5 implies
that G is abelian.
Finally, an abelian group of order pq, with p < q primes, is necessarily cyclic:
indeed it must contain elements g, h of order p, g, respectively (for example by
Cauchy's theorem), and then \gh\ pq by Exercise II. 1.14.
=
For example, this statement 'classifies' all groups of order 15, 33, 35, 51,
such groups are necessarily cyclic.
...:
The argument given in the proof is rather 'high-brow', as it involves the
automorphism group of H\that is precisely why we gave it. For low-brow alternatives,
see
Exercise 2.18
or
Remark 5.4.
The condition q ^ 1 modp in Claim 2.16 is clearly necessary: indeed, IS3I
is the product of two distinct primes, and yet S3 is not cyclic. The argument
in the
proof shows that if G
=
=
2-3
given
\pq\,with p
of order p, then G is cyclic. If q
=
< q prime, and G has a normal subgroup
lmodp, it can be shown that there is in fact
a unique noncommutative group of order pq up to isomorphism:
the reader will
work this out after learning about semidirect products (Exercise 5.12). But we are
in fact already in the position of obtaining rather sophisticated information about
this group, even without knowing its construction in general (Exercise 2.19).
For fun, let's tackle the
case
in which p
=
2.
IV. Groups, second encounter
202
Claim 2.17.
Let q be
order 2q. Then G
=
odd prime, and let G be
the dihedral group.
an
D2q,
a
noncommutative
group
of
Cauchy's theorem, 3y G G such that y has order q. By the third Sylow
theorem, (y) is the unique subgroup of order q in G (and is therefore normal).
Proof. By
Since G is not commutative and in particular it is not cyclic, it has no elements of
order 2g; therefore, every element in the complement of (y) has order 2; let x be
any such element.
The conjugate
xyx~l
yr for
=
xyx~l oiyhyx
some r
is
element of order g,
1.
an
between 0 and q
so
xyx~l
(y). Thus,
G
Now observe that
(yr)r
since
\x\
=
2.
=
(xyx~l)r
~l
Therefore, yr
=
xyrx~l
=
by Corollary
0 <
r < q
=
x2y(x~1)2
(r-l)(r
+
r
=
l or
r
=
=
Therefore
=
q
r
=
q
(r
1)
(r
or q
+
1);
since
1.
1, then xyx~l
y; that is, xy
yx. But
Exercise II. 1.14), and G is cyclic, a contradiction.
If r
y
l)
II. 1.11. Since q is prime, this says that q
1, it follows that
=
implies
e, which
g|(r2-l)
=
=
then the order of xy is
2q (by
1, and we have established the relations
( x2
<
yq
=
e,
=
e,
=
\yx
xyq~1.
These
are the relations satisfied by generators x,y of
verified in Exercise II.2.5; the statement follows.
D2Q,
as
the reader
hopefully
Claim 2.17 yields a classification of groups of order 2g, for q an odd prime: such
be either abelian (and hence cyclic, by the usual considerations) or
3, we recover the result of Exercise 1.13:
isomorphic to a dihedral group. For q
a group must
=
every noncommutative group of order 6 is isomorphic to Dq
=
S3.
Exercises
2.1. > Prove Claim 2.2.
[§2.1]
2.2. > Let G be a group.
every
automorphism
A
subgroup
H of G is characteristic if
<p(H)
C H for
cp of G.
Prove that characteristic
subgroups
are
normal.
Let H C K C G, with H characteristic in K and K normal in G. Prove that H
is normal in G.
Let G, K be groups, and assume that G contains
to K. Prove that H is normal in G.
a
single subgroup
H isomorphic
Exercises
203
Let K be
are
a
normal subgroup of
a
relatively prime. Prove that
finite group G, and assume that
K is characteristic in G.
and \G/K\
\K\
[§2.1, §2.4, 2.13, §3.3]
2.3. Prove that
some
abelian group G is simple if and only if G
a nonzero
positive prime integer
2.5.
Let G be a
(up
to
simple
for
simple if and only if its only homomorphic images
homomorphism G
Gf) are the trivial
groups G' such that there is an onto
group and G itself
Z/pZ
p.
2.4. > Prove that a group G is
(i.e.,
=
^
isomorphism). [§3.2]
group, and assume cp
:
G
G' is
nontrivial group
a
homomorphism. Prove that cp is injective.
2.6. Prove that there
In
are no simple groups of order 4, 8, 9, 16, 25, 27, 32, or 49.
prove that no p-group of order > p2 is simple.
fact,
2.7. Prove that there
simple
are no
groups of order 6, 10, 14, 15, 20, 21, 22, 26,
28, 33, 34, 35, 38, 39, 42, 44, 46, 51, 52, 55, 57, or 58.
(Hint: Example 2.4.)
2.8. Let G be a finite group, p a prime integer, and let N be the intersection of
the p-Sylow subgroups of G. Prove that N is a normal p-subgroup of G and that
every normal p-subgroup of G is contained in N. (In other words,
with respect to the property of being a homomorphic image of G of
for
some
G/N
is final
order \G\/pa
a.)
2.9.
Let P be a p-Sylow subgroup of a finite group G, and let H C G be a psubgroup. Assume H C NG(P). Prove that H C P. (Hint: P is normal in NG(P),
so PH is a subgroup of NG(P) by Proposition II.8.11, and \PH/P\
\H/(PnH)\.
-i
=
Show that this implies that PH is a p-group, and hence PH
maximal p-subgroup of G. Deduce that H C P.) [2.10]
2.10.
-i
Let P be
conjugation
on
the
a
p-Sylow subgroup of a finite group G, and act with P by
set of p-Sylow subgroups of G. Show that P is the unique fixed
(Hint:
2.11. > Use the second
together
P since P is
a
point of this action.
an
=
Use Exercise
2.9.) [2.11]
Sylow theorem, Corollary 1.14, and Exercise
Sylow theorem. [§2.4]
2.10 to paste
alternative proof of the third
2.12. Let P be a p-Sylow subgroup of a finite group G, and let H C
lmodp.
subgroup containing the normalizer NG(P). Prove that [G : H]
G be a
=
2.13.
-i
Let P be
a
p-Sylow subgroup of
Prove that if P is normal in
Exercise
G, then
a
finite group G.
it is in fact characteristic in G
(cf.
2.2).
Let H C G be
a
subgroup containing the Sylow subgroup
P.
Assume P is
normal in H and H is normal in G. Prove that P is normal in G.
Prove that
NG(NG(P))
=
P.
[3.12]
2.14. Prove that there are no
simple
groups of order 18, 40, 45, 50, or 54.
IV. Groups, second encounter
204
all groups of order n < 15, n ^ 8,12: that is, produce a list of
nonisomorphic groups such that every group of order n^8,12,n<15is isomorphic
2.15.
Classify
to one group in the list.
2.16.
-i
Let G be
noncommutative group of order 8.
a
Prove that G contains elements of order 4 and
Let y be
x
0 (y),
elements of order 8.
no
element of order 4. Prove that G is generated by y and by
an
such that x2
=
e or
x2
=
an
element
y2.
In either case, G
{e,y,y2,y3,x,yx,y2x,y3x}. Prove that the multiplication
table of G is determined by whether x2
e or x2
y2, and by the value of xy.
=
=
Prove that necessarily xy
right by
=
y3x. (Hint:
=
To eliminate xy
=
y2x, multiply
on
the
y.)
Prove that G ^ D8
or
G ^ Q8-
[6.2, VII.6.6]
division ring (Definition III. 1.13), and assume
that R is necessarily commutative (hence, a field), as follows:
2.17.
-i
Let R be
a
The group of units of R has order 63. Prove it has
of order 9. (Sylow.)
a
\R\
=
commutative
64. Prove
subgroup G
Prove that R is the only sub-division ring of R containing G.
Prove that the set of elements of R commuting with every element of G is
(Cf. Exercise III.2.10.)
a
Conclude that G is contained in the center of R. Recall that the center of R is
a
sub-division ring of R containing G.
sub-division ring of R
(cf.
Exercise
III.2.9),
and conclude that R is commutative.
Like Exercise III.2.11, this is a particular case of
according to which every finite division ring is a field.
a
theorem of Wedderburn,
[VII.5.16]
2.18. > Give an alternative proof of Claim 2.16 as follows: use the third Sylow
theorem to count the number of elements of order p and q in G; use this to show
that there are elements in G of order neither 1 nor p nor q\ deduce that G is cyclic.
[§2.5]
2.19. > Let G be
Show that
q
=
a
noncommutative group of order pq, where p < q
are
primes.
1 mod p.
Show that the center of G is trivial.
subgroups of G.
Find the number of elements of each possible order
Draw the lattice of
in G.
Find the number and size of the conjugacy classes in G.
[§2-5]
2.20. How many elements of order 7
2.21. Let p < q <
r
that G is not simple.
are
there in
a
simple
be prime integers, and let G be
group of order 168?
a group
of order pqr. Prove
3. Composition series and
2.22. Let G be
a
205
solvability
finite group, n
\G\,and p be a prime divisor of n. Assume that
n that is congruent to 1 modulo p is 1. Prove that G is simple.
=
the only divisor of
2.23.
-i
Let
Np
denote the number of p-Sylow
subgroups of a
group
G. Prove that
if G is simple, then |G| divides Np\ for all primes p in the factorization of |G|. More
generally, prove that if G is simple and H is a subgroup of G of index iV, then |G|
divides N\. (Hint:
Example 2.15.
Exercise
This problem capitalizes
II.9.12.)
on
the idea behind
[2.25]
2.24. > Prove that there are no noncommutative
simple groups of order less
possible order for a
than 60.
If you have sufficient stamina, prove that the next
noncommutative
simple
group is 168.
(Don't
feel too bad if you have to cheat and look up
few particularly troublesome orders >
2.25.
-i
Assume that G is
a
simple
a
60.) [§4.4]
group of order 60.
Sylow's theorems and simple numerology to prove that G has either five
fifteen 2-Sylow subgroups, accounting for fifteen elements of order 2 or 4.
(Exercise 2.23 will likely be helpful.)
Use
or
If there
are fifteen 2-Sylow subgroups, prove that there exists an element g G G
of order 2 contained in at least two of them. Prove that the centralizer of g has
index 5.
Conclude that every simple
group10
of order 60 contains
a
subgroup of index
5.
[4.22]
3.
Composition
series and
solvability
We have claimed that simple groups (in the sense of Definition 2.3) are the 'basic
constituents' of all finite groups. Among other things, the material in this section
will
(partially) justify
this claim.
3.1. The Jordan-Holder theorem. A series of subgroups Gi of
decreasing
sequence of
G
The length of
a
a group
G is
a
subgroups starting from G:
=
G0 2 Gi
D
G2 2
.
series is the number of strict inclusions.
A series is normal if
G^+i is normal in Gi for all i. We will be interested in the
length of a normal series in G; if finite, we will denote this number11 by
£{G). The number £{G) is a measure of how far G is from being simple. Indeed,
1 if and only if G is nontrivial and
£(G) 0 if and only if G is trivial, and £(G)
simple: for a simple nontrivial group, the only maximal normal series is
maximal
=
=
GD{e}.
The reader will prove later (Exercise 4.22) that there is in fact only one simple group of
order 60 up to isomorphism and that this group contains exactly five 2-Sylow subgroups. The
result obtained here will be needed to establish this fact.
There does not appear to be
a
standard notation for this concept.
IV. Groups, second encounter
206
Definition 3.1. A composition series for G is
G
=
Go 2 Gi 2 G2 2
such that the successive quotients
G^/Gi+i
normal series
a
2 Gn
are
{e}
=
simple.
j
It is clear (by induction on the order) that finite groups have composition series,
while infinite groups do not necessarily have one (Exercise 3.3). It is also clear that
if a normal series has maximal length £(G), then it is a composition series. What is
conceivably, there could exist maximal normal
lengths (the longest ones having length £{G)). For example, why
not clear is that the converse holds:
series of different
can't there be
a
finite group G with
G
£{G)
G1
2
3 and two different composition series
=
G2
2
&}
2
and
G
(that
is:
finite group G with
is simple)?
a
that G/G[
2
G[
2
£(G)
3 and
=
{e}
simple normal subgroup G[ such
a
Part of the content of the Jordan-Holder theorem is that
happen. In fact, the theorem is much
series have the same length, but they
precise:
also have the
more
(luckily)
this cannot
only do all composition
same quotients (appearing,
not
however, in possibly different orders).
Theorem 3.2 (Jordan-Holder). Let G be
G
G
=
=
Gi/Gi+i, H[
=
and let
G0^G1DG2D...DGn
for G.
G'i/G'^
Then
agree
m
(up
=
to
n,
{e},
=
G'0DG[DG2D---DG'rn
=
be two composition series
Hi
a group,
{e}
=
and the lists
of quotient
isomorphism) after
a
groups
permutation of
the indices.
Proof. Let
G
(*)
=
Go 2 Gi 2 G2 2
2 Gn
a composition series. Argue by induction on n: if
there is nothing to prove. Assume n > 0, and let
be
G
(**)
=
n
G'0DG'1DG'2D---DG'm
{e}
=
=
=
0, then G is trivial, and
{e}
be another composition series for G. If G\
G[, then the result follows from the
1 < n.
induction hypothesis, since G\ has a composition series of length n
=
We may then
assume
G\ ^ G[. Note that GiG^
(Exercise 3.5), and G\ C GiG[; but there
between G\ and G since GjG\ is simple.
in G
Let K
=
G1nG'lJ
=
G: indeed, GiG^ is normal
normal subgroups
are no proper
and let
K 2 Ai 2 K2 2
2 Kr
=
{e}
3. Composition series and
207
solvability
composition series for K. By Proposition II.8.11 (the "second isomorphism
theorem"),
be
a
Gl
^
=
G1nG[
K
are
simple. Therefore,
we
G^Gi
G[
have two
G 2 Gi
D
=
Gi
d
K
G[
new
K
^_
G
^
Gi
composition series for G:
2 Ki 2
2 &}
'••
G 2 G[ 2 K 2 #i 2
2 {e}
which only differ at the first step. These two series trivially have the
and the same quotients (the first two quotients get switched from one
same
length
series to the
other).
Now
as
claim that the first of these two series has the
we
the series
same
(*). Indeed,
Gi 2
K 2 Kx 2 K2 2
2 Kr
=
{e}
composition series for G\: by the induction hypothesis, it
length and quotients as the composition series
is
length and quotients
a
must have the same
Gi2G2D-OG„={e};
verifying
By
n
our
the
claim
in
n
particular, r
2).
token, applying the induction hypothesis to
same
(and
note
that,
=
the series of
length
1, i.e.,
G[ 2
K 2 Kx 2 K2 2
shows that the second series has the
same
2 ^n-2
=
{e},
length and quotients
as
(**),
and the
statement follows.
3.2. Composition factors; Schreier's theorem. Two normal series are
equivalent if they have the same length and the same quotients (up to order).
Jordan-Holder theorem shows that any two maximal finite series of
a group
The
are
equivalent. That is, the (isomorphism classes of the) quotients of a composition
series depend only on the group, not on the chosen series. These are the
composition factors of the group. They form a multiset12 of simple groups: the 'basic
constituents' of our loose comment back in §2.
It is clear that two isomorphic groups must have the same composition factors.
Unfortunately, it is not possible to reconstruct a group from its composition factors
alone (Exercise 3.4). One has to take into account the way the simple groups are
'glued' together;
we
will
come
back to this point in
§5.2.
The intuition that the composition factors of a group are its basic constituents
is reinforced by the following fact: if G is a group with a composition series, then
the composition factors of every normal subgroup N oi G are composition factors
of G and the remaining ones are the composition factors of the quotient G/N.
See §1.2.2 for a reminder on multisets: they are sets of elements counted with multiplicity.
For example, the composition factors of Z/4Z form the multiset consisting of two copies of Z/2Z.
IV. Groups, second encounter
208
Example 3.3. Let G
=
Z/6Z
=
Then
{[0], [1], [2], [3], [4], [5]}.
{[0],[1],[2],[3],[4],[5]}2{[0],[3]}2{[0]}
is
a
composition series for G; the quotients
(normal) subgroup
N
{[0], [2], [4]}
=
are
Z/3Z, Z/2Z, respectively.
The
'turns off' the second factor: indeed, intersecting
the series with N gives
{[0],[2],[4]}D{[0]}
series with composition factor
a
Z/3Z.
turns off the first factor: keeping in mind
{[0]
a
+
TV, [1]
N}
+
=
{[0]
series with lone composition factor
=
{[0]},
On the other hand, 'mod-ing out by NJ
[3] + N [1] + iV, etc., we find
=
+
TV, [1]
+
TV}
D
{[0]
+
TV},
Z/2Z.
j
This phenomenon holds in complete generality:
Proposition 3.4. Let G be a group, and let N be a normal subgroup of G. Then
G has a composition series if and only if both N and G/N have composition series.
Further, if this is the
case, then
e(G)
factors of G
and the composition
N and
=
e(N)
consist
+
e(G/N),
the collection
of
of composition factors of
of G/N.
G/N has a composition series, the subgroups appearing in it correspond
subgroups of G containing iV, with isomorphic quotients, by Proposition II.8.10
(the "third isomorphism theorem"). Thus, if both G/N and N have composition
series, juxtaposing them produces a composition series for G, with the stated
consequence on composition factors.
Proof. If
to
The
converse
is
a
a
G
{e}
=
Go 2 Gi 2 G2 2
and that iV is
sequence
a normal subgroup of G.
of subgroups of the latter:
N
such that
Gi+i
=
2 Gn
=
Intersecting the series with iV gives
GnNDG1nND...D{e}DN
=
a
{e}
Gi fl N, for all i. We claim that this becomes a
repetitions are eliminated. Indeed, this follows once
fl iV is normal in
composition series for iV
we
composition series
little trickier. Assume that G has
once
establish that
GjDN
Gi+i
is either trivial
be
omitted)
G).
factors of
or
n N
Gi N, and the corresponding inclusion may
G^+i
to
isomorphic
Gi/G^+i (hence simple, and one of the composition
(so
To
that
see
fl iV
=
fl
this, consider the homomorphism
GinN^Gi^-^-:
3. Composition series and
209
solvability
clearly G*+i fl iV; therefore (by the first isomorphism theorem)
injective homomorphisms
the kernel is
an
GiDN
Gi+i
is normal
As for
obtain
Gi+i
subgroup of Gi/Gi+\. Now, this subgroup
Gi/Gi+i is simple; our claim follows.
of subgroups from a composition series for G:
fl
G/N,
a sequence
a
and
GDG1NDG1NDD{jo}N
~
~
N
such that
(Gi+iN)/N
~
N
=
~
N
is normal in
have
d
fl N
N)/{Gi+\fl N) with
(because N is normal in G)
identifying (Gi
we
N
(GiN)/N.
x G/Nh
As above,
we
have to check that
(GjN)/N
(Gi+1N)/N
is either trivial
Gi/Gi+i. By the third isomorphism theorem, this
(GiN)/(Gi+iN). This time, consider the homomorphism
isomorphic
or
quotient is isomorphic to
to
°iN
Gi^GiN
Gi+1N
this is surjective (check!), and the subgroup Gi+i of the source is sent to the identity
element in the target; hence (by Theorem II.7.12) there is an onto homomorphism
Gi
d
+ N
—»
Gi+\
Since
Gi/Gi+i
isomorphic
is
to it
simple,
it follows that
(Exercise 2.4),
as
.
Gi+i
(Gi
+ N
+
N)/(Gi+i
+
N)
is either trivial
or
needed.
we have shown that if G has a composition series and iV is normal
in G, then both iV and G/N have composition series. The first part of the argument
yields the statement on lengths and composition factors, concluding the proof.
Summarizing,
One nice consequence of the Jordan-Holder theorem is the following
observation. A series is a refinement of another series if all terms of the first appear in the
second.
Proposition 3.5.
Any
equivalent refinements.
Proof. Refine the
theorem.
two normal series
series to
a
of
a
finite
group
ending with {e} admit
composition series; then apply the Jordan-Holder
In fact, Schreier's theorem
asserts that this holds for all groups
(while
the
argument given here only works for groups admitting a composition series, e.g.,
groups). Proving this in general is reasonably straightforward, from judicious
finite
applications of
the second isomorphism theorem
(cf.
Exercise
3.7).
IV. Groups, second encounter
210
3.3. The commutator
been
a
while since
derived series, and solvability. It has
a universal object; here is one.
For any
subgroup,
have encountered
we
A
group G, consider the category whose objects are group homomorphisms a : G
from G to a commutative group and whose morphisms a
are
the
reader
fi
(as
should expect) commutative diagrams
A
where
?—>B
cp is a
homomorphism.
Does this category have an initial object? That is, given
exist
a
a group
G, does there
which is universal with respect to the property of being
a commutative group
homomorphic image of G?
Yes.
Such a group may well be thought of as the closest 'commutative approximation'
of the given group G. To verify that this universal object exists, we introduce the
following important notion. (The diligent reader has begun exploring this territory
already, in Exercise
II.7.11.)
Definition 3.6. Let G be
a group.
The commutator subgroup of G is the subgroup
generated by all elements
ghg~lh-1
with g,h E G.
j
The element
ghg~lh~l
is often denoted
[g,h]
and is called the commutator of
g and h. Thus, g, h commute with each other if and only if
In the
[G,G];
same
this is
a
[g, h]
=
e.
notational style, the commutator subgroup of G should be denoted
bit heavy, and the common shorthand for it is G', which offers
the possibility of iterating the notation. Thus, G" may be used to denote the
commutator subgroup of the commutator subgroup of G, and G^ denotes the i-th
iterate. We will adopt this notation in this subsection for convenience, but not
elsewhere in this book
(as
we want to
be able to 'prime' any letter
we
wish, for
any
reason).
First
we
record the
Lemma 3.7. Let cp
have
:
following trivial,
G\
G<i be
a group
<P(\9M)
and
ip(G[)
C
but useful, remark:
=
homomorphism. Then Vg, h
G
G\ we
Vp(9),<p{h)]
G'2.
This simple observation makes the key properties of the commutator subgroup
essentially immediate (cf. Exercise II.7.11):
Proposition
G'
3.8. Let G' be the commutator
is normal in
G;
the quotient G/G'
is commutative;
subgroup of G. Then
3. Composition series and
if
:
a
A is
G
a
211
solvability
homomorphism of G
to a commutative group, then G' C
ker a;
the natural
Proof. These
(cf.
projection G
are
is universal in the sense
G/G'
explained above.
all easy consequences of Lemma 3.7:
Lemma 3.7, the commutator subgroup is characteristic, hence normal
—By
Exercise 2.2).
Lemma 3.7, the commutator of any two cosets gG', KG' is the coset of the
—By
[g,/i]; hence it is the identity in G/G'. As noted above, this implies
that G/G' is commutative.
commutator
—Let
a : G
a(G') CA'
=
{e}:
A be
a
homomorphism
that is, G' C ker
to a commutative group.
By Lemma 3.7,
a.
—The
universality follows from the previous point and from the universal
property of quotients (Theorem II.7.12).
Taking
successive commutators of
a group
produces
a
descending
sequence of
subgroups,
G
which is 'normal' in the
sense
Definition 3.9. Let G be
G'
D
D
G"
D
indicated in
G'"
D
,
§3.1.
The derived series of G is the sequence of
a group.
subgroups
G D G' D G" D G'" D
j
.
The derived series may or may not end with the identity of G. For example,
if G is commutative, then the sequence gets there right away:
GDG'
however,
=
{e};
if G is simple and noncommutative, then it gets stuck at the first step:
G
(indeed, G' is
simple).
normal and ^
{e}
=
as
G'
=
G"
=
G is noncommutative; but then G'
=
G since
G is
Definition 3.10. A group is solvable if its derived series terminates with the
identity,
j
For example, abelian groups
are
solvable.
The importance of this notion will be most apparent in the relatively distant
future because of a brilliant application of Galois theory (§VII.7.4). But we can
already appreciate it in the way it relates to the material we just covered. A normal
series is abelian, resp., cyclic, if all quotients are abelian, resp., cyclic13.
Proposition
3.11. For
a
finite
group
G, the following
are
equivalent:
(i) All composition factors of G are cyclic.
(ii) G admits a cyclic series ending in {e}.
Thus,
a
composition series should be called 'simple'; to
our
knowledge, it is
not.
212
(Hi)
G admits
(iv)
G is solvable.
Proof,
=>
groups are
(ii)
(iii)
=>
we
(iii)
=>
are
is also trivial, since the derived series is abelian
abelian series.
(by
the second
3.8).
only have
to prove
G
an
{e}.
a
point in Proposition
Thus,
encounter
trivial, (iii) ==> (i) is obtained by refining an
series
composition
(keeping in mind that the simple abelian
cyclic p-groups).
(i)
(iv)
abelian series ending in
an
abelian series to
be
Groups, second
IV.
=
(iii)
==>
Go 2 Gi 2 G2
Then
we
(iv).
D
For this, let
D
Gn
=
{e}
claim that G^ C d for all i, where G^ denotes
the i-th 'iterated' commutator subgroup.
This
G'
C
can
be verified by induction.
Gi, by the third point
fact that
Gi/Gi+i
For i
=
1, G/G\ is commutative; thus
Proposition 3.8. Assuming we know G« C Gu the
implies G\ C G^+i, and hence
in
is abelian
G<i+1)HG(i))'CG;cGi+i,
as
claimed.
In particular we obtain that G^ C Gn
terminates at {e}, as needed.
Example 3.12. All p-groups
p-group are
Corollary
simple
(what
p-groups
3.13. Let N be
and only if both N and
are
a
G/N
=
{e}:
that is, the derived series
solvable.
Indeed, the composition factors of
else could they be?), hence cyclic.
normal subgroup
solvable.
of
a group
a
j
G. Then G is solvable if
are
Proof. This follows immediately from Proposition 3.4 and the formulation of
solvability in terms of composition factors given in Proposition 3.11.
The Feit-Thompson theorem asserts that every finite group of odd order is
solvable. This result is many orders of magnitude beyond the scope of this book:
the original 1963 proof runs about 250 pages.
Exercises
213
Exercises
3.1. Prove that Z has normal series of
3.2. Let G be
a
finite
cyclic
in terms of
Compute £(G)
group.
is not
arbitrary lengths. (Thus, £(Z)
finite.)
Generalize to
|G|.
finite solvable groups.
3.3.
>
Prove that every finite group has
not have a
3.4. > Find an example of
factors. [§3.2]
3.5.
>
a
composition series. Prove that Z does
composition series. [§3.1]
Show that if H, K
subgroup of G.
two
nonisomorphic
groups with the same
normal subgroups of
are
a group
composition
G, then HK
is
a
normal
[§3.1]
a composition series if and only if both G\ and G<i
and
how
the
do,
explain
corresponding composition factors are related.
3.6. Prove that G\ x G<i has
3.7. > Locate and understand a proof of (the general form of) Schreier's theorem
that does not use the Jordan-Holder theorem. Then obtain an alternative proof of
the Jordan-Holder theorem, using Schreier's. [§3.2]
3.8. > Prove Lemma 3.7.
[§3.3]
a nontrivial p-group.
Construct explicitly an abelian series for G,
using the fact that the center of a nontrivial p-group is nontrivial (Corollary 1.9).
This gives an alternative proof of the fact that p-groups are solvable (Example 3.12).
3.9. Let G be
3.10.
-i
Let G be
Define inductively an increasing sequence Zq
{e} C
as follows: for i > 1, Z{ is the subgroup of G
a group.
=
Z1CZ2O" of subgroups of G
(as
corresponding
in
Proposition II.8.9)
Prove that each Zi is normal in G,
A group is14 nilpotent if Zm
=
G for
to the center of
so
that this definition makes
p-groups are
Prove that nilpotent
sense.
some m.
Prove that G is nilpotent if and only if
Prove that
G/Zi-i.
G/Z(G)
is
nilpotent.
nilpotent.
groups are solvable.
Find a solvable group that is not
nilpotent.
[3.11, 3.12, 5.1]
3.11.
Exercise
-1
Let H be
3.10).
a
nontrivial normal subgroup of
Prove that H intersects
smallest index such that 3h
commutator
[g, h].)
Since p-groups
^
are
e, h G
a
nilpotent
Z(G) nontrivially. (Hint:
HC\Zr. Contemplate
a
group G
Let
r
>
(cf.
1 be the
well-chosen
nilpotent, this strengthens the result of Exercise 1.9.
[3.14]
There are many alternative characterizations for this notion that are equivalent to the one
given here but not too trivially so.
214
IV.
3.12. Let H be a proper
subgroup of
a
Prove that H C
finite
nilpotent
is nontrivial.
Groups, second
encounter
group G (cf. Exercise 3.10).
First dispose of the case in
Ng(H). (Hint: Z(G)
Z(G\ and then use induction to deal with the case
does contain Z(G).) Deduce that every Sylow subgroup of a finite
which H does not contain
in which H
3.13.
For
-i
normal15. (Use Exercise 2.13.)
group is
nilpotent
a group
G, let G^ denote the iterated commutator,
that each G^ is characteristic
3.14. Let H be
nontrivial normal subgroup of
a
Prove that H contains
Let
(Hint:
r
(hence normal)
(cf.
an
§3.3.
Prove
[3.14]
solvable group G.
nontrivial commutative subgroup that is normal in G.
H fi G^ is nontrivial. Prove
be the largest index such that K
=
use
Exercise 3.13 to show it is normal in
example showing that H need
Exercise
a
in
as
a
that K is commutative, and
Find
in G.
not intersect the center of G
G.)
nontrivially
3.11).
integers, and let G be a group of order p2q. Prove that G
particular case of Burnside 's theorem: for p, q primes, every
3.15. Let p, q be prime
is solvable.
is
a
paqb
is
(This
group of order
solvable.)
3.16. > Prove that every group of order < 120 and
3.17. Prove that the
^
60 is solvable.
Feit-Thompson theorem is equivalent
simple group has even order.
[§4.4, §VII.7.4]
to the assertion that
every noncommutative finite
4.
The
symmetric
group
4.1. Cycle notation. It is time to give a second look at symmetric groups. Recall
that Sn denotes the group of permutations (i.e., automorphisms in Set) of the set
{1,..., n}.
In
§11.2
we
denoted elements of Sn in
a
straightforward
but inconvenient
way:
G
_
~
(I
\S
234567
8\
127534
6)
would stand for the element in Ss sending 1 to 8, 2 to 1, etc.
There is clearly "too much" information here (the first row should be implicit),
same time it seems hard to find out anything interesting about a
permutation from this notation. For example, can the reader say anything about the
and at the
conjugates of
then try
in 5s? For maximal enlightenment, try
after
again
absorbing the material in §4.2. In
a
to do Exercise 4.1 now, and
short,
we
should be able to
do better.
As often is the case, thinking in terms of actions helps. By its very definition,
on the set
{1,... ,n}; so does every subgroup of Sn. Given a
the group Sn acts
permutation
{1,..., n}.
a G
Sn, consider the cyclic
group
The orbits of this action form
a
(a) generated by
a
and its action
partition of {1,..., n}; therefore,
'This property characterizes finite nilpotent groups; cf. Exercise 5.1.
on
every
4. The symmetric group
215
Sn determines a partition of {1,..., n}. For example, the element
above splits {1,..., 8} into three orbits:
g G
{1,2,3,6,8},
{4,7},
a G
Ss given
{5}.
The action of (a) is transitive
on each orbit.
This means that one can get from
any element of the orbit to any other element and then back to the original one by
applying
a
times. In the
enough
example,
1i-»8i-»6i-»3i-»2i-»1,
5
4i-»7i-»4,
(nontrivial) cycle is an element of Sn with
{1,..., n}, the notation
Definition 4.1. A
5.
•-?
exactly
one
nontrivial
orbit. For distinct ai,..., ar in
(a\(i2...ar)
denotes the cycle in Sn with nontrivial orbit
0>l
'
0*2
'
{ai,..., ar}, acting
0>r
'
Cbl-
length of the cycle. A cycle of length
In this case, r is the
as
is called
r
an
r-cycle,
The identity is considered a cycle of length 1 in a trivial way and is denoted
as well be denoted by (i) for any i).
j
by
(1) (and could just
Note that
(aid2
...
ar)
(d2
=
...
ara\) according
to the notation introduced
cycle, but a nontrivial cycle only
determines the notation 'up to a cyclic permutation'.
Two cycles are disjoint if their nontrivial orbits are. The following observation
deserves to be highlighted, but it does not seem to deserve a proof:
in Definition 4.1:
Lemma 4.2.
cycles,
commute.
Disjoint cycles
The next one
Lemma 4.3.
the notation determines the
gives
Every a
G
us
the alternative notation
Sn,
a
^
we were
looking for.
product of disjoint nontrivial
factors.
e, can be written as a
in a unique way up to permutations
of
the
we have seen, every a G Sn determines a partition of
{1,... ,n} into
orbits under the action of (a). If a ^ e, then (a) has nontrivial orbits. As a acts
as a cycle on each orbit, it follows that a may be written as a product of cycles.
Proof. As
The proof of the uniqueness
The cycle notation for
a
G
product of disjoint cycles found
example,
is left to the reader
Sn is the (essentially) unique expression of a as a
in Lemma 4.3 (or (1) for a
e). In our running
=
<t
=
(18632)(47),
and keep in mind that this expression is unique
a
and all of these
=
are
(63218)(47)
(Exercise 4.2).
=
(74)(21863)
=
'the same' cycle notation for
cum grano
salis:
(32186)(74)
a.
we
could write
IV. Groups, second encounter
216
4.2. Type and
conjugacy classes in Sn. The cycle
features—such
as the not-too-unique uniqueness
annoying
or
notation has
pointed out
obviously
a moment ago,
the fact that
(123)
element of S3 just as well as of Siooooo* and only the context can tell.
it
is invaluable as it gives easy access to quite a bit of important
However,
information on a given permutation. In fact, much of this information is carried already
can
be
an
by something even simpler than the cycle decomposition.
A partition of an integer n > 0 is a nonincreasing16 sequence of positive integers
whose sum is n. It is easy to enumerate partitions for small values of n. For example,
5 has 7 distinct partitions:
5=1+1+1+1+1
=2+1+1+1
=2+2+1
=3+1+1
The partition Ai > A2 >
=
3 + 2
=
4 + 1
=
5.
> Ar may be denoted
[Ai,..., ArJ;
for example, the fourth partition listed above would be denoted [3,1,1]. A nicer
'visual' representation is by means of the corresponding Young (or Ferrers) diagram,
obtained by stacking Ai boxes on top of A2 boxes on top of A3 boxes on top of
For example, the diagrams corresponding to the
[1,1,1,1,1] [2,1,1,1]
[2,2,1]
[3,1,1]
a G
orbits of the action of (a)
{1,..., n}.
It is
hopefully
clear
(from
partitions listed above
[3,2]
Sn is the partition of
Definition 4.4. The type of
on
seven
[4,1]
n
are
[5]
given by the sizes of the
the argument proving Lemma
j
4.3)
that the type of
Sn is simply given by the lengths of the cycles in the decomposition of a as
the product of disjoint cycles, together with as many l's as needed. In our running
example,
a
(18632)(47) G S8
a
G
=
has type [5,2,1]:
'Of
course this choice is
arbitrary, and nondecreasing sequences would do just
as well.
4. The symmetric group
217
ED
The main
why 'types'
reason
introduced is
are
a consequence
of the following
simple observation.
Lemma 4.5. Let
r G
Sn, and let (a\...ar) be
t(cl\
The
ai G
{1,..., n};
act on the
ar)r~l
recall that
in
notation for
by checking that both sides
Proof. This is verified
For example, for 1 < i <
it should; the other
products
in groups.
act in the same way on
{1,..., n}.
r
(aiT~l)(r(ai... ar)r~l)
as
for the action of the permutation r_1 on
§11.2.1 that we would let our permutations
agreed
right, for consistency with the usual
we
cycle. Then
(a\T~l...arr~l).
=
a\T~l stands
notation
funny
...
a
cases are
Oi(oi... ar)r~l
=
=
ai+ir~l
left to the reader.
By the usual trick of judiciously inserting identity factors t~xt, this formula
for computing conjugates extends immediately to any product of cycles:
r(ai... ar)
(&i
...
6s)r_1
This holds whether the cycles
disjoint cycles remain disjoint
the
following important
Proposition 4.6.
the
same
=
(air-1
...
arr~l)
{b\r~l
...
bsr~l).
disjoint or not. However, since r is a bijection,
after conjugation. This is essentially all there is to
are
observation:
Two elements
of Sn
conjugate in Sn if and only if they have
are
type.
Proof. The 'only if part of this statement follows immediately from the preceding
considerations: conjugating a permutation yields a permutation of the same type.
As for the 'if part, suppose
<7i
=
a2
=
(oi... ar)(bi... bs)
(ci... ct)
and
permutations with the
are two
> t
bj
=
(so
b'jT,
so G\
(a[...a'r)(b'1...b's).-.(c'1...c't)
the type is
...,
and a2
Cfc
are
=
c'kr
same
[r, 5,..., t]).
type, written in cycle notation, with
Let
r > s >
be any permutation such that a^
a^r,
k. Then Lemma 4.5 implies g<i
tg\t~x,
r
for all i, j, ...,
as needed.
=
=
conjugate,
Example 4.7. In 5s,
(18632)(47)
must be
tells
us
and
conjugate, since they have the
(12345)(67)
same
type. The proof of Proposition 4.6
that
t(18632)(47)t"1
=
(12345)(67)
IV. Groups, second encounter
218
for
'12
3
4
5
6
7
^18
6
3
2
4
7
5
this may be checked by hand in a second. Running this check, and
especially staring at the second row in r vis-a-vis the cycle notation of the first
j
permutation, should clarify everything.
and of
course
Summarizing, the type (or
everything about conjugation in Sn.
Corollary 4.8.
partitions of n.
The number
For example, there
alikes drawn above.
are
the corresponding Young
of conjugacy
diagram)
tells
us
classes in Sn equals the number
7 conjugacy classes in S5, indexed
by the
of
Tetris look-
It is also
reasonably straightforward to compute the number of elements in each
conjugacy class, in terms of the type. For example, in order to count the number
of permutations of type [2,2,1] in S5, note that there are 5!
120 ways to fill the
corresponding Young diagram with the numbers 1,..., 5:
=
ai
a2
a3
a4
a5
that is, 120 ways to write
a
permutation
as a
product of
two
2-cycles:
(aia2)(a3a4);
but switching a\
a2 and as
a±, as
same permutation. Therefore there are
<-?
<-?
well as
switching the
two
cycles, gives the
120
15
2-2-2
permutations of type [2,2,1]. Performing this computation for all Young diagrams
for 55 gives us the size of each conjugacy class, that is, the class formula (cf. §1.2)
for 55:
120
Example
4.9. There
=
1 + 10 + 15 + 20 + 20 + 30 + 24.
are no
normal subgroups of size 30 in S5.
Indeed, normal subgroups are unions of conjugacy classes (§1.3); since the
1
29 cannot be written as a sum of the
identity is in every subgroup and 30
=
numbers appearing in the class formula for S5, there is
no
such subgroup.
j
This observation will be dwarfed by much stronger results that we will prove
soon (such as Theorem 4.20, Corollary 4.21); but it is remarkable that such precise
statements
can
be established with
so
little work.
4. The symmetric group
219
Transpositions, parity, and the alternating
4.3.
group. For
n >
1, consider
the polynomials
An
]^[
=
(Xi-Xj) eZ[Xi,...,Xn],
l<i<j<n
that
is,
Ai
=
A2
=xi -x2,
A3
A4
We
=
=
l,
(xi
(xi
-
-
with any
can act
x2)(xi
x2)(xi
a G
Sn
-
-
xs)(x2
x3)(xi
-
-
x3),
x4)(x2
x3)(x2
-
-
x4)(x3
-
x4),
An, by permuting the indices according
on
Ana
JJ
:=
to a:
(xiCT-xiCT).
l<i<j<n
For
example,
A4(1234)
(x2
=
-
x3)(x2
-
x4)(x2
-
xi)(x3
-
x4)(x3
-
Xi){x±
-
x{)
=
-A4.
In general, it is clear that Ana is still the product of all binomials (xi —Xj),
where
the factors are permuted and some factors may change sign in the process. Hence,
Ana
±An, where the factor ±1 depends on a.
=
Definition 4.10. The sign of
by the action of a on An:
a
permutation
Ana
We say that
a
permutation is
Note that Va,
r G
Sn
we
even
=
multiplication17,
( l)ar
we see
=
is determined
(-l)CTAn.
j
have
=
(Ana)r
:
as
Viewing { —1,+1}
(—1)CT(
—1)T.
a group
under
that the 'sign' function
c:Sn-{-l,+l},
is
5n, denoted (-1)CT,
if its sign is +1 and odd if its sign is —1.
An(ar)
it follows that
a G
e(a):=(-iy
homomorphism.
Here is a different viewpoint on this sign function. A transposition is
length 2. Every permutation is a product of transpositions:
a
Lemma 4.11.
cycle of
Transpositions generate Sn.
by Lemma
transpositions, and indeed
Proof. Indeed,
4.3 it suffices to show that every
(ai... ar)
as may
a
be checked by
applying18
=
(aia2)(aia3)
cycle
a
product of
(aiar),
both sides to every element of
This is the group of units in Z; of course it is isomorphic to C<2.Don't forget that permutations act on the right.
is
{1,..., n}.
IV. Groups, second encounter
220
given permutation may be written as a product of transpositions
However, whether an even number of transpositions or an
odd number is needed only depends on a. Indeed,
Of
course a
in many different ways.
Lemma
resp.,
4.12.
Let
a
odd, according
=
t\ -rr
be a
product of transpositions.
Then
a
is even,
to whether r is even, resp., odd.
immediately from the facts that e is a homomorphism and the
of
a
indeed, (ij) acts on An by permuting its factors and
transposition is —1:
sign
the
of
one
factor (namely, the factor ±(xi
changing
sign
exactly
Xj)).
Proof. This follows
Definition 4.13. The alternating group
even
permutations
a G
The alternating group is
indeed, An
{1,..., n},
denoted An, consists of all
j
a
normal subgroup of 5n, and
[on
for n > 2:
on
Sn.
=
kere, and
:
An\
=
2
is surjective for
e
n >
2.
belongs to An in terms of the
(by
4.11) a cycle is even, resp., odd,
if it has odd, resp., even, length19. Pictorially, a permutation a G Sn is even if and
only if n and the number of rows in the Young diagram of a have the same parity.
Take n
5 for example:
It is very easy to tell whether
a
permutation
a
the argument proving Lemma
type of a. Indeed
=
The starred types correspond to
corresponding conjugacy classes,
even
up the sizes of the
permutations. Adding
1 + 15 + 20 + 24
=
60,
confirms that A*, has index 2 in S*,.
However, do
that these
not read too much into this
computation:
we
are
not
claiming
the sizes of the conjugacy classes in A§, only that these are the
classes
in S§ making up the normal subgroup A§. Conjugacy in An is
conjugacy
rather interesting and ties in nicely with the issue of the simplicity of An.
are
Conjugacy in An\ simplicity of An and solvability
[o~]sn, resp., [cr]^n, the conjugacy class of an even permutation
Clearly [o-]ati ^ Msn; we proceed to compare these two sets.
4.4.
Lemma 4.14. Let
is
half
the size
n>2, and let
a G
of [o~]sn, according
An. Then [ct]ati
=
[&]sn
to whether the centralizer
of Sn. Denote
a
or
in
5n,
resp.,
the size
Zsn(o~)
by
An.
of [ct]ati
is not or is
contained in An.
Note the unfortunate terminology clash. For example, 3-cycles
are
even
permutations.
4. The symmetric group
Proof.
(Cf.
Exercise
221
Note that
1.16.)
ZAn(<r)=AnnZSn(<T):
this follows immediately from the definition of centralizer (Definition 1.6). Now
recall that the centralizer of a is its stabilizer under conjugation, and therefore the
size of the conjugacy class of
If
ZSri(cr)
C
a
equals the index of
An, then ZAri(a)
[Sn : Z5»]
=
[Sn : ZAn(a)]
=
=
ZSti(g),
its centralizer.
that
so
[Sn : An][An : ZAri(a)}
=
2
[An : ZAti(g)};
therefore, [cr]An is half the size of [cr]sn in this case.
If Zsn(o~) 2 An, then note that AnZsri(o~)
5n: indeed, AnZsri(o~) is a
of
is
cf.
Sn (because An
normal;
Proposition II.8.11), and it properly contains
subgroup
=
An, so it must equal Sn as An has index 2 in Sn. Applying the second isomorphism
theorem (Proposition II.8.11) gives
[An
so
to
=
[An : An
the classes have the
[c]ati
to
ZAri(a)}
:
=
=
size. Since
same
Msn> completing
Z5»]
n
[^s» : ZSn(°)]
[cr]Ari
C
[Sn '• ZSrA°)l
=
in any case, it follows that
[a]sTl
the proof.
Therefore, conjugacy classes of even permutations either are preserved from Sn
An or they split into two distinct, equal-sized classes. We are now in a position
give precise conditions determining which happens when.
Proposition 4.15. Let a G An, n>2. Then the conjugacy class of a in Sn splits
into two conjugacy classes in An precisely if the type of a consists of distinct odd
numbers.
Proof. By Lemma 4.14, we have to verify that Zsn((J) 1S contained in An precisely
when the stated condition is satisfied; that is, we have to show that
a
precisely when the type of
Write
g
in
cycle
=
r must
(including cycles
of length
(oi... ax)(bi... 6M)
=
(aiT_1
.
.
Assume that A, \x...,v
by
r is even
=>
1):
(ci... c„),
that
(Lemma 4.5)
TGT~l
to~t~1
consists of distinct odd numbers.
notation
g
and recall
a
=
preserve each
a\T~l)(biT~l
.
are
cycle
.
.
.
6MT_1)
.
.
.
(CiT_1
odd and distinct. If tgt~x
in a,
as
all cycle lengths
r(oi... a\)r~l
=
are
=
.
.
.
CVT~X).
g, then
conjugation
distinct:
etc.,
(oi... aA),
that is,
(air-1... aXT~l)
This means that r acts as a
(oi... oa),
cyclic permutation
in the same way as a power of
t =
=
on
etc.
(e.g.)
ai,..., a\ and therefore
{a\...a\). It follows that
(oi... ax)r(bi... b^)s
(ci... cv)1
222
IV.
Groups, second
encounter
Since all cycles have odd lengths, each cycle is an even
r
and
must
then be even as it is a product of even permutations.
permutation;
This proves that Zsn (&) Q An if the stated condition holds.
for suitable r, s,...,£.
Conversely,
assume
that the stated condition does not hold: that is, either some
even length or all have odd length but
of the cycles in the cycle decomposition have
two of the cycles have the same length.
In the first case, let r be an
Note that rar~l
a: indeed, r
than
r.
Since
Zsn(&) 2 -Am
has
r
even
length, then
needed.
as
A
assume
=
/x, and consider the
permutation
r =
conjugating by
As
a.
with itself and with all cycles in a other
it is odd as a permutation: this shows that
commutes
In the second case, without loss of generality
odd
cycle decomposition of
in the
even-length cycle
=
r
r
(aibi)(a2b2)'"(axbx)
simply interchanges the first
two
Zsn(o~) 2 ^n,
is odd, this again shows that
:
in a; hence rar~l
cycles
and
we are
=
a.
done.
Example 4.16. Looking again at A5, we have noted in §4.3 that the types of
the even permutations in S5 are [1,1,1,1,1], [2,2,1], [3,1,1], and [5]. By
Proposition 4.15 the conjugacy classes corresponding to the first three types are
in A5, while the last one splits.
Therefore there
A5
are
exactly
5 conjugacy classes in
preserved
A5, and the class formula for
is
60
=
1 + 15 + 20 + 12 + 12.
j
circle of thought begun in the first section of
this chapter. Along the way, the reader has hopefully checked that every simple
group of order < 60 is commutative (Exercise 2.24); and we now see why 60 is
Finally!
We
can now
complete
a
special:
Corollary 4.17.
order 60.
The
alternating
group
A5
is a
simple
noncommutative group
of
Proof. A normal subgroup of A5 is necessarily the union of conjugacy classes,
contains the identity, and has order equal to a divisor of 60 (by Lagrange's theorem).
The divisors of 60 other than 1 and 60 are
2, 3 4, 5
,
,
6
,
10, 12, 15 20, 30;
,
counting the elements other than the identity would give
one
of
1, 2, 3,4, 5, 9, 11, 14, 19, 29
as a sum
of numbers ^ 1 from the class formula for A5. But this simply does not
happen.
The reader will check that Aq is simple, by the same method (Exercise 4.21).
case that all groups An, n > 5, are simple (and noncommutative),
It is in fact the
implying that Sn
is not solvable for
n >
5, and these facts
view of applications to Galois theory (cf., e.g., Corollary
is trivial, As
Z/3Z is simple and abelian, and A± is not
=
rather important in
VII.7.16). Note that A2
are
simple (Exercise 2.24).
4. The symmetric group
The
it is the
(Can
223
alternating group A$ is also called the icosahedral (rotation)
group of symmetries of an icosahedron obtained through
verify20
the reader
this
group:
indeed,
rigid motions.
fact?)
The simplicity of An for n > 5 may be established by studying 3-cycles. First
of all, it is natural to wonder whether every even permutation may be written as a
product of 3-cycles, and this is indeed so:
Lemma 4.18.
Proof. Since
The
alternating
every even
group
An
permutation is
is
a
generated by 3-cycles.
product of
an even
number of 2-cycles,
it suffices to show that every product of two 2-cycles may be written
3-cycles. Therefore, consider
as
product of
product
a
(ab)(cd)
with
a
y^ b,
If
d.
y^
c
(ab)
(cd), then this product is the
{a, &}, {c, d} have exactly one element
=
nothing to prove.
If
may assume c
and observe
a
=
(ab)(ad)
If
{a, &}, {c, d}
are
=
(abc)(adc),
done.
we are
Now
(abd).
disjoint, then
(ab)(cd)
and
=
identity, and there is
in common, then we
we can
capitalize
Claim 4.19. Let
contains all
n
>
5.
on our
If
a
study of conjugacy
normal subgroup
in An:
of An
contains a
3-cycle, then
it
3-cycles.
Proof. Normal subgroups are unions of conjugacy classes, so we just need to verify
that 3-cycles form a conjugacy class in An, for n > 5. But they do in 5n, and the
type of a 3-cycle is [3,1,1,... ] for n > 5; hence the conjugacy class does not split
in An, by Proposition 4.15.
The general
statement now follows
Theorem 4.20.
Proof. We have
n
=
6. For
n >
The
alternating
easily by tying
group An is
already checked this for
6, let iV be
a
n
=
up loose ends:
simple for
n >
5.
5, and the reader has checked it for
nontrivial normal subgroup of An\ we will show that
An, by proving that iV contains 3-cycles.
necessarily N
Let r G iV, r 7^ (1), and let a G An be a 3-cycle. Since the
=
trivial (Exercise 4.14) and 3-cycles generate An,
we may assume
center of
that
r
and
An is
a
do
not commute, that is, the commutator
[r, a]
=
r(ar~1a~1)
=
(rar~1)a~1
It is good practice to start with smaller examples:
group is isomorphic to A4.
for instance, the tetrahedral rotation
224
Groups, second
IV.
encounter
identity. This element is in N (as is evident from the first expression,
normal) and is a product of two 3-cycles (as is evident from the second
expression, since the conjugate of a 3-cycle is a 3-cycle).
Therefore, replacing r by [r, a] if necessary, we may assume that r G iV is
is not the
as
a
N is
nonidentity permutation acting
TC{l,,,,,n}
with
\T\
=
6. Now
on
6 elements:
<
we may
view Aq
that is,
as a
on
a
subset of
a set
subgroup of An, by letting
subgroup NdAq of Aq is then normal (because N is normal) and
nontrivial (because t e N D Aq and r ^ (1)). Since Aq is simple (Exercise 4.21),
this implies N D Aq
Aq. In particular, N contains 3-cycles.
By Claim 4.19, this implies that N contains all 3-cycles. By Lemma 4.18, it
follows that N
An, as needed.
it act
on
T. The
=
=
Corollary 4.21. For
Proof. Since
n >
5, the group Sn is
An is simple, the
not solvable.
sequence
Sn^AnD {(1)}
is
a
composition series for Sn. It follows that the composition factors of
Z/2Z
on are
and An. By Proposition 3.11, Sn is not solvable.
In particular, S*, is a nonsolvable group of order 120. This is in fact the smallest
order of a nonsolvable group; cf. Exercise 3.16.
Exercises
4.1.
>
Compute the number of elements
'12
in
in the conjugacy class of
3
4
5
6
7
12
7
5
3
4
6;
Ss. [§4.1]
4.2. >
Suppose
(oi... ar)(bi ...bs)'-(ci...ct)
are two
(Hint:
=
(di... du)(ei... ev)
products of disjoint cycles. Prove that the factors
The two corresponding partitions of
4.3. Assume
a
has type
What is |a |? What
[Ai,..., Ar]
can you say
{1,..., n}
and that the A^'s
about
|a|,
agree up to order.
must
are
(/i... fw)
agree.) [§4.1]
pairwise relatively prime.
without the additional hypothesis
the numbers A^?
4.4. Make
sense
of the 'Taylor series' of the infinite product
11111
(1-x)
(1-x2)
(1-x3)
(1-x4)
(1-x5)
Prove that the coefficient of xn in this series is the number of partitions of
4.5. Find the class formula for
5n,
n <
6.
n.
on
Exercises
225
4.6. Let N be
normal subgroup of S±. Prove that
a
\N\ 1, 4, 12,
=
or
24.
generated by (12) and (12... n).
(Hint: It is enough to get all transpositions. What is the conjugate of (12) by
(12...n)?) [4.9, §VII.7.5]
4.7. > Prove that Sn is
4.8.
For n > 1, prove that the subgroup H of Sn consisting of permutations
1
is isomorphic to 5n_i. Prove that there are no proper subgroups of Sn
fixing
-i
properly containing
H.
[VII.7.17]
4.7, 54 is generated by (12) and (1234). Prove that (13) and
of D% in S4. Prove that every subgroup of S4 of order 8 is
conjugate to ((13), (1234)). Prove there are exactly 3 such subgroups. For all n > 3
prove that Sn contains a copy of the dihedral group D271, and find generators for it.
By Exercise
(1234) generate a
4.9.
4.10.
copy
Prove that there
-1
More
generally, find
permutation of given type in Sn.
are
a
exactly (n
1)! n-cycles
in on.
formula for the size of the conjugacy class of
a
[4.11]
prime integer. Compute the number of p-Sylow subgroups of Sp.
Exercise
(Use
4.10.) Use this result and Sylow's third theorem to prove again the
if
implication in Wilson's theorem (cf. Exercise II.4.16.)
'only
4.11. Let p be a
4.12. > A
subgroup G of Sn
is transitive if the induced action of G
on
{1,..., n}
is transitive.
Prove that if G C Sn is transitive, then
|G|
is
a
multiple of
n.
List the transitive subgroups of 53.
Prove that the
following subgroups of S4
are
all transitive:
((1234)) C4 and its conjugates,
((12)(34),(13)(24))^C2xC2,
((12)(34), (1234)) ^ D8 and its conjugates,
-
=
-
-
A4, and 54.
With a bit of stamina, you
-
can prove
that these
are
the only transitive subgroups
ofS4.
[§VII.7.5]
4.13.
(If
determinants.) Prove that the sign of a permutation a,
4.10, equals the determinant of the matrix Ma defined in
you know about
as defined in Definition
Exercise II.2.1.
4.14. > Prove that the center of
4.15.
is
Justify
An is trivial for
the 'pictorial' recipe given in
§4.3
n >
4.
[§4.4]
to decide whether a
permutation
even.
4.16. The number of conjugacy classes in
An,
n >
2, is (allegedly)
1,3,4,5,7,9,14,18,24,31,43,....
Check the first several numbers in this list by finding the class formulas for the
corresponding alternating groups.
IV. Groups, second encounter
226
4.17. >
Find the class formula for A4.
Use it to prove that A± has
4.18. For
Prove that
no
subgroup of order
6.
[§11.8.5]
>
5, let H be a proper subgroup of An. Prove that [An
An does have a subgroup of index n for all n > 3.
n
4.19. Prove that there
Construct21
are no
nontrivial actions of An
nontrivial action of A±
2?
action of A4 on a set S with |5|
a
on
a set
on any set
5, \S\
=
3.
S with
Is there
a
> n.
H]
:
|5|
< n.
nontrivial
=
4.20.
five
-1
Find all fifteen elements of order 2 in A5, and prove that A*, has exactly
2-Sylow subgroups. [4.22]
Aq is simple, by using its class formula (as is done for A*,
of
proof
Corollary 4.17). [§4.4]
4.21. > Prove that
in the
that A$ is the only simple group of order 60, up to isomorphism.
Exercise
2.25, a simple group G of order 60 contains a subgroup of index 5.
(Hint: By
Use this fact to construct a homomorphism G
S5, and prove that the image of
4.22.
-1
Verify
this homomorphism must be A5.) Note that A5 has exactly five 2-Sylow subgroups;
cf. Exercise 4.20. Thus, the other possibility contemplated in Exercise 2.25 does
not
5.
occur.
[2.25]
Products of groups
We already know that products exist in Grp (see §11.3.4); here we analyze this notion
further and explore variations on the same theme, with an eye towards the question
of determining the information needed to reconstruct
factors.
a group
5.1. The direct product. Recall from §11.3.4 that the
groups H, K is the group supported on the set H x K, with
from its composition
(direct) product
of two
operation defined
componentwise. We have checked (Proposition II.3.4) that the direct product satisfies
the universal property defining products in the category Grp.
There are situations in which the direct product of two subgroups iV, H of a
group G may be realized as a subgroup of G. Recall (Proposition II.8.11) that if
one of the subgroups is normal, then the subset NH of G is in fact a subgroup of
G. The relation between NH and N x H depends on how N and H intersect in
G,
take
so we
a
look at this intersection.
The 'commutator'
generated by all
Lemma 5.1.
[A, B]
commutators
Let
of two subsets A, B of G
[a, b] with a e A, b e B.
N, H be normal subgroups of
[N,H]
You
sides
on a
a group
(see §3.3)
is the
subgroup
G. Then
CNDH.
can think algebraically if you want; if you prefer geometry, visualize pairs of opposite
tetrahedron.
5. Products of groups
Proof. It suffices to
227
this
verify
[n, ft]
generators; that is, it suffices to check that
on
n(hn-lh-1)
=
=
(n/in"1)^-1
e N D H
N, ft G H. But the first expression and the normality of N show that
N\the second expression and the normality of H show that [n, H] G H.
for all n G
[n, ft]
G
5.2. Let
Corollary
N, H be normal subgroups of a
group
G. Assume Nf)H
=
{e}.
Then N, H commute with each other:
(Vn
Proof. By Lemma 5.1,
In fact, under the
Proposition
[N, H]
same
5.3. Let
nh
N) (Vft eH)
G
=
{e}
if Nf)H
hypothesis
more
=
=
{e};
hn.
the result follows immediately.
is true:
N, H be normal subgroups of a
group
G, such that Nf)H
=
ThenNH^NxH.
{e}.
Proof. Consider the function
<p
defined
by (p(n, ft)
=
iV
:
x
H
NH
-?
nft. Under the stated hypothesis, ip is
a group
homomorphism:
indeed
<p((nuhi)
(n2,ft2))
=
=
=
since iV, H
commute
<p((nin2,hih2))
niri2hih2
n\h\n<ih<2,
5.2
by Corollary
=
<p((nuhi)) '<p((n2,h2)).
The homomorphism cp is surjective by definition of NH. To
consider its kernel:
ker (p
If nft
=
same
token for ft,
(p is
e, then n e N and n
inject ive.
Thus (p is
an
we
{(n, h)
=
=
ft-1
conclude ft
isomorphism,
=
as
G
\nh
e N x H
H; thus
e; hence
n
=
=
(n, ft)
e
=
verify
it is injective,
e}.
since N f)H
=
{e}. Using
the identity in N
x
the
H, proving
needed.
Remark 5.4. This result gives an alternative argument for the proof of Claim 2.16:
if |G|
pq, with p < q prime integers, and G contains normal subgroups i7, K of
=
is the case if q ^ 1 modp, by Sylow), then H C\K
{e}
H x K. As \HK\ \G\ pq,
and
then
5.3 shows HK
Proposition
necessarily,
HxK
this proves G
Z/pZ x Z/qZ. Finally, (1,1) has order pq in this group,
so G is cyclic, with the same conclusion we obtained in Claim 2.16.
j
order
p, g,
respectively (as
=
=
=
=
=
=
IV. Groups, second encounter
228
5.2. Exact sequences of groups; extension problem. Of course, the
hypothesis that both subgroups iV, H are normal is necessary for the result of
Proposition 5.3: for example, the permutations (123) and (12) generate subgroups
iV,
H of
53 meeting only
{e}, and N is normal in S3, but S3 NH is not isomorphic to
the direct product of N and H. It is natural to examine this more general situation.
at
Let iV, H be
assumptions
subgroups of
H)
on
description of the
=
a group
G, with N normal (but with
and such that N f) H
=
{e};
G
assume
=
no
NH. We
a
are
priori
after
a
structure of G in terms of the structure of N and H.
notationally convenient to use the language of exact sequences, introduced
§111.7.1. A (short) exact sequence of groups is a sequence of groups
group homomorphisms
It is
for modules in
and
>N—^G
1
>1
>H
where ip is surjective and (p identifies N with ker^. In other words
isomorphism theorem), use (p to identify N with a subgroup of G; then
is exact if N is normal in G and
ip induces
the first
(by
the
H.
sequence
isomorphism G/N
The reader should pause a moment and check that if G, N, H are abelian, then
this notion matches precisely the notion of short exact sequence of abelian groups
an
(i.e., Z-modules) from §111.7.1; a notational difference is that here the trivial group
is denoted22 '1' rather than '0'.
Of
course
there always is
is
a very
and
(n, e#)
special
>NxH
>N
1
map n E N to
an exact sequence
case:
(n, ft)
:
However, keep in mind that this
example mentioned above, there also is an
G N x H to ft.
reiterate the
to
>1
>H
exact sequence
>
1
yet S3 £ C3
x
C3
>
S3
>
C2
,
C2.
Definition 5.5. Let iV, H be groups. A group G is
there is
> 1
an exact sequence
an
extension
of
H
by
N if
of groups
1
>iV
>G
>H
>1.
j
The extension problem aims to describe all extensions of two given groups,
For example, there are two extensions of C2 by C3: namely
up to isomorphism.
Cq
=
Cs
are no
x
C2 and S3;
we
will
soon
be able to
verify that,
up to
isomorphism, there
other extensions.
The extension problem is the 'second half of the classification problem: the
first half consists of determining all simple groups, and the second half consists of
figuring out how these can be put together to construct any group23. For example,
This is not unreasonable, since groups are more often written 'multiplicatively' rather than
so the identity element is more likely to be denoted 1 rather than 0.
As mentioned earlier, the first half has been settled, although the complexity of the work
leading to its solution justifies some skepticism concerning the absolute correctness of the proof.
The status of the second half is, as far as we know, (even) murkier.
additively,
5. Products of groups
229
if
G
is
a
=
Go 2 G1
D
D
G2
G3 2 G4
composition series, with (simple) quotients Hi
of Hq by
extension
an
extension
of H\ by
an
=
=
extension
{e}
Gi/Gi+i,
then G is
an
of H2 by H3: knowing
the
composition factors of G and the extension process, it should in principle be possible
to reconstruct G.
We
going to 'solve' the extension problem
subgroup of G, intersecting N at {e}.
are
H is also
a
in the
particular
case in
which
Definition 5.6. An exact sequence of groups
1
>N
>1
>H
>G
split if H
the corresponding extension) is said to
subgroup of G, so that N D H
{e}.
(or
may be identified with a
=
We encountered this terminology in §111.7.2 for modules, thus for abelian groups.
Note that the notion examined there appears to be more restrictive than
Definition 5.6, since it requires G to be isomorphic to a direct product N x H. This
apparent mismatch evaporates because of Proposition 5.3: in the abelian case,
every split extension (according to Definition 5.6) is in fact a direct product.
Of
are
Exercise
course
usually
split extensions are anyway very special, since quotients of a group G
isomorphic to subgroups of G, even in the abelian case (cf.
not
5.4).
a normal subgroup of a group
NH and Nf)H
{e}. Then G is
Lemma 5.7. Let N be
of G such that G
=
=
Proof. We have to construct
let N
we
an exact sequence
>
1
G, and let H be a subgroup
split extension of H by N.
a
> G
N
> 1 ;
H
G be the inclusion map, and
we
prove
that
G/N
=
H.
For this,
consider the composition
a:H^G^
Then
n E
is surjective: indeed, since G
N and h E H, and then
a
gN
Further,
as
ker
a
=
{h
e H
=
nhN
hN
=
=
=
G/N.
NH, \/g G
h(h-lnh)N
N}
=
N fi H
=
=
hN
{e};
G we have g
=
=
nh for
some
a(h).
therefore
a
is also injective,
needed.
To recap, if in the situation of Lemma 5.7 we also require that H be normal in
G, then G is necessarily isomorphic to the 'trivial' extension N x H: this is what
we have proved in Proposition 5.3. We are seeking to describe the extension 'even
if H is not normal in G.
IV. Groups, second encounter
230
Internal/semidirect products.
5.3.
The attentive reader should have noticed
that the key to Proposition 5.3 is really Corollary 5.2: if both N and H are normal
and NnH
{e}, then N and H commute with each other. This is what ultimately
=
causes
the extension NH to be trivial.
then every
subgroup
conjugation
H of G acts
determines
a
Now, recall that
N
H
:
7
ft
AutGrp(iV),
-?
^
as soon as
in fact
by conjugation:
homomorphism
on
(cf.
N is normal,
Exercise 1.21)
7/i-
N acts by 7^(71) := hnh~l.)
(Explicitly, for h G H the automorphism 7^ : N
The subgroups H and N commute precisely when 7 is trivial. Corollary 5.2 shows
that if N and H are both normal and N D H
{e}, then 7 is indeed trivial.
=
This is the crucial remark. The next several considerations may be summarized
a subgroup of G, N fi H
{e} and G NH,
follows: if N is normal in G, H is
as
then the extension G of H by N
7
:
H
AutQrp(N).
=
=
conjugation action
following triviality, which
may be reconstructed from the
The reader is advised to stare at the
is the motivating observation for the general discussion:
(*)
n1h1n2h2
(Vni,n2 eN),(\/huh2 G H)
=
(n1(h1n2h^1)) (hih2).
This says that if we know the conjugation action of H on iV, then we can recover
the operation in G from this information and from the operations in iV and H.
Here is the general discussion. It is natural to abstract the situation and begin
with any two groups N, H and an arbitrary homomorphism24
0:H^AutGrp(N),
Define
let
an
operation
•#on the set N x H as follows: for ni,n2 G iV and
(ni,hi)
This will look
more
Lemma 5.8.
The
h^0h.
99
reasonable
resulting
(n2,h2)
once
:=
it is
structure
(N
hi, h2
G
i7,
(ni6hl(n2),hih2).
compared with (*)!
H, •#) is
x
a group, with
identity element
(ejv,etf).
Proof. The reader should carefully verify this.
(ni,hi)99(0h-i(n^l),h^1)
and similarly in the
=
reverse
Definition 5.9. The group
denoted by iV x# H.
For example, inverses exist because
(ni9hl(9h-i(n^1)), hihf:)
=
(nin^l,eH)
=
(eN,eH)
order.
(N
x
i7, •#)is
a
semidirect product of iV and H and is
j
For example, the ordinary direct product is a semidirect product and
the trivial map. If the reader feels a little uneasy about giving
corresponds to 9
one name (semidirect product) for a whole host of different groups supported on
=
the Cartesian product, welcome to the club.
The
reason
why
we are not
In fact, it gets
denoting the image of h by 0
as
0(h)
worse
still: it is not
is that this is an
for the image of n
N obtained by applying
the automorphism corresponding to h. The alternative 0h(n) looks a little easier to parse.
automorphism of N and we dislike the notation
0(h)(n)
5. Products of groups
231
omit '0' from the notation and
uncommon to
write N
simply
xi
H for
a
semidirect
product25.
In any case, the notation •#is too
heavy to carry around, so we generally revert
simple juxtaposition of elements in order to denote multiplication
back to the usual
in N xi 6» H.
The
following proposition
checks that semidirect products
are
split
extensions:
5.10. Let
Proposition
morphism; let G
G
contains
=
N, H be
groups, and let 0
H
:
be
AutQrp(N)
a
homo-
corresponding semidirect product. Then
N >\g H be the
isomorphic copies of N and H;
H is a surjective homomorphism, with kernel N;
the natural projection G
thus N is normal in G, and the sequence
1
is
>1
>H
(split) exact;
iVHtf
G
>N x\eH
>N
=
=
{eG};
NH;
the homomorphism 0
n E
N
we
is realized
by conjugation
in G: that is,
for
h G H and
have
hnh~l
Oh(n)
in G.
Proof. The functions iV
G, H
-?
G defined for
-?
n G
iV, h
G H
by
n i—?
h i—?
(n, e#),
(ejv,h)
are manifestly injective homomorphisms, allowing us
corresponding subgroups of G. It is clear that N D H
(n,eH)9Q (eNjh)
shows that G
=
i7 defined
me
H with the
{ec},
and
(n,/i)
ft
i—?
surjective homomorphism, with kernel iV; therefore iV is normal
(ejv, ft)
as
{(ejv, e#)}
=
by
(n, ft)
a
identify N,
NH.
The projection G
is
=
to
=
(n, en)
^
(e^, ft)-1
=
(0h(n), ft)
^
(eN, ft-1)
=
in G.
Finally,
(0h(n),eH),
claimed in the last point.
Our original goal of 'reconstructing' a given split extension of
group iV is a sort of converse to this proposition. More precisely,
This is actually OK, if N and H
case
are
the implicit action is just conjugation.
given
as
subgroups of
a group
a common group
H
by
a
G, in which
IV. Groups, second encounter
232
Proposition 5.11. Let N, H be subgroups of a
NH. Let 7
Assume that N f) H
{e}, and G
h
n
E
E
H,
N,
conjugation: for
group
:
H
bijection. We need
to
=
=
lh(n)
=
G, with N normal in G.
AutQrp(N) be defined by
hnh~l.
Then G^N y\1H.
Proof. Define
a
function
ip:N
(p(n, h)
by
=
nh\ this is clearly
phism, and indeed
(Vn
(p((nuhi)
a
G
iV), (V/i
«7
(n2,/i2))
verify
that cp is
a
homomor-
H):
G
=
=
=
=
as
y\1H^G
(p((ni7hl(n2),hih2))
<p{{nl(hln2h^l),hlh2))
ni/iin2(/if1/ii)/i2 (nihi)(ri2h2)
=
(p((ni,hi))(p((n2,h2))
needed.
When realized 'within'
product
is sometime called
a group, as
an
in the previous proposition, the semidirect
internal product.
Remark 5.12. If N and H commute, then the conjugation action of H on iV is
trivial; therefore 7 is the trivial map, and the semidirect product iV xi7 H is the
direct product N xH. Thus, Proposition 5.11
in this case.
Example
C2: if C3
5.13. The
=
{e, £/, £/2},
automorphism
recovers
the result of Proposition 5.3
j
group of
C3 is isomorphic
then the two automorphisms of C3
f
e 1—?
e,
id: <
y^y,
0 :
i2~y\
<
to the
cyclic
group
are
e 1—?
e,
y^y2
,y2
*-+y-
are two homomorphisms C2
Autcrp(C'3): the trivial map, and
isomorphism sending the identity to id and the nonidentity element to a. The
semidirect product corresponding to the trivial map is the direct product C3 x C2
Cq; the other semidirect product C3 xi C2 is isomorphic to S3. This can of course
Therefore, there
the
=
be checked by hand (and you should do it, for
from Proposition 5.11, since iV
((123)), H
=
fun); but it also follows immediately
((12)) C S3 satisfy the hypotheses
=
of this result.
j
The reader should contemplate carefully the slightly more general case of
dihedral groups (Exercise 5.11); this enriches (and in a sense explains) the discussion
presented in Claim 2.17.
In fact, semidirect products shed light on all groups of order pq, for p < q
primes; the reader should be able to complete the classification of these groups
lmodp, then there is exactly one such nonbegun in §2.5.2 and show that if q
=
commutative group up to
isomorphism (Exercise 5.12).
Exercises
233
The reader would in fact be well-advised to try to
semidirect products to
subgroup N is found (typically
groups of small order: if a nontrivial normal
classify
by applying Sylow's theorems),
with
some
use
luck the classification is reduced to the
study of possible homomorphisms from known groups to AutQrp(N) and
carried out. See Exercise 5.15 for a further illustration of this technique.
can
be
Exercises
5.1.
Let G be
>
Assume
all Pi
a
are
finite group, and let Pi,..., Pr be its nontrivial
normal in G.
Prove that G
=
Prove that G
is
on
|G|.
Together
if each of
P\ x
Pr. (Induction
x
nilpotent. (Hint: Mod
What is the center of
a
out
Sylow subgroups.
Proposition 5.3.)
the
center, and work by induction
by
on r; use
direct product of
groups?)
with Exercise 3.10, this shows that a finite group is
its Sylow subgroups is normal. [3.12, §6.1]
nilpotent if and only
by N. Prove that the composition factors of G
the collection of the composition factors of H and those of N.
5.2. Let G be an extension of H
are
5.3. Let
G
be
a
G0 2 Gi
=
D
normal series. Show how to 'connect'
of groups, involving only
{e}, G,
D
{e}
Gr
=
to G
{e}
by
and the quotients Hi
means
=
of
r exact sequences
Gi/Gi+\.
5.4. > Prove that the sequence
> Z
0
is exact but does not split.
5.5. In
Proposition III.7.5
0
[§5.2]
we
> M
0
—^
> Z/2Z
Z
have
seen
that if
an exact sequence
—^
> N/(ip(M))
N
of abelian groups splits, then ip has
split sequences of groups ?
a
0
left-inverse. Is this necessarily the
for
case
5.6. Prove Lemma 5.8.
5.7. Let iV be
that
a
a group,
and let
may be realized as
containing iV
as a
a
:
iV
conjugation,
iV be
in the
an
sense
normal subgroup and such that
automorphism of N. Prove
that there exists
a(n)
=
gng~l
for
a group
some g G
5.8. Prove that any semidirect product of two solvable groups is solvable.
that semidirect products of nilpotent groups need not be nilpotent.
5.9.
>
Prove that if G
=
iV
x
H is commutative, then G^N
x
H.
[§6.1]
G
G.
Show
IV. Groups, second encounter
234
Let N be
5.10.
and
|G/iV|
\H\ |G/iV|.
> For
a
finite group G, and assume that \N\
is a subgroup H in G such that
all
n
>
a
semidirect product of N and H.
0 express D2n
as
semidirect product Cn xi<9 C2, finding 0
a
[§5.3]
explicitly.
5.12. >
normal subgroup of
Prove that G is
=
5.11.
a
relatively prime. Assume there
are
Classify
then either G is
groups G of order pq, with p < q
cyclic
or q
=
lmodp and there
is
of
prime: show that if |G|
pq,
exactly one isomorphism class
(You will likely have to use the
=
noncommutative groups of order pq in this case.
fact that AutGrp(Cq) ^ Cq-i if q is prime; cf. Exercise
II.4.15.) [§2.5, §5.3]
N >\g H be a semidirect product, and let K be the subgroup of G
to
ker 0 C H. Prove that K is the kernel of the action of G on the
corresponding
set G/H of left-cosets of H. [5.14]
5.13.
Let G
-1
=
5.14. Recall that
S3
isomorphism. Prove that
(C2
=
x
AutGrp(C2
C2) x, Ss
C2) (Exercise II.4.13).
x
Let
1
be this
S4. (Hint: Exercise 5.13.)
=
5.15. > Let G be a group of order 28.
Prove that G contains a normal
subgroup
iV of order 7.
isomorphism, the only groups of order 4 are
C4 and C2
C2. Prove that there are two homomorphisms C4
Autcrp(^)
and two homomorphisms C2 x C2
Autcrp(^) up to the choice of generators
Recall
(or
prove
again) that,
up to
x
for the sources.
Conclude that there are four groups of order 28 up to
direct
products C4
x
C7, C2
x
C2
x
C7, and
isomorphism: the
two
two noncommutative groups.
Prove that the two noncommutative groups are
D2% and C2
x
D14.
[§5.3]
5.16. Prove that the quaternionic group Qg (cf. Exercise III. 1.12)
as a semidirect
product of
two nontrivial
cannot be written
subgroups.
5.17. Prove that the multiplicative group H* of nonzero quaternions (cf.
Exercise III. 1.12) is isomorphic to a semidirect product SU2(C) x R+. (Hint:
Exercise III.2.5.) Is this semidirect product in fact direct?
6. Finite abelian groups
We will end this chapter by treating in
finite abelian groups mentioned in
6.1.
some
detail the classification theorem for
§11.6.3.
Classification of finite abelian groups. Now that we have acquired more
with products, we are in a position to classify all finite abelian groups26.
familiarity
In due time
classify
'Of
all
(Proposition VI.2.11,
finitely generated
course
Exercise
abelian groups:
VI.2.19)
as
we
will in fact be able to
mentioned in Example II.6.3, all
fancier semidirect products will not be needed here; cf. Exercise 5.9.
6. Finite abelian groups
235
In particular, this is the
such groups are products of cyclic groups
abelian groups: this is what we prove in this section.
.
case
for finite
we exclusively deal with abelian groups, we revert to the
notations: thus the operation will be denoted +; the identity will
be 0; direct products will be called direct sums (and denoted 0); and so on.
Since in this section
abelian
style of
First of all,
has been with
we
will
congeal
into
in
one
form
another since at least
us
Lemma 6.1. Let G be
or
an
simple observation that
statement a
as
far back as28 Exercise II.4.9.
abelian group, and let H, K be subgroups such that
H + K
H 0 if.
an
\K\are relatively prime. Then
Proof.
explicit
\H\,
=
By Lagrange's theorem (Corollary II.8.14), HDK
{0}. Since subgroups
are automatically normal, the statement follows from
=
of abelian groups
Proposition 5.3.
Now let G be a
of G
finite
Sylow subgroups of G
abelian group. For each prime p, the
automatically normal
is unique, since it is
are p-groups
p-Sylow subgroup
in G. Since the distinct nontrivial
for different primes p, Lemma 6.1 immediately
implies the following result.
Corollary 6.2. Every finite abelian group is the direct
sum
of its
nontrivial
Sylow
subgroups.
(The diligent reader knew already that this had to be the case, since abelian groups
are nilpotent; cf. Exercise 5.1.)
Thus, we already know that every finite abelian
group is a direct sum of p-groups, and our main task amounts to classifying abelian
p-groups for a fixed prime p. This is somewhat technical; we will get there by a
seemingly roundabout path.
Lemma 6.3. Let G be
an
abelian p-group, and let g G G be
an
element
of maximal
order. Then the exact sequence
0
>
(g)
a
subgroup
>
G
>
G/(g)
> 0
splits.
Put otherwise, there
G/(g)
is
L of G such that L maps isomorphically to
(g) f)L {0} and (g)+L G.
via the canonical projection, that is, such that
Note that it will follow that G
=
(g)
The main technicality needed
particular case:
0
=
=
L, by Proposition 5.3.
in order to prove this lemma is the
following
Lemma 6.4. Let p be a prime integer and r > 1. Let G be a noncyclic abelian
group of order pr+1, and let g G G be an element of order pr. Then there exists an
element h G G, h
0 (g),
such that
\h\
=
p.
more general result is that of modules over Euclidean
principal ideal domains.
In fact, this observation will really find its most appropriate resting place when we prove
The natural context to prove this
rings
or even
the Chinese Remainder Theorem, Theorem V.6.1.
IV. Groups, second encounter
236
special case of Lemma 6.3 in the sense that, with notation as in
necessarily (ft)
G/(g), and in fact G
(g) 0 (ft) (and the reader
is warmly encouraged to understand this before proceeding!). That is, we can split
off the 'large' cyclic subgroup (g) as a direct summand of G, provided that G is
Lemma 6.4 is
a
the statement,
=
=
not cyclic and not much larger than (g). Lemma 6.3 claims that this can be done
whenever (g) is a maximal cyclic subgroup of G. We will be able to prove this more
general
easily
statement
once
the particular
case
is settled.
(g) by K, and let ft' be any element of G, ft' 0 K.
K
in
is
normal
since G is abelian; the quotient group G/K has
G
subgroup
order p. Since ft' 0 K, the coset ft' + K has order p in G/K', that is, ph' G K. Let
Proof of Lemma 6.4. Denote
The
k
=
ph'.
\k\divides pr; hence it is a power of p. Also
and G would be cyclic, contrary to the hypothesis.
Note that
pr+1
\h'\=
=
=
(pg); thus,
k
=
Then let ft
mpg for some
ft'
=
showing that \h\
Proof of Lemma 6.3.
requires
=
(since
ph'
ft' 0
p(mg)
-
K),
k
=
and
-
k
=
0,
p, as stated.
=
we
will
on the order of G; the case |G|
p°
that G is nontrivial and that the statement
induction
Argue by
proof. Thus
no
Z.
m G
mg: ft ^ 0
ph
1
otherwise
\k\ ps for some s < r; k generates a subgroup (k) of the cyclic
K, of order ps. By Proposition II.6.11, (k)
{pr~sg). Since s < r, (k) C
Therefore
group
\k\^ pr,
assume
=
=
is true for every p-group smaller than G.
element of maximal order, say pr, and denote by K the
generated by g; this subgroup is normal, as G is abelian. If G K,
Let g G G be
(g)
subgroup
an
=
trivially. If not, G/K is a nontrivial p-group, and hence it
contains an element of order p by Cauchy's theorem (Theorem 2.1). This element
generates a subgroup of order p in G/K, corresponding to a subgroup G' of G of
order pr+1, containing K. This subgroup is not cyclic (otherwise the order of g is
then the
statement holds
not
maximal).
is
That is, we are in the situation of Lemma 6.4: hence we can conclude that there
element h G G' (and hence h G G) with h^L K and \h\ p. Let H
(h) C G
an
=
be the
subgroup generated by ft, and
note that K D H
=
=
{0}.
Now work modulo H. The quotient group G/H has smaller size than G, and
H
+
g
generates a cyclic subgroup K'
(K + H)/H K/(KnH) K of maximal
order in G/H. By the induction hypothesis, there is a subgroup L' of G/H such
=
that K'
+ L'
=
G/H
and K' fi V
=
=
{Og/h}-
=
This subgroup L' corresponds to
a
L of G containing H.
subgroup
Now
we
claim that
(i)
K + L
=
G and
(ii)
K D L
=
{0}. Indeed,
we
have the
following:
(i)
For any
a
G
mg + £ + H
a G
K + L
as
G, there
(since
exist mg + H G
if7 + V
needed.
=
G/H).
K'\ £
+ H G V such that
This implies
a
mg
G
a +
H
=
L, and hence
6. Finite abelian groups
(ii)
If
a G
a G
(i)
and
KnL, then
K n i7
=
(ii) imply
237
H G K'C\Lf
a +
{0}, forcing
the lemma,
as
a
=
0,
=
as
{Og/h},
and hence
a G
H. In particular,
needed.
observed in the comments
following
the statement.
Now we are ready to state the classification theorem; the proof is quite
straightforward after all this preparation work. We first give the statement in a somewhat
coarse form, as a corollary of the previous considerations:
Corollary
6.5. Let G be
a
abelian group.
finite
groups, which may be assumed to be
Proof. As noted in
of the Sylow
cyclic
Then G is
a
direct
sum
of cyclic
p-groups.
Corollary 6.2, G
theorems).
is a direct sum of p-groups (as a consequence
We claim that every abelian p-group P is a direct sum of
cyclic p-groups.
To establish this, argue by induction on \P\.There is nothing to prove if P is
trivial. If P is not trivial, let g be an element of P of maximal order. By Lemma 6.3
P
for
some
=
{g)®P'
subgroup P' of P\by the induction hypothesis P'
is
a
direct
sum
of cyclic
p-groups, concluding the proof.
6.2. Invariant factors and elementary divisors. Here is a more precise version
common to state the result in two equivalent
of the classification theorem. It is
forms.
Theorem 6.6. Let G be
a
finite
nontrivial abelian group.
Then
there exist prime integers pi,... ,pr and positive integers n^ such that
there exist positive integers 1 < d\ |
G=
aiZ
Further, these decompositions
are
| ds
such that
\G\ d\
=
\G\
=
ds and
0---0-T-ZT
dslj
uniquely determined by G.
The first form
Corollary 6.5,
so
is nothing but a more explicit version of the statement of
already been proven. We will explain how to obtain the second
first. The uniqueness statement29 is left to the reader (Exercise 6.1).
it has
form from the
The prime powers appearing in the first form of Theorem 6.6 are called the
elementary divisors of G; the integers d{ appearing in the second form are called
invariant factors. To go from elementary divisors to invariant factors, collect the
Of
permutation
course the
'uniqueness'
statement
only holds up
to trivial
of the factors. The claim is that the factors themselves
that two direct sums of either form given in the statement
match.
are
are
manipulation such
as a
determined by G, in the
sense
isomorphic only if their factors
IV. Groups, second encounter
238
elementary divisors in a table, listing (for example) prime powers according to
increasing primes in the horizontal direction and decreasing exponents in the vertical
direction; then the invariant factors are obtained as products of the factors in each
row:
dr
=
dr-l
=
dr-2
=
IpT1
pT
pT
\p112
pT
pT
\P113
pT
pT
Conversely, given the invariant factors d{, obtain the rows of this table by
factoring d{ into prime powers: the condition d\ |
| dr guarantees that these will
be decreasing.
Repeated applications of Lemma 6.1 show that if d
primes pi and positive rti (as is the case in each row of the
=
z
z
^
for
p™1 -p™r
table), then
distinct
z
~dZ~p^Z^'"^p^Z'
proving that the
two
decompositions given
This will likely be much clearer
Example 6.7. Here
29160
=
23
36
are
once
in Theorem 6.6
indeed equivalent.
are
the reader works through
the two decompositions for
a
(random)
2Z
and here is the
few examples.
group of order
5:
~
I
a
2Z
3Z
3Z
32Z
corresponding table of
5Zy/
32Z
invariant
90
=
2
32
18
=
2
32
6
=
2
3
3
=
~
6Z
\3Z
~
18Z
factors/elementary
90Z
divisors:
5
3
6.8. There are exactly 6 isomorphism classes of abelian groups of
23 32 5; the six possible tables of elementary divisors
Indeed, 360
are shown below. In terms of invariant factors, the six distinct abelian groups of
order 360 (up to isomorphism, by the uniqueness part of Theorem 6.6) are therefore
Example
order 360.
=
z
z
360Z'
z
z
3Z0
120Z'
z
2Z0
180Z'
z
z
6Z06OZ'
z
z
z
2Z02Z09OZ'
z
2Z
z
0
6Z
z
0
30Z'
6. Finite abelian groups
23
360=
32
23
120=
239
3=
3
5
180=
22
2=
2
5
3
32
60=
22
3
6=
2
3
5
5
90=
2
2=
2
2=
2
32
5
5
30=
2
3
6=
2
3
2=
2
6.3. Application: Finite subgroups of multiplicative groups of fields. Any
classification theorem is useful in that it potentially reduces the proof of general
facts to explicit verifications. Here is
one
example illustrating this strategy:
a finite abelian group, and assume that for every integer
elements g G G such that ng
0 is at most n. Then G is cyclic.
Lemma 6.9. Let G be
the number
of
n
=
The reader should try to prove this 'by hand', to appreciate the fact that
it is not entirely trivial. It does become essentially immediate once we take the
classification of finite abelian groups into account. Indeed, by Theorem 6.6
Z
Z
G-(^)0'"e(4)
for
some
positive integers
for all g G G
Therefore 5
=
(so
set F* of
s >
1, then |G|
>
ds), contradicting
ds and dsg
to a
nonzero
elements of
particularly
nice
proof of the following important
back in Example II.4.6. Recall that the
commutative group under multiplication.
a
field F is
a
=
f(a)
=
n
linear factors, this shows31 that if
n distinct elements a G F.
f(x)
G
F[x]
-
a)
can
has degree n, then
0 for at most
Theorem 6.10. Let F be
group
0
we ran across
Also recall (Example III.4.7) that a polynomial f(x) G F[x] is divisible by (x
if and only if /(a)
0; since a nonzero polynomial of degree n over a field
have at most
=
the hypothesis.
1; that is, G is cyclic.
weak form of which30
a
But if
that the order of g divides
Lemma 6.9 is the key
fact,
| ds.
1 < d\ |
(F, •). Then
afield,
and let G be
a
finite subgroup of the
multiplicative
G is cyclic.
Proof. By the considerations preceding the statement, for every n there are at
1
most n elements a G F such that an
0, that is, at most n elements a G G
-
such that an
=
1. Lemma 6.9
=
implies then that G
is
cyclic.
The diligent reader has proved that particular case in Exercise II.4.11. The proof hinted
at in that exercise upgrades easily to the general case presented here. My point is not that the
classification theorem is necessary in order to prove statements such as Theorem 6.10; our point
is that it makes such statements nearly evident.
Unique factorization in
F[x]
formally later; cf. Lemma V.5.1.
is secretly needed here.
We will deal with this issue more
IV. Groups, second encounter
240
As
a
(very) particular
is the fact
to in
multiplicative
Example II.4.6.
case, the
((Z/pZ)*, •)is cyclic: this
group
pointed
Preview of coming attractions: Finitely generated (as opposed to just finite)
abelian groups are also direct sums of cyclic groups. The only difference between
the classification of finitely generated abelian groups and the classification of finite
abelian groups explored here is the possible presence of a 'free' factor Z0r in the
decomposition. The reader will first prove this fact in Exercise VI.2.19, as a
consequence of 'Gaussian elimination over integral domains', and then recover it again
as
a
over
particular case of the classification theorem for finitely generated modules
Neither Gaussian elimination nor the very general
PIDs, Theorem VI.5.6.
harder to prove than the particular case of finite abelian
worked
out by hand in this section—a
common benefit of finding
groups laboriously
the right general point of view is that, as a rule, proofs simplify. Technical work
Theorem VI.5.6
are any
such
as that performed in order to prove Lemma 6.3 is absorbed into the work
necessary to build up the more general apparatus; the need for such technicalities
evaporates
in the process.
Exercises
6.1. > Prove that the decomposition of a finite abelian group G as a direct sum
of cyclic p-groups is unique. (Hint: The prime factorization of |G| determines the
primes, so it suffices to show that if
z
z
®.
prVL
with ri >
G defined by g
-?
6.2.
a
Prove that Z(G)
Classify
6.5. Let p be
of abelian
6.6.
/T\
z
C^
/T\
z
.
.
.
/T\
pSlZ
> sn, then
m
£>S-Z
and r^
Si for all i. Do this by
the image of the homomorphism
n
=
group pG obtained
=
as
pg.) [§6.2, §VI.5.3, VI.5.12]
Complete the classification of
6.3. Let G be
6.4.
^
.
pTrnZ
> rm and si >
induction, by considering the
G
.
groups of order 8
(cf.
Exercise
noncommutative group of order p3, where p is
Z/pZ and G/Z(G) Z/pZ x Z/pZ.
=
2.16).
a
prime integer.
=
abelian groups of order 400.
a
prime integer. Prove that the number of distinct isomorphism classes
pr equals the number of partitions of the integer r.
groups of order
> How many abelian groups of order 1024
are
there,
up to
isomorphism?
[§11.6.3]
6.7.
p
:
G
-i
Let p > 0 be a prime integer, G a finite abelian group, and denote
G the homomorphism defined by p(g)
pg.
Let A be a finite abelian group such that
pA
Z/pZ.
Prove that pkerp and
by
=
p(cokerp)
are
both 0.
=
0. Prove that A
=
Z/pZ 0
0
241
Exercises
Prove that ker p
=
coker p.
Prove that every subgroup of G of order p is contained in ker p and that every
subgroup of G of index p contains im p.
Prove that the number of subgroups of G of order p equals the number of
subgroups of G of index p.
[6.8]
6.8.
Let G be
-i
ri2 >
(mi
)•
a
finite abelian p-group, with elementary divisors pni,..., pUr {n\ >
a subgroup H with invariant divisors pmi,... ,pm,s
Prove that G has
and only if s < r and rrii < rti for i
1,..., s. (Hint: One
For the other, with notation as in Exercise 6.7, compare
H and G to establish s < r; this also proves the statement if all ni
1.
> ?7i2 >
•)if
=
direction is immediate.
kerp for
For the
=
general
case use
induction, noting that if G
=
0^Z/pniZ,
then
p(G)
=
e.z/p^z.)
Prove that the
description holds for the homomorphic images of G. [6.9]
same
a subgroup of a finite abelian group G. Prove that G contains a
isomorphic to G/H. (Reduce to the case of p-groups; then use Exercise 6.8.)
6.9. Let H be
subgroup
Show that both hypotheses 'finite' and 'abelian'
has a unique subgroup of order 2.)
are
needed for this result.
(Hint:
Qs
6.10. The dual of a finite group G is the abelian group Gv
where C* is the multiplicative group of C.
Prove that the image of every
1 for
of polynomials xn
Prove that if G is
a
cyclic groups; then
a G
HomGrp(G,C*),
:=
Gv consists of roots of I in C, that is, roots
some n.
finite abelian group, then G
Gv. (Hint: First prove this for
use the classification theorem to generalize to the arbitrary
=
case.)
In Example VIII.6.5
6.11.
we
will encounter another notion of 'dual' of
Use the classification theorem for finite abelian groups
classify
all finite modules
over
the ring
6.13.
G, H, K be finite abelian
(Theorem 6.6)
to
Z/nZ.
Prove that if p is prime, all finite modules
6.12. Let
a group.
over
Z/pZ
are
groups such that G 0 H
=
free32.
G 0 K. Prove that
Let G, H be finite abelian groups such that, for all positive integers n, G
H. (Note:
and H have the same number of elements of order n. Prove that G
-i
=
The 'abelian' hypothesis is
necessary! C4
x
C4 and Qs
x
C2
are
nonisomorphic
groups both with 1 element of order 1, 3 elements of order 2, and 12 elements of
order
4.) [§11.4.3]
6.14. Let G be
a
order p. Prove that
You are welcome
finite abelian p-group, and assume G has only
cyclic. (This is in some sense a converse to
G is
to try to prove it 'by hand', but
will simplify the argument considerably.)
'As
we will see in
use
subgroup of
Proposition II.6.11.
one
of the classification theorem
Proposition VI.4.10, this property characterizes fields.
242
6.15.
IV.
Groups, second
encounter
Let G be a finite abelian group, and let a G G be an element of maximal
order in G.
reproduces
Prove that the order of every b G G divides
the result of Exercise
6.16. Let G be
an
\a\. (This essentially
II.1.15.)
abelian group of order n, and assume that G has at most
n. Prove that G is cyclic.
subgroup of order d for all d
one
Chapter
Irreducibility
V
and
factorization in
integral
domains
We
attention back to rings and analyze several useful classes of
One guiding theme in this chapter is the issue of factorization:
move our
domains.
integral
we
will
address the problem of existence and uniqueness of factorizations of elements in
or k[x] (for k
ring, abstracting good factorization properties of rings such as Z
field) to whole classes of integral domains. The reader may want
following picture
a
a
to associate the
with the first part of this chapter:
Integral domains
Blanket assumption: all rings considered in this chapter will be commutative1. In
fact, most of the special classes of rings we will consider will be integral domains,
Also,
recall that all our rings have 1; cf. Definition III.l.l.
243
244
that is,
Definition
1.
Irreducibility and factorization
V.
commutative rings with 1 and with
no
nonzero
integral domains
in
zero-divisors
(cf.
III.1.10).
Chain conditions and existence of factorizations
1.1. Noetherian rings revisited. Let R be a commutative ring. Recall that R
is said to be Noetherian if every ideal of R is finitely generated (Definition III.4.2).
In fact, this is a special case of the corresponding definition for modules: a
over a ring R is Noetherian if every submodule of M is finitely generated
(Definition III.6.6). In §111.6.4 we have verified that this condition is preserved
module M
through exact sequences: if M, N, P
0
is
an exact sequence
and P
are
Noetherian
fact is that every
R-modules and
are
>N
>M
>P
>0
of R-modules, then M is Noetherian if and only if both N
(Proposition III.6.7). An easy and useful consequence of this
finitely generated
module
over a
Noetherian ring is Noetherian
(Corollary III.6.8).
The Noetherian condition may be expressed in alternative ways, and it is useful
some familiarity with them.
to acquire
Proposition 1.1. Let R be
the following are equivalent:
(1) M
is
a commutative
Noetherian; that is,
(2) Every ascending
chain
every submodule
of submodules of M
Ni
is a chain
ring, and let M be
C
N2
of submodules of M,
(3) Every nonempty family of
C
N3
of M
is
of
R-module. Then
finitely generated.
stabilizes; that is,
if
C
then 3i such that Ni
submodules
an
M has
a
=
Ni+i
=
Ni+2
maximal element w.r.t.
inclusion.
The second condition listed here is called the ascending chain condition (a.c.c.)
for submodules. For M
R, Proposition 1.1 tells us (among other things) that a
=
ring is Noetherian if and only if the ascending chain condition holds for its ideals.
Proof. (1)
=>
(2):
Assume that M is Noetherian, and let
#1 C JV2 C JV3 C
be
a
chain of submodules of M. Consider the union
N
=
\jNi:
i
the reader will
Since M is Noetherian, N is
N => Uk
Now Uk
Ni for some i\
,nr are contained
by picking the largest such i, we see that 3i such that all ni,
in Ni. But then N C Ni, and since Ni C A^+i C
are all contained in N
Ni,
verify that
finitely generated, say N
N is
=
a
submodule of M.
(ni,... ,nr).
=
it follows that Ni
(2)
=>
=
Ni+i
=
Ni+2
=
...
as
(3): Arguing contrapositively,
submodules that does not have
a
needed.
assume
that M admits
maximal element. Construct
an
a
family & of
infinite ascending
1. Chain conditions and existence of factorizations
245
&\since N\ is not maximal in ^", there
element N2 of & such that N\ C JV2; since N$ is not maximal in ^", there
element N$ of & such that N2 C JV3; etc. The chain
chain as follows: let N\ be any element of
exists
an
exists
an
iViCJV2CJV3C...
does not stabilize, showing that
(3)
Assume
(1):
=>
(2) does not
(3) holds, and let
hold.
iV be
a
submodule of M.
Then the
family & of finitely generated subsets of N is nonempty (as (0) G &)\hence it has
N:
a maximal element N'. Say that N'
(m,... ,nr). Now we claim that N'
indeed, let n G iV; the submodule (ni,..., nr, n) is finitely generated, and therefore
it is in 3?\ as it contains Nf and Nf is maximal, necessarily (ni,..., nr, n)
iV; in
n
as
needed.
G
iV',
particular
=
=
=
This shows that N
this implies that
=
N' is finitely generated, and since N C M
arbitrary,
was
M is Noetherian.
Noetherian rings are a very useful and flexible class of rings. In §111.6.5 we
mentioned the important fact that every finite-type algebra over a Noetherian ring is
Noetherian. 'Finite-type (commutative) algebra' is just a fancy name for a quotient
of
a
polynomial ring (§111.6.5),
Theorem 1.2. Let R be
ring
R[x\,...,xn].
generated
Noetherian ring, and let J be an ideal
Then the ring R[x\,...,xn]/J is Noetherian.
modules
a
over
them to be Noetherian
as
this is what the fact states:
finite-type R algebras
Note that
as
so
rings (that is,
as
as
of the
polynomial
(in general) very far from being finitely
(cf. again §111.6.5), so it would be foolish to expect
R
are
R-modules. The fact that they turn out to be Noetherian
over themselves) provides us with a huge class of
modules
examples of Noetherian rings, among which are the rings of (classical) algebraic
geometry and number theory. Thus, entire fields of mathematics are a little more
manageable thanks to Theorem 1.2.
The proof of this deep fact is surprisingly easy. By Exercise 1.1, it suffices to
prove that
R Noetherian =>
and
an
R[x\,...,xn] Noetherian;
immediate induction reduces the statement to the
which carries a
Lemma 1.3
distinguished
(Hilbert's
basis
theorem).
R Noetherian =>
Proof. Assume R is Noetherian, and let I be
that I is finitely generated.
Recall that if
called the leading
A
=
{0}
f(x)
=
ad,xd
+
ad-\xd~l -\
coefficient of f(x). Consider
U
It is clear that A is
{a
an
generated. Thus there
G R
I
a
is
a
an
ideal of
h ao G
the
R[x]
R[x],
R[x]
following
leading coefficient of
ideal of R
exist
following particular
case,
name:
an
Noetherian.
We have to prove
and ad
^ 0,
then ad is
subset of R:
element of
/}.
(Exercise 1.6); since R is Noetherian, A is finitely
elements f\{x),...,fr{x) G / whose leading coefficients
ai,..., ar generate A as an ideal of R.
Irreducibility and factorization
V.
246
Now let d{ be the degree of fi(x), and let d be the
degrees. Consider the sub-i?-module
M
in
integral domains
maximum among these
(l,x,x2,...,xd-1) QR[x],
=
that is, the R-module consisting of polynomials of degree < d. Since M is finitely
generated as a module over R, it is Noetherian as an R-module (by Corollary III.6.8).
Therefore,
the submodule
MnJ
of M is finitely generated
i?,
over
say
by gi(x),... ,gs(x)
G I.
Claim 1.4.
I
=
(/lW,
.
.
.
,
fr(x),9l(x),
,9s(x))'
This claim implies the statement of the theorem. To prove the claim,
need
in /.
to prove the C
If
>
dega(x)
3bi,..., br
we only
inclusion; to this end, let a(x) G I be an arbitrary polynomial
d, let a be the leading coefficient of a(x). Then a G A, so
G R such that
a
Letting
e
=
dega(x),
so
has degree <
-
G
R[x]
and
we are
a(x)
=
-
hx^^Mx)
brxe-drfr(x)
-
pi(x)fi(x)
pi(x)fi(x)
pr(x)fr(x)
=
cigi(x)
+
+
+
pr(x)fr(x)
+
csgs(x)
+
csgs(x),
+
Cigi(x)
+
(A(x),..., fr{x),9i{x),.. .,98{x)),
the proof of Claim 1.4, hence of Lemma 1.3, hence of Theorem 1.2.
a,b E R. We say that a divides 6,
of a, if b G (a), that is,
or
that
is
a
b
(3c eR),
the notation
a
Two elements a, b
are associates
if
a
=
a
(commutative) ring,
divisor of 6,
or
that b is
a
and let
multiple
ac.
b.
are associates
Lemma 1.5. Let a, b be
b
polynomials
done, since this verifies that
1.2. Prime and irreducible elements. Let R be
use
finite list of
pr(x)fr(x)
G
We
a
But this places this element in M fi /; therefore 3c\,...,cs G R
pi(x)fi(x)
completing
obtain
we
such that
a(x)
a(x)
+ brar.
Iterating this procedure,
e.
has degree < d.
such that
+
b\a\
that e> di for all i, this says that
a(x)
Pi(x),... ,Pr(x)
=
nonzero
and only
if
a
=
if
(a)
=
elements
ub, for
u
(6),
of
a
an
that is, if
a
b and b
a.
integral domain R. Then
unit in R.
a
and
1. Chain conditions and existence of factorizations
3c, d
Proof. Assume a and b are associates. Then
b
therefore
a
=
bd
=
=
Since cancellation
is
c
The
a
unit,
converse
G R such that
bd;
=
acd, i.e.,
o(l
Thus
a
ac,
247
by
as
cd)
-
=
0.
elements hold in integral domains, this implies cd
nonzero
=
1.
needed.
is left to the reader.
Incidentally, here the reader sees why it is convenient to restrict our attention
integral domains: for this argument to work, a must be a non-zero-divisor.
For example, in Z/6Z the classes [2]6, [4]6 of 2 and 4 are associates according to
our definition, yet [4]6 cannot be written as [2]6 times a unit.
Away from the
comfortable environment of integral domains, even harmless-looking statements
to
such as Lemma 1.5 may fail.
generalize directly the corresponding notions in Z.
explore
going
analogs of other common notions in Z, such as 'primality'
and 'irreducibility', in more general integral domains.
The notions reviewed above
We
to
are
Definition
1.6. Let R be
An element
a G
R is prime if the ideal
(a)
is prime; that is,
a
is not
a
unit
(cf. Proposition III.4.11)
and
An element
a G
a
are
=
be =>
is
(b
b
a\c).
or
a
is not
a
unit
unit and
a
or c
is
a
unit).
useful alternative ways to think about the notion of 'irreducible':
a
=
be implies that
a
a
=
bc implies that
(a)
is
an
=
associate of b
(b)
or
(a)
C
(a)
is maximal among proper
=>
(b)
=
(a)
a
only if
is irreducible if and
(b)
(a |
R is irreducible if
a
There
be =>
|
a
nonunit
integral domain.
an
or
(b)
(a)
=
=
or
of c;
(c) (Lemma 1.5);
(1) (Exercise 1.12);
principal ideals (just rephrasing the previous
point!).
is important to realize that primality and irreducibility are not equivalent;
is somewhat counterintuitive since they are equivalent in Z, as the reader
It
this
should
verify2 (Exercise 1.13).
What is true in general is that prime is stronger
than irreducible:
Lemma 1.7. Let R be
element. Then
a is
an
integral domain, and let
This fact will be fully explained by the general theory,
right away.
a
G
R be
a
nonzero
prime
irreducible.
so
the reader should work this out
V.
248
Irreducibility and factorization
in
integral domains
Proof. Since (a) is prime, (a) ^ (1); hence a is not a unit. If a
6c, then
be
a G (a); therefore 6 G (a) or c G (a) since (a) is prime.
Assuming without
=
=
loss of generality b G
(a) C (6): hence (a)
=
We will
soon see
(a),
(6),
we
have
that is,
C
(6)
a
and 6
(a).
are
On the other hand
associates,
under what circumstances the
as
a
be implies
=
needed.
converse statement
holds.
1.3. Factorization into irreducibles; domains with factorizations.
Definition 1.8. Let R be
such that
r
=
qi
an
integral domain. An element
r G
R has
a
factorization
into irreducibles if there exist irreducible elements <7i,...,#n
(or decomposition)
-
qn.
This factorization is unique if the elements qi are determined
by
r up to
order
and associates, that is, if whenever
is another factorization of
qi after
r
(possibly) shuffling
into irreducibles, then
the factors.
m
=
n
and
q[
is
an
associate of
j
Definition 1.9. An integral domain R is a domain with factorizations (or
'factorizations exist in R') if every nonzero, nonunit element r G R has a factorization
into irreducibles.
j
Definition 1.10.
An integral domain R is
domain (abbreviated UFD), if every
factorization into irreducibles.
factorial,
or
a
unique factorization
unique
nonzero, nonunit element r G R has a
j
The terminology introduced in Definition 1.9 does not appear to be too
standard; by contrast, UFDs are famous.
We will study the unique factorization condition in §2. For now, it seems
worth spending a little time contemplating the mere existence of factorizations.
Interestingly, this condition is implied by an ascending chain condition, for a special
class of ideals.
Proposition 1.11. Let R be an integral domain, and let r be a nonzero,
element of R. Assume that every ascending chain of principal ideals
nonunit
(r)C(n)C(r2)C(r3)C...
stabilizes.
Then
r
has
Proof. Assume that
particular,
(r)
C
(n),
r
a
r
factorization
does not have
a
factorization into irreducible elements. In
is itself not irreducible; thus 3r*i,si G R such that r
riSi and
C
If
both
have
factorizations
into
si
r*i,
irreducibles, the
(r)
($i).
=
product of these factorizations gives
(e.g.)
into irreducibles.
a
factorization of r; thus
we may assume
that
r*i does not have a factorization into irreducibles. Therefore we have
(r)
and r*i does not have
a
c
(n)
factorization; iterating this argument
increasing chain
(r)C(n)C(r2)C(r3)C...,
constructs an
infinitely
Exercises
249
contradicting
hypothesis.
our
Thus, factorizations exist in integral domains in which the ascending chain
condition holds for principal ideals. This has the following immediate and convenient
consequence:
Corollary
Proof.
1.12. Let R be a Noetherian domain.
Then
factorizations
exist in R.
By Proposition 1.1, Noetherian domains satisfy the ascending chain
condition for all ideals.
Corollary 1.12 verifies part of the picture presented at
chapter: the class of Noetherian domains is contained in the
factorization.
This inclusion is proper:
beginning of the
the
class of domains with
for instance, the standard
example of
a
non-Noetherian ring,
Z[xi,x2,x3,...],
does have factorizations.
(Example III.6.5)
Z[#i,£2,...] involves
Indeed, every given polynomial / G
so it belongs to a subring
variables3,
only finitely many
Z[#i,
,xn] isomorphic to an ordinary polynomial ring over Z; further, this subring contains every divisor of /. It follows easily (cf. Exercise 1.15) that the
ascending chain condition for principal ideals holds in Z[x\,X2,xs,...], because it holds
in Z[xi,... ,xn] (since this ring is Noetherian, by Hilbert's basis theorem).
Exercises
Remember that in this section all rings
1.1.
>
Let R be
Noetherian ring.
a
1.2.
Prove that if
theorem.)
1.3. Let A; be
/,
a
define
mapping t to
Prove that
/.
taken to be commutative.
Noetherian ring, and let / be
an
ideal of R. Prove that
R[x]
is Noetherian,
so
is R.
is
a
(This
is
a
'converse' to Hilbert's
field, and let / G k[x], f 0 k. For every subring R of k[x] containing
a homomorphism ip :
R by extending the identity on k and
k[t]
This makes every such R
k[x]
is
finitely generated
a
as a
k[^-algebra (Example III.5.6).
k[^-module.
Prove that every subring R
as
Prove that every subring of
k[x] containing
above is finitely generated
A; is
a
1.5. Determine for which sets S the power set ring
as a
k [^-module.
Noetherian ring.
1.4. Let R be the ring of real-valued continuous functions
Prove that R is not Noetherian.
Exercise
R/I
[§1.1]
basis
k and
are
£?(S)
on
the interval
is Noetherian.
(Cf.
III.3.16.)
'Remember that polynomials
are
[0,1].
finite linear combinations of monomials; cf. §111.1.3.
V.
250
1.6. > Let / be
an
ideal of
R[x],
Theorem 1.2. Prove that A is
1.7.
Prove that if R is
(cf. §111.1.3)
a
Irreducibility and factorization
an
in
integral domains
and let A C R be the set defined in the proof of
ideal of R.
[§1.1]
Noetherian ring, then the ring of power series
(Hint: The order of a power series Yl^o aiX%
is also Noetherian.
R[[x]]
1S ^ne
smallest i for which a^ ^ 0; the dominant coefficient is then a^. Let Ai C Rbe the
set of dominant coefficients of series of order i in /, together with 0. Prove that
•.This
an ideal of R and Aq C Ai C A<i C
sequence stabilizes since i? is
Noetherian, and each Ai is finitely generated for the same reason. Now adapt the
proof of Lemma 1.3.)
Ai is
1.8. Prove that every ideal in
ideals.
(Hint:
Let & be the
a
Noetherian ring R contains a finite product of prime
ideals that do not contain finite products of
family of
prime ideals. If & is nonempty, it has a maximal element M since R is Noetherian.
Since M G &, M is not itself prime, so 3a, b G R s.t. a 0 M, b 0 M, yet ab G M.
What's wrong with this?)
1.9.
Let R be a commutative ring, and let / C R be a proper ideal. The reader
will prove in Exercise 3.12 that the set of prime ideals containing / has minimal
elements (the minimal primes of I). Prove that if R is Noetherian, then the set
-i
of minimal primes of I is finite. (Hint: Let & be the family of ideals that do not
have finitely many minimal primes. If 3? ^ 0, note that 3? must have a maximal
/, and / is not prime itself. Find ideals Ji, J2 strictly larger than /, such
that J1J2 C /, and deduce a contradiction.) [VIA 10]
element
1.10.
-1
By Proposition 1.1,
a
ring R is Noetherian if and only if it satisfies the
for ideals. A ring is Artinian if it satisfies the d.c.c. (descending chain
condition) for ideals. Prove that if R is Artinian and / C R is an ideal, then R/I
a.c.c.
Artinian. Prove that if R is
Let
for
r G
i?,
some n.
is, prime
r
^
Artinian integral domain, then it is
0. The ideals (rn) form a descending sequence; hence
Therefore
ideals
are
)
an
1.12. > Let R be an
(a)
(rn)
=
is
(Hint:
(rn+1)
(that
rings4). [2.11]
equivalence relation.
integral domain. Prove that a G R is irreducible if and only
principal ideals of R. [§1.2, §2.3]
is maximal among proper
1.13. > Prove that prime <^=> irreducible in Z.
1.14. For a, b in a commutative
if and
field.
Prove that Artinian rings have Krull dimension 0
maximal in Artinian
1.11. Prove that the 'associate' relation is an
if
a
only if the class of b
in
ring i?,
R/(a)
[§1.2, §2.3]
prove that the class of a in
R/(b)
is prime
is prime.
1.15. > Identify S
Z[#i,... ,xn] in the natural way with a subring of the
polynomial ring in countably infinitely many variables R
rL[xi,X2,x$,...]. Prove
that if / G S and (/) C (g) in i?, then g G S as well. Conclude that the
ascending chain condition for principal ideals holds in i?, and hence R is a domain with
=
=
factorizations.
[§1.3, §4.3]
One can prove that Artinian rings are necessarily Noetherian; in fact, a ring is Artinian if
and only if it is Noetherian and has Krull dimension 0. Thus, the d.c.c. implies the a.c.c, while
the a.c.c. implies the d.c.c. if and only if all prime ideals are maximal.
2.
UFDs, PIDs, Euclidean domains
251
1.16. Let
Z[xi,x2,x3,...}
(xi -xi,x2-xi,...y
Does the ascending chain condition for principal ideals hold in Rl
1.17.
>
Consider the subring of C:
Z[V^5]
:=
{a
Prove that this ring is isomorphic to
Prove that it is
Define
a
N(zw)
a
a, b G
Z[t]/(t2
5).
Z}.
Noetherian integral domain.
'norm' N
on
Z[v^5] by setting N(a
N(z)N(w). (Cf.
=
triVb |
+
Exercise
+
biy/5)
=
a2
562. Prove that
+
III.4.10.)
are ±1. (Use the preceding point.)
Z[-\/—5]
i\/h, 1 iV§ are all irreducible nonassociate
Prove that the units in
Prove that 2, 3, 1 +
elements of
Z[>/=5).
element listed in the preceding point is prime. (Prove that the
rings obtained by mod-ing out the ideals generated by these elements are not
Prove that
no
integral domains.)
Prove that Z[v^5]
is not
a
UFD.
[§2.2, 2.18, 6.14]
2.
UFDs, PIDs, Euclidean domains
2.1.
R
is
integral domain
Irreducible factors and greatest common divisor. An
a
UFD if factorizations exist in R and
Thus, in
a
UFD all elements
(other
unique in the
are
than 0 and the
sense
units)
of Definition 1.8.
determine
a
multiset
multiplicity'; cf. §1.1.1) of irreducible factors, determined
(a
to
the
associate
relation.
We can also agree that units have no factors; that is,
up
the corresponding multiset is 0.
set of elements 'with
The
following
UFDs, such
as
trivial remark is at the root of most elementary facts about
the characterization of Theorem 2.5:
Lemma 2.1. Let R be
(a)
Q
(b)
multiset
of
<^=>
a
UFD, and let a,b,c be
elements
of irreducible factors of
factors of a;
the multiset
irreducible
a and b are associates
(that is, (a)
=
the irreducible factors of a product be
of b and of c.
a
nonzero
(b))
are
<^=>
of R.
b is contained in the
the two multisets
the collection
Then
of all
coincide;
irreducible
factors
The proof is left to the reader (Exercise 2.1). The advantage of working in
UFD resides in the fact that ring-theoretic statements about elements of the
ring often reduce to straightforward set-theoretic statements about multisets of
irreducible elements, by means of Lemma 2.1.
V.
252
One important
divisors.
Irreducibility and factorization
in
integral domains
instance of this mechanism is the existence of greatest common
(at least for integers) in previous
We have liberally used this notion
chapters;
appreciate it from
now we can
Definition 2.2. Let R be
is a
greatest
(d)
is the smallest
common
an
divisor
a
technical perspective.
integral domain, and let a,b
abbreviated
(often
principal ideal
In other words, d is
a more
gcd of
c
a
e R. An element d G R
and b if
(a, b)
C
(d)
and
in R with this property.
a
|
'gcd')
of
and b if d
a, c
|
|
a, d
b =>
c
|
| 6,
j
and
d.
This definition is immediately extended to any finite number of elements.
Note that greatest
defined uniquely by this
and 6, so is every associate of d. Thus,
the notation 'gcd(a,6)' should only be used for the associate class formed by all
greatest common divisors of a, b. Of course, language is often (harmlessly) abused
on this point. For example, the fact that we can talk about the greatest common
prescription: if d is
a
greatest
divisors
common
common
are not
divisor of
a
divisor of two integers is due to the fact that in Z there is
distinguished element in each class
one).
Also note that greatest common
a
of associate
a
convenient way to choose
integers (that is, the nonnegative
divisors need not exist
(cf.
Exercise
2.5);
but
they do exist in UFDs:
UFD, and let
Lemma 2.3. Let R be
a
have
divisor.
greatest
a
Proof. We
where
u
and
can
common
a, b be nonzero elements
of R.
Then a, b
write
v are
units, the elements
<& are
irreducible,
<& is not an associate of qj
for i ^ j, and c^ > 0, /% > 0 (so that the multisets of irreducible factors of a,
resp., 6, consist of those qi for which ai > 0, resp., Pi > 0; the units u, v are
included since the irreducible factors are only defined up to the associate relation).
We claim that
A
a
_
min(ai,/3i)
ql
-
-
-
rnin(ar,/3r)
qr
gcd of a and b. Indeed, d is clearly a divisor of a and 6; and if c also divides
and 6, then the multiset of factors of c must be contained in both multisets of
factors for a and b (by Lemma 2.1); that is,
is a
a
c
with
w a
(again by
=
unit and ji < c^, ji < Pi.
Lemma 2.1), as needed.
wql1
q1/
This implies
ji <
min(a^,/^),
and hence
c
d
course the argument given in the proof generalizes one of the standard
to
compute greatest common divisors in Z: find smallest exponents in prime
ways
factorizations. But note that this is not the only way to compute the gcd in Z;
In fact, greatest common divisors
we will come back to this point in a moment.
Of
in Z have properties that should not be expected in
a more
general domain: for
UFDs, PIDs, Euclidean domains
2.
253
example, the result of Exercise II.2.13 does not generalize to arbitrary domains
(and not even to arbitrary UFDs), as the reader will check in Exercise 2.4.
2.2. Characterization of UFDs. It is easy to construct integral domains where
The diligent reader has already analyzed one example
unique factorization fails.
(Exercise 1.17); for another,
in the domain5
C[x,y,z,w]
(xw yz)
the
(classes
of
another; since
the)
xw
into irreducibles:
elements x, y, z, w are irreducible and not associates of one
xw has two distinct factorizations
0 in i?, the element r
yz
r
Note that this
factorizations do exist in
=
=
=
xw
=
yz.
ring is Noetherian, by Theorem
R). Thus,
there
1.2
(and
in
particular
Noetherian integral domains that
are
are not
UFDs.
Also note that this ring provides an example in which the converse to Lemma 1.7
(the class of) x is irreducible, but the quotient
does not hold: indeed,
fC[x,y,z,w] //A
(xw-yz) J
)
is not
is,
an
integral domain (because
^
y
C[x,y,z,w]
(x,xw-yz)
^ 0,
z
^ 0,
=
C[x,y,z,w]
(x,yz)
and yet yz
=
0 in this
ring);
that
is not prime.
x
fact, and maybe
a little surprisingly, the issue of unique factorization is
linked
with
the
relation between primality and irreducibility. Indeed,
inextricably
the 'converse' to Lemma 1.7 does hold in UFDs:
In
Lemma 2.4. Let R be
is
a
UFD, and let
be
a
an
irreducible element
of R.
Then
a
prime.
Proof. The element
a
is not
a
unit, by definition of irreducible. Assume be
G
(a):
and by Lemma 2.1 the irreducible factors of a, that is, a itself,
must be among the factors of b or of c. We have b G (a) in the first case and c G (a)
in the second. This shows that (a) is a prime ideal, as needed.
thus
(be)
C
In fact,
(a),
more
is true. Provided that the
ideals holds, then UFDs
are
ascending chain condition for principal
characterized by the equivalence between irreducibility
and primality.
Theorem 2.5. An integral domain R is
the
for principal
a. c. c.
(
=>
an
)
are
of R
Assume that R is
prime. To
ascending chain
elements of R
UFD if and only if
ideals holds in R and
every irreducible element
Proof.
a
a
is prime.
UFD. Lemma 2.4 shows that irreducible
prove that the a.c.c. for
principal ideals holds, consider
(ri)C(r2)C(r3)C....
the
In algebraic geometry, this is the ring of a 'quadric cone in A4'. The vertex of this cone
is a singular point, and this has to do with the fact that R is not a UFD.
origin)
(at
Irreducibility and factorization
V.
254
in
integral domains
By Lemma 2.1, this chain determines a corresponding descending chain of multisets
of irreducible factors. A descending chain of finite multisets clearly stabilizes, and it
follows
(
<<=
ducibles
to
Lemma 2.1
(by
Now
)
assume
that
(r^)
=
(ri+i)
that R satisfies the
=
(ri+2)
a.c.c.
=
for large enough i.
for principal ideals and irreexist in i?; we have
implies that factorizations
1.11
prime. Proposition
are
Let #i,..., gm and
verify uniqueness.
and
again)
q[,..., q'n
be irreducible elements of i?,
assume
qi-'qrn=q'l'-q'nThen
for
q[
some
q'n
Therefore
q2 by uq'2,
(qi)
is
prime ideal (by hypothesis); thus q\
a
(qi)
we may assume to
=
u
we
find
qi'-qm
Repeating
G
be 1 after changing the order of the factors.
uq\ for some u G R. Since q[ is irreducible and q\ is not a unit,
is a unit. Thus qi and q[ are associates. Canceling qi and replacing
q[
necessarily
and
(#i),
G
i, which
=
q,2'-q'n'
this process matches all factors
because otherwise we will obtain 1
fact that irreducibles
=
n,
by one. It is clear that m
product of irreducibles, contradicting the
a
one
=
units.
are not
pleasant statement, but it does not make it particularly easy to
check whether a given ring is in fact a UFD. The situation is not unlike that with
Noetherian rings: the a.c.c. is a sharp equivalent formulation of the Noetherian
condition, but in practice one relies more often on other tools, such as Hilbert's
basis theorem, in order to establish that a given ring is Noetherian. Does an analog
of Hilbert's basis theorem hold for unique factorization domains?
Theorem 2.5 is
to
a
Answering this question requires
it in §4.
some
preparatory work, and
we
will
come
back
2.3. PID
=>
UFD. There are simple ways to produce examples of UFDs.
Recall (from §111.4) that a principal ideal domain (abbreviated PID) is an integral
domain in which every ideal is principal. In §111.4 we have observed that
PIDs
are
Z and
Noetherian.
k[x] (where
A; is
a
field)
are
PIDs.
If R is
a PID and a,b E R, then d is a greatest common divisor for a and b if
and only if (a, 6)
(d). In particular, if d is a greatest common divisor of a
and b in a PID i?, then d is a linear combination of a and b: 3r, s G R such
=
that d
=
ra-\-sb.
If / is a nonzero ideal in a
We
now
add
one
Proposition
PID, then / is prime if and only if it is maximal.
important point
2.6.
If R
is a
to this list:
PID, then
it is a UFD.
Noetherian, the a.c.c. holds in R (for all
ideals, hence in particular for principal ideals); we verify that irreducible elements
are prime in i?, which implies that R is a UFD, by Theorem 2.5.
Proof. Let R be a PID. Since PIDs are
2.
UFDs, PIDs, Euclidean domains
Let
a
G
R be
that either b
D
(a,6)
(a);
or
as
c
irreducible element, and
(a). If b G (a), we are
an
is in
i? is
This implies that
PID, 3d
a
(a, 6)
by irreducible elements
=
(d)
are
e R such that
(1),
=
because
a
assume
be G
done; thus,
(d)
(a);
(a,6),
=
sb
ra +
c
=
1.
rac +
=
b
have to show
0 (a).
and hence
(cf.
Then
(d) 2 (a)-
is irreducible and ideals
Multiplying by
sbc G
we
assume
maximal among principal ideals
Hence 3r,s G R such that
as
255
generated
Exercise
1.12).
c, we obtain that
(a)
needed.
In particular, k[x] is a UFD if A; is a field, and Z is a UFD—which
hopefully
will not come as a big surprise to our readers6. Note that at this point we have
recovered the fact stated in Exercise 1.13 in full glory.
Proposition 2.6 justifies another feature of the picture given at the beginning
of this chapter: the class of UFDs is contained in the class of PIDs. We will soon
see that the inclusion is proper, that is, that there are UFDs which are not PIDs;
for example, Z[x] is
Z[x]
as we
will
are not
soon
is
a
unique factorization domain
be able to prove
Noetherian,
discussed after
(Exercise 2.12), yet
not a PID
as
more
In fact, there
(Theorem 4.14).
represented
in the picture;
are
UFDs which
however, these examples
material has been developed
are
best
(Example 4.18).
already get feel for the gap between UFD and PID, by
the
fact
contemplating
(recalled above and essentially tautological) that, in a PID,
common
divisors
of a, b are linear combinations of a and b. This is a
greatest
The reader
can
a
strong requirement: for example, it characterizes PIDs among Noetherian
domains, as the reader will check (Exercise 2.7); it does not hold in general UFDs
very
(Exercise 2.4).
2.4.
A; is
properties of Zand k [x] (where
special than PIDs: they are Euclidean
Euclidean domain => PID. The excellent
a
field)
make these rings
even
more
domains.
Informally,
Euclidean domains
with remainder': this is the
case
are
rings
is that in both Z and
integer
n
and
in which
for Z and for
k[x],
k[x] one can define a notion
degf(x) for a polynomial f(x). In
the size of the 'remainder' in
a
as
one may
perform
§111.4.
a
'division
observed in
of 'size' of
an
The point
element: \n\
for an
both cases, one has control over
division. The definition of Euclidean domain simply
abstracts this mechanism.
For the purpose of this discussion7,
Z^°.
v : R
{0} -?
a
valuation
on an
integral domain R
is any
function
This fact is known as the fundamental theorem of arithmetic.
Entire libraries have been written on the subject of valuations, studying a more precise
notion than what is needed here.
Irreducibility and factorization
V.
256
Definition
2.7.
satisfying
A Euclidean valuation
following property8: for
the
on
all
integral domain R
an
R and all
a G
integral domains
in
is
valuation
a
b G R there exist
nonzero
q,r G R such that
a
with either r
admits
=
0
or
v(r)
qb
+ r,
An integral domain R is
v(b).
<
=
k[x]
The
r
is the remainder. Division
provide examples,
that Z
so
Euclidean domains.
are
Proposition 2.8. Let R be
and
Euclidean domain if it
j
We say that q is the quotient of the division and
with remainder in Z and in k[x] (where A; is a field)
and
a
Euclidean valuation.
a
Euclidean domain. Then R is
a
a
modeled after the instances encountered for Z
proof is
the reader has
k[x] (which
hopefully
PID.
(Proposition III.4.4)
III.4.4).
worked out in Exercise
Proof. Let I be
If I
an ideal of i?; we have to prove that I is principal.
{0},
there is nothing to show; therefore, assume / ^ {0}. The valuation maps the
nonzero elements of / to a subset of Z-°; let b E I be an element with the smallest
valuation. Then
C
(b)
clearly
/,
For this, let
=
we
we
claim that /
only need
a G
/ and
in
some g,r
i?, with
r
verify
(b).
division
with
remainder:
apply
by the minimality of v(b)
Therefore
r
=
=
0
or
Proposition 2.8 justifies
v(r)
<
r
a
=
v(b).
qb
a
=
qb
one more
(6),
G
is proper, as
which is not
More precisely,
either
(a, b)
=
a
as
domains
as
(b) (that is,
/,
picture
our
at the
in the picture.
v(r)
<
v(b).
beginning of the
principal ideal
Producing
can
an
explicit
easy, but the gap
in fact be described very sharply: PID
satisfying
a)
have
we cannot
needed.
Euclidean domain is not
or
have
G I :
a
so
weaker requirement than 'division
'Dedekind-Hasse valuation' is
b divides
we
is contained in the class of
suggested
a
between PIDs and Euclidean domains
may be characterized
with remainder'.
Since
But
feature of
chapter: the class of Euclidean domains
domains.
This inclusion
example of a PID
needed.
qb-\-r
=
among nonzero elements of
0, showing that
as
that I Q
a
for
therefore / is principal,
(6);
=
to
there exists
a
valuation
r G
(a, b)
v
such that Va, 6,
such that
v(r)
<
v(b).
This latter condition amounts to requiring that there exist q,s G R such that
as
bq + r with v(r) < v(b); hence a Euclidean valuation (for which we may in fact
=
choose
s
domain
=
is
1)
a
is
a
PID if and
For example, this
It is not
Dedekind-Hasse valuation. It is not hard to show that
can
only if it
admits
a
Dedekind-Hasse valuation
be used to show that the ring
uncommon to also
Z[(l
require that v(ab) > v(b) for all
needed in the considerations that follow, and cf. Exercise 2.15.
+
an
is a
\J—19)/2]
nonzero a,
b
integral
(Exercise 2.21).
PID: the
R; but this is not
UFDs, PIDs, Euclidean domains
2.
norm
257
considered in Exercise 2.18 in order to prove that this ring is not a Euclidean
a Dedekind-Hasse valuation9.
Thus, this ring gives an
domain turns out to be
example of
a
PID that is not
Euclidean domain.
a
giving them their
algorithm computing greatest common
the Euclidean algorithm. As Euclidean domains are PIDs, and hence UFDs,
One excellent feature of Euclidean domains, and the
one
names, is the presence of an effective
divisors:
know that they do have greatest common divisors. However, the 'algorithm'
obtained by distilling the proof of Lemma 2.3 is highly impractical: if we had to
factor two integers a, b in order to compute their gcd, this would make it essentially
we
impossible (with current technologies and factorization algorithms) for integers
a few hundred digits. The Euclidean algorithm bypasses the necessity of
of
factorization: greatest common divisors of thousand-digit integers may be computed in
fraction of a second.
a
The key lemma
which the algorithm is based is the
on
following
trivial general
fact:
Lemma 2.9. Let
a
=
=
a
Proof. Indeed,
r
proving (a, b)
(6,r).
C
ring R. Then (a, b)
+ r in a
bq
bq
G
(a, 6), proving (6,r)
C
=
(6, r).
and
(a, 6);
a
=
bq
+ r G
(b,r),
In particular,
(Vc£R),
M)C(c)
^
(6,r)C(c);
that is, the set of common divisors of a, b and the set of
coincide. Therefore,
Corollary 2.10. Assume
a
gcd, and
Of course
a
=
bq-\-r. Then
gcd(a,6)
in this case
'gcd(a, b)
=
=
a, b have a
common
divisors of 6,
gcd if and only ifb,
r
r
have
gcd(6,r).
gcd(6, r)'
means
that the two classes of associate elements
coincide.
These considerations hold
Euclidean domain. Then
over
the remainders
r.
over any
integral domain;
assume now
that R is
a
division with remainder to gain some control
two
elements
Given
a, b in i?, with b ^ 0, we can apply
we can use
division with remainder repeatedly:
a
=
b
=
bqi +n,
riq2 + r2,
n =r2q2 + rs,
as
long
as
the remainder r^ is
Claim 2.11.
This process terminates: that is, r^
This boils down to
readers.
nonzero.
a
case-by-case analysis, which
=
we are
0
for
some
N.
happily leaving
to our most patient
Proof. Each line
would have
in the table is
infinite
an
a
decreasing
v(b)
of
Irreducibility and factorization
V.
258
division with remainder. If
nonnegative integers, which is
>
v(r2)
>
v(rs)
nonsense.
a
=
r0qi +n,
b
=
nq2+r2,
r\
7^
=
r2q2 + r3,
r^-3
=
rN-2qN_2 + r^-i,
rN_2
=
rjv-itfjv-i
follows: letting tq
as
=
6,
0.
Proposition 2.12.
Proof.
no Ti were zero, we
>
Thus the table of divisions with remainders must be
with vn-i
integral domains
sequence
v(ri)
>
in
With notation
as
above,
r^v-i is a
gcd of a,
b.
By Corollary 2.10,
gcd(a,6) =gcd(6,ri) =gcd(rur2)
=
=
gcd(rN_2,rN_i).
But rN-2
rN-iqN-i gives rN-2 e (rN-i); hence (rN-2,rN-i)
(rN-i).
Therefore rjv-i is a gcd for rjv-2 and r^v-i, hence for a and 6, as needed.
=
=
The ring of integers and the polynomial ring over a field are both Euclidean
Fields are Euclidean domains (as represented in the picture at the
domains.
beginning of the chapter), but
not for a very
interesting
the remainder of the
reason:
division by a nonzero element in a field is always zero,
as a 'Euclidean valuation' for trivial reasons.
so every
function qualifies
We will study another interesting Euclidean domain later in this chapter
(§6.2).
Exercises
2.1. > Prove Lemma 2.1.
2.2. Let R be a
[§2.1]
UFD, and let
1. Prove that a divides
a,
6,
c
be elements of R such that
a
be and
gcd(a, b)
=
c.
positive integer. Prove that there is a one-to-one correspondence
preserving multiplicities between the irreducible factors of n (as an integer) and
2.3. Let
n
be
a
the composition factors of Z/nZ (as a group).
may be used to prove that Z is a UFD.)
2.4. > Consider the elements x,y in
yet 1 is not
a
linear combination of
Z[x,y].
x
and y.
(In fact,
the Jordan-Holder theorem
Prove that 1 is
(Cf.
Exercise
a
gcd of
x
and y, and
II.2.13.) [§2.1, §2.3]
Exercises
259
2.5. > Let R be the
a2t2
degree 1: a0 +
subring of Z[t] consisting of polynomials with
-\
h
Prove that R is indeed
of
no
term
an
integral
adtd.
subring of Z[£], and conclude that R
a
is
domain.
List all
common
divisors of t5 and t6 in R.
Prove that t5 and t6 have
no
gcd
in R.
[§2-1]
2.6. Let R be
a
domain with the property that the intersection of any
ideals in R is necessarily
principal
Show that greatest
Show that UFDs
2.7.
common
satisfy
a
family of
principal ideal.
divisors exist in R.
this property.
Noetherian domain, and assume that for all nonzero a, b in i?,
the greatest common divisors of a and b are linear combinations of a and b. Prove
that R is a PID. [§2.3]
Let R be
>
2.8. Let R be
a
a
UFD, and let I ^ (0) be
an
ideal of R. Prove that every descending
chain of principal ideals containing I must stabilize.
2.9.
The height of a prime ideal P in a ring R is (if finite) the maximum length h
C Ph
P in R. (Thus, the Krull dimension
of a chain of prime ideals Po C Pi C
of i?, if finite, is the maximum height of a prime ideal in R.) Prove that if R is a
UFD, then every prime ideal of height 1 in R is principal. [2.10]
-i
=
2.10.
-i
nonzero
It is
a consequence
element in
Assuming this,
domain R is
a
a
of
a
theorem known
as
KrulVs Hauptidealsatz that every
a prime ideal of height 1.
2.9, and conclude that a Noetherian
Noetherian domain is contained in
prove a converse to Exercise
UFD if and only if every prime ideal of height 1 in R is principal.
[4.16]
2.11.
Let R be a
artinian ring
(cf.
PID, and let / be
Exercise
2.12. > Prove that if
2.13. > For
and only if
R[x]
is
a
PID, then R is
a,b,c positive integers with
a
b. Prove that xa
Prove that if A; is a
v
such that
v(ab)
>
v(b)
=
for all
Show that
R/I
that the d.c.c. holds in
is
an
R/I.
[§2.3, §VI.7.1]
1, prove that ca
1 divides cb
1 if
1 in
ad +
field, then k[[x]]
2.15. > Prove that if R is a Euclidean
valuation
c >
field.
a
1 divides xb
For the interesting implications, write b
into account.) [§VII.5.1, VII.5.13]
2.14. >
ideal of R.
a nonzero
1.10), by proving explicitly
is
r
a
Z[x] if and only if a b. (Hint:
with 0 < r < a, and take 'size'
Euclidean domain.
[§4.3]
domain, then R admits a Euclidean
a,b e R. (Hint: Since R is a Euclidean
Definition 2.7. For a ^ 0, let v(a) be the
nonzero
domain, it admits a valuation v as in
minimum of all v(ab) as b G i?, 6^0. To
see
that R is
a
Euclidean domain with
well, let a, b be nonzero in i?, with b a; choose g, r so that a
bq+r,
with v(r) minimal; assume that v(r) > v(b), and get a contradiction.) [§2.4, 2.16]
respect to
v as
=
Irreducibility and factorization
V.
260
2.16.
Let R be
v(ab)
>
v(b)
Euclidean domain with Euclidean valuation v\ assume that
nonzero a,b e R (cf. Exercise 2.15).
Prove that associate
a
same
2.17.
a
Let R be
-i
either r
0
Euclidean domain that is not
y/d,
unit.
or r a
2.18. > For
Q and
case d
valuation and that units have minimum valuation.
nonunit element
=
an
c
in R such that Va G
integer d, denote by
with
norm
Let 5
(1
=
+
N defined
iy/l9)/2,
Prove that the smallest values of
> 5 if b
bS)
Prove that the units in
If
Z[S]
c G
divide 2
or
3 in
2.19.
(fe*,-)
:
a +
±
following subring of Q(V—19):
i^™\a,bez\.
N(z)
for
z
=
a +
bS G
Z[S]
are
0, 1, 4, 5. Prove
are ±1.
in Exercise 2.17, prove that
specified
=
=
(Z, +)
Z[S]
0.
>/z19)/2]
A discrete valuation
-i
a
qc + r and
=
—19.
=
=
+
a
the smallest subfleld of C containing
Z[5], and conclude that c ±2 or c
Jg G Z[5] such that 5 qc + r with c
Conclude that Z[(l
R with
G
bl +
satisfies the condition
Now show that
v
Q(Vd)
and consider the
+
+
field. Prove that there exists
in Exercise III.4.10. See Exercise 1.17 for the
as
Z[6\:=L
that N(a
a
i?, 3g, r
[2.18]
in this problem, you will take d
—5;
=
integral domains
for all
elements have the
nonzero,
in
such that
is not
=
±2, ±3 and
Euclidean domain.
a
on
a
field A; is
v(a
+
b)
>
a
c must
±3.
r
=
0, ±1.
[§2.4, 6.14]
homomorphism of abelian
groups
for all a,b e k* such that
min(v(a),v(b))
bek*.
Prove that the set R
Prove that R
is
a
:=
{a
G £;*
>
| v(a)
0}
U
{0}
is
subring of fe.
a
Euclidean domain.
Rings arising in this fashion are called discrete valuation rings, abbreviated DVR.
They arise naturally in number theory and algebraic geometry. Note that the
Krull dimension of
correspond
to
a
DVR is 1
particularly
(Example III.4.14);
nice points
Prove that the ring of rational numbers
integer p is a DVR.
in
algebraic geometry, DVRs
'curve'.
on a
a/b
with b not divisible by
a
fixed prime
[2.20, VIII.1.19]
2.20.
-i
As
seen
in Exercise 2.19, DVRs
must be PIDs. Check this directly,
for
some
k > 1.
=
2.21. > Prove that an
are
Euclidean domains. In particular, they
follows. Let R be
a
DVR, and let t G R
ideal, then I
1. Prove that if I C R is any nonzero
v(t)
(The element
element such that
as
t is called
integral domain
is
a
'local parameter' of
a
PID if and
only if it
be an
=
(tk)
R.) [4.13, VII.2.18]
admits
a
Dedekind-
(Hint: For the <^^ implication, adapt the argument in
=>
let v(a) be the size of the multiset of irreducible factors
Hasse valuation.
Proposition 2.8; for
[§2-4]
,
of
a.)
3. Intermezzo: Zorn's lemma
2.22.
a
a
Suppose R
-i
261
Compute d
2.23.
such that d
=
2.24. > Prove that there
about pi
than 2,000 years
2.25.
Variation
-i
a +
R is
also
find a, b
2922476045110123 6.
infinitely many prime integers. (Hint: Assume by
complete list of all positive prime integers. What
pjy + 1? This argument was already known to Euclid,
are
Pi,... ,Pjv is a
can you say
more
a
gcd(5504227617645696,2922476045110123). Further,
=
5504227617645696
contradiction that
integral domains, and assume that
gcd for a and b in R. Prove that d is
C S is an inclusion of
PID. Let a,b e R, and let d G R be
gcd for a and b in 5. [5.2]
on
ago.) [2.25, §5.2, 5.11]
the theme of Euclid from Exercise 2.24: Let
f(x)
G
Z[x]
be
1. Prove that infinitely many primes
polynomial such that /(0)
divide the numbers /(n), as n ranges in Z. (If pi,... ,pw were a complete list of
primes dividing the numbers /(n), what could you say about f(pi -p^a;)?)
1 is unnecessary.
Once you are happy with this, show that the hypothesis /(0)
a nonconstant
=
=
-p^ax). Finally,
(If /(0) a 7^ 0, consider f(pi
about 0.) [VII.5.18]
=
note that there is
nothing special
3. Intermezzo: Zorn's lemma
3.1.
Set
theory, reprise. We leave ring theory for
a moment
and take
a
little
detour to contemplate an issue from set theory. As remarked at the very outset,
only naive set theory is used in this book; all set-theoretic operations we have used
so
far
are
nothing
more
than
formalization of intuitive ideas regarding collections
a
occasionally need to refer to a less 'intuitively obvious'
set-theoretic statement: for example, this statement is needed in order to show that
every ideal in a ring is contained in a maximal ideal (Proposition 3.5).
of objects. However,
we
will
An order relation
This set-theoretic fact is Zorn's lemma.
on
a set
Z is
relation ¦< which is reflexive, transitive, and antisymmetric: the first two terms
familiar to the reader, and the third means that
(Va, b
G
a<b and b <a ==>
Z),
a
=
a
are
b.
Typical prototypes are the < relation on Z or the inclusion relation C among subsets
of a given set. We use a -< b to denote a ^ b and b ^ a.
A pair
(Z, ^), consisting
of
a set
Z and
an
order relation ^
on
Z, is called
poset, for partially ordered set. The qualifier 'partially' is not necessary, but
convenient, as it reminds us that if a, b G Z, then it is not necessarily the case
a
that
a
^ b
in general
or
b ^
a:
(while (Z,
for example, C does not satisfy this additional requirement
<) does). An order is total if it does satisfy this additional
requirement. A totally ordered set is not called
but rather a chain.
An element
m
of
a
a
'toset',
poset Z is maximal if nothing
as
comes
the order:
(Va
G
Z),
m -< a
=>
m
=
would
a.
seem
reasonable,
properly 'after'
it in
Irreducibility and factorization
V.
262
in
integral domains
For example, maximal ideals in a ring are maximal elements in the set of proper
ideals, that is, ideals ^ (1) (by Proposition III.4.11). An upper bound for a subset
S of a poset Z is an element u G Z coming after every element of S:
(Vo€5),
a<u.
(or, rather, 'least')
Notions of 'smallest'
'largest'
or
are
defined in the evident
way.
Posets may
example, the
or may not
family of finite
All this terminology
Lemma 3.1
in Z has
that
have maximal elements, upper bounds, etc.: for
an infinite set does not have maximal elements.
subsets of
comes
(Zorn's lemma).
an upper
together
Let Z be
in the
a
following
statement:
nonempty poset. Assume that every chain
bound in Z; then there exists
a
maximal element in Z.
The status of Zorn's lemma is peculiar: on the one hand, it is complex enough
no one we know of finds it 'intuitively clear'; on the other hand, it turns out to
choice, which does look reasonable to most
'well-ordering theorem', which we find intuitively unreasonable.
be logically
equivalent10
people, and
to the
to the axiom of
We will not prove all these equivalences (the diligent reader will prove them in the
exercises and should in any case have no trouble locating more detailed proofs);
but we will attempt to describe the situation in general terms and explain why the
unreasonable
statement
The axiom
a set
Z, then
of
implies the other
two.
family of disjoint nonempty subsets of
by selecting one element x from each X G 3?.
choice states that if & is
we can
form
a new set
a
This may sound reasonable, but it raises rather subtle points: for example, it can
be shown that this axiom is independent of the other axioms of (Zermelo-Frankel)
set theory.
Also, it has disturbingly counterintuitive consequences, such as the
Banach-Tarski
paradox11.
The subtlety of the axiom of choice boils down to the following question: how
does one choose the element x? This is not controversial if Z (and hence &) is
finite, but, not
surprisingly,
it becomes
A suitable order relation
on
an
Z would
issue if Z is infinite.
come
in
handy here: if
nonempty subset of Z has
¦< is an order
least element, then we
be the least element of X for each X G &. We say that Z
relation on Z such that every
could simply let x
is well-ordered by ^,
or
that -< is
a
well-ordering
on
a
Z, if this is the
case.
(The
abbreviation woset is also used in the literature, but not very often.) For example,
the set Z>0 of positive numbers is well-ordered by <; this fact is called the well-
ordering principle.
Thus, the statement
set
of the axiom of choice is
Z>0 of positive numbers. It may
seem
is not well-ordered by <; however, it takes
Equivalent in the
sense
that each
can
be
really 'clear' if the
somewhat less
a moment
derived
so
set Z is the
for the set Z, since Z
(Exercise 3.4)
to construct a
from the other together with the other
axioms of 'Zermelo-Frankel' set theory.
One can subdivide a solid ball of radius 1 into finitely many pieces, then reassemble them
after rotations and translations in 3-space and obtain two balls of radius 1.
3. Intermezzo: Zorn's lemma
different
relation
on
263
Z, making Z
also completely transparent for Z
a
=
well-ordered set. Thus, the axiom of choice is
or any countable set (like Q) for that matter.
Z
The well-ordering principle is at the basis of proofs by induction: in fact (cf.
3.5) it is equivalent to the so-called principle of induction. More is true,
Exercise
however.
induction'
If
is any woset, then
(Z, :<)
on
Let S C Z be
a
following 'principle of
consider the
subset such that Va G Z
((V6
then S
we can
Z:
G
Z),
6 -<
a
=>
beS)
=>
S;
a G
Z.
=
That is, Z is the only set S with the property that a G S if all b -< a are in 5.
(Note that the least element a of Z is automatically in 5, since the condition on
all b -< a is vacuously true in this case.)
The reader will recognize that for Z
=
Z>0 with the relation <, this is the
ordinary principle of induction.
Claim 3.2. Let Z be any woset. Then the principle of induction holds for Z.
Proof. Let S
C Z be a subset with the
property given above, and
assume
S C Z.
Then the complement T of S in Z is nonempty; hence it has a least element a.
Now if b -< a, then necessarily b e S (otherwise b G T, contradicting the fact that a
is the least element in T). But the property of S then forces a G 5, contradiction.
Therefore
a
T is empty; i.e., S
=
Z.
This is remarkable, since it extends induction to uncountable
is available.
sets12, provided
well-ordering
For example, if we had a well-ordering on R, then we could prove a statement
about all reals 'by induction'. Many people find this somewhat counterintuitive,
and hence so seems the following amazing claim:
Theorem 3.3
(Well-ordering theorem). Every
set admits a
well-ordering.
As argued above, this implies that the axiom of choice holds on every set: if you
accept the well-ordering theorem, the statement of the axiom of choice becomes as
evident for every set as it is for the set of positive integers. The catch is of course
that the axiom of choice is used in the proof of the well-ordering theorem; in fact,
the statements are equivalent.
In any case, the statement of the well-ordering theorem is easy to absorb (even
if perhaps less 'intuitively clear' than the axiom of choice). The well-ordering
theorem is equivalent to Zorn's lemma, since they are both equivalent to the axiom
of choice. The good news is that the derivation of Zorn's lemma from the well-
ordering theorem
is
reasonably straightforward:
Well-ordering theorem => Zorn's lemma. Let (Z, <)
that every chain in Z has
Even beyond; in the
induction.
an upper
be
a
nonempty poset such
bound in Z. By the well-ordering theorem, there
context of ordinals this induction
principle is called transfinite
is
Irreducibility and factorization
V.
264
well-ordering13
a
^
Z. Define
on
a
in
integral domains
function / from Z to the power set of Z
as
follows:
(
/(a)
=
{
totally ordered by <;
({a}
U
(J f(b))
is
0 if ({a}
U
(J /(6))
is no*
if
{a}
totally ordered by
<.
by ^ implies that / is defined for every a G Z.
Indeed, let T be the set of elements of Z for which / is not defined; if T ^ 0,
T has a least element a w.r.t. ^; but then f(b) is defined for all b -< a, and the
The fact that Z is well-ordered
prescription given for /(a) defines /
Let S
=
UaGZ /(a)-
b -< a, then both
(say)
at a, a contradiction.
ft *s c^ear that ^ is totally ordered by <: if a, 6 G 5, and
a
and 6 belong to
{a}
U
|J /(&),
b^a
which is
totally ordered by
We claim that S is
If S
Claim 3.4.
C
S'
<
by
construction.
maximal totally ordered subset14 of Z\
a
C Z and
5; is fo£a% ordered 6y <, then S
=
S'.
Indeed, let T be the complement of S in S'. If T is nonempty, let
a G
T and
observe that
Wu(J/W
is
totally ordered by <, since it is a subset of [a] U S C 5;. But
a G 5, a contradiction.
Since 5 is a chain, by the hypothesis of Zorn's lemma S has
then
/(a)
=
{a},
that is,
an upper
bound,
be maximal in Z w.r.t. <, verifying the statement
of Zorn's lemma. Indeed, if m! > m, then S U {m'} is totally ordered; hence
m since ra is an
S
S U {mf} by the claim. This means m! G 5, and hence m!
It is
m.
now
clear that
m must
=
=
bound for S.
upper
Zorn's lemma is the key to several basic results in algebra, the first of which we
about to encounter; the reader will sample a few more results in the exercises.
The reader will also encounter equally basic applications of Zorn's lemma (or of
are
other manifestations of the axiom of
theorem
3.2.
m
of
in
choice)
in other fields: for
topology and the Hahn-Banach theorem
example Tychonoff's
analysis.
in functional
Application: Existence of maximal ideals. Recall (§111.4.3) that an ideal
ring R is maximal if and only if R/m is a field, if and only if no other ideal
a
stands between / and R
inclusion, in the
=
family of
(1),
Watch out:
and X
we
are
(about
m
is maximal with respect to
proper ideals of R. It is not obvious from this definition
that maximal ideals exist, but
something)
that is, if and only if
they do:
considering
two
orderings
which we know something
on
Z:
<
(about
which we want to say
already).
The existence of maximal totally ordered subsets is known as the Hausdorff maximal
principle; it is also equivalent to the axiom of choice.
Exercises
265
Proposition 3.5. Let J ^
there exists a maximal ideal
be
(1)
m
family of
of
a commutative
ring R.
Then
I.
by applying condition (3) from
containing I. The argument for
This is immediate if R is Noetherian,
Proposition 1.1 to the
ideal
a proper
of R containing
proper ideals of R
arbitrary rings is
In fact, this is the first
a classic application of Zorn's lemma.
application given by M. Zorn in his 1935 article introducing a 'maximum principle'
(now known as Zorn's lemma). The result had earlier been proven by Krull, using
the well-ordering theorem.
Proof. The set J^ of proper ideals of R containing I is ordered by inclusion. Then
let ^ be a chain of proper ideals, and consider
U:={JJ.
We claim that U is
a proper ideal containing 7; hence it is an upper bound for ^
in J?. This proves that every chain in J? has an upper bound, and it follows that
J? has maximal elements, by Zorn's lemma.
To verify our claim, it is clear that U contains I and that it is
an
ideal
(for
example, if a, b G £/, then 3J e J? such that a,b e J; hence a±b G J, and therefore
a±b G U). We have to check that U is proper. But if U
(1), then 1 G J for some
J e J?, contradicting the fact that J consists of proper ideals.
=
Note that this argument relies crucially
on
the fact that R has
unit 1, and indeed the stated fact is not true in
(Exercise
3.9). Also,
the
is known to be
use
a multiplicative
general for rings without 1
of Zorn's lemma is not just
equivalent
to the axiom of
a
convenient trick; the statement
choice, by work of W. Hodges.
Exercises
3.1. Prove that every
3.2. Prove that
a
well-ordering
totally ordered
set
is total.
is
(Z, ^)
a woset
if and only if every descending
chain
z\h z2 h z3 >z
in Z stabilizes.
3.3. Prove that the axiom of choice is equivalent to the statement that
function is surjective if and only if it has a right-inverse (cf. Exercise 1.2.2).
3.4. > Construct
can
explicitly
be well-ordered,
3.5. >
even
is well-ordered,
Vn G Z>0 there
a
well-ordering on Z. Explain why you know that Q
performing an explicit construction. [§3.1]
well-ordering
assume
are no
set-
without
(ordinary) principle
Prove that the
statement that < is
a
a
on
Z>0.
of induction is equivalent to the
(To prove by induction that
(Z>0,<)
it is known that 1 is the least element of Z>0 and that
integers between
n
and
n +
1.) [§3.1]
Irreducibility and factorization
V.
266
3.6. In this exercise
assume
in
integral domains
the truth of Zorn's lemma and the conventional set-
theoretic constructions; you will be proving the well-ordering theorem.
Let Z be
nonempty set, and let 2f be the set of pairs (5, <) consisting of a
a well-ordering < on S. Note that 3f is not empty (singletons
a
subset S of Z and of
can
be
Define
well-ordered).
a
relation ¦< on 2f by prescribing
(S,<)±(T,<')
if and only if S C T, < is the restriction of <' to 5, and every element of S precedes
S w.r.t. <'.
every element of T
Prove that ¦< is
order relation in 3f.
an
Prove that every chain in 3? has
Use Zorn's lemma to obtain
Thus every set admits
3.7. In this exercise
a
a
maximal element
well-ordering,
assume
bound in 3?.
an upper
as
in 3f. Prove that M
(M, <)
=
Z.
stated in Theorem 3.3.
the truth of the axiom of choice and the conventional
set-theoretic constructions; you will be proving the well-ordering theorem15.
Let Z be
7(5) 0
is a
nonempty set.
a
Use the axiom of choice to choose
S for each proper subset S C Z. Call
well-ordering
5, and for
on
every a e
Show how to begin constructing
begin in the
Define
an
a
S,
a
a
an
pair (S,<) a j-woset if S
^({b G S,b < a}).
element
C
Z,
<
=
7-woset, and show that all 7-wosets must
same way.
ordering
7-wosets by prescribing that ([/, <")
on
¦<
if and only if
(T, <;)
U C T and <" is the restriction of <'.
Prove that if
For two
(17, <")
-<
(T, <;),
then
7(J7)
G T.
7-wosets (S,<) and (T, <;), prove that there is
both w.r.t. ¦<.
(U, <") preceding
(Note:
There is
no
need to
a
maximal 7-woset
Zorn's lemma!)
use
Prove that the maximal 7-woset found in the previous point in fact equals
(T, <;). Thus,
or
Prove that there
¦< is a total
is
a
maximal 7-woset
need not and should not be
Prove that M
=
a
well-ordering,
3.8. Prove that every nontrivial
subgroup. Prove that (Q, +) has
3.9.
>
as
w.r.t. ¦<.
finitely generated
no
Zorn's lemma
group has a maximal proper
maximal proper subgroup.
Consider the rng (= ring without 1; cf. §111.1.1)
endowed with the trivial multiplication qr
-1
(Again,
stated in Theorem 3.3.
(Q, +)
that this
3.10.
(M, <)
invoked.)
Z.
Thus every set admits
group
(5, <)
ordering.
consisting
=
rng has no maximal ideals.
of the abelian
0 for all g,
r G
Q. Prove
[§3.2]
As shown in Exercise III.4.17, every maximal ideal in the ring of continuous
real-valued functions
vanishing at
a
on a
compact topological space K consists of the functions
point of K.
'We learned this proof from notes of Dan Grayson.
4.
Unique factorization
in
267
polynomial rings
are maximal ideals in the ring of continuous real-valued
the real line that do not correspond to points of the real line in the same
fashion. (Hint: Produce a proper ideal that is not contained in any maximal ideal
corresponding to a point, and apply Proposition 3.5.) [III.4.17]
Prove that there
functions
on
3.11. Prove that a UFD R is a PID if and only if every nonzero prime ideal in R is
maximal. (Hint: One direction is Proposition III.4.13. For the other, assume that
every nonzero prime ideal in a UFD R is maximal, and prove that every maximal
ideal in R is principal; then use Proposition 3.5 to relate arbitrary ideals to maximal
ideals, and prove that every ideal of R is principal.)
commutative ring, and let / C R be a proper ideal. Prove that
the set of prime ideals containing / has minimal elements. (These are the minimal
3.12.
-i
Let R be
primes of
3.13.
Let
-i
0
r
a
J.) [1.9]
Let R be
commutative ring, and let N be its nilradical
a
(Exercise III.3.12).
N.
Consider the
family & of ideals
of R that intersect the ideal
(r) only
at 0. Prove
that & has maximal elements.
Let I be a maximal element of &. Prove that I is
Conclude
r
$
N =>
r
is not in the intersection of all prime ideals of R.
Together with Exercise III.4.18,
ring R equals the intersection of
3.14.
-i
r
this shows that the nilradical of
all prime ideals of R.
The Jacobson radical of
maximal ideals in R.
that
prime.
(Thus,
a
commutative
[III.4.18, VII.2.8]
commutative ring R is the intersection of the
the Jacobson radical contains the nilradical.) Prove
a
is in the Jacobson radical if and
only if
1 + rs is invertible for every
s G
R.
[VI.3.8]
a (commutative) ring R is Noetherian if every ideal of R is finitely
Assume
the seemingly weaker condition that every prime ideal of R is
generated.
Let
& be the family of ideals that are not finitely generated in
finitely generated.
3.15. Recall that
R. You will prove &
=
0.
If & 7^ 0, prove that it has
Prove that
R/I
Prove that there
are
Give
of
a structure
Prove that
I/J1J2
Prove that
/ is
Thus,
a
a
maximal element /.
is Noetherian.
ideals Ji, J^ containing /, such that J\J<i Q /•
R/I
is
a
module to
I/J1J2
and
JiJ\/J\
finitely generated R/I-module.
finitely generated, thereby reaching
a
ring is Noetherian if and only if its prime ideals
contradiction.
are
finitely generated.
4. Unique factorization in polynomial rings
We
regular programming and study unique factorization in
will finally establish the fact (already hinted at) that R[x] is
now return to
polynomial
rings;
if R is
a
we
UFD.
a
UFD
V.
268
Irreducibility and factorization
in
integral domains
Among the necessary preparatory work we also discuss the field of fractions of
integral domain; this is a particular instance of the process of localization of a
ring (or a module) at a multiplicative subset, which the diligent reader will explore
an
a
little
in the exercises.
more
Primitivity and content; Gauss's lemma. One
4.1.
issue that
have
we
encountered already, and will encounter again, is the description of the ideals of
the polynomial ring R[x], given enough information about R. For example, we
have proved that the ideals of k[x] are principal if A; is a field, and Hilbert's basis
theorem shows that all ideals of R[x]
(Exercise
An
III.4.4 and Lemma
even more
ideal of
naive observation is
:=
Lemma 4.1. Let R be
simply that
{a0
+ aix H
h
~
The proof of this lemma is
theorem and is left to the reader
Corollary
If I
4.2.
is
a
a
every ideal I of R generates an
(R/I)[x],
We will
use
an
G
R[x] | Vi, o»
ideal
of R.
G
/}.
Then
I[i'
standard application of the first isomorphism
(Exercise 4.1).
prime ideal of R, then IR[x] is prime in R[x].
Proof. If I is prime in i?, then R/I is
and therefore
adxd
ring, and let I be
a
IR[x]
IR[x]
this fact in
an
is prime in
integral domain; hence
following definitions
application will be to UFDs.
is
i?[x]//i?[x]
=
R[x].
can
be studied for every commutative ring;
Definition 4.3. Let R be a commutative
/
so
a moment.
The
a
are
R[x]:
IR[x]
be
finitely generated if all ideals of R
are
1.3).
=
ao + a\xH
our
main
ring, and let
V
ad,xd
G
R[x]
polynomial.
/
is very primitive if for all prime ideals p of i?,
/
is primitive if for all
/ 0 pi?[x].
ideals
of
i?, / 0 pi?[x].
principal prime
p
j
The notion of 'primitive' polynomial is standard, but it is not usually
presented in this way; the standard definition is the equivalent formulation given in
Lemma 4.5. As for 'very primitive', this term is essentially a joke—our
excuse for
bringing up this notion is that it is very natural and that unfortunately some
blur the distinction between this notion and the notion of 'primitive'. We
hope to ward off any possible confusion by being rather explicit on this point. Very
primitive polynomials are primitive, but the converse does not hold in general, even
references
for UFDs
(cf. Exercise 4.3).
Perhaps the most important
remark.
fact about 'primitivity' is the
following
easy
4.
Unique factorization
in
Lemma 4.4. Let R be
a commutative
fg is primitive
Proof. This is
269
polynomial rings
<^=>
an easy consequence
ring. Then for f,g£ R[x]
both
and g
f
are
primitive.
of Corollary 4.2:
fg primitive <^=> Vp prime and principal in R, fg 0
Vp prime and principal
<^=>
/
<^=>
since
R, f 0 pR[x] and
g
0 pR[x]
is primitive and g is primitive
is prime if p is
pR[x]
in
pR[x]
a
prime ideal.
The analogous equivalence holds for very primitive polynomials (Exercise 4.4).
Lemma 4.4 will be ultimately responsible for the fact that R[x] is a UFD if R is
If R is a UFD, the notion
a UFD, which is our main objective in this section.
of 'primitive' has the following interpretation; we accompany it with the 'very
primitive'
Lemma
as
case, for
comparison.
4.5. Let R be a commutative ring and
f
=
a$ + a\x+
+
adXd
G
R[x]
above.
f
if and only if (ao,..., ad)
(1).
UFD, then f is primitive if and only if gcd(ao,..., a<j)
is very primitive
If R
is
a
=
=
1.
(1), then no prime ideal can contain all coefficients a^,
is
/
very primitive. Conversely, if / is very primitive, then the
coefficients of / are not all contained in any one prime ideal, and in particular they
are not all contained in any maximal ideal (since maximal ideals are prime). Thus,
Proof. If
(ao,...,ad)
=
and it follows that
(ao,..., ad)
=
(1)
in this case, since every proper ideal is contained in
a
maximal
ideal by Proposition 3.5. This proves the first point.
For the second point note that, in a UFD, gcd(ao,..., ad) ^ 1 if and only if
exists an irreducible element q G R such that (ao,... ,ad) C (q). As (q) is
there
then prime
(by
Lemma
2.4),
the second point follows
as
well.
With this in mind, in order to capitalize on Lemma 4.4, it is convenient to give
the gcd of the coefficients of a polynomial. We now assume that R is a
a name to
UFD, since this is the
case
Definition 4.6. Let R be
denoted
cont/,
is the
of greatest interest in the applications.
a
gcd of
UFD. The content of
its coefficients.
a nonzero
polynomial /
G
R[x],
j
the content of a polynomial is only defined up to units. We find this
distasteful.
What is uniquely determined is the principal ideal generated
ambiguity
the
content
of
which
we will denote
by
/,
Of
course
(cont/)
:
is primitive precisely when (cont/)
(1). We will take a somewhat stubborn
and
deal
with
ideals
approach
principal
throughout, rather with individual (but only
defined up to unit) elements. The reader should keep in mind that ideals may be
thus
/
multiplied (cf. §111.4.1), and if
is the principal ideal (ab).
=
(a), (b)
are
principal ideals, then their product (a)(b)
V.
270
Irreducibility and factorization
The proof of the following remarks
is left to the reader
=
(cont/)(/),
if (f)
integral domains
is immediate from the definition; hence it
(Exercise 4.5):
Lemma 4.7. Let R be
(/)
in
(c)(#)>
UFD, and let f
a
where
with
c e
G
R[x].
Then
is primitive;
f
R and g primitive, then
(c)
=
(cont/).
In view of its consequences, the following fact is rather profound, and it
therefore deserves a name. Some references call the result of Lemma 4.4, or its main
consequence
Gauss's lemma. We prefer to
(Theorem 4.14),
use
this
name
for the
following statement.
Proposition 4.8
Let R be
(Gauss's lemma).
(cont/p)
Proof. This follows easily from
(fg)
=
our
=
a
UFD, and let f,g
R[x\. Then
(cont/)(contp).
preparatory work. Write
((cont/)(/)) ((contg)(g))
=
(cont/)(contp) (fg),
/, g primitive. By Lemma 4.4, fg is primitive; by Lemma 4.7 it follows that
(cont/)(contp) is the content of fg, as needed.
with
Note the
following
immediate consequence:
Corollary 4.9. Let R be
(cont/)
4.2.
C
a
UFD, and let f,gE R[x]. Assume (f)
C
(g).
Then
(contp).
integral domain. Gauss's lemma is the key
if
observation
that
is a UFD, then so is R[x], However, we need
R
important
The field of fractions of an
to the
one more
tool before
we can prove
this fact; this is
one
instance of the important
process of localization.
In the case we need for our immediate goal, the process starts from any integral
domain and produces a field, in precisely the same way the field Q of rational
numbers may be obtained from the integral domain Z. This construction is known
as
field of fractions (or field of quotients) of the integral domain R. It satisfies
following beautiful universal property.
Given an integral domain R, consider the category & whose objects are pairs
the
the
(i,K),
where K is
a
field and i
:
R
^->
K is
an
injective ring homomorphism. Morphisms
defined in the style of every analogous construction we have encountered: that is,
L
morphism (i,K)
(j, L) is determined by a homomorphism of fields a : K
are
a
making the following diagram
V
commute.
Unique factorization
4.
in
271
polynomial rings
Note that a, as every homomorphism of fields, is necessarily injective
III.3.10); this is compatible with the requirement that i : R ^-> K be injective
to begin with. The injectivity of R ^-> K also forces R to be an integral domain,
(Exercise
since
subrings of fields
Definition
4.10. The
are
necessarily integral domains.
field of fractions K(R) of R
is
an
initial object of the
category St.
j
Thus, K(R) is the 'smallest field containing R\
The usual caveats apply to any such definition: the initial object of & carries
not only the information of a field K (the field of fractions proper), but also of a
specific realization of R as a subring of K\this distinction is blurred in common use
of the language. Also, of course this prescription only defines the field of fractions
up to isomorphism (as is true of any universal object; cf. Proposition 1.5.4).
Finally, just stating the universal property does not guarantee that the soughtfor initial object exists. Thus, our next task is the construction of an initial object
for 8%. Fortunately, the construction is essentially straightforward: we just need to
formalize a notion of 'fraction' of elements of R.
Consider the set R x (R*) of pairs (a,r) of elements of i?, where r ^ 0. The
pair (a,r) will be associated with a 'fraction' ^; the reader would be well-advised
to put this book away now and carry out the construction of K(R) on his/her own,
profiting from this major hint.
Here is the construction. Denote by
with respect to the
following equivalence
(a,r)~(6, s)
The reader will
verify
the equivalence class of
-
(a, r)
G Rx
(R*)
relation:
<^=>
that this is indeed
as
-
br
=
0.
equivalence relation. As
an
a
set, K(R)
is defined by
K(R):=
{-|aei?,rei?,r^o}.
It is clear that these 'fractions' behave much
if
s
7^
as
a
rs
r
ordinary fractions.
0: indeed,
(as)r
by associativity
=
a(rs)
and commutativity in i?, and this shows
(as,rs)
as
as
~
(a,r)
needed.
We define operations
on
K(R)
follows:
as
a
b
r
s
as +
_
rs
a
b
ab
r
s
rs
br
For example,
Of
Irreducibility and factorization
V.
272
course one must
the reader.
guarantees that
Now
that these operations
verify
^
rs
0 if
claim that
we
^
r
an
0
is made into
K(R)
are
integral domain
and s ^ 0.
The fact that R is
a
The
a
(H)
(bt
+
"
r
'
r(st)
st
ab
a c
r s
r
a(bt)
r(st)
This is
a(cs)
r(st)
a
following
amounts to the
ab
ac
rs
rt
t'
element is the fraction y (or in fact
multiplicative identity is the fraction y (or in fact
zero
r s
rs
1
s r
rs
1
°
for any
^
nonzero
for any
^ 7^ jjj (that is, for all r ^ 0), every nonzero element
promised.
An 'inclusion map' i : R^> K(R) is defined by
for all
as
is also left to
is used in these definitions: it
field by these operations.
a(bt + cs)
cs)
integral domains
well-defined; this
straightforward verification; for example, distributivity
computation:
;i
in
r
in
K(R)
has
R); the
R). Since
G
nonzero r G
an
inverse,
a
a^
Y'
immediately checked that this is an injective ring homomorphism. It is common
this map to identify R with its isomorphic copy inside K(R) and simply
view R as a subring of K(R).
it is
to
use
Claim
4.11.
Proof. Let
j
(i,K(R))
:
R
We need to define
^->
an
is initial in 8%.
L be any injective ring homomorphism from R to a field L.
L so that the diagram
induced homomorphism j : K(R)
K—-—>L
j
it
commutes, and
forced upon
Thus j
us:
we must
show that j is unique. Now, the definition of j is in fact
if j exists
as a
homomorphism, then necessarily
is indeed unique, if it exists. On the other hand, the prescription
does define
a
function
K(R)
L: indeed, if
as
=
br
(a,r)
~
(6,5),
then
4.
Unique factorization
in
273
polynomial rings
in i?, hence
j(a)j(s) =j(b)j(r)
in
L, and (note that j(r), j(s)
in L since r,
are nonzero
s are nonzero
in i? and j is
injective)
rtaWr)-1^)^)-1,
showing
that the proposed j is well-defined. The reader will
verify
that it is
a
ring
homomorphism, concluding the proof of the claim.
4.12. With the notation introduced
above, K(Z)
Q.
The universal property implies immediately that F ^-> K(F) is
Example
if
F is itself
a
famous field:
=
an
field. Thus, the construction adds nothing to Q, R, C,
a
If R is any
integral domain,
so
that
R[x]
is also
an
isomorphism
Z/pZ,
etc.
integral domain, K(R[x])
is
Definition 4.13. The field of rational functions with coefficients in R is the field
of fractions of the ring R[x]. This field is denoted R(x).
j
Elements of
R(x)
are
fractions of polynomials
q(x)
with
p(x),q(x)
'function' R
which
q(a)
element in
=
R[x]
G
and
R given
by
q(x) ^
a i->
0.
|£4
The term
function
is not defined for all
0, that is); and the function itself does
R(x) (cf.
Exercise
is inaccurate, since the
a G
i?
(not
III.2.7).
j
R UFD ==> R[x] UFD. We are now in a position to prove
Hilbert's basis theorem for unique factorization domains, that is,
4.3.
Theorem 4.14. Let R be
a
UFD; then R[x]
example, this result (and
For
Z[#i,..., xn]
and
for those for
not suffice to determine the
k[xi,..., xn] (for
an
k
a
the
analogue of
is a UFD.
immediate
field)
are
induction) shows that the rings
UFDs. Theorem 4.14 is also often
called Gauss's lemma.
As
a measure
power series ring
of how delicate the statement of Theorem 4.14 is, note that the
is not necessarily a UFD if R is a UFD; examples of this
R[[x]]
phenomenon are however not easy to construct. Of course k[[x]] is a UFD if A; is
field, since k[[x]] is a Euclidean domain in this case (Exercise 2.14).
Theorem 2.5, in order to prove Theorem 4.14, we have to verify that
satisfies the a.c.c. for principal ideals and that every irreducible element in
By
is prime, provided that
questions to matters in
as
we
know, K[x]
R
is a UFD
domain => PID => UFD,
The
between
following
R[x]
and
is itself
K[x],
as
a
R[x]
R[x]
UFD. The general idea is to reduce these
K(R) is the field of fractions of R:
where K
(in
a
=
fact it is
shown in
a
Euclidean domain, and Euclidean
§2).
lemma captures the most crucial ingredient of the interaction
K[x\:
Irreducibility and factorization
V.
274
Lemma 4.15. Let R be
f,g
nonzero
and denote by
C
(contp)
c
(g)K
UFD, and let K
a
K(R)
=
be its
integral domains
field of fractions.
For
denote by (f), (g) the principal ideals fR[x], gR[x] in R[x],
{q)k the principal ideals fK[x], gK[x] in K[x\. Assume
R[x],
(/)#-,
G
in
and
(cont/)
U)k-
Then(g)C(f).
Proof. Since
(g)x
C
(/)k,
where h G
fh,
we nave 9
K[x],
Write h
=
where
^ft,
a,b E R and ft G R[x] is a primitive polynomial: this can be done by collecting
common denominators in ft, then applying the first point of Lemma 4.7. We then
have
bg
R[x]. By
in
Gauss's lemma and since ft is primitive,
(a cont/)
and since
(contp)
C
an
(cont/) by hypothesis,
some c e
we
C
obtain
(bcont/).
integral domain and (cont/) ^ (0), this implies
a
for
(b cont g)\
=
(acont/)
Since R is
afh
=
R. But then ft
=
|ft
=
=
eft G
be
i?[x],
and g
=
fh
e
(/);
that is,
(g)
C
(/)
D
needed.
in
i?[x],
of
The first application is the following description of the irreducible elements
R[x]\ this will also be used in the proof of Theorem 4.14 and is independently
as
interesting.
Proposition 4.16. Let R be a UFD, and let K be its field offractions. Let f G R[x]
be a nonconstant, irreducible polynomial. Then f is irreducible as an element of
K[x).
Proof. First note that
and
/
/ is primitive: otherwise
would not be irreducible.
Next, assume /
gh, with g, ft G
unit in K[x\. Let c,deK such that
=
a
g
and
g, ft are primitive
thus
(contgh)
=
(1)
=
=
K[x]\ we
in
polynomials
cd
7^
0 is
a
unit in K.
By Lemma
ideals of
R[x]\ that is,
this implies that either g
verifying
that
/
have to prove that either g
or
ft is
dh,
R[x]. By
/
or
=
ugh with
ft is
is irreducible in
a
=
4.15
(/)
as
=
could factor out its content,
Lemma 4.4, gh is also primitive;
(cont/); further,
U)k
as
h
cg,
we
=
obtain
(gh)
R[x] a unit. As / is
R[x]. But then g or ft
u G
unit in
K[x],
(gh)K
we
irreducible in
were
units in
R[x],
K[x],
4.
Unique factorization
We have
so one may
in
275
polynomial rings
always found this fact almost
counterintuitive: K is
expect that it should be 'easier' to factor polynomials
'larger' than i?,
over
K than
over
R. Proposition 4.16 tells us that this is not the case: with due attention to special
cases, irreducibility in R[x] is 'the same as' irreducibility in K[x\. To be precise,
Corollary 4.17. Let R be a UFD and K the field of fractions of R. Let f G R[x]
be a nonconstant polynomial. Then f is irreducible in R[x] if and only if it is
irreducible in K[x] and primitive.
The proof amounts to tying up loose ends, and
(Exercise
we
leave it to the reader
4.21).
We
can now prove
the main result of this section. We will make systematic
use
of the characterization of UFDs found in Theorem 2.5.
Proof of Theorem 4.14. We begin by verifying the
in R[x]. Let
a.c.c.
for principal ideals
(/i)c(/2)c(/3)c...
ascending chain of principal ideals of R[x]. By Corollary 4.9, this induces
ascending chain of principal ideals
be
an
an
(cont/J C (cont/2) C (cont/3) C
in i?; since R is a UFD, this chain stabilizes: that is,
(cont/.)
(cont/.+1) for16
i ^> 0. On the other hand, with notation as in Lemma 4.15 we have
=
(/ikc^cf/^c...
as a sequence
this
of ideals in
sequence stabilizes.
K[x\, since K[x]
Therefore, (fi)K
By Lemma 4.15, (f)
ideals stabilizes, as needed.
=
Next,
we
consider
an
(/i+i)
is
=
UFD
a
(/i+i)k
(because
it is
a
PID; cf. §2.3)
for i > 0.
for i ^> 0; that is, the given chain of principal
irreducible element
/ of R[x].
prime ideal; by Theorem 2.5 it then follows that R[x] is
We
verify that (/) is a
UFD, as stated.
R as R is a UFD, and it
a
If / is irreducible and constant, then / is prime in
follows that / is prime in R[x] (by Corollary 4.2). Thus we may assume that /
nonconstant and irreducible (and in particular primitive) in R[x].
By Proposition 4.16, / is irreducible as an element
(f)K is prime in K[x], Consider the composition
of
K[x]\since K[x]
is
a
is
PID,
\J)K
We claim that kerp
(/). Indeed, the inclusion D is trivial; for the other inclusion,
note that p(g)
0 implies that g is divisible by / in K[x\: that is, (g)x Q (/)#"> and
=
=
we
have
(contp)
obtain that
we
(g)
C
C
(cont/)
(/), i.e.,
find that p induces
an
since
(cont/)
=
(1)
as
in
g is divisible
is primitive. By Lemma 4.15, we
R[x], as needed. Since kerp (/),
/
by /
infective homomorphism
R[x]
=
K[x]
77T^(7k*
16'For
i > 0' is shorthand for
(3N
>
0) (Vt
>
N)
...,
that is, 'for all sufficiently large i.
V.
276
Irreducibility and factorization
Since the ring
(J)k
the ring
R[x],
on
on the right is an integral domain (as
the left. This proves that (/) is prime in
if R is
Summarizing,
UFD, then factorization
a
in
R[x]
in
integral domains
is prime in if[#]),
and we are done.
is 'the
same
so
is
as'
factorization in the polynomial ring K[x] over the field of quotients of R. If f(x) G R[x],
then f(x) has a prime factorization in K[x] for the simpler reason that K[x] is a
PID, hence
UFD; but if R itself is a UFD, then we know that each of the factors
R[x] to begin with (cf. Exercise 4.23).
a
may be assumed to be in
4.18. As mentioned already, Theorem 4.14 implies that several
rings, such as Z[#i,... ,xn] or C[#i,... ,xn], are UFDs; for example Z[x] is a
UFD, as announced in §2.3. Further, arguing as in Exercise 1.15 to reduce to the
Example
important
case
an
of finitely many indeterminates, it follows that Z[xi,^2,...] is a UFD: this is
a non-Noetherian UFD, promised a while back (and illustrating the
example of
last missing feature of the picture presented at the beginning of the
chapter).
j
Exercises
4.1.
>
4.2.
Let R be a
Prove Lemma 4.1.
ring, and let / be
maximal in i?, then
4.3. > Let R be
[§4.1]
a
IR[x]
an
ideal of R.
is maximal in
PID, and let /
G
R[x].
is very primitive. Prove that this is not
Prove
or
disprove that if /
is
R[x],
Prove that / is primitive if and only if it
necessarily the case in an arbitrary UFD.
[§4-1]
ring, and let f,g
4.4. > Let R be a commutative
fg
is very primitive ^=> both
/
G
and g
R[x\. Prove
are very
that
primitive.
[§4-1]
4.5. > Prove Lemma 4.7.
4.6. Let R be
Prove that
a
[§4.1]
PID, and let K be its field of fractions.
every element c G K can be written as a finite sum
i
where the pi
are
&
nonassociate irreducible elements in i?, Vi > 0, and a^pi
are
relatively prime.
If 5Zi"%"
Pi
Pi
=
Qi, n
YsiJ ~^i
are
two
sucn
expressions,
prove that
<?/
=
Relate this
s^ and o»
=
to
reshuffling)
6» modp[\
to the process of
when you took calculus.
(up
integration by 'partial fractions'
you learned about
Exercises
277
4.7. > A subset S of
tively closed) if (i)
of pairs (a, s) with
a
commutative ring R is
(ii) s,t
1 G S and
R,
a G
(a, 5)
s G
5
as
(a',*')
-
a
=>
e S
multiplicative subset (or multiplicas£ G 5. Define a relation on the set
follows:
<^>
(3£G 5), £(s'a
-
sa')
0.
=
Note that if R is an integral domain and S
R
{0}, then 5 is a multiplicative
subset, and the relation agrees with the relation introduced in §4.2.
=
Prove that the relation
~
is
an
equivalence relation.
Denote by ^ the equivalence class of (a, s), and define the
such 'fractions'
on
these operations
as
are
introduced in the special
well-defined.
the
ones
same
operations +,
of §4.2. Prove that
case
The set S~XR of fractions, endowed with the operations +, •,is the localization17
of R at the multiplicative subset S. Prove that 5_1i? is a commutative ring and
a 1—?
defines a
y
that the function
Prove that
£(s)
is invertible for every
Prove that R
that
7(5)
ring homomorphism £ : R
s G
S~lR.
S.
S~lR is initial among ring homomorphisms /
5 G S.
i?' such
R
:
is invertible in R' for every
Prove that S~lR is
Prove that S~lR
an
integral domain if R
is the zero-ring if and
is
an
only if
integral domain.
0 G 5.
[4.8, 4.9, 4.11, 4.15, VII.2.16, VIII.1.4, VIII.2.5, VIII.2.6, VIII.2.12, §IX.9.1]
4.8.
multiplicative subset of a commutative ring i?, as in Exercise 4.7.
on the set of pairs (m, s), where m G M
R-module M, define a relation
Let S be
-1
For every
and
a
~
S:
s G
(m,s)
~
(ra',57)
<^>
(3*
G
5), ^/ra
-
sra;)
=
0.
an equivalence relation, and define an S~1R-module structure
S~lM of equivalence classes, compatible with the i?-module structure
The module S~lM is the localization of M at S. [4.9, 4.11, 4.14, VIII.1.4,
Prove that this is
on the set
on M.
VIII.2.5, VIII.2.6]
4.9.
Let S be
-1
a
multiplicative subset of
a
commutative ring i?, and consider the
localization operation introduced in Exercises 4.7 and 4.8.
Prove that if / is
an
ideal of R such that IDS
=
0, then18
Ie
:=
5-1/ is
a proper
ideal of S-XR.
If £
of
:
S_1R is the natural homomorphism, prove that if J is
:=
£~l( J) is an ideal of R such that J n S 0.
R
S~lR,
then Jc
J, while (Ie)c
{a e R\(3s e S) sa e I}.
that
example showing
(Ie)c need not equal /, even if / fi S
Prove that (Jc)e
Find
an
Let S
=
a proper
ideal
=
=
{l,£,x2,... }
=
in R
=
C[x,y\.
What is
(Ie)c
for /
=
=
0. (Hint:
(xy)l)
[4.10, 4.14]
The terminology is motivated by applications to algebraic geometry; see Exercise VII.2.17.
The superscript e stands for 'extension'
the superscript
c in the next
(of
the ideal from
point stands for 'contraction'.
a smaller
ring
to a
larger ring);
V.
278
4.10.
gives
With notation
-i
as
from S and the set of ideals
disjoint from S.) [4.16]
(Notation
-i
assignment p i—?
5_1p
prime ideals of R disjoint
S_1R. (Prove that (pe)c
p if p is a prime ideal
of
set oi
=
in Exercise 4.7 and
as
integral domains
in
in Exercise 4.9, prove that the
inclusion-preserving bijection between the
an
4.11.
Irreducibility and factorization
4.8.)
A ring is said to be local if it has
a
single maximal ideal.
Let R be a commutative
set S
=
R
denoted
ring, and let p be a prime ideal of R. Prove that the
p is multiplicatively closed. The localizations S~1R, S~lM are then
i?p, Mp.
an inclusion-preserving bijection between the prime ideals
of Rp and the prime ideals of R contained in p. Deduce that Rp is a local ring19.
Prove that there is
[4.12, 4.13, VI.5.5, VII.2.17, VIII.2.21]
4.12.
be
an
Af
(Notation
-i
as
in Exercise
Let R be
4.11.)
i?-module. Prove that the following
=
Mp
a
commutative ring, and let M
equivalent20:
0.
=
Mm
are
=
0 for every prime ideal p.
0 for every maximal ideal
m.
(Hint: For the interesting implication, suppose that M ^ 0; then the
| (Vra G M),rm 0} is proper. By Proposition 3.5, it is contained in
ideal m. What can you say about Mm?) [VIII.1.26, VIII.2.21]
R
=
4.13.
-i
Let k be
a
field, and let
v
be
a
discrete valuation
corresponding DVR, with local parameter t
Prove that R is local
(see
(Exercise 4.11), with
m is invertible.)
Exercise
on
k.
ideal
a
{r
G
maximal
Let R be the
2.20).
maximal ideal
m
=
(t). (Hint:
Note
that every element of R
Prove that k is the field of fractions of R.
a PID, and let p be a prime ideal in A. Prove that the localization
(p), define a valuation on the field
Ap (cf. Exercise 4.11) is a DVR. (Hint: If p
Now let A be
=
of fractions of A in terms of 'divisibility by
p\)
[VII.2.18]
We have the following picture in mind associated with the two operations
R/P, Rp
for
a
prime ideal P:
Spec R/P
p*
Spec/?
Spec Rp
Can you make any sense of this?
The way to think of this type of results is that a module M over R is zero if and only if it
is zero 'at every point of Spec J?'. Working in the localization Mp amounts to looking 'near' the
point p of Spec R; this is what is local about localization. As in this exercise,
'global' features by looking locally at every point.
one can
often detect
Exercises
279
4.14. With notation
as
Ne and N
in Exercise 4.8, define operations N
for submodules N C M, N C
5_1M, respectively, analogously
defined in Exercise 4.9. Prove that (Nc)e
=
Nc
i—?
i—
to the
operations
N. Prove that every localization of
a
Noetherian module is Noetherian.
In particular, all localizations S~XR of
4.15.
Exercise
-i
Let R be
a
UFD, and let S be
a
a
Noetherian ring
are
Noetherian.
multiplicatively closed subset of R (cf.
4.7).
Prove that if q is irreducible in i?, then
S~lR.
Prove that if
a/s
is irreducible in
S~1R,
q/1
then
is either irreducible
a/s
is
an
or
associate of
unit in
a
q/1
for
some
irreducible element q of R.
Prove that 5_1i? is also
a
UFD.
[4.16]
4.16. Let R be
a
Noetherian integral domain, and let
element. Consider the multiplicatively closed subset S
s
=
G
i?,
s
^ 0,
{1, s, s2,... }.
be
prime
a
Prove that
R is a UFD if and only if S~lR is a UFD. (Hint: By Exercise 2.10, it suffices to
show that every prime of height 1 is principal. Use Exercise 4.10 to relate prime
ideals in R
prime ideals in the localization.)
On the basis of results such as this and of Exercise 4.15, one might suspect
that being factorial is a local property, that is, that R is a UFD if and only if Rp
to
a UFD for all primes p, if and only if Rm is a UFD for all maximals m.
This
is regrettably not the case. A ring R is locally factorial if Rm is a UFD for all
maximal ideals m; factorial implies locally factorial by Exercise 4.15, but locally
is
factorial rings that
are not
factorial do exist.
> Let F be a field, and recall the notion of characteristic of a ring
(Definition III.3.7); the characteristic of a field is either 0 or a prime integer
(Exercise III.3.14.)
4.17.
Show that F has characteristic 0 if and only if it contains
F has characteristic p if and only if it contains
Show that
(in
the prime
subfield of
both
cases)
a copy
a copy
of the field
of Q and that
Z/pZ.
this determines the smallest subfield of F; it is called
F.
[§5.2, §VII.1.1]
4.18.
are
-i
Let R be
an
integral domain. Prove that the invertible elements
as constant polynomials.
[4.20]
in
R[x]
the units of i?, viewed
a G R in a ring is said to be nilpotent if an
0 for some
nilpotent, then 1 + a is a unit in R. [VI.7.11, §VII.2.3]
4.19. > An element
Prove that if
4.20.
a
is
=
Generalize the result of Exercise 4.18
as
follows:
let R be
h adXd G R[x]; prove that / is
a$ + a\X-\
ring, and let /
if
in
is
a
unit
and
R
ao
ai,..., a^ are nilpotent. (Hint: If
only
=
is the inverse of /, show by induction that
that ad is
nilpotent.)
al^~lbe-i
=
a
bo
a
0.
commutative
unit in
+
n >
R[x]
b\X-\
if and
h
bexe
0 for all i > 0, and deduce
V.
280
Irreducibility and factorization
4.21. > Establish the characterization of irreducible
in Corollary 4.17.
4.22. Let k be a
field, and let /, g
/ and g have a nontrivial
common factor in k[x,y].
a
be two
common
polynomials
factor in
f(x)
as a
product of factors
Deduce that if
then
a(x),fi(x)
4.24.
are
=
a
UFD given
k(x)[y],
k[x, y]
=
k[x] [y].
Prove that
then they have
G
R[x],
that there exists
and
a
c
a
nontrivial
assume
G
f(x)
=
K such that
(ca(x))(c-l(3(x))
in
R[x\.
a(x)/3(x)
f(x) G R[x] is monic and a(x) G K[x] is monic,
both in R[x] and fi(x) is also monic. [§4.3, 4.24, §VII.5.2]
=
In the same situation as in Exercise
coefficient of
in
UFD, K its field of fractions, f(x)
a(x)P(x) with a(x), (3{x) in K[x\. Prove
ca(x) G R[x], c~l(3(x) G R[x], so that
splits f(x)
over a
polynomials
[§4.3]
if
4.23. > Let R be
integral domains
in
4.23,
prove that the
product of
any
with any coefficient of (3 lies in R.
4.25. Prove Fermat's last theorem
for polynomials:
the equation
/, g, h not all constant. (Hint: First, prove
tn
/,
relatively prime. Next, the polynomial 1
factorizes in C[t] as U7=i(l
for
deduce
that
e2™/n;
C
fn
CO
U7=i(h C$0Use unique factorization in C[t] to conclude that each of the factors h
Qg is an
n-th power. Now let h
h
h
cn
is
where
the
n > 2
an,
bn,
g
Qg
C^g
(this
this
to
obtain
a
relation
where
Use
+
A,
hypothesis enters).
(Xa)n (/x6)n
(^c)n,
has
no
solutions in
that
C[t]
for
n >
2 and
g, h may be assumed to be
~
=
=
=
=
~
=
=
//,
v are
suitable complex numbers. What's wrong with
this?)
pattern of proof would work in any environment where unique
factorization is available; if adjoining to Z an n-th root of 1 lead to a unique factorization
The
same
domain, the full-fledged Fermat's last theorem would be
as easy to prove as
indicated in this exercise. This is not the case, a fact famously missed by G. Lame as
he announced a 'proof of Fermat's last theorem to the Paris Academy on March 1,
1847.
5.
Irreducibility
of
polynomials
Our work in §4.3, especially Corollary 4.17, links the irreducibility of
over a UFD with their irreducibility over the corresponding field of fractions.
For example, irreducibility of polynomials in Z[x] is 'essentially the same' as
polynomials
irreducibility in Q[x]. This can be useful in both directions, provided we have ways
to effectively verify irreducibility of polynomials over one kind of ring or the other.
We collect here several remarks aimed at establishing whether a given polynomial
is irreducible and briefly discuss the notion of algebraically closed field.
5.
Irreducibility of polynomials
281
5.1. Roots and reducibility. Let R be a ring and
a root if f(a)
0. Recall (Example III.4.7) that
is
=
/
a
G
R[x].
An element
polynomial f(x)
G
a G
R
R[x]
is
by (x a) if and only if a is a root of /. More generally, we say that a is a
root of / with multiplicity r if (x
a)r divides / and (x a)r+1 does not divide r.
The reader is probably familiar with this notion from experience with calculus. For
divisible
-
-
-
example, the graph of polynomials with roots of multiplicities 1, 2, and 3 at 0 look
(near the origin), respectively, like
V
Lemma 5.1.
degree
n.
Let R be
an
Then the number
integral domain, and let f G R[x] be a polynomial of
of roots of f, counted with multiplicity, is at most n.
Proof. The number of roots of
viewed
/
byK.
of
as a
polynomial
/
over
in R is less than
or
equal
to the number of roots
the field of fractions K of R\ so
we may
replace R
Now, K[x] is a UFD, and the roots of / correspond to the irreducible factors
/ of degree 1. Since the product of all irreducible factors of / has degree n, the
number of factors of degree 1 can be at most n, as claimed.
of
It is worth
remembering the trick of replacing
an
integral domain R by
its field
of fractions K, used in this proof: one is often able to use convenient properties
due to the fact that K is a field (the fact that K[x] is a UFD, in this case), even
if R is very far from
itself need
not be a
satisfying
such properties
(R[x\ may
not be a
UFD, since R
UFD).
Also note that the statement of Lemma 5.1 may fail if R is not an
For example, the degree 2 polynomial x2 + x has four roots over
domain.
integral
Z/6Z.
V.
282
Irreducibility and factorization
in
integral domains
The
it
innocent Lemma 5.1 has important applications: for example, we used
(without much fanfare) in the proof of Theorem IV.6.10. Also, recall that a
polynomial /
R[x]
G
determines
an
'evaluation functions'
(cf. Example III.2.3)
checked in Exercise III.2.7 that in general
i?, namely
f(r).
the function does not determine the polynomial; but polynomials are determined
by the corresponding functions over infinite integral domains:
r i—?
The reader
R
Corollary 5.2. Let R be an infinite integral domain, and let f,g G R[x] be
polynomials. Then f
g if and only if the evaluation functions r i-> f(r), r i->
=
g(r)
agree.
Proof.
f
Indeed, the
two functions agree if and
g\ but a nonzero
polynomial
over
only if
every a G R is a root of
R cannot have infinitely many roots, by
Lemma 5.1.
Clearly, an irreducible polynomial of degree > 2
holds for degree 2 and 3, over a field:
can
have
The
no roots.
converse
Proposition 5.3. Let k be a field. A polynomial
irreducible if and only if it has no roots.
f
k[x] of degree
G
2
or
3 is
Proof. Exercise 5.5.
Example
5.4. Let F2 be the field
The polynomial
/(0)
/(l)
(because F2[£]
=
f(t)
t2 +
=
Z/2Z.
t + 1 G
1. Therefore the ideal
=
is
construction of
a
a
PID: don't
(t2
F2[f]
+1 +
is irreducible, since it has no roots:
1) is prime in F2[£], hence maximal
forget Proposition III.4.13).
This gives
a
one-second
field with four elements:
F2[*]
(t2+t + l)'
(hopefully) constructed this field in Exercise III. 1.11, by the tiresome
process of searching by hand for a suitable multiplication table. The reader will
now have no difficulty constructing much larger examples (cf. Exercise 5.6).
j
The reader
Proposition 5.3 is pleasant and can help to decide irreducibility over more
general rings: for example, a primitive polynomial / G Z[x] of degree 2 or 3 is
irreducible if and only if it has no roots in Q. Indeed, / is irreducible in Z[x]
if and only if it is irreducible in Q[x] (by Corollary 4.17).
1
Ax2
(2x \){2x+ 1) is primitive and reducible in
=
e.g.
no
for
roots. Also, keep in
polynomials of degree > 4:
integer
Q[x],
but it has
no
mind that the statement of
for
example, x4
+ 2x2 + 1
=
However, note that
Z[x], although it has
Proposition 5.3
(x2 + l)2
may fail
is reducible in
rational roots.
Looking for rational roots of a polynomial in Z[x] is in
endeavor, due to the following observation (which holds over
often called the 'rational root test'.
Proposition 5.5. Let R be
f(x)
a
=
UFD, and let K be
a0 + a±x H
h
its
anxn
principle
every
field of fractions.
G
R[x],
a
finite
UFD).
Let
This is
Irreducibility of polynomials
5.
and let
q
|
c
=
|
G K be a root
283
of f,
with p,q G R,
gcd(p,q)
1.
=
Then p
ao and
an in R.
Proof.
By hypothesis,
pn
p
a0 + Oi- H
q
that
=
^an~r
qn
0;
is,
a0gn
+
aipq71'1
+
+
anpn
=
0.
Therefore
a0gn
=
-p(aign-1
+
+
---
anpn-1),
proving that
p | (ao#n). Since the gcd of p and g is one, this implies that the
multiset of factors of p is contained in the multiset of irreducible factors of ao, that
is, p
|
o0.
An
entirely similar argument
Looking for rational
5.6.
Example
proves that
3
-
2x +
roots of the
3x2
-
is therefore reduced to trying fractions
|
+
polynomial
3x4
2x5
-
±1,±2, p
±1,±3. As it
and
it
follows that it
possibilities,
with q
is the only root found among these
rational
root of the polynomial.
only
=
=
j
closed fields. A homomorphism of
Adding roots; algebraically
5.2.
2xs
|
happens,
is the
q\an.
fields
i:k^F
is necessarily injective (cf. Exercise III.3.10): indeed, its kernel is a proper ideal
of /c, and the only proper ideal of a field is (0). In this situation we say that F
(or more properly the homomorphism i) is an extension of k. Abusing language a
little,
we
then think of k
case as soon as
as
contained in F: k C F. Keep in mind that this is the
F.
there is any ring homomorphism k
standard procedure for constructing extensions in which a given
polynomial acquires a root; we have encountered an instance of this process in
Example III.4.8. In fact, the resulting field is almost universal with respect to this
There is
a
requirement.
Proposition
5.7.
Let k be
a
field,
and let
f(t)
be
k[t]
G
a
nonzero
irreducible
polynomial. Then
'
(/w)
is a
field,
endowed with
composition k
f(x)
G
^
k[x]
a
natural homomorphism i
k[x]
F) realizing
C
has
F[x]
a root in
:
it as an extension
F, namely the
k
coset
F
^->
of k.
(obtained
Further,
oft;
as
the
Irreducibility and factorization
V.
284
ifkQKis
phism j
:
f has
diagram
any extension in which
F
K such that the
a
in
integral domains
root, then there exists
a
homomor-
>K
k^
i\
W
F
3
commutes.
Proof. Since k is
Proposition
F
a
is
field, k[t]
a
PID; hence (/(£)) is
III.4.13. Therefore F is indeed
by underlining,
we
k[t], by
k[t]/(f(t))
maximal ideal of
=
have
/(£)
as
a
field. Denoting cosets in
a
=
/W=°>
claimed.
To
f(u)
=
verify
the second part of the statement, suppose k C K is an extension and
u G K. This means that the evaluation homomorphism
0, with
k[t]
e :
defined
by e(g(t))
:=
g(u)
vanishes at
property of quotients gives
a
/(£);
K
-?
hence
(f(t))
C
ker(e),
and the universal
unique homomorphism
D
satisfying the stated requirement.
We have stopped short of claiming that the extension constructed in
Proposition 5.7 is universal with respect to the requirement of containing a root of /
because the homomorphism j appearing in the statement is not unique: in fact,
the proof shows that there are as many such homomorphisms as there are roots of
/ in the larger field K. We could say that F is versal, meaning 'universal without
uni(queness)'. We would have full universality if we included the information of the
root of /; we will come back to this construction in Chapter VII, when we analyze
field
extensions in greater
Example
is
5.8. For k
(isomorphic to)
=
depth.
R and
C: this
was
f(x)
=
x2+l,
checked
the field constructed in Proposition 5.7
carefully
in
Example III.4.8.
Similarly, Q[£]/(£2
2) produces a field containing Q and in which there is a
'square root of 2'. There are two embeddings of this field in R, because R contains
two distinct square roots of 2:
±y/2.
j
/ G k[x] acquires a root in the
Proposition 5.7 implies that / has a linear (i.e., degree 1)
F; in particular, if deg(/) > 1, then / is no longer irreducible over F.
The fact that the irreducible polynomial
extension F constructed in
factor
over
Given any polynomial
which / factors completely
in which this
/
as a
already happens
hence it is given
a name.
G
k[x],
it is easy to construct
an
extension of k in
product of linear factors (Exercise 5.13). The case
nonzero / is very important, and
in k itself for every
5.
Irreducibility of polynomials
algebraically closed if all irreducible polynomials
Definition 5.9. A field k is
k[x]
285
have degree 1.
in
j
We have encountered this notion in passing, back in Example III.4.14
III.4.21). The following lemma is left to the reader:
(and
Exercise
Lemma 5.10.
f
A
k is algebraically closed
field
k[x] factors completely
G
nonconstant
polynomial f
G
as
k[x]
a
if and only if every polynomial
linear
product of
factors, if and only if every
has
a root in
k.
In other words, a field k is algebraically closed if the only polynomial equations
with coefficients in k and no solutions are equations 'of degree 0', such as 1
0.
=
The result known
Nullstellensatz
as
this result in Chapter III) is
the pillars of (old-fashioned)
in
a vast
due to David Hilbert;
(also
we
have mentioned
generalization of this observation and
algebraic geometry.
We will
come
one
of
back to all of this
§VII.2.
Is any of the finite fields we have encountered (such as Z/pZ, for p prime)
closed? No. Adapting Euclid's argument proving the infinitude of
algebraically
prime integers
reveals that algebraically closed fields
(Exercise 2.24)
Proposition 5.11. Let k be
an
algebraically closed field. Then k
is
are
infinite.
infinite.
Proof. By contradiction, assume that k is algebraically closed and finite; let the
elements of k be ci,..., cjy.
Then there
ci),..., (x
are
cn).
N irreducible monic
exactly
polynomials
in
k[x], namely (x
Consider the polynomial
f(x)
=
(x-ci)
(x-cN)
+ l:
for all c G k we have
/(c)
since
c
equals
one
=
(c-ci)
of the q's. Thus
(c-cN)
f(x)
is
+ l
=
a nonconstant
1^0,
polynomial with
no
roots,
contradicting Lemma 5.10.
Since finite fields are not algebraically closed, the reader may suspect that
algebraically closed fields necessarily have characteristic 0 (cf. Exercise 4.17). This
is not so: there are algebraically closed fields of any characteristic. In fact, as we will
see in due time, every field F may be embedded into an algebraically closed field;
the smallest such extension is called the 'algebraic closure' of F and is denoted F.
Thus, for example, Z/2Z is an algebraically closed field of characteristic 2.
The algebraic closure Q is a countable subfield of C. The extension Q C Q,
and especially its 'Galois group', cf. §VII.6, is considered one of the most important
objects of study in mathematics (cf. the discussion following Corollary VII.7.6).
We will construct the algebraic closure of
5.3. Irreducibility in
completely over C. Indeed,
a
field rather explicitly, in §VII.2.1.
C[x], R[#], Q[x], Every polynomial /
Theorem 5.12. C is algebraically closed.
G
C[x]
factors
Gauss
(which
(we
a
sketch of such
a
is,
\z\
find
fundamental theorem of algebra require more than we
one in §VIL6, after we have seen a little Galois
will encounter
little complex analysis makes the statement nearly trivial. Here
argument. Let / G C[x] be a nonconstant polynomial; the
an
task (cf. Lemma 5.10)
we can
integral domains
the
as
'Algebraic' proofs of the
theory); curiously,
in
providing the first proof of this fundamental theorem
fundamental theorem of algebra.)
is credited with
is indeed known
know at this point
is
Irreducibility and factorization
V.
286
an r
G
consists of proving that / has
R large enough that \f(z)\>
a root
|/(0)|
in C. Whatever
for all
z
on
/(0)
the circle
r
(since / is nonconstant, lim^oo \f(z)\ +00). The disk \z\< r is compact,
the continuous function \f(z)\
has a minimum on it; by the choice of r, it must
=
so
=
be somewhere in the interior of the
Exercise
(cf.
principle
say at
disk,
then implies that
5.16)
z
f(a)
=
=
The minimum modulus
a.
0, q.e.d.
By Theorem 5.12, irreducibility of polynomials in C[x] is
be:
polynomial /
a nonconstant
Every /
G
C[x]
G
C[x]
5.13.
1 and the
-
simple:
R[x] of degree > 3 is reducible.
polynomials in R[x] are precisely the polynomials
G
quadratic polynomials
f
with b2
as
Every polynomial f
The nonconstant irreducible
of degree
=
ax2
+ bx + c
Aac < 0.
Proof. Let
/
G
l[x]
be
a nonconstant
/
=
polynomial:
a0 + a\xH
with all ai G R. By Theorem 5.12,
/
has
Applying complex conjugation
a0 + a±z H
h
an~zn
this says that ~z is also
either
z
of / in
in
7^
=
R[x];
=
=
0.
noting that
z\-+~z and
h a^~zn
of /. There
r was
anzn
root z:
are two
=
a*
=
a* since a* G
a0 + a±z H
R,
V anzn
=
0
:
possibilities:
real to begin with, and hence
(x
r)
is
a
factor
or
In this
(x
case
x2
divides
z
anxn,
complex
h
a^ + aTz H
a root
~z, that is,
2, and then
C[x],
=
V
a
a0 + a\zH
z
simple as it can
if
only it has degree 1.
as
of degree > 2 is reducible.
Over R, the situation is almost
Proposition
is irreducible if and
z) and (x
(since C[x]
-
(z
+
~z)x
~z)
is
a
are
nonassociate irreducible factors of
/
UFD!)
+ z~z
=
(x
-
z) (x
-
~z)
/.
Actually,
Gauss's first proof apparently had a gap; the first rigorous proof is due to Argand.
Later, Gauss fixed the gap in his proof and provided several other proofs.
5.
Irreducibility of polynomials
Since both
z + ~z
nonconstant
/
and
zz
are
has
R[x]
G
an
287
both real numbers, this analysis shows that every
irreducible factor of degree < 2, proving the first
statement.
The second statement then follows immediately from Proposition 5.3 and the
fact that the quadratic polynomials in R[x] with no real roots are precisely the
polynomials ax2 + bx + c with b2 Aac < 0.
-
The
following
remark is
an
immediate consequence of Proposition 5.13 and is
otherwise evident from elementary calculus
Corollary
5.14.
Every polynomial f
G
(Exercise 5.17):
R[#] of odd
degree has
a
real root.
Summarizing, the issue of irreducibility of polynomials in C[x] or R[x] is very
straightforward. By contrast, irreducibility in Q[x] is a complicated business: as we
have seen (Proposition 4.16), it is just as subtle as irreducibility in Z[x\. There is
no sharp characterization of irreducibility in Z\x\\
but the following simple-minded
remarks
Eisenstein's criterion, discussed in
(and
§5.4)
suffice in many interesting
cases.
One simple-minded remark is that if (p
is reducible in i?, then
a
=
be, then (p(a)
Of
(p(c)
course
=
is
likely
(p(a)
(p(b)(p(c).
S is
R
:
a
to be reducible in S.
this is not quite right: for example because
may be units in S. But with due care for
special
homomorphism, and a
This is just because if
(p(a) and/or (p(b) and/or
cases
this observation may
(p(a) is irreducible
be used to great effect, especially in the contrapositive form: if
in 5, then a is (likely...) irreducible in R.
Here is
formalizing these remarks, for R Z[x\. The natural
a surjective homomorphism tt : Z[x]
Z/pZ
Z/pZ[x\:
is
obtained
from
quite simply, 7r(/)
/ by reading all coefficients modulo p; we will
write '/modp' for the result of this operation.
a
simple
statement
projection Z
=
induces
Proposition 5.15. Let
f
G
Z[x]
integer. Assume fmodp has the
Then
is irreducible in
f
be
a
primitive polynomial, and let p be
same
degree
as
f
and is irreducible in
a
prime
Z/pZ[x\.
Z[x\.
Proof.
Argue contrapositively: if / is primitive and reducible in Z[x] and deg/
n, then f
d, deg/i
e, d + e
n, and both d, e, positive. But
gh with degg
then the same can be said of /modp, so fmodp is also reducible.
=
=
=
The
hypothesis
cases' mentioned in
Sx2
+ Sx +
yet (2x
+
1)
=
=
is necessary,
the discussion preceding
on
degrees
precisely to take care of the 'special
the statement. For example, (2x3 +
1) equals (x2
is
a
+ x + 1) mod 2, and the latter is irreducible in Z/2Z[x];
factor of the former. The point is of course that 2x + 1 is a unit
mod 2.
Warning: As
we
will
see
in
a
fairly
distant future
(Example VII.5.3),
there
are irreducible polynomials in Z[x] which are reducible modulo all primes!
So
Proposition 5.15 cannot be turned into a characterization of irreducibility in Z[x],
It is however rather useful in producing examples. For instance,
V.
288
Corollary
5.16.
There
are
Irreducibility and factorization
irreducible polynomials in
Z[x]
in
and
integral domains
Q[x] of arbitrarily
large degree.
Proof. By Proposition 4.16, the statement for Z[x] implies the one for Q[x], By
Proposition 5.15, it suffices to verify that there are irreducible polynomials in
of arbitrarily large degree, for any prime integer p.
of Exercise 5.11.
This is
Z/pZ[x]
case
5.4. Eisenstein's criterion. We
giving this result
are
in
a
a
particular
separate subsection
of the fact that it is very famous; it is also very simple-minded,
only
like the considerations leading up to Proposition 5.15, and further it is not really a
on account
'criterion',
in the
that it also does not give
sense
a
characterization of irreducibility.
The result is usually stated for Z, but it holds for every ring R.
Proposition 5.17. Let R be
a
(commutative) ring,
and let p be
a
prime ideal of
R. Let
f
be
a
polynomial, and
an
Then
f
V
for
i
=
0,..., n
=
R[x]
0p2.
is not the
product of polynomials of degree
Argue by contradiction. Assume /
h
deg less than n
deg /; write
=
gh
R[x].
< n in
in
R[x],
with both d
=
degg and
=
g
and
G
1;
Proof.
e
anxn
that
assume
0 p;
ai G p
a0
ao + a\xH
=
note that
=
b0
+
b±x
necessarily d
H
h
bdxd,
where / denotes / modulo p,
=
gh
=
g_
e >
=
0, this implies bo
=
boco
G
c0 + c\XH
/
h
cexe,
modulo p: thus
m(R/p)[x],
etc.
By hypothesis, /
anxn modulo
factors
of
integral domain,
/ must also
But then ao
=
> 0 and e > 0. Consider
l
Since d > 0,
h
p, where an
^
0 in
be monomials: that
bdxd,
h
=
R/p.
Since
R/p
is
an
is, necessarily
cexe.
G p, Co G p.
p2, contradicting
the hypothesis.
For example, x4 + 2x2 + 2 must be irreducible in Z[x] (and hence in Q[x]): apply
Eisenstein's criterion with R
Z, p
(2). More exciting applications involve whole
classes of polynomials:
=
=
Example 5.18. For all n and all primes p, the polynomial xn
p is irreducible
in Z[x], This follows immediately from Eisenstein's criterion and gives an
alternative proof of Corollary 5.16.
j
Exercises
289
Example 5.19. This is
criterion. Let p be a prime
probably the most
integer, and let
f(x)
These polynomials
are
=
x2
1 + x +
famous
+
called cyclotomic;
+
we
application of Eisenstein's
xp~l
G
Z[x].
will encounter them again in
§VII.5.2.
It may not look like it, but Eisenstein's criterion may be used to prove that
is irreducible. The trick (which the reader should endeavor to remember) is
f(x)
x + 1.
apply the shift x —>
only if f(x + 1) is; thus we are
to
f{x
+
1)
=
1 +
It is
hopefully
clear that
(x
+
1)
+
(x
+
l)2
is irreducible. This is better than it looks: since
+
+
f(x)
=
(x
For p prime and fc
There is nothing
to
1,... ,p
=
+
l)p~l
(xp
(p)
Eisenstein's criterion proves that this is irreducible, since
Claim 5.20.
is irreducible if and
f(x)
reduced to showing that
1, p divides
l)/(x
=
1),
p and
©•
this, because
\k)
k\(p-k)\
and p divides the numerator and does not divide the denominator; the claim follows
as p is prime. It may be worthwhile noting that this fact does not hold if p is not
prime: for
example,
4 does not divide
(2).
j
Exercises
5.1.
if
-1
f(a)
Let
=
f(x)
f(a)
G
C[x]
=
=
prove that a is a root of
f(r~l\a)
value of the k-th derivative of
if and only if
/
gcd(f(x)J'(x)) ^
=
0 and
/
f^r\a)
at a. Deduce that
1.
if and only
denotes
the
f{k\a)
with multiplicity
=
0, where
f(x)
G
C[x]
r
has multiple roots
[5.2]
C, and let f(x) be an irreducible polynomial in F[x].
multiple roots in C. (Use Exercises 2.22 and 5.1.)
5.2. Let F be a subfleld of
Prove that
f(x)
has
no
h a2x2
a0x2ri + a2x2n~2 -\
ring, and let f(x)
polynomial only involving even powers of x. Prove that if g(x) is
5.3. Let R be
so
is
a
=
+ a0 e
a
R[x]
factor of
be
a
/(#),
g(—x).
5.4. Show that xA + x2 + 1 is reducible in
without finding its
5.5. > Prove
(complex)
Z[x\. Prove
that it has
no
rational roots,
roots.
Proposition 5.3. [§5.1]
5.6. > Construct fields with 27 elements and with 121 elements.
[§5.1]
V.
290
5.7. Let R be
integral domain, and let f(x) G R[x]
by its value at any d +
an
Prove that f(x)
Irreducibility and factorization
is determined
be
a
integral domains
in
polynomial of degree d.
1 distinct elements of R.
field, and let ao,..., Q>d be distinct elements of K. Given any
K, construct explicitly a polynomial f(x) G K[x] of degree d
such that /(ao)
bo,..., f(ad)
bd, and show that this polynomial is unique.
First
solve
the
problem
assuming that only one b{ is not equal to zero.) This
(Hint:
5.8.
Let K be
-i
elements
a
in
bo,... ,bd
=
=
process is called Lagrange interpolation.
5.9.
-i
Pretend you
Exercise
over
5.8)
Q[x].
to
give
a
can
[5.9]
factor integers, and then
finite
algorithm
to factor
Use your algorithm to factor
use
Lagrange interpolation (cf.
polynomials22
(x
l)(x
2){x
with integer coefficients
S)(x
4)
+ 1.
[5.10]
1 is irreducible in
Q[x]
in
F[x]
of arbitrarily high degree. (Hint: Exercise 2.24. Note that Example 5.16 takes
of Z/pZ, but what about all other finite fields?) [§5.3]
care
5.10. Prove that the
for all
n >
1.
polynomial (x l)(x 2)
(x n)
along the lines of Exercise 5.9.)
(Hint:
-
-
-
-
Think
5.11. > Let F be a finite field. Prove that there are irreducible
5.12.
Prove that
applying the
linear
polynomial
in
5.13.
>
Let k be
a
construction in
k[x] produces
polynomials
Proposition 5.7
to an irreducible
field isomorphic to k.
a
field, and let / G k[x] be any polynomial. Prove that there is an
/ factors completely as a product of linear terms. [§5.2,
extension k C F in which
§VI.7.3]
5.14.
How many different
embeddings of the field Q[£]/(£3
2)
are
there in R?
How many in C?
5.15. Prove Lemma 5.10.
5.16. > If you know about the 'maximum modulus principle' in complex analysis:
formulate and prove the 'minimum modulus principle' used in the sketch of the
proof of the fundamental theorem of algebra. [§5.3]
5.17. > Let
theorem
to
/ G R[#] be a polynomial of odd degree. Use the intermediate value
give an 'algebra-free' proof of the fact that / has real roots. [§5.3,
§VII.7.1]
5.18. Let / G Z[x] be a cubic polynomial such that /(0) and
with odd leading coefficient. Prove that / is irreducible in Q[x].
5.19. Give
a
proof of the fact that y/2
is not rational
5.20. Prove that x6 + 4x3 + 1 is irreducible
5.21. Prove that 1 +
5.22.
Let R be
a
x +
x2 -\
UFD, and let
are
odd and
by using Eisenstein's
criterion.
by using Eisenstein's
+ xn~l is reducible
a
G
R be
/(l)
an
over
Z if
n
criterion.
is not prime.
element that is not divisible by
Prove that xn
a is
the square of some irreducible element in its factorization.
irreducible for every integer n > 1.
'It is in fact much harder to factor integers than integer polynomials.
6. Further remarks and examples
5.23. Decide whether
y5
+ x
y3
+
291
x3y
C[x, y, z, w]/(xw
5.24. Prove that
+ x is reducible or irreducible in
yz)
is
integral domain, by using Eisenstein's
§2.2, but we did not have the patience
an
used this ring as an example in
to prove that it was a domain back then!)
criterion.
(We
Further remarks and
6.
C[x, y],
examples
Chinese remainder theorem. Suppose all you know about an integer is
its class modulo several numbers; can you reconstruct the integer? Also, if you
are given arbitrary classes in Z/nZ for several integers n, can you find an N G Z
6.1.
satisfying
all these congruences
simultaneously?
If an integer N satisfies given
multiple of n\ n^ to N produces
an integer satisfying the same congruences (so N cannot be entirely reconstructed
1 mod 2 and N
from the given data); and there is no integer N such that N
2 mod 4, so there are plenty of congruences which cannot be simultaneously satisfied.
The
answer
is
no
in both cases, for trivial
congruences modulo ni,..., n^, then
adding
reasons.
any
=
=
The Chinese remainder theorem (CRT) sharpens these questions so that they
in fact be answered affirmatively, and (in its modern form) it generalizes them
a much broader setting. Its statement is most pleasant for PIDs; it is very
can
to
impressive23
even
in the limited context of Z.
Protoversions of the theorem have
already appeared here and there in this book, e.g., Lemma
We will first give the more general version, which is in
R be any commutative ring.
Theorem
6.1. Let
Ji,..., J*, be ideals of R such that U
+
IV.6.1.
some way
Ij
=
simpler. Let
(1) for
all i
^ j.
Then the natural homomorphism
R
_
(f
is surjective and induces
an
R
:
R
x
>
x
h
h
isomorphism
R
~
h-'h
R
R
h
h
by the canonical projections
R/Ij and the universal property of products; the homomorphism <p is induced
The 'natural' homomorphism (p is determined
R
by
virtue of the universal property of quotients, since I\
-
Ik
C
Ij
for all j, hence
h'"Ik Q ker(p. In fact, clearly
ker (p
=
11
n
fl
Ik;
the second part of the statement of Theorem 6.1 follows immediately from the
first part, the 'first isomorphism theorem', and the following (independently
interesting) observation:
so
Allegedly,
the CRT.
a
large number of 'Putnam' problems
can
be
solved
by clever applications of
Irreducibility and factorization
V.
292
of R
Lemma 6.2. Let Ji,... ,/& be ideals
Then h
fl Ik.
Ik
h fl
such that U +
Ij
integral domains
in
=
(1) for
all i
^ j.
=
Proof. A simple induction reduces the general statement to the case k
2.
Therefore, assume / and J are ideals of i?, such that J+J
(1). The inclusion IJ C ID J
=
=
holds for all ideals /, J,
If / + J
if
the task amounts to proving If) J C IJ when /+J
then there exist elements
(1),
=
so
I, b
a
J such that
b
a +
=
=
(1).
1. But
J fl J, then
r
r
because
ra
e
IJ
as
r
=
1
r
=
J and
e
r(a
a
+
b)
=
I, while rb
e
rb G IJ
ra +
:
e IJ as r e I and b e J.
The
statement follows.
Thus,
under the
to set up.
only need to prove the first part of Theorem 6.1: the surjectivity of <p
given hypotheses. The proof of this is a simple exercise but may be tricky
The following remark streamlines it considerably:
we
Lemma 6.3.
Let /i,...,/^ be ideals
of R
such that U + Ik
(1) for
all i
=
ak G
U.
1, there is nothing
to
=
l,...,/c-l. T/ien(Ji •••/*_!)+Jfc(1).
=
Proof. By hypothesis, for i
1,..., k
=
1 there exists
Ik such that 1
ai G
Then
(1
-
oi)
(1
-
afc_i)
h
G
4-1,
and
1
because it is
a
Jfc,
G
Argue by
induction
k. For k
on
=
the statement is known for fewer ideals. Thus,
that the natural projection induces an isomorphism
For k >
assume
(1 -Oi)---(l -afc_i)
combination of ai,..., a^-i G Ik-
Proof of Theorem 6.1.
show.
-
1,
assume
Jrt
Jrt
h-'h-i
h
it
x
and all that
we
we may
x
h-i
have left to prove is that the natural homomorphism
R
R
R
y
h'-h-i
is surjective. By Lemma 6.3,
of two ideals.
(I\
h-i)
+ h
h
=
(1);
thus
we are
reduced to the
case
Let then 7, J be ideals of a commutative ring R, such that I + J
ri mod I and r
ri,rj G R; we have to verify that 3r G R such that r
=
=
Since I + J
=
(1),
there
are a
I, b
J such that
a +
6
=
1. Let
then
as a G
as b G
r
=
r
=
arj +
(1
-
a)ri
=
ri +
a(rj
-
r/)
=
ri
mod I
rj
mod J
7, and
J,
as
(1
-
6)rj
+
bri
=rj +
b(ri
needed, and completing the proof.
-
rj)
=
r
(1),
=
=
and let
rj mod J.
arj +
6rj:
6. Further remarks and examples
In
293
PID, the CRT takes the following form:
a
6.4.
Corollary
gcd(a^, dj)
=
1
Let R be
all i
for
PID, and let
a
Let
^ j.
a
=
a\
R
(f
defined by
This
(a)
r +
(r
i->
+
Then the
ak.
R
x
-—-
'
(a)
x
(oi)
(ai),..., r
+
function
R
—>
:
ai,...,ak G R be elements such that
(ofc)
(a/-))
is an
isomorphism.
immediate consequence of Theorem 6.1, since (in a PID!) gcd(a, b)
(1) as ideals. This is not the case for arbitrary UFDs, and
indeed the natural map
1 if and
is
an
=
only if (a, b)
=
AA
(2)
(x)
1. But the kernel of this
is not surjective (check this!), even though gcd(2,x)
in
is
and
this
is
what
can
be
map
expected
general from the CRT over UFDs
(2a;);
=
(Exercise 6.6).
As Z is
PID, Corollary 6.4 holds for Z and gives the promised
a
answer to a
revised version of the questions posed at the beginning of this subsection. In fact,
tracing the proof in this case gives an effective procedure to solve simultaneous
congruences over Z
the argument).
To
this
see
and let
n
=
prime24, and
(or
in fact
over any
explicitly, let
more
-n/-; for each
ni
Euclidean domain,
as
will be apparent from
ni,... ,rik be
i, let
=
rrii
pairwise relatively prime integers,
n/rii. Then rii and rrii are relatively
the Euclidean algorithm to explicitly find integers a^,^
we can use
such that
aiUi + biirii
The numbers qi
=
=
1.
birrii have the property that
qi
=
1
mod rii,
qi =0
mod rij
\/j^
i
(why?). These integers may be used to solve any given system of congruences
modulo ni,..., rik'. indeed, if r\,...,r^ G Z are given, then
N
:=riqi +
+ rkqk
h 7v H
h rk
satisfies
N
=
ri
0 H
0
=
n
mod m
for all i.
The same
Exercise 6.7 for
an
procedure
example.
can
be applied
over any
Euclidean domain R;
see
:This is a particular case of Lemma 6.3 and is clear anyway from gcd considerations.
V.
294
Irreducibility and factorization
in
integral domains
integers. The rings Z and k[x] (where k is a field) may be the
of
Euclidean domains known to our reader. The next most famous
only examples
is
the
example
ring of Gaussian integers; this is a very pretty ring, and it has
6.2.
Gaussian
elementary but striking applications in number theory (one instance of which
will
may
define this ring
Z[t]
and
we are
going
The notation
Proposition
5.7): Z[i]
x2 + 1, that is,
to
verify
Z[i]
is
that
Z[i]
as
Z[x]
(*2 + l)'
:=
is
a
Euclidean domain.
justified by the discussion in §5.2 (cf. especially
ring containing Z and a
1. The natural embedding
root i of
may be viewed as the 'smallest'
a square
Z[x]
C
"
(x2
+
1)
subring of C; tracing this embedding shows that
Z[i]
Thus,
=
are
+ bi
of
(Example III.4.8)
we
realize
Z[i]
as
could equivalently define
eC\a,beZ}.
as consisting of the complex numbers whose
this
integers;
may be referred to as the 'integer lattice'
the reader should think of
real and imaginary parts
inC:
{a
root
R[x]
(x2 + 1)
and the identification of the rightmost ring with C
a
we
in
§6.3).
Abstractly, we
see
-2
Z[i]
%
This picture is particularly compelling, because it allows us to 'visualize' principal
ideals in Z[i]: simple properties of complex multiplication show that the multiples
of a fixed w G Z[i] form a regular lattice superimposed on Z[i\. For example, the
fattened dots in the picture
6. Further remarks and examples
295
represent the ideal (—2i) in Z[i\: an enlarged, tilted lattice superimposed
the integer lattice in C. The reader should stop reading now and make sure
understand why this works out so neatly.
A Gaussian integer
is
N(a
Geometrically, this
The
norm
of
a
+
complex number, and
bi)
(a
:=
Gaussian integer is
a
The
multiplicative in the
function
sense
N is
bi)
=
a2
Proof. The multiplicativity is
properties of complex conjugation:
N(zw)
=
an
a
+
to
a norm
b2.
a +
bi.
nonnegative integer; thus N is
a
function
Z^°.
-?
Euclidean valuation
a
that \/z,w G
on
Z[i\;further,
N is
Z[i]
N(zw)
that N is
-
such it has
is the square of the distance from the origin to
Z[t]
Lemma 6.5.
bi)(a
+
as
on
=
N(z)N(w).
immediate consequence of the elementary
(zw)(~zw)
=
(zl)(ww)
=
N(z)N(w).
Euclidean valuation, we have to show how to perform a 'division
Z[i\. It is not hard, but it is a bit messy, to do this algebraically;
will
we
attempt to convince the reader that this is in fact visually evident (and
then the reader can have fun producing the needed algebraic computations).
To
see
a
with remainder' in
Let z,w e Z[i], and assume w ^ 0. The ideal (w) is a lattice superimposed
Z[i\. The given z is either one of the vertices of this lattice (in which case z is a
multiple of w, so that the division z/w can be performed in Z[i], with remainder 0)
on
of the 'boxes' of the lattice. In the latter case, pick any of the
vertices of that box, that is, a multiple qw of it;, and let r
z
qw. The situation
or
it sits inside
one
=
may look
as
follows:
V.
296
Then
we
Irreducibility and factorization
in
integral domains
have obtained
z
qw + r,
=
length of the segment between qw and z.
box, and the square of the size of the box
and the norm of r is the square of the
Since this segment is contained in a
is
N(w),
we
have achieved
N(r)
completing
<
N(w),
the proof.
As we know, all sorts of amazing properties hold for the ring of Gaussian
integers as a consequence of Lemma 6.5: Z[i] is a PID and a UFD; irreducible
elements are prime in Z[i]; greatest common divisors exist and may be found by
applying the Euclidean algorithm, etc. Any one of these facts would seem fairly
challenging in itself, but they are all immediate consequences of the existence of a
Euclidean valuation and of the general considerations in §2.
The fact that the
norm
is
multiplicative simplifies further the analysis of the
ring Z[i\.
Lemma 6.6.
Proof. If
u
The units
is
a
ofZ[i]
are
Z[i], then
N(uv)
N(l)
±1, ±i.
unit in
then
N(u)N(v)
This
implies N(u)
=
=
=
=
there exists
1
v
G
Z[i]
by multiplicativity,
1, and the only elements in Z[i] with
such that
so
N(u)
uv
is
norm 1 are
a
=
1.
But
unit in Z.
±1, ±i. The
statement follows.
Lemma 6.7. Let q G
such that
N(q)
=
p or
Z[i] be a prime element. Then there is
N(q) p2.
a
prime integer p
G Z
=
Proof. Since q is not a unit, N(q) ^ 1 (by Lemma 6.6). Thus N(q) is a nontrivial
product of (integer) primes, and since q is prime in Z[i] D Z, q must divide one
prime integer factors of N(q); let p be this integer prime. But then q | p
p2.
Z[i], and it follows (by multiplicativity of the norm) that N(q) | N(p)
Since N(q) ^ 1, the only possibilities are N(q)
p and N(q)
p2, as claimed.
of the
in
=
=
=
6. Further remarks and examples
297
Both possibilities presented in Lemma 6.7 occur, and studying this dichotomy
be key to the result in §6.3. But we can already work out several cases
further will
'by hand':
Example
6.8. The prime
integer 3 is
a
prime element of Z[i]; this
can
be verified
by proving that 3 is irreducible in Z[i] (since Z[i] is a UFD). For this purpose, note
that since N(S)
9, the norm of a factor of 3 would have to be a divisor of 9,
that is, 1, 3, or 9. Gaussian integers with norm 1 are units, and those with norm
=
9
are
have
associates of 3
norm
equal
(Exercise 6.10);
to 3.
thus
a
nontrivial factor of 3 would necessarily
Z[i]\ hence 3 is indeed
But there are no such elements in
irreducible.
The prime integer 5 is not
a
prime element of Z[i]: running through the
same
argument as we just did for 3, we find that a factor of 5 should have norm 5, and
2 + i does. In fact, 5
(2 + i)(2 i) is a prime factorization of 5 in Z[i].
=
Visually, what happens is that the lattice generated by 3 in Z[i] cannot be
refined further into a tighter lattice, while the one generated by 5 does admit a
refinement:
A
®
®
®
®
®
®
®
®
»"
®
<§>
?
S)
^^
®
/
..®-''#
<§>
®
®
®
®
®
®
®
®
®
Irreducibility and factorization
V.
298
(the circles represent complex numbers with
in
integral domains
3, 5, respectively).
norm
j
The reader is encouraged to work out several other examples and to try to figure
out what property of an integer prime makes it split in Z[i] (like 5 and unlike 3).
But this should be done before reading on!
6.3. Fermat's theorem
of squares. Once armed with our knowledge
little further teaches us something new about Z—
on sums
about rings, playing with Z[i]
a
beautiful and famous example of the
generalization over brute force.
this is
a
success
achieved by judicious
use
of
complete the circle of thought begun with Lemma 6.7. Say that
prime integer p splits in Z[i] if it is not a prime element of Z[i]\ we have seen in
Example 6.8 that 5 splits in Z[i], while 3 does not.
First, let
us
a
Z[i] if
Lemma 6.9. A positive integer prime p G Z splits in
sum
of
and only
if
it is the
two squares in Z.
Proof. First assume that p
=
a2 + b2, with
p
(a
=
+
a, b G Z. Then
bi)(a
-
bi)
a2 + b2
p ^ 1, so neither of the two factors is a unit
Z[i], and N(a bi)
Z[i] (Lemma 6.6). Thus p is not irreducible, hence not prime, in Z[i],
Conversely, assume that p is not irreducible in Z[i\: then it has an irreducible
factor q G Z[i], which is not an associate of p. Since q | p, by multipUcativity of
the norm we have N(q) N(p)
p2, and hence N(q)
p since q and p are not
associates (thus N(q) ^ p2) and q is not a unit (thus N(q) ^ 1). If q
a + bi, we
in
±
=
=
in
=
=
=
find
p
verifying
that p is the
Thus, 2 splits
5
(as
=
sum
2
l2
=
+
=
N(q)=a2+b2J
of two squares and completing the proof.
l2 +
l2),
22,
13
3,
7,
and
=
22
so
+
do
32,
17
19,
23,
=
l2
+
42,
while
remain prime if viewed
as
11,
elements of
Z[i],
The next puzzle is as follows: what else distinguishes the primes in the first
list from the primes in the second list? Again, the reader who does not know the
already should pause and
conjecture! Do not read the
conjecture and then prove
on your own25!
answer
try to come up with a
that
next statement before trying this
The point we are trying to make is that these are not difficult facts, and a conscientious
reader really is in the position of discovering and proving them on his or her own. This is extremely
remarkable: the theorem we are heading towards was stated by Fermat, without proof, in 1640,
and had to wait about one hundred years before being proven rigorously (by Euler, using 'infinite
descent'). Proofs using Gaussian integers are due to Dedekind and had to wait another hundred
years. The moral is, of course, that the modern algebraic apparatus acts
of our skills.
as a
tremendous amplifier
6. Further remarks and examples
Lemma 6.10.
299
A positive odd prime integer p splits in
Z[i] if
and only
if
it is
congruent to 1 modulo 4.
Proof. The
Z[i]/(p)
is
an
question is whether p is prime as an element of Z[i], that is, whether
integral domain. But we have isomorphisms
Z[t]
(p)
Z[x]/(x2
(p)
1)
+
Z[x]
(p,*2 + l)
Z[x]/(p)
(*2 + l)
Z/pZ[x]
(x2 + l)
by virtue of the usual isomorphism theorems (including an appearance of Lemma 4.1)
and taking some liberties with the language (for example, (p) means three different
things, hopefully self-clarifying through the context). Therefore,
p
in
splits
Z[i]
and we are reduced to
is not
an
integral domain
1)
is not
^=>
Z[i]/(p)
^=>
Z/pZ[x]/(x2
^=>
x2
+ 1 is not irreducible in
^=>
x2
+ 1 has a root in
^=>
there is an
verifying
+
integer
an
integral domain
Z/pZ[x]
Z/pZ
n
such that n2
1 mod p,
=
that this last condition is equivalent to p
=
1 mod
provided that p is an odd prime.
For this purpose, recall (Theorem IV.6.10) that the multiplicative group G of
Z/pZ is cyclic; let g G G be a generator of G. Since p is assumed to be odd, p 1
is an even number, say 2£: thus g has order |G|
2£. Also, denote the class of an
integer n mod (p) by n. Since g is a generator of G, for every integer n there is an
integer m such that n
grn.
The class ^1 generates the unique subgroup of order 2 of G (unique by the
4,
=
=
classification of subgroups of
same, we have
a
cyclic
group,
Proposition II.6.11);
since
g*
does the
0*==!.
Therefore, with
n
=
grn,
we see
that n2
=
1
modp if and only if g2rn
=
g£,
that
is, if and only if
2m
Summarizing,
3n G Z such that n2
=
=
e mod 21
lmodp if and only if 3m
G Z such that
£mod2£. Now this is clearly the case if and only if £ is even, that is, if and
1 mod 4; and we
2£ is a multiple of 4, that is, if and only if p
only if (p
1)
2m
=
=
=
are done.
Putting together Lemma
6.9 and Lemma 6.10 gives the
following beautiful
number-theoretic statement:
Theorem 6.11
and only
if p
=
Remark 6.12.
(Fermat).
A positive odd prime p G Z is
a sum
of
two squares
if
1 mod4.
Lagrange proved (in 1770) that
every
positive integer is the
sum
of
four squares. One way to prove this is not unlike the proof given for Theorem 6.11:
it boils down to analyzing the splitting of (integer) primes in a ring; the role of Z[i]
is taken here by
a
ring of 'integral quaternions'.
j
in
integral domains
pause and note that there is no mention of
UFDs, Euclidean
V.
300
The reader should
Irreducibility and factorization
groups, etc., in the statement of Theorem 6.11,
domains, complex numbers, cyclic
although a large selection of these tools
used in its
was
This is of
proof.
course
the
dream of the generalizer: to set up an abstract machinery making interesting facts
that would be extremely mysterious or that would require
(nearly) evident—facts
cleverness
to be understood without that machinery.
Fermat-grade
Exercises
6.1. Generalize the CRT for two ideals,
commutative ring i?; prove that there is
o
as
follows. Let /, J be ideals in
of i?-modules
>JL-
>r-^^x^
/
J
>mj
>o
/ + J
(Also, explain why
where (p is the natural map.
Theorem 6.1, for k
2.)
a
an exact sequence
this implies the first part of
=
6.2. Let R be
a
Prove that R ^
commutative ring, and let
R/(a)
x
R/(l
-
a G
R be
an
element such that a2
=
a.
a).
Show that the multiplication in R endows the ideal (a) with a ring structure,
a as the identity26.
Prove that (a)
R/(l a) as rings. Prove that R
with
(a)
x
-
=
(1
a)
6.3. Recall
Let R be
a
as
=
rings.
(Exercise 3.15)
that
a
ring R is called Boolean if a2
finite Boolean ring; prove that R ^
Z/2Z
x
x
a
=
for all
a G
R.
Z/2Z.
a finite commutative ring, and let p be the smallest prime dividing
Let
ii,... ,ifc be proper ideals such that U + Ij
\R\.
(1) for i ^ j. Prove that
<
<
k <
Prove
\R\.(Hint:
l^"1 |/i|
|/fc|
(\R\/p)k.)
6.4. Let R be
=
logp
6.5. Show that the map
6.6. > Let R be
a
Z[x]
Z[x]/(2)
Z[x]/(x)
x
is not surjective.
UFD.
Let a,b e R such that
gcd(a,6)
=
1. Prove that
Under the hypotheses of Corollary 6.4
prove that the function (p is
(a)
fl
(b)
(ab).
=
(but only assuming
that R is
a
UFD)
injective.
[§6-1]
6.7.
Find
>
a
polynomial /
G
Q[x]
such that
/
=
1
mod(x2 + l)
and
/
=
xmodx100.
[§6-1]
6.8.
factorization.
positive integer and n
p^1 -p%T its
By the classification theorem for finite abelian groups (or,
-i
Let
n
G
Z be
a
This is an extremely unusual situation.
=
prime
in fact,
simpler
Note that this ring (a) is not a subring of R if
(a) and R differ.
a / 1 according to Definition III.2.5, since the identities in
Exercises
301
considerations; cf. Exercise
II.4.9)
z
z
z
(n)
(p?1)
(Prr)
abelian groups.
as
Use the CRT to prove that this is in fact
ring isomorphism.
a
Prove that
z y
/
z
\*
/
z
(recall that (Z/nZ)* denotes the group of units of Z/nZ).
Recall (Exercise II.6.14) that Euler's <j)-function </>(n) denotes the number of
positive integers < n that are relatively prime to n. Prove that
(!>{n)=val1-\vi-l)--Var-\Vr-l).
[II.2.15, VII.5.19]
6.9. Let / be
ideal of
an
Z[i\. Prove
that if
Z[i]. Show
(z) and N(z)
6.10. > Let z,w e
Show that if
w G
that
=
z
Z[i]/I
and
z
6.11. Prove that the irreducible elements in
integer primes congruent to 3
prime congruent to 1 mod 4.
6.12.
-i
Z[i]
|e
^$^|
-
Why
\^-q\<\-
<
and
Z[>/2]
Z[V2]
is
a
-
^^|
multiplicative: A^(z^)
{a
=
+
(Hint:
g^a
=
:
Z[>/2]
->
to
Let
w
z
=
!%+%>
+
=
a+
bi,
integer
=
c+di
^in<^ integers
e +
i/.
e^/
Prove that
N(z)N(w). (Cf.
by N(a
Exercise
(Hint:
a
C C.
2).
Z[t]/(t2
Z defined
Follow the
is
Z}
a, 6 G
Z[y/2]
Z[V2]
Prove that
1 + i; the
an
\,
has infinitely many units.
Prove that
[§6.2]
bi with a2 + b2
and set g
<
6^2 |
ring, isomorphic
Prove that the function iV
valuation.
|/
a ±
N(w).
=
does this do the job?) [6.13]
Consider the set
Prove that
=
associates.
w are
are, up to associates:
Prove Lemma 6.5 without any 'visual' aid.
w ^ 0. Then z/w
such that
-.
and
4; and the elements
mod
be Gaussian integers, with
6.13.
associates, then N(z)
w are
then
N(w),
is finite.
+
6^2)
=
a2
-
2b2 is
III.4.10.)
Euclidean domain, by using the absolute value of iV
same
steps
as
in Exercise
as
6.12.)
[6.14]
6.14.
the
Working
norm
N(a
as
+
in Exercise 6.13, prove that
by/^2)
=
a2 +
is a Euclidean
l\\f—2]
domain.
(Use
2b2.)
Vd)/2]
If you are particularly adventurous, prove that Z[(l +
is also a Euclidean
—11.
domain27 for d
can
still
use
the
norm
defined
—3,
—7,
by N(a+bVd)
(You
=
a2
=
db2\ note that this is still an
'You
are
integer
probably wondering why
Q(Vd);
we
on
Z[(l
+
switched from
V/d)/2],
Z[Vd]
are the 'ring of integers' in
the form they take depends
study is a cornerstone of algebraic number theory.
on
to
if d
Z[(l
=
+
1 mod
4.)
\fd)/2\.These
rings
the class of d modulo 4. Their
V.
302
The five values d
Z[Vd],
Z[(l
resp.,
the ring Z[(l
is
a
PID
Stark around
6.15. Give
integer
is still
a
PID
(cf. §2.4
and Exercise 2.18 for d
elementary proof (using modular arithmetic) of the fact that if
an
n
be
Prove that
a +
integer
m
n
a
and
n are two
integers both of which
also be written
ran can
sum
the
as
sum
an
of two squares.
be written
can
as sums
of two squares.
positive integer.
is
a sum
of two squares if and only if it is the
norm
of
Gaussian
a
bi.
factoring a2 + b2 in Z and a + bi in Z[i], prove that n is a
only if each integer prime factor p of n such that p
if and
with
Q(Vd)
by Alan Baker and Harold
is not even a UFD, as you
l\\f—5]
yourself
6.16. Prove that if
6.17. Let
the
-19);
proven
is congruent to 3 modulo 4, then it is not the
n
=
other negative values for which the ring of integers in
Also, keep in mind that
in Exercise 1.17.
1966.
of two squares, then
By
=
are no
have proved all by
integral domains
are the only ones for which
—1,-2,
resp., —3,-7,-11,
is
Euclidean.
For
the
values
d
-19, -43, -67, -163,
>/d)/2],
conjectured by Gauss and only
was
in
=
Vd)/2]
+
fact that there
+
Irreducibility and factorization
an even power
in
sum of two squares
=
3 mod 4 appears
n.
One ingredient in the proof of Lagrange's theorem on four squares is the
following result, which can be proven by completely elementary means. Let p > 0
be an odd prime integer. Then there exists an integer n, 0 < n < p, such that np
6.18.
-i
may be written as 1 +
a2 + b2 for
a2,
Prove that the numbers
two
0 <
integers
a
<
a, b. Prove this
result,
l)/2, represent (p
(p
+
as
follows:
l)/2
distinct
congruence classes mod p.
Prove the
same
for numbers of the form
b2,
1
0 <b <
(p
l)/2.
Now conclude, using the pigeon-hole principle.
[6.21]
6.19.
§ (1
-i
Let I C EI be the set of quaternions
+ i +
j
+
k)
+ bi +
cj
Prove that I is
a
Prove that the
norm
an
+ dk with a,
(cf.
(noncommutative) subring
of the ring of quaternions.
N(w) (Exercise III.2.5)
N(wi)N(w2).
integer and N(wiW2)
Exercise III. 1.12) of the form
b,c,de Z.
of
an
integral quaternion
Prove I has exactly 24 units in I: ±1, ±i, ±j, ±/c, and
Prove that every
a,b,c,d G Z.
w G
I is
an
associate of
an
element
The ring I is called the ring of integral quaternions.
6.20.
a
-i
Let I be
as
w G
I is
=
|(±1 ± i ± j ± k).
a +
bi + cj + dk G I with
[6.20, 6.21]
in Exercise 6.19. Prove that I shares most
good properties of
Euclidean domain, notwithstanding the fact that it is noncommutative.
Let 2,w G I, with
N(r)
<
w
^
0.
Prove that 3q,r e I such that z
qw + r, with
little tricky; don't feel too bad if you have to cheat
N(w). (This is a
somewhere.)
and look it up
=
Exercises
303
Prove that every left-ideal in I is of the form Iw for
Prove that every z,w G I, not both zero, have
d in I, of the form az + (3w for a, (3 G I.
a
L
some w G
'greatest
right-divisor'
common
[6.21]
6.21. Prove
Lagrange's theorem
four squares. Use notation
on
as
in Exercises 6.19
and 6.20.
Let
I and
z G
n G
in I is 1 if and
Z. Prove that the greatest
only if (N(z),m)
=
1 in Z.
common
(If
az +
right-divisor of
pn
=
z
and
1, then N(a)N(z)
n
=
where (3 is obtained by changing the signs of the
coefficients of i, j, k. Expand, and deduce that (N(z),n) | 1.)
N(l—
Pn)
=
(1
fin)(I
-
Pn),
odd prime integer p,
For
an
z
l+ai+bj such that p | N(z).
=
that
Say
is not
that
a
w G
unit and not
an
Exercise 6.18 to obtain
use
Prove that
z
and p have
an
integral quaternion
right-divisor
a common
associate of p.
I is irreducible \iw
=
aP implies that either
a or
(3
is
a
unit. Prove
that integer primes are not irreducible in L Deduce that every positive prime
integer is the norm of some integral quaternion.
Prove that
every
positive integer is the
norm
of
some
integral quaternion.
the last point of Exercise 6.19 to deduce that every positive integer
may be written as the sum of four perfect squares.
Finally,
use
Chapter
Linear
VI
algebra
In several branches of science, 'algebra' means 'linear algebra': the study of vector
spaces and linear maps, that is, of the category of modules over a ring R in the very
special case in which R
fields, such as R or C).
is
a
field (and often
one
restricts attention to very
special
of the main themes in this chapter. However, we will stress
that much of what can be done over a field can in fact be done over less special rings.
In fact, we will argue that working out the general theory over integral domains
This will be
one
produces invaluable tools when
fields: the paramount example
field, which will be obtained as
corollary of the classification of finitely generated modules over a PID.
Throughout the main body of this chapter (but not necessarily in the exercises)
we are
working
over
being canonical forms for matrices with entries in
a
R will denote
integral domain;
an
most of the
commutative1 rings without major
a
theory
can
be extended to arbitrary
difficulty.
1. Free modules revisited
1.1.
i?-Mod. For
generalities
on
modules,
see
§111.5.
A module
over
R is
an
abelian
group M, endowed with an action of R. The action of r G R on ra G M is denoted
rm: there is a notational bias towards left-modules (but the distinction between
left- and right-modules will be immaterial here as R is commutative). The defining
axioms of a module tell us that for all r*i,r2,r G R and ra,rai,7712 G M,
(n.
Ira
+
=
r(rai
r2)m
ra
+
=
and
7712)
rim + r2ra,
(ri^ra
=
=
ri(r2ra),
rrrti + rra,2.
The hypothesis of commutativity is very convenient, as it allows us to identify the notion of
left- and right-modules, and the fact that integral domains have fields of fractions simplifies many
arguments. The reader should keep in mind that many results in this chapter can be extended to
more
general rings.
305
VT. Linear
306
Modules
over
a
ring R form the category i?-Mod, which
we
algebra
encountered in
Chapter III. This category reflects subtle and important properties of R, and
are
going
to attempt to uncover some of these features. As a first
we
approximation,
look at the 'full subcategory' (cf. Exercise 1.3.8) of R-Mod whose objects are
free modules and in which morphisms are ordinary morphisms in i?-Mod, that is,
we
i?-linear
group homomorphisms.
The goal of the first few sections of this chapter is to give a very explicit
description of this subcategory: in the case of finitely generated free modules, this
can
will
be done by means of matrices with entries in R. Later in the chapter, we
see that matrices may also be used to describe important classes of nonfree
modules.
1.2. Linear independence and bases. The reader is invited to review the
free i?-modules given in §111.6.3: FR(S) denotes an i?-module containing
definition of
a given set S and universal with respect to the existence of a set-map from S. We
proved (Claim III.6.3) that the module R®s with 'one component for each element
of S" gives an explicit realization of FR(S).
The main point of this subsection will be that, for reasonable rings i?, the
can be recovered2 'abstractly' from the free module FR(S). For example, it
will follow easily that i?m
Rn if and only if m
n; this is the first indication
set S
=
=
that the category of
(finitely generated)
free modules does indeed admit
a
simple
description.
Our main tool will be the famous concepts of linearly independent subsets and
bases. It is easy to be imprecise in defining these notions. In order to avoid obvious
traps, we will give the definitions for indexed sets (cf. §1.2.2), that is, for functions
M from a (nonempty) indexing set / to a given module M. The reader
i : I —>
should think of
i
as a
selection of elements of M, allowing for the possibility that
a G I may not all be distinct.
the elements ma G M corresponding to
Recall that for all sets I there is
function i
:
I
M determines
—>
a
a
canonical injection j
:
I
FR(I) and
:
FR(I)
—>
unique R-module homomorphism
cp
any
M
making the diagram
M
FR(I) —^
3
I
commute: this is
Definition
1.1. We say that the indexed set i
(f is
injective;
surjective.
Put in
S
=
a
{ma}a£i
'This is
precisely the universal property satisfied by FR(I).
not
i is
:
I
M is
linearly dependent otherwise. We
linearly independent if
say that i generates M if (p is
j
slightly messier, but perhaps
more
common,
way,
an
indexed set
of elements of M is linearly independent if the only vanishing linear
necessarily the
case
if R is not commutative.
1. Free modules revisited
307
combination
^] rama
=
0
ot€l
is obtained
indexed
by choosing
sums are
finite
=
0 Va G /; S is
linearly dependent otherwise. The
set generates M if every element of M may be written as
some choice of ra.
a
ra
(As
defined in
sum.
assuming that
a
a
notational
warning/reminder, keep
module; therefore,
in this context the notation
When writing Ylaei rn°i m an ordinary
0 for all but finitely many a e I.)
ma
module,
one
f°r
only finite
^2
stands for
is
implicitly
=
indexed sets in Definition 1.1 takes
Using
Y^aei rama
in mind that
care
of obvious particular
cases
such
M is not
ra
0: Hi : I
being linearly dependent since 1 ra + (—1)
itself injective, then (p is surely not injective (by the commutativity of the diagram
and the injectivity of j), so that i is linearly dependent in this case. Because of
M amounts to a (special)
this fact, the datum of a linearly independent i : I
choice of distinct elements of M; the temptation to identify the elements of / with
as ra
and
ra
=
—>
—>
the corresponding elements of M is historically irresistible and essentially harmless.
Thus, it is common to speak about linearly dependent/independent subsets of M.
We will conform
carefully
to this common
practice; the reader should
parse the statements
and correct obvious imprecisions that may arise.
A simple application of Zorn's lemma shows that every module has maximal
linearly independent subsets. In fact, it gives the following conveniently stronger
statement:
Lemma 1.2.
Let M be an
subset. Then there
R-module, and let S C M be a linearly independent
linearly independent subset of M containing
exists a maximal
S.
family 5? of linearly independent subsets of M containing 5,
ordered by inclusion. Since S is linearly independent, 5? ^ 0. By Zorn's lemma, it
suffices to verify that every chain in 5? has an upper bound. Indeed, the union of
a chain of linearly independent subsets containing S is also linearly independent:
because any relation of linear dependence only involves finitely many elements and
Proof. Consider the
these elements would all belong to
one
subset in the chain.
Remark 1.3. This statement is in fact known to be equivalent to the axiom of
choice; therefore, the use of Zorn's lemma in one form or another cannot be
bypassed,
j
Note that the singleton {2} C Z is a 'maximal linearly independent subset' of
it does not generate Z. In general this is an additional requirement, leading
to the definition of a basis.
Z, but
Definition 1.4. An indexed set B
M is
a
basis if it generates M and is linearly
independent.
Again,
one
of all b G B
finite
(or
j
often talks of bases
are
as
'subsets' of the module M; since the images
is rather harmless. When B is
necessarily distinct elements, this
at any rate
countable),
the extra information carried by the indexed set
VT. Linear
308
can
be encoded by ordering the elements of B; to emphasize this,
one
algebra
talks about
ordered bases.
are necessarily maximal linear independent subsets and minimal
this holds over every ring. What will make modules over afield, i.e.,
subsets;
generating
vector spaces, so special is that the converse will also hold.
In any case, only very special modules admit bases:
Bases
Lemma 1.5.
An R-module M is
B C M is
basis
a
if
and only
if
and only if it admits a basis. In fact,
M is an
the natural homomorphism R®D
free if
isomorphism.
Proof. This is immediate from Definition 1.1: if B C M is linearly independent
M is injective
and generates M, then the corresponding homomorphism R®D
M is an isomorphism, then B is identified
and surjective. Conversely, if (p : R®D
with
a
subset of M which generates it
independent
By
(because
(p is
(because
(p is
surjective)
and is linearly
injective).
Lemma 1.5, the choice of
a
basis B of
a
free module M amounts to the
M; this will be an important observation in due
isomorphism R®D
time (e.g., in §2.2).
Once a basis B has been chosen for a free module M, then every element m G M
choice of
can
an
=
be written uniquely
as a
linear combination
m
2_] rbb
=
beB
with rh G R.
As
are 0 in any such
1.3. Vector
always, remember that all but finitely
expression.
spaces. Lemma 1.5 is all that is needed to prove the fundamental
observation that modules over a
a
field k
are
field
are
necessarily free. Recall that modules
over
(Example III.5.5). Elements of a vector space
(surprise, surprise) vectors, while elements of the field are called3 scalars.
are
called
many of the coefficients r&
called k-vector spaces
By Lemma 1.5, proving that vector spaces are free modules amounts
that they admit bases; Lemma 1.2 reduces the matter to the following:
Lemma 1.6.
Let R
=
k be
a
Again,
and let V be
field,
maximal linearly independent subset
ofV;
a
then B is
a
this should be contrasted with the situation
linearly independent subset of Z, but it is not
a
Proof. Let
is not
v G V, v 0 B.
Then B
maximality of B; therefore, there exist
U
{v}
k-vector space.
basis ofV.
over
cibi
H
proving
Let B be
rings: {2} is
a
a
maximal
basis.
linearly independent, by the
Co,..., ct G k and
(distinct) &i,..., bt
such that
c0v +
to
\-ctbt
=
0,
This terminology is also often used for free modules over any ring.
G B
1. Free modules revisited
with not all Co,..., c*
309
equal
Now,
to 0.
c$
^
0: otherwise
dependence relation among elements of B. Since k is
v
proving that
v
=
(-c^c^bi
-\
h
a
we
field,
would get
cq is a
a
linear
unit; but then
(-CQlct)bt,
is in the span of B. It follows that B generates V,
needed.
as
Summarizing,
Proposition 1.7. Let R
be a linearly independent
k be
=
set
of
a
field, and
of V.
let V be
k-vector space. Let S
a basis B of V
a
Then there exists
vectors
containing S.
In particular, V is
free
as a
k-module.
Proof. Put Lemma 1.2, Lemma 1.5, and Lemma 1.6 together.
We could also contemplate this situation from the 'mirror' point of view of
generating sets:
Lemma 1.8. Let R
=
minimal generating set
Every
set
k be
a
for V;
and let V be
field,
then B is
generating V contains
a
a
basis
basis
a
k-vector space.
Let B be
a
ofV.
ofV.
Proof. Exercise 1.6.
Lemma 1.8 also fails
fields
a
(but
not over
on more
general rings)
general rings (Exercise 1.5).
a
subset B of
maximal linearly independent subset
a vector space
^=> it is
a
is
To reiterate, over
a basis ^=> it is
minimal generating set.
1.4. Recovering B from FR(B). We are ready for the 'reconstruction' of a set
(up to a bijection!) from the corresponding free module FR(B). This is the
B
result
justifying the notion of
rank of a free module. Again,
dimension of
we prove a
a vector space, or, more
generally, the
somewhat stronger statement.
Proposition 1.9. Let R be an integral domain,
Let B be a maximal linearly independent subset
and let
M be
of M,
and let S be
a
free R-module.
a linearly
independent subset. Then4" \S\< \B\.
In particular, any two maximal linearly independent subsets
over an
integral domain have the
same
assume
that R
=
k is
a
a
free
module
cardinality.
Proof. By taking fields of fractions, the general case over
easily reduced to the case of vector spaces over a field; see
then
of
field and M
=
V is
a
an
integral domain
is
Exercise 1.7. We may
/c-vector space.
We have to prove that there is an injective map j : S ^-> B, and this can be
done by an inductive process, replacing elements of B by elements of S 'one-byone'. For this, let < be a well-ordering on 5, let v G 5, and assume we have defined
j for all
w
G
S with
w
<
v.
Let B' be the set obtained from B by replacing all
Here, |A| denotes the cardinality of the set A, a notion with which the reader is hopefully
familiar. The reader will not lose much by only considering the case in which B, S are finite sets;
but the fact is true for 'infinite-dimensional spaces' as well, as the argument shows.
VT. Linear
310
j(w) by
w, for w < v, and assume
independent subset of V. Then
j(v) 7^ j(w)
f°r aU
we
(Transflnite)
is
as
induction
a
maximal linearly
B may be defined so that
w < v\
the set B" obtained from B' by replacing
independent subset.
5,
B' is still
(inductively) that
claim that j(v) G
algebra
(Claim V.3.2)
j(v) by
v
is still
a
maximal linearly
then shows that j is defined and injective
on
needed.
To verify our claim, since B' is
linearly dependent (as an indexed
a
maximal linearly independent set, B' U {v}
so that there exists a linear dependence
set5),
relation
(*)
cqV +
cibi
H
h
ctbt
=
0
zero and the bi distinct in B'. Necessarily cq ^ 0 (because
B' is linearly independent); also, necessarily not all the bi with a ^ 0 are elements
of S (because S is linearly independent). Without loss of generality we may then
with not all q equal to
assume
we set
that c\ ^ 0 and bi G B'
j(v)
=
All that is left
by
in B' is
v
a
^ j(w) for
now
is the verification that the set B" obtained
maximal linearly independent subset. But by using
v
this is
S. This guarantees that bi
an easy consequence
subset. Further details
are
=
-%lcibi
(*)
to write
c^ctbu
of the fact that B' is
a
maximal linearly independent
left to the reader.
1.11. Let R be an
FR(A)
Proof.
w < v,
by replacing b\
Example 1.10. An uncountable subset of C[x] is necessarily linearly
Indeed, C[x] has a countable basis over C: for example, {1, x, x2, x3,...
Corollary
all
b\.
^
integral domain, and let A,
FR(B)
^^
there is a
B be sets.
dependent.
}.
j
Then
bisection A^B.
Exercise 1.8.
Remark 1.12. We have learned in Lemma 1.2 that
linearly independent subset S to
Proposition 1.9 shows that we
can
in fact do this
maximal linearly independent subset.
Remark 1.13. As
domain, then i?m
the 'IBN
=
we can
'complete'
every
proof of
elements
of
a given
by borrowing
a maximal one. The argument used in the
j
particular case of Corollary 1.11, we see that if R is an integral
n. This says that integral domains satisfy
Rn if and only if m
Basis Number) property'.
a
=
(Invariant
Strange as it may
seem, this obvious-looking fact does not hold over arbitrary
for
the
example,
rings:
ring of endomorphisms of an infinite-dimensional vector
space does not satisfy the IBN property. On the other hand, integral domains are a
bit of an overshoot: all commutative rings satisfy the IBN property (Exercise 1.11).
This allows for the possibility that v
the process we are about to describe gives
B' 'already'. In this case, the reader can check that
j(v)
=
v.
1. Free modules revisited
311
One way to think about this is that the category of finitely generated free
modules over (say) an integral domain is 'classified' by Z-°: up to isomorphisms,
there is exactly one finitely generated free module for any nonnegative integer. The
task of describing this category then amounts to describing the homomorphisms
between objects corresponding to two given nonnegative integers; this will be done
in
§2.1.
j
byproduct of the result of Proposition 1.9,
important definition.
As
a
we can now
give the following
integral domain. The rank of a free R-module M,
denoted rk# M, is the cardinality of a maximal linearly independent subset of M.
The rank of a vector space is called the dimension, denoted dim/- V.
j
Definition
1.14. Let R be an
This definition will in fact be adopted for
modules when the time comes, in §5.3.
more
general finitely generated
Finite-dimensional vector spaces over a fixed field form
spaces are free modules (Proposition 1.7), Corollary 1.11
dimensional vector spaces are
a
category. Since vector
implies that two
if
if
and
isomorphic
only they have the same
finite-
dimension.
subscripts i?, k
are often omitted, if the context permits. But note that, for
the
2,
example, viewing
complex numbers as a real vector space, we have dim^ C
The
=
while dimcC
=
1. So some care is warranted.
Proposition 1.9 tells us that every linearly independent subset S of a free Rmodule M must have cardinality lower than rk# M. Similarly, every generating set
must have cardinality higher than the rank. Indeed,
Proposition 1.15. Let R be an integral domain, and let M be a free R-module;
assume that M is generated by S: M
(S). Then S contains a maximal linearly
independent subset of M.
=
Proof.
space.
By
Exercise 1.7
Use Zorn's
we may assume
that R is
a
field and M
V is
a vector
linearly independent subset B C S which is
S. Arguing as in the proof of Lemma 1.6 shows that
it follows that B generates V. Thus B is a basis, and
lemma to obtain
a
maximal among subsets of
S is in the span of £?, and
hence a maximal linearly independent subset of V,
as
needed.
Remark 1.16. We have used again the trick of switching from
to its field of fractions.
=
an
integral domain
The second part of the argument would not work over
arbitrary integral domain, since maximal linearly independent subsets over an
integral domain are not generating sets in general.
Another standard method to reduce questions about modules over arbitrary
commutative rings to vector spaces is to mod out by a maximal ideal; cf. Exercise 1.9
and following.
j
an
VT. Linear
312
algebra
Exercises
1.1.
and
Prove that R and C
-i
(C, +)
1.2.
-i
are
isomorphic
are
as
isomorphic
Q-vector spaces. (In particular, (R, +)
as
groups.) [II.4.4]
Prove that the sets listed in Exercise III. 1.4
compute their dimensions.
1.3. Prove that
SU2(C)
=
all R-vector spaces, and
are
[1.3]
S03(R)
R-vector spaces.
as
(This
particularly interesting, from the dimension computation of
these two spaces may be viewed
as
the tangent spaces to
is immediate, and not
Exercise 1.2. However,
SU2OC),
/; the surjective homomorphism SU2(C)
SOs(R) you
Exercise II.8.9 induces a more 'meaningful' isomorphism SU2(C)
at
find this
1.4.
SOs(R),
S03(R). Can
you
isomorphism?)
A Lie bracket on V is an
Let V be a vector space over a field k.
[-,.]: V
resp.,
constructed in
V
x
^
operation
Fsuch that
(\/u,v,w G V),(Vo,6€fc),
[au
+
bv, w]
=
a[u, w]
+
(VveV), [v,v]=0,
(Wu,v,w G V), [[u, v],w]
and
(This
b[v, w],
+
axiom is called the Jacobi
bracket is called
a
[w, au + bv]
[[v,iy],w]
a
[[iy,w],v]
+
a[w, u]
=
+
b[w, v],
0.
A vector space endowed with
identity.)
Lie algebra. Define
=
category of Lie algebras
over a
a
Lie
given field.
Prove the following:
In
a
Lie algebra V,
[u,v]
=
for
—[v,u]
all u,
v G
V.
wv defines a Lie
/c-algebra (Definition III.5.7), then [v,w] := vw
bracket on V, so that V is a Lie algebra in a natural way.
This makes g[n(R), gin(C) into Lie algebras. The sets listed in Exercise III. 1.4
are all Lie algebras, with respect to a Lie bracket induced from gl.
If V is
a
-
su2(C) and so3(R)
1.5. > Let R be
an
are
isomorphic
as
Lie algebras
integral domain. Prove
Every linearly independent subset of
a
or
over
R.
disprove the following:
free i?-module may be completed to
a
basis.
Every generating subset of
a
free R-module contains
a
basis.
[§1.3]
1.6. > Prove Lemma 1.8.
1.7. > Let R be
an
[§1.3]
integral domain, and let
M
=
R®A be
a
free R-module. Let K
K®A in the evident
be the field of fractions of i?, and view M as a subset of V
C
M is linearly independent in M (over R) if and
way. Prove that a subset S
=
only if it is linearly independent in V
(over K).
Conclude that the rank of M
(as
Exercises
313
R-module) equals
an
generates M
over
1.8. > Deduce
the dimension of V
i?, then it generates V
Corollary
1.11 from
(as
a
K-vector
K. Is the
over
Prove that if S
space).
converse
true?
[§1.4]
Proposition 1.9. [§1.4]
1.9. > Let R be a commutative ring, and let M be an R-module. Let m be a
maximal ideal in i?, such that mM
0 (that is, rm
0 for all r G m, m G M).
Define in a natural way a vector space structure over R/m on M. [§1.4]
=
-i
be
F/mF
a commutative ring, and let F
R®D be a free module over R.
maximal ideal of i?, and let k
be
the quotient field. Prove that
R/m
k®D as k-vector spaces. [1.11]
Let R be
1.10.
Let
m
=
=
a
=
=
1.11. > Prove that commutative rings satisfy the IBN property.
Proposition V.3.5 and Exercise 1.10.) [§1.4]
1.12. Let V be a vector space over a field
endomorphisms
Prove that
(cf.
Exercise
Endfc_vect(V
©
/c, and let R
III.5.9). (Note
V)
=
RA
as an
1.13.
Let A be
-i
Prove that A
=
an
general.)
i?-module.
=
/ceN.
/ceN.)
abelian group such that EndAb(^4) is a field of characteristic 0.
carries a Q-vector space structure; what
be?) [IX.2.13]
Let V be
-i
be its ring of
that R is not commutative in
Q. (Hint: Prove that A
must its dimension
1.14.
=
Endfc_vect(V)
=
Prove that R does not satisfy the IBN property if V
(Note that V^V®Vi£V
(Use
a
finite-dimensional vector space, and let ip
:
V
V be
homomorphism of vector spaces. Prove that there is an integer n such that
ker<^n+1 kenp71 and im<^n+1 lm(pn.
Show that both claims may fail if V has infinite dimension. [1.15]
a
=
=
question of Exercise 1.14 for free i?-modules F of finite rank,
F be an i?-module
integral domain that is not a field. Let ip : F
1.15. Consider the
where R is
an
homomorphism.
What property of R immediately guarantees that ker
Show that there is
(pn for all
1.16.
is
a
i?-module homomorphism cp
a
module
decreasing
in which all quotients
a
Prove
a
a
over a
ring R. A finite composition
=
M0
Mi/Mi+i
D
are
M1
n ^>
<^n+1
0?
C
series for M
(if it
D
D
Mm
=
(0)
simple i?-modules (cf. Exercise III.5.4). The
The composition factors are
Mi/Mi+i.
Jordan-Holder theorem for modules: any two finite composition series
same length and the same (multiset of) composition factors.
module have the
(Adapt
ker (pn for
F such that im
series is the number of strict inclusions.
the quotients
of
F
sequence of submodules
M
length of
:
=
0.
Let M be
-i
exists)
n >
an
<^n+1
the proof of Theorem
IV.3.2.)
VT. Linear
314
We
say that M has
length
This notion is well-defined
m
if M admits
algebra
finite composition series of length m.
of the result you just proved. [1.17,
a
as a consequence
1.18, 3.20, 7.15]
1.17. Prove that
over
k
Exercise
case
its
equals
a k-vector space V has finite length as a module
if and only if it is finite-dimensional and that in this
its dimension.
1.16)
1.18. Let M be
R-module of finite length
an
Exercise
(cf.
m
Prove that every submodule N oi M has finite length
of Proposition IV.3.4.)
Prove that the 'descending chain condition'
(Use induction on the length.)
Prove that if R is
an
integral domain that
(d.c.c.)
is not
a
(cf.
length
1.16).
n < m.
the proof
(Adapt
for submodules holds in M.
field and F is
a
free R-module,
then F has finite length if and only if it is the 0-module.
a field, and let f(x) G
k[x] be any polynomial. Prove that there
multiple of f(x) in which all exponents of nonzero monomials are prime
1.19. Let k be
exists
a
integers. (Example: for f(x)
(1 +
x
+x
){2x
=
1 + x5 +
x6,
—x
—x
—x
+x
+x
+x
=
(Hint: k[x]/(f(x))
is
a
2x2
-
)
x3
finite-dimensional /c-vector
+
x5
+
2x7
+
2a11
-
x13
+
x17'.)
space.)
Let A, B be sets.
Prove that the free groups F(A), F(B) (§11.5) are
if
if
B. (For the interesting direction:
and
there
is a bijection A
isomorphic
only
^
^
remember that F(A)
F(B) => Fab(A)
Fab(B), by Exercise II.7.12). This
1.20.
-.
=
extends the result of Exercise II.7.13 to possibly infinite sets A, B.
2.
Homomorphisms
of free modules, I
2.1. Matrices. As pointed out in §1, Corollary 1.11 amounts to
of free modules
set
A
over
(determined
isomorphism
is
an
[II.5.10]
integral domain: if
up to a
bijection)
precisely the
same
F is
a
such that F
=
thing
as
a
classification
free module, then there is
i?eA. The choice of such
the choice of
a
This is simultaneously good and bad news. The good
it must be possible to 'understand'6
basis of F
news
a
an
(Lemma 1.5).
is that if i<\,F2
are
free, then
KomR(FuF2)
entirely in terms
That is, this set
corresponding sets A\,A2 such that i*\ R®Al, F2
morphisms in i?-Mod may be identified with
of the
of
=
=
R®A2.
Hom^(i?eAl,i?eA2).
The bad
news
is that this identification
is not 'canonical', because it
depends
on
For notational simplicity we will denote
common, and
no
confusion is likely.
KomR(FuF2)
=
Hom^(i?eAl,i?eA2)
i?eAl,
the chosen isomorphisms F\
Horn#_Mod(-M\N) by Hom#(M, iV);
=
this is very
Homomorphisms of free modules, I
2.
315
F2
R®A<1, that is, on the choice of bases. We will therefore have to do
to deal with this ambiguity (in §2.2).
=
some
work
The point of this subsection is to deal with the good news, that is, describe
Hom#(i?eAl,i?eA2). This can be done in a particularly convenient way when the
free modules
Also
are
the
finitely generated,
case we are
going
to
analyze
more
carefully.
of the good features of the category R-M06
(from §111.5.2)
is that the set of morphisms Hom#(M, N) between two R-modules is itself an
i?-module, in a natural way. The task is then to describe as explicitly as possible7
recall
that
one
Hom^(i?n,i?m)
as an
R-module, for
every choice of ra,n G Z-°.
This will be done
by
means
of matrices with entries in R. We trust that the
reader is familiar with the general notion of an ra x n matrix; we have occasionally
used matrices in examples given in previous chapters, and they have showed up
in several exercises. An
elements of R. It is
rows
and
n
matrix with entries in R is
ra x n
these elements
common to arrange
r2\
Vij)i=lym"
¦>rn
j=l,---
=
with rij G R. For any
I
,n
ran
consisting of
ra
given
r12
rln\
r22
r2n
;
;
ra, n the set
;
^ra2
TmnJ
M.m,n(R)
of
ra x n
matrices with entries
abelian group under entrywise addition8
("ij)
(cf. Example
+
II. 1.5); this is in fact
(*>ij)
an
:=
(aij+bij)
R-module, under the action
r(ai:j)
for
choice of
I
V ml
an
a
columns:
Mi
in R is
simply
as an array
(rai:j)
:=
R. From this point of view, the set of
the i?-module Rmn.
r G
ra x n
matrices is
simply
a copy
of
However, there are other interesting operations on these sets. If A
(a^) is
matrix and B
is
a p x n matrix, then one may define the product
(bkj)
of A and B as
=
an ra x p
=
p
A- B
=
(oifc) (bkj)
:=
(22aikbkj)',
k=l
this operation is (clearly) distributive with respect to addition and compatible with
the R-module structure. It is just a tiny bit messier to check that it is associative,
in the sense that if A, B, C are matrices of size, respectively, m x p, p x q, and
q
x n,
then
(A.B).C
=
A.(B-C).
We are now writing Rn for what was denoted R®n in previous chapters.
abuse of language, but it makes for easier typesetting.
write
This is a slight
Note that we will be dropping the extra subscripts giving the range of the indices and will
(r^) rather than (nj)i=iy... >m: these subscripts are an eyesore, and the context usually
j l,--- ,n
=
makes the information redundant.
VT. Linear
316
The reader should have
(Exercise
no
algebra
trouble reconstructing the proof of this fact
2.2).
In particular,
have
we
a
binary operation
on
the set
matrices, and this operation is associative, distributive
M.n(R)
of square
n x n-
w.r.t. +, and admits the
identity element
(the identity matrix,
in fact
an
denoted
/l
0
0\
0
1
0
\o o
In). That
i)
...
is,
M.n(R)
is
in very
i?-algebra (Exercise 2.3). Except
a ring (Example III. 1.6) and
special cases (such as n
1)
=
this ring is not commutative.
are
Matrices of type n x 1 are called column n-vectors; matrices of type 1 x m
called row m-vectors, for evident reasons. An element of a free R-module Rn
is nothing but the choice of n elements of i?, and we
into row or column vectors if we please; the standard
as
can arrange
choice
these elements
is to represent them
column vectors:
V2
€Rn.
\VJ
We will denote by e, the elements of the 'standard basis' of Rn:
0
ei
0
=
e2
=
V
W
so
6n
w
that
/v
\»n
The elements Vj G R
are
3
=
the 'components' of
Interpreting elements of Rn
1
v.
column vectors, we can act on Rn with anmxn
matrix, by left-multiplication: if A
(a^) is an m x n matrix and v G Rn is a
column vector, the product is a column vector in i?m:
as
=
A-v
/flu
Q>12
a\n\ M\
&21
^22
CL2n
I
+ ai2v2 -\
Q>\\V\
Q>2\V\+ a22^2 +
V2
h ainVn
'
'
'
+ CL2nVn
eR"
=
\flml
Lemma 2.1.
The
For
function
R-modules.
Ojmn/
dm2
all
(p
m x n
:
Rn
\VnJ
\CLmlVl +
am2^2 H
+ amnVn/
matrices A with entries in R:
R171 defined by
<p(v)
=
A
v
is a
homomorphism of
2.
Homomorphisms of free modules, I
317
Every R-module homomorphism Rn
unique
m x n
i?m
is determined in this way
by
a
matrix.
point follows immediately from the elementary properties of
matrix multiplication recalled above: Vr, s G i?, Vv, w G Rn
Proof. The first
(p(rv
as
sw)
+
=
A
(rv
+
sw)
rA
=
v
+ sA
w
=
np(v)
+
s(p(w)
needed.
For the second point, let (p
i?m be
Rn
:
a^ be the i-th component of ^(e^),
so
(p(ej)
a
homomorphism of i?-modules; let
that
=
\amj/
Then A
is
(a^)
=
/
an m x n
matrix, and Vv
anvi + ai2v2 +
-
-
a2lVi + a22V2 ~\
A-w
G i?n with
components Vj,
fa>ijVj\
+ ainvn
JL
\-0>2nVn
a2jVj
E
=
3
\Q>rnlVl +
as
am2^2 H
h
0>rnnVn)
=
1
\amjVj/
needed. The homomorphism induced by a nonzero matrix is manifestly nontrivimplies that the matrix A associated to a homomorphism cp is uniquely
ial; this
determined by
(p.
The reader should attempt to remember the recipe associating a matrix A to
homomorphism cp as in the proof of Lemma 2.1: the j-th column of A is simply
the column vector (p(ej). The collection of these vectors determines (p—since
a
a
homomorphism
is determined
by
its action
on a set
of generators.
Lemma 2.1 yields the promised explicit description of
Corollary
2.2. The
correspondence introduced
HomjR(i?n,i?m):
in Lemma 2.1 gives an
isomorphism
of R-modules
MmAR)
=
^omR{RnJRrn).
Proof. The reader will check that the correspondence is
of R-modules; this is enough, by Exercise III.5.12.
This is very
good
category; that is,
a
bijective homomorphism
forget that i?-Mod is
there
is a function9
morphisms. Therefore,
news, and it gets even better. Don't
we can compose
Hom^(i?p,i?m)
x
RomR(Rn,Rp)
a
Hom^(i?n,i?m),
-?
Here we am reversing the order of the Horn sets on the left w.r.t. the convention used in
Definition 1.3.1.
VT. Linear
318
mapping (ip, ip)
to
(poip-,
on
the other hand, the matrix product gives
(R)
mapping (A, B)
to A
B. That
X
is,
MPtn(R)
we
have
a
Mmtn(R)>
diagram
>Mm,n(R)
X
Mp,n(R)
Homfl(i^,i?m)
x
RomR(Rn,Rr)
Lemma 2.3.
This diagram commutes.
Hom^(i?n,i?m)
the isomorphisms obtained in
(induced by)
are
function
"?
Mm,p(R)
where the vertical maps
Corollary 2.2.
a
algebra
That is, the matrix corresponding to
composition (poip is the product of the matrices corresponding
to (p and
a
ip.
Proof. This follows immediately from the associativity of matrix multiplication:
for v e Rn and A e Xm,p(i?), B e MPtn(R),
A-(B-v)
that
=
(A-B)-
v;
is, the successive action of B and A is equivalent
Here is a summary of the situation.
construct
a
'toy category'
as
follows:
to the action of A
B.
integral domain i?,
Given an
we
can
let Z-° be the set of objects, and we
to be the set of matrices A/fm,n(i?), with
we
define the set of morphisms Hom(n,ra)
composition given by the matrix product
(cf.
Exercise
1.3.6).
Then
we
have verified
that this toy category is 'essentially the same as' the category of finitely generated
free R-modules and R-module homomorphisms10.
For example, by virtue of Proposition 1.7, if R
k is a field, then i?-Mod
/c-Vect consists exclusively of free /c-modules. The toy category we just described
is a faithful snapshot of the category of finite-dimensional k-vector spaces.
=
2.2. Change of basis. It is time to deal with the bad
are
only defined
universal property.
a
Free modules
a
Writing
up to
isomorphism,
free module
as
a
as
=
news.
is any structure
direct
sum
R®A
satisfying
amounts to
making the choice of one specific realization. As we pointed out at the end of §1.2,
by Lemma 1.5 this choice is equivalent to the choice of a basis of the module.
The price to pay for representing homomorphisms of free modules by matrices
is that this correspondence relies on the choice of a basis: we say that it is not
canonical. It is essential to be able to keep track of this choice. For finitely generated
free modules, this also boils down to the action of
Let F be a free
finitely generated,
module, and choose
so
that A, B
are
a
two bases
matrix,
as we
A, B for F;
finite sets, and further
proceed
assume
to see.
that F is
\A\ \B\(= rkF) by
=
The reader should not take this statement too seriously: we have identified together all
free modules isomorphic to a given one, which is a rather drastic operation. We will take a more
formal look at this situation in Example VIII. 1.8.
Homomorphisms of free modules, I
2.
Proposition 1.9. The
two bases
319
correspond
to two
¦*F,
R®A
isomorphisms
R®D
4>
*F.
Then
7p
R®A
Q(f
-+R®B
which
to
a
matrix
as seen in Lemma 2.1;
isomorphism,
corresponds
call this matrix N%. Explicitly, the j-th column of N% is the image of
element of A, viewed as a column vector in R®D. That is, letting11
is
an
A
we
have
N%
=
(n^-);=i,.
B
(ai,...,ar),
=
,
=
(bu
we may
the
j-th
>br),
with
tl>
-l
0(P(aj)
=J2niihi'
2=1
This equality is written in R®D; in practice one views the basis vectors a^, b2vectors of F and would therefore simply write
as
r
2=1
that is, the corresponding equality in F.
Definition 2.4. The matrix
N%
is called the matrix
Since it represents an isomorphism,
necessarily an invertible matrix.
To
our
knowledge there
is
no
the
of
the change
matrix of the
established convention
of
basis.
change of basis
on
j
is
the notation for this
matrix, and in any case we would find it futile to try to remember any such
convention. Any choice of notation will lead to a pleasant 'calculus': for example,
the
definition given above implies immediately N£
and
(N^)~x
N^N^ N%- In
our experience, any such manipulation is immediately clarified by writing
isomorphisms out explicitly, so no convention is necessary.
For example, let's work out the action of a change of basis on the matrix
=
representation of a homomophism a : F
taking care of the needed manipulations is
R®A
=
G of two free modules.
The
diagram
R®c
V>c
R®D
R®B
Let
Mg
N%
a
be the matrix of
l
o
1/%
ip
o
<p, as
above, and let Mq be the matrix for
p.
a basis is convenient, for example because it allows
'j-th element' of the basis. Hence we have the 'tuple' notation.
Ordering the elements of
about the
=
l
us
to
talk
VT. Linear
320
Choose the basis A for
the matrix P
F and C for G. Then the matrix representing
corresponding
algebra
a
will be
to
p-loaov:R®A^R®c.
If
on
a
by the
the other hand
matrix
we
choose the basis B for F and D for G, then
Q corresponding
we
represent
to
<j-loao^:R®B
^
R®D.
Now,
a_1otto^(^o p-1) oao((po {v%)-1)
=
^c°
{p~l oao(p)o (i/^)"1,
and by Lemma 2.3 this shows
Q
=
Mg
P
(JVf )-1 =Mg -P-Ng.
This is hardly surprising, of course: starting from a
combination of b's, N£ converts it into a combination of a's;
under
a as a
Q's job,
as
combination of c's; and
Mq
expressed as a
by giving its image
into d's. That accomplishes
vector
P
converts it
acts on it
it should.
Summarizing
this discussion,
Proposition 2.5. Let
a : F
G be a homomorphism of finitely generated free
modules, and let P be a matrix representing it with respect to any choice of bases
for F and G. Then the matrices representing a with respect to any other choice of
bases are all and only the matrices of the form
M'P-N,
where M and N
are
invertible matrices.
Definition 2.6. Two matrices P, Q G
the
same
Mm,n(R)
homomorphism of free modules Rn
are
Pm
equivalent if they represent
up to a choice of basis.
j
This is manifestly an equivalence relation. Properly speaking, the 'abstract'
G is not represented by one matrix as much as by the whole
homomorphism a : F
equivalence class with respect to this relation. Proposition 2.5 gives a computational
interpretation of equivalence of matrices: P and Q are equivalent if and only if there
are
invertible M and N such that Q
=
MPN.
2.3. Elementary operations and Gaussian elimination. The idea is now to
on Proposition 2.5, as follows: given a homomorphism a : F
G between
capitalize
'special' bases in F and G so that the matrix of a takes a
particularly convenient form. That is, look for a particularly convenient matrix in
each equivalence class with respect to the relation introduced in Definition 2.6.
two free modules, find
For
this,
we
can
matrix P and then
start
with random bases in F and G, representing a by a
and left by invertible
(by Proposition 2.5) multiply P on the right
matrices, in order to bring P into whatever form is best suited for
There is
Consider the
performed
an even more concrete way to
three 'elementary
matrix P:
following
on a
our
needs.
deal with equivalence computationally.
that can be
(row/column) operations'
2.
Homomorphisms of free modules, I
switch
two rows
add to
one row
multiply
(by
two
2.7.
Two matrices
a sequence
a
multiple of another
(or column)
one row
P
by
of P;
columns)
(resp., column)
all entries in
Proposition
obtained from
Proof. To
(or
321
of P by
a
row
(resp., column);
unit of R.
P, Q G .A/fm,n(P) are equivalent if Q
of elementary operations.
may be
that elementary operations produce equivalent matrices, it suffices
Proposition 2.5) to express them as multiplications on the left or right12 by
see
invertible matrices. Indeed, these operations may be performed by suitably
multiplying by the matrices obtained from the identity matrix by performing the
operation. For example, multiplying
interchanges
on
the left by
/l
0
0
0\
0
0
0
1
0
0
10
\0
1
0
the second and fourth
same
row
of
\0
0
a
0/
4 x
n
matrix; multiplying
on
the right
by
1/
adds to the third column of
a m x 3 matrix the c-multiple of the first column. The
will formalize this discussion and prove that all these matrices are
indeed invertible (Exercise 2.5).
diligent reader
The matrices corresponding to the elementary operations are called elementary
Linear algebra over arbitrary rings would be a great deal simpler if
Proposition 2.7 were an 'if and only if statement. This boils down to the question
matrices.
of whether every invertible matrix may be written as a product of elementary
matrices, that is, whether elementary matrices generate the 'general linear group'.
Definition
is the group
2.8. The n-th
of units in
general linear
.A/fn(P),
group over the
ring P, denoted GLn(P),
that is, the group of invertible
nxn matrices
with
entries in P.
j
Brief mentions of this notion have occurred already; cf. Example II. 1.5 and
several exercises in previous chapters.
The elementary matrices are elements of GLn(P); in fact, the inverse of
elementary matrix is (of course) again an elementary matrix. The following
an
surely known to the reader, in one form or another; in view of the foregoing
considerations, it says that the relation introduced in Definition 2.6 is under good
control over fields.
observation is
Multiplying
columns.
on
the left acts
on
the
rows
of the matrix; multiplying
on
the right acts
on its
VT. Linear
322
Proposition
2.9. Let R
=
k be
a
field,
and letn>0 be
an
algebra
integer. Then GLn(/c)
is generated by elementary matrices.
equivalent
of elementary operations.
two matrices are
Thus,
a sequence
Proof. Let A
=
(a^)
be
over a
field if
and only
if they
are
linked by
invertible matrix. In particular, some entry in
a row switch if necessary, we may
an n x n
the first column of A is nonzero; by performing
assume
an
that an is
nonzero.
Multiplying the first
row
by
a^1,
we may assume
that
1:
=
/
1
ai2
o>in\
0>21
&22
air
Q>nn/
of the first
(—a2i)-multiple
&n2
\&nl
Adding
to the second row the
After performing the analogous operation
nonzero
on
all rows,
row
clears the
we may assume
(2,1) entry.
that the only
entry in the first column is the (1,1) entry:
/I
ttl2
a>i2
0>ln\
0
a22
0>ln
\0
an2
annJ
...
of the first column
Similarly, adding to the second column the (—ai2)-multiple
(1,2) entry. Performing this operation on all columns reduces the matrix
clears the
to the form
A
o
o
o
«22
O-ln
1
0
0
Q>nn/
\0
(clearly invertible) (n 1)
an2
where A' denotes a
x
(n
1) matrix13.
Repeating the process on A' and on subsequent smaller matrices reduces A to
the identity matrix In. In other words, In may be obtained from A by a sequence
of elementary operations:
In
where M and iV
are
a
M
A
products of elementary
A
is itself
=
=
TV,
matrices. But then
M~l -N~l
product of elementary matrices, yielding the
statement.
N M; thus, the
that, with notation as in the preceding proof, A~l
explained in the proof may be used to compute the inverse of a matrix (also
Note
process
cf. Exercise
=
3.5).
When applied only to the rows of a matrix, the simplification of a matrix by
of elementary operations is called Gaussian elimination. This corresponds
means
The vertical and horizontal lines alert the reader
to the fact that the sectors of the matrix
are themselves matrices; this notation is called a block matrix.
2.
Homomorphisms of free modules, I
multiplying the given matrix
and it suffices in order to reduce
to
(Exercise
on
323
the
left by
a
product of elementary matrices,
identity
any square invertible matrix to the
2.15).
We will sloppily call 'Gaussian elimination' the more drastic process including
column operations as well as row operations; this is in line with our focus on
equivalence rather than 'row-equivalence'. Applied
process yields the following:
Proposition
2.10.
Over
a
field,
m x n
every
to any
matrix is
rectangular matrix, this
equivalent
to a matrix of
the form
'
(where
r <
min(ra,n)
'
Ir
0
0
0
and '0' stands for null matrices of appropriate
sizes).
Different matrices of the type displayed in Proposition 2.10
are mequivalent
rank
cf.
describes all
2.10
considerations;
(for example by
§3.3). Thus, Proposition
classes
of
matrices
over
a
field
and
shows
that
for
equivalence
any given ra, n there
are
in fact
2.4.
only finitely
(over
many such classes
a
field!).
Gaussian elimination over Euclidean domains. With due care,
Gaussian elimination may be
should suffice
performed over every Euclidean
general case: let
domain. A 2
x 2
example
to illustrate the
(: ^H*<*>.
for a Euclidean domain i?, with Euclidean valuation N. After switching rows and/or
columns if necessary, we may assume that N(a) is the minimum of the valuations
of all entries in the matrices. Division with remainder gives
b
with r
=
0
or
N(r)
<
N(a). Adding
=
aq + r
to the second column the
(-g)-multiple
of the
first produces the matrix
(a
\c
If
r
the
^ 0,
so
that
N(r)
<
N(a),
we
r
d
-
qcj
begin again and shuffle
rows
and columns
so
that
(1,1) entry
has minimum valuation. This process may be repeated, but after a
finite number of steps the (1,2) entry will have to vanish: because valuations are
nonnegative integers and
at each iteration the valuation of the
Trivial variations of the
producing
a
same
(1,1) entry decreases.
procedure will clear the (2,1) entry
as
well,
matrix
(J?)Now
(this
with
no
is the cleverest
part)
we
claim that
remainder. Indeed, otherwise
we can
we may assume
add the second
that
e
divides
row to
/
the first,
in
i?,
VT. Linear
324
and
Again,
start all over with this new matrix.
(1,1) entry,
be to decrease the valuation of the
must
reach the condition
The reader will have
end result is the
e
algebra
the effect of all the operations will
after a final number of steps we
so
f.
some
fun describing this process in the general
remark:
case.
The
following useful
Proposition 2.11. Let R be a Euclidean domain, and let P G
is equivalent to a matrix of the form14
Mm,n(R)-
Then P
'
'di
0
0
0
dr
0
o
0
o,
| dr.
with d\ |
This is called the Smith normal
form of
the matrix.
Proposition 2.11 suffices to prove a weak form of one of our main goals (a
classification theorem for finitely generated modules over PIDs) over Euclidean domains;
this will hopefully be clear in a little while, but the reader can gain the needed
insight right away by contemplating Proposition 2.11 vis-a-vis the classification of
finite abelian groups
Remark 2.12. As
in the
(Exercise 2.19).
a consequence
proof of Proposition 2.9,
of the preceding considerations and arguing as
that GLn(i?) is generated by elementary
we see
a Euclidean domain. The reader may think that some cleverness in
elimination may extend this to more general rings, but this will
Gaussian
handling
not go too far: there are examples of PIDs R for which GLn(i?) is not generated
matrices if R is
by elementary
matrices.
On the other hand,
some
cleverness does manage to produce a Smith normal
a PID: the presence of a good gcd suffices to
form for any matrix with entries in
adapt the procedure sketched above (but
and column
operations).
We will take
a
one may need more than elementary row
direct approach to this question in §5.
j
Exercises
2.1. Prove that the subset of M.2(R) consisting of matrices of the form
CO
is
a group
under matrix multiplication and is isomorphic to
2.2. > Prove that matrix
:Here r <
min(m,n),
multiplication
is associative.
the bottom-right 0 stands for
a
null
(i?, +).
[§2.1]
(m
r)
x
(n
r) matrix,
etc.
Exercises
325
2.3. > Prove that both
M.n(R) and Hom#(i?n, Rn) are R-algebras in a natural way
Hom#(i?n,i?n)
Mn(R) of Corollary 2.2 is an isomorphism of
and the bijection
=
i?-algebras. (Cf. Exercise III.5.9.) In particular, if the matrix M corresponds to
the homomorphism cp : Rn
i?n, then M is invertible in Mn(R) if and only if cp
is an isomorphism. [§2.1, §3.2, §6.1]
—>
2.4. Prove
Corollary
2.5. > Give
2.6.
-i
its
a
2.2.
formal argument proving Proposition 2.7.
A matrix with entries in
nonzero rows are
the leftmost
leftmost
all above the
zero rows
form
if
and
entry of each row is 1, and it is strictly
entry of the row above it.
The matrix is further in reduced
The leftmost
echelon
row
nonzero
nonzero
the leftmost
field is in
a
[§2.3]
nonzero
nonzero
row
entry of each
entries in
a
to the
right of the
echelon form if
row
is the
matrix in
row
only
nonzero
echelon form
entry in its column.
are
called pivots.
Prove that any matrix with entries in a field can be brought into reduced
echelon form by a sequence of elementary operations on rows. (This is what is
more
2.7.
properly called Gaussian elimination.) [2.7, 2.9]
-i
Let M be
(Exercise 2.6).
a
matrix with entries in
Prove that if
a
a row vector r
field and in reduced
is
a
row
linear combination
echelon form
YlaiYi
°f the
of M, then a* equals the component of r at the position corresponding
to the pivot on the i-th. row of M. Deduce that the nonzero rows of M are linearly
nonzero rows
independent. [2.9]
2.8.
P.
PN for an invertible matrix
Two matrices M, N are row-equivalent if M
Prove that this is indeed an equivalence relation, and that two matrices with
=
-i
row-equivalent if and only if
entries in a field are
other by
2.9.
-i
a sequence
Let k be
a
of elementary operations
one may
on rows.
be obtained from the
[2.9, 2.12]
field, and consider row-equivalence (Exercise 2.8)
on
the set of rax
Prove that each equivalence class contains exactly one matrix
in reduced row echelon form (Exercise 2.6). (Hint: To prove uniqueness, argue by
contradiction. Let M, N be different row-equivalent reduced row echelon matrices;
n matrices
.A/fm,n(/c).
that they have the minimum number of columns with this property. If the
leftmost column at which M and N differ is the /c-th column, use the minimality
assume
to prove that M, N may be assumed to be of the form
Use Exercise 2.7 to obtain
h-i
*
0
*
a
contradiction.)
The unique matrix in reduced
row
echelon form that is row-equivalent to
given matrix M is called the reduced echelon form of M. [2.11]
a
VT. Linear
326
2.10. > The
row space
of
a
algebra
matrix M is the span of its rows; the column space of
row-equivalent matrices have the same row
M is the span of its column. Prove that
space
and isomorphic column spaces.
2.11. Let k be a field and M G
spanned by the
form of M
2.12.
-i
(cf.
[2.12, §3.3]
.A/fm,n(/c).
Prove that the dimension of the space
of M equals the number of
Exercise 2.9).
rows
Let k be
a
nonzero rows
field, and consider row-equivalence
on
in the reduced echelon
.A/fm,n(/c) (Exercise 2.8).
By Exercise 2.10, row-equivalent matrices have the same row space. Prove that,
conversely, there is exactly one row-equivalence class in .A/fm,n(/c) for each subspace
of kn of dimension <
2.13.
a
-i
m.
[2.13, 2.14]
The set of subspaces of given dimension in a fixed vector space is called
In Exercise 2.12 you have constructed a bijection between the
Grassmannian.
Grassmannian of r-dimensional subspaces of kn and the
set of reduced row echelon
matrices with n columns and r nonzero rows.
For r
1, the Grassmannian is called the projective space. For a vector space V,
the corresponding projective space FV is the set of 'lines' (1-dimensional subspaces)
in V. For V
/cn, FV may be denoted P£-1, and the field k may be omitted if it is
=
=
clear from the context. Show that
U k1 U
/c°,
P£_1
may be written as a union kn~l U kn~2 U
and describe each of these subsets 'geometrically'.
1
Thus, Pn_1 is the union of n 'cells'15, the largest one having dimension n
for
the
choice
of
all
be
written
Grassmannians
may
(accounting
notation). Similarly,
as
unions of cells. These
are
called Schubert cells.
Prove that the Grassmannian of
(n
l)-dimensional subspaces of kn admits a
P£_1. (This phenomenon will be
VIII.5.17]
cell decomposition entirely analogous to that of
explained in Exercise VIII.5.17.) [VII.2.20, VIII.4.7,
2.14. > Show that the Grassmannian
Gr^(2,4)
of 2-dimensional subspaces of /c4 is
(Use Exercise 2.12; list
the union of 6 Schubert cells: /c4 U ks U k2 U k2 U k1 U k°.
all the possible reduced echelon
forms.) [VIII.4.8]
2.15. > Prove that a square matrix with entries in a field is invertible if and
if it is equivalent to the identity, if and only if it is row-equivalent
if and only if its reduced echelon form is the identity. [§2.3, 3.5]
2.16. Prove
Proposition 2.10.
2.17. Prove
Proposition
2.18.
Suppose
a :
Z3
to the
only
identity,
2.11.
Z2 is represented by the matrix
/ -6
12
18\
V-15
36
54;
with respect to the standard bases. Find bases of Z3, Z2 with respect to which
is given by a matrix of the form obtained in Proposition 2.11.
'Here,
a 'cell' is
simply
a
subset endowed with
a
natural bijection with k^ for
some
I.
a
3. Homomorphisms of free modules, II
2.19. >
Prove
prove the more
3.
Corollary IV.6.5 again as a corollary of Proposition 2.11.
general fact that every finitely generated abelian group is
of cyclic groups.
sum
327
In
a
fact,
direct
[§11.6.3, §IV.6.1, §IV.6.3, §2.4, §4.1, §4.3]
Homomorphisms
of free modules, II
The work in §2 accomplishes the goal of describing
Hom#(F, G),
where F and G
free i?-modules of finite rank; as we have seen, this can be done most explicitly
if, for example, R is a field or a Euclidean domain. We are not quite done, though:
even over fields, it is important to 'understand the answer'—that
is, examine the
are
classification of
homomorphisms
into
finitely
many
classes, highlighted
at the end
the frequent appearance of invertible matrices makes it necessary
to develop criteria to tell whether a given matrix is or is not in the general linear
group. We begin by pointing out a straightforward application of the classification
of
§2.3. Also,
of homomorphisms.
3.1. Solving systems of linear equations. A moment's thought reveals that
Gaussian elimination, through the auspices of statements such as Proposition 2.9,
gives us a tool to solve systems of linear equations over a field or a Euclidean
domain16. This is another point which does not
or memory
tools
as
a
should take
gymnastic: writing things
carefully
a typical situation, suppose
careful treatment
care
of producing
need be. To describe
a>nXi H
is
seem to warrant a
out
system of
m
equations in
n
1- ainXn
=
h
unknowns, with a^, bi
in
a
Euclidean domain R.
We want to find all solutions #i,..., xn in R.
fan
With A
oin\
and
=
\bn
*/
'solving for
(xi
(b\
b
=
x
this amounts to
=
\En/
x' in the matrix equation
Ax
=
b.
n and the matrix A is square and invertible, then the solutions
Of course, if m
obtained simply as
=
are
x
=
A-1b.
Here A~l may be obtained through Gaussian elimination (cf. Proposition 2.9)
through determinants (§3.2).
Of
field case,
systems over arbitrary integral domains R by reducing
by embedding R in its field of fractions.
course one can deal with
or
to the
VI. Linear algebra
328
Even without requiring anything special of the 'matrix of coefficients' A,
and column operations will take it to the standard form presented in
Proposition 2.11; that is, it will yield invertible matrices M, N such that
M-A-N
/ di
0
o
V
0
0/
row
=
o
with notation
procedure:
and N
as in Proposition 2.11.
Gaussian elimination is a constructive
watching yourself as you switch/combine rows or columns will produce
explicitly. Now, letting y
(yj) and c Mb, the system
=
M
=
(M A N)
y
=
c
solves itself:
( diyi
t*
yv
0
yj
=
d~1Cj
ci
t*
=
cr+i
only if dj
has solutions if and
case
=
for j
the reader will check
0 for j > r; moreover, in
Cj for all j and Cj
and
is
1,... ,r,
arbitrary for j > r. This yields y,
yj
that x
all
solutions to the original system.
Ny gives
=
=
this
and
=
Such arguments
can be packaged into convenient explicit procedures to solve
of
linear
equations. Again, it seems futile to list (or try to remember) any
systems
such procedure; it seems more important to know what is behind such techniques,
so as to
be able to
come up
with
one
when needed.
In any case, the reader should be able to justify such recipes rigorously. The
most famous one is possibly Cramer's rule (Proposition 3.6) which relies on
determinants and is,
incidentally, essentially useless
in practice for any decent-size
problem.
a : F
G be a homomorphism of free R-modules
A
and
let
be
the
matrix
rank,
representing a with respect to a choice
The determinant. Let
3.2.
of the
same
of bases for F and G. As
a consequence
isomorphism if and only if A
as a
of Lemma 2.3
(cf.
Exercise
2.3),
a
is
an
that is, if and only if it is invertible
matrix with entries in R. This may be detected by computing the determinant
is
a
unit in
M.n(R),
of A
Definition
3.1. Let A
=
(a^)
G
M.n(R)
be
a square
matrix. Then the determinant
of A is the element
n
det(A)
=
Yl (-!)" II °*M
R-
3. Homomorphisms of free modules, II
329
Here Sn denotes the symmetric group on {1,... n}, and we write17 a(i) for
the action of a G Sn on i G {1,..., n}; (-1)CT is the sign of a permutation
(Definition IV.4.10, Lemma IV.4.12).
The reader is surely familiar with determinants, at least over fields. They
satisfy a number of remarkable properties; here is a selection:
,
—The
determinant of
matrix A
a
equals the determinant of
its transpose A1
=
defined by setting
(alj),
a\j
for all i and j
detA*=
(that is,
the
rows
of A1
=
are
aji
the columns of
A). Indeed,
£(-irn<4(i)=£(-i)<7n°-wi= Et-^rK-M
i=l
aGSri
i=l
creSri
2=1
creSri
by the commutativity of the product in R; and a-1 ranges over all permutations
of {1,... n} as a does the same, and with the same sign, so the right-most term
,
equals det A.
—If
two rows
is
enough
or
columns of
to check this for
matrix A agree, then det A
a square
matching columns; the
the previous observation. If columns j and
a
and the transposition
%
(jf),
so
det A
for
are
rows
0. Indeed, it
follows by applying
=
equal, the contribution
to
to the contribution due to the
=
0.
A
—Suppose
(a,ij) and B (bij) agree on all but at most one row: a^ b^
for
all
and
some fixed k. Let
/c,
7^
j
c^ := a^
b^ for i ^ /c, Ckj := Cbkj + frfcj*
=
if
of A
Sn is equal and opposite in sign
det A due to a a G
product of
f
case
=
=
=
and let C
:=
Then
(c^).
det(C)=det(A)+det(B).
This follows immediately from Definition 3.1 and distributivity. Applying this
observation to the transpose matrices gives an analogous statement for matrices
differing
at most
along
We will record
a
column.
officially
more
the effect of elementary operations
on
determinants:
Lemma 3.2. Let A be
a square matrix
with entries in
Let A! be obtained from A by switching
det(A')
=
(column).
Then
A by adding to
det(A') det (A).
from
Then
Let A' be obtained
ce R.
integral domain R.
two rows or two columns.
a row
(column)
a
multiple of another
=
from
det(A')
=
A by multiplying
a
row
(column) by
an
element18
cdet(A).
effect of
elementary operation on det A is the same
multiplying det A by the determinant of the corresponding elementary matrix.
In other words, the
that
Then
-det(A).
Let A! be obtained
row
an
an
as
Consistency with previous encounters with the symmetric group (e.g., §IV.4) would demand
ia; but then we would end up with things like iau(T\ which are very hard to parse.
For an elementary operation, c should be a unit; this restriction is not necessary here.
we write
VT. Linear
330
algebra
Proof. These
are all essentially immediate from Definition 3.1.
For example,
in
two
columns
amounts
to
each
<j
the
definition
switching
correcting
by a fixed
m
the
of
all
contributions
to
the
the
definition.
The
transposition, changing
sign
^2
third point is immediate from distributivity. Combining the third operation and
the two remarks preceding the statement yields the second point. Details are left
to the reader (Exercise 3.2).
a
This observation simplifies the theory of determinants drastically. If R
.A/fn(/c), Gaussian elimination (Proposition 2.10) shows
=
k is
field and P G
A
where
r < n
and Ei,
Ej
det(A)
(In particular,
det A
^
E1...Ea.'
=
elementary
are
=
0
Ir
'
0
|
0
E[ '-E'b,
/
matrices. Then Lemma 3.2 gives that
Y[ det(Ei) JJ det(£$) det
only if
r
=
n.)
l
0
0
0
Useful facts about the determinant follow
from this remark. For example,
Proposition 3.3. Let R be
A
square matrix A G
The determinant is
a commutative
M.n{R)
a
ring.
is invertible
if
and only
homomorphism19 GLn(R)
if det A
is
a unit in
(#*,•)•More
R.
generally,
forA,BeMn(R),
det(A'B)=det(A)det(B).
Proof for R
=
a
field. If R
=
k is
a
field,
we can use
the considerations
immediately preceding the statement. The first point is reduced to the
matrix
case
of
a
block
'
0
0
0
0 if and only if the linear
isomorphism. In particular, for all A, B
we have det( AB)
0 if and only if AB is not an isomorphism, if and only if A or B
is not an isomorphism, if and only if det(A)
0 or det(B)
0. So we only need
to check the homomorphism property for invertible matrices. These are products
of elementary matrices 'on the nose', and the homomorphism property then follows
from Lemma 3.2.
for which it is immediate. In fact, this shows that det A
map kn
kn
corresponding
=
to A is not an
=
=
=
Before giving the (easy) extension to the case of arbitrary commutative rings,
it is helpful to note the following explicit formulas. A 'submatrix' obtained from a
given matrix A by removing a number of rows and columns is called a minor of A\
more
properly, this
term refers to the determinants
'Recall that (R*, •)denotes the groups of units of R.
of square
submatrices obtained
3. Homomorphisms of free modules, II
the cofactors of A are the (n
1) x (n
A
will
More
for
we
let
sign.
precisely,
(a^)
in these ways. If A G
of
A, corrected by
a
331
M.n(R),
/
AM :=(-l)i+''det
for all
for
i
aij-i
aij+i
O'i-ll
O'i-lj-l
O'i-lj+l
tti—ln
tti+ii
cii+ij-i
ai+ij+i
0>i+ln
With notation as
=
all j
=
Proof. This is
1,..., n, det(A)
l,..., n, det(A)
a
minors
din
Q>n
Q>nn /
Q>nj+l
Lemma 3.4.
1)
=
above,
=
=
Y^j=i o>ijA^\
YJl=i a^A™.
simple (if slightly messy) induction
on n,
which
we
leave to the
diligent reader.
Of
Lemma 3.4 is simply stating the famous strategy computing a
expanding it with respect to your favorite row or column. This works
course
determinant
by
wonderfully for
totally useless for large ones, since the
needed
to
it
computations
apply grows as the factorial of the size of the
matrix. From a computational point of view, it makes much better sense to apply
Gaussian elimination and use the considerations preceding Proposition 3.3.
very small matrices and is
number of
But Lemma 3.4 has the
Corollary
3.5. Let R be
^(ii)
following important implication:
a commutative
A(n^\
ring and A
(A^
G
Mn(R)-
A(n1^
A
n)
\AV
Note the switch
is called the
Proof.
adjoint
j^{nn) J
\A(ln)
Then
=
det(A)In.
J^{nn)
in the role of i and j in the matrix of cofactors. This matrix
matrix of A.
Along the diagonal of the right-hand side,
one is evaluating (for example)
this is
a restatement
of Lemma 3.4.
Off the diagonal,
X>;^W)
3
=
1
for i' ^ i. By Lemma 3.4 this is the same as the determinant of the matrix obtained
by replacing the i-th row with the i'-th row; the resulting matrix has two equal rows,
so its determinant is 0, as needed.
VT. Linear
332
In particular,
its determinant:
A~l
this holds
3.5 proves that
Corollary
computation of cofactors is
likely
a
a
:
:
\4(ln)
^W
commutative ring,
over any
elimination is
I
det(A)~l
=
invert
we can
detA is
as soon as
'expensive',
better alternative
in any
so
least
(at
given
over
matrix if
a
algebra
we can
invert
In practice the
unit.
concrete case Gaussian
cf. Exercise 3.5.
fields);
But this formula for the inverse has good theoretical significance. For example,
in
we are now
a
position
complete the proof of Proposition 3.3.
to
Proof of Proposition 3.3
the second and from what
A e
admits
Mn(R)
if A admits
an
an
for commutative
have just seen.
inverse A-1 e Mn(R) if
we
inverse A~l G
then
M.n{R),
det^det^"1)
rings. The first point follows from
Indeed, we have checked that
det(A) is a unit in R; conversely,
detiAA'1)
=
=
det(Jn)
=
1
by the second statement, so that det(A~x) is the inverse of det(A) in R.
Thus, we just need to verify the second point, that is, the homomorphism
property of determinants. In order to verify this over every (commutative) ring,
it suffices to verify the 'universal' identity obtained by writing out the claimed
equality, for matrices with indeterminate entries.
For example, for
n
=
2 the
statement is
detf*1
X2)det(Vl
yA=det(Xiyi+X2V3
\xs x±J
\y3 y±J
\x3y1 + £42/3
*i0* +
W)
x3y2 + x^VaJ
which translates into the identity
(xix±
-
x2X3)(yiy4
-
2/22/3)
since this identity holds
=
(xiyi
+
x2yz){xzy2
+
x±y±)
-
(xiy2
+
x2y4)(x3yi
+
£42/3);
Z[#i,..., 2/4], it must hold in any commutative ring, for
choice
of
x\,...,2/4: indeed, Z is initial in Ring.
any
Now, we
particular it
in
have verified that the homomorphism property holds over
over the field of fractions of
Z[xn,... ,xnn,yu,...,
holds
follows that it does hold in
Z[xn,..., xnn, 2/n,..., y<nn],
and
we are
fields;
in
2/nn]-
It
done.
The 'universal identity' argument extending the result from fields to arbitrary
commutative rings is a useful device, and the reader is invited to contemplate it
carefully.
As
an
application of determinants (and especially cofactors)
case of a system of n equations in n unknowns
we
can
now
go
back to the special
A
cf. §3.1,
in the
Proposition
in which detA is
case
3.6
matrix obtained
(Cramer's rule).
a
x
=
b,
unit.
Assume
det(A)
is a unit, and let
by replacing the j-th column of A by the column
xj=det(A)-1det(A^).
A^ be the
vector b.
Then
3. Homomorphisms of free modules, II
Using Lemma 3.4, expand
Proof.
333
det(A^)
with respect to the j-th column:
n
2=1
Therefore
(xA
:
A~xb
=
(A™
AW
/h
\A^ln^
A^nn\
\bn/
det(A)-1
=
\XnJ
/det(A)-1det(A(1))\
det(A)-1
=
\E?=i^(in)&i.
\det(A)-ldet(A^)J
which gives the statement.
D
each equivalence class of
3.3. Rank and nullity. According to Proposition 2.10,
matrices over a field has a representative of the type
0
0
The
two different m x n matrices of this type are
why
reason
0
surely inequivalent
V and W free modules, vector spaces in this case) is
a
matrix
of
this
form, then r is the dimension of the image of a.
represented by
r cannot represent the same a.
matrices
with
different
Therefore,
is that if
V
:
a
W
(with
This integer r is called the rank of the matrix and deserves some attention. We
will discuss it for matrices with entries in a field, leaving generalizations to more
general rings
to the reader
The column
(rows)
of P.
(row)
(for
now at
least).
space of a matrix P over a field k is the span of the columns
The column
(row)
rank of P is the dimension of the column
(row)
space of P.
Proposition
3.7.
The
row
rank
of a
matrix over a
field
k equals its column rank.
Proof. Equivalent matrices have the same ranks. Indeed, let P G
row space of P consists of all row vectors
(fli
obtained
(wi
as
each vi ranges in k.
wm)
(wi
dm)
=
W,
(vi
-"
=
If Q
vm) Af_1;
^)'Q= (Vl
{vi
=
dimension,
as
the
Vm)'P
MPN, with M and iV invertible, let
then
vrn)M-1(MPN)
=
This shows that multiplication on the right by iV maps the
phically, since iV is invertible) to the row space of Q; thus
same
A/(m,n(/c);
claimed. Minimal variations
on
the
the column ranks of P and Q agree. (Cf. Exercise 2.10.)
(a1
am)
row space
of P
the
same
N.
(isomor-
two spaces have the
argument show that
VT. Linear
334
Applying this observation and Proposition
row
rank
=
r
2.10 reduces the question to matrices
'
'
for which
algebra
Ir
0
0
0
column rank, proving the statement.
=
The result upgrades easily to matrices
(applying the usual trick of embedding R
In view of Proposition 3.7,
Definition 3.8. Let M G
we can
over
arbitrary integral domain R
an
in its field of
fractions).
simply talk about the rank of
A/(m,n(/c) be a matrix over a field
(or, equivalently, row) space.
a
matrix:
k. The rank of M is
the dimension of its column
One way to rephrase Proposition 2.10 is that matrices
to
up
equivalence by their rank.
j
over a
field
are
classified
The foregoing considerations translate nicely in more abstract terms for a linear
W between finite-dimensional vector spaces over a field k. Using the
map a : V
convenient language introduced in §111.7.1, note that each a determines an exact
sequence
of vector spaces
0
>kera
>ima
>V
>0
.
Definition 3.9. The rank of a, denoted rka, is the dimension of ima. The nullity
of a is dim(kera:).
j
Claim 3.10. Let
a :
V
W be
a
linear map
of finite-dimensional
vector spaces.
Then
(rank
Proof. Let
by
n
an n x m
=
dimF and
of
m
a)
=
of
(nullity
+
a)
=
dim V.
dim W. By Proposition 2.10
we can
represent
a
matrix of the form
^
'
Ir
0
0
0
From this representation it is immediate that rka
with the stated consequence.
=
r
and the nullity of
a
is
n
r,
Summarizing, rka equals the (column) rank of any matrix P representing a;
similarly, the nullity of a equals 'dimF minus the (row) rank' of P. Claim 3.10 is
the abstract version of the equality of row rank and column rank.
3.4. Euler characteristic and the Grothendieck group. Against our best
efforts, we cannot resist extending these simple observations to more general
complexes. Claim
3.10 may be reformulated
as
follows:
Proposition 3.11. Let
0
be
a
short exact sequence
>V
>U
>W
>0
of finite-dimensional
vector spaces.
dim(V)
dim(W).
=
dim(J7)
+
Then
3. Homomorphisms of free modules, II
Equivalently, this
Consider then
V.
a
amounts to the relation
> Vn-i
VN
(cf. §111.7.1). Thus, cti-i
requirement that im(a^+i)
defined
dim(V/C7)
complex of finite-dimensional
0
:
335
o
=
oti
C
dim(F)
=
dim(C/).
vector spaces and linear maps:
>
> V0
Vi
> 0
0 for all i.
This condition is equivalent to the
recall that the homology of this complex is
ker(a^);
the collection of spaces
as
Ht(V.)
The complex is exact if
im(a^+i)
=
=
J^2iL.
im(ai+i)
ker(a^)
for all i, that is, if
Hi(V9)
=
0 for all i.
Definition 3.12. The Euler characteristic of V, is the integer
i
The original motivation for the introduction of this number is topological: with
suitable positions, this Euler characteristic equals the Euler characteristic obtained
by triangulating a manifold and then computing the number of vertices of the
triangulation, minus the number of edges, plus the number of faces, etc.
The following simple result is then
generalization of Proposition 3.11:
Proposition 3.13.
With notation
straightforward (and
a
very
useful)
above,
as
N
X(V.)
=£(-!)* dim(JJ4(V;)).
i=0
In particular,
ifV.
Proof. There is
is exact, then
nothing
Proposition 3.11 if N
1
rr
0
:
=
to show for N
0.
=
0, and the result follows directly from
a complex
(Exercise 3.15). Arguing by induction, given
OiN-1
OiJSJ
_
,r
V.
=
x(Y*)
Oil
Vjv_i
VN
OL\
Vi
> 0,
V0
that the result is known for 'shorter' complexes. Consider then the
we may assume
truncation
V.
0
:
> VN-i
>
> Vi
> V0
> 0
Then
x(V.)=x(K)
+
(-l)Ndim(VN),
and
Hi(V.)
=
Hi(Vi)
for 0 <
% <
N
-
2,
while
HN-!(V:)
=
ker(<*N-i)>
HN-!(V.)
=
ker^N~'\
HN(V.)
By Proposition 3.11 (cf. Claim 3.10),
dim(V/v)
=
dim(im(ajv))
+
dim(ker(o:^))
=
ker(a^).
VT. Linear
336
algebra
and
dim(H]y-i(V9))
dim(ker(a:jv-i))
=
dim(im(o:^));
therefore
dim^-i^.'))
Putting
-
dim(VN)
=
dim(HN^(V.))
-
dim(HN(V.)).
all of this together with the induction hypothesis,
N-l
x(K)=E(-1)idim(^(y.'))
i=0
gives
x(V.)=X(V:)
+
(-l)Ndim(VN)
N-l
J2 (-1)* dim(ff4(K)) + W dMVN)
=
i=0
N-2
^(-l)Mim(^(K)) + (-l)iV-1(dim(^iv_1(K))-dim(Fiv))
=
N-2
^(-l)Mim(^(F0) + (-l)^"1(dim(^iV-i(F0)-dim(^iv(F.)))
=
N
2=0
as
needed.
In terms of the topological motivation recalled above, Proposition 3.13 tells us
that the Euler characteristic of a manifold may be computed as the alternating sum
of the ranks of its homology, that is, of its Betti numbers.
Having come this far, we cannot refrain from mentioning the next, equally
simple-minded, generalization. The reader has surely noticed that the only tool
used in the proof of Proposition 3.13 was the 'additivity' property of dimension,
established in Proposition 3.11: if
0
>V
>U
>0
>W
is exact, then
dim(V)
Proposition
3.13 is
a
formal
=
dim(J7)
consequence
of
+
dim(W).
this
one
property of dim.
we can reinterpret what we have just done in the following
Consider the category k-Vecr of finite-dimensional k-vector spaces.
With this in mind,
curious way.
Each object
V of fc-Vect/ determines an isomorphism class [V]. Let F(k-Vecr) be
the free abelian group on the set of these isomorphism classes; further, let E be the
subgroup generated by the elements
[V]
-
[U]
-
[W]
for all short exact sequences
0
>U
>V
>W
5-0
3. Homomorphisms of free modules, II
in /c-Vect
.
337
The quotient group
r^/z
f^
v,
K(k-VecV)
:=
F(k-Vectf)
—-—
-
E
is called the Grothendieck group of the category
fc-Vect/.
by V in the Grothendieck group is still denoted
[V].
a
The element determined
More generally, a Grothendieck group may be defined for any category admitting
notion of exact sequence.
Every complex V9 determines
Xk{V.)
an
element in
£(-l)4[VS]
:=
K(k-Vecr ), namely
tf
(fc-Vect').
i
Claim 3.14.
With notation
as
above,
Xk 'is an Euler characteristic
in
we
have the
following:
in the sense that it
satisfies
the
formula given
Proposition 3.13:
i
is a 'universal Euler characteristic7, in the following sense. Let G be an
abelian group, and let 5 be a function associating an element ofG to each finitedimensional vector space, such that S(V)
V and S(V/U)
5(V) ifV
Xk
=
S(V)
S(U).
For V9
a
=
=
complex, define
XG(v.)
=
Y^(-m(vi).
i
Then 5 induces
a
(unique)
group
homomorphism
K(k-Vectf)
mapping
Xk(V.)
In particular, 5
=
to
xg(V.).
dim induces
a group
K(k-Vectf)
such that
Xk(V»)
This is in
fact
an
G
-?
homomorphism
Z
-?
x(^«)isomorphism.
The best way to convince the reader that this impressive claim is completely
trivial is to leave its proof to the reader (Exercise 3.16). The first point is proved
by adapting the proof of Proposition 3.13; the second point is a harmless mixture
of universal properties; the third point follows from the second; and the last point
follows from the fact that
dim(fc)
=
1 and the IBN property.
point is, in fact, rather anticlimactic: if the impressively abstract
Grothendieck group turns out to just be a copy of the integers, why bother
defining it? The answer is, of course, that this only stresses how special the category
fc-Vect/ is. The definition of the Grothendieck group can be given in any context in
The last
VT. Linear
338
algebra
which complexes and a notion of exactness are available (for example, in the
category of finitely generated20 modules over any ring). The formal arguments proving
Claim 3.14 will go through in any such context and provide
of 'universal Euler characteristic'.
We will
come
with
us
a
useful notion
back to complexes and homology in Chapter IX.
Exercises
3.1. Use Gaussian elimination to find all integer solutions of the system of equations
7x- S6y + 12z
J
-Sx + 42y
-
Uz
proof of Lemma
3.2. > Provide details for the
=
=
3.2.
1,
2.
[§3.2]
3.3. Redo Exercise II.8.8.
3.4. Formalize the discussion of 'universal identities':
is it true that if
properties
an
identity holds
in
commutative ring i?, for every choice of Xi
every
by what cocktail of universal
Z[#i,... ,xr], then it holds
G i?? (Is the commutativity
over
of R
necessary?)
3.5. > Let A be
consider the
n x
an
n
square
invertible matrix with entries in
a
field, and
(2n) matrix B
(A\In) obtained by placing the identity matrix
to the side of A. Perform elementary row operations on B so as to reduce A to In
(cf. Exercise 2.15). Prove that this transforms B into (Jn|A_1).
(This
is
n x
a
=
much
using determinants
3.6.
-i
Let R be
R-module. Let A
more
as
in
efficient way to compute the inverse of
=
(mi,... ,mr)
a
Mr(R)
be
a
matrix such that A
:
3.7.
be
-i
an
that
=
0 for all me M.
Let R be
a
(Hint: Multiply by
(1
b)M
=
:
.
Prove that
\0/
adjoint.) [3.7]
commutative ring, M a finitely generated i?-module, and let J
M. Prove that there exists an element b G J such
ideal of R. Assume JM
+
the
finitely generated
=
\mrJ
det(A)m
by
/0\
fmA
G
matrix than
§3.2.) [§2.3, §3.2]
commutative ring and M
a
a
0.
(Let
=
mi,... ,mr be generators for M.
/mA
B with entries in J such that
.
\m7
Find an r x r matrix
/miN
Then
use
Exercise
3.6.) [3.8,
\mr
VIII. 1.18]
We cannot define a
Note that some finiteness condition is likely to be necessary.
'Grothendieck group of fc-Vect' because the isomorphism classes of objects in fc-Vect do not form
a set: there is one for each cardinal number, and cardinal numbers do not form a set.
Exercises
339
Let R be
3.8.
-i
J be
an
a commutative ring, M be a finitely generated i?-module, and let
ideal of R contained in the Jacobson radical of R (Exercise V.3.14). Prove
that M
=
0
^=>
JM
=
M.
(Use
Exercise 3.7.
This is
Nakayama's lemma,
a
result with important applications in commutative algebra and algebraic geometry.
A particular case was given as Exercise III.5.16.) [III.5.16, 3.9, 5.5]
Let R be a commutative local ring, that is, a ring with a single maximal
ideal m, and let M, N be finitely generated R-modules. Prove that if M
mM+iV,
N. (Apply Nakayama's lemma, that is, Exercise 3.8, to M/N. Note
then M
3.9.
-i
=
=
that the Jacobson radical of R is
3.10.
Let R be
-i
a
module. Note that
m.) [3.10]
commutative local ring, and let M be a finitely generated Ris a finite-dimensional vector space over the field i?/m;
M/mM
let mi,... ,mr G M be elements whose cosets mod mM form
Prove that mi,...,mr generate M.
(Show
form of
3.11.
that
(mi,... ,mr) + mM
Exercise 3.9.) [5.5, VIII.2.24]
Explain how
=
a
basis of
M/mM.
M; then apply Nakayama's lemma
in the
to use Gaussian elimination to find bases for the row space and
the column space of a matrix over a field.
3.12.
Let R be
-i
an
integral domain, and let
are linearly dependent
M G
that the columns of M
3.13. Let k be
a
field. Prove that
a
3.14.
M is the smallest
with
m < n.
Prove
A/fm,n(/c) has rank < r if and only
AAr,n(k) such that M PQ. (Thus the
matrix M G
A/(m,r(/c), Q
such integer.)
if there exist matrices P G
rank of
Mm,n(R),
[5.6]
R.
over
G
=
Proposition 3.11 to the case of finitely generated free modules
domain.
integral
(Embed the integral domain in its field of fractions.)
Generalize
over any
3.15. > Prove
Proposition 3.13 for the
3.16. > Prove Claim 3.14.
case
N
=
1.
[§3.4]
[§3.4]
3.17. Extend the definition of Grothendieck group of vector spaces given in §3.4 to
(possibly infinite) dimension, and prove
the category of vector spaces of countable
that it is the trivial group.
Ab^p be the category of finitely generated abelian groups. Define
Grothendieck group of this category in the style of the construction of if
Z.
and prove that
3.18. Let
a
(/c-Ved/),
K(Abfd)
=
3.19.
Let Ab/ be the category of finite abelian groups. Prove that assigning to
every finite abelian group its order extends to a homomorphism from the Grothendieck
-i
group
K(Abf)
3.20. Let
over
a
to the
itMVIod^
multiplicative
group
(Q*, •)•
[3.20]
be the category of modules of finite length (cf. Exercise 1.16)
an abelian group, and let 5 be a function assigning an
ring R. Let G be
element of G to every simple i?-module. Prove that 5 extends to
from the Grothendieck
group of
il-Mod'
Explain why Exercise 3.19 is
a
a
homomorphism
to G.
particular
case
of this observation.
VT. Linear
340
algebra
1 G Z for every simple module M shows
(For another example, letting S(M)
that length itself extends to a homomorphism from the Grothendieck group of
=
R-Modf
4.
to
Z.)
Presentations and resolutions
After this excursion into the idyllic world of free modules, we can come back to earth
and see if we have learned something that may be useful for more general situations.
Modules
over a
field
necessarily free (Proposition 1.7),
are
not so for modules over
In fact, this property will turn out to be
more
general rings.
fields
(Proposition 4.10).
a
characterization of
It is important that we develop some understanding of nonfree modules. In this
section we will see that homomorphisms of free modules carry enough information
to allow us to deal with many nonfree modules.
4.1. Torsion. There
the most spectacular
are
one
several ways in which a module M may fail to be free:
is that M may have torsion.
Definition 4.1. Let M be
if
an
R-module. An element
m G
M is
a torsion
linearly dependent, that is, if
element
3r G i?, r ^ 0, such that rm
0. The
subset of torsion elements of M is denoted Totr(M). A module M is torsion-free
is
{m}
if Tor r(M)
=
{0}.
A torsion module is
=
module M in which every element is
a
torsion element.
a
j
no uncertainty about the base
ring.
A commutative ring is torsion-free as a module over itself if and only if it
is an integral domain; this is a good reason to limit the discussion to integral
domains in this chapter. Also, the reader will check (Exercise 4.1) that if R is an
integral domain, then Tor(M) is a submodule of M. Equally easy is the following
The subscript R is usually omitted if there is
observation:
Lemma 4.2. Submodules and direct
Free modules
over an
sums
integral domain
are
of torsion-free
torsion-free.
modules
are
torsion-free.
Proof. The first statement is immediate; the second follows from the first, since
an integral domain is torsion-free as a module over itself.
Lemma 4.2 gives
in
an
module
which
a
good
integral domain R
a
R1).
are
source
of torsion-free modules:
torsion-free
In fact, ideals provide
module may fail to be free.
us
(because they
are
for example, ideals
submodules of the free
with examples of another mechanism in
Example 4.3. Let R
(2,x). Then / is not a free i?-module.
Z[x], and let /
More generally, let / be any nonprincipal ideal of an integral domain i?; then / is
a torsion-free module which is not free.
=
=
4. Presentations and resolutions
Indeed, if /
Proposition 1.9
rank
its rank would have to be 1 at most,
basis for / would be
(a
1 over
free, then
were
341
thus
itself);
a
by
linearly independent subset of i?, and R has
element would suffice to generate /, and / would be
one
principal.
j
What is behind this example is a characterization of PIDs in terms of 'torsiona rank-1 free module' (Exercise 4.3). This is a facet of the main
free submodules of
result towards which
modules
PIDs
we are
heading, that is, the classification of finitely generated
The gist of this classification is that finitely
generated modules over a PID can be decomposed into cyclic modules. We have
essentially proved this fact already for modules over Euclidean domains (it follows
over
(Theorem 5.6).
from Proposition 2.11; see Exercise 2.19), and we have looked in great detail at the
particular case of Z-modules, a.k.a. abelian groups (§IV.6); we are almost ready to
deal with the general case of PIDs.
Definition 4.4. An i?-module M is cyclic if it is generated by
if M
R/I for some ideal / of R.
a
singleton, that is,
=
j
The equivalence in the definition is hopefully clear to our reader, as an
of the first isomorphism theorem for modules (Corollary III.5.16).
immediate consequence
If not, go back and
Exercise III.6.16.
(re)do
Cyclic modules are witness to the difference between fields and more general
rings: over a field /c, a cyclic module is just a 1-dimensional vector space, that is a
'copy of /c'; over more general rings, cyclic modules may be very interesting (think
of the many hours spent contemplating cyclic groups). In fact, we can tell that a
ring is a field by just looking at its cyclic modules:
Lemma 4.5. Let R be
torsion-free.
Proof. Let
Then R is
c G
i?,
c
integral domain. Assume that
field.
an
a
^ 0;
then M
=
R/(c)
is
a
every
cyclic R-module
cyclic module. Note that Tor(M)
is
=
M: indeed, the class of 1 generates R/(c) and belongs to Tor(M) since c 1 is 0
mod (c) and c ^ 0. However, by hypothesis M is torsion-free; that is, Tor(M)
=
{0}.
Therefore M
This shows
=
Tor(M)
R/(c)
unit. Thus every
is the
nonzero
is the
zero
zero
module.
R-module; that is, (c)
(1). Therefore,
a unit, proving that R is a field.
=
c
is
a
element of R is
a simple-minded illustration of the fact that we can study a ring R
the
module
structure over R, that is, the category .R-Mod, and that
by studying
Lemma 4.5 is
we may not even
need to look at the whole of i?-Mod to be able to draw strong
conclusions about R.
4.2. Finitely presented modules and free resolutions. The 'right' way to
think of a cyclic i?-module M is as a module which admits an epimorphism from i?,
VT. Linear
342
viewing the latter
as
the free rank-1 R-module
:
M
R1
algebra
> 0.
fact that M is surjected upon by a free R-module is nothing special. In fact,
every module M admits such an epimorphism:
The
M
R®A
0
M
provided that we are willing to take A large enough; if we are desperate, A
will surely do. This is immediate from the universal property of free modules; if
the reader does not agree, it is time to go back and review §111.6.3. What makes
cyclic modules special is that A can be chosen to be a singleton.
We are now going to focus on a case which is also special, but not quite as
special as cyclic modules: finitely generated modules are modules for which we can
=
a finite set (cf. §111.6.4). Thus,
from
a finite-rank free module:
epimorphism
choose A to be
we
will
assume
that M admits
an
M
0
i?m —^—>
for
some
integer
m.
The image by
tt
of the
m vectors
in
a
basis of R71 is
a set
of
generators for M.
Finitely generated modules are much easier to handle than arbitrary modules.
For example, an ideal of R can tell us whether a finitely generated module is torsion.
Definition 4.6. The annihilator of
Ann^(M)
:=
{r
an
i?-module M is
G
R\VmG M,rm
=
0}.
The subscript is usually omitted. The reader will check
Ann(M) is an ideal of R and that if M is a finitely generated
integral domain, then
M is torsion if and
only if Ann(M) ^
j
(Exercise 4.4)
module and R
that
is
an
0.
finitely generated modules. It
comfortably large collection of such
We would like to develop tools to deal with
turns out that matrices allow us to describe a
modules.
Definition 4.7. An R-module M is
finitely presented
if for
some
positive integers
m, n there is an exact sequence
Rn
Such
a sequence
is called
a
—^
> M
Rm
> 0.
presentation of M.
j
In other words, finitely presented modules are cokernels (cf. III.6.2) of homomorphisms between finitely generated free modules. Everything about M must
be encoded in the homomorphism <p; therefore, we should be able to describe the
module M by studying the matrix corresponding to (p.
There is
modules, but
on
a gap between finitely presented modules and finitely generated
reasonable rings the two notions coincide:
In context the exactness of a sequence of .R-modules will be understood, so the displayed
sequence is a way to denote the fact that there exists a surjective homomorphism of .R-modules
from R to M; cf. Example III.7.2. Also note the convention of denoting R by R1 when it is viewed
as a
module
over
itself.
4. Presentations and resolutions
If R
finitely presented.
Lemma 4.8.
Proof. If M is
a
343
finitely generated R-module
is a Noetherian ring, then every
finitely generated module, there
is
is
an exact sequence
M
0
i?m —^—>
for
some
(Corollary
Since R is Noetherian, i?m is Noetherian as an i?-module
Thus ker7r is finitely generated; that is, there is an exact sequence
m.
III.6.8).
Rn
for
some n.
Once
Putting together the
we
presentation,
we
an exact
two sequences
>0
gives
a
presentation of M.
have gone one step to obtain generators and two steps to get
should hit upon the idea to keep going:
Definition 4.9. A resolution of
is
^ker?r
an
a
i?-module M by finitely generated free modules
complex
>
R™3
R1712
i?mi
i?m°
M
0.
j
Iterating the argument proving Lemma 4.8 shows that if R is Noetherian,
every finitely generated module has a resolution as in Definition 4.9.
It is
an exact
important conceptual step
complex of free modules
an
...
±
R1713
>
to realize that M may be studied
R1712
>
R1711
>
then
by studying
R1710
resolving M, that is, such that M is the cokernel of the last map. The i?m° piece
keeps track of the generators of M; i?mi accounts for the relations among these
records relations among the relations; and so on.
generators; R™2
Developing this idea
in full
generality would take
us too
far for
now:
for
would have to deal with the fact that every module admits many different
example,
resolutions (for example, we can bump up every rrii by one by direct-summing each
we
complex with a copy of R1, sent to itself by the maps in the complex).
carefully later on, in Chapter IX.
However, we can already learn something by considering coarse questions, such
term in the
We will do this very
A priori, there is no reason
as 'how long' a resolution can be.
resolutions to be 'finite', that is, such that rrii
0 for i ^> 0.
conditions tell us something special about the base ring R.
=
The first natural question of this type is, for which rings R
to
expect
a
free
Such finiteness
is it the
case
that
every finitely generated R-module M has a free resolution 'of length 0', that is,
stopping at mo? That would mean that there is an exact sequence
0
> Rm°
> M
> 0.
Therefore, M itself must be free. What does this say about R?
Proposition 4.10. Let R be
every
an
finitely generated R-module
integral domain. Then R
is free.
is a
field if
and only
if
VT. Linear
344
Proof. If R
algebra
field, then every R-module is free, by Proposition 1.7. For the
finitely generated R-module is free; in particular, every
in
module
is
free;
particular, every cyclic module is torsion-free. But then R
cyclic
is a field, by Lemma 4.5.
is
a
converse, assume that every
The next natural question
admit free resolutions of length
concerns
terms, that is, to require that for every
beginning of
rings for which finitely generated modules
phrase the question in stronger
finitely generated R-module M and every
1. It is convenient to
free resolution
a
Rm0 _^_> M
the resolution
>
0,
be completed to a length 1 free resolution. This would amount
that
there exist an integer rai and an i?-module homomorphism
demanding
such
that
the sequence
Rmi
Rm°
can
to
-?
0
> 0
M
i?m° —^
i?mi
Equivalently, this condition requires that the module ker7r of relations
necessarily be free.
is exact.
among the tuq generators
Claim
4.11.
Let R be an
integral domain satisfying this property.
Then R
is
aPID.
Proof. Let I be an ideal of
have
an
i?, and apply the condition
to M
=
R/I.
Since
we
epimorphism
Ri^L^R/I
>o,
the condition says that ker-zr is free; that is, I is free. Since I is
a
free submodule
of i?, which is free of rank 1, / must be free of rank < 1 by Proposition 1.9.
Therefore I is generated by one element, as needed.
The classification result for finitely generated modules
which
we
keep bringing
up, will
essentially be
over
a converse to
PIDs
(Theorem 5.6),
Claim 4.11: the
mysterious condition requiring free resolutions of finitely generated modules to have
length at most 1 turns out to be a characterization of PIDs, just as the length 0
condition is
work this
a
characterization of fields
out in
(as proved
in
Proposition 4.10). We will
§5.2.
Reading a presentation. Let us return to the brilliant idea of studying
finitely presented module M by studying a homomorphism of free modules
4.3.
(*)
ip
:
Rn
a
i?m
coker</?. As we know, we can describe cp completely by considering
matrix A representing it, and therefore we can describe any finitely presented
module by giving a matrix corresponding to (a homomorphism corresponding to) it.
such that M
a
=
4. Presentations and resolutions
345
In many cases, judicious use of the material developed in §2 allows
determine the module M explicitly. For example, take
us to
a homomorphism Z2
Z3, hence to a Z-module, that is,
abelian
The
reader
should
G.
finitely generated
group
figure out what G is more
explicitly (in terms of the classification of §IV.6; cf. Exercise 2.19) before reading
on. In the rest of this section we will simply tie up loose ends into a more concrete
recipe to perform these operations.
this matrix corresponds to
a
Incidentally,
operations
on
a
modules
number of software packages
(say,
over
polynomial rings);
perform sophisticated
personal favorite is Macaulay2.
can
a
packages rely precisely on the correspondence between modules and matrices:
with due care, every operation on modules (such as direct sums, tensors, quotients,
etc.) can be executed on the corresponding matrices. For example,
These
Lemma 4.12. Let A, B be matrices with entries in an integral domain R, and let
M, N denote the corresponding R-modules. Then M 0 N corresponds to the block
matrix
'
A
0
0
B
Proof. This follows immediately from Exercise 4.16.
Coming back to
(*),
have chosen for Rn or
note that the module M cannot know which bases we
i?m; that is, M
=
coker^ really depends
on
the
homomorphism (p, not on the specific matrix representation we have chosen for (p. This is an
issue that we have already encountered, and treated rather thoroughly, in §2.2 and
following: 'equivalent' matrices represent the same homomorphism and hence the
same
module.
In the context
we are
two matrices A, B represent the
P, Q such that B
PAQ.
exploring
now,
module M
same
if
Proposition 2.5 tells
us
that
there exist invertible matrices
=
But this
have
case
is not the whole story. Two different homomorphisms <^i, cp2 may
isomorphic cokernels, even if they act between different modules: the extreme
being any isomorphism
whose cokernel
Therefore,
if
a
is 0
(regardless
matrix A'
of the isomorphism and
corresponds
to a module
no
matter
what
m
M, then (by Lemma 4.12)
is).
so
does the block matrix
'
Ir
0
0
A'
i
where Ir is the
identity matrix (and r is any
could be replaced here by any invertible matrix.
The
r x r
following proposition attempts
nonnegative
integer);
to formalize these observations.
in
fact, Ir
VT. Linear
346
algebra
Proposition 4.13. Let A be a matrix with entries in an integral domain R, and
let B be obtained from A by any sequence of the following operations:
switch two
add
rows or two
to one row
multiply
columns;
(resp., column)
all entries in
one row
multiple of another
a
(or column) by
the only nonzero entry in
column containing that entry.
if
a unit is
Then B represents the
same
R-module
a row
A,
as
a unit
row
of R;
(or column),
up to
(resp., column);
remove
the
row
and
isomorphism.
Proof. The first three operations are the 'elementary operations' of §2.3, and they
transform a matrix into an equivalent one (by Proposition 2.7); as observed above,
this does not affect the corresponding module, up to isomorphism.
As for the fourth operation, if u is a unit and the only nonzero entry in (say) a
row, then by applications of the second elementary operation we may assume that
u
is also the
assume
that
only
u
nonzero
is in fact the
entry in its column; without loss of generality
we may
of the matrix; that is, the matrix is in block
(1,1) entry
form:
0
A
=
o
x
But then A and A' represent the
module,
same
as
needed.
Example 4.14. The matrix with integer entries
an abelian group G.
second column, obtaining
'l
3>
2
3
Subtract three times the first column from the
determines
'1
0
2
-3
K5
the
(1,1) entry
remove
the first
is
a
row
unit and the
-6,
only
nonzero
entry in the first row,
so
we can
and column:
'-3>
-6,
now
change the sign, and subtract
twice the first
row
from the second, leaving
'3>
Therefore G is isomorphic to the cokernel of the homomorphism
(p:Z—>Z0Z
mapping 1 to (3,0). This homomorphism is injective and identifies Z with the
subgroup 3Z 0 0 of the target. Therefore
^
G ^ coker cp
^
zez
z
^
0 Z.
j
——-
3Z 0 0
3Z
Exercises
347
elimination, the 'algorithm' implicitly described in
over Euclidean domains (e.g., over the polynomial
By virtue of Gaussian
Proposition 4.13 will work without fail
ring in
variable
that it will
identify the finitely
explicit direct sum of cyclic
in
4.14.
as
This
is
too
much
to
modules,
Example
expect over more general rings,
since in general elementary transformations do not generate GL\ cf. Remark 2.12.
one
over
field),
a
in the
generated module corresponding to
a
sense
matrix with
an
Exercises
4.1. > Prove that if R is
is
a
integral domain
integral domain and M is an R-module, then Tor(M)
example showing that the hypothesis that R is an
an
submodule of M. Give
an
is necessary.
[§4.1]
integral domain i?, and let N be a torsion-free
is torsion-free. In particular, Hom#(M, R) is
fact again; see Proposition VIII.5.16.) [§VIII.5.5]
4.2. > Let M be a module over an
module.
Prove that
Hom#(M, N)
(We will run
torsion-free.
4.3. > Prove that
into this
integral domain R
an
4.4. > Let R be a commutative
Prove that
Ann(M)
is
ring and M
an
Ann(M)
=
0.
a
only if
PID if and
every submodule
(Of
course
an
i?-module.
ideal of R.
an
integral domain and
if and only if Ann(M) ± 0.
Give an example of a torsion
If R is
is
[§4.1, 5.13]
of R itself is free.
M is
finitely generated,
module M
over
prove that M is torsion
integral domain, such that
an
this example cannot be finitely
generated!)
[§4.2, §5.3]
4.5.
-i
Let M be
of
R/I (viewed
over a commutative ring R. Prove that an ideal I of R
element of M if and only if M contains an isomorphic copy
module
a
is the annihilator of
an
as an
R-module).
The associated primes of M are the prime ideals among the ideals Ann(m),
for m G M. The set of the associated primes of a module M is denoted Assr(M).
Note that every prime in Assr(M) contains Ann#(M). [4.6, 4.7, 5.16]
4.6.
of
-i
Let M be
ideals
Ann(m),
a
module
as
over a
commutative ring i?, and consider the family
the nonzero elements of M. Prove that the
m ranges over
family are prime ideals of R. Conclude that if R is
then
Noetherian,
AssR(M) ^ 0 (cf. Exercise 4.5). [4.7, 4.9]
maximal elements in this
4.7.
Let R be a commutative Noetherian ring, and
module over R. Prove that M admits a finite series
-i
M
in which all quotients
=
M0
Mi/Mi+i
(Hint: Use Exercises 4.5 and 4.6
D
are
Mi
D
D
of the form
Mm
R/p
let M be
=
a
finitely generated
(0)
for
some
to show that M contains an
prime ideal p of R.
isomorphic copy M'
VT. Linear
348
of R/pi for
prime pi. Then do the same with M/M', producing an M" D M'
i?/p2 for some prime p2. Why must this process stop after
some
such that M" jM'
finitely
many
over
=
steps?) [4.8]
4.8. Let R be
module
algebra
commutative Noetherian ring, and let M be
a
R. Prove that every prime in
Assr(M)
finitely generated
primes
a
appears in the list of
produced by the procedure presented in Exercise 4.7. (If p is an associated prime,
then M contains an isomorphic copy N of R/p. With notation as in the hint in
Exercise 4.7, prove that either pi
isomorphically to a copy of R/p in
=
In particular, if M is
is finite.
a
p
or
N fi M'
M/M';
=
0. In the latter case, N maps
iterate the
finitely generated module
reasoning.)
over a
Noetherian ring, then
Ass(M)
4.9. Let M be
a
module
union of all annihilators of
primes of M.
as
nonzero
Exercise
(Use
commutative Noetherian ring R. Prove that the
elements of M equals the union of all associated
over a
4.6.)
Deduce that the union of the associated primes of a Noetherian ring R
a module over itself) equals the set of zero-divisors of R.
4.10. Let R be
primes of
commutative Noetherian ring. One can prove that the minimal
Ann(M) (cf. Exercise V.1.9) are in Ass(M). Assuming this, prove that
a
the intersection of the associated
over
itself) equals
primes of a Noetherian ring R (viewed
4.12. Let p be a
=
as a
module
the nilradical of R.
4.11. Review the notion of presentation
notion of presentation introduced in §4.2.
let R
(viewed
prime ideal of
k[xi,... ,xn]/p.
a
of
a group,
(§11.8.2),
polynomial ring k[xi,... ,xn]
and relate it to the
over a
Prove that every finitely generated module
field /c, and
R has a
over
finite presentation.
tuple (ai, a2,..., an) of elements of R is a
regular sequence if a\ is a non-zero-divisor in i?, a2 is a non-zero-divisor modulo22
(ai), as is a non-zero-divisor modulo (ai,a2), and so on.
For a, b in R, consider the following complex of i?-modules:
4.13.
-i
Let R be
(*)
a
commutative ring. A
tR^^ReR^^R^^j^G)
0
>0
tt is the canonical projection, di(r, s)
ra + sb, and d2(£)
otherwise, d\ and d2 correspond, respectively, to the matrices
where
=
<«»).
Prove that this is indeed
Prove that if
The
(a, b)
complex (*)
regular sequence,
module
is
a
a
complex, for
regular
every a and b.
sequence, this
complex
complex of (a, b).
complex provides us with
is exact.
Thus, when (a, b)
a
class of <X2 in
R/(ai)
is a non-zero-divisor in
is
a
free resolution of the
R/(a,b). [4.14, 5.4, VIII.4.22]
!That is, the
Put
(bt, —at).
(.«)¦
is called the Koszul
the Koszul
=
R/(ai).
5. Classification of finitely
4.14.
generated modules
PIDs
over
349
A Koszul complex may be defined for any sequence ai,..., an of elements
2 seen in Exercise 4.13 and the case n
commutative ring R. The case n
3
reviewed here will hopefully suffice to get a gist of the general construction; the
of
-i
a
=
general
case
=
will be given in Exercise VIII.4.22.
Let a,b,c e R. Consider the
following complex:
—^R^^
>R—^RQRQR
—^RqR^R
(^y
0
where
tt
is the canonical projection and the matrices for di, cfe, ds are,
/0
(a
c)
b
[
,
\b
Prove that this is indeed
Prove that if
(a, 6, c)
Koszul complexes
geometry. [VIII.4.22]
is
a
are very
a
free resolution of Z
a
4.16.
>
Let cp
:
a
a
0/
respectively,
/a\
I —&
J
,
-
c/
every a,
sequence, this
regular
&,
c.
complex
is exact.
important in commutative algebra and algebraic
ring R
=
Z[x,y],
where
x
and y act by 0.
R.
over
i?m and ip
Rn
0
complex, for
4.15. > View Z as a module over the
Find
-b\
-c
-c
>0
i?9 be two i?-module homomorphisms,
i?p
:
and let
y? e V
be the morphism induced
:
#n © Rp
direct
on
coker(<^
0
sums.
ip)
=
-+
Rm © ^
Prove that
coker cp 0 coker ip.
[VIII.4.21]
4.17. Determine
(as
a
better known
entity)
1 + 3x
2x
1 + 2x
l-\-2x-x2
X2
X
over
5.
the polynomial ring
Classification of
the module represented by the matrix
k[x]
over a
3x\
2x\
J
X
field.
finitely generated
modules
over
PIDs
It is finally time to prove the classification theorem for finitely generated modules
over arbitrary PIDs. We have already proved this statement in the special case of
finite Z-modules
and the diligent reader has worked out a proof in the less
special case of finitely generated modules over Euclidean domains, in Exercise 2.19.
Now we go for the real thing.
(§IV.6),
VI. Linear
350
Submodules of free modules. Recall
5.1.
algebra
(Lemma 4.2, Example 4.3)
that
a
arbitrary integral domain R is necessarily
torsion-free but need not be free. For example, the ideal /
(x,y) of R k[x,y]
k
a
for
is
torsion-free
as
an
but
not free: for this
field,
i?-module,
(with
example)
to become really, really evident, it is helpful to rename x
b when these
a, y
are viewed as elements of / and observe that a and b are not linearly independent
submodule of
a
free module
over
an
=
=
=
k[x, y],
over
since ya
xb
0.
=
On the other hand, submodules of
free: simply because
=
a
free module
over a
every module over a field is free
field
are
automatically
(Proposition 1.7).
It is
reasonable to expect that 'some property in between' being a field and being a
friendly UFD such as k[x,y] will guarantee that a submodule of a free module is
free. We will now prove that this property is precisely that of being a principal
ideal domain.
Proposition 5.1.
and let M C F be
PID, let F be
Let R be a
a submodule.
We will actually prove
Then M is
a more
finitely generated free
free.
a
precise result, in view of the full
the classification theorem: we will show that there is a basis
elements a\,...,am
of R (with
2/l
form
a
basis
that there
=
m
module
<n)
over
R,
statement of
(#i,... ,xn) of F
and
such that
QlXXi
,
ym
=
of M. That is, not only do we prove
'compatible' bases of F and M.
&mXm
that M is free, but
we
also show
are
In order to do this, we may of course assume that M ^ 0: otherwise there is
nothing to prove. Most of the work will then go into showing that if M ^ 0, we can
split one direct summand off M; iterating this process will prove the proposition23.
This is where the PID condition is used, so we will single out the main technical
point into the following statement.
Lemma 5.2. Let R be a
let M C F be
submodules F'
PID, let F be
finitely generated free module over R, and
R, x e F, y e M, and
F' n M, and
M, such that y
ax, M'
a nonzero submodule.
C F and M' C
F=
a
Then there exist a e
=
(x)eF',
M
=
=
(y)®M'.
The reader would find it instructive to pause a moment and try to imagine how
the PID hypothesis may enter into the proof of this lemma. The PID hypothesis is
a hypothesis on i?, so the question is, where is the special copy of R to which we
can apply it? Don't read on until you have spent a moment thinking about this.
It
is tempting to look for this copy of R among the 'factors' of F
example, we could map F to R by projecting onto the first component:
7r(ri,...,rn)
=
Rn: for
=n.
But there is nothing special about the first component of i?n; in fact there is
nothing special about the particular basis chosen for F, that is, the particular
In a loose sense, this is the same strategy we employed in the proof of the classification
result for finite abelian groups in §IV.6.
generated modules
5. Classification of finitely
PIDs
over
351
representation of F as Rn. Therefore, this does not look too promising. The way
out of this bind is to democratically map F to R in every possible way: we consider
all homomorphisms (p : F
R. This is a set that does not depend on any extra
so
it
is
more
to
choice,
likely
carry the information we need.
cp(M) is a submodule of i?, that is, an ideal of R; so we have a
the PID hypothesis with profit. In fact, the much weaker fact that
R is Noetherian guarantees already that there must be some homomorphism a for
For each cp,
chance to
which
use
is maximal among all ideals
a(M)
of such
homomorphism
choices. The fact that R is
clearly
PID tells
a
This copy of i?, that is, the target
<p(M).
special for
a, is
a
a
us
good
reason,
independent of inessential
principal, and then we are
that
a(M)
is
submodule of i?, that is,
is
in business.
Here is the formal argument:
Proof. For all (p G
Hom#(F,i?), <p(M)
a
an
ideal.
family of all these ideals is nonempty, and PIDs are Noetherian; therefore (by
Proposition V.1.1) there exists a maximal element in the family, say a(M), for a
R. The fact that M ^ 0 implies immediately that some
homomorphism a : M
The
ip(M) 7^
a(M) ^
(for example,
0
take for cp the projection to
a
suitable factor of
Rn);
Since R is
a
PID, a(M) is principal: a(M)
(a)
=
for
^
element y G M such that
a(Af),
elements a, y mentioned in the statement.
a
there exists
G
We claim that
a
an
divides
a
and
a.
i?,
a
0. Since
These
are
the
let b be
a
b is simply a gcd of
consider the homomorphism
PID; of
course
=
+ sip. Since a G
b
therefore
-0(M);
a(y)
=
and let r,s e R such that b
ra + s(p(y);
we
have
(6),
a(M) C (b). On the other hand
(p(y)),
:= rot
a
some a G
Hom#(F, R). Indeed,
for all (p G
(p(y)
generator of (a, ip(y)) (which exists since R is
ip
hence
0.
ra +
s(p(y)
=
(ra
+
s(p)(y)
=
ip(y)
G
^(M);
ip(M). It follows that a(M) C ^(Af), and by maximality a(M)
(a)
(6), and in particular a divides ip(y), as claimed.
C
(b)
hence
Let y
=
=
(si,..., sn)
=
=
element of F
as an
=
Rn. Each s^ is the image of y by
a
i? (that is, the i-th projection), so a divides all of them by
homomorphism F
what we just proved. Therefore 3r*i,..., rn G R such that Si
ar^ let
=
x
=
(ri,...,rn)
G F.
By construction, y
integral domain and
This is the element x mentioned in the statement.
Further,
a
implies
a(y)
a(x)
=
=
Finally,
a(ax)
aa(x);
=
since R is
an
=
a
ax.
^ 0,
this
verify
the
1.
let F'
we
stated direct
First,
=
=
ker
a
and M'
=
F' D M, and
we can
proceed
sums.
every
z G
F may be written
z
=
a(z)x
as
+
(z
a(z)x);
by linearity
a(z
a(z)x)
=
a(z)
a(z)a(x)
=
a(z)
a(z)
=
0,
to
VI. Linear
352
that is,
rx
a(z)x G kera.
0
a(rx)
z
F'
G
=>
implies that
This
=>
=
ra(x)
F
(x)
=
=>
=0
r
+ F'.
=
0:
algebra
On the other
that is,
(x)
hand,
n F'
0.
=
Therefore
as
a
claimed
Exercise
(cf.
Second, if
(2)
z
G
ca, we have
=
5.1).
M, then
a(z)x
z
and this leads
as
=
a(z)x
-
splitting
cy;
z as
z-cy e Mf)F'
=
G
above,
=
a(M)
(a). Writing
M',
=
(y)
0
M',
the proof.
Once Lemma 5.2 is established, the proof of Proposition 5.1 is
Proof of Proposition
toMCF produces
an
5.1. If M
=
0,
=
mere
busywork:
done. If not, applying Lemma 5.2
submodule M^ C M such that
we are
element y\ G M and
Af
If M^
=
we note
before to
M
concluding
a(z): indeed, a(z)
divides
a
cax
=
a
(yi)0Af(1).
0, we are done; otherwise, apply Lemma 5.2 again (to M^
G M^ and M^ C M^ so that
C
=
obtain y2
F)
to
MHi/iletWeK2)).
This process may be continued, producing elements yi,..., 2/m G M such that
M
so
long
as
=
(yi>0...0(ym)0M<m>,
the module M^ is
However, by Proposition 1.9
nonzero.
we
know
since 2/1,... ,2/m are linearly independent in F. It follows that the process
must stop; that is, M^
0 for some m <n. That is,
m < n,
=
M=
is
free,
as
(yi)0---©(ym)
needed.
Note that the proof has not used part of the result of Lemma 5.2, that is, the
fact that the 'factor' (y) of M is a submodule of a corresponding factor (x) of F.
This is needed in order to
upgrade Proposition
5.1
along the lines mentioned after
its statement. Here is that stronger statement:
Corollary
and let M
nonzero
of M.
5.3. Let R be
a
PID, let F be
C F be a submodule.
elements ai,..., am
Further,
of R (m
we may assume a\
over
module
over
R,
(#i,... ,xn) of F and
(ai#i,..., amxm) is a basis
a2
<
|
such that
n)
|
am.
compared with Proposition
PIDs'; cf. Remark 2.12.
This statement should be
'Smith normal form
finitely generated free
a
Then there exist a basis
2.11:
it amounts to
a
5. Classification of finitely
Proof. Now that
we
generated modules
over
know that submodules of
a
PIDs
353
free module
are
free,
we see
that
the submodule F' C F produced in Lemma 5.2 is free. The first part of the
statement then follows from Lemma 5.2, by an inductive argument analogous to
the proof of Proposition 5.1, and is left to the reader.
The most delicate part of the statement is the divisibility condition. By
induction it suffices to prove a\ a<i, and for this we refer back to the proof of Lemma 5.2:
(ai) is maximal among the ideals <p(M) of i?, as (p ranges over all homomorphisms
F
R.
Since
<p(yi)
Consider then
=
Therefore a<i
The
<p(a,iXi)
y>{aix<2)
=
any24 homomorphism
=
=
1.
<p{x{)
<p{x2)
(ai)
(p(M); by maximality (ai)
<p(M).
D
<p(M)
(ai)> proving a\ a<i as needed.
^(2/2)
£
logic of the argument given
ip such that
=
=
C
ai, we have
=
in this section is somewhat
it takes
tricky:
Lemma 5.2 to prove Proposition 5.1; but then it takes Proposition 5.1 (proving
that F' is free) to revisit Lemma 5.2 and squeeze from it the stronger Corollary 5.3.
5.2. PIDs and resolutions. Proposition 5.1 allows
in §4.2.
us to
complete the circle of
ideas begun
Proposition 5.4. Let R be an integral domain. Then R is a PID if
for every finitely generated R-module M and every epimorphism
Rm0
there exist
a
free R-module R1711
and
a
_^?_>
M
homomorphism
and only
if
> 0,
tti : R1711
i?m° such that
the sequence
0
> i?mi
—^->
M
Rm° -^—>
> 0
is exact.
Proof. The fact that the stated condition implies that R is
in Claim 4.11. For the converse, let 7ro : -Rm°
M be an
ker7To
is free
tti : i?mi
PID
proved
epimorphism; then
by Proposition 5.1; the result follows by choosing any isomorphism
a
was
ker(TTo).
-?
The content of Proposition 5.4 is possibly simpler than its awkward formulation:
Proposition 5.4 is simply a characterization for PIDs analogous to the
characterization of fields worked out in Proposition 4.10. This is another example of the fact
that we can study a ring R by looking at how R-M06 is put together.
The reader
well imagine the 'length-n' version of the condition appearing
we may ask for the class of integral domains R with the property
that for all finitely generated modules M and all partial free resolutions of M,
can
in these results:
7Ti
T^ri-l
ftm^!
>
7Tn
>
Rm0 _?_> M
> 0
,
In order to define a homomorphism on a free module F, one may define it on a basis of F
and then extend it by linearity; there is no restriction on the possible choices for the images of
basis elements. The reader who is not sure about this should review the universal property of free
modules!
VI. Linear
354
there
exists
a
homomorphism
> i?m-
0
7rn : i?mn
i?771™-1such that
7Tn-l
7Tn
algebra
7Ti
y
Rmri-i
7To
y Rrn0
> M
is exact. We have proved that this condition characterizes fields for
l.
for n
> Q
n
=
0 and PIDs
=
dejd vu, as we have already encountered
precisely for fields and 1 for PIDs: the Krull dimension of a
ring; cf. Example III.4.14. Of course this is not a coincidence, but the full relation
between the Krull dimension and the bound on the length of free resolutions is
The alert reader should now have a
a
notion that is 0
beyond the scope of this book25. 'Most' rings (even rings with finite Krull dimension)
do not satisfy any such bound for all modules. The local rings satisfying the flniteness
condition described above for
geometry
to 'smooth'
points
on
some n
in the
correspond
varieties of dimension <
language of algebraic
n.
5.3. The classification theorem. The notion of the rank of
(Definition
1.14)
a
free module
extends naturally to every finitely generated module M
over an
integral
domain R.
Definition 5.5. Let R be an integral domain. The rank rk M of a finitely generated
i?-module M is the maximum number of linearly independent elements in M.
j
It should be clear that this number is finite
Theorem 5.6. Let R be
the
following
a
PID, and let M be
finitely generated R-module.
Then
hold:
There exist distinct prime ideals
an
a
(Exercise 5.6).
(</i),..., (qn)
R, positive integers r^, and
G
isomorphism
MsKkM
•(©(£>
There exist nonzero, nonunit ideals (ai),...,
2 (a>m)t and an isomorphism
(#2) 2
M^RrkM~
These decompositions
are
R
"
(ttl)
(am) of R,
such that
{a\)
2
R
"
(0>m)'
unique (in the evident sense).
After all our preparatory work, this statement practically proves itself. We
will describe the arguments in general terms, leaving the details to the enterprising
reader.
The two forms taken by the theorem go under the name of invariant factors
and elementary divisors, as in the case of abelian groups. As in the case of abelian
groups, the equivalence between the two formulations amounts to careful
bookkeeping, and
we
The
exercises to
will not prove it
formally;
see
§IV.6.2 for
most persistent of readers will take a more
§IX.8.
a
reminder of how to go from
sophisticated look
at these bounds in the
5. Classification of finitely
one to
generated modules
over
PIDs
355
the other. The role played by Lemma IV.6.1 in that discussion is taken
over
by the Chinese remainder theorem, Theorem V.6.1, in this more general setting.
As the two formulations are equivalent, it suffices to prove the existence and
uniqueness of the
decompositions given
The existence is
a
in Theorem 5.6 for any
direct consequence of the results in
is an epimorphism
one
of the two.
let M be
§5.1. Indeed,
a
finitely generated module; thus there
->M
Rn
where
ker
n
->0
is the number of generators of M. Apply Corollary 5.3 to the submodule
a basis (#i,..., xn) of Rn and nonzero elements <2i,..., <2m
Rn: there exist
7r C
of R such that
(a\X\,...,
amxm)
is
a
is, M is presented by the
That
basis of ker tt and further a\ |
n x m
|
am.
matrix
/ai
0\
\0
0/
By Proposition 4.13, we may assume that ai,..., am are not units: if any one of
them is, the corresponding row and column may be omitted from the matrix (and
n corrected accordingly). It follows that
M ^
with ai I
is proved.
I
am nonzero
R
R
(ttl)
(ttm)
e
nonunits,
as
i?(n-™)
prescribed by Theorem 5.6, and the
existence
As for the uniqueness of the representations, if
M
with T
a
torsion
module26,
then
r
=
^
Rr 0 T,
rkM and T ^
Tor^(M) (Exercise 5.10). Thus,
the 'free part' and the 'torsion part' in the decompositions given in Theorem 5.6
are determined uniquely.
This reduces the uniqueness question to the structure of torsion modules:
assuming that
R
e (p?n
i>3
(*)
R
TX
e
(qkSk*Y
with pi, qk irreducible elements of i?, the task is to show that the range of the
indices is the same and, up to reordering, (pi)
{qi) and r^
sij for all i, j.
=
Since
nonzero
a
finitely generated module
(Exercise 4.4),
'Recall that this
is torsion if and
it is reasonable that
means
=
Ann(M)
that every element of T is
only if
will be of
a torsion
its annihilator is
some use
element; cf. §4.1.
here.
VI. Linear
356
Lemma 5.7. Let M be
0).
Then
Ann(M)
a torsion
module, expressed
(am). Further,
=
as in
the prime ideals
Theorem 5.6
(qi)
algebra
(with rk M
=
precisely the prime
are
ideals of R containing Ann(M).
Proof. By hypothesis
R
M 2*
R
~
(ai)
with ai
|
|
am. If r G
Ann(M),
0
In particular r
For the reverse
=
=
0 modulo
inclusion,
(am)'
then
r(l,...,l)
that is,
(am);
assume r
G
decomposition, write y
(yi,... ,2/m),
have r^
0 for all i; therefore ry
0,
=
=
=
(r,...,r).
r
G
Thus
(am).
Ann(M)
C
(am).
(am) and y G M. Identifying M with its
with yi G R/(di). Since r G (am) C (a*), we
and
=
Ann(M)
r G
as
needed.
For the second part of the statement, tracing the equivalence between the two
decompositions in Theorem 5.6 shows that the q^s are precisely the irreducible
factors of
am, so the assertion follows from the first part.
By Lemma 5.7, the
(up
coincide
to inessential
{pi}, {qk} of irreducibles appearing in (*) must
units): the ideals they generate are precisely the prime
sets
ideals containing Ann(M). The fact that the sets coincide also follows from the
observation that the isomorphism in (*) must match like primes, in the sense that
an
element of
a
factor
R
in the
same
source must
prime
uniqueness
land in
pi\this follows
combination of quotients in the target by powers of the
by comparing annihilators (cf. Exercise 5.11). Therefore,
a
is reduced to the
case
of
a
single irreducible
q G R: it suffices to show
that if
R
(5-1)®
with r\>
Z
R
Z
%^)_(<^)
> rm and si >
> sn, then
(<?s")'
m
=
n
and ri
=
Si for all i.
This
reproduces precisely the key step the reader used in the proof of uniqueness
for the particular case of abelian groups, in Exercise IV.6.1. Of course we will not
spoil the reader's fun by giving any more details this time (Exercise 5.12).
situation
Remark 5.8. To summarize, the information carried by a torsion module over a
PID is equivalent to the choice of a selection of nonzero, nonunit ideals (ai),..., (am)
such that
(oi)
We have
seen
that
(am)
D
(o2)
2
2
( am)'
is in fact the annihilator ideal of the module; it would be
nice if we had a similarly explicit description of all invariants (ai),..., (am). As
preview of coming attractions, we could call the ideal
a
(oi •••am)
In situations in which we can compute
both the annihilator and the characteristic ideal, comparing them may lead to
the characteristic ideal of the module27.
Warning:
This does not seem to be standard terminology.
Exercises
357
strong conclusions about the module. For example, a torsion module is
and only if its annihilator and characteristic ideals coincide.
cyclic if
Further, the prime ideals appearing in Theorem 5.6 have been characterized
the prime ideals containing the annihilator ideal. They may just as well be
characterized as those prime ideals containing the characteristic ideal, as the reader
as
will check
(Exercise 5.15).
Finally, note that from
point of view it is immediate that the characteristic
We will run into this fact again, in an
j
important application (Theorem 6.11).
ideal
this
is contained in the annihilator ideal.
Exercises
5.1.
M
Let N, P be submodules of a module M, such that N D P
{0} and
N + P. Prove that M
N 0 P. (This is a word-for-word repetition of
>
=
=
=
Proposition IV.5.3 for
5.2.
M is
Complete the proof of Corollary
5.4. Let R be
b
integral domain, and let
torsion if and only if rkM
M be
Let R be an
Prove that
5.3.
modules.) [§5.1]
and
£ (a),
an
=
a
finitely generated i?-module.
0.
5.3.
integral domain, and
R/(a), R/(a,b)
are
assume that a,b e R
both integral domains.
are
such that
a
^ 0,
Prove that the Krull dimension of R is at least 2.
Prove that if R satisfies the finiteness condition discussed in
then
You
n >
can
concrete
§5.2 for
some
n,
2.
this second point by appealing to Proposition 5.4. For a more
argument, you should look for an R-module admitting a free resolution of
prove
length 2 which cannot be shortened.
Prove that
(a, b)
is
a
regular
Prove that the i?-module
Can you
see
sequence in R
R/(a,b)
has
a
(Exercise 4.13).
free resolution of length exactly 2.
how to construct analogous situations with
n >
3 elements ai,..., an?
5.5.
Recall (Exercise V.4.11) that a commutative ring is local if it has a single
maximal ideal m. Let R be a local ring, and let M be a direct summand of a finitely
generated free i?-module: that is, there exists an i?-module N such that M 0 N is
-i
a
free i?-module.
Choose elements mi,..., mr G M whose cosets mod mM are a basis of M/mM
the field R/xn. By Nakayama's lemma, M
(mi,..., mr)
as a vector space over
=
(Exercise 3.10).
Obtain
a
surjective homomorphism
n :
F
=
i?0r
M.
VI. Linear
358
Show that
splits, giving
7r
an
isomorphism
F
=
algebra
(Apply Exercise III.6.9
M0ker it.
to the surjective homomorphism n and the free module M 0 N to obtain
F; then use Proposition III.7.5.)
splitting M
Show ker 7r/m ker 7r
ker7r
=
=
0. Use
Nakayama's lemma (Exercise 3.8)
a
to deduce that
0.
Conclude that M ^ F is in fact free.
Summarizing,
free i?-module
[VIII.2.24, VIII.6.8, VIII.6.11]
local ring, every direct summand of a finitely generated28
Using the terminology we will introduce in Chapter VIII, we
over a
is free.
would say that 'projective modules over local rings are free'. This result has strong
implications in algebraic geometry, since it underlies the notion of vector bundle.
Contrast this fact with Proposition 5.1, which shows that,
finitely generated free module is free.
over a
PID,
every
submodule of a
integral domain, and let M
(mi,..., mr)
generated module. Prove that rkM < r. (Use Exercise 3.12.) [§5.3]
5.6. > Let R be
5.7. Let R be
an
integral domain, and let
an
Prove that rkM
5.8. Let R be
=
=
integral domain, and let
an
=
r
finitely generated module
over
R.
a
a
=
may be chosen so that 0
N
M
N/M
-? -? -?
0 splits.
-?
integral domain, and let
0
be
finitely
finitely generated module over
Rr, such that
free submodule N
M be
if and only if M has
M/N is torsion.
If R is a PID, then N
an
a
a
rk(M/Tor(M)).
R. Prove that rkM
5.9. Let R be
M be
be
an exact sequence
> Mi
> M2
> M3
> 0
of finitely generated i?-modules. Prove that rkM2
=
rkMi
+
rkM3.
a homomorphism from the Grothendieck
finitely generated i?-modules to Z (cf. §3.4).
Deduce that 'rank' defines
the category of
5.10. > Let R be
with T
r
=
a
an
group of
Rr 0 T,
integral domain, M an i?-module, and assume M
directly (that is, without using Theorem 5.6) that
=
torsion module. Prove
rkM and T ^
TovR(M). [§5.3]
integral domain, let M, iV be R-modules, and let ip : M
homomorphism. For m G M, show that Ann((ra)) C Ann((<p(ra))). [§5.3]
5.11. > Let R be an
be
a
5.12.
>
Complete the proof of uniqueness
Exercise IV.6.1 may be
in Theorem 5.6.
(The
N
hint in
helpful.) [§5.3]
finitely generated module over an integral domain R.
Prove that if R is a PID, then M is torsion-free if and only if it is free. Prove
that this property characterizes PIDs. (Cf. Exercise 4.3.)
5.13. Let M be
example of a finitely generated module
isomorphic to a direct sum of cyclic modules.
5.14. Give an
is not
a
over an
integral domain which
The finite rank hypothesis is actually unnecessary, but the proof is harder without this
condition.
6. Linear transformations of
a
free module
359
5.15. > Prove that the prime ideals appearing in the elementary divisor version of
the classification theorem for a torsion module M over a PID are the prime ideals
containing the characteristic ideal of M,
as
defined in Remark 5.8.
[§5.3]
5.16. Prove that the prime ideals appearing in the elementary divisor version of
the classification theorem for a module M over a PID are the associated primes
of M,
as
defined in Exercise 4.5.
5.17. Let R be
a
PID. Prove that the Grothendieck group
(cf. §3.4)
of the category
of finitely generated i?-modules is isomorphic to Z.
Linear transformations of
6.
a
free module
One beautiful application of the classification theorem for finitely generated
modules
over
PIDs is the determination of 'special' forms for matrices of linear maps of
itself.
a vector space to
Several fundamental concepts are associated with the general notion of a linear
map from a vector space to itself, and we will review these concepts in this section.
Not
any
surprisingly, much of the discussion
integral domain i?, and we will stay
may be carried out for free modules over
at this level of
generality
section. Theorem 5.6 will be used with great profit when R is
a
in
(most of)
field,
the
in the next
section.
6.1. Endomorphisms and similarity. Let R be an integral domain. We have
considered in some detail the module Hom#(F, G) of R-module homomorphisms
between two free modules.
For example, we have argued that describing such
for
homomorphisms
finitely generated free modules F, G amounts to describing
matrices with entries in i?, up to 'equivalence': the equivalence relation introduced
in
§2.2
We
accounts for the
now
(arbitrary)
shift the focus
that is, the R-module
choice of bases for F and G.
little and consider the special case in which F
G,
Endj^F) of endomorphisms a of a fixed free R-module F:
a
=
End#(F) is in fact an R-algebra: the operation of composition makes
ring, compatibly with the i?-module structure (cf. Exercise 2.3).
Note that
a
From the point of view championed in
End#(F)
Hom#(F, F)
§2,
it
the two copies of F appearing in
unrelated; they are just (any) isomorphic
given rank. There are circumstances, however, in which
we need to really choose one representative F and stick with it: that is, view a as
acting from a selected free module F to itself, not just to an isomorphic copy of
itself. In this situation we also say that a is a linear transformation of F, or an
representatives of
a
operator
=
free module of
on
are
a
F.
From this different point of view it makes sense to compare elements of F
we apply a; for example, we could ask whether for some v G F
'before and after'
VI. Linear
360
we may
have
a(v)
=
Av for
some
M of F may be sent to itself by
F.
a with the identity29 / : F -?
can
A G i?, or more generally
In other words, we can
a.
whether
compare
a
algebra
submodule
the action of
In terms of matrix representations (in the finite rank case) the description of a
be carried out as we did in §2, but with one interesting twist. In §2.2 we dealt
homomorphism changes when we change
identifying source and target,
with how the matrix representation of a
bases in the source and in the target. As
choose the
we must
same
basis for both
we are now
source
and target. This leads to
a
different
notion of equivalence of matrices:
Definition 6.1. Two square matrices A, B G Mn(R) are similar if they represent
F of a free rank-n module F to itself, up to the
the same homomorphism F
choice of a basis for F.
j
new
This is clearly an equivalence
notion is readily obtained:
6.2.
Proposition
exists
an
Two matrices
relation. The analog of Proposition 2.5 for this
A,Be M.n{R)
are
similar
and only
if
if
there
invertible matrix P such that
B
=
PAP~\
The reader who has really understood Proposition 2.5 will not need any detailed
proof of this statement: it should be apparent from staring at the butterfly diagram
Rn
Rn
J
a
Rn
which
we
are
choosing the
Rn
essentially copying from §2.2.
same
basis for
The difference here
track of the change of basis are
respect to the choice of basis dictated by (p (resp.,
matrix of the 'top' (resp., bottom) composition
(f-1
(resp., -0-1
B
=
o a o
ip).
is that
we
are
and target; hence the two triangles keeping
the same. Suppose A (resp., B) represents a with
source
o a o
(p
:
Rn
ip)\ that
is, A
(resp., B)
is the
Rn
-?
If P is the matrix representing the change of basis it, then
PAP~l simply because the diagram commutes:
ip~l
o a o
ip
=
(jr
o
(p-1)
Proposition 6.2 suggests
a
o a o
((p 7r-1)
o
=
tt o
(<^-1
o a o
cp)
o
7r-1.
useful equivalence relation among endomorphisms:
Definition 6.3. Two i?-module homomorphisms of
a
free module F to itself,
a,(3:F^F,
The existence of the identity is
handy to be able to invoke it.
one
of the axioms of
a
category; every
now and
then, it is
6. Linear transformations of
are
similar if there exists
a
free module
361
automorphism
an
ft
=
7T O a
tt :
F
F such that
07T_1.
J
In the finite rank case, similar endomorphisms are represented by similar
matrices, and two endomorphisms a, ft are similar if and only if they may be represented
by the same matrix by choosing appropriate (possibly different) bases on F. The
interesting invariants determined
surprisingly)
by similar endomorphisms
will turn out
(not
to be the same.
From a
study: the
group-theoretic viewpoint, similarity
GL(F)
End#(F) by conjugation,
group
=
Autr(F)
is
an
eminently natural
of automorphisms of
a
notion to
free module acts
on
and two endomorphisms a, ft are similar if and only if
in
are
the
same
orbit
under this action. If the reader prefers matrices, the
they
of
invert
ible
nxn
matrices with entries in R acts by conjugation on
group GLn(i?)
the module
Mn(R)
only if they
are
of square nxn matrices, and two matrices are similar if and
same orbit.
in the
Natural questions arise in this context. For example, we should look for ways
to distinguish different orbits: given two endomorphisms a, ft, can we effectively
decide whether a and ft are similar? One way to approach this question is to
determine invariants which
can
distinguish different orbits.
Better still, we can
a given similarity
look for 'special' representatives within each orbit, that is, of
class. In other words, given an endomorphism a : F
F, find
a
basis of F with
respect to which the matrix description of a has a particular, predictable
Then a, ft are similar if these distinguished representatives coincide.
This is what
friendly
case
we
eventually going
are
of vector spaces
over a
to squeeze out of Theorem
shape.
5.6, in the
fixed field.
6.2. The characteristic and minimal polynomials of an endomorphism.
Let a G End#(F) be an endomorphism of a free R-module; henceforth we are often
tacitly going to
assume
that F is finitely generated
(this
is necessary anyway
as
aiming to translate what we do into the language of matrices). We want
to identify invariants of the similarity class of a, that is, quantities that will not
change if we replace a with a similar linear map ft G End#(F).
we are
For example, the determinant is such
Definition
6.4. Let
a G
End#(F).
A is the matrix representing
a
a
quantity:
The determinant of
a
is det
a :=
det A, where
with respect to any choice of basis of F.
j
Of course there is something to check here, that is, that det A does not depend
the choice of basis. But we know (Proposition 6.2) that A, B represent the
same endomorphism a if and only if there exists an invertible matrix P such that
on
B
=
PAP~l;
then
det(B)
by Proposition
basis.
3.3.
=
detiPAP-1)
=
det(P)
det (A)
det(P_1)
=
det(A)
Thus the determinant is indeed independent of the choice of
same argument shows that if a and ft are similar linear
Essentially the
transformations, then det (a)
=
det(/3) (Exercise 6.3).
VI. Linear
362
By Proposition 3.3,
is
not
a
field, det a ^
0
0
if
linear transformation
is invertible if and
a
a
6.5. Let
Proposition
Then det a ^
a
field, this of course means simply det a ^
says something interesting:
unit in R. If R is
a
a
be
and only
a
if
a
linear
transformation of
a
algebra
only if det a
0. Even if R is
free R-module
F
Rn.
=
is infective.
K, and view
Proof. Embed R in its field of fractions
a as a
linear transformation
of Kn\ note that the determinant of a is the same whether it is computed over R
or over K. Then a is injective as a linear transformation Rn
Rn if and only if it
is injective as a linear transformation Kn
Kn, if and only if it is invertible as a
linear transformation Kn
Of
even
Kn, if and only if det
a
^
0.
integral domains other than fields a may fail to be surjective
care is required even over fields, if we abandon the hypothesis
modules are finitely generated (cf. Exercises 6.4 and 6.5).
course over
if det
a
that the free
and
^ 0;
Another quantity that is invariant under similarity is the trace. The trace of
matrix A
(a^) G Mn(R) is
a
=
square
n
:=^Oii,
tr(A)
2=1
that is, the
of its diagonal entries.
sum
Definition 6.6. Let
where A
a
G
The trace of
End#(F).
is the matrix representing
Again, we have to check that
following computation.
a
is defined to be tra
:=
tr
with respect to any choice of basis of F.
a
A,
j
this is independent of the choice of basis; the key
is the
Lemma 6.7. Let
Proof. Let A
=
A, B
(a^),
G
B
Mn(R).
=
(bij).
Then
tr(AB)
Then AB
n
tr(AB)
=
=
tr(BA).
(Y^k=i aikbkj)\ hence
n
^2^2aikbki.
=
2=1 fc = l
This expression is symmetric in A, £?,
With this understood, if B
tr(B)
=
tv((PA)p-1)
showing (by Proposition 6.2)
Again,
=
=
so
it must
PAP'1,
=
tv((p-1P)A)
that similar matrices have the
same
=
same
tr(A),
trace,
as
needed.
token shows that similar linear
same trace.
The trace and determinant of
in fact just two of
then
tviP-^PA))
the reader will check that the
transformations have the
equal tr(BA).
a sequence
of
n
an endomorphism a of a rank-n free module are
invariants, which can be nicely collected together
into the characteristic polynomial of
a.
6. Linear transformations of
free module
a
363
Definition 6.8. Let F be
a free R-module, and let a G End#(F).
Denote by /
F. The characteristic polynomial of a is the polynomial
the identity map F
Pa(t) :=det(tl -a)
Proposition
6.9. Let F be
coefficient oft71-1
R[i\.
free R-module of rank
a
The characteristic polynomial
The
G
Pa(t) is
Pa(t) equals
in
a monic
j
n, and let
a G
End#(F).
polynomial of degree
n.
tr(a).
The constant term
If a
and (3
are
of Pa(t) equals (-l)ndet(o:).
similar, then Pa(t)
Pp(t).
=
Proof. The first point is immediate, and the third is checked by setting t
0. To
be
a matrix representing a with respect
verify the second assertion, let A
(a^)
to any basis for F, so that
=
=
ft
-
an
-oi2
t
-a2i
Pa(t)
=
-
-oin
a22
-&2n
det I
dni
t
—Q>ni
annJ
Expanding
the determinant according to Definition 3.1 (or in any other way), we
that the only contributions to the coefficient of tn~l come from the diagonal
entries, in the form
see
n
^t'-t
(-an) -t'-t,
2=1
and the statement follows.
Finally,
such that P
assume
=
that
7r o a o
and (3 are similar.
and hence
a
7r_1,
tl
are
similar
-
(as endomorphisms
must have the
same
p
of
=
7T O
(tl
-R[£]n). By
determinant, and this
-
a)
Then there exists
O
an
invertible
tt
7T"1
Exercise 6.3 these two transformations
proves the fourth
point.
By Proposition 6.9, all coefficients in the characteristic polynomial
tn
are
invariant under
only
ones
-
(tva)tn-1
similarity;
that have special
as
far
+
as we
+
(-l)n(deta)
know,
trace and determinant are the
names.
Determinants, traces, and
more
generally the characteristic polynomial
can
show very quickly that two linear transformations are not similar; but do they tell
us unfailingly when two transformations are similar? In general, the answer is no,
even over fields. For example, the two matrices
'-(J!)- HI
both have characteristic polynomial (t
l)2, but they are not similar. Indeed, I is
I for
the identity and clearly only the identity is similar to the identity: PIP~l
all invertible matrices P.
=
VI. Linear
364
Therefore, the equivalence relation defined by prescribing that two
polynomial is coarser than similarity.
This hints
transformations have the same characteristic
that there must be other interesting quantities associated with
transformation and invariant under similarity.
We
a
algebra
linear
thoroughly in a short while (at least
already gain
insight by contemplating another kind
fields),
of 'polynomial' information, which also turns out to be invariant under similarity.
As EndjR(F) is an iJ-algebra, we can evaluate every polynomial
are
going
but
over
to understand this much more
we can
some
f(t)
at any a G
=
rmtm
+
rwx*™-1
+
+ r0 G
R[t]
Endfl(F):
f(a)
=
rmam
+
+
r^a™"1
...
+ r0 e
EndR(F).
In other words, we can perform these operations in the ring End#(F);
multiplication by r G R amounts to composition with rl G Endj^F), and ak stands for the
o a of a with itself.
/c-fold composition a o
The set of polynomials such that
f(a)
=
0 is
an
ideal of
annihilator ideal of
Lemma 6.10.
If a
R[t] (Exercise 6.7),
and (3
are
similar, then J^ot
Proof. By hypothesis there exists
(3k
=
(noaon~l)k
we see
that for all
which
we
will denote J?a and call the
a.
=
=
invertible
an
tt
(7roQ:o7r_1) (7roo:o7r_1)
f(t)
o
G
R[t]
we
such that (3
o
-o
=
tt o a o
(7roo:o7r_1)
7r_1. As
7roak o7r_1,
=
have
/(/?)=
It follows immediately that f(a)
^p-
=
7TO
/(<*)
0 <^=>
0
/(/?)
7T"1.
=
0, which is the
statement.
Going back to the simple example shown above, the polynomial t 1 is in the
annihilator ideal of the identity, while it is not in the annihilator ideal of the matrix
An optimistic reader might now guess that two linear transformations are
similar if and only if both their characteristic polynomials and annihilator ideals
coincide. This is unfortunatley not the case in general, but we promise that
situation will be considerably clarified in short order (cf. Exercise 7.3).
In any case, even the simple example given above allows us to point
remarkable fact.
Note that the
(common)
characteristic polynomial
(t
the
out a
l)2
of I
and A annihilates both:
This is not
a
coincidence:
Pa(t)
G
ya for all linear transformations
a.
That is,
6. Linear transformations of
free module
a
365
Theorem 6.11
the linear
(Cayley-Hamilton). Let Pa(t)
transformation a G Endj^F). Then
P«(a)
This beautiful observation
rule30,
in the form of
can
be the characteristic polynomial
of
0.
=
be proved directly by judicious use of Cramer's
Exercise 6.9. In any case, the Cayley-
Corollary 3.5; cf.
Hamilton theorem will become essentially evident once we connect these linear
algebra considerations with the classification theorem for finitely generated modules
PID; the adventurous reader can already look back
why the Cayley-Hamilton theorem is obvious.
at Remark 5.8 and
over a
out
figure
we cannot expect too much of R[t], and it
priori concerning ya. However, consider the field of
fractions K of R (§V.4.2); viewing a as an element of End^(i^n) (that is, viewing
If R is
the entries of
have
arbitrary integral domain,
an
hard to say something
seems
an
a
a
matrix representation of
annihilator ideal
^
J^i
'over
a as
elements of K, rather than
i?),
a
will
K\and it is clear that
The advantage of considering J^i * C K[t]
has a (unique) monic generator.
JSt
is that
K[t]
is
a
PID, and it follows that
a free i?-module, and let a G End#(F).
Let K be
The minimal polynomial of a is the monic generator
Definition 6.12. Let F be
the field of fractions of R.
ma(t)eK[t] oijiK).
j
With this terminology, the Cayley-Hamilton theorem amounts to the assertion
that the minimal polynomial divides the characteristic polynomial: ma(t) Pa(t).
Of
the situation is simplified, at least from an expository point of view,
field: then K
R, ma(t) G R[t], and ya
(ma(t)). This is one
we will eventually assume that R is a field.
course
if R is itself
reason
why
a
=
=
6.3. Eigenvalues, eigenvectors, eigenspaces.
Definition 6.13. Let F be
a free R-module, and let a
transformation of F. A scalar A G R is an eigenvalue for
be
G
End#(F)
a
if there exists
a
v
linear
G
F,
v^O, such that
a(v)
=
Av.
j
For example, 0 is an eigenvalue for a precisely when a has a nontrivial kernel.
The notion of eigenvalue is one of the most important in linear algebra, if not in
algebra, if not in mathematics, if not in the whole of science. The set of eigenvalues
of a linear transformation is called its spectrum. Spectra of operators show up
everywhere, from number theory to differential equations to quantum mechanics.
The spectrum of a ring, encountered briefly in §111.4.3, was so named because it
may be interpreted
(in important motivating contexts)
as a scalar and cannot
spectrum in the
sense
0. Unfortunately
det(al a)
det(0)
det(t/ a) of the characteristic polynomial, t is assumed to
be replaced by a. Applying Cramer's rule circumvents this obstacle.
Here is an even more direct
this does not work: in the definition
act
as a
'proof: Pol((x)
=
=
=
VI. Linear
366
of Definition
6.13. The 'spectrum of the
hydrogen atom'
is also
a
algebra
spectrum in this
sense.
It is
for if (3
hopefully
immediate that similar transformations have the
7r_1 and a(v)
7r o a o
=
(3(ir(v))
as
7r(v) ^
If
0 if
F is
=
7r_1(7r(v))
finitely generated,
same
spectrum:
Av, then
=
(a(v))
n o
=
7r(Av)
this shows that every eigenvalue of
^ 0,
v
7r o a o
=
then
we
have the
a
=
is
A7r(v);
an
following useful
eigenvalue of (3.
translation of the
notion of eigenvalue:
Lemma 6.14. Let F be
the set
a
of eigenvalues of
finitely generated R-module, and
is precisely the set of roots in
a
a G
R
of
End#(F).
Then
the characteristic
Pa(t).
polynomial
Proof. This is
A
is
a
an
straightforward
eigenvalue for
a
consequence of
Proposition 6.5:
<^=>
3v
^
0 such that
a(v)
<^>
3v
^
0 such that
(XI
<^=> XI
as
let
a
<^>
det(AJ
^
Pa(A)
=
-
AJ(v)
a)(v)
=
0
is not injective
-
=
a)
=
0
0
claimed.
It is
an
evident consequence of Lemma 6.14 that eigenvalue considerations
on the base ring R.
depend very much
Example 6.15. The matrix
(0
-1>
has no eigenvalues over R, while it has eigenvalues over C: indeed, the characteristic
polynomial t2 + 1 has no real roots and two complex roots. The reader should
observe that, as a linear transformation of the real plane R2, this matrix corresponds
to a 90° counterclockwise rotation; the reason why this transformation has no (real)
eigenvalues
Example
is that
no
6.16. An
direction in the plane is preserved through
example with
a
a
90° rotation,
j
different flavor is given by the matrix
2; hence it has eigenvalues over R, but
Q. Geometrically, the corresponding transformation flips the plane about
line; but that line has irrational slope, so it contains no nonzero vectors with
This has characteristic polynomial t2
not over
a
rational components.
j
As another benefit of Lemma 6.14,
we may now
introduce the
following
algebraic multiplicity of an eigenvalue of a linear
finitely generated free module is its multiplicity as a root
notion:
Definition 6.17. The
transformation
a
of
a
characteristic polynomial of
a.
of the
j
6. Linear transformations of
a
For example, the identity
multiplicity equal
free module
on
367
F has the
single eigenvalue 1, with (algebraic)
to the rank of F.
algebraic multiplicities is bounded by the degree of the
polynomial, that is, the dimension of the space. In particular,
The sum of the
characteristic
Corollary
at most
6.18.
If
transformation
n.
The number
of eigenvalues of
linear
a
transformation of Rn is
then every linear
the base ring R is an algebraically closed field,
has exactly n eigenvalues (counted with algebraic
multiplicity).
Proof. Immediate from Lemmas 6.14, V.5.1, and V.5.10.
There is a different notion of multiplicity of
the corresponding31 eigenspace may be.
Definition 6.19.
i?-module F. Then
Let A be an
eigenvalue of
a nonzero v G
F is
Av, that is, if
eigenvalue A, if a(v)
is the eigenspace corresponding to A.
=
an
v G
an
eigenvalue, related
linear transformation
a
eigenvector for
ker(A7
-
a).
a,
to how
big
of
free
a
corresponding
The submodule
a
to the
ker(A7
-
a)
j
Definition 6.20. The geometric multiplicity of
an
eigenvalue A
is the rank of its
eigenspace.
j
This is clearly invariant under similarity (Exercise 6.14). Geometric and
algebraic multiplicities do not necessarily coincide: for example, the matrix
GO
has characteristic polynomial (t
l)2, and hence the single eigenvalue 1, with
2.
for
a vector v
However,
algebraic multiplicity
=
A
\0
equals lv if and only if v2
eigenvalue 1 consists of the
1\
(vA
l) \v2)
=
Qj1),
fvi
+v2
V
v2
0: that is, the eigenspace corresponding to the single
and has dimension 1. Thus, the
geometric multiplicity of the eigenvalue is 1 in this example.
=
span of the vector
(J)
Contemplating this example will convince the reader that the geometric
an eigenvalue is always less than its algebraic multiplicity. In the neatest
multiplicity of
possible situation, an operator
eigenvalues in the base ring i?,
a on a
free module F of rank
n
may have all its
n
and each algebraic multiplicity may agree with the
corresponding geometric multiplicity. If this is the case, F may then be expressed
direct sum of the eigenspaces, producing the so-called spectral decomposition of
F determined by a (cf. Exercise 6.15 for a concrete instance of this situation). The
action of a on F is then completely transparent, as it amounts to simply applying
as a
a
(possibly)
different scaling factor
on
each piece of the spectral decomposition.
6lli we were serious about pursuing the theory over arbitrary
compelled to call this object the eigenmodule of a. However, we
our attention to the case in which the base
ring is
a
field,
integral domains, we
eventually going
are
so we will not
bother.
would feel
to restrict
VI. Linear
368
If A
admits
is
a
a
matrix representing
a G
End#(F)
spectral decomposition if and only if A
/Ai
A
where Ai,..., An
are
=
with respect to any basis, then
is similar to a diagonal matrix:
0
0
0
A2
0
\0
0
\J
...
a.
In this
case
we
say
diagonalizable. A moment's thought reveals that the columns of
of V consisting of eigenvectors of
Once more,
bring Theorem
that A
P
are
(or a)
then
a
is
basis
a.
promise that the situation will become (even) clearer
we
a
y-l
P
the eigenvalues of
algebra
once we
5.6 into the picture.
Exercises
In these exercises, R denotes
6.1. Let k be
an
an
integral domain.
infinite field, and let
Prove that there
are
finitely
Prove that there
are
infinitely
det(a
(3)
be any positive integer.
equivalence classes of
matrices in
M.n(k).
similarity classes of
matrices in
M.n(k).
many
free i?-module of rank n, and let a,/3 G End#(F).
det(a)det(/3). Prove that a is invertible if and only
6.2. Let F be
o
many
n
=
Prove that
a
if det(a)
is
invertible.
6.3. > Prove that if
det(/3)
and
tr(a)
6.4. > Let F be
=
a
a
and (3
are
tr(/3). (Do
similar in the
sense
of Definition 6.3, then
this without using Proposition
a
is not
=
6.9!) [§6.2]
finitely generated free R-module, and let a be a linear
an example of an injective a which is not surjective; in
surjective precisely when det R is not a unit. [§6.2]
transformation of F. Give
prove that
det(a)
fact,
a field, and view k[t] as a vector space over k in the evident way.
example of a /c-linear transformation k[t]
k[t] which is injective but not
surjective; give an example of a linear transformation which is surjective but not
6.5. > Let k be
Give
an
injective.
[§6.2, §VII.4.1]
same characteristic polynomial if and
if
have
the
same
trace
and
determinant.
Find two 3x3 matrices with
only
they
the same trace and determinant but different characteristic polynomials.
6.6. Prove that two 2x2 matrices have the
6.7.
>
Let
a G
End#(F)
be
that the set of polynomials
6.8. > Let A G
Mn(R)
A and A1 have the
[§7-2]
be
same
a
linear transformation
f(t)
G
R[t]
a square
such that
on a
f(a)
=
free i?-module F. Prove
0 is
an
ideal of
R[i\. [§6.2]
matrix, and let A1 be its transpose. Prove that
characteristic polynomial and the
same
annihilator ideals.
Exercises
369
Cayley-Hamilton theorem, as follows. Recall that every square
adjoint matrix, which we will denote adj(M), and that we proved
that
(Corollary 3.5)
adj(M) M
det(M) I. Applying this to M tl A (with
6.9. > Prove the
matrix M has
an
=
A
a
matrix realization of
(*)
a G
=
End#(F)) gives
Bd](tI-A)'(tI-A)
Prove that there exist matrices Bk
then
use
(*)
to obtain
Pa(A)
-
=
M.n{R)
=
Pa(t)'L
such that
adj(£/
A)
=
Y^k=o Bktk',
0, proving the Cayley-Hamilton theorem. [§6.2]
6.10. > Let Fi, F2 be free i?-modules of finite rank, and let ai, resp., 0:2, be linear
transformations of Fi, resp., F2. Let F
F\0 F2, and let a
ct\ 0 0:2 be the
=
linear transformation of F
Prove that Pa(t)
=
Pai(t)Pa2(t).
multiplicative under direct
=
to ct\ on
restricting
F\ and
0:2 on
F2.
That is, the characteristic polynomial is
sums.
an example showing that the minimal polynomial
under direct sums.
Find
is not
multiplicative
[6.11, §7.2]
6.11.
-1
Let
be
a
a
linear transformation of
a
finite-dimensional vector space V,
subspace, that is, such that a{V{) C Vlt Let ai be
the restriction of a to Vi, and let V2
V/V\. Prove that a induces a linear
transformation 0:2 on V2, and (in the same vein as Exercise 6.10) show that Pa(t)
Pai(t)Pa2(t). Also, prove that trar traj + tro£, for all r > 0. [6.12]
and let V\ be
an
invariant
=
=
=
6.12. Let
a
be
a
linear transformation of
a
finite-dimensional C-vector space V.
Prove the identity of formal power series with coefficients in C:
W^)=exp(Strar7)(Hint:
of
a.
The left-hand side is essentially the inverse of the characteristic polynomial
Use Exercise 6.11 to show that both sides are multiplicative with respect
to exact sequences
0—>Vi V
Use this and the fact that
algebraically closed
(why?)
a
V2
0, where Vi is
an
invariant
subspace.
admits nontrivial invariant subspaces since C is
to reduce to the case dimF
=
1, for which the identity is
elementary calculus exercise.)
With due care, the identity can be stated and proved (in the same way)
arbitrary fields of characteristic 0. It is an ingredient in the cohomological
interpretation of the 'Weil conjectures'.
an
6.13. Let A be
a square
matrix with integer entries. Prove that if A is
eigenvalue of A, then in fact A G Z.
a
over
rational
(Hint: Proposition V.5.5.)
6.14. > Let A be an eigenvalue of two similar transformations a, (3. Prove that the
geometric multiplicities of A with respect to a and (3 coincide. [§6.3]
a be a linear transformation on a free i?-module F, and let vi,..., vn
be eigenvectors corresponding to pairwise distinct eigenvalues Ai,...,An. Prove
that vi,..., vn are linearly independent. (Hint: If not, there is a shortest linear
6.15. > Let
VI. Linear
370
combination riv^
of
a on
0 with all rj G iJ, fj ^ 0.
+
+ rmVjm
this linear combination with the product by A^.)
=
It follows that if F has rank
a
spectral decomposition of
6.16.
-i
F.
n
and
a
elements
(viewing
v
G
V
column
as
Compare the
distinct eigenvalues, then
n
a
action
induces
[§6.3, VII.6.14]
The standard inner product
V
on
(v, w)
W
has
algebra
:=
=
Rn is the map 7x7->R defined
v*
by
w
vectors).
The standard hermitian product
on
Cn is the map l^x^^C defined by
=
(v, w)
:=
v*
w,
M, M^ stands for the matrix obtained by taking the complex
conjugates of the entries of the transpose Ml.
where for any matrix
These products satisfy evident linearity properties: for example, for A
G C and
v,w G W
(Av,w)
Prove32
that
on
(Vv,w
Likewise,
G
A(v,w).
JAn(M) belongs
matrix M G
a
the standard inner product
=
to
On(R)
if and only if it preserves
Rn:
Rn)
prove that a matrix M G
(Mv,Mw)
=
(v,w).
JAn(C) belongs to Un(C)
on Cn. [6.18]
if and only if it
preserves the standard hermitian product
We say that two vectors v, w of Rn or Cn are orthogonal if (v, w)
0.
v-1
v
w
v.
of
is
the
set
of
vectors
that
are
to
complement
orthogonal
orthogonal
6.17.
The
=
-i
Prove that if v
^
0 in V
=
Rn
or
Cn, then v-1
is
a
subspace of V of dimension
n
1.
[7.16, VIII.5.15]
6.18.
Let V
-i
Exercise 6.16.
=
Rn, endowed with the standard inner product defined
A set of distinct vectors vi,...,vr is orthonormal if
(v^,v^)
in
=
0 for
j. Geometrically, this means that each vector has length 1,
and different vectors are orthogonal. The same terminology may be used in Cn,
i
7^ j
and 1 for i
=
product.
M
if
Prove that
and only if the columns of M
G On(R)
only if the rows of M are orthonormal33.
w.r.t. the standard hermitian
Formulate and
prove an
analogous
statement for
called the unitary group. Note that, for
of
norm
6.19.
-i
n
=
are
orthonormal, if and
Un(C). (The
group
Un(C)
is
1, it consists of the complex numbers
1.) [6.19, 7.16, VIII.5.9]
Let vi,..., vr form
an
orthonormal set of vectors in Rn.
See Exercise II.6.1 for the definitions of On(K) and Un(C). We trust that basic facts on
products are not new to the reader. Among these, recall that for v, w Mn, (v, v) equals the
square of the length of v, and (v, w) equals the product of the lengths of v and w and the cosine
of the angle formed by v and w. Thus, the reader is essentially asked to verify that M
On(K) if
and only if M preserves lengths and angles, and to check an analogous statement in the complex
environment.
The group On(K) is called the orthogonal group; orthonormal group would probably be a
inner
more
appropriate terminology.
7. Canonical forms
371
Prove that vi,...,vr are linearly
basis of the space V they span.
Let
w
=
independent;
so
they form
+ arvr be a vector of V. Prove that ai
aivi +
More generally, prove that if
w
G
orthogonal projection wy of
orthogonal to all vectors of V.)
the
=
an
orthonormal
(v^, w).
Rn, then (v^,w) is the component of
(That is,
w onto V.
prove that
w
v^ in
wy
is
For reasons such as these, it is convenient to work with orthonormal bases. Note
that, by Exercise 6.18, a matrix is in On(R) if and only if its columns form an
orthonormal basis of Rn. Again, the reader should formulate parallel statements
for hermitian products in Cn.
[6.20]
6.20. The Gram-Schmidt process takes as input a choice of linearly independent
vectors vi,..., vr of Rn and returns vectors wi,..., wr spanning the same space V
0 for i ^ j. Scaling each w^ by its
spanned by v1,...,vf and such that (w^,w^)
length, this yields an orthonormal basis for V.
=
Here is how the Gram-Schmidt process works:
Wi
:= vi.
For k > 1, Wfc
Exercise
(Cf.
6.21.
-i
:= v/j.
v*. onto
(wi,..., Wfc_i).
Prove that this process accomplishes the stated goal.
6.19.)
A matrix M G
symmetric if and
M.
Xn(Rn) is symmetric if Ml
only if (Vv,w G Cn), (Mv,w)
(v,Mw).
A matrix M G
and
orthogonal projection of
only if (Vv,w
=
Prove that M is
=
Mn(Cn) is hermitian if M"1"
Cn), (Mv,w)
(v,Mw).
G
=
M. Prove that M is hermitian if
=
In both cases, one may say that M is self-adjoint; this means that shuttling it
one side of the product to the other does not change the result of the operation.
from
A hermitian matrix with real entries is symmetric. It is in fact useful to think
as particular cases of hermitian matrices. [6.22]
of real symmetric matrices
Prove that the eigenvalues of a hermitian matrix (Exercise 6.21) are real.
Also, prove that if v, w are eigenvectors of a hermitian matrix, corresponding to
different eigenvalues, then (v,w)
0. (Thus, eigenvectors with distinct eigenvalues
6.22.
-i
=
for
a
real symmetric matrix
7.
Canonical forms
are
orthogonal.) [7.20]
7.1. Linear transformations of free modules; actions of polynomial rings.
use Theorem 5.6 to better understand the similarity relation
We have promised to
VI. Linear
372
and
algebra
special forms for square matrices with entries in a field. We are now
ready
good on our promise.
The questions the reader should be anxiously asking are, where is the PID?
Where is the finitely generated module? Why would a classification theorem for
to obtain
to make
these things have anything to tell
us
about matrices?
Once these questions
are
answered, everything else will follow easily. In a sense, this is the one issue that
we keep in our mind concerning all these considerations: once we remember how to
get an interesting module out of a linear transformation of vector spaces, the rest
is essentially an immediate consequence of standard notions.
Once more, while the main application will be to vector spaces and matrices
field, we do not need to preoccupy ourselves with specializing the situation
over a
before
our
we
get to the key point. Therefore, let's keep working for
a
while longer
on
given integral domain34 R.
Claim
giving
7.1.
an
Giving
linear
a
R[t]-module
transformation
structure on
on a
free R-module
F, compatible with
F is the
same as
its R-module structure.
Here R[t] is the polynomial ring in one indeterminate t over the ring R. The
claim is much less impressive than it sounds; in fact, it is completely tautological.
Giving
an
compatible with the i?-module
of
homomorphism
i?-algebras
i?[£]-module
the same as
giving
a
structure on F
(f
R[t]
:
structure is
Endfl(F)
-?
(this is an insignificant upgrade of the very definition of module from §111.5.1).
By the universal property satisfied by polynomial rings (§111.2.2, especially
Example III.2.3), giving ip as an extension of the basic i?-algebra structure of End#(F) is
just the same as specifying the image <p(t), that is, choosing an element of End#(F).
This is precisely what the claim says.
use of a certain language may obfuscate the matter,
simple. Given a linear transformation a of a free module F, we can
define the action of a polynomial
Our propensity for the
which is very
/(*)
on
F
as
follows: for every
/(*)(v)
rmtm
=
v e
:=
F,
+
rwf"-1
+
^.^"-'(v)
where
ak denotes the /c-fold composition of
this in
§6.2. Conversely,
R[t]
F, that is, 'multiplication by V:
a :
F
+ r0
R[t]
set
rmam(v)
an
+
action of
a
+
+ r0v,
with itself; we have already run into
F determines in particular a map
on
£v;
v i—?
the compatibility with the i?-module structure
linear transformation of F.
Again,
on
F tells
us
precisely that
a
is
a
this is the content of Claim 7.1.
Tautological as it is, Claim 7.1 packs a good amount
the R[t] module knows everything about similarity:
of information.
Indeed,
In fact, the requirements that R be an integral domain and that F be free are immaterial
here; cf. Exercise 111.5.11.
7. Canonical forms
373
Lemma 7.2. Let a, (3 be linear
corresponding R[t]-module
are
transformations of a free R-module F. Then the
isomorphic if and only if a and (3
structures on F are
similar.
Proof. Denote by Fa,
Claim 7.1.
Assume first that
transformation
a
F
n :
Fp
the two
that is,
7T o a
(3
o 7r.
We
view
can
7T O a O
=
tt as an
Fa
an
by
(3
a,
as
per
invertible i?-linear
7T_1;
i?-linear map
Fp-
—>
i?[£]-linear: indeed, multiplication by
We claim that it is
F
on
and (3 are similar. Then there exists
F such that
(3
=
defined
i?[£]-modules
t is a in Fa and
in
Fp,
F#
are
/?
so
7r(£v) 7r o a(v) (3 7r(v) ^7r(v).
invertible i?[£]-linear map Fa
F#, proving
i?[£]-modules.
=
Thus
is
7r
isomorphic
The
reverse
=
—>
an
as
converse
implication
is obtained
and is left to the reader
Corollary
o
=
7.3. There is
that Fa and
essentially by running this argument
(Exercise 7.1).
correspondence between the similarity classes
F and the isomorphism classes of
R-module
free
a one-to-one
of R-linear transformations of
R[t]-module structures on F.
a
Of course the same statement holds for square matrices with entries in
the finite rank
in
i?,
in
case.
Corollary 7.3 is the tool we were looking for, bridging between classifying
similarity classes of linear transformations or matrices and classifying modules.
task now becomes that of translating the notions we have encountered in §6
the module language and seeing if this teaches us anything new.
The
into
In any case, since we have classified finitely generated modules over PIDs,
we will be able to classify similarity classes of transformations of
it is clear that
finite-rank free modules—provided
that R[t] is a PID. This is where the additional
that
be
a
field
enters
the discussion (cf. Exercise V.2.12).
R
hypothesis
7.2.
k[^-modules
simply
a
and the rational canonical form. It is
case
in which R
=
k is
a
field
giving V
time to
F
so
=
V is
=
By the preceding discussion, choosing
as
finally
and F has finite rank;
finite-dimensional vector space. Let n
dim V.
specialize to the
a
k[^-module
structure
a
linear transformation of V is the
(compatible
with its vector space
same
structure);
similar linear transformations correspond to isomorphic k[^-modules. Then V is
a finitely generated module over
k[t], and k[t] is a PID since k is a field
(Exercise III.4.4 and
§V.2); therefore,
we are
precisely
in the situation covered
by the
classification theorem.
Even before
spelling
out the result of
applying Theorem 5.6,
of its slogan version: every finitely generated module
over a
we can
PID is
a
make
direct
use
sum
VI. Linear
374
of cyclic modules. This tells
that
us
algebra
understand all linear transformations
we can
of finite-dimensional vector spaces, up to similarity, if we understand the linear
transformation given by multiplication by t on a cyclic k[^-module
k[t]
V
where f(t)
is
a nonconstant
monic
/(0
It is worthwhile
obtaining
polynomial:
*n +
=
+
---
+ ro.
matrix representation of this poster
good
a
rn-i*n-1
case.
We
choose the basis
1,
t,
tn~l
-..,
of V (cf. Proposition III.4.6). Recall that the columns of the matrix corresponding
to a transformation consist of the images of the chosen basis (cf. the comments
preceding Corollary 2.2). Since multiplication by t on V acts as
1 ^t,
t^t2,
n-l
>tn
n-l
=
-rn_i*'
r0,
the matrix corresponding to this linear transformation is
/0
0
0
0
-r0
1
0
0
o
-n
0
1
0
0
-r2
0
-rn-2
-rn-i/
Definition 7.4. This is called the companion matrix of the polynomial
denoted Cf(t).
/(£),
j
Theorem 5.6 tells
us (among other things) that every linear transformation
matrix representation into blocks, each of which is the companion matrix
polynomial. Here is the statement of Theorem 5.6 in this context:
admits
of
a
a
field, and let V be a finite-dimensional vector space. Let
transformation on V, and endow V with the corresponding k[t]-module
Theorem 7.5. Let k be
a be a linear
structure,
as in
a
Claim 7.1. Then the following hold:
There exist distinct monic irreducible polynomials
positive integers r^ such that
^
=
0
m
as
k[t]-modules.
k[t\
fait)'**)
pi(t),... ,ps(t)
G
k[t]
and
7. Canonical forms
375
There exist monic nonconstant
fi(t)
fm(t)
V
as
=
k[t]
(AW)
e
k[t]
such that
k[t]
Um{t))
k[t]-modules.
Via these isomorphisms, the action
the
polynomials fi(t),...,fm(t)
and
of a
onV corresponds to multiplication by t.
Further, two linear transformations a, (3 are similar if and only if they have
collections of invariants Pi(t)Tij ('elementary divisors7), fi(t) ('invariant
same
factors').
Proof. Since dim V is finite, V is finitely generated as a /c-module and a fortiori as
a
/c[£]-module. The two isomorphisms are then obtained by applying Theorem 5.6.
All the relevant polynomials may be chosen to be monic since every polynomial
field is the associate of a (unique) monic polynomial (Exercise III.4.7). The
over a
fact that the action of a corresponds to multiplication by t is precisely what
defines the corresponding k[^-module structure on V. The statement about similar
transformations follows from Corollary 7.3.
the question raised in §6.1: we now have
list of invariants which describe completely the similarity class of a linear
Theorem 7.5
answers
(over fields)
transformation. Further, these invariants provide
with
a
special
Definition 7.6. The rational canonical form of
a
linear transformation
of
a
us
a
matrix representation
given linear transformation.
vector space
a
of
a
V is the block matrix
/ ch(t)
where
fi(t),
...,
fm(t)
are
cfm(t)
J
the invariant factors of
a.
The rational canonical form of
a square
matrix is
j
(of course)
canonical form of the corresponding linear transformation. The
the rational
following
statement is an
immediate consequence of Theorem 7.5.
Corollary 7.7. Every linear transformation admits a rational canonical form. Two
linear transformations have the same rational canonical form if and only if they are
similar.
Remark 7.8. The 'rational' in rational canonical form has nothing to do with
Q; it is meant to remind the reader that this form can be found without leaving
the base field.
The other canonical form
we
will encounter will have entries in
possibly larger field, where the characteristic polynomial of the transformation
factors completely.
j
a
VI. Linear
376
algebra
explicitly another wish expressed at the end of §6.1, that
distinguished representative in each similarity class of matrices/linear
transformations. The rational canonical form is such a representative.
Corollary
is, to find
The
7.7 fulfills
one
next obvious
question is how the invariants we just found relate
produce invariants of similar transformations in §6.
to our
more naive attempts to
Proposition 7.9. Let fi(t) |
transformation a on a vector space
| fm(t)
be the invariant
factors of
a
linear
V. Then the minimal polynomial ma(t) equals
and the characteristic polynomial Pa(t) equals the product fi(t)
fm(t).
fm(t),
-
Proof. Tracing definitions, (ma(t)) is the annihilator ideal of V when this is viewed
as a k[^-module via a (as in Claim 7.1). Therefore the equality of ma(t) and fm(t)
is
a restatement
of Lemma 5.7.
Concerning the characteristic polynomial, by Exercise 6.10 it suffices to prove
the statement in the cyclic case: that is, it suffices to prove that if f(t) is a monic
polynomial, then f(t) equals the characteristic polynomial of the companion
C/(t). Explicitly,
be
a
monic
polynomial; then the task
amounts to
showing that
0
0
0
r0
-1
t
0
0
n
0
-1
t
0
Tl
(t
det
This
can
be done by
Corollary
7.10
matrix
let
fit)0
0
0
t
\o
0
0
-1
an easy
rn-2
t + rn.
l/
induction and is left to the reader
(Cayley-Hamilton).
The minimal polynomial
(Exercise 7.2).
of a
linear
transformation divides its characteristic polynomial.
Proof. This has
Finding
now
become evident,
as
promised
Claim 7.1. As
k[t]
(cf.
a
Euclidean domain, we know that this can be done by
§2.4). In practice, the process consists of compiling the
is in fact
Gaussian elimination
§6.2.
given matrix A amounts to working
specific k[^-module corresponding to A as in
the rational canonical form of
out the classification theorem for the
in
a
information obtained while diagonalizing tl A over k[t] by elementary operations.
The reader is invited to either produce an original algorithm or at least look it up
in
a
standard reference to
see
what is involved.
Major shortcuts may come to our aid. For example, if the minimal and
characteristic polynomials coincide, then one knows a priori that the corresponding
module is cyclic (cf. Remark 5.8), and it follows that the rational canonical form is
simply the companion matrix of the characteristic polynomial.
7. Canonical forms
377
In any case, the strength of a canonical form rests in the fact that it allows us
reduce general facts to specific, standard cases. For example, the following
statement is somewhat mysterious if all one knows are the definitions, but it becomes
to
essentially trivial with
Proposition
a
pinch of rational canonical forms:
7.11. Let A G
be
Mn(k)
a square matrix.
Then A is similar to its
transpose.
Proof. If B is similar to A and
we can prove
that B is similar to its transpose
then A is similar to its transpose Atm. because B
At
Therefore,
=
=
PAP-1,
Bl
=
£?*,
QBQ~l give
(PtQP)A(PtQP)-\
it suffices to prove the statement for matrices in rational canonical form.
Further, to prove the statement for a block matrix, it clearly suffices to prove it
for each block; so we may assume that A is the companion matrix C of a polynomial
f(t). Since the characteristic and minimal polynomials of the transpose Cl coincide
with those of C (Exercise 6.8), they are both equal to f(t). It follows that the
rational canonical form of Cl is again the companion matrix to
and C are similar, as needed.
/(£);
therefore Cl
7.3. Jordan canonical form. We have derived the rational canonical form from
the invariant factors of the module corresponding to a linear transformation. A
useful alternative can be obtained from the elementary divisors, at least in the case
in which the characteristic polynomial factors completely over the field k. If this is
not the case, one can enlarge k so as to include all the roots of the characteristic
polynomial: the reader proved this in Exercise V.5.13. The price to pay will be that
of a linear transformation a G Endk(V) may be a matrix
larger than k. In any case, whether two transformations are
independent of the base field (Exercise 7.4), so this does not affect
the Jordan canonical form
with entries in a field
similar
or not
is
the issue at hand.
a G Endfc(V), obtain the elementary
k[^-module, as in Theorem 7.5:
Given
corresponding
divisor decomposition of the
k[t\
(P«(*)ry)'
V*©
It is
an immediate consequence of Proposition 7.9 and the equivalence between
the elementary divisor and invariant factor formulations that the characteristic
polynomial Pa(t) equals the product
rN*)r
Lemma 7.12. Assume that the characteristic
polynomial Pa(t) factors completely;
that is,
s
Pa(t)
=
l[(t
-
Xi)™*
2=1
where \i,i
=
1,...
,s, are the distinct
multiplicities; cf. $6.3). Then Pi(t)
=
eigenvalues of a (and
(t
\i), and
rrii
=
mi are their
V r^.
algebraic
VI. Linear
378
algebra
In this situation, the minimal polynomial of a equals
s
rna(t)
=
Y[(t-\i)m*Xj{rij}.
2=1
Proof. The first statement follows from uniqueness of factorizations. The
statement about the minimal polynomial is immediate from Proposition 7.9 and the
bookkeeping giving the equivalence of the two formulations in Theorem 7.5.
The elementary divisor decomposition splits V into a different collection of
cyclic modules than the decomposition in invariant factors: the basic cyclic bricks
are now of the form
k[t]/(p(t)r) for a monic prime p{t); by Lemma 7.12, assuming
that the characteristic polynomial factors completely
over
k, they
in fact of the
are
form
k[t]
((X-X)r)
for
some
A G k
(which equals
eigenvalue of a) and
an
r
>
0.
Just
as we
did for
'companion matrices', we now look for a basis with respect to which a (=
multiplication by t; keep in mind the fine print in Theorem 7.5) has a particularly
matrix representation. This time we choose the basis
(t-\y-\(t-xy-2,
Multiplication by
f (t
(t
-
-
t on V acts
xy-1 ^t(t
xy-2 ^t(t
i>->t
Therefore, with respect
=
-
-
(in
the first line,
xy-1
xy-2
=
=
(t-\)+
x(t
-
(«-a)°
-..,
the fact that
use
xy-1
(t- xy-1
=
+
+
(t- xy
x(t
-
=
simple
i.
(t X)r
x(t
-
=
0 in
V)
as
to
A,
xy-\
xy-2,
\.
to this basis the linear transformation has matrix
A
1
0
0
0\
0
0
0
A
1
\0
0
0
0
A/
Definition 7.13. This matrix is the Jordan block of size
r
corresponding
denoted J\tr.
We
can
j
put several blocks together for
a
given A: for (r^)
=
(ri,... ,rg),
let
/ JA,ri
yA,(rj)
V
we get new distinguished representatives of the
given linear transformation:
With this notation in hand,
similarity class of
a
J*,re J
7. Canonical forms
379
Definition 7.14. The Jordan canonical
vector space
form of
linear transformation
a
a
of
a
V is the block matrix
(j.Ai,(n7)
|
where
(t
\i)Tij are
| JAa,(rai) /
the elementary divisors of
Lemma
(cf.
a
7.12).
j
This canonical form is 'a little less canonical' than the rational canonical form,
sense that while the rational canonical form is really unique, some ambiguity
in the
is left in the Jordan canonical form:
there is
way to choose the order
special
no
of the different blocks any more than there is a way to
order35 the eigenvalues
Ai,..., As.
Therefore, the analog of Corollary 7.7 should state that two linear
only if they have the same Jordan canonical form up to a
reordering of the blocks. As we have seen, every linear transformation a G Endk(V)
admits a Jordan canonical form if Pa(t) factors completely over h; for example, we
can always find a Jordan canonical form with entries in the algebraic closure of k.
transformations are similar if and
Example 7.15. One use of the Jordan canonical form is the enumeration of all
possible similarity classes of transformations with given eigenvalues. For example,
are 5 similarity classes of linear transformations with a single eigenvalue A
with algebraic multiplicity 4, over a 4-dimensional vector space: indeed, there are 5
different ways to stack together Jordan blocks corresponding to the same eigenvalue,
within a 4 x 4 square matrix:
there
/
A
0
0
0
/A
1
0
°
0
A
0
0
0
A
0
0
0
0
0
0
0
0
A
0
v°
0
0
x
J
^
/A
1
0
0
0
A
0
0
A
0
0
0
A
1
0
A/
0
0
0
A /
(X
1
0
(X
1
0
0
0
A
0
0
A
1
0
0
0
0
0
A
1
o
o
0
0
0
A /
0
o
A
/
algebraic and
eigenvalue. The algebraic multiplicity of A as an
The Jordan canonical form clarifies the difference between
geometric
multiplicities of
an
is (of course) the number of times A appears in
Recall that the geometric multiplicity of A equals
the dimension of the eigenspace of A.
eigenvalue of
a
linear transformation
the Jordan canonical form of
Proposition 7.16.
a
a.
The geometric multiplicity of A
the number of Jordan blocks corresponding
Of course if the
ground field is Q or M, then we can
an option on fields like C.
order. However, this is not
as
an
eigenvalue of a equals
form of a.
to A in the Jordan canonical
order
the A^'s in, for example, increasing
VI. Linear algebra
380
Proof. As the geometric multiplicity is clearly additive in direct sums, it suffices
to show that the geometric multiplicity of A for the transformation corresponding
to a
single Jordan block
J
/A
1
0
0\
0
A
0
0
0
0
A
1
\0
0
0
A/
=
is 1.
AA
Let
be
v
an
eigenvector corresponding
to A. Then
W
(Vl\
v2
/Xvi
A^2
(vA
v2
=
J
v2\
+ ^3
=
\vr)
\vr)
+
A^r
/
yielding
V2
That is, the eigenspace of A is
=
=
Vr
=
0.
generated by
A\
o
W
and has dimension 1,
as
needed.
7.4. Diagonalizability. A linear transformation of
izable if it can be represented by a diagonal matrix.
a vector space
V is diagonal-
The question of whether a given linear transformation is or is not diagonalizable
is interesting and important: as we pointed out already in §6.3, when a G End/c(F)
is diagonalizable, then the vector space V admits a corresponding spectral
decomposition:
more
a
is
diagonalizable if and only if V admits
could be said
on
this important theme, but
a
basis of eigenvectors of
we are
giving useful criteria for diagonalizability.
For example, the diagonal matrix representing
a
already
will
a.
Much
in the position of
necessarily be
a
Jordan
canonical form for a, consisting of blocks of size 1. Proposition 7.16 tells us that
this can be detected by the difference between algebraic and geometric multiplicity,
so we
get the following characterization of diagonalizable transformations:
Corollary 7.17. Assume the characteristic polynomial of
completely
over
k.
Then
algebraic multiplicities of all
Another
view of the
a
its
a
G
Endk(V) factors
diagonalizable if and only if the geometric and
eigenvalues coincide.
is
same
result is the
following:
Exercises
381
Proposition 7.18. Assume the characteristic
completely over k. Then a is diagonalizable if
of a
has
no
multiple
polynomial of a G Endk(V) factors
and only if the minimal polynomial
roots.
equivalent to having all Jordan blocks of size 1
Therefore, if the characteristic polynomial of a
factors completely, then a is diagonalizable if and only if all exponents r^ appearing
in Theorem 7.5 equal 1. By the second part of Lemma 7.12, this is equivalent to
the requirement that the minimal polynomial of a have no multiple roots.
Again, diagonalizability
is
in the Jordan canonical form of
a.
Proof.
is
a
diagonalizable
in these statements is necessary in order to
a matrix
factorizability
The condition of
guarantee that
Jordan canonical form exists. Not surprisingly, whether
or not depends on the ground field. For example,
is not diagonalizable
over
R
(cf. Example 6.15),
while it is diagonalizable
Endowing the vector space V with additional structure
important classes of operators, for which
product) singles
out
The reader
will get
(such
as an
over
C.
inner
obtain precise and
delicate results on the existence of spectral decompositions. For example, this can
be used to prove that real symmetric matrices are diagonalizable (Exercise 7.20.)
a taste
one can
of these 'spectral theorems' in the exercises.
Exercises
As above, k denotes
7.1. >
a
field.
Complete the proof of Lemma
7.2. > Prove that the characteristic
7.2.
[§7.1]
polynomial of the companion
matrix of
a
monic
polynomial f(t) equals f(t). [§7.2]
7.3. > Prove that two linear transformations of
similar if and only if they have the
Is this true in dimension 4? [§6.2]
same
a vector space
of dimension < 3
are
characteristic and minimal polynomials.
a field, and let K be a field containing k.
Two square matrices
B
in
be
viewed
as
matrices
with
entries
the
G
A,
larger field K. Prove
Mn(k) may
that A and B are similar over k if and only if they are similar over K. [§7.3]
7.4. > Let k be
7.5. Find the rational canonical form of
/Ai
0
I
are
A2
.
V0
assuming that the Aj
0
all distinct.
a
0\
0
'
0
matrix
diagonal
I
Xr)
'
VI. Linear
382
algebra
7.6. Let
/ 6
A
-10
-10
3
-5
-6
\-l
2
3
=
Compute the characteristic polynomial of A.
Find the minimal polynomial of A (use the Cayley-Hamilton theorem!).
Find the invariant factors of A.
Find the rational canonical form of A.
7.7. Let V be a k-vector space of dimension n, and let a G Endfc(V). Prove that
the minimal and characteristic polynomials of a coincide if and only if there is a
vector v G V such that
v,
is
a
q(v),
-..,
q""1^)
basis of V.
7.8. Let V be a k-vector space of dimension n, and let a G Endfc(V). Prove that the
characteristic polynomial Pa(t) divides a power of the minimal polynomial ma(t).
similarity classes of linear transformations
7.9. What is the number of distinct
n-dimensional vector space with
multiplicity n?
an
one
on
fixed eigenvalue A with algebraic
7.10. Classify all square matrices A G Mn(k) such that A2
action of such matrices 'geometrically'.
=
A,
up to
similarity.
Describe the
7.11. A square matrix A G
some
Mn(k)
is
nilpotent (cf. Exercise V.4.19) if Ak
0 for
integer k.
Characterize nilpotent matrices
Prove that if Ak
than
n
(=
=
0 for
7.12.
-i
some
the size of the
Prove that the trace of
Let V be
a
a
in terms of their Jordan canonical form.
integer /c, then Ak
=
that
a
induces
a
0 for
integer k
some
no
larger
matrix).
nilpotent
matrix is 0.
finite-dimensional k-vector space, and let
diagonalizable linear transformation. Assume that W
so
=
linear transformation
a G
Endk(V)
C V is an invariant
a\w G Endfc(W).
Prove that
be
a
subspace,
a\w is
also
diagonalizable. (Use Proposition 7.18.) [7.14]
7.13.
Let R be an integral domain. Assume that A G M.n(R)
with distinct eigenvalues. Let B G Mn(R) be such that AB
-i
=
is
diagonalizable,
BA.
Prove that
B is also diagonalizable, and in fact it is diagonal w.r.t. a basis of eigenvectors
of A. (If P is such that PAP'1 is diagonal, note that PAP'1 and PBP~l also
commute.) [7.14]
7.14. Prove that
'commuting transformations
may be
simultaneously diagonalized',
finite-dimensional vector space, and let a, (3 G
End/c(F) be diagonalizable transformations. Assume that a(3 Pa. Prove that V
has a basis consisting of eigenvectors of both a and (3. (Argue as in Exercise 7.13
in the
following
sense.
Let V be
a
=
to reduce to the
case
in which V is
an
eigenspace for
a; then use Exercise
7.12.)
Exercises
383
7.15. A complete flag of subspaces of a
V of dimension
vector space
n
is
a sequence
of nested subspaces
0
with dim V{
=
i. In other
V0
=
C
words,
V1
a
C
C
Fn_!
complete flag
C
is
a
Fn
F
=
composition series in the
sense
of Exercise 1.16.
Let V be
a
finite-dimensional vector space over an algebraically closed field.
transformation a of V preserves a complete flag: that is,
Prove that every linear
there is a complete flag
Find
a
as
a(Vi)
C
Vi.
linear transformation of R2 that does not preserve
(Schur form) Let
Un(C) such that
7.16.
P G
above and such that
A G
£
Mn(C)
0
A
=
Prove that there exists
.
*
*
*\
A2
*
*
/Ai
a
complete flag.
a
unitary matrix
P-\
P
0
0
\0
0
An-i
0
*
\n)
So not only is A similar to an upper triangular matrix (this is already guaranteed
by the Jordan canonical form), but a matrix of change of basis P 'triangularizing'
the matrix A can in fact be chosen to be in Un(C). (Argue inductively on n. Since
C is algebraically closed, A has at least
(vi> vi)
=
and keep
one
eigenvector vi, and we may assume
complement v^-; cf. Exercise 6.17,
1- For the induction step, consider the
in mind Exercise
6.18.)
The upper triangular matrix is
a
Schur form for A. It is
7.17.
A matrix M G Mn(C) is normal if MM^
matrices and hermitian matrices are both normal.
-i
Prove that triangular normal matrices
7.18.
-i
are
=
(of course)
not
unique.
M^M. Note that unitary
diagonal. [7.18]
Prove the spectral theorem for normal operators: if M is a normal matrix,
an orthonormal basis of eigenvectors of M.
Therefore, normal
then there exists
operators
are
diagonalizable; they determine a spectral decomposition of the vector
they act. (Consider the Schur form of M; cf. Exercise 7.16. Use
space on which
Exercise
7.17.) [7.19, 7.20]
7.19. Prove that
a
matrix M G
orthonormal basis of eigenvectors.
M.n(C) is normal if and only if it
(Exercise 7.18 gives one direction;
admits
an
prove the
converse.)
7.20. > Prove that real symmetric matrices are diagonalizable and admit an
orthonormal basis of eigenvectors. (This is an immediate consequence of Exercise 7.18
and Exercise 6.22.) [§7.4]
Chapter
VII
Fields
The minuscule amount of algebra we have developed so far allows us to scratch the
surface of the unfathomable subject of field theory. Fields are of basic importance
in many subjects—number
theory and algebraic geometry
come immediately to
surprising that their study has been developed to a particularly
high level of sophistication. While our profoundly limited knowledge will prevent us
from even hinting at this sophistication, even a cursory overview of the basic notions
mind—and
it is not
will allow
us to
deal with remarkable applications, such
We will also get
as
the famous problems of
glimpse of the beautiful
of
an
of
the
Galois
theory of field extensions, the
subject
theory,
amazing interplay
of
and
solvability
polynomials,
group theory.
constructibility of geometric figures.
a
1. Field extensions, I
We first deal with several basic notions in field theory, mostly inspired by easy linear
algebra considerations. The keywords here are finite, simple, finitely generated,
algebraic.
1.1.
study of fields is (of course) the study of the category
III. 1.14), with ring homomorphisms as morphisms. The
is to understand what fields are and above all what they are in relation to one
Basic definitions. The
Fid of fields
task
(Definition
another.
The place to start is a reminder from elementary ring theory1: every ring
homomorphisms from a field to a nonzero ring is injective. Indeed, the kernel of
a ring homomorphism to a nonzero ring is a proper ideal (because a homomorphism maps 1 to 1 by definition and 1 ^ 0 in nonzero rings), and the only proper
ideal in a field is (0). In particular, every ring homomorphism of fields is injective
(fields
rings by definition!);
(cf. Proposition III.2.4).
are nonzero
The last time
we made
this point
was
every
morphism
in Fid is
a
monomorphism
back in §V.5.2.
385
VII. Fields
386
K between two fields identifies the first with a
Thus, every morphism k
subfield of the second. In other words, one can view K as a particular way to
enlarge /c, that is, as an extension of k. Field theory is first and foremost the
study of field extensions. We will denote a field extension by k C K (which is less
than optimal, because there may be many ways to embed a field into another);
other popular choices are K/k (which we do not like, as it suggests a quotienting
operation) and
K
k
(which
will avoid most of the time,
we
as
it is hard to
typeset).
field k is its characteristic; cf. Exercise V.4.17. We
have a unique ring homomorphisms i : Z
k (Z is initial in Ring: 1 must go to 1
by definition of homomorphism, and this fixes the value of i(n) for all n G Z); the
The coarsest invariant of
a
characteristic of /c, char/c, is defined to be the nonnegative generator of the ideal
ker r, that is, char k
0 if i is injective, and char k
p > 0 if ker i
(p) ^ (0).
=
=
=
Put otherwise, either n 1 is only 0 in k for n
0, in which case char/c
1
n
0 for some n ^ 0; char k
p > 0 is then the smallest positive integer
=
or
=
-
=
=
0,
for
0 in k. Since fields are integral domains, the image i(Z) must be an
integral domain; hence ker i must be a prime ideal. Therefore, the characteristic of
a field is either 0 or a prime number.
If k C K is an extension, then char/c
char if (Exercise 1.1). Thus, we could
define and study categories Fldo, Fldp of fields of a given characteristic, without
losing any information: these categories live within Fid, without interacting with
each other2. Further, each of these categories has an initial object: Q is initial
which
p
-
1
=
=
in Fldo, and3 ¥p := Z/pZ in Fldp; the reader has checked this easy fact in
Exercise V.4.17, and now is an excellent time to contemplate it again. In each case,
the initial object is called the prime subfield of the given field. Thus, every field is
in
a
canonical way an extension of its prime subfield: studying fields really means
extensions. To a large extent, the 'small' field k will be fixed in what
studying field
follows, and our main
with the evident (and
an
object of study
by
now
will be the category Fid/- of extensions of /c,
reader) notions of morphisms.
familiar to the
The first general remark concerning field extensions is that the larger field
algebra, and hence a vector space over the smaller one (by the very definition
algebra; cf. Example
III.5.6).
Definition 1.1. A field extension k C F is
dimension dim F
=
n as a vector space over
finite, of degree
n, if F has
k. The extension is
The degree of a finite extension k C F is denoted by
oo if the extension is infinite).
[F : k]
[F
infinite
:
(finite)
otherwise.
k] (and
we
write
=
They do interact in other ways, however—for
example, because Z is initial in Ring,
maps to all
3We
(rather
is
of
j
so
Z
fields.
are
going
than 'just'
to denote the field
as a
group).
Z/pZ by ¥p
in this chapter, to stress its role as a field
1. Field extensions, I
387
§V.5.2 we have encountered a prototypical example of finite field extension:
procedure starting from an irreducible polynomial f(x) G k[x] with coefficients
in a field and producing an extension K of k in which f(t) has a root. Explicitly,
In
a
is such
an
extension
The quotient is
(Proposition V.5.7).
a
field because
k[t]
is
because of the
PID; hence irreducible elements generate maximal ideals—prime
UFD property (cf. Theorem V.2.5) and then maximal because nonzero prime ideals
a
are
maximal in
a
(cf. Proposition III.4.13). The coset of t is a root of f(x)
an element of
K[x\. The degree of the extension k C K
the polynomial f(x) (this is hopefully clear at this point and
PID
when this is viewed
as
equals the degree of
was checked carefully in
The reader
Proposition III.4.6).
perhaps all finite extensions are of this type.
This is unfortunately not the case, but as it happens, it is the case in a large
class of examples. As we often do, we promise that the situation will clarify itself
considerably once we have accumulated more general knowledge; in this case, the
may wonder whether
relevant fact will be
Also recall that
Proposition 5.19.
we proved that these
extensions
are
almost universal with
respect to the problem of extending k so that the polynomial f(t) acquires a root.
We pointed out, however, that the uni in universal is missing; that is, if k C F is
an
extension of k in which
f(t)
has
a
root, there may be many different ways to
put K in between:
kCKCF.
Such questions
central to the study of extensions,
look at this situation.
are
so we
begin by giving
1.2. Simple extensions. Let k C F be a field extension, and let a
smallest subfield of F containing both k and a is denoted k(a); that is,
intersection of all subfields of F containing k and
a
G
second
F.
k(a)
The
is the
a.
Definition 1.2. A field extension k C F is simple if there exists
such that F
k(a).
an
element
a G
=
The extensions recalled above
the coset of t, then K
k(a):
then it must contain (the coset
=
are
of this kind: if K
=
a
j
k[t]/(f(t))
indeed,
of) every polynomial expression
if
F
and
a
denotes
subfield of K contains the coset of t,
in t, and hence it
must be the whole of K.
We have
always found the
notation
suggests that all such extensions are in some
to the field k(t) of rational functions in
k(a)
somewhat unfortunate, since it
way
isomorphic and possibly all isomorphic
one
indeterminate t
(cf. Example V.4.12).
This is not true, although it is clear that every element of k(a) may be written as
a rational function in a with coefficients in k (Exercise 1.3). In any case, it is easy
to classify simple extensions: they are either isomorphic to k(t) or they are of the
prototypical kind recalled above. Here is the precise statement.
Proposition 1.3. Let k C
map
e :
k[t]
k(a) be
k(a), defined by f(t)
a
i->
simple
/(a).
extension.
Then
we
Consider the evaluation
have the
following:
VII. Fields
388
if and only if k C k(a) is an infinite extension.
isomorphic to the field of rational functions k(t).
is injective
e
k(a)
is
and only
e
is not injective
a
unique monic irreducible
[k(a) k]
:
if
if
k C
In this case,
finite. In this case there exists
polynomial p(t) G k[t] of degree n
k(a)
nonconstant
is
=
such that
k(a)
M-
-
Via this isomorphism,
the monic
a corresponds to the coset oft.
The polynomial p(t)
0 in k(a).
polynomial of smallest degree in k[t] such that p(a)
p(t) appearing
The polynomial
minimal
Of
polynomial of
in the first part of the statement is called the
k.
a over
the minimal polynomial of an element a of a ('large') field depends
2
For example, y/2 G C has minimal polynomial t2
course
('small') field k.
Q, but t y/2 over R.
the base
on
is
=
over
-
Proof. Let F
=
k(a). By the
'first isomorphism theorem', the image of e
:
F
k[t]
isomorphic to k[t]/ker(e). Since F is an integral domain, so is k[t]/ker(e); hence
ker(e) is a prime ideal in k[i\.
—Assume
0; that is, e is an injective map from the integral domain
ker(e)
F.
to
the
field
the
universal property of fields of fractions (cf. §V.4.2), e
By
k[t]
is
=
extends to
a
unique homomorphism
k(t)
The
(isomorphic) image
of
k(t)
in F is
F.
-?
field containing k and a; hence it equals F
a
by definition of simple extension.
Since
are
(that is, the images
the powers 1, t, t2,
therefore the extension k C F is infinite in this
is injective, the powers a0
all distinct and linearly independent
e
l,a, a2, a3,
=
linearly independent
over
/c);
over
k
...
(because
e(t1))
...
are
case.
(p(t)) for a unique monic irreducible nonconstant
polynomial p(t), which has smallest degree (cf. Exercise III.4.4!) among all nonzero
polynomials in ker(e). As (p(t)) is then maximal in k[t], the image of e is a subfield
—If
ker(e) ^
0,
then
ker(e)
=
of F containing a
e(t). By definition of simple extension, F
that is, the induced homomorphism
=
=
the image of e;
(p(t))
is
an
isomorphism.
In this
case
particular the extension is finite,
a
[F
as
:
k]
=
degp(£),
as
recalled in
§1.1,
and in
claimed.
The alert reader will have noticed that the proof of Proposition 1.3 is essentially
rehash of the argument proving the 'versality' part of Proposition V.5.7.
Example 1.4. Consider the extension Q C R.
1. Field extensions, I
The
389
polynomial x2 2 G Q[x] has roots in R: therefore,
a homomorphism (hence a field extension)
by Proposition V.5.7
there exists
{t2
'
2)
-
(the coset of) t is a root a of x2 2. Proposition 1.3 simply
identifies the image of this homomorphism with Q(a) C R.
such that the image of
This is hopefully crystal clear; however, note that even this simple example
shows that the induced morphism e is not unique (hence the 'lack of uni'): because
2 in R. Concretely, there are two possible
there are more than one root of x2
choices for a: a
and
a
The choice of a determines the evaluation
+V2
—y/2.
-
=
map
e
=
2)
Q[t]/(t2
and therefore the specific realization of
as a
subfleld of R.
The reader may be misled by one feature of this example: clearly Q(a/2)
and this may seem to compensate for the lack of uniqueness lamented
Q(—^/2),
=
above. The morphism may not be unique, but the image of the morphism surely
is? No. The reader will check that there are three distinct subflelds of C isomorphic
to
Q[t]/(t3
Ay,
2) (Exercise 1.5).
-
there's the rub. One of
our
main
goals
in this
chapter will be
to
single
out
a condition on an extension k C F that guarantees that no matter how we embed F
in a
larger extension, the images of these (possibly
many
different) embeddings
will
all coincide. Up to further technicalities, this is what makes an extension Galois.
Q C Q(\/2) will be a Galois extension, while Q C Q(\/2) will not be a Galois
Thus,
extension. But Galois extensions will have to wait until §6, and the reader
this issue aside for
One way to
can
put
now.
recover
data. This leads to the
j
uniqueness is
following
incorporate the choice of the
to
root in the
(uni)versality statement, adapted
refinement of the
for future applications.
Proposition 1.5. Let i
extensions.
Letpi(t)
resp., ol<2,- Let i :
k\
G
:
k\ C F\
ki[t],
&2 be
resp.,
an
k\{a\),k<i C F2 £2(0:2) be two finite simple
P2(t) A^], be the minimal polynomials of a\,
=
=
isomorphism, such that4
i(pi(t))=P2(t).
Then there
that j(ai)
Proof.
exists a unique
=
isomorphism j
:
F2 agreeing with i
F\
on
k\ and such
oil-
Since every element of
k\(oc\)is
a
linear combination of powers of a\with
on k\ (which agrees with i) and by
isomorphism j as in the statement is
coefficients in £4, j is determined by its action
j(cti), which is prescribed
to be 0:2- Thus an
uniquely determined.
To
see
induces
an
that the isomorphism j exists, note that
ki[t]
Of
R[t]
as
i maps
Pi(t)
to
P2(t),
it
isomorphism
course
any
homomorphism of rings /
S[t] sending
—*¦
k2[t]
:
R
S induces
—*
t to t, named in the same way
by
a
a harmless
unique ring homomorphism
abuse of language.
VII. Fields
390
Composing
with the isomorphisms found in Proposition 1.3 gives j:
.
as
7
,
ki[t]
~
s
~
k2[t]
~
D
needed.
Thus, isomorphisms lift uniquely to simple extensions,
polynomials. We
of extensions:
minimal
may say that
so long as they preserve
extends
and
draw
the following diagram
i,
j
fci(ai) ^^^/c2(^2)
h
>k2
We will be particularly interested in considering isomorphisms of a field to
itself, subject to the condition of fixing a specified subfield k (that is, extending the
identity on /c), that is, automorphisms in Fldfc. Explicitly, for k C F an extension,
we will analyze isomorphisms j : F -^ F such that Vc G k,j(c)
c.
=
We will denote the group of such automorphisms by Autfc(F). This is probably
the most important object of study in this chapter, so we should make its definition
official:
Definition 1.6. Let k C F be
the extension, denoted
such that j\k idfc.
a
Autfc(F),
field extension. The group of automorphisms of
F
automorphisms j : F
is the group of field
=
Corollary 1.7. Le£ k C F
the minimal polynomial of a
roots
ofp(x)
in
F;
in
j
=
k(a)
over
be
a
k. Then
simple finite extension, and let p(x) be
| Autfc(F)| equals the number of distinct
particular,
\Autk(F)\<[F:k],
with equality if and only if p(x) factors
polynomials.
over
F
as a
Proof. Let j
product of distinct linear
G Autfc(F). Since every element of F is
with coefficients in /c, and j extends the identity on
polynomial expression in a
/c, j is determined by j(a).
a
Now
Pti(<*))=j(p(a))=j(0)=0:
is
therefore, j(a)
necessarily a root of p(x). This shows that | Autfc(F)|
than the number of roots of p(x).
is
no
larger
On the other hand, by Proposition 1.5 every choice of a root ofp(£) determines
lift of the identity to an element j G Autfc(F), establishing the other inequality.
a
Corollary 1.7 is our first hint of the powerful interaction between group theory
and the theory of field extensions; this theme will haunt us throughout the chapter.
The proof of Corollary
polynomial p(t)
irreducible
1.7
G
really
k[t]
in
bijection between the roots of
larger field F and the group Autfc(F).
sets up a
a
particularly optimistic reader may then hope that the roots of
p(t)
come
an
A
endowed
1. Field extensions, I
391
with a group structure, but this is
the choice of
However,
and
one root a,
have established that
we
hoping for
no one root
if F
too much: the
bijection depends
on
is 'more beautiful' than the others.
oip(t)
k(a) is
a
=
simple
extension
ofk,
then the
group Aut/c(F)
faithfully and transitively on the set of roots of p(t) in F. In
other words, Autk(F) can be identified with a certain subgroup of the symmetric
acts
group
acting
the set of roots of
on
p(t).
More generally (Exercise 1.6) the automorphisms of any extension act on roots
of polynomials. One conclusion we can draw from these preliminary considerations
is that the analysis of Autk(F) is substantially simplified
k(a) and further if the minimal polynomial p(x) of
extension
distinct linear terms in F.
attention
on
is of this type.
be
(The
fully appreciated
1.3. Finite and
Proposition 1.3
It will not
provides
with
k, of degree
n
simple
degp(x)
will focus
our
precisely if
it
if
n
=
dichotomy encountered
in
of the most important definitions in field theory:
[k(a)
a
:
field extension, and let a G F. Then a is algebraic
k] is finite; a is transcendantal over k otherwise.
The extension k C F is algebraic if every
a G
F is
algebraic
over
k.
j
By Proposition 1.3, a G F is algebraic over k if and only if there exists a
0. The minimal polynomial of
polynomial f(x) G k[x] such that f(a)
is the monic polynomial of smallest degree satisfying this condition; as we have
nonzero
a
we
extension is Galois
an
extensions. The
one
Definition 1.8. Let k C F be
over
F is a
reader is invited to remember this fact, but its import will not
until we develop substantially more material.)
algebraic
us
C
factors into
surprise that
come as a
such extensions in later sections:
if k
a
=
seen, it is necessarily irreducible.
a is algebraic over /c, then every
polynomial with coefficients in k.
Also note that if
be written as a
Finite extensions
are
<
k(a)
may in fact
necessarily algebraic-
Lemma 1.9. Let k C F be
k, of degree
element of
finite
a
extension.
Then every
a G
F is
algebraic
over
[F : k}.
Proof. Since k C
k(a)
bydimkF=[F:k}.
C
F, the dimension of k(a)
as a
/c-vector space is bounded
D
Concretely, if k C F is finite and a G F, then the powers l,a, a2... are
necessarily linearly dependent; and any nontrivial linear dependence relation among
them provides us with a nonzero polynomial f(x) G k[x] such that f(a)
0.
=
The literal
will
converse
of the claim that 'finite extensions
in
example
understanding the situation requires
we
see an
algebraic' is not true;
However, something along these lines holds;
a moment.
a more
careful look at finiteness conditions.
First of all, compositions of finite extensions
degree behaves nicely with respect
Proposition 1.10. Let k
only
if
to this
are
are
field extensions.
finite. In this case,
[F:k]
=
finite extensions, and the
operation:
C E C F be
both k C E and E C F
are
[F:E][E:k\.
Then k C F is
finite if and
VII. Fields
392
Proof. If F is finite-dimensional
and
as a vector space over
dependence relation of elements of
/c, then
so
is its
subspace E;
the field k gives one over
the larger field E. It follows that if k C F is finite, then so are k C E and E C F.
any linear
Conversely
basis for F
assume
k C E and E C F
/c, and let (/i,..., /n) be
ran products
over
show that the
are
a
F
over
both finite.
basis for F
Let
over
(ei,..., em)
F.
be
a
It will suffice to
(ei/i,ei/2,...,em/n)
form
a
basis for F
over
k.
Let g G F. Then 3di,.. .,dn e E such that
n
since the elements
k, 3cij,..., cm^-
fj
span F over E.
G /c such that
Since F is
spanned by the elements
e2- over
Mj
m
dj
=
y
^
Cij ei.
2=1
It follows that
m
9
=
n
^2^2cijeifj
:
i=l j = l
therefore the products
eifj
span F over k.
(This
suffices to prove that k C F is
finite.)
To
for
A^
verify
that these elements
are
linearly independent,
assume
G k. Then we have
^(^A^ei)/,- =0,
3
implying ^^A^e^
over
=
0 for all j, since the elements
E. As the elements e2-
equal 0,
as
i
are
linearly independent
fj
over
linearly independent
/c, this shows that all A^
are
needed.
The formula given in the statement should look familiar to the reader: glance
§11.8.5. Of course this is not accidental; the reader should take it as
at the end of
another hint that the world of groups and the world of fields have deep
case of groups, we draw the immediate (but powerful) consequence,
interaction. As in the
reminiscent of
Lagrange's theorem:
Corollary 1.11.
field (that is, k C
Let k C F be a
E C
F).
finite extension,
Then both
[E k]
:
As with Lagrange's theorem, judicious
mysterious statements melt into trivialities.
and
use
and let E be
[F E]
:
divide
an
intermediate
[F k\.
:
of this result will make otherwise
1. Field extensions, I
393
Example 1.12. Let k C F be a
element over /c, of odd order. Then
in
a2,
field
we
k(a2)
is intermediate between k and
k C
what
can
we
say
polynomial t2
a
G
F be
algebraic
polynomial
an
may be written as a
/c(a2),
=
k(a2)
C
k(a):
/c(a);
about the degree d of A;(a) over /c(a2)? Since a satisfies the
k(a2)[t], d < 2. On the other hand, d divides [k(a) : /c]
a2 G
by Corollary 1.11,
in
a
with coefficients in k.
Indeed,
/c(a)
extension, and let
claim that
and
[k(a)
is odd,
/c]
:
and in particular
a G
/c(a2),
so
d
^
2.
Therefore d
=
which is the claim.
1, proving
j
Here is something else that should evoke fond memories for the reader: back
we had encountered an important distinction between finite algebras
§111.6.5
and algebras of finite type. An algebra is finite over the base ring if it is finitely
generated as a module, that is, if it admits an onto homomorphism (of modules)
from a finitely generated free module; a commutative algebra is of finite type if
it admits an onto homomorphism (of algebras) from a polynomial ring in finitely
many variables.
Something along the same lines is going to occur here. An extension k C F is
finite if and only if dim/- F is finite, that is, if and only if F is a finite /c-algebra.
The other finiteness condition takes the following form.
A field extension k C F is finitely generated if there exist
ai,..., an G F such that
Definition 1.13.
F
=
k(ai)(a2)...(an).
j
Remark 1.14. This is not literally the same as saying that F is a finite-type
/c-algebra: even in the case of simple extensions, if a is transcendental, then the
natural evaluation map e : k[t]
k(a) of Proposition 1.3 is not onto. However,
this is
an
epimorphism of rings
in the
categorical
sense, as the reader will check
(Exercise 1.17); thus, the 'categorical spirit' behind the
preserved in this context.
finite-type
condition is
an interesting question: what if a field extension k C F is
finite-type /c-algebra? This turns out to be an important issue, and we will
back to it in §2.2.
j
This subtlety raises
really
come
a
We will
use
the notation
k(ai,a2,---,an)
For k C F and ai,..., an G F, the elements
of the field
are all the elements of F which may be written
as rational functions in the c^'s, with coefficients in k (cf. Exercise 1.3).
Put
for the field
k{a{){a2)... (ctn).
F
k(ai,... ,an)
=
otherwise, k(ai,..., an) is the smallest subfield of F containing fc(ai),..., k(an)
(it is the composite of these subfields). It is clear that the order of the elements c^
is irrelevant.
for
Coming back to the issue of
finitely generated extensions.
finite
vs.
algebraic, these
two notions do coincide
VII. Fields
394
Proposition 1.15. Let k C F
extension.
Then the
(i) k
finite
these conditions
=
be
k(a\,...,an)
finitely generated field
a
equivalent:
extension,
extension.
algebraic
an
Each ct{ is algebraic
(Hi)
If
C F is a
k C F is
(ii)
following
are
are
over
k.
then
satisfied,
[F
:
<
k]
product of the degrees of
the
ct{
over k.
Proof.
need
Lemma 1.9 shows that
to prove that
(iii)
(i),
==>
(i)
(ii); (ii)
==>
(iii) trivially. Thus,
==>
and to bound the degree of F
over
we
only
k in the process.
Assume that each oti is algebraic over /c, and let di be the degree of oti over k.
C
k{ai) is finite, of degree di. By Exercise 1.16, each extension
By definition, k
C
*;(«!,...,Qi_i)
is finite, of
<
degree
k
di. Applying Proposition 1.10
C
C
k(a±)
While rather
C
k(ai, a2)
proves that k C F is finite and
to us: it says
*;(«!,...,Qi)
<
[F : /c]
C
d\
that if
a
and (3
k(au
dn,
straightforward, Proposition
(in particular)
are
to the
as
...,
composition of extensions
an)
=
F
D
needed.
always seemed remarkable
algebraic over a field /c, then so are
1.15 has
a ± (3, aft, ctf3~l, and any other rational function of a and (3. If all we knew were
the simple-minded definition of 'algebraic' (that is, 'root of a polynomial'), this
would seem rather mysterious: given a polynomial f(x) of which a is a root and a
polynomial g(x) of which (3 is a root, how do we construct a polynomial h(x) such
that (for example) h(a
+
(3)
=
0? The
answer
is that
we
do not need to perform
any such construction, by virtue of Proposition 1.15.
One immediate consequence of this observation is that the set of algebraic
elements of any extension forms a field:
Corollary 1.16. Let k C F be
E
=
{a
field
a
G F
|
a
extension. Let
is
algebraic
over
k}.
Then E is
a
Example
Q; then Q
1.17. Let
are
is
is
algebraic.
Note that Q C Q is not
field.
a
QCCbe the set of complex numbers that
field, by Corollary 1.16, and the extension Q C Q
algebraic
over
(tautologically)
finite extension, because in it there are elements
indeed, there exist irreducible polynomials in Q[x]
a
of arbitrarily high degree over Q:
of arbitrarily high degree, as we observed in
Corollary V.5.16.
Elements of Q are called algebraic numbers; the complex numbers in the
complement of Q are transcendental over Q (by definition); they are simply called
transcendental numbers.
Elementary cardinality considerations show that Q
is
countable; thus, the set of transcendental numbers is uncountable. Surprisingly,
it may be substantially difficult to prove that a given number is transcendental:
for example, the fact that e is
Charles Hermite. The number
transcendental
n
was
is transcendental
only proved around 1870, by
(Lindemann,
ca.
1880);
en is
1. Field extensions, I
transcendental
395
(Gelfond-Schneider, 1934).
No
one
knows whether 7re,
tt
+ e, or 7re
are transcendental.
j
Another consequence of Proposition 1.15 is the important fact that
compositions of algebraic extensions are algebraic (whether finitely generated or not):
Corollary
1.18.
and only if both k
Proof. If k
over E, and
Let k C E C F be
Then k C F is
field extensions.
algebraic.
algebraic if
C E and E C F are
C F is
algebraic, then
every element of E is
algebraic over k, hence
k; thus E C F and k C E are
every element of F is
algebraic
over
algebraic.
a
Conversely, assume k C E and E C F are both
is algebraic over £, there exists a polynomial
f(x)
such that
subfleld
=
xn +
en_ixn_1
0. This implies that
f(a)
/c(eo,..., en_i) C j£; therefore,
=
/c(e0,..., en_i)
is
a
C
algebraic, and let
+ e0 G
+
in fact
is
a
a G
F. Since
£[x]
'already' algebraic
over
the
/c(e0,..., en_i, a)
finite extension. On the other hand,
k C
/c(e0,...,en_i)
finite extension by Proposition 1.15, since each ei is in E and therefore algebraic
over k. By Proposition 1.10
is
a
/c C
is finite. This
implies that
a
is
/c(e0,...,en_i,a)
algebraic
over
/c,
as
needed, by Lemma 1.9.
We will essentially deal only with finitely generated extensions, and
Proposition 1.15 will simplify our work considerably. Also, since finitely generated
extensions are compositions of simple extensions, the reader should expect that we
give
a
will
careful look at automorphisms of such extensions.
It may in fact
come as a
extensions turn out to be
simple
surprise that, in many cases, finitely generated
to begin with; we will prove a precise statement
(Proposition
(but interesting)
this effect in due time
contemplate easy
serve as an
Example
encouragement to look at many
1.19. Consider the extension
Proposition
—By
1.15
we
to
5.19). The reader already has enough tools to
examples, such as the following. This should
Q
more.
C
Q(V% y/3).
know that this is
a
finite
(hence algebraic) extension,
of degree at most 4.
—Thus
any five elements in
consider
powers of
1,
must be
y/2
(>/2
+
+
Q(V% y/S)
must be
linearly dependent
over
Q. We
y/3:
>/3),
(V^+v/3)2,
linearly dependent. Thus, there
(v^+v/3)3,
(V^+v/3)4
must be rational numbers qo,..., #3 such
that
(V2 + yfef
+
q3(V2 + V^)3
+
q2(V2 + VS)2
+
qi(V2 + y/3)
+ q0
=
0.
VII. Fields
396
arithmetic shows that this relation
—Elementary
q2
=
-10,
=
#3
0: that is,
\/2+ y/3
/(*)
In
fact,
it is
is
a root
t4
=
-
is satisfied for qo
=
1, qi
=
0,
of the polynomial
10t2
easily checked that f(t) vanishes
+ 1.
at all four combinations
±V2±Vs.
—It
follows that
f(t)
=
(t- (-V2
y/3))(t
-
(-V2 + y/3))(t
-
and that
Thus
f(t) is irreducible over Q (no
y/2 + y/3 has degree 4 over Q.
the composition of
—Consider
-
(>/2
proper factor of
V3))(t
-
coefficients).
extensions
C
Q(>/2, >/3).
[Q(>/2 + VS) Q]
<
[Q(V% >/3) Q].
C
{y/2 + V3))
has rational
f(t)
Q(>/2 + >/3)
Q
-
By Corollary 1.11,
4
On the other hand
:
=
we
know that
:
[Q(a/2, VS) Q]
:
<
4; thus
we can
conclude that
this degree is exactly 4, and it follows that
Q(v/2,V/3)=Q(V/2 + V/3):
the extension
was
typical example,
simple
begin with. (In due
to
not a contrived
—Staring
now at
time
we
will prove that this is
a
one.)
the composition
QCQ(V/2)CQ(V/2,v/3),
Proposition 1.10 tells
us
that
[Q(V% VS) Q(v/2)]
:
=
2. That is,
a
side-effect of the
3 must be irreducible
computation carried out above is that the polynomial t2
over Q(<s/2); this is not surprising, but note that this method is very different from
anything
so
we
encountered back in
§V.5.
—The
fact that the other roots of the minimal polynomial of y/2 + y/3 'looked
much like' y/2 + V3 is not surprising. Indeed, since Q(a/2) C Q(V% y/3) is
simple extension, we know (cf. Proposition 1.5 and following) that there is an
automorphism of Q(V% y/S) fixing Q(a/2) (and hence Q) and swapping y/3 and
VS. Similarly, there is an automorphism fixing Q(y/3) and swapping y/2 and
a
y/2. These automorphisms must act on the set of roots of f(t) (Exercise 1.6):
applying them and their composition to y/2 + y/3 produces the other three roots of
—In
fact, at this point
we
know that G
=
AutQ(Q(<\/2,y/S))
must consist
(by Corollary 1.7) and has at least two elements of order 2 (both
automorphisms found above have order 2). This is enough to conclude that G is
not a cyclic group, and hence it must be isomorphic to the group Z/2Z x Z/2Z.
of 4 elements
j
Exercises
397
Exercises
1.1. > Prove that if k C K is
the category Fid has
1.2. Define
1.3.
field extension, then char/c
[§1.1]
field extension, and let
a
a
F.
G
Prove that the field
consists of all the elements of F which may be written as
with coefficients in k. Why does this not give (in general)
k(t)
char if. Prove that
=
the category Fldfc of extensions of k.
carefully
Let k C F be
>
a
initial object.
no
a
k(a)
rational function in a,
an onto
homomorphism
fc(a)? [§1.2, §1.3]
-
k(a) be a simple extension, with a transcendental over k. Let E be a
k(a) properly containing k. Prove that k(a) is a finite extension of E.
1.4. Let k C
subfield of
1.5.
(Cf. Example 1.4.)
>
Prove that there is exactly
Prove that there
are
one
subfield of R isomorphic to
Q[t]/(t2 2).
to Q[t]/(t3
2).
exactly three subfields of C isomorphic
-
From a 'topological' point of view, one of these copies of Q[£]/(£3
different from the other two: it is not dense in C, but the others are.
1.6. > Let k C F be
that Autfc(F)
a
field extension, and let
acts on the set of roots of
action need not be transitive
1.7. Let k C F be
a
or
faithful.
f(x)
k[x]
G
be
a
Let
field extension, and let
polynomial. Prove
a G
F be
algebraic
over
=
f(x)
G
Prove that
k[x].
Let
f(x)
be the roots of
k[x] be
f(x). For
G
Assume that oli G k
over
k.
f(a)
=
0 if and
-i
an
polynomial
a
subset I C
only for
I
=
is the minimal
sense
0 and I
f(x).
polynomial of
a
certain
of Definition VI.6.12.
field k of degree d, and let ai,..., ad
{1,..., d}, denote by ai the sum ^2ieI ol%.
over a
=
{1,..., d}.
Prove that
f(x)
is irreducible
a
finite field. Prove that the order
Let k be
a
field.
is
\k\
a power
Prove that the ring of square
of
n x n
a
prime integer.
matrices
Mn(k)
isomorphic copy of every extension of k of degree < n. (Hint: If k C F
extension of degree n and a G F, then 'multiplication by a' is a /c-linear
contains
is
a
a
only if p(x)
[7.14]
1.9. Let k be
1.10.
k.
p(x) G k[x] is an irreducible monic polynomial such that p(a) 0; prove
p(x) is the minimal polynomial of a over /c, in the sense of Proposition 1.3.
/c-linear transformation of F, in the
-i
[§1.2]
f(x).
[§1.2, §1.3]
Show that the minimal polynomial of
1.8.
looks very
Provide examples showing that this
Suppose
that
2)
an
transformation of
F.) [5.20]
Let k C F be
a finite field extension, and let p(x) be the characteristic
polynomial of the /c-linear transformation of F given by multiplication by a. Prove
that p(a)
0.
1.11.
-i
=
VII. Fields
398
a polynomial satisfied by an element of an
satisfied by \/2+ y/3 over Q, and compare
Example 1.19. [1.12]
This gives an effective way to find
extension. Use it to find a polynomial
this method with the one used in
1.12.
Let k C F be
-i
^cf(«),
multiplication
by
a
a finite field extension, and let a G F.
The norm of a,
is the determinant of the linear transformation of F given by
(cf. Exercise 1.11, Definition VI.6.4).
Prove that the
norm
is
multiplicative: for a,/3GF,
NkcF(<*P)
NkcF(a)NkcF(P)-
=
Compute the norm of a complex number viewed as an element of the extension
(and marvel at the excellent choice of terminology). Do the same for elements
R C C
of
extension
an
Q(Vd)
of Q, where d is
1.13.
Define the trace
-i
an
integer that is
not a square, and compare
[1.13, 1.14, 1.15, 4.19, 6.18, VIII.1.5]
the result with Exercise III.4.10.
of
tikCF^)
element
an
a
of
a
finite extension F of
a
field k by following the lead of Exercise 1.12. Prove that the trace is additive:
trkQF(a
+
ft)
for a, ft G F. Compute the trace of
an
integer that is
not a square.
trkQF(a)
=
+
element of
an
trkCF(ft)
an
extension
C
Q
Q(Vd),
[1.14, 1.15, 4.19, VIII.1.5]
1.14.
Let k C k(a) be a simple algebraic extension, and let xd-\-a(i-iXd~1-\
be the minimal polynomial of a over k. Prove that
-i
trfccfc(a)(a)
(Cf.
Exercises 1.12 and
1.15.
Let k C F be
-i
=
for d
Nkck(a)(a)
and
-dd-i
=
\-a0
(-l)da0.
1.13.) [4.19]
a
finite extension, and let
a G
F. Assume
[F
k(a)]
:
=
r.
Prove that
trfcCF^)
(Cf.
a
=
and
rtrkck(a)(a)
NkcF(a)
Nkck(a)(a)r.
=
1.13.) (Hint: If /i,...,/r is a basis of F over k(a) and
is a basis of F over k. The matrix
/c, then {fia?)i=\,...,r
Exercises 1.12 and
has degree d
over
j=i\-\d-i
corresponding to multiplication by
square blocks.) [4.19, 4.21]
1.16. > Let k C L C F be
then L C
L(a)
is finite and
1.17. > Let k C F
=
a
with respect to this basis consists of
fields, and let
[L(a) L]
:
k(ai,... ,an)
<
be
a G
F. If k C
k(a)
is
a
r
identical
finite extension,
[k(a) k\. [§1.3]
:
a
finitely generated
extension.
Prove that
the evaluation map
k[ti,...,tn]
is an
1.18.
epimorphism of rings (although
-i
Let R be
a
->
ring sandwiched between
of k. Prove that R is
Is it necessary to
a
tn-+
F,
it need not be
a
on
onto). [§1.3]
field k and
an
algebraic
field.
assume
that the extension is
algebraic? [1.19]
extension F
Exercises
399
a field extension of degree p, a prime integer.
Prove that
F
in F. (Use
of
k
and
contained
subrings
properly containing
properly
1.19. Let k C F be
there
are no
Exercise
1.18.)
tf2
prime integer, and let a
polynomial of degree < p. Prove that a
1.20. Let p be
constant
in
g(a)
a
G R. Let
=
may be
g(x)
G
Q[x]
expressed
be any
as a
non-
polynomial
with rational coefficients.
Prove that
analogous
an
1.21. Let k C F be
a
y/2
statement for
is false.
field extension, and let E be the intermediate field consisting
are algebraic over k. For a G F, prove that a is algebraic
of the elements of F which
only ii
E if and
over
E. Deduce that
a
1.22. Let k C F be a field
d, e,
of (3
resp. Assume
over
1.23.
k. Prove
d,
e are
is
algebraically closed.
extension, and let a G F, /? G F be algebraic, of degree
relatively prime, and let p(x) be the minimal polynomial
is irreducible
p(x)
Q
Express y/2 explicitly
as
over
k(a).
polynomial function
a
in
y/2
+
V3
with rational
coefficients.
1.24. Generalize the situation examined in Example 1.19: let k be a field of
characteristic 7^ 2, and let a, b G k be elements that are not squares in k; prove that
k(y/a,Vb)
=
k(^/a + Vb).
Prove that
resp., is not,
1.25.
-.
k{^/a, y/b)
a square
Let f
has degree 2, resp., 4,
V^+V^.
:=
Find the minimal polynomial of £
V^2
Prove that y 2
Prove that \/2 >/2
-
By Proposition
Q. Find the
Prove that
k according to whether ab is,
over
in k.
over
Q, and show that Q(£) has degree
is another root of the minimal
G
Q(f). (Hint: (a
1.5, sending £
to
+
\/2 V2
-
matrix of this
AutQ(Q(£))
is
6) (a
-
6)
defines
4
over
automorphism of Q(£)
over
polynomial of £.
=
an
a2
-
automorphism
cyclic of order 4.
w.r.t. the basis
b2.)
1,£,£2,£3.
[6.6]
1.26.
be
a
-i
algebras from the
is
field extension, let / be an indexing set, and let {c^}^/
This choice determines a homomorphism ip of kpolynomial ring k[I] on the set I to F (the polynomial ring
Let k C F be
a
choice of elements of F.
free commutative /c-algebra; cf. Proposition
a
polynomial
/(#!,... ,xn)
G
k[xi,... ,xn]
Prove that ai,... ,an
assignment t\i—>•
ai,...,tn
isomorphism)
[1.27]
an
are
such that
We say that {ai}iei is
For example, distinct elements
k if there is no nonzero
III.6.4).
algebraically independent over k if (p is injective.
ai,..., an of F are algebraically independent over
/(ai,..., an)
=
0.
algebraically independent if and only if the
homomorphism of /c-algebras (and hence
an defines a
i—>•
from the field of rational functions
k(ti,..., tn)
to
fc(ai,..., an).
VII. Fields
400
1.27.
With notation and terminology as in Exercise 1.26, the indexed set {c^}^/
transcendence basis for F over k if it is a maximal algebraically independent
set in F.
is
-i
a
Prove that
{c^}^/
is
a
transcendence basis for F
algebraically independent and F is algebraic
Prove that transcendence bases exist.
over
k have the
over
(Mimic the proof of Proposition VI.1.9. Don't feel
only with the case of finite transcendence bases.)
a
k if and only if it is
(Zorn)
Prove that any two transcendence bases for F
The cardinality of
over
k({ai}i^i).
cardinality.
prefer to deal
same
too bad if you
transcendence basis is called the transcendence degree of F
/c, denoted tr.deg^cF- [1-28, 1.29, 2.19]
over
1.28.
-i
Let k C E C F be field extensions. Prove that
only if both tr.deg^c^ and tr.deg£CF (see
tr.degfcCF
=
Exercise
1.27)
tr.degfcCF
are
is finite if and
finite and in this
case
tr.degfccs +tr.deg£CF
[7.3]
1.29.
basis
An extension k C F is purely transcendental if it admits
{ai}i£i (see Exercise 1.27) such that F k({ai}i^i).
-i
a
transcendence
=
Prove that any field extension / C F may be decomposed as a purely
extension followed by an algebraic extension. (Not all field extensions
transcendental
may be decomposed
as an
algebraic
extension followed
by
a
purely transcendental
extension.) [1.30]
k(a) be a simple extension, with a transcendental
k(a) properly containing k. Prove that tr.deg^c^
1.30. Let k C
a
subfield of
over
=
k. Let E be
1.
Luroth's theorem asserts that in this situation k C E is itself a
transcendental
extension of k\ that is, it is
simple
purely transcendental (Exercise 1.29).
2. Algebraic closure, Nullstellensatz, and
a
little
algebraic geometry
One of the most important extensions of a field k is its algebraic closure k C /c; this
mentioned already in §V.5.2, and we can now prove that k exists and is unique
(up to isomorphism). Once this circle of ideas is approached, the temptation to say
was
a
few words about the important result known as Hubert's Nullstellensatz, at the
a small digression into algebraic geometry, will simply be irresistible.
cost of
2.1. Algebraic closure. Recall (Definition V.5.9) that a field K is algebraically
closed if all irreducible polynomials in K[x] have degree 1, that is, if every
polynomial in
factors completely
(Exercise
every maximal ideal in
K[x]
III.4.21)
Now that
again,
as
we
follows.
have
a
little
as a
more
product of linear
K[x]
terms.
is of the form
vocabulary,
(x
Equivalently
c),
we can recast
for
c G
K.
this definition yet
2.
and
Algebraic closure, Nullstellensatz,
Lemma 2.1. For
K
is
The
following
little algebraic geometry
are
401
equivalent:
algebraically closed.
K has
If K
the
afield K,
a
no
nontrivial algebraic extensions.
C L is any extension and
proof is
a
a G
L is
algebraic
straightforward application of the
over
K, then
definitions and
a
a G
K.
good
exercise
(Exercise 2.1).
Definition 2.2. An algebraic closure of
such that k is algebraically closed.
a
field k is
an
algebraic
extension k C k
j
Of course, every polynomial f(x) G k[x] will split into linear factors over k: so
does every polynomial in the much larger k[x], as k is algebraically closed. The
requirement that k C k be algebraic ensures that no other intermediate field L,
may be algebraically closed: indeed, L C k would be a nontrivial algebraic
extension, contradicting Lemma 2.1. In fact, the only subfleld of k containing all roots
of all nonconstant polynomials in k[x] is k itself (Exercise 2.2); thus, the algebraic
a field is 'as small as possible' subject to this requirement. Not
surprisingly, k turns out to be unique up to isomorphism, so that we can speak of the
algebraic closure of k.
closure of
Theorem
Every field k admits
isomorphism.
2.3.
unique up to
an
algebraic closure k
k; this
C
extension is
Concerning existence, the idea is to construct 'by hand' a huge extension K of
polynomial f(x) G k[x] factors completely. The elements of K which
algebraic over k will form an algebraic closure of k.
The construction is done in steps, each step including 'one more root' of all
k where every
are
nonconstant polynomials in
Lemma 2.4. Let k be
a
k[x\. Here
Proof.
in
(This
construction is
bijection with the
a
formalization of this step.
Then there exists
field.
every nonconstant polynomial
is
f(x)
G
k[x]
an
has at least
apparently due
to Emil
extension
k C K such that
one root in
Artin.)
K.
Consider
a set
polynomials f(x)
k[x],
{tf}
k\SF\be the corresponding polynomial ring5 in all the indeterminates tf.
I C k\ST\be the ideal generated by all polynomials /(£/).
set of nonconstant monic
let
Then I is
a proper
ideal. Indeed, otherwise
we
G
2?
=
and
Let
could write
n
(*)
i
=
!>•/<(</<),
2=1
where a2- G
an
k\ST\. We claim that this cannot be done: indeed, we can construct
extension KF where the polynomials fi(x),..., fn(%) have roots ai,..., an,
Here is another rare occasion where we need to consider polynomial rings with possibly
infinitely many indeterminates. An element of k[J7] is simply an ordinary polynomial with coefficients
in k, involving a finite number of indeterminates from S".
VII. Fields
402
respectively (apply Proposition V.5.7
and plug in tfi
oti, obtaining
n
times);
view
(*)
as
an
in
identity
F[^],
=
n
1
=
n
XI
ai
'
h^
5Z ai
=
2=1
which is
2
=
'
0
=
°'
1
nonsense.
Since I is proper, it is contained in
Thus, we obtain a field extension
a
maximal ideal
m
(Proposition V.3.5).
m
by construction every nonconstant monic (and hence every
f(x) has a root in K, namely the coset of tf.
The field constructed in Lemma 2.4 contains at least
nonconstant) polynomial
one root
of each
nonconstant polynomial f(x) G k[x], but this is not good enough: we are seeking a field
which contains all roots of such polynomials. Equivalently, not only should f(x)
have (at least) one linear factor x
a in K[x], but we need the quotient polynomial
f(x)/(x
-
a)
K[x]
G
to have a linear factor and the
have one, etc. This prompts
us to
consider
a
quotient by that
new
factor to
whole chain of extensions
where K\ is obtained from k by applying Lemma 2.4, K2 is similarly obtained
from K\,etc.
Now consider the union L of this chain6. For every two a,b e L, there is an i
Ki\ we can define a + 6, a b in L by adopting their definition in
such that a, b G
Ki, and the result does not depend
field.
the choice of i. It follows easily that L is
a
The field L is algebraically closed.
Claim 2.5.
Proof. If
on
f(x)
G
L[x]
is
a nonconstant
hence f(x) has a root in Ki+i
has a root in L, as needed.
C L. That
polynomial, then f(x) G Ki for some i;
is, every nonconstant polynomial in L[x]
Proof of the existence of algebraic closures. The existence of algebraic
closures is now completely transparent, because of the following simple observation:
Lemma 2.6. Let k C L be
k
Then k is
The
an
:=
a
field extension,
{a
a is
construction reviewed above
This is
algebraic
over
k}.
algebraic closure of k.
L containing any given field /c,
§VIII.1.4.
G L
with L algebraically closed. Let
an
so
provides
us
with
the lemma is all
example of direct limit,
a
notion
which
we
we
an
algebraically closed field
need to prove.
will
introduce
more
formally in
2.
and
Algebraic closure, Nullstellensatz,
a
little algebraic geometry
403
By Corollary 1.16, k is a field, and the extension k C k is tautologically
To verify that k is algebraically closed, let a be algebraic over k; then
algebraic.
kCkC k(a)
is a composition of algebraic extensions, so k C k(a) is algebraic (Corollary 1.18),
and in particular a is algebraic over k. But then a G /c, by definition of the latter.
It follows that k is algebraically closed, by Lemma 2.1.
Remark 2.7. Zorn's lemma
sneakily used
was
in this
proof (we used the
existence
of maximal ideals, which relies upon Zorn's lemma), and it is natural to wonder
whether the existence of algebraic closures may be another statement equivalent to
the axiom of choice.
this is not the case7.
Apparently,
j
deal with uniqueness. It may be tempting to try to set things up
that algebraic closures would end up being solutions to a universal
problem; this would guarantee uniqueness by abstract nonsense. However, we run
into the same obstacle encountered in Proposition V.5.7: morphisms between
Next,
in such
we
a way
extensions depend on choices of roots of polynomials, so that algebraic closures
not 'universal' in the most straightforward sense of the term.
Therefore,
a
bit of independent work is needed.
Lemma 2.8. Let k C L be
afield extension,
be any algebraic extension. Then there exists
As
pointed
are
out
above,
i is
by
no means
Proof. This argument also relies
with L algebraically closed. Let k C F
L.
a morphism of extensions i : F
unique!
Zorn's lemma. Consider the set Z of homo-
on
morphisms
%K
where K is
an
:
K
L
-?
intermediate field, k C K C F, and iK restricts to the identity on
: k C L defines an element of Z. We give
/c; Z is nonempty, since the extension ik
a
poset structure to Z by defining
iK
< %k>
iK' restricts to iK on K. To verify that every chain C in Z has
bound in Z, let Kc be the union of the sources of all iK £ C {Kc is clearly
if a G Kc, define iKc{a) t° be i/<-(a), where iK is any element of C such
if K C K' C F and
an upper
a
field);
that
a
a G
clearly independent of the chosen K and defines
L restricting to the identity on k. This homomorphism is
K. This prescription is
homomorphism Kc
an upper
bound for C.
By Zorn's lemma, Z admits
intermediate field k C G C F. Let H
We claim that G
there
is
a
=
maximal element ic, corresponding to
ic{G) be the image of G in k.
F: this will prove the statement, because it will
L extending the identity on k.
homomorphism ip : F
=
Arguing by contradiction,
the
a
extension G C
G{a).
assume
Since
a
G
that there exists
F is
algebraic
an
imply that
F
G, and consider
k, it is algebraic over G;
an a G
over
Allegedly, the existence of algebraic closures is a consequence of the compactness theorem
for first-order logic, which is known to be weaker than the axiom of choice.
VII. Fields
404
thus, it is
a root
of
an
irreducible polynomial
p(x)
G
G[x].
Consider the induced
homomorphism
iG
and let
We
are
fields
G[x]
H[x],
^
is an irreducible polynomial over H, and it has
the hypothesis that L is algebraically closed!).
in the situation considered in Proposition 1.5: we have an isomorphism of
h(x)
(3
a root
:
%g
=
in L
G
iG(g(x)). Then h(x)
(this is where we use
H;
we
irreducible polynomials
G(a), H{(3)\ and a, (3 are roots of
iG(g(x)). By Proposition 1.5, %g lifts to an
have simple extensions
g(x), h(x)
=
isomorphism
H(fi) C L
iG(a) : G(a)
sending a to (3. This contradicts the maximality of %g\
the argument.
-
hence G
=
F, concluding
Proof of uniqueness of algebraic closures. Let k C k,k C k\betwo algebraic
closures of /c; we have to prove that there exists an isomorphism k\ k extending
the identity on k.
Since k C k\ is algebraic and /c is algebraically closed, by Lemma 2.8 there
a homomorphism i : k\—>
k extending the identity on k\this homomorphism
exists
is trivially injective, since k\ is a field. It is also surjective, by Lemma 2.1:
otherwise k\ C k is a nontrivial algebraic extension, contradicting the fact that k\ is
algebraically closed. Therefore i is an isomorphism, as needed.
2.2. The Nullstellensatz. If K is
maximal ideal in
has
K[x]
is of the form
an
(x c),
algebraically closed field, then
for
c G
K
(Exercise III.4.21).
every
This statement
straightforward-looking generalization to polynomial rings in more indeterif K is algebraically closed, then every maximal ideal in K[x\,...,xn] is
the
,xn
of
form (xi cu
cn).
this
statement
is
more
Proving
challenging than it may seem from the looks of
a
minates:
-
-
...
it; it is one facet of the famous theorem known as Hubert's Nullstellensatz ("theorem
the position of zeros"). The Nullstellensatz has made a few cameo appearances
earlier (see Example III.4.15 and §111.6.5 in particular), and we will spend a little
on
on it now. We will not prove the Nullstellensatz in its natural generality, that
for
all fields; this would take us too far. Actually, there are reasonably short
is,
proofs of the theorem in this generality, but the short arguments we have run into
time
end up
replacing
a
wider
(and useful)
find these arguments particularly
context with
insightful
or
ingenious cleverness;
we
do not
memorable.
By contrast, the theorem has a very simple and memorable proof if we make
the further assumption that the field is uncountable. Therefore, the reader will have
complete proof of the theorem (in the form given below, which does not assume
algebraically closed) for fields such as R and C but will have to trust
on faith regarding other extremely important fields such as Q.
a
the field to be
The
applications of the Nullstellensatz involve basic definitions in
algebraic geometry, and we will get a small taste of this in the next section; but the
theorem itself is best understood in the context of the considerations on 'finiteness'
mentioned in §1.3; see especially Remark 1.14. We pointed out that if k C F
most vivid
2.
Algebraic closure, Nullstellensatz,
is finitely generated
as
(that is, 'finite-type')
extensions F that
question.
it so:
It is
Theorem 2.9
is
a
and
a
little algebraic geometry
field extension, then F need not be finitely generated
k-algebra, and we raised the question of understanding
finite-type /c-algebras. The following result answers this
a
as a
are
favorite version of the Nullstellensatz; therefore
our
405
Then k C F is
§111.6.5,
an
algebra k
S
=
As
(and
if S
mentioned above,
somewhat
F
As
pointed
(as an algebra),
k[t] is an obvious
may very well be of finite type
without being finite (i.e., finitely generated as a module): S
example. The content of Theorem 2.9 is that the distinction between
and finite disappears
name
afield extension, and assume that
finite (hence algebraic) extension.
a
The reader should pause and savor this statement for a moment.
out in
will
Let k C F be
(Nullstellensatz).
finite-type k-algebra.
we
is a
field.
we
will
finite-type
only
prove this statement under an additional
unnatural) assumption,
which reduces the argument to elementary
linear algebra.
Proof for uncountable fields. Assume that k is uncountable.
Let k C F be
algebra
over
k;
in
field extension, and assume that F is finitely generated as an
particular, it is finitely generated as a field extension. We have
a
to prove that k C F is
it is algebraic, that is,
a
finite extension,
every a G F
or
that
Now, by hypothesis
equivalently (by Proposition 1.15) that
k is the root of
F is the quotient of
a
a nonzero
polynomial ring
/c[xi,...,xn]
over
f(x)
G
k[x\.
k:
»F ;
implies that there is a countable basis of F as a vector
this is the case for k[x\,...,xn]. Consider then the set
this
k, because
space over
\a-cjc
this is
an uncountable subset of F\therefore it is linearly dependent over k
(Proposition VI. 1.9). That is, there exist distinct ci,..., cm G k and nonzero coefficients
Ai,..., Am G k, such that
a
a
c\
Expressing the left-hand side with
Cn
a common
denominator gives
9(ol)
for
f(x) ^ 0 in k[x] and g(a) ^ 0 (Exercise 2.4).
nonzero f(x) G k[x], and we are done.
This implies
f(a)
=
0 for
a
Regardless of its proof, the form of the Nullstellensatz given in Theorem 2.9
does not look much like the other results we have mentioned as associated with it.
For
one
thing,
it is not too clear
why Theorem
the 'position of zeros'. The result deserves
a
2.9 should be called
little
more
attention.
a
theorem
on
VII. Fields
406
The form stated at the
beginning of this
section is obtained
Corollary 2.10. Let K be an algebraically closed
Then I is maximal if and only if
..., xn].
as
field, and let
follows:
I be
an
ideal
of
K[x\,
1
for
=
\X\ C\,
.
.
.
,
Xn
Cn),
ci,...,Cn G K.
Proof. For c\,...,cn
K[xu...,xn]
C\,
yX\
is
(Exercise III.4.12)
let
m
be
a
a
.
.
.
,
field; therefore (#i
maximal ideal in
„K
Xn
Cn)
is maximal.
cn)
ci,..., xn
K[xi,... ,xn],
that the quotient is
so
a
Conversely
field and the
natural map
_^
%=
K[xu...,xn]
m
is
a
The field L is finitely generated as a if-algebra; therefore
algebraic extension by Theorem 2.9. Since K is algebraically closed,
field extension.
K C L is
an
this implies that
L
K
=
(Lemma 2.1); therefore,
(p:
such that
m
=
Let q
ker<p.
(x\
but
(x\
Cn)
ci,..., xn
:=
-
K[xi,...,xn]
tp(xi).
a
surjective homomorphism
K
-?
Then
c\,...,xn
is maximal,
there is
-
so
Cn)
C ker ip
this implies
=
m
m;
=
{x\
ci,..., xn
cn),
as
needed.
Note how very
which
may
(trivially)
false the statement of Corollary 2.10 is over fields
in R[x]. No such silliness
algebraically closed: (x2 + 1) is maximal
over C, for any number of variables.
are not
occur
algebraic geometry. Corollary 2.10 is one of the main
'classical'
why
algebraic geometry developed over C rather than fields such
as R or Q: it may be interpreted as saying that if K is algebraically closed, then
there is a natural bisection between the points of the product space Kn and the
2.3. A little affine
reasons
ring K[x\,...,xn]- This is the beginning of
a fruitful dictionary
into
and
It
is
worthwhile
translating geometry
algebra
conversely.
formalizing this
statement and exploring other 'geometric' consequences of the Nullstellensatz.
maximal ideals in the
field, A^ denotes the affine
n-tuples of elements of K:
For K
set of
a
A7}c
Elements of
A^
are
=
space of dimension n over
K, that is, the
{(cu...,cn)\cieK}.
called points.
Objection: Why
don't
we
just
use
'ifn' for this
object?
Because the latter
already stands for the standard n-dimensional vector space over K\ so it carries
with it a certain baggage: the tuple (0,..., 0) is special, so are the vector subspaces
of Kn, etc.
any other
no point of
A^ is 'special'; we can carry any point to
simple translation. Similarly, although it is useful to consider
By constrast,
point by
a
2.
and
Algebraic closure, Nullstellensatz,
a
little algebraic geometry
407
affine subspaces, these (unlike vector subspaces) are not required to go through the
origin. Affine geometry and linear algebra are different in many respects.
We have already established that, by the Nullstellensatz, if K is algebraically
closed, then there is a natural bijection between the points of A^ and the maximal
ideals of the polynomial ring K[x\,...,xn]. Concretely, the bijection works like
this: the point p
(ci,..., cn) corresponds to the set of polynomials
=
{/(£)
S(p)
where
f(p)
K[xu. ..,xn}\f(p)
¦¦=
is the result of
applying the evaluation
/(xi,...,xn)
This of course is
G K.
nothing but the homomorphism
(f
defined
0},
map:
/(ci,...,cn)
i->
=
by prescribing
:
K[xi,...,xn]
q; the set
J^(p)
i—?
Xi
is
K
-?
simply kenp, which
is the maximal
ideal
\X\ C\,
.
.
.
,
Xn
Cn).
field; the content of
correspondence p
J^(p)
is
that
maximal
ideal
of
2.10
Corollary
every
K[xi,... ,xn] corresponds to a point
of A^ in this way, if K is algebraically closed.
It is natural to upgrade the correspondence as follows (again, over arbitrary
fields): for every subset S C AJ-, consider the ideal
may be defined over every
This
i—?
S(S)
5, f(p)
K\xu...,xn\ Vp
{f(x)
:=
=
0},
that is, the set of polynomials 'vanishing along 5'; it is immediately checked that
this is indeed an ideal of K[xi,... ,xn]. We can also consider a correspondence
in the
reverse
direction, from ideals of K[xi,... ,xn]
to subsets of
AJ-,
defined by
setting for every ideal /,
r(/):={p
Thus, y(I)
as
f
(c1,...,c„)€A^|V/€/,/(c1,...,c„)
0}.
=
=
is the set of
e I: the set of
solutions of all the polynomial equations
out in AJ-' by the polynomials in /.
common
points 'cut
/
=
0
In its most elementary manifestation, the dictionary mentioned in the beginning
of the section consists precisely of the pair of correspondences
{subsets
of
A^}
>
<
y
{ideals
The set
is
y(I) is often (somewhat improperly)
(also improperly) the ideal of S.
The function V
K[xi,...,xn],
can
be defined
whether A is
an
ideal
(using
or
the
in
K[x\,...,xn]}
called the variety
same
.
of I,
prescription)
not; it is clear that Y{A)
=
while
*#{S)
for any set A C
y(I) where / is
the ideal generated by A (Exercise 2.5). Hilbert's basis theorem (Theorem V.1.2)
tells us something rather interesting: K[xi,..., xn] is Noetherian when K is a field;
hence
every ideal / is
generated by
a
finite number of elements.
ideal / there exist polynomials /i,..., fr such that
r(/)
=
(abusing
r(/1,...,/P).
Thus, for every
notation
a
little)
VII. Fields
408
In other words, if a set may be defined by any set of polynomial equations, it may
in fact be defined by a finite set of polynomial equations8.
We
Y, J^
behave
well
with
to
unions
and
inclusions,
respect
they
reasonably
(they
will.
the
reader
should
feel
free
to
research
the
at
But
intersections, etc.);
topic
will pass in silence many simple-minded properties of the functions
reverse
we want to focus the reader's attention on the fact that
bisections; this is a problem, if
'geometry' and 'algebra'.
by
we
really
(of course) ^,
want to construct a
J
are not
dictionary between
For example, there are (in general) lots of subsets of A^ which are not cut out
of polynomial equations, and there are lots of ideals in K[x\,...,xn] which
a set
are not
obtained
*#{S)
as
for any subset S C
AJ-.
We will leave to the reader the task of finding examples of the first phenomenon
(Exercise 2.6). The way around it is to restrict our attention to subsets of A^ which
are
of the form
y{I)
for
some
ideal /.
Definition 2.11. An
exists
an
ideal I C
(affine) algebraic set is a subset
K[xi,..., xn] for which S ^(/).
S C
A^
such that there
=
j
This definition may seem like a cheap patch: we do not really know what the
image of Y is, so we label it in some way and proceed. This is not entirely true—the
family of algebraic subsets of a given A^ has a solid, recognizable structure: it is
the
family of
closed subsets in
a
topology
on
A^ (Exercise 2.7);
this topology is
called the Zariski topology.
Studying affine algebraic sets is the business of affine algebraic geometry. The
'geometric' properties of these sets in terms of 'algebraic'
of
properties
corresponding ideals or other algebraic entities associated with them.
aim is to understand
Example 2.12. Here
are
pictures of
two
algebraic subsets of
*
The first is Y((y
-
x2));
A^:
/
the second is
Y((y2
-
x3)).
What feature of the ideal
(y2 -x3) responsible for the 'cusp' at (0,0) in the second picture? The reader
will find out in any course in elementary algebraic geometry.
j
is
The fact that J is not surjective leads to interesting considerations. The
an ideal that is not the ideal of any set is (x2) in K[x\: because
standard example of
We distinctly remember being surprised by this fact the first time
we ran into it.
2.
Algebraic closure, Nullstellensatz,
and
a
little algebraic geometry
wherever x2 vanishes, so does x; hence if x2 G
well. This motivates the following definitions.
Definition 2.13. Let / be
ideal in
an
a
*#{S)
for
a set
5, then
409
x G
*#{S)
commutative ring R. The radical
of I
as
is
the ideal
Vl:={reR\3k>0,rkel}.
An ideal / is
a
radical ideal if /
=
\fl.
j
\/lis indeed an ideal; it may
prime ideals containing / (Exercise 2.8).
The reader should check that
the intersection of all
Example
2.14. An element of a commutative
in the radical of the ideal
(0) (cf.
Exercise
be characterized as
ring R is nilpotent if and only if it is
The reader has encountered this
V.4.19).
ideal, called the nilradical of i?, in the exercises, beginning with Exercise III.3.12.
Prime ideals p are clearly radical: because fk G p =>
\/pQ P (the other inclusion holds trivially for all ideals).
field, and
of K[xi,...,xn].
Lemma 2.15. Let K be
is a radical ideal
Proof. The inclusion
satisfied.
To
integer k
verify
a
J[S)
fk
e
subset
a
G p, and therefore
j
ofA1^.
Then the ideal
yJJ[S) holds for every ideal, so
yJJ[S) C J{S), let / G y/S(S).
C
the inclusion
> 0 such that
let S be
/
J?(S);
it is
J^(S)
trivially
Then there is
an
that is,
0.
(VpGS),
f(p)k
(VpeS),
/(p)=o,
=
But then
proving /
G
^{S\
as
needed.
In view of these considerations, for any field we can refine the correspondence
A^ and ideals of K[x\,...,xn] as follows:
between subsets of
{algebraic
subsets of
A^}
<
>
{radical
ideals in
K[xi,..., xn]}
;
as we have argued, it is necessary to do so if we want to have a good dictionary. In
the rest of the section, J? and Y will be taken as acting between these two sets.
tightened the correspondence substantially, the situation is still
idyllic. Over arbitrary fields, the function Y is now surjective by the very
definition of an affine algebraic subset (cf. Exercise 2.9); but it is not necessarily
injective. An immediate counterexample is offered over K
Rby the ideal (x2 + l):
While we have
less than
=
this is maximal, hence prime, hence radical, and
r(x2 +1)
Here is where the Nullstellensatz
=
0
=
comes to our
r(i).
aid, and here is where
we see
Y of that
ideal).
the Nullstellensatz has to do with the 'zeros' of
an
ideal
(=
what
Proposition 2.16 (Weak Nullstellensatz). Let K be an algebraically closed field,
and let I C K[xi,..., xn] be an ideal. Then y(I)
0 if and only if I
(1).
=
=
VII. Fields
410
Proof. If I
=
Conversely,
a
then
(1),
maximal ideal
assume
Y(I)
that J
^ (1). By Proposition V.3.5,
I is then contained in
Since K is algebraically closed, by Corollary 2.10
m.
m
for
0 by definition.
=
if;
some c\,...,cn G
so
=
(xi
-ci,...,xn
V/(#i,..., xn)
have
we
-cn)
G 7 there exist gi,...,gn
^
[#i,
xn]
•,
such that
n
/(xi,...,xn)
^2g(xi,...,xn)(xi -a).
=
2=1
In particular,
n
/(ci,...,cn)
=
^5f(ci,...,cn)(ci -a)
=0
:
2=1
that is,
(ci,..., Cn)
^(7).
G
y(I) ^ 0
This proves
if I
^ (1),
and
we are
done.
This is encouraging, but it does not seem to really prove that Y is injective.
As it happens, however, it does: we can derive from the 'weak' NuUstellensatz the
following stronger result, which does imply injectivity:
Proposition 2.17 (Strong NuUstellensatz). Let K be
and let I C K[xi,..., xn] be an ideal. Then
f {?{!))
=
an
algebraically closed field,
V7.
This implies that the composition /of is the identity on the set of
radical ideals; that is, Y has a left-inverse, and therefore it is injective (cf.
Proposition 1.2.1!).
Proof. Note that
y(y(I))
is
a
radical ideal
immediately from the definitions that / C
/(f (/));
Vicy/s(y(i))
We have to
verify
the
reverse
inclusion:
(by
Lemma
2.15).
It follows
therefore
s(y(i)).
=
assume
that
/
G
J^(f(/));
that is,
assume
that
f(p)
for all p G A^ such that \/g G
for which fm G L
I,g(p)
=
0;
=
0
we
have to show that there exists
an m
The argument is not difficult after the fact, but it is fiendishly clever9. Let y
be
an
extra
variable, and consider the ideal J generated by / and
K[xi,...,xn,y],
More explicitly,
I
(since K[xi,... ,xn]
resp.,
is Noetherian,
=
7
fy
in
(9U-">9r)
finitely
many generators
/(#i,...,£n) G K[xi,...
F(xi,...,xn,y) G K[xi,...,xn,y], and
gi\X\f
1
assume
¦En)! resp.,
,
suffice);
we can
view
xn\, as polynomials Gi (xi,..., xn,
y),
let
J=(Gu...,Gr,l-Fy).
This is called the Rabinowitsch trick and dates back to 1929. It was published in a one-page,
eighteen-line article in the Mathematische Annalen.
2.
and
Algebraic closure, Nullstellensatz,
As
K[x\,...,xn, y] has n +
We claim y(J)
0.
variables, the ideal J defines
1
411
little algebraic geometry
a
a
subset
V(J)
A^1*1.
in
=
Indeed, assume on the contrary that y(J) ^ 0, and let p
A^+1 be a point of G T( J). Then for i 1,..., r
(ai,..., an, 6)
=
in
=
Gi(ai,...,an,y)
but this
=
0;
means
Pi(oi,...,on)
for i
=
1,...,
r; that
is, (ai,..., an)
^(/).
G
=
0
Since / vanishes at all points of
y(I),
this implies
/(oi,...,on)
=
0,
and then
(1 -Fy)(au...,
But this contradicts the
Now
conclude J
use
=
6)
on,
1
=
/(oi,..., an)6
-
assumption that
p G
=1-0
=
1.
^(J). Therefore, y( J)
=
0.
the weak Nullstellensatz: since K is algebraically closed, we can
there exist polynomials Hi(x\,...,xn,y), i
l,...,r,
(1). Therefore,
=
and L(xi,... ,xn,y), such that
r
^2Hi(xi,... ,xn,y)Gi(xi,... ,xn,y) + L(xi,...,xn,y)(l
We are not done
being clever: the
of rational functions
F(xi,...,xn,y)y)
=
1.
equality in the field
polynomial ring, which we can do,
next step is to view this
K rather than in the
over
-
since the latter is a subring of the former. We can then plug in y
1/F and still
get an equality, with the advantage of killing the last summand on the left-hand
=
side:
Further,
{-*i
(Gi
first
is just the
n
name
variables),
%11
'
'
'
i
^W) j-p )
Qix^l")
'
'
'
i
%n)
of gi in the larger polynomial ring and only depends
on
the
while
/
1\
_
hj(xu...,xn)
l\XU~"Xn'Fj~ f(xu...yxn)^
where hi G
1,...,
r.
K[xi,...,xn]
Expressing
and
in terms of
m
is
integer large enough to work for all i
denominator, we can rewrite the identity
an
=
a common
as
feiffi
H
1-
hr9r
_
fm
or
/m
=
higi
We have proved this identity
polynomials, it holds in K[x\,...,xn].
this proves / G V% and we are done.
+
+
hrgr.
in the field of rational
functions;
The right-hand side is
as
an
it
only involves
element of /,
so
412
a
VII. Fields
The addition of the variable y in the proof looks more reasonable (and less like
is an ordinary
trick) if one views it geometrically. The variety Y(l xy) in
A^
hyperbola:
The projection (x,y)
*—>
x
may be used to
identify y(l
of the origin (that is, y{x)) in the "x-axis"
is a way to realize the Zariski open set A^
A^1*1
of dimension
xy)
with the complement
(viewed
A^-). Similarly, Y(l fy)
y(f) as a Zariski closed subset in a
as
higher.
The proof of Proposition 2.17 hinges on this fact, and this fact is also
important building block in the basic set-up of algebraic geometry, since it can
space
one
used to show that the Zariski topology has a basis consisting of
isomorphic in a suitable sense to) affine algebraic sets.
We will officially record the conclusion
functions Y, J^\
Corollary 2.18. Let K be
an
we were
(sets
which
an
be
are
able to establish concerning the
algebraically closed field. Then for
any
n
>
0 the
functions
{algebraic
are inverses
of
subsets
of A^}
>
<
y
{radical
ideals in
K[x\,...,xn]}
each other.
Proof. Proposition 2.17 shows that J^o y is the identity on radical ideals, so Y is
injective, and Y is surjective by definition of affine algebraic set. It follows that
Y is
a
bijection and J?
is its inverse.
is algebraically closed, then studying sets defined by
polynomial equations in a space Kn is 'the same thing as' studying radical ideals in a
polynomial ring K[x\,...,xn].
This correspondence is actually realized even more effectively at the level of
if-algebras. We say that a ring is reduced if it has no nonzero nilpotents; then R/I
is reduced if and only if J is a radical ideal (Exercise 2.8).
Summarizing:
If K
2.
and
Algebraic closure, NuUstellensatz,
Definition 2.19. Let S C
A^
be
a
little algebraic geometry
algebraic
an
set. The coordinate
413
ring of S is the
quotient10
K[S]
Thus,
the coordinate ring of
algebra of finite type. The reader
ring,
as
K[xi,...,Xn]
:-
J{S)
algebraic
an
will establish
set is a
a
reduced,
commutative K-
'concrete' interpretation of this
the ring of 'polynomial functions' on 5, in Exercise 2.12. One way to think
(in its manifestation as Corollary 2.18) is that, if K is
about the NuUstellensatz
algebraically closed, then
every reduced commutative if-algebra of finite type R
be
realized
as
the
coordinate
may
ring of an amne algebraic set 5; the points of S
in
a
natural
to
the
maximal ideals of R (Exercise 2.14).
correspond
way
In fact, in basic
algebraic geometry one defines the category of amne algebraic
correspondence: morphisms of algebraic sets
are defined so that they match homomorphisms of the corresponding (reduced,
finite-type) if-algebras. (The reader will encounter a more precise definition in
Example VIII.1.9.)
This dictionary allows us to translate every 'geometric' feature of an algebraic
set (such as dimension, smoothness, etc.) into corresponding 'algebraic' features of
the corresponding coordinate rings. Once these key algebraic features have been
identified, then one can throw away the restriction of working on reduced finitetype algebras over an algebraically closed field and try to 'do geometry' on any (say
sets over a field K in terms of this
commutative, Noetherian) ring. For example, it turns out that PIDs correspond
to (certain) smooth curves in the afflne geometric world; and then one can try to
think of Z as a 'smooth curve' (although Z is not a reduced finite-type if-algebra
over any field K\) and try to use theorems inspired by the geometry of curves to
understand features of Z—that
is, use geometry to do number
theory11.
Another direction in which these simple considerations may be generalized is
by 'gluing' afflne algebraic sets into manifold-like objects: sets which may not
be afflne algebraic sets 'globally' but may be covered by afflne algebraic sets. A
concrete way to do this is by working in projective space rather than afflne space;
the globalization process can then be carried out in a straightforward way at the
algebraic level, where
it translates into the
study of 'graded' rings and modules—
notions for which, regretfully, we will find no room in this book (except for a
glance, in §VIII.4.3). In the second half of the twentieth century a more abstract
more
viewpoint, championed by Alexander Grothendieck and others, has proven to be
extremely effective; it has led to the intense study of schemes, the current language
of choice in algebraic geometry. We have encountered the simplest kind of schemes
when
we
introduced the spectrum of
a
ring R, Speci?, back
in
§111.4.3.
The notation 'K[5]' is fairly standard, although it may lead to confusion with the usual
polynomial rings; hopefully the context takes care of this. It makes sense to incorporate the name
of the field in the notation, since one could in principle change the base field while keeping the
'same' equations encoded in the ideal
<#(S).
This is of course a drastic oversimplification.
414
VII. Fields
Exercises
2.1. > Prove Lemma 2.1.
2.2. > Let k C k be an
[§2.1]
algebraic closure, and let
that every polynomial f(x) G
L[x]. Prove that L k. [§2.1]
C
k[x]
L[x]
L be
factors
an
as a
intermediate field. Assume
product of linear
terms in
=
2.3. Prove that if /c is
2.4. > Let k be
be
nonzero
(This
a
a
countable field, then
field, let
so
is k.
ci,..., cm G /c be distinct
elements, and let Ai,..., Am
elements of k. Prove that
fact is used in the proof of Theorem
2.5. > Let K be a
field, let A be
generated by A. Prove that
Y(A)
a
2.9.) [§2.2]
subset of if [#i,...
=
Y(I)
in
and let I be the ideal
,xn],
A£. [§2.3]
2.6. > Let If be your favorite infinite field. Find
examples of subsets S
C
A^
which
for any ideal I C if [#i,... ,xn]. Prove that if if is a
finite field, then every subset S C A^ equals V(I) for some ideal I C if [#i,..., xn\.
cannot be realized
as
V(I)
[§2-3]
2.7. > Let if be
subsets of A^
a
nonnegative integer. Prove that the set
family of closed sets of a topology on AJ-. [§2.3]
field and
is the
2.8. > With notation
Prove that the set
as
vT
n a
of
algebraic
in Definition 2.13:
is
an
ideal of R.
vT corresponds to the nilradical of R/I via the correspondence
R/I and ideals of R containing /.
Prove that \flis in fact the intersection of all prime ideals of R containing
(Cf. Exercise V.3.13.)
Prove that
between ideals of
Prove that I is radical if and only if
R/I
is reduced.
Exercise
(Cf.
/.
III.3.13.)
[§2.3]
2.9. > Prove that every affine
algebraic
2.10. Prove that every ideal in
2.11. Assume
algebra which
2.12.
a
a
A^
equals y(I) for
Noetherian ring contains
field is not algebraically closed.
Find
is not the coordinate ring of any affine
> Let if be an infinite field.
set S C
set
A
a
a
radical ideal I.
a power
of its radical.
reduced
finite-type
algebraic
polynomial function
[§2.3]
if-
set.
on an
affine algebraic
(the evaluation function of) a polynomial
if [#i,... ,xn]. Polynomial functions on an algebraic S manifestly
is the restriction to S of
/(#!,... ,xn) G
form a ring and in fact a if-algebra. Prove that this if-algebra
the coordinate ring of 5. [§2.3, §VIII. 1.3, §VIII.2.3]
is
isomorphic
to
Exercises
415
2.13. Let K be
if-algebra of
space
an
algebraically closed field. Prove that every reduced commutative
an algebraic set S in some affine
finite type is the coordinate ring of
AJ-.
algebraically closed field if, the points of an algebraic
ring K[S] of 5, in such a
2.14. > Prove
over an
set S
the maximal ideals of the coordinate
way
/
G
that,
correspond to
that if p corresponds to the maximal ideal mp, then the value of the function
K[S]
2.15.
-i
at p
equals the
Let K be
coset of
in
/
K[S]/mp
^ if.
[§2.3, VIII.1.8]
algebraically closed field. An algebraic subset S of A^
an
is
the union of two algebraic subsets properly
contained in it. Prove that S is irreducible if and only if its ideal *#{S) is prime, if
and only if its coordinate ring K[S] is an integral domain.
irreducible if it canned be written
An irreducible algebraic
example) y(xy)
algebraic sets
2.16.
set is 'all in one
in the affine
are
called
> Let K be
an
as
plane
A2^
(affine algebraic)
a
a
function
if[A^]
Irreducible affine
[2.18]
varieties.
algebraically closed field.
K{x\,...,xn) is the field of fractions of
function
piece', like A^ itself, and unlike (for
with coordinates x, y.
The field of rational functions
=
K [#i,...
,xn];
every rational
^
(with G ^ 0 and F, G relatively prime) may be viewed as defining
the open set A^
y{G)\ we say that a is 'defined' for all points in
=
on
complement of Y(G).
Let G e if [#i, ...,xn] be irreducible. The set of rational functions that are
defined in the complement of V{G) is a subring of if (#i,... ,xn). Prove that this
the
subring may be identified with the localization (Exercise V.4.7) of
multiplicative set {1, G, G2, G3,... }. (Use the Nullstellensatz.)
The
K[AJ-]
at the
considerations may be carried out for any irreducible algebraic set 5,
field of 'rational functions' K(S) the field of fractions of the integral
same
as
adopting
domain if [5].
[2.17,2.19, §6.3]
algebraically closed field, and let m be a maximal ideal of
K[x\,...,xn], corresponding to a point p of AJ-. A germ of a function at p is
determined by an open set containing p and a function defined on that open set; in
2.17.
-i
Let K be
an
(dealing with rational functions and where the open set may be taken
to be the complement of a function that does not vanish at p) this is the same
information as a rational function defined at p, in the sense of Exercise 2.16.
our context
Show how to identify the ring of germs with the localization K [A^]m
in Exercise
(defined
V.4.11).
As in Exercise 2.16, the same discussion can be carried out for any algebraic
set. This is the origin of the name 'localization': localizing the coordinate ring of a
variety V at the maximal ideal corresponding to a point p amounts to considering
only functions defined in a neighborhood of p, thus studying V 'locally', 'near p\
[V.4.7]
2.18.
-i
Let K be
x2, C2 : y2
=
in Example
x3 in
2.12).
an
algebraically closed field. Consider the
A2K (pictures
two 'curves'
of the real points of these algebraic sets
C\ :
are
y
=
shown
VII. Fields
416
Prove that
K[Ci]
=
K[t]
=
if[A^],
while
K[C2]
subring K[t2,t3} of K[t] consisting of polynomials
(Note
f(x) + g(x)y + h(x, y)(y2
h(x,y).)
are
both irreducible
Prove that
is
UFD, while K[C2] is
K[Ci]
a2t2
+
+
adtd
with
-
Show that Ci, C2
a
a0 +
that every polynomial in K[x, y] may be written as
x3) for uniquely determined polynomials /(#), g(x),
£-coefficient.
zero
may be identified with the
(cf.
Show that the Krull dimension of both if
Exercise
2.15).
not.
[Ci]
and K [C2] is 1.
(This
is
why these
prime
sets would be called 'curves'. You may use the fact that maximal chains of
ideals in K[x, y] have length 2.)
The origin (0,0) is in both Ci, C2 and corresponds
to the maximal ideals mi,
K[C2], generated by the classes of x and y.
Prove that the localization K[Ci]mi is a DVR (Exercise V.4.13). Prove that the
localization -ftT[Cf2]m2 1S n°t a DVR. (Note that the relation y2
x3 still holds
in this ring; prove that K[C2]m2 is not a UFD.)
K[Ci],
resp., rri2, in
resp.,
=
The fact that
a
maximal ideal
a curve
such
K[C2]m2
as
local parameter, that is, a single generator for its
V.2.20), is a good algebraic translation of the fact that
DVR admits
(cf.
Exercise
a
C\has a single, smooth branch through (0,0). The maximal ideal of
generated by just one element, as the reader may verify. [2.19]
cannot be
2.19. Prove that the fields of rational functions
and C2 of Exercise 2.18
are
(Exercise 2.16)
of the
curves
isomorphic and both have transcendence degree
C\
1 over k
(cf.
Exercise
it
reason why we should think of C\ and C2 as 'curves'. In fact,
be proven that the Krull dimension of the coordinate ring of a variety equals
transcendence degree of its field of rational functions. This is a consequence of
1.27).
This is another
can
the
Noether's normalization theorem,
2.20.
a cornerstone
of commutative algebra.
Recall from Exercise VI.2.13 that P^ denotes the 'projective space'
parametrizing lines in the vector space Kn+l. Every such line consists of multiples of
a nonzero vector (co,..., cn) G ifn+1, so that
P^ may be identified with the
quotient
-i
(in
relation
the set-theoretic
~
sense
of
§1.1.5)
of ifn+1
{(0,..., 0)} by
the equivalence
defined by
(c0,..., Cn)
~
(cq, ...,cn)
(3A
<^>
G
if*), (cq, ...,cn)
=
(Ac0,..., Acn).
The 'point' in P^ determined by the vector (co,..., cn) is denoted (co :...: cn);
these are the 'projective coordinates'12 of the point. Note that there is no 'point'
(0:...:0).
Prove that the function A^
P^ defined by
(ci,...,cn)
is
a
bijection. This function
similar functions, prove that
This is
the point,
so
by the point.
are not
(l:ci :...:cn)
is used to realize
P^
can
A^
be covered with
as
n
a
subset of
+1 copies of
PJ-. By using
AJ-, and relate
abuse of language. Keep in mind that the c$'s are not determined by
'coordinates' in any strict sense. Their ratios are, however, determined
a convenient
they
\-+
3. Geometric impossibilities
417
this fact to the cell decomposition obtained in Exercise VI.2.13.
out carefully the case n
2.) [2.21, VIII.4.8]
(Suggestion:
Work
=
2.21.
-i
notation
Let
as
F(xo,... ,xn)
G
be
a
homogeneous polynomial. With
is well-defined: it does not
cn)
P^
(co,..., cn) chosen for the
subset ofP£:
point (co
K[xo,... ,xn]
in Exercise 2.20, prove that the condition
G
:...:
r(F)
:=
points
{(co
(co
:...:
:...:
cn)
G
Prove that this 'projective algebraic set'
cn).
We
lF(co,... ,cn)
can
P£ | F(c0,..., cn)
can
=
0' for
a
the representative
then define the following
depend
on
=
be covered with
0}.
n +
1 affine
algebraic
sets.
'projective algebraic geometry' can be developed along
path taken in this section for affine algebraic geometry,
using 'homogenous ideals' (that is, ideals generated by homogeneous polynomials;
The basic definitions in
essentially the
same
§VIII.4.3) rather than ordinary ideals. This problem shows one way to relate
projective and affine algebraic sets, in one template example. [VIII.4.8, VIII.4.11]
see
3. Geometric impossibilities
Very simple considerations
in field
theory dispose easily of a whole class of geometric
a very long time—problems
which
in
solved
the
recent
past.
only
relatively
problems which had seemed unapproachable for
preoccupied the Greeks and were
These problems have to do
with the construction of certain
geometric
with 'straightedge and compass', that is, subject to certain very strict rules.
The practical utility of these constructions would appear to be absolute zero, but
figures
the intellectual satisfaction of being able to thoroughly understand matters which
stumped extremely intelligent people for a very long time is well worth the little
side trip.
by straightedge and compass. We begin with two points
O, P in the ordinary, real plane. You are allowed to mark ('construct') more points
and other geometric figures in the plane, but only according to the following rules:
3.1. Constructions
If you have constructed two points A, B, then you
them (using your straightedge).
If you have constructed two points A, B, then you
center at A and containing B (using your compass).
You
can
can
can
draw the line joining
draw the circle with
mark any number of points of intersection of any two distinct lines,
or circles that you have drawn already.
line and circle,
Performing
circles;
these actions leads to complex (and beautiful) collections of lines and
that we have 'constructed' a geometric figure if that figure appears
we say
as a subset of the whole
For example, here is
two
of its vertices:
picture.
a
recipe
to construct an
equilateral triangle with O, P
as
VII. Fields
418
(1)
draw the circle with center O, through P;
(2)
draw the circle with center P, through O;
(3)
let Q be any of the two points of intersection of the two circles;
(4)
draw the lines through O, P; O, Q; and P, Q.
Of
course
It
O, P, Q
are
vertices of
an
equilateral triangle.
with a specific sequence of basic moves
pleasant
a
or
constructing
given figure
geometric configuration. The enterprising reader
could try to produce the vertices of a pentagon by a straightedge-and-compass
construction, without looking it up and before we learn enough to thwart pure
is
exercise to try to
a
come up
geometric intuition.
The general question is to decide whether a figure can or cannot be constructed
this way. Three problems of this kind became famous in antiquity:
trisecting angles;
squaring circles;
doubling cubes.
For example,
one assumes one
and asks whether
for
some
has already constructed two lines forming an angle 6
lines forming 0/3. This is trivially possible
one can construct two
constructible 9:
we
just constructed
able to trisect the angle 7r; but
The
answer
equals the
side of
a
is
no.
can
Similarly,
an
angle of 7r/3,
so
evidently we
angles?
were
this be done for all constructible
it is not
possible
to construct a square whose area
(constructible) circle, and it is not possible to construct the
cube whose volume is twice as large as a cube with given (constructible)
area
of
a
side.
Field theory will allow us to establish all this13. Later we will return to these
constructions and study the question of the constructibility of regular polygons
with
given side: it is very
squares, hexagons; it is not
a
easy to construct
However, for the impossibility of squaring circles
we are
taking
on
faith.
equilateral triangles (we just did),
too hard to construct pentagons; but the question
we will use
the transcendence of n, which
3. Geometric impossibilities
of which
polygons
are
419
constructible and which
are not
connects
beautifully
with
deeper questions in field theory.
The three problems listed above have an illustrious history; they apparently
date back to at least 414 B.C., if one is to take this fragment from Aristophanes'
"The birds"
as an
indication:
METON: By measuring with
the circle becomes
The problems depend
For example,
one can
a
a square
on
straightedge there
with
a court
within.14
the precise restrictions imposed
on
the constructions.
angles if one is allowed to mark points on the
ruler); in fact, this was already known to Archimedes
trisect
(thus making it a
3.13).
Translating these geometric
straightedge
(Exercise
feasibility of
a
problems into algebra requires establishing the
few basic constructions.
one can construct a line containing a given point A and
given line £ (thus, £ contains at least another constructible point B). For
this, draw the circle with center A and containing B;
First of all,
perpendicular
to a
—if
this circle only intersects £
perpendicular to £:
at
£?, then the line through A and B is
the circle intersects £
—otherwise,
at a second
point C (whether A is
on
£
or
not):
and once C is determined, the line through A and perpendicular to £ may be found
by joining the two points of intersection D, E of the two circles with centers at £?,
resp., C, and containing C, resp., B:
14We thank Matilde Marcolli for the translation. For those who
dp-dcoi ^exprjaco xavovi 7ipoaxi'dei<; 'iva
6 xuxXoc yevTQTai aoi xexpaycovoc xav ^xeacoi ayopa
are less
ignorant than
we are:
VII. Fields
420
to a
Applying this
given line:
construction twice gives the line through
a
point A and parallel
4
which is often handy.
Judicious application of these operations allows
system into the picture.
reference
construct
point
perpendicular Cartesian
(1,0)
on
us to
bring
a
Cartesian
Starting with the initial two points O, P,
axes
centered at O, with P marking
we can
(say)
the
the 'x-axis':
—4-
and constructing a
X
(x, 0), Y (0,
=
two
points X
=
=
point A
y)
(x,0)
=
(x,y)
is
equivalent
and Y'
=
(y,0):
constructing its projections
equivalent to constructing the
to
onto the axes, and in fact it is
421
3. Geometric impossibilities
It follows that determining which figures are constructible by straightedge and
compass is equivalent to determining which numbers may be realized as coordinates
of a constructible point.
Definition 3.1. A real number
is constructible if the point
compass (assuming O
(0,0) and P
will denote by ^r C R the set of constructible real numbers.
with
r
straightedge and
Also,
we can
=
=
(r, 0) is constructible
(1,0), as above). We
j
the real plane with C, placing O at 0 and P at 1, and
identify
we
x+iy is constructible if the point (x, y) is constructible by straightedge
and compass. We will denote by ^c Q C the set of constructible complex numbers.
Summarizing the foregoing discussion, we have proved:
say
that
z
=
Lemma 3.2. A point
if
x +
iy
G
(x, y)
is constructible
^c, if and only if x,y
The next obvious
by straightedge and
compass
if and
only
G ^r.
question is what kind of
structure these sets of constructible
numbers carry, and this prompts us to look at a few more basic
straightedge-and-
compass constructions.
Lemma 3.3.
Likewise, ^c is
The subset
a
^r
C
subfield ofC,
R
of constructible numbers
fact ^c
^kW-
and in
Proof. The set *4 C R is nonempty,
need
B
R.
a
subfield of
so
in order to show it is
a
field, we only
by a nonzero
to show that it is closed with respect to subtraction and division
constructible number
see
is
=
(cf.
Exercise
II.6.2).
The reader will check that ^r is closed under subtraction (Exercise 3.2). To
that ^r is closed under division, let a,b e ^r, with 6^0. Since A
(0, a) and
(6,0) are constructible by hypothesis, we can construct C as the y-intersect
=
=
of the line through P
=
(1,0)
and parallel to the line through A and £?,
as
in the
following picture15:
In the picture we are assuming a > 0 and b > 1; the reader will check that other possibilities
lead to the same conclusion.
422
VII. Fields
triangles AOB and COP
Since the
that
are
similar,
we see
that C
=
(0,a/6);
it follows
is constructible.
a/b
a subfleld of C is an immediate consequence of the fact that
^r is a field, and the proof of this is left to the enjoyment of the reader (Exercise 3.7).
The fact that ^c
^k(i) is a restatement of the fact that x + iy G ^c if and only
The fact that ^c is
if
x
and y
are
in ^fc.
We may view ^ and ^c as extensions of Q, drawing a bridge between constructibility by straightedge and compass and field theory: we will be able to
understand constructibility of geometric figures if we can understand the field extensions
we can
Happily,
understand these extensions!
3.2. Constructible numbers and quadratic extensions. Our goal is to prove
following amazingly explicit description of ^ (immediately implying one for
the
%)•
Theorem 3.4. Let 7 G R. Then 7 G ^r
5\,...,5k such that \/j 1,..., k
i/
and on/y
2/
£/iere e:ris£ rea/ numbers
=
u...,6j):Q(6u...,6j-1)]=2
and 7 G
Q(5i,... ,4).
In other words, 7 G R is constructible if and only if it
a sequence of (real) quadratic extensions over Q.
can
be placed in the top
Since ^c
^k(0» tne
same statement holds for constructible complex numbers, including i in the list of
5j (or simply allowing the 5j to be complex numbers; cf. Exercise 3.9).
The proof of this theorem is as explicit as one can possibly hope. From a given
straightedge-and-compass construction of a point (x, y) one can obtain an explicit
field of
sequence
=
x and y belong to the extension
from any element 7 of any such extension one can obtain
construction of a point (7,0) by straightedge and compass.
S\,...,5k
as
in the statement, such that
Q(5i,..., 5k)', conversely,
an
explicit
'geometry to algebra' direction. A configuration
of points, lines, and circles obtained by a straightedge-and-compass construction
Proof. Let's first argue in the
3. Geometric impossibilities
423
by the coordinates of the points and the equations of the lines
and circles. Suppose that at one stage in a given construction all coordinates of all
points and all coefficients in the equations of lines and circles belong to a field F\
we will say that the configuration is defined over F.
may be described
Then we claim that for every object constructed at the next stage, there exists
number 5 G R, of degree at most 2 over F, such that the new configuration is
defined over F(5). The 'only if part of the theorem follows by induction on the
a
number of steps in the construction, since at the beginning the configuration
is, the pair of points O (0,0), P (1,0)) is defined over Q.
=
(that
=
Verifying our claim amounts to verifying it for
straightedge-and-compass constructions. The reader
the basic operations defining
will check (Exercise 3.5) that
the point of intersection of two lines defined over F has coordinates in F and that
lines and circles determined by points with coordinates in F are defined over F. So
5
=
1 works in all these cases.
C, assume that £ is not parallel to
entirely analogous otherwise) and that it does meet C;
For the intersection of a line £ and a circle
the 2/-axis
(the argument
is
let
y
=
mx + r
be the equation of £ and let
x2 +
y?
+ ax +
by
+ c
=
0
We are assuming that a,b,c,m,r G F.
Then the xpoints of intersection of £ and C are the solutions of the equation
be the equation of C.
coordinates of the
x2 +
(mx
+
r)2
b(mx
+ ax +
+
r)
+ c
=
0.
The 'quadratic formula' shows that these coordinates belong to the field
where D is the discriminant of this polynomial: explicitly,
D
=
(2mr
+ bm +
a)2
4(ra2
-
+
l)(r2
+ br +
c),
but this is unimportant. What is important is that D G F; hence 5
our
F(y/~D),
=
\fl)satisfies
requirement.
For the intersection of
(distinct)
two
circles defined
over
F, nothing
new
is
needed: if
x2
+
y2
+ aix +
biy
+ ci
=
0,
x2 + y2 + a2x + b2y + c2
0
the
two
shows
that
their points of intersection
circles, subtracting
equations
coincide with the points of intersection of a circle and a line:
=
are two
x2 +
(oi
with the
same
conclusion
y2
-
as
+ axx +
a2)x
+
(6i
biy
-
+ cx
b2y)
in the previous
+
=
0,
(ci
-
c2)
=
0,
case.
This completes the verification of the 'only if part of the theorem.
To prove that every element of an extension as stated is constructible, again
argue by induction: it suffices to show that if (i) 5 G R, (ii) all elements of F are
constructible, and (iii)
r
=
52
e
F, then 5 is constructible (note that in order
to
424
VII. Fields
construct
an
element of degree 2
the discriminant of its minimal
we can
picture
'take square roots' by
(if
a
over
F, it suffices
to construct the square root of
polynomial). Therefore,
all
straightedge-and-compass
we
have to show is that
construction. Here is the
1):
r >
(r, 0) is constructible, so is B (—r,
0); C is the midpoint of the segment BP
are
Exercise
constructible,
(midpoints
3.1); the circle with center C and containing
P intersects the positive y-axis at a point Q
(0,5), and elementary geometry
If A
=
=
=
shows that 52
=
Therefore 5 is constructible, concluding the proof of the theorem.
r.
For example,
one can construct an
53°
cos.
=
angle of 3°: allegedly
hy/3 + i)J5 + y/h + 4(>/6
V2)(V$
lb
-
-
o
and this
expression shows that
53°
cos.
that is
(by
Theorem
constructed,
Here
1),
so
is the
is another
3.4),
cos
e
Q(V% >/3, V5)(\£+^);
3° is constructible. Of
course once
A
=
(cos 0,0)
angle 9:
example of
a
'constructive' application of Theorem 3.4:
Example 3.5. Regular pentagons
are
constructible.
is
3. Geometric impossibilities
425
Indeed, it suffices to construct the point A
cos(27r/5) satisfies
7
(cos(27r/5),0),
=
and it
so
happens that
=
472
(*)
(in fact,
+
27
7 is half of the inverse of the
-
1
=
golden
0
ratio:
7
^~l).
=
It follows that
7
Q(V/5), and hence it is constructible by Theorem 3.4. In fact, the proof shows
how to construct y/b, so the reader should now have no difficulty producing a
straightedge-and-compass
We will
come
construction of
a
regular pentagon (Exercise 3.6).
back to constructions of regular polygons
once we
thoroughly (cf. §7.2). By the way,
out that 7
cos(27r/5) should satisfy the identity (*)? This
complex number arithmetic; see Exercise 3.11.
field theory
a
little
more
=
3.3. Famous impossibilities. Theorem 3.4 easily settles
in §3.1, via the following immediate consequence:
j
have studied
how does one
figure
is easy modulo
some
the three problems
mentioned
Corollary 3.6. Let 7 G % be
a
constructible number. Then
[Q(7) : Q]
is a power
of2.
Proof. By Lemma 3.3 and Theorem 3.4, there exist 5\...,5k Gl such that
7eQ((51,...,4,i),
and each Sj has degree < 2
Proposition 1.10 shows that
over
Q(5\,...,Sj-i). Repeated application
of
[Q{51,...,5kji):Q]
is
a power
of 2, and since
QCQ(7)CQ(Ji,...,ffc,i),
the statement follows from
In
Corollary
1.11.
particular, constructible real numbers
must
satisfy
the
Consider the problem of trisecting an angle. We know
angle of 60° (this is a subproduct of the construction of an
Constructing an angle of 20° is equivalent (Exercise 3.10)
complex number 7 on the unit circle and with argument tt/9:
to
same
we
condition.
can
construct
an
equilateral triangle).
constructing the
VII. Fields
426
60°
Now
79
=
1; that is, 7 satisfies the polynomial
*9 + l
It does not
satisfy t3
=
(*3
l)(*6-*3
+
+ 1; therefore its minimal
t6
-
ts
However, this polynomial is irreducible
Eisenstein's criterion), so
+
l).
polynomial
over
Q
is
a
factor of
+ 1.
over
[Q(7) : Q]
Q (substitute
=
t
1, and apply
t
1—?
6.
By Corollary 3.6,
which cannot be trisected.
7 is not constructible. Therefore there exist constructible
Similarly,
ity of
angles
cubes cannot be doubled because that would imply the constructibil-
\/2, which has degree 3
over
Q, contradicting Corollary 3.6.
Squaring circles amounts to constructing 7r, and tt is not even algebraic
(although, as we have mentioned, the proof of this fact is not elementary), so this is
also not possible.
As another example, constructing
a
7-th
(complex)
root of
a
regular 7-gon would
amount to
constructing
1, £:
By definition, £ satisfies
t7 -i
=
(t-i)(t6
+
t5
+
--
+ t +
£ 7^ 1, £
satisfy the cyclotomic polynomial t6 -\
(Example V.5.19); hence again we find that £ has degree 6
iy,
+ 1. This is irreducible
must
as
implies that the regular 7-gon
Of course '7' is not too
polynomial of degree
p
Q, and Corollary 3.6
straightedge and compass.
positive prime integer, the cyclotomic
over
cannot be constructed with
special: if p
1 is irreducible
is a
(Example
V.5.19
again);
hence
[Q(Cp):Q]=p-i,
where
£p
is the
complex p-th
root of 1 with argument
reveals that if p is prime, then the
regular
2n/p. Therefore, Corollary
p-gon can be constructed
a power of 2. This is even more restrictive than it looks at
first,
only if
since if p
p
=
2k
3.6
1 is
+ 1
Exercises
427
is prime, then
form22 +1
necessarily k
are
is itself
a power
of 2
Primes of the
(Exercise 3.15).
called Fermat primes:
3, 5, 17, 257, 65537
(the 'next one', 232 + 1 4294967297 641 6700417, is not prime; in fact, no one
knows if there are any other Fermat primes, and yet some people conjecture that
=
there
=
infinitely many).
These considerations alone do
are
regular
us
if p is a Fermat prime, then the
straightedge and compass; but they do tell
24 + 1
would be p
17, and it just so happens
not tell us that
p-gon can be constructed with
that the next
case to
consider
=
=
that
\f\7 1
2tt
-
+
_
\/2^34+ 6^17+ ^2(^17
-
l)>/l7- y/Vf
-
8^2^17
+ vT? + y/2 \/\7
-
y/Vf
by Gauss
at age 19. Therefore (by Theorem 3.4) the 17-gon is constructible
and
compass. As we have announced already, the situation will be
by straightedge
further clarified soon, allowing us to bypass heavy-duty trigonometry.
as noted
Exercises
A, B are constructible, then the midpoint of the segment AB
is also constructible. Prove that if two lines £\,£2 are constructible and not parallel,
3.1. > Prove that if
then the two lines bisecting the angles formed by £\ and £2
are
also constructible.
[§3-2]
3.2. > Prove that if a, b
3.3. Find
an
constructible numbers, then
are
explicit straightedge-and-compass
so
is
b.
a
construction for the
[§3.1]
product of two
real numbers.
3.4. Show how to square a
triangle by straightedge and
compass.
3.5. > Let F be a subfleld of R.
(xa,Va), B
(xb,Vb) be two points in
Prove that the line through A, B is defined over F
Let A
=
=
R2, with xa,VAj %B,yi3 F(that is, it admits an equation
with coefficients in F).
Prove that the circle with
Let ^1,^2 be two distinct,
£if]£2' Prove that x,y
center at A and
nonparallel lines
containing B is defined
in
R2,
defined
over
over
F.
F, and let (x, y)
=
e F.
[§3.2]
3.6. > Devise
[§3.2]
a way to construct a
regular pentagon with straightedge and
compass.
VII. Fields
428
3.7. > Identify the (real) plane with C, and place O, P at 0,1 G C. Let ^c be the
set of all constructible points, viewed as a subset of C. Prove that ^c is a subfield
ofC. [§3.1]
3.8. For 5 G C, 5 ^ 0, let 9s be the argument of 5 (that is, the angle formed by
the line through 0 and 5 with the real axis). Prove that 5 G ^c if and only if \S\,
cos 0$, sin 9s are all constructible real numbers.
Let 71,... ,7fc G ^c be constructible complex numbers, and let if be the
3.9.
>
field
Q(7i,... ,7/c)
Prove that 5 G
^c. Let 5 be
C
complex number such that [K(5)
a
:
K]
=
2.
%•
Deduce that there
are no
irreducible polynomials of degree 2
^c and that
over
^c is the smallest subfield of C with this property. [§3.2]
3.10.
>
Prove that if two lines in any constructible configuration form
of 9, then the line through O forming
angle
an
angle of 9 clockwise with respect
an
to the
line OP is constructible.
angle 9
Deduce that an
cos
9 is constructible.
3.11.
g
>
Verify
sin(27r/5),
=
that 7
cos(27r/5)
=
note that z
3.12. Prove that the
is constructible
anywhere
plane if and only if
in the
[§3.3]
=
satisfies the relation
angles of
1° and 2°
(*) given
in
§3.2.
(If
1.) [§3.2]
7 + icr satisfies z5
are not
constructible.
(Hint:
Given what
know at this point, you only need to recall that there exist trigonometric formulas
for the sum of two angles; the exact shape of these formulas is not important.) For
what integers n is the angle n° constructible?
we
3.13. > Prove that
a
=
|
in the
following picture:
O
This says that angles
can
be trisected if
we
1
allow the
P
use
of
straightedge with markings (here, the
construction works if
distance of 1
this construction
on
the
ruler). Apparently,
was
a
'ruler', that is,
we can
mark
a
a
fixed
known to Archimedes.
[§3-1]
3.14. Prove that the
regular 9-gon
is not constructible.
3.15. > Prove that if 2k + 1 is prime, then k is
4.
Field
extensions,
It is time to continue
here
are
a power
of 2.
[§3.3]
II
our survey
of different flavors of field extensions. The keywords
splitting fields, normal, separable.
4. Field extensions, II
429
Splitting fields
4.1.
and normal extensions. In
algebraic closure k of any given field k:
§2
we
have constructed the
every polynomial in
k[x]
factors
as
a
product of linear terms (that is, 'splits') in k[x], and k is the 'smallest' extension
of k satisfying this property.
Here is an analogous, but more modest, requirement: given a subset & C k[x]
of polynomials, construct an extension k C F such that every polynomial in 3?
splits as a product of linear terms over F, and require F to be as small as
possible with this property. We then call F the splitting field for 3?. In practice we
will only be interested in the case in which & is a finite collection of
polynomials /i(x),... ,/r(x); then requiring that each fi(x) splits over F is equivalent to
requiring that the product fi(x)
fr{x) splits over F.
That is, for
it is not restrictive to
our purposes
single polynomial f(x)
G
assume
that & consists of
a
k[x\.
Definition 4.1. Let k be
The splitting field for f(x)
a
field, and let f(x)
k is
over
an
G
k[x]
be
a
polynomial of degree d.
extension F of k such that
d
=cY[(x-ai)
f(x)
2=1
splits
f(x)
in
F[x],
and further F
k(a\,...,a^)
=
is
generated
over
k by the roots of
in F.
a splitting field exists: given f(x) G
k[x], the subfield
k by the roots of f(x) in k satisfies the requirements in
have written the splitting field, and this is justified by the
Note that it is clear that
F C k generated
over
Definition 4.1. But
we
uniqueness part of the
this extension can be.
following
Lemma 4.2. Let k be
f(x)
over
In
f(x)
=
a
field,
basic observation, which also evaluates 'how big'
and let
f(x)
k[x\. Then the splitting field F for
[F : k] < (deg/)!.
of fields and g(x) G k'[x] is such that
G
k is unique up to isomorphism, and
fact, if t : k'
t(g(x)),
k is any isomorphism
then
t
extends to
k' to any splitting field of
f(x)
isomorphism of
an
over
any
splitting field of g(x)
over
k.
Proof. We first construct explicitly a splitting field and obtain the bound on the
degree mentioned in the statement. Then we prove the second part, which implies
uniqueness up to isomorphism.
a splitting field and the given bound on the degree are an
of
our
basic
simple extension found in Proposition V.5.7. Arguing
easy application
assume
the
inductively,
splitting field has been constructed and the bound has
The construction of
been proved for all fields and all polynomials of degree
irreducible factor of f(t) over /c; then
is
an
(the
extension of
coset a of
t)
degree degq
<
and therefore
deg/,
a
in which
linear factor
(deg/
q(x) (and
x
a.
1).
hence
Let
f(x))
q(t)
has
The polynomial
be any
a root
h(x)
:=
VII. Fields
430
f(x)/(x a)
h(x), and
e
F'[x\ has
(deg/
degree
1);
therefore
a
splitting field F
exists for
[F:F']<(deg/-l)!.
The factors of
f(x)
follows that F is
over
are
(x
and the factors of
h(x),
which
are
linear; it
=
[F: Ff][Ff : k]
<
(deg/)(deg/
1)!
-
=
(deg/)!
stated.
To prove that isomorphisms
fields
as
k'
/c
a
a)
splitting field for f(x) and
a
[F:k]
as
F
C /c.
Since the
morphism of
roots of
g(x)
extensions
in
G,
we
are
isomorphic
i :
/c
G
over
/c'. Since G is generated
=
f(x)
k(ai,...,ad)
=
t(g(x))
C
over
k' by the
fc,
in k. Therefore L
:=
is
t(G)
to
i
L).
By definition, Q(i) is the splitting field of x2
field
for the same polynomial, over R.
splitting
Example
the
t
/c extend to isomorphisms of splitting
isomorphism and splitting field G. Applying this observation
and to the identity k
k proves the statement (as both G and F
independent of the chosen
to the given
k'
have
the roots of
are
:
a
L(G)
where the c^'s
t
splitting field for g(x) and consider the composition
extension /c' C G is algebraic, by Lemma 2.8 there exists
stated, let G be
4.3.
+ 1 over
Q, and C
is
j
1 over Q is generated by £ := e27rl/8:
Example 4.4. The splitting field F of x8
1
the
roots
of
x8
are
all
the
roots
of 1, and all of them are powers
8-th
indeed,
ofC:
In fact, £ is
F
Q(C) is
=
a root
of the polynomial x4 + 1, which is irreducible
'already' the splitting field of x4
+ 1. The
degree of
over
F
Q; therefore
Q is
over
Q]=4,
way less than the bounds
8!, 4! obtained
in Lemma 4.2.
splitting field (even) better,
(? is in F, and
y/2 C + C7;
Q(i, y/2). Conversely, ( ^(l + i) e Q(i, y/2).
Therefore, the splitting field of x4 +1 (a.k.a. the splitting field of x8 1) is Q(i, y/2).
Analyzing Q C Q(i, y/2) (as we did for the extension in Example 1.19) shows that
its group of automorphisms over Q is Z/2Z x Z/2Z.
To understand this
so
is
=
thus F contains
note that i
=
=
j
4. Field extensions, II
431
Example 4.5. Variation
of x4 + 1 we consider x4
1.
on the theme: x4
1: this
x4-l
=
The situation
polynomial factors
(x_i)(x
and it follows that the splitting field is the
+
i)(x2
same as
+
changes if instead
Q,
over
i)5
for x2 + 1, that is, just
Q(i).
j
Example 4.6. Variation on the theme: x4 + 2. The overoptimistic reader may
1 over
now hope that the difference between the splitting fields of x4 + 1 vs. x4
Q is just due to the fact that the first polynomial is irreducible over Q and the
second is not. This example will nip any such guess in the bud. With notation as
in Example 4.4, the roots of x4 + 2 are
Therefore, with K
=
Q( v^C v^C3, v^C5, v^C7)
K C
On the other hand,
y/2
Q(C, </2)
=
=
the splitting field of x4 + 2,
Q(t, V% </2)
(^/2C)3/(v/2C3)
G
(^2C)/C
=
Q(t, </2).
K\hence
^2
£
^(1 i)GK;
conclusion is that the splitting field of x4 + 2 equals K
=
hence
+
computation
of x4
+ 2 is
(Exercise 4.3)
certainly
not
=
shows
[K : Q]
isomorphic
%
(^2C)2/>/2
=
if. Therefore,
=
to the
=
Q(t, </2)
Q(i, \/2). A
8 and in particular the
splitting field of x4
C
K; hence
if, and the
G
simple degree
splitting field
+ 1.
j
Examples such as these have always left me with the impression that field
theory must be rather mysterious: innocent-looking variations on the parameters
of a problem may cause dramatic changes in the corresponding field extensions. On
plus side, this hints that field theory can indeed be a very precise tool in (for
example) the study of polynomials.
Splitting fields will play an important role in the rest of the story. They are
even more special than they may appear to be at first: it turns out that not only do
they split the given polynomial, but they also automatically split any irreducible
polynomial which dares touch them with a root. This makes splitting fields normal
the
extensions:
Definition 4.7. A field extension k C F is normal if for every irreducible
polynomial f(x) G /c[x], f(x) has a root in F if and only if f(x) splits as a product
linear factors
over
Theorem 4.8. A
splitting field of
F.
of
j
field extension k C F is finite
polynomial f(x) G k[x\.
and normal
if and
only
if F
is the
some
Proof. Assume k C F is finite and normal.
Then F is
finitely generated:
F
=
k(ai,... ,ar) with c^ algebraic over k. Let Pi(t) be the minimal polynomial of c^
over k. As F is normal over /c, each Pi(t) splits completely over F, and hence so
f(t) Pi(t) -pr(t). It follows that F is the splitting field of f(x).
Conversely, assume that F is a splitting field for a polynomial f(x) G /c[x], and
let p(x) G k[x] be an irreducible polynomial, such that F contains a root a of p(x).
Viewing F as a subfield of the algebraic closure /c, let (3 G k be any other root of
p(x); we are going to show that (3 G F. This will prove that F contains all roots of
p(x), implying that k C F is normal, and k C F is finite by Lemma 4.2.
does
=
-
-
VII. Fields
432
By Proposition 1.5, there
exists
an
isomorphism
t :
k{(3) extending the
F{(3) C fc, viewed
k(a)
k and sending an/j. We also consider the subfleld
identity
as an extension of k(/3). Putting everything in one diagram:
on
Fc
/c(a)c
<^ I
W
> k
>F(W
>k
join the
two copies of k on the right by the identity map, the
does
not commute: i sends a G k in the top row to (3 G k in the
corresponding diagram
bottom row.) Now observe that F may be viewed as the splitting field of f(x) over
k(a): indeed, it contains all the roots of f(x) and is generated over k (and hence
(Note:
If
we
k(a)) by these roots. By the same token, F((3) is the splitting field for f(x)
k{(3). By Lemma 4.2, t extends to an isomorphism i' : F
F{(3). Since t
over
over
/c, i' is an isomorphism of k-vector spaces; in particular,
dim/- F{fi) (keep in mind that splitting fields are finite extensions).
restricts to the
dim/-
F
=
identity
Now consider the
given by
on
different
/c-linear map of k-vector spaces i
kCFC
Since F and
an
:
F{p) simply
F
the inclusion within k:
.F(/?)
F(P)
k-vector spaces of the
are
C k.
same
finite dimension,
i must also be
isomorphism. In other words,
[F((3)
that is, /?
:
F]
=
1;
G F as needed.
Remark 4.9. This
D
argument is somewhat delicate. With notation
as
in the
proof,
the diagram
/c(a)c
> F
r
r
fc(/?)C
is commutative, while the
>F(J3)
diagram
/c(a)c
>F
*
r
fcC9)C
is not commutative if
(3 ^
a.
Indeed,
>F(J3)
i
sends
a to
/?, while
i sends
key point in the argument is the observation that if there exists
between
V
draw this conclusion; cf. Exercise VI.6.5.
The
isomorphism
injective linear map
V,W, then every
isomorphism. Finite dimensionality is necessary
two finite-dimensional vector spaces
W must be an
a to a.
one
in order to
j
Example 4.10. If a complex root of an irreducible polynomial p(x) G Q[x] may
be expressed as a polynomial in i and v^2 with rational coefficients, then all roots
4. Field extensions, II
433
p(x) may be expressed likewise in terms of i and y/2. Indeed, we have checked
(Example 4.6) that Q(i, \/2)is a splitting field over Q; hence it is a normal extension
of
ofQ.
j
Separable polynomials. Our intuition may lead us to think that if a
as a product of linear factors and we are not 'purposely' repeating
4.2.
polynomial factors
of the factors
one
(as
in
(x
l)2(x
2)),
then these will be distinct. For example,
surely irreducible polynomials necessarily split
as
products of distinct factors
in
an
algebraic closure, right? Wrong.
4.11. Let p be a
Example
over
prime, and consider the field ¥p(t) of rational functions
Then the polynomial
Wp.
xp
is irreducible:
in
Fp[£]),
in
an
by Eisenstein's
hence in
criterion it is irreducible in
Fp (£)[#] by Proposition
modestly
a
L[x];
that is,
has multiplicity p
u
u
is prime
of this polynomial
Fp [£][#] (since (t)
be
a root
¥p(t) (for example
xp-t
in
V.4.16. Let
L could be the algebraic closure of ¥p(t),
field
for
the
splitting
polynomial). Then (Exercise 4.8)
extension L of
or more
-te¥p(t)[x]
=
(x- u)p
as a root
of
f(x).
In other words, the minimal polynomial over ¥p(t) of u G L vanishes p times
at u, and there is nothing to do about this: no smaller power of (x
u) than
in
xp
t
has
coefficients
smaller
would
a
nontrivial
power
give
(x u)p
¥p(t) (a
factor of xp
t is irreducible).
j
t, and xp
=
We have always found this example difficult to visualize, because of intuition
developed in characteristic 0, and as we will see, no such pathology can occur in
characteristic 0.
Definition 4.12.
Let k be
a
field.
A polynomial f(x) G k[x] is separable if it
is inseparable if it has multiple
multiple factors over its splitting field; f(x)
factors over its splitting field.
has
no
j
t in Example 4.11 is inseparable. We suppose the
Thus, the polynomial xp
terminology reflects the fact that we cannot 'separate' its roots: they come clumped
together like quarks in a proton, and we cannot take them apart.
Of
course
Definition 4.12 does not depend
on
the chosen splitting field, since
isomorphism (Lemma 4.2). In fact, to detect separability,
we can use any field in which the polynomial splits as a product of linear factors
(by Exercise 4.1). The first, somewhat surprising, observation about separability is
that we can in fact detect it without leaving the field of coefficients of f(x). This
fact uses a notion borrowed from calculus: for a polynomial
these
are
unique
up to
f(x)
we
denote by
f'{x)
=
a0 + aix +
a2x2
-\
h
anxn,
the 'derivative'
f'(x)
=
a\+
2a<ix H
h nanxn~x.
VII. Fields
434
Of
course
this is
arbitrary fields.
purely formal operation; no limiting process is
Still, all expected properties of derivatives hold,
should check: for example,
Lemma 4.13. Let k be
only if f(x) and f'{x)
(fg)'
divisor of
+
fg'
field, and let f(x)
relatively prime.
a
are
and
f(x)
fg
=
By definition, f(x) and f'{x)
common
at work over
a
are
the reader
usual.
as
G
as
Then
k[x\.
f(x)
is
separable if and
relatively prime precisely when the greatest
Note that if k C F, the gcd of f(x) and
is 1.
f'{x)
is the same whether it is considered in k[x] or in F[x]: for example because it
be computed by applying the Euclidean algorithm (§V.2.4), and this proceeds
in exactly the same way whether it is performed in k[x] or in F[x],
f'{x)
can
Proof. First
a
assume
that
is not
f(x)
separable. Then f(x) has
a
root in
multiple
splitting field F; that is,
f(x)
for
some a G
F, g(x)
F[x],
G
f'{x)
and it follows that
(x
gcd(/(a:),/'(*)) ^
1 in
are not
m(x
a) is
F[x];
m >
-
(x- a)mg(x)
2. Therefore
ar~lg{x)
(x
-
<*)"*</(*),
factor of
f'(x) in F[x\. Therefore
k[x]; that is, /(*), f'(x)
and
f(x)
gcd(/(x), f'(x)) ^
a common
hence
+
1 in
relatively prime.
Conversely,
gcd(/(x),/'(#)) ^ 1,
assume
irreducible factor
we
=
and
=
(x
a)
in the
so
f(x)
and
f'{x)
have
algebraic closure k of k. Write f(x)
a common
(x
=
a)h(x);
have
f\x) h(x)
=
and it follows that
x
a
divides
h(x)
+
(x-a)h\x),
since it divides
f'{x).
But then
(x-a)2\f(x);
hence
f(x)
is
inseparable.
For example, the polynomial xp
t of
Example
4.11 may be
inseparable without invoking splitting fields: the derivative of xp
characteristic p, and gcd(xp
t,0) xp t ^ I.
seen to
t equals
be
pxp~x
=
0 in
=
This
example captures
Lemma 4.14. Let k be
polynomial. Then f'(x)
a
=
one
of the key features of inseparability:
field,
and let
f(x)
G
k[x]
be
an
inseparable irreducible
0.
f(x) is inseparable, f(x) and f'{x) have a common irreducible factor
q(x) by Lemma 4.13; but as f(x) is itself irreducible, q(x) must be an associate
of f(x), and in particular its degree is larger than the degree of f'{x) if f'{x) ^ 0.
As q(x) | f'(x), the only option is that f'{x) =0.
Proof. Since
This
already tells
characteristic 0:
us
that irreducible polynomials
in characteristic 0, the derivative of
are
a
necessarily separable in
polynomial
nonconstant
4. Field extensions, II
435
clearly cannot vanish. In fact, Lemma 4.14 gives
inseparable, irreducible polynomials must look like.
us
a
precise picture of what
If
n
f(x)
=
^ aiX%
is irreducible and inseparable, then the characteristic of the field
prime p, and by Lemma 4.14 we must have
must be a
positive
n
/,(a:)
^taia:i-1=0;
=
i=0
0 for all i. Now iai
0 is automatic if i is a multiple of p, and it
0 for all the indices i which are not multiples of p. Therefore, the only
implies ai
nonvanishing coefficients in f(x) must be those corresponding to indices which are
multiples of p:
that is, iai
=
=
=
f(x)
therefore, f(x)
=
apxp
a0 +
must in fact be a
+
a2px2p
polynomial
picture further by working out Exercise 4.13 and
This leads
we
us to a
have not yet
+ ...;
(The reader
following.)
in xp.
precise result linking separability and
a
will refine this
flavor of fields that
into.
run
Definition 4.15. Let k be
homomorphism is the map k
a
field of characteristic p > 0.
x i—?
xp.
The Frobenius
k defined by
j
The Frobenius homomorphism may not look like a homomorphism of rings, but
(Exercise 4.8). It must be injective, as is every nontrivial ring homomorphism
from a field; but it is not necessarily surjective.
it is
Definition
4.16. A field k is
perfect
if char k
=
0
or
if char k > 0 and the Frobenius
homomorphism is surjective.
Proposition
j
a field. Then k is perfect if and only if all irreducible
separable.
4.17. Let k be
polynomials in
k[x]
are
Proof. We will prove that irreducible polynomials over a perfect field
leaving the other implication to the reader (Exercise 4.12).
are
We have already noted that irreducible polynomials are separable
zero. In positive characteristic p, we have observed that
characteristic
separable,
over
an
inseparable irreducible polynomial must be of the form
m
f(x)
=
J2*i-(xpY-
Since Frobenius is surjective, there exist b{ such that bP
m
six)
=
y, w*
i=0
where
g(x)
=
polynomial.
E(6^)p
i=0
/
=
m
E fc**
\i=0
a{. Thus
\P
=
s(*)p>
J
and keeping in mind that the Frobenius map is a
contradicts the irreducibility of /(#), so there is no such
^2b{Xl
homomorphism. But this
m
=
=
fields of
VII. Fields
436
Corollary 4.18. Finite
are
polynomials
fields
separable.
Proof. The Frobenius
perfect. Therefore,
over
finite
fields,
irreducible
injective (because it is a homomorphism of fields), so
by the pigeon-hole principle. Therefore finite fields
and
the
second
perfect,
part of the statement follows from Proposition 4.17.
it is surjective
are
are
map is
finite fields,
over
The reader now sees why we have chosen the somewhat unusual field ¥p(t)
Example 4.11: we need a field of positive characteristic, but no example of
irreducible inseparable polynomial can be found over Fp, by Corollary 4.18; to
in
example,
concoct an
in
an
we
(that is,
element
need to enlarge the field so that it is infinite and to throw
t) for which there is no p-th root, that is, which is not in
the image of Frobenius.
Separable extensions and embeddings in algebraic closures. The
terminology examined in the previous section extends to the language of field
extensions. If k C F is an extension and a G F, we say that a is separable over k if the
minimal polynomial of a over k is separable; a is inseparable otherwise.
4.3.
Definition
separable
4.19. An
algebraic field
extension k C F is
separable if
every
F is
a G
k.
over
j
With this terminology, Proposition 4.17 may be restated in the
following
more
impressive form:
Proposition 4.20. A field k
is
perfect if
and only
if
every
algebraic
extension
of
k is separable.
In particular, algebraic extensions of Q (or any field of characteristic
of every finite field are necessarily separable.
zero)
and
The separability condition is exceedingly convenient, and we will essentially
adopt it henceforth: all extensions we will seriously consider will be separable. One
convenient feature of separable extensions is the following alternative description
of the separability condition.
We have
embedded in
be done
F
k
(Lemma 2.8)
seen
an
that every algebraic extension k C F may be
algebraic closure k
in many different ways. If
extending the identity
on
fc, and we pointed out that this can in general
finite, the number of different homomorphisms
C
k is denoted by
[F:k]8.
Definition 4.21. This is the separable degree of F
over
k.
j
It is clear that this number is independent of the chosen algebraic closure k C k.
What does it have to do with
Lemma 4.22. Let k C
the number
[k(a) : k]s
<
of
separability?
simple algebraic extension. Then [k(a) : k]s equals
of the minimal polynomial of a. In particular,
with
equality
if and only if a is separable over k.
k],
k(a)
be
a
distinct roots in k
[k(a)
:
4. Field extensions, II
437
a rehash of the proof of
k
Corollary
extending idfc the image t(a), which
k(a)
must be a root of the minimal polynomial of a. This correspondence is injective,
since t(a) determines t (as t extends the identity on k). To see it is surjective, let
Proof. The
proof
is
essentially (and
1.7. Associate with each
not
by coincidence)
i:
(3 G k be any other root, and consider the extension k{(3) C /c; by Proposition 1.5
there is an isomorphism k(a)
k(/3) sending a to /?, and composing with the
C
D
k defines the corresponding l, as needed.
embedding k((3)
Thus, the 'separable degree' does detect separability, for simple extensions.
Further, it is multiplicative over successive extensions:
Lemma 4.23. Let k C E C F be
and only
if
both
[F E]s, [E k]s
:
:
algebraic extensions. Then [F
finite, and in this case
:
k]s
is
finite if
are
[F:k}s
=
[F:E}s[E:k}s.
Proof. Different embeddings of E into k extend to different embeddings of F into /c,
k extending the identity on E extend
by Lemma 2.8; and embeddings of F into E
a fortiori the identity on k. Therefore, if any of
[F : E]s, [E : k]s is infinite, then
=
so
is
[F : k]s.
implication, it suffices to prove the stated degree formula. But
every embedding
extending the identity on k may be obtained in two steps:
first extend the identity to an embedding E C /c, which can be done in [E : k]s ways;
E to an embedding F C E
and then extend the chosen embedding E C k
k,
which can be done in [F : E]s ways. There are precisely [F : E]S[E : k]s ways to
do this, as stated.
For the
converse
F C k
=
Lemmas 4.22 and 4.23 allow
in the
(i)
F
(ii) k
=
k(a\,...,ar),
C F is
(iii) [F:k]8
separability of finite
algebraic closure:
us to recast
light of counting embeddings
Proposition 4.24. Let k C
the following are equivalent:
=
F be a
in
an
finite
extension.
where each ai is separable
Then
over
extensions
[F : k]s
<
entirely
[F : k],
and
k;
separable;
=
[F:k].
Proof. Since F is finite
over
/c, it is finitely generated. Let F
=
k(ai,..., ar).
Then using Lemma 4.23, Lemma 4.22, and Proposition 1.10,
[F : k]8
=
<
=
[fe(ai,...,ar_i)(ar) : k(au
...
,ar_i)]s
[fe(ai,..., ar_i)(ar) : k(au
...
,ar-i)]
[fe(ai) : k]8
[Kai) : *]
[F:k}.
This proves the stated inequality.
(i) => (iii): If each oti is separable over fc, then it is separable over the field
fc(ai,..., Oii-\) (Exercise 4.15), so the inequality is an equality by Lemma 4.22.
VII. Fields
438
(iii)
Assume
(ii):
=>
[F : k]a
=
and let
[F : k],
a G
F. We have fc C
fc(a)
C
F;
hence by Lemma 4.23
[F
:
fc(a)]fl[fc(a) : k]s
Since both separable degrees
=
are
[F : k]s
[F : k]
=
less than
or
[F : k(a)][k(a) : A;].
=
equal
plain counterparts, the
to their
equality implies
[k(a) : k]8
that
a
is
separable, by Lemma
proving
separable according to Definition 4.19.
(ii)
=>
(i)
[k(a) : fc],
=
Therefore the extension k C F is
4.22.
is immediate from Definition 4.19, since finite extensions
are
finitely
generated.
For example, if a is separable over fc, then every (3 G k(a) is separable. This
seem rather mysterious from the definition alone: why should the fact that
the minimal polynomial of a has distinct roots in k imply the same for the minimal
would
polynomial of (31 It does, by virtue of Proposition 4.24.
Remark 4.25. While
divides
[F k]s
-^sep over
that
:
field
we
have only proved
in
[F k]:
:
fact, [F
:
k]s
[F : k]s
is the
<
[F : k],
degree of
a
one can
in fact prove
certain intermediate
k. The diligent reader will prove this in Exercise 4.18.
Exercises
4.1.
>
Let k be
over
field, f(x)
a
k. Let k C K be
an
and let F be the splitting field for
k[x],
G
extension such that
K. Prove that there is
a
f(x) splits
homomorphism
F
f(x)
over
product of linear factors
extending the identity on k.
as a
K
[§4-2]
splitting field of x6
4.2. Describe the
4.3.
over
>
same
for x4 + 4.
Q (cf. Example 4.6). [§4.1]
4.5. > Let F be a
a
Q. Do the
Find the order of the automorphism group of the splitting field of x4 + 2
4.4. Prove that the field
be
+ x3 + 1 over
factor of
Q(v//2)
is not the
splitting field for
f(x).
a
splitting field of
any
polynomial
over
Q.
polynomial f(x) G k[x], and let g(x) G k[x]
a unique copy of the splitting field of
Prove that F contains
g(x). [§5.1]
4.6.
Let k C Fi, k C F<i be two finite extensions, viewed
as
embedded in the
algebraic closure k oik. Assume that i<\ and F2 are splitting fields of polynomials
in k[x\. Prove that the intersection F\fl F2 and the composite F\F2 (the
smallest subfield of k
(Theorem
4.8 is
containing both F\ and F2)
likely going to be helpful.)
4.7. > Let k C F
=
k(a)
be
a
are
simple algebraic
both also splitting fields
=
k.
extension. Prove that F is normal
k if and only if for every algebraic extension F C K and every
a(F) F. [§6.1]
over
over
a G
Autk(K),
Exercises
439
4.8. > Let p be
that16 (a + by
a
Using the
4.9.
prime, and let k be
a
field of characteristic p. For a, 6 G K, prove
[§4.2, §5.1, §5.2]
a? + V>.
=
notion of 'derivative' given in
§4.2,
prove that
(fg)'
=
fg + fg'
for
all polynomials /, g.
4.10. Let k C F be
not divide
4.11.
is the
a
finite extension in characteristic p > 0. Assume that p does
Prove that k C F is separable.
[F : k].
> Let p be a prime integer. Prove that the Frobenius homomorphism
identity. (Hint: Fermat.) [§5.1]
on
¥p
field, and assume that k is not perfect. Prove that there are
irreducible
inseparable
p and u G fc, how many
polynomials in k[x\. (If char/c
4.12. > Let k be a
=
roots does xv
-
have in
u
fc?) [§4.2]
field of positive characteristic p, and let f(x) be an irreducible
Prove
that
there exist an integer d and a separable irreducible
polynomial.
4.13. > Let k be
a
polynomial fsep(x) such that
f(x)
=
fsep(xP").
inseparable degree of f(x). If /(#) is the minimal
polynomial of an algebraic element a, the inseparable degree of a is defined to
be the inseparable degree of f(x). Prove that a is inseparable if and only if its
The number
pd
is called the
inseparable degree
The picture to
polynomial
of
is > p.
keep
in mind is
Exercise
4.14.
-i
element
a are
distributed into
as
follows:
the roots of the minimal
deg/sep 'clumps',
each collecting a number of
coincident roots equal to the inseparable degree of a. We say that a is 'purely
inseparable' if there is only one clump, that is, if all roots of f(x) coincide (see
f(x)
4.14). [§4.2, 4.14, 4.18]
Let k C F be
a
G
F is
algebraic extension, in
purely inseparable over k if ap
positive characteristic p.
an
G k for some17 d > 0.
extension is defined to be purely inseparable if every
over k.
a G
F is
An
The
purely inseparable
Prove that a is purely inseparable if and only if [k(a) : k]s
1, if and only if
its degree equals its inseparability degree (Exercise 4.13). [4.13, 4.17]
=
4.15.
>
Let k C F be
an
algebraic extension, and let
For every intermediate field k C E C F, prove that
a
separable over k.
separable over E. [§4.3]
a G
is
F be
Let k C E C F be algebraic field extensions, and assume that /c C E is
separable. Prove that if a G F is separable over E, then k C E(a) is a separable
extension. (Reduce to the case of finite extensions.)
4.16.
-i
Deduce that the set of elements of F which
intermediate field
For F
=
separable over k form an
0 Fsep is inseparable over k.
k, ksep is called the separable closure of k. [4.17, 4.18]
Fsep,
such that every element
a G
are
F,
a
This is sometimes referred to as the freshman's dream, for painful reasons that are likely
all too familiar to the reader.
Note the slightly annoying clash of notation: elements of k are not inseparable, yet they
are purely inseparable according to this definition.
VII. Fields
440
4.17.
Let k C F be
an algebraic extension, in positive characteristic.
in Exercises 4.14 and 4.16, prove that the extension Fsep C F is
inseparable. Prove that an extension k C F is purely inseparable if and
-i
notation
Fsep
=
as
k.
With
purely
only if
[4.18]
4.18. > Let k C F be
inseparable degree [F
Prove that
[k(a)
a
k]i
:
finite extension, in positive characteristic.
to be the
k]i equals
:
quotient [F
Define the
k]/[F k]s.
:
:
the inseparable degree of a,
as
defined in
Exercise 4.13.
Prove that the inseparable degree is multiplicative:
extensions, then [F : k]i
[F : E]i[E : k]{.
if k C E C F
are
finite
=
finite extension is purely inseparable if and only if its inseparable
degree equals its degree.
Prove that
a
With notation
as
[F Fsep]. (Use
:
In particular,
in Exercise 4.16, prove that
Exercise
and
[Fsep : k]
=
[F : k]i
=
4.17.)
divides
[F : k]s
[F : k]s
[F : fc];
the inseparable degree
[F : k]i
is
an
integer.
[§4-3]
4.19.
-i
Let k C F be
embeddings of
finite separable extension, and let ^i,..., id be the distinct
extending id/-. For a G F, prove that the norm NkCF(&)
a
F in fc
(cf. Exercise 1.12) equals Yli=iLi(a) and its trace tr^cF^) (Exercise 1.13) equals
E*=i <*(<*)• (Hint- Exercises 1.14 and 1.15.) [4.21, 4.22, 6.15]
4.20.
all
a G
4.21.
-i
Let k C F be
a
finite
-i
=
1 and
trkcF(ct
-
a
(a))
a G
=
0.
Let k C F C F be finite separable extensions, and let
NkcF(a)
=
and
NkcE(NECF(a))
(Hint:
Use Exercise 4.19: if d
into k
lifting idfc
to
separable extension, and let
Autfc(F), NkcF(a/cr(a))
=
[E
:
k]
and
trfccF(o:)
e
must divide into d groups of e
=
[F
:
=
F. Prove that for
[6.16, 6.19]
a G
F. Prove that
trkcE(trECF(a)).
J5],
the de embeddings of F
each, according
to their restriction
F.)
This 'transitivity' of norm and trace extends the result of Exercise 1.15 to
separable extensions. The separability restriction is actually unnecessary; cf.
Exercise 4.22. [4.22]
4.22. Generalize Exercises 4.19—4.21
to all finite extensions k C F.
raise to power
5.
Field
[F : k]i\ for
extensions,
the trace, multiply by
(For
the norm,
[F : k]{.)
III
The material in §4 provides us with the main tools needed to tackle several key
examples. In this section we study finite and cyclotomic fields, and we return to
the question of when
a
finite
extension is in fact
simple (cf. Example 1.19).
5. Field extensions, III
441
5.1. Finite fields. Let F be
(§1.1)
that F may be viewed
a
finite
as an
field, and let
p be its characteristic. We know
extension
FpCF
of
Wp
it is
Z/pZ;
=
let d
isomorphic
to
F^
as a vector space,
The general question
to every power of a
Since F has dimension d
[F : Wp].
=
we pose
prime, and
we
as a vector space over
and in particular
|F|
=
pd
is whether there exist fields of
aim to
classifying
all fields of
a
is
a power
Fp,
of p.
cardinality equal
given cardinality.
The conclusion
we will reach is as neat as it can be: for every prime power q
exactly one field F with q elements, up to isomorphism.
Also recall (Remark III. 1.16) that, by a theorem of Wedderburn, every
there
exists
finite
division ring is in fact a finite field. Thus, relaxing the hypothesis of commutativity
in this subsection would not lead to a different classification.
Theorem 5.1. Let q
=
of the polynomial xq
let F be
Conversely,
for
xq
x
-
over
pd
x
¥p.
be
a power
over
¥p
is a
of a prime integer p. Then the splitting field
field with precisely q elements.
field with exactly q elements; then F is
The polynomial xq
x is separable over
¥p.
a
a
splitting field
-
Proof. Let F be the splitting field of xq
x
of f(x)
xq
x in F.
Since f'{x)
qxq~l
=
over
1
=
¥p.
Let E be the set of roots
1
=
(as
q
0 in
=
1; hence (Lemma 4.13) f(x) is separable, and E
p),
(f(x),f'(x))
consists of precisely q elements. We claim that £ is a field, and it follows that
characteristic
have
we
=
E
F. Indeed, F is generated
of F containing E is F itself.
=
To
see
that E is
a
field, let a,b
(a-b)q
(using Exercise 4.8;
p
2). If 6^0,
by the
note that
=
roots of
e E. Then aq
aq +
(-l)q
=
(-l)qbq
hence the smallest subfleld
/(#);
a
=
=
and bq
=
6; it follows that
a-b
-1 if p is odd and
(-l)q
=
+1
=
—1
if
=
(ab~l)q
aq(bq)~l =ab~l.
Thus E is closed under subtraction and division by a nonzero element, proving that
E is a field and concluding the proof of the first statement.
=
To prove the second statement, let F be a field with exactly q elements. The
1
nonzero elements of F form a group under multiplication, consisting of q
elements; therefore, the
(multiplicative)
(Example II.8.15). Therefore,
a
of
is,
course
0q
0
=
all elements of
^
0 => aq~l
order of every
=
1
=> aq
-
nonzero a
a
Corollary 5.2. For every prime power q there exists
of order q, up to isomorphism.
F divides q
1
0;
=
0. In other words, the polynomial xq
F!); it follows that F is a splitting field
G
x
for
one
has q roots in F (that
xq
x, as stated.
and only
one
finite field
Proof. This follows immediately from Theorem 5.1 and the uniqueness of splitting
fields (Lemma 4.2).
442
VII. Fields
Since there is
exactly
one
isomorphism class of fields of order
q for any
given
power q, we can devise a notation for a field of order q; it makes sense to
prime
adopt18 ¥q.
field of
This is called the Galois
order q.
Example 5.3. Let p be a prime integer. Then we claim that the polynomial x4 +1
is reducible19 over ¥p (and therefore over every finite field).
Since x4 + 1
assume
the
8
that p is
(x
=
l)4
+
in
the statement holds for p
F2[x],
odd prime. Then
an
we
=
2.
Thus,
claim that x4 +1 divides xp
we may
Indeed,
—x.
hence
(Exercise II.2.11);
(Exercise V.2.13); hence
square of every odd number is congruent to 1 mod 8
| (p2
-
1);
hence x8
-
(x4
1 divides xp
+
1) | (x8
_1
1) |
-
-
1
(xp2-x
-
1) |
(xp2
-
x).
It follows that x4 + 1 factors completely in the splitting field of xp
in ¥p2. If a is a root of x4 + 1 in ¥p2, we have the extensions
x, that is,
FpCFp(a)CFp2;
therefore
or
2
over
showing
(Corollary 1.11) [Fp(a) : ¥p] divides [Fp2 : ¥p]
¥p. But then its minimal polynomial is a factor
=
2. That is,
of degree 1
has degree 1
2 of
(x4 + 1),
that the latter is reducible.
j
Theorem 5.1 has many interesting consequences, and
rest of this subsection.
Corollary 5.4. Let p be
a
prime, and let d
< e be
we
sample
a
few in the
positive integers. Then there is
and only if d e. Further, if d e, then there is
¥pd
such extension, in the sense that ¥pe contains a unique copy of ¥pd.
C
an extension
one
a
or
All
extensions
Proof. If there
is
¥pe if
C
¥pd
an
¥pe
are
extension
simple.
as
[¥pe ¥p] by Corollary
Conversely, assume that d\e.
divides
that
stated, then ¥p C ¥pd C Fp«; hence
precisely that d | e.
[¥pd
:
¥p]
1.11. This says
:
pe_l
we see
exactly
pd
-
(Exercise V.2.13).
1 divides
=
pe
-
As
(p*_l)((p<')$-l+...
+
l))
1, and consequently xp
~1
-
1 divides
xpe'~l
-
1
Therefore
(xpd
-x) |
By Theorem 5.1, ¥pe is a splitting field
contains a unique copy of the splitting
(xpe -x).
for the second polynomial. It follows that it
field for the first polynomial (Exercise 4.5),
that is, of ¥pd.
of
a
For the last statement, recall that the multiplicative group of nonzero elements
finite field is necessarily cyclic (Theorem IV.6.10). If a G ¥p* is a generator of
this group, then
so
¥pd
C
¥pe
8Keep
is
a
will generate
¥pe
over any
subfield; if d
e, this says
¥pe
=
¥pd (a),
simple.
in mind that
¥q
is not the ring
Z/qZ
unless q is prime: if q is composite, then
Z/qZ
is not an integral domain.
Of course x4 + 1 is irreducible over Z. This example shows that Proposition V.5.15 cannot
be turned into an 'if and only if statement.
5. Field extensions, III
These results
can
443
be translated into rather precise information
over a finite field. For example,
on
the structure
>
1 there exist
of the polynomial ring
Corollary 5.5. Let F be
irreducible
finite field. Then for
polynomials of degree n in F[x\.
Proof. We know F
is
extension
an
¥pd
=
C
¥pd for some prime p and some d
Fp<*n, generated by an element
polynomial of
and it follows that the minimal
polynomial of degree
In fact,
our
in
n
an
5.6. Let F
=
the factorization of xql
of degree d,
polynomials
as
F[x]
a
over
> 1.
a.
F
n
By Corollary
Then
=
¥pd
[¥pdn
is
5.4 there
¥pd]
:
=
n,
irreducible
an
.
analysis of
factorizations, leading to
polynomials in ¥q[x]:
Corollary
all integers
a
d ranges
extensions of finite fields tells us about explicit
inductive algorithm for the computation of irreducible
¥q
be
x
in
Proof. By Theorem 5.1,
over
¥q F.
¥qn
and let
finite field,
F[x]
consists
of
n
be
a
positive integer. Then
all irreducible monic polynomials
of n. In particular, all these
the positive divisors
over
factor completely
a
in
¥qn.
is the
splitting field of a:9'
x over
and hence
Fp,
=
polynomial of degree d, then F[x]/(f(x))
F(a)
d of F, that is, an isomorphic copy of F^d. By Corollary 5.4,
| n, then there is an embedding of F^d in ¥qn. But then a must be a root of
x is a multiple of /(#), as this is the minimal polynomial
x, and hence xq
If
is
an
if d
xq
of
is
monic irreducible
a
=
This proves that every irreducible polynomial of degree d
a.
xqU
f(x)
extension of degree
-
n
is
a
factor of
x.
l
Conversely,
a
d
of
=
f(x);
we
if
f(x)
is
an
irreducible factor of xq
have the extensions F
dega. It follows that d
n,
=
C
¥q
x, then
C
¥q(a)
F^n,
¥qn contains a root
¥q(a)
¥qd for
and
=
again by Corollary 5.4.
The picture we are trying to convey is the following: the qn roots of xq
x
into disjoint subsets, with each subset collecting the roots of each and every
irreducible polynomial of degree d n in F[x].
clump
2: F2
contemplate the case q
Z/2Z.
l: the polynomial x2
n
x factors as the product of x and (x
1) (which
could write as (x + 1) just as well, since we are working over F2). These are all
Example
5.7. Let's
=
=
we
-
the irreducible polynomials of degree 1
n
=
=
-
2: the
polynomial x4
-
over
x must
F2.
factor
as
the product of all irreducible
polynomials of degree 1 and 2; in fact
x4
-
x
=
x(x
-
l)(x2
and the conclusion is that there is exactly
over
F2, namely x2
+ x + 1.
one
+ x +
1),
irreducible polynomial of degree 2
444
VII. Fields
n
=
x by x(x
1) is a polynomial of degree 6, which
of
the
two
irreducible
product
polynomials of degree 3 over
3: the quotient of x8
must therefore be the
F2. It takes
a moment to
find them:
x3+x2
It also follows that
an
F8
x3+£
+ l,
+ l.
may be realized in two ways as a
quotient of ¥2[x] modulo
irreducible polynomial:
F2M
(X3+X2 + 1)
F2[s]
(X3+X + 1)'
We know that these two fields must be isomorphic because of Theorem 5.1; it is
instructive to find an explicit isomorphism between them (Exercise 5.3).
n
=
n
=
4 and 5: these
cases are
left to the enjoyment of the reader
6: the factorization of x64
x must
include x,
x
1, the
(Exercise 5.4).
one
irreducible
polynomial of degree 2, and the two irreducible polynomials of degree 3 found above,
and that leaves room for 9 polynomials of degree 6, which must be all and only the
irreducible polynomials of degree 6 over F2. So the 64 elements of ¥$4 cluster as
follows:
The two dotted rectangles delimit the (unique)
these intersect in the (unique) copy of F2.
Again, F64
copies
of F4 and F8 contained in F64;
by quotienting F2[x] by the ideal generated by any
polynomials of degree 6; this gives 9 'different' realizations of this
may be realized
of the irreducible
field.
j
nothing, but it may help us focus
going
regarding finite fields.
Note that the effect of any automorphism of ¥$4 must be to scramble elements in
each sector represented in the picture, without mixing elements in different sectors
and without interchanging sectors. Indeed, every automorphism of an extension
The picture drawn above for F64
last element of information
on one
sends roots of
Since
be much
for
prime
to extract
irreducible polynomial to roots of the
extensions of finite fields
us to
a
an
means next to
we are
same
polynomial.
previous work allows
simple extensions,
precise. Restricting our attention to the extensions ¥p C ¥pd,
know these can be realized as simple extensions by an element
are
our
more
p, we
with minimal polynomial of degree d.
This polynomial is necessarily separable
5. Field extensions, III
(Fp
is
perfect),
so
445
Corollary
1.7
immediately gives
the size of the automorphism
us
group:
\Aut¥p(¥pd)\=d.
But what is this
Proposition
group?
5.8.
is
AutFp(Fpd)
cyclic, generated by the Frobenius isomorphism.
Proof. Let
xp. The
(p be the Frobenius homomorphism ¥pd
Fp<*: <p(x)
is
an
on
a
finite
field
homomorphism
isomorphism
(Corollary 4.18) and
Frobenius
=
restricts to the identity
¥p (Exercise 4.11),
on
priori the size of this group, all
Let
e
=
-
(Lemma V.5.1). Equivalently, d
Since
AutFp(Fpd).
so cp G
know
we
a
have to show is that the order of cp is d.
id; hence xp*
\(p\. Then (pe
(nonzero) polynomial xpe x has pd
=
words, the
we
< e,
for all
x
=
roots in
x
Fp<*;
G
In other
¥pd.
this implies
pd
<
pe
yielding
|AutFp(Fpd)|<M.
But then these
two numbers must be
divides the order of the group
Two last comments
on
equal,
since the order of
an
element always
(Example II.8.15).
finite fields
in order:
are
—For
n large enough, the stupendously large (but finite) field
¥pn\ works
approximation of the algebraic closure ¥p: indeed, this field contains roots
polynomials of degree < n in Fp[x], by Corollary 5.6.
This observation
can
be turned into the explicit construction of
5.4 there are extensions
an
as an
of all
algebraic
closure of ¥p: by Corollary
¥p
C
¥p2
C
C
Fp6
and the union of this chain of fields20 gives
Fp4!
a copy
C
.
of
.
¥p.
category, which (by Corollary 5.4) should strongly remind
the reader of the 'toy' category encountered in Exercise 1.5.6. There is an
—Finite
fields form
a
difference, however: in that category every object has exactly one
automorphism, while we have just checked that finite fields may have many automorphisms.
Challenge: The reader has encountered a very similar situation in a different (but
important
related)
context. Where?
Cyclotomic polynomials and fields. In the last several
5.2.
often encountered roots of 1 in C; the extensions they generate
Let
thus,
n
the
be
a
sections
are very
n roots
of the polynomial xn
Pictorially, the
subgroup of order
roots of 1 are
n
1 in C
placed
n
at the vertices of a
regular
n-gon centered
at 0, with one vertex at 1.
more
e27"/n;
distinct powers of £n;
of the multiplicative group of C, which we
are
the
have
important.
positive integer. We will denote by £n the complex number
they form a cyclic
will denote \xn.
A
we
formal construction requires the notion of direct limit; cf. §VIII. 1.4.
VII. Fields
446
\Cn
0
A primitive n-th root of 1 is a generator of \xn. Thus, £n is primitive; from our
work on cyclic groups, we know (Corollary II.2.5) that (J£ is a primitive n-th root
of 1 if and
roots of
only if m
is
relatively prime
n
*»(*)==
£ primitive
is
a
particular, there
ton. In
are
</>(n) primitive
1, where </> is Euler's totient function (cf.