NOTATIONS
The number after each entry refers to a page where the symbol is explained in the text.
Sets and Functions
cEB
c
is an element of the set B, 509
crJ,B
c
is not an element of the set B, 509
0
Empty set [or null set], 510
B�C
Bis a subset of C, 510
B-C
Relative complement of set C in set B, 511
BnC
Intersection of sets Band C, 511
nA,
le/
BUC
UAi
le/
BXC
f : B-+ C
f(b)
Intersection of the sets A1 with i E /, 511
Union of sets Band C, 511
Union of the sets A1 with iEI, 511
Cartesian product of sets Band C, 512
Function [or mapping] from set B to set C, 512
Image ofb under the function/:B-+C, or the value of /atb, 512
Identity map on the set B, 512
Composite function of f:B...+C and g:C-+D, 512-513
Image of the function/:fi....+.C, which is a subset of C, 517
Important Sets
Nonnegative integers, 523
Integers , 3
Rational Numbers, 49, 191
Real Numbers, 45, 191
Complex numbers, 49, 191
0*,n*,C*
O**, IR**
Nonzero elements of
Positive elements of
0, R, C respectively, 178, 192
Q, IR respectively, 178, 192
Integers
bIa
(a,b)
(a1> "2· ... , a,J
[a , b]
b divides
a
[orb is a factor of
a], 9
Greatest common divisor (gcd) of
a
and b, 10
Greatest common divisor (gcd) of a1, ":!·
Least common multiple (lcm) of
a
•
.
.
, a,,, 16
andb , 16
�2012'=-l..Hmlq.All ...... -IUJ ..... �..- .. .......- .. -.,, ..,...0oo ..-....... ------ .. _..-...--�).-- ...
__ ..,._,..__ ... ____ ....,. ...... c.g.,,.i...mo,--1111rigll< ........ - ...... ..,-lt..._.,..,...._.....,lt.
[ a 1o a,.,
. • . '
a,,]
a ==h(modn)
Least common multiple (lcm) of a., a,.,
. . •
, a,,, 1 6
a is congruent to b modulo n , 25
[a] or [a],,
Congruence class of a modulo n, 27, 28
z,,
Set of congruence classes modulo n, 30
Rings and Ideals
lR
M(IR)
M(Z),M(O),
M(C),M(Zn)
0
M(R)
Ring of 2 X 2 matrices over the real numbers n, 46
Ring of 2 X 2 matrices over Z, 0, C, Z" respectively, 48
z.ero matrix in M(lll), 4 7
Ring of2 X 2 matrices over a commutative ring R with identity, 48
R=S
Ring R is isomorphic to ring S, 72
(c)
Principal ideal generated by c, 144
(ch c2, ... , cJ
a==
Multiplicative identity element in a ring with identity, 44
b(modl)
a+I
Rjl
J+J
/J
Z[Vd]
Z[i] or Z[v'=I]
C)z[x]
N:Z[W)-+ Z
F(x)
Ideal generated by c11 c2,
a is
• • •
, c1,, 145
congruent to b modulo the ideal I, 145
Coset [congruence class] of a modulo the ideal/, 147
Quotient ring [or factor ring] of the ring R by the ideal I, 147, 154
Sum of ideals I and J (which is also an ideal), 149
Product of ideals I and J (which is also an ideal), 150
The subring {r + sv'd I d, r, s E Z} of C, 322
Ring ofGaussian integers, 322
Ring of polynomials in O[x] whose constant term is an integer, 336
Norm function, 346
Field of quotients [or field of rational functions] of the polynomial ring
F[x] over the field F, 358
Polynomials
R[x]
deg/(x)
f(x)lg(x)
f(x) == g(x)(mod p(x))
(f(x)] or [/(x)]p(x)
F[x]jp(x)
Ring of polynomials with coefficients in the ring R, 86
Degree of the poly nomial/(x), 88
f(x) divides [or is a factor of] g(x), 96
f(x) is congruent to g(x) modulo p(x), 125
Congruence class [or residue class] off(x) modulo p(x), 126
Ring of congruence classes modulo p(x), 128, 131
List continues on inside back cover.
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ABSTRACT
ALGEBRA
An Introduction
THIRD EDITION
THOMAS W. HUNGERFORD
Saint Louis University
�I�
BROOKS/COLE
CENGAGE Learning·
Australia• Brazil •japan• Korea•Mexico• Singapore• Spain •United IClngdom •United States
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•
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BROOKS/COLE
CENGAGE LearningAbstract Algebm: An
Introduction, Third Edition
Thomas H. Huneerford
Publisher/Executive Editor:
Richard Stratton
C> 2014, Brooks/Cole, Cengage Learning
ALL RIGHTS RESERVED. No part of this work cavered by the copyright
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1 2 3 4 5 6 7 16 15 14 13 12
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__ ..,._.,. __ ... _.,_lbo_......,....,_..Clol'8'LNmlnl---daM ..____ ..,_if__,_,o!gbll_,.....it.
Dedicated to the memory of
Vincent 0. McBrien
and
Raymond J. Swords, S.J.
College of the Holy Cross
�2012�Lomliq.All--llf"1 ... 0.<q0od._«..,..._lo_«m)*t.0..IO-...... ._--.-- .. -- -...--�).--kao
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TABLE OF
CONTENTS
Preface
ix
To the Instructor
To the Student
xii
xiv
Thematic Table of Contents for the Core Course
xvi
Part 1 The Core Course
CHAPTER 1
CHAPTER 2
CHAPTER 3
CHAPTER 4
Arithmetic inZRevisited
3
1.1
The Division Algorithm
1.2
Divisibility
3
1.3
Primes and Unique Factorization
9
17
Conuruence inZandModularArithmetic
25
2.1
Congruence and Congruence Classes
25
2.2
Modular Arithmetic
2.3
The Structure of Z,, (p Prime) and Zn
Rings
32
37
43
3.1
Definition and Examples of Rings
3.2
Basic Properties of Rings
3.3
Isomorphisms and Homomorphisms
Arithmetic in f[x]
44
59
70
85
4.1
Polynomial Arithmetic and the Division Algorithm
4.2
Divisibility in f[x]
4.3
lrreducibles and Unique Factorization
86
95
100
v
CqrJJQlll:t012C...�Allllla"'._..., ..... <Djlild._ar..,..._.,_arlaJ1111.llmto_dpll....,tinl_.-_ .. __.,..__�,).-..-...
_... ..,_.... __ ....-..., _ ... _.....,...,_c.g.,1.am1ag---riglltto....... --.. ..,-11_11jjjU_.....,. ...
vi
Table of Contents
4.4
Polynomial Functi ons, Roots, and Reducibility
4.5* lrreducibillty in O[x]
4.6* lrreducibillty in IR[x] and C[x]
CH A PH R 5
CH A PH R 6
120
5.1
Congruence in F[x] and Congruence Classes
5.2
Congruence-Class Arithmetic
5.3
The Structure of F{x]/(p(x)) When p(x) Is Irreducible
6.1
Ideals and Congruence
6.2
Quotient Rings and Homomorphisms
7.1
135
141
Ideals and Quotient Rings
Groups
125
130
141
152
Rf/When /Is Prime or Maximal
162
169
Definition and Examples of Groups
169
7.1.A Definition and Examples of Groups
7.2
Basic Properties of Groups
7.3
Subgroups
7.4
Isomorphisms and Homomorphisms
183
196
203
214
7.5* The Symmetric and Alternating Groups
c H A PT E R 8
125
Congruence in f[x] and Congruence-Glass Arithmetic
6.3* The Structure of
CH A PH R 7
105
112
Normal Subgroups and Quotient Groups
227
237
8.1
Congruence and Lagrange's Theorem
8.2
Normal Subgroups
8.3
Quotient Groups
8.4
Quotient Groups and Homomorphisms
237
248
255
8.5* The Simplicity of An
263
273
Part 2 Advanced Topics
CHAPTER 9
Topics in Group Theory
279
281
9.1
Direct Products
9.2
Finite Abelian Groups
9.3
The Sylow Theorems
281
289
298
9.4
Conjugacy and the Proof of the SylowTheorems
9.5
The Structure of Finite Groups
304
312
*Sections in the Core Course marked ., may be omitted or postponed. See the beginning of each
such section tor specifics.
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�--mJ"� ..,..,_.._aot_....,-.dk1.bl�._-...,m-._�l...Mmiog---a..nghtv...,....�cua1mt•..,.._:if..-...-i:agtu�...-.:it.
Table of Contents
CHAPTER 1 O
Arithmetic in Integral Domains
Euclidean Domains
10.2
Principal Ideal Domains and Unique
322
332
10.3
Factorization of Quadratic Integers
10.4
The Field of Quotients of an Integral Domain
353
10.5
Unique Factorization in Polynomial Domains
359
Field Extensions
11.1
11.3
Vector Spaces
365
Separability
394
399
GaloisTheory
382
388
11.6 Finite Fields
12.1
376
Algebraic Extensions
11.4 Splitting Fields
11.5
344
365
11.2 Simple Extensions
CHAPTER 12
321
10.1
Factorization Domains
CHAPTER 11
407
The Galois Group
407
12.2 The Fundamental Theorem of Galois Theory
12.3 Solvability by Radicals
415
423
Part 3 Excursions and Applications
CHAPTER 13
vii
Public-Key Cryptography
435
437
Prerequisite: Section 2.3
CHAPTER 14
The Chinese RemainderTheorem
443
14.1 Proof of the Chinese Remainder Theorem
Prerequisites: Section 2.1, Appendix C
14.2
443
Applications of the Chinese Remainder Theorem
Prerequisite: Section 3.1
14.3 The Chinese RemainderTheorem for Rings
450
453
Prerequisite: Section 6.2
CHAPTER 1 5
Geometric Constructions
459
Prerequisites: Sections 4.1, 4.4, and
CHAPTER 16
Algebraic CodingTheory
16.1
Linear Codes
4.5
471
471
Prerequisites: Section 7.4, Appendix F
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�--mJ"nw--l� ... lllll---.O,-dlK.1.'1»�._._....,.n-..�i.....q:IMK9ma.ftgbtm-...,.,..�eroa11m•..,.1111mo�....:Dgbl.l�
...
...... :it.
viii
Table of Contents
16.2 DecodingTechniques
4S3
Prerequisite; Section 8.4
16.3 BCH Codes
492
Prerequisite: Section 11.6
Part 4 Append ices
499
A. Logic and Proof
500
B. Sets and Functions
509
C. Well Ordering and Induction
D. Equivalence Relations
531
E. The Binomial Theorem
537
F. Matrix Algebra
6. Polynomials
Bibliography
523
540
545
553
Answers and Suggestions for Selected Odd-Numbered
Exercises
Index
556
589
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�--mJ"��-douflEl�dltd... � ...... ..,._,..�i....iog--a.ftgbt1D--��-..,.tlal9if....:ligb.l�...-.
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PREFACE
This book is intended for a first undergraduate course in modem abstract algebra.
Linear algebra is not a prerequisite. The flexible design makes the text suitable for
courses of various lengths and different levels of mathematical sophistication, in­
cluding (but not limited
to) a traditional abstract algebra course, or one with a more
applied flavor, or a course for prospective secondary school teachers. As in previous
editions, the emphasis is on clarity of exposition and the goal is to produce a book that
an average student can read with minimal outside assistance.
New in the Third Edition
Groups Fir st Option Those who believe (as I do) that covering rings before groups
is the better pedagogical approach
to abstract algebra can use this edition exactly as
they used the previous ones.
Nevertheless, anecdotal evidence indicates that some instructors have used the sec­
ond edition for a "groups first" course, which presumably means that they liked other
aspects of the book enough that they were willing to take on the burden of adapting it to
their needs. To make life easier for them (and for anyone else who prefers "groups first")
It is now possible (though not necessary) to use this text for
a course that covers groups before rings.
See the TO THE INSTRUCTOR section for details.
Much of the rewriting needed to make this option feasible also benefits the "rings
first" users. A number of them have suggested that complete proofs were needed in
parts of the group theory chapters instead of directions that said in effect "adapt the
proof of the analogous theorem for rings". The full proofs are now there.
Proofs for Beginners Many students entering a first abstract algebra course have
had little (or no) experience in reading and writing proofs. To assist such students (and
better prepared students as well), a number of proofs (especially in Chapters 1and2)
have been rewritten and expanded. They are broken into several steps, each of which
is carefully explained and proved in detail. Such proofs take up more space, but I think
it's worth it if they provide better understanding.
So that students can better concentrate on the essential topics, various items from
number theory that play no role in the remainder of the book have been eliminated
from Chapters 1 and2 (though some remain as exercises).
ix
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dmoed.'lblf:q.....-l&d.mmiat..... llld...-.n,.6d... IJ'Mllll:..,..�Cmg91...m.tg ...... 'lllll:righttD111mJN��-..,m.:if.--.-��:Nqllirll:it.
x
Preface
More Examples and Exercises In the core course (Chapters
1-8), there
are 35%
more examples than in the previous edition and 13% more exercises. Some older exer­
cises have been replaced, so 18% of the exercises are new. The entire text has about 350
examples
and 1600 exercises. For easier reference, the examples are now numbered.
Coverage The breadth of coverage in this edition is substantially the same as in
the preceding ones, with one minor exception. The chapter on Lattices and Boolean
Algebra (which apparently was rarely used) has been eliminated. However, it is avail­
able at our website (www.CengageBrain.com) for those who want to use it.
The coverage of groups is much the same as before, but the first group theory chapter
in the second edition (the longest one in the book by far) has been divided into two chap­
ters of more manageable size. This arrangement has the added advantage of making the
parallel development of integers, polynomials, groups, and rings more apparent.
Endpapers The endpapers now provide a useful catalog of symbols and notations.
Website The website (www. CengageBrain.com) provides several downloadable
programs for TI graphing calculators that make otherwise lengthy calculations in
Chapters 1 and 14 quite easy. It also contains a chapter on Lattires and Boolean
Algebra, whose prerequisites are Chapter 3 and Appendices A and B .
Continuing Features
Thematic Development The Core Course (Chapters
1-8) is organized around two
themes: Arithmetic and Congruence. The themes are developed for integers (Chapters 1
and2),polynomials (Chapters 4and 5),rings(Chapters3 and6),andgroups(Chapters 7
and
8).
See the Thematic Table of Contents in the TO THE STUDENT section for a
fuller picture.
Congruence The Congruence theme is strongly emphasized hi the development of
quotient rings and quotient groups. Consequently, students can see more clearly that
ideals, normal subgroups, quotient rings, and quotient groups are simply an extension
of familiar concepts in the integers, rather than an unmotivated mystery.
Useful Appendices These contain prerequisite material (e.g., logic, proof, sets,
functions, and induction) and optional material that some instructors may wish to
introduce (e.g., equivalence relations and the Binomial Theorem).
Acknowledgments
This edition has benefited from the comments of many students and mathematicians
over the years, and particularly from the reviewers for this edition. My warm thanks to
Ross Abraham, South Dakota State University
George DeRise, Thomas Nelson Community College
Kimberly Blee, California State University, Sacramento
Sherry Ettlich, Southern Oregon University
Lenny Jones, Shippensburg University
Anton Kaul, California Polytechnic University, San Luis Obispo
Wojciech K.omornicki, Hamline University
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dl9mBd.--mJ"��dl-.ool�.dllw;.1_blOftdl.'-uiag...,._._Cmgqe�---rlgbtm--�c:oi-.•..,.--il......_.�� ......:it.
Preface
xi
Ronald Merritt, Athens State University
Bogdan Nita, Montclair State University
Tara Smith, University of Cincinnati
It is a particular pleasure to acknowledge the invaluable assistance of the Cengage
staff, especially Molly Taylor, Shaylin Walsh, Cathy Brooks, and Alex Gontar. I also
want to express my appreciation to my copyeditor, Martha Williams, whose thorough
reading of the manuscript significantly improved the final text. Charo Khanna and the
MPS Limited production staff did an excellent job.
John Oprea (Cleveland State University), Greg Marks (Saint Louis University),
and David Leep (University of Kentucky) provided assistance on several points, for
which I am grateful.
Finally, a very special thank you to my wife Mary Alice for her patience, under­
standing, and support during the preparation of this revision.
T.W.H.
OlpJ!ripl.2012c..p.,�Al...-a-..ct..V.,-autt.-ggpW.-....S.w�iD ...... erkaJ)ld.0.lo-�...... -1bkd.pat;r�_,.a..�m...&•lkx*.ndiltll'�•).liidbmW...,...bM
clmiimd..... mJ"���ad�dlK:l--�...-...,m-..��--a.ftgbtm--..�ooa11m•..,.tilll9:1f���...-.:it.
T 0 THE INSTRUCTOR
Here are some items that will assist you in making up your syllabus.
Course Planning
Using the chart on the opposite page, the Table of Contents (in which optional sections
are marked), and the chapter introductions, you can easily plan courses of varying length,
emphasis, and order of topics. If you plan to cover groups before rings, please note that
Section 7.1 should be replaced by Section 7.1. A (which appears immediately after 7.1).
Appendices
Appendix A (Logic and Proof) is a prerequisite for the entire text. Prerequisites for
various parts of the text are in Appendices B-F. Depending on the preparation of
your students and your syllabus, you may want to incorporate some of this material
into your course. Note the following.
•
Appendix B (Sets and Functions): The middle part (Cartesian
products and binary operations) is first used in Section 3.1 [7.1.A].* The last
five pages (injective and surjective functions) are first used in Section 3.3 [7.4].
•
Appendix C (Induction): Ordinary induction (Theorem C.1) is first used
in Section 4.4. Complete Induction (Theorem C.2) is first used in Section 4.1
[9.2]. The equivalence of induction and well-ordering (Theorem C.4) is not
needed in the body of the text.
•
Appendix D (Equivalence Relations): Important examples of
equivalence relations are presented in Sections 2.1, 5.1, 6.1, and 8.1, but the
formal definition is not needed until Section 10.4 [9.4].
•
Appendix E (The Binomial Theorem): This is used only in Section
•
Appendix F (Matrix Algebra): This is a prerequisite for Chapter 16 but
11.6 and occasional exercises earlier.
is not needed by students who have had a linear algebra course.
Finally, Appendix G presents a formal development of polynomials and indetermi­
nates. I personally think it's a bit much for beginners, but some people like it.
Exercises
The exercises in Group A involve routine calculations or short straightforward proofs.
Those in Group B require a reasonable amount of thought, but the vast majority
should be
accessible
to most students. Group C consists of difficult exercises.
Answers (or hints) for more than half of the odd-numbered exercises are given
at the end of the book. Answers for the remaining exercises are in the Instructor's
Manual available to adopters of the text.
xii
"The section numbers in brackets are for groups-first courses.
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...
1111*"it.
To the Instructor
xiii
CHAPTER INfERDEPENDENCE*
I.
Arithmetic
inZ
13.
Public-Key
Cryptography
15.
Geometric
Constructions
4.
....<See Note bek1w)
Arithmetic
inF(x]
s.
Congnience
inF[x]
14.3
The CRT
for Rings
8.
Normal
6.
------1 Ideals &
Subgroups
&Quotient
Groups
Quotient
Rings
16.1, 16.2
Algebraic
Coding
'TheoJy
10.
Arithmetic
in Integral
Domains
NOTE: To go quickly from Chapter 3 to Chapter 6, first cover Section 4.1 (except the
proof of the Di vision Algorithm), then proceed to Chapter 6. If you plan to cover
Chapter 11, however, you will need to cover Chapter 4 first.
•A solid arrowA--->B means thatA is a prerequisite for B; a dashed arrow A-...8 means that B depends
only on parts of A (see the Table of Contents for specifics). For the dotted arrow S ··>6, see the Note
at the bottom of the chart.
.......
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fa:ml.beBom:.ndkir�1).HdlmUl.:Nvillwt..
...... mJ"nppNH9d�
...ua;,dlld.1t1e�a-mag�c.pge��-rightlu
��-..,.m..if......_._zigtus:wtrietliuas
k.
......
.......
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TO THE STUDENT
Overview
This book begins with grade-school arithmetic and the algebra of polynomials from
high school (from a more advanced viewpoint, of course). In later chapters of the
book, you will
see
how these familiar topics fit into a larger framework of abstract
algebraic systems. This presentation is organized around these two themes:
Arithmetic You will see how the familiar properties of division, remainders, factor­
ization, and primes in the integers carry over to polynomials, and then to more general
algebraic systems.
Congruence You may be familiar with "clock arithmetic".* This is
an example
of
congruence and leads to new finite arithmetic systems that provide a model for what
can be done for polynomials and other algebraic systems. Congruence and the related
concept of a quotient object are the keys to understanding abstract algebra.
Proofs
The emphasis in this course, much more than in high-school algebra, is on the rigor­
ous logical development of the subject. If you have had little experience with reading
or writing proofs, you would do well to read Appendix A, which summarizes the basic
rules of logic and the proof techniques that are used throughout the book.
You should first concentrate on understanding the proofs in the text (which is quite
different from constructing a proof yourself). Just as you can appreciate a new build­
ing without being an architect or a contractor, you can verify the validity of proofs
presented by others,
in
even if you caKt see how anyone ever thought of doing it this way
thefirst place.
Begin by skimming through the proof to get an idea of its general outline before
worrying about the details in each step. It's easier to understand an argument if you
know approximately where it's headed. Then go back to the beginning and read the
proof carefully, line by line. If it says "such and such is true by Theorem 5.18", check
to see just what Theorem 5.18 says and be sure you understand why it applies here. If
you get stuck, take that part on faith and finish the rest of the proof. Then go back and
see if you can figure out the sticky point .
*When the hour hand of a clock moves 3 hours or 15 hours from 12, it ends in the same position, so
3 15 on the clock. If the hour hand starts at 12 and moves B hours, then moves an additional
9 hours, it finishes at 5; so B + 9 5 on the clock.
=
=
xiv
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To the Student
xv
When you're really stuck, ask your instructor. He or she will welcome questions that
arise from a serious effort on your part.
Exercises
Mathematics is not a spectator sport. You can't expect to learn mathematics without
doing mathematics, any more th an you could learn to swim without getting in the
water. That's why there are so many exercises in this book.
The exercises in group A are usually straightforward. If you can't do almost all of
them, you don't really understand the material. The exercises in group B often require
a reasonable amount of thought-and for most of us, some trial and error as well. But
the vast majority of them are within your grasp. The exercises in group C
are
usually
difficult ... a good test for strong students.
Many exercises will ask you to prove something. As you build up your skill in un­
derstanding the proofs of others (as discussed above), you will find it easier to make
proofs of your own. The proofs that you will be asked to provide will usually be much
simpler than proofs in the text (which can, nevertheless, serve as models).
Answers (or hints) for more th an half of the odd-numbered exercises are given at
the back of the book.
Keeping It All Straight
1-8), students often have trouble seeing how the various
Thematic Table of Contents on the next two
arranged according to the themes of arithmetic and congruence, so you can
In the Core Course (Chapters
topics tie together, or even if they do. The
pages is
see how things fit together.
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THEMATIC TABLE OF
CONTENTS FOR THE
CORE COURSE
TOPICS ...
INTEGERS
P OLYNOMIALS
THEMET
ARITHMETIC
1. Arithmetic in Z Revisited
4. Arithmetic in
Flxl
Division Algorithm
1.1 The Division Algorithm
4.1 Polynomial Arithmetic
and the Division Algorithm
Divisibility
1.2 Divisibility
4.2 Divisibility in F[x]
Primes and
Factorization
1.3 Primes and Unique
Factorization
4.3 Irreducibles and Unique
Factorization
Primality Testing
1.3 Theorem 1.10
4.4 Polynomial Functions,
Roots, and Reducibility
4.5 Irreducibility in O[x]
4.6 Irreducibility in
CONGRUENCE
2. Congruence in Z and
R[x] and qx]
5. Congruence in Ff xi and Congruence
Modular Arithmetic
Cl� Arithmetic
2.1 Congruence and
Congruence Classes
5.1 Congruence in F[x] and
Congruence Classes
Congruence-C/Q11s
Arithmetic
2.2 Modular Arithmetic
5.2 Congruenoe-Oass Arithmetic
Quotient Structures
2.3 The Structure of z,
When p Is Prime
5.3 The Structure of F[x]/p(x)
When p(x) Is Irreducible
Congruence
OTHER
Isomorphism and
Homomorphism
xvi
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__ ODJ_.... __ ... _..,. _ .. _ .....,. ..,_ c.g,.u....,. -- ... ridi<"'...... - -• ..,-11..-.-..... -...-.11.
Thematic Table of Contents for the Core Course
xvii
Directions: Reading from left to right across these two pages shows how the theme
or
subtheme in the left-hand column is developed in the four algebraic systems listed i n the
top row. Each vertical column shows how the themes are carried out for the system listed
at the top of the column.
RINGS*
GROUPS*
3. Rings
7. Groups
7. l Definition and Examples of Groups
3.1 Rin�
7.5 The Symmetric and Alternating Groups
7.2 Basic Properties of Groups
3.2 Basic Properties of Rings
7.3 Subgroups
6. Ideals and Quotient Rings
8. Normal Subgroups and Quotient Groups
6.1 Ideals and Congruence
8.1 Congruence
8.2 Normal Subgroups
8.5 The Simplicity of An
6.2 Quotient Rings and
Homomorphisms
8.3 Quotien t Groups
8.4 Quotient Groups and Homomorphisms
6.3 The Structure of R//When Ils
Prime or Maximal
7.4 Isomorphisms and Homomorphisms
3.3 Isomorphisms and
Homomorphisms
*In the Arithmetic Theme, the sections of Chapters 3 (Rings) and
subthemes (as do the sections
you will see in Chapter
of Chapters 1
10 (Arithmetic
8 (Groups) do
not correspond to the individual
and 4). For integral domains, however, there is a correspondence, as
in Integral Domains).
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1
CHAPTER
Arithmetic in "11._ Revisited
Algebra grew out of arithmetic and depends heavily on it. So we begin our study of
abstract algebra with a review of those facts from arithmetic that are used frequently
in the rest of this book and provide a model for much of the work we do. We stress
primarily the underlying pattern and properties rather than methods of computation.
Nevertheless, the fundamental concepts are ones that you have seen before.
•
The Division Algorithm
Our starting point is the set of all integers Z
=
{O, ±1, ±2, . . . } . We assume that you
are familiar with the arithmetic of integers and with the usual order relation (<) on
the set Z. We also assume the
WELL-ORDERING AXIOM Every nonempty subset of the set of nonnegatiVe
integers contains a smallest element.
If you think of the nonnegative integers laid out on the usual number line, it is
intuitively plausible that
each subset
contains an element that lies to the left of all the
other elements in the subset-that is the smallest element. On the other hand, the Well­
Ordering Axiom does not hold in the set Z of all integers (there is no smallest negative
integer). Nor does it hold in the set of all nonnegative rational numbers (the subset of
all positive rationals does not contain a smallest element because, for any positive ratio­
nal number r, there is always a smaller positive rational-for instance, r/2).
NOTE: The rest of this chapter and the next require Theorem
1.1,
which
is stated below. Unfortunately, its proof is a bit more complicated than
is desirable at the beginning of the course, since some readers may not
have seen many (or any) formal mathematical proofs. To alleviate this
3
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4
Chapter 1
Arithmetic in Z Revisited
situation, we shall first look at the origins of Theorem 1.1 and explain the
idea of its proof. Unless you have a strong mathematical background, we
suggest that you read this additional material carefully before beginning
the proof.
To ease the beginner's way, the proof itself will be broken into several
steps and given in more detail than is customary in most books. However,
because the proof does not show how the theorem is actually used in prac­
tice, some instructors may wish to postpone the proof until the class has
more experience in proving results. In any case, all students should at least
read the outline of the proof
Steps
(its
first three lines and the statements of
1-4).
So here we go. Consider the following grade-school division problem:
Quotient
---+
Divisor
�·
Divicknd
Check:
11
7
---+
+--- Quotient
77
+5
82
12
R£mainckr
11
X7 +--- DiviSor
+--- R£mainder
+--- Divicknd
5
The division process stops when we reach a remainder that is less than the divisor.
All the essential facts are contained in the checking procedure, which may be verbally
summarized like this:
dividend = (divisor) (quotient) + (remainder).
Here is a formal statement of this idea, in which the dividend is denoted by
divisor by b, the quotient by q, and the remainder by
Theorem 1.1
> 0. Then
a=bq+r
1.1
there exist unique integers q and r such
and
Os r<b.
allows the possibility that the dividend
quires that the remainder
r
a
might be negative but re­
must not only be less than the divisor b but also must be
nonnegative. To see why this last requirement is necessary, suppose a
by b = 3, so that
the
The Division Algorithm
Let a, b be integers with b
that
Theorem
a,
r:
-14 = 3q + r.
If
we
= -14 is divided
only require that the remainder be less than
the divisor 3, then there are many possibilities for the quotient
q
and remainder r,
including these three:
-14 = 3(-3) + (-5),
with
< 3
[Here
q = -3 and r = -5.]
-14 = 3(-4) + ( 2)
with -2 < 3
[Here
q = -4 and r = -2.]
with 1
[Here
q=
-
-14 = 3(-5) + 1,
,
-5
< 3
-5 and r
= l.].
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1.1
The Division Algorithm
5
When the remainder is also required to be nonnegative as in Theorem 1.1, then there
is exactly one quotient q and one remainder
r,
namely, q
=
-5 and
r =
1, as will be
shown in the proof.
The fundamental idea underlying the proof of Theorem 1.1 is that division is just
repeated subtraction. For example, the division of 82 by 7 is just a shorthand method
for repeatedly subtracting 7:
82
-7
75 +--82- 7·l
40
-7
68 +--82-7
·
-7
33 +---- 82 - 7. 7
2
-7
-7
61+--82-7. 3
26 +---- 82-7. 8
-7
-7
5 4 +--82- 7·4
19 +---- 82- 7. 9
-7
-7
47 +--82-7
·
12 +---- 82-7. 10
5
-7
-7
5
40+--82- 7·6
+---- 82- 7
•
11
The subtractions continue until you reach a nonnegative number less than 7 (in this
case
5).
The number 5 is the remainder, and the
nwnber of
multiples of 7 that were
subtracted (namely, 11, as shown at the right of the subtractions) is the quotient.
In the preceding example we looked at the numbers
82- 7
·
1,
82- 7
·
2,
82- 7
·
3, and so on.
In other words, we looked at numbers of the form 82
found the smallest nonnegative one (namely,
-
7x for
5). In the proof
x""'
1, 2, 3,
.
.
.
and
of Theorem 1.1 we shall
do something very similar.
Proof of Theorem 1.1* ... Let a and b be fixed integers with b > 0. Consider the sets
of all integers of the form
a-bx,
Note that
where
x is an integer
and
a
-bx� 0.
x may be any integer-positive, negative, or 0---but a -bx must
be nonnegative. There are four main steps in the proof, as indicated below.
Step I
Show that Sis nonempty byfinding a valU£ for x such that a -bx� 0.
Proof of Step I: We first
show that
a
+ b la I � 0. Sinceb is a positive
integer by hypothesis, we must have
b�l
bja]
<'!:
Jaj
bJal �-a
[Multiply both sides of the precedinginequality by Jaj.]
[Because lal
<?:-aby
the defmition of absolute value.]
a+ bjaj� 0.
•for an alternate proof by induction of part of the theorem, see Example 2 in Appendix C.
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6
Chapter 1
Arithmetic in l Revisited
Now let x= -!al. Then
a - bx= a - b (-lal ) =a+ bla[2: 0.
Hence, a
-
bx is in Swhen x= -!al, which means that Sis nonempty.
Step 2 Find qand r such that a
=
bq + rand r
�
0.
Proof of Step 2: By the Well-Ordering Axiom, S contains a smallest
element-call it r. Since r E S, we know that r 2: 0 and t=a - bx fo r
some x, say x=q. Thus,
r = a
-
bq and r 2: 0,
a= bq + r and r
or, equivalently,
?.:
0.
Step 3 Show that r<b.
Proof of Step3: We shall use a "proof by contradiction" (which is
explained on page 506 of Appendix A). We want to show that r<b.
So suppose, on the contrary, that r
2:
b. Then r - b2:: 0, so that
0 sr - b=(a - bq) - b= a - b(q + 1).
Since a - b(q + 1) is nonnegative, it is an element of Shy definition . But
since bis positive, it is certainly true that r
-
b<r. Thus
a - b(q+ 1) = r - b<r.
The last inequality states that a - b(q + 1)-which is an element of
S-is less than r, the smallest element of S. This is a contradiction.
So our assumption that r
;;:::
bis false, and we conclude that r<b.
Therefore, we have found integers q and r such that
a=bq+r
0 sr<b.
and
Step 4 Show that r and q are the only numbers with these properties (that's what
"unique" means in the statement of the theorem).
Proof of Step 4: To prove uniqueness, we suppose that there are integers
q1 and r1 such that a= bq1 + r1 and 0 sr1<b, and prove that q1=q
and r1 =r.
Since a=bq+r and a= bq1 + rh
bq + r
=
we
have
bq1 + r1
so that
(*)
b(q - q1)=r1 - r.
Furthermore,
Osr<b
0 sr1< b.
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1.1
The Division Algorithm
7
Multiplying the first inequality by -1 (and reversing the direction of the
inequality), we obtain
-b< -rs O
0
s r1 <b.
Adding these two inequalities produces
-b <
r
1 -r<b
-b< b(q
-1 < q
-
q1) < b
q1 <1
[By Equation(*)]
[Divide each term by b.]
q1 is an integer (because q and q1 are integers) and the only
q - q1 = 0 and q = q1•
Substituting q - q1 = 0 in Equation(*) shows that r1 - r = 0 and
hence r = r1• Thus the quotient and remainder are unique, and the
But q -
integer strictly between -1 and 1 is 0. Therefore
proof is complete.
•
When both the dividend a and the divisor bin a division problem are positive, then
the quotient and remainder are easily found either by long division
(as
on
page 4) or
with a calculator when the integers involved are larger.
EXAMPLE 1
Suppose a
= 4327 is divided by b = 281. Entering a/b in a calculator produces
15.39857
··.The integer to the left of the decimal point (15 here) is the quo­
·
tient q and the remainder is
r =a - bq = 4327 - 281 15 = 112.
•
These calculations are shown on the graphing calculator screen in Figure 1.
4327/281
15.39857651
4327-281*15
112
FIGURE1
When the dividend
a
is negative, a slightly different procedure is needed so that the
remainder will be nonnegative.
*The symbol• indicates the end of a proof.
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8
Chapter 1
Arithmetic in 7L Revisited
EXAMPLE 2
Suppose a= -7432 is divided b y b = 453. Entering a/b in a calculator pro­
duces -16.40 618
• ·
·.In this case the quotient q is not -16; instead,
q = (the integer to the left of the decimal point) -1 = -16 - 1= -17.
(Without this adjustment, you will end up with a negative remainder.) Now,
as
usual,
r =a - bq
=
-7432 - 453
·
(-17)
=
269 .
The preceding calculations are summarized in the calculator screen in Figure 2.
-74321'453
-16.40618102
-7432-453*( -17)
269
FIGURE2
• Exercises
A. In Exercises 1 and2,find the quotient q and remainder r when a is divided by b,
without using technology. Check your answers.
1. (a) a:=: 17;b=:4
2. (a) a
""'
-51; b= 6
(b) a= O; b
=
19
(b) a= 302; b= 19
(c) a = -17; b= 4
(c) a= 2000; b= 17
In Exercises 3 and 4, use a calculator tofind the quotient q and remainder r when
a is divided by b.
3. (a) a= 517; b= 83
(b) a= -612; b = 74
(c) a= 7,965,532; b= 127
4. (a) a= 8,12 6,493; b= 541
(b) a= -9,217,645; b= 617
(c) a= 171,819,920;b = 4321
5. Let a be any integer and let b and e be positive integers.. Suppose that when
a is divided by b, the quotient is q and the remainder is r, so that
a = bq + r
and
0 s r < b.
If ae is divided by be, show that the quotient is q and the remainder is re.
B. 6. Let a, b, e, and q be as in Exercise 5. Suppose that when q is divided by e, the
quotient is k. Prove that when a is divided by be, then the quotient is also k.
7. Prove that the square of any integer a is either of the form 3k or of the
form 3k + 1 for some integer k. [Hint: By the Division Algorithm, a must
be of the form 3q or 3q + 1 or 3q + 2.]
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1.2
Divisibility
9
8. Use the Division Algorithm to prove that every odd integer is either of the
form 4k + 1 or of the form 4k + 3 for some integer k.
9. Prove that the cube of any integer a has to be exactly one of these forms:9k
or9k + 1or9k + 8 for some integer k.
[Hint: Adapt the hint in Exercise 7,
and cube a in each case.]
IO. Let
n be a positive integer. Prove that a and cleave the same remainder when
n if and only if a c = nk for some integer k.
divided by
-
11. Prove the following version of the Division Algorithm, which holds for both
positive and negative divisors.
Extended Division Algorithm: Let a andb be integers with b :# 0. Then there
exist unique integers q a11d rsuch that a= hq + randO s r < JbJ.
[Hint: Apply Theorem1.1when a is divided by
(b > 0 and b< O).]
•
lb 1- Then consider two cases
Divisibility
An important case of division occurs when the remainder is 0, that is, when the divisor
is a factor of the dividend. Here is a formal definition:
Definition
Leta and b be integers with
b:;:. 0. We say that b divides a (or that b is a divi­
of a) if a =be for some integer c. In symbols, "b
divides a" is written b Ia and "b d oes not divide a" is written b .ta.
sor of a, or that b i s a factor
EXAMPLE 1
= 3 8, but 3 .( 17. Negative divisors are allowed: -6 [ 54
(-6)(-9), but-6.((-13).
3 f 24 because 24
because 54
=
•
EXAMPLE 2
Every nonzero integer b divides 0 because 0
have I la because a= 1
Remark
If
=b
•
0. For every integer a, we
·a.
b divides a, then a = be for some
c.
-a= b(-c), so that
-a is also a divisor of a.
Hence
b I (-a). An analogous argument shows that every divisor of
Therefore
a
and
-a
hal·e the same divisors.
Supposea:¢:0andb I a. Then a= be, so that laf= lbl le!. Consequently,
lbl s laf. This last inequality is equivalent to
Jal s b s lal. Therefore
Remark
0s
-
(i) every divisor of the nonzero integer a is less than or equal to I a I;
(ii) a nonzero integer has only finitely many dMsors.
...
..
..
....
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10
Chapter 1
Arithmetic in 1 Revisited
All the divisors of the integer 12 are
l,
-1, 2, -2, 3, -3, 4, -4, 6, -�. 12, -12.
Similarly, all the divisors of 30 are
1, -1, 2, -2, 3, -3, 5, -5, 6, ...,5, 10, -10, 15, �15, 30, -30.
The common divisors of 12 and 30 are the numbers that divide both 12 and 30, that
is, the numbers that appear on both of the preceding lists:
l,
-
I, 2, -2, 3, -3, 6, -6.
The largest of these common divisors, namely 6, is called the "greatest common
divisor" of 12 and 30. This is an example of the following definition.
Definition
Leta and b be integers, not both O. The greatest common divisor (gcd) of
bis the largest integer d that divides both a and b. In other words,
a and
dis the gcd
of a and b provided that
(1) dla and dlb;
(2) ifcjaand clb, then cs d.
The greatest common divisor of a and b is usually denoted (a, b).
If a and
b
are not both 0, then their gcd exists and is unique. The reason is that
a nonzero integer has only finitely many divisors, and so there are only a finite num­
ber of common divisors. Hence there must be a unique largest one. Furthermore, the
greatest common divisor of a and b satisfies the inequality
(a,
because I is
a
b) ??.
l
common divisor of a and b.
EXAMPLE 3
(12, 30)
=
6, as shown above. The only common divisors of 10 and 21 are 1 and
-1. Hence (10, 21) = 1. Two integers whose greatest common divisor is 1, such
as
10 and 21, are said to be relatively prime.
EXAMPLE 4
The common divisors of an integer a and 0 are just the divisors of a. If a > 0,
then the largest divisor of a is clearly a itself. Hence, if a > 0, then (a, 0) = a.
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1.2
Divisibility
11
Listing all the divisors of two integers in order to find their gcd can be quite time
consuming. However, the Euclidean Algorithm
(Exercise
15) is a relatively quick
method for finding gcd's by hand. You can also use technology.
Technology Tip: To find a gcd on a Tl-graphing calculator, select "gcd" in the
NUM submenu of the MATH menu.
We have seen that 6 =(12, 30). A little arithmetic shows that something else is true
here: 6 is
a
linear combination of 12 and 30. For instance,
6 = 12(-2) + 30(1)
6 = 12(8) + 30(-3).
and
You can readily find other integers u and
v
such that 6 = 12u + 30v. The following
theorem shows that the same thing is possible for any greatest common divisor.
Theorem 1.2
Let
a
and b be integers, not both 0, and let d be their greatest common divi­
sor. Then there exist (not necessarily unique) integers u and v such that
d =au + bv.
CAUTION:
Read the theorem carefully. The fact that d =au+ bv does
not imply that d =(a, b). See Exercise 25.
For the benefit of inexperienced readers, the proofs of Theorem 1.2 and
Corollary 1.3 will be broken into several steps. The basic idea of the proof of
Theorem 1.2 is to look at all possible linear combinations of a and b and find one
that is equal to d.
Proof of Theorem 1.2 ... 1..et s be the set of all linear combinations of a and b, that is
S = {am+hnlm,n E
Step
Z}.
I Find the smallest positive element of S.
Proof of Step 1: Note that d1- + b2 = aa + bb is in Sand d1- + b2 ;;?: 0.
Since a and bare not both 0, Ql + b2 must be positive. Therefore S
contains positive integers and hence must contain a smallest positive
integer by the Well-Ordering Axiom. Let t denote this smallest positive
element of S. By the definition of S, we know that t =au + bv for
some integers u and v.
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12
Chapter 1
Arithmetic in l Revisited
Step 2
Prove that t is the gcd of a and b, that is, t =d
Proof of Step 2: We must prove that t satisfies the two conditions in the
definition of the gcd:
(1) tla and tlb;
(2)
If
cla and c jb,
then
c :St.
Proof of (1): By the Division Algorithm, there are integers q and r
such that a = tq + r, with 0s r < t. Consequently,
r =a - tq,
r =a - (au + bv)q =a - aqu - bvq,
r =a(l - qu)
+
b(-vq):
Thus r is a linear combination of a and b, and hence r E S. Since
r < t (the smallest positive element of S), we know that r is not
positive. Since r 2!: 0, the only possibility is that r =0. Therefore,
a = tq + r = tq + 0 = tq, so that t Ia. A similar argument shows
that t I b. Hence, t is a common divisor of a and b.
Proof of (2): Let c be any other common divisor of a and b, so that
c I a and c Ib. Then a =ck and b =cs for some integers k ands.
Consequently,
t =au + bv =(ck)u + (cs)v
= c(ku + sv).
The first and last terms of this equation show that
c It. Hence,
cs It !by the second Remark on page 9. But tis positive, so It I =t.
Thus cs t.
This shows that t is the greatest common divisor d and completes
the proof of the theorem.
Technology Tip: To find the
•
gcd of a and b and express it in the form au + bu on
a TI calculator, download the GCD program on our website (www.CengageBrain
.com). Figure
is
1 shows the result when you enter a =2579 and b = 4321: The gcd
1 and you can easily verify that 2579 826 + 4321 ( -493) =1.
•
•
AU+BV"'6CD•
LI=
""'
1
826
-493
Don&
FIGUREt
To do the same thing with Maple, use the command
igcdex(a, b, 's', 't');.
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dlremad.'lmm,-��._
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..
1.2
Divisibility
13
Corollary 1.3
Let a and b be integers, not both 0, and let d be a positive integer. Then dis the
greatest common divisor of a and b if and only if d satisfies these conditions:
(i) di a and
di b;
(ii) if cla and
Proof• T he proof
clb, then cld.
of an
"if and only if" statement requires two
steps
(see page 507 in Appendix A).
Step I
Prove: If d= (a, b), then d satisfies conditions (i) and (ii).
Proof of Step 1: If d=(a, b), then by the definition of the gcd, d divides
both a and b. So d satisfies condition (i).
To verify that d satisfies condition (ii), suppose that c isan integer such
thatc laand clb. Then a=crand b=cs for some integers rand s, by the
definition of "divides". By Theorem 1.2 there are integers uand v such that
d=au+bv
d= (cr)u + (cs)u
[Because a = er and b = cs.]
d= c(ru + sv)
[Factor c out of both terms.]
But this last equation says that c Id. T herefore,
Step 2
d satisfies condition (ii).
Prove: If dis a positive integer that satisfies conditions (i) and (ii), then
d=(a,b).
Proof of Step 2: To
prove that
d= (a, b),
we must show that
d satisfies
the requirements of the definition of the gcd, namely,
(1)
d la and d lb;
(2) If
c I a and c I b,
then
cs d.
since requirement (1) and condition (i) are
d satisfies requirement (2), suppose c is an inte­
ger that divides both a and b, then c I dby condition (ii). Consequently,
by the second Remark on page 9, cs l dl. But dis positive, so ldl = d.
Thus, cs d. T herefore, d satisfies requirement (2) and, hence, dis the
gcd of a and b. •
Obviously
d satisfies (1)
identical. To prove that
T he answer to the following question will be needed on several occasions. If
then under what conditions is it true that a I b or
a
Ic? It is cer tainly not
a I be,
always true, as
this example shows:
613. 4,
Note that
6 has a
When a divisor of
but
and
6.f4.
nontrivial factor in common with 3 and another in common with 4.
be has no common factors (except
±1) with either b or
c,
then there
is a useful answer to the question.
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...
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14
Chapter 1
Arithmetic in l Revisited
Theorem 1.4
If albc and
{a, b) =1, then a le.
Proof� Since (a, b) =1, Theorem 1.2 shows that au+ bv=1 for some integers
u and v. Multiplying this equation by c shows that acu+ bcv=
a jbc, so that be= ar for some r. Therefore
c=acu+ bcv = acu + (ar)v=a(cu+
c.
But
)
ro .
The first and last parts of this equation show that a I c.
•
• Exercises
1.
Findthe greatest common divisors. You should be able to do parts (a)-(c) by
hand, but technology is OK for the rest.
(a)
(56, 72)
(b) (24, 138)
(c) (112, 57)
(d)
(143, 231)
(e) (306, 657)
(f) {272, 1479)
(g) (4144, 7696)
(h) (12378, 3054)
2. Prove that bj a if andonly if (-b) la.
3. If a I b and b I c, prove that a I c.
4. (a) Ifalbandalc,provethatal(b+c).
(b) Ifa I banda I c, prove thata I(hr+ ct) forany r, t
E Z.
5. If a andbarenonzero integers such that alb and b I a, prove that a= ±b.
6. If
a I b and c I d, prove that ac I bd.
7. If a < 0, find (a, 0).
8. Prove that (n,
n+
1) =1 for every integer n.
9. If a I c and b I c, must ab divide c? Justify your answer.
10. If (a, 0)= 1, what can a possibly be?
11. If n E Z, what are the possible values of
(a) (n, n
+ 2)
(b)
(n,n + 6)
12. Suppose that (a, b)= 1and(a, c) = 1. Are any of the following statements
false? Justify your answers.
(a)
(ab, a) =1
(b)
(b, c) =1
(c) (ab,
c) =1
13. Suppose that a, b, q, and r are integers such that a= bq+ r. Prove each of the
following statements.
(a)
Every common divisor
c
of
a
and bis also a common divisor of
b and r.
[Hint: For some integers sand t, we have a=cs and b=ct. S ubstit ute
these results into a= bq+ r, and show that c Ir.]
......
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fmnb•Bom:.ndlat�l).BdlmiM._...._
-...ed.
-.�-i:mi111111*-'GE1�.tkl.1t1e�._.....
o.pg.i...marg--.t111ftaht
1
1D_,,,,.�ca.-.111:..,...._w...._._.:dibb�...-. ..
...
.......
1.2
Divis I bl lity
15
(b) Every common divisor of b and r is also a common divisor of a and b.
(c) (a, b) = (b, r).
14. Find the smallest positive integer in the given set. [Hint: Theorem
(a) {6u + 15vlu,v
(b) {12r + 11slr,s
E Z}
1.2.]
E Z}
15. The Euclidean Algorithm is an efficient way to find (a, b) for any positive
integers a and b. It only requires you to apply the Division Algorithm
several times until you reach the gcd, as illustrated here for (524, 148).
(a) Verify that the following statements are correct.
524 "" l;l8·3 ,80
;
,
148, = ,, 80·1 + ,68
-�
,•'
0 :s; 80
<
148
0 :S 68< 80
80 =, , 68:3 + ,12
0 :S 12<68
68
os 8<12
=
,,,.12:.5 + 8
,,.:,.
'
ti= ,.8:i+4
,
8
=
os4<8
[The diviSor in each line becomes
the dividend in the next line,
and the remainder in each line
becomes the divisor in the next line.]
[As shown in part (b), the last
remainder, namely 4,
is the gcd (a, b).]
nonzero
4·2 + 0
(b) Use part (a) and Exercises 13 and Example 4 to prove that
(524, 148) = (148, 80) = (80, 68) = (68, 12)=(12, 8)=(8, 4)=(4, 0)=4.
Use the Euclidean Algorithm to find
(d) (322, 148)
(c) (1003, 456)
(e) (5858, 1436)
The equations in part (a) can be used to express the gcd 4 as a linear
combination of 524 and 148 as follows. First, rearrange the first 5 equations in
part (a), as shown below.
80=524 - 148·3
68 = 148 - 80
12 = 80 - 68·3
8=68-12·5
4=12-8
(1)
(2)
(3)
(4)
�
(f) Equation (1) expresses 80 as a linear combination of 524 and 148. Use this
fact and Equation (2) to write 68 as a linear combination of 524 and 148.
(g) Use Equation (1), part (f), and Equation (3) to write 12
combination of 524 and 148.
as
a linear
(b) Use parts (f) and (g) to write 8 as a linear combination of 524 and 148.
(i) Use parts (g) and (h) to write the gcd 4 as a linear combination of 524 and
148, as desired.
(j) Use the method described in parts (t)-(i) to express the gcd in part (c) as a
linear combination of 1003 and 456.
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16
Chapter 1
Arithmetic in l Revisited
B.16. If (a, b) = d, prove that
(� %)
integers rands (Why?). So
(r, s)
.
=
1. [Hint: a= dr and b= ds for some
a/d =rand b/d =sand you must prove that
(a, b) and divide the resulting equation by d.]
= 1. Apply Theorem 1.2 to
17. Suppose (a, b) = 1. If a Ic and b I c, prove that a b I c. [Hint: c = ht (Why?), so
albt. Use Theorem 1.4.]
18. If c > 0, prove that (ca, ch) = c(a, b). [Hint: Let (a, b) = d and (ca, ch)
Show that cd Ik and k Ied. See Exercise 5.]
=
k.
19. If al(b + c) and (b, c) = 1, prove that (a, b) = 1 =(a, c).
20. Prove that (a, b)
=
(a, b +at) for every t
E
Z.
21. Prove that (a, (b, c)) = ((a, b), c).
22. If (a, c) = 1 and (b, c) = 1, prove that (ab, c) = 1.
23. Use induction to show that if (a, b) = 1, then (a, Ii') = 1 for all n 2!: 1. *
24. Let a, b, c
Z. Prove that the equation
E
ax +by == c has integer solutions if
(a, b) I c.
and only if
25. (a) If a, b, u, v E Z are such that
(b) Show by example that if
au
+bv = 1, prove that (a, b) = 1.
au+ bv = d > 1, then (a, b) may not bed.
26. If a I c and b I c and (a, b) = d, prove that ab Ied.
27. If c lab and (c, a)= d, prove that cldb.
28. Prove that a positive integer is divisible by 3 if and only if the sum of its digits
is divisible by 3. [Hint: 103 = 999 +1 and similarly for other powers of 10.]
29. Prove that a positive integer is divisible by 9 if and only if the sum of its digits
is divisible by 9. [See Exercise 28.]
30. If al! a2,
, an are integers, not all zero, then their greatest common
divisor (gcd) is the largest integer d such that d I a1for every i. Prove that
there exist integers u1 such that d = a1u1 + a2u2 +
+ anu,.. [Hint: Adapt
the proof of Theorem 1. 2.]
•
•
•
·
•
·
31. The least common multiple (lcm) of nonzero integers a1, �
, ak is the
smallest positive integer m such that a1lm for i = 1, 2,
, k and is denoted
[a1> �
, ak1.
• • •
•
...
•• •
(a)
•
Find each of the following: [6,
10], [4, 5, 6, 10], (20, 42], and [2, 3, 14, 36, 42].
a1 I t for i = 1, 2,
, k, prove that
, ak] It. [Hint: Denote [ai. a2,
, ak] by m. By the Division
Algorithm, t = mq +r, with 0 s r < m. Show that a1 Ir for i = 1, 2, ... , k.
(b) If t is an integer such that
[ai. a2,
•
•
•
...
•
•
•
Since m is the smallest positive integer with this property, what can you
conclude about r?]
*Induction is discussed in Appendix C.
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1.3
Primes and Unique Factorization
17
32. Let a and b be integers, not both 0, and let tbe a positive integer. Prove that tis
the least oommon multiple of a and b if and only if t satisfies these conditions:
(i)
(ii)
a
] t and b I t;
If a [ c and b Jc, then t I c.
C. 33. If a> 0 and b > 0, prove that
[a, b]
=
(:.�) .([a, b]
is defined in Exercise 31.)
34. Prove that
(a) (a, b)l(a + b, a - b);
•
(b)
if
a is odd and bis even, then (a, b) =(a+ b, a
(c)
if a and bare odd, then
2(a, b)
=
- b);
(a + b, a - b).
Primes and Unique Factorization
Every nonzero integer n except
:!:l
has at least four
distinct divisors, namely 1, -1, n, -n.
Integers that have only these four divisors play a crucial role.
Definition
An lntegerp is said to be prime if p * 0, ±1 and the only divisors {jf pare
·
±1 and±p.
EXAMPLE 1
3, -5, 7, -11, 13, and -17 are prime, but 15 is not (because 15 has divisors
other than :!:: 1 and :!:: 15, such as 3 and 5). The integer 4567 is prime, but prov­
ing this fact from the definition requires a tedious check of all its possible divi­
sors. Fortunately, there are more efficient methods for determining whether an
integer is prime, one of which is discussed at the end of this section.
It is not difficult to show that there
are
infinitely many distinct primes (Exercise 32).
Because an integer p has the same divisors as
p
p, we see that
-
is prime if and only if -p is prime.
If p and q are both prime and p I q, then p must be one of 1, -1, q,
prime, p =fo :!:: 1. Hence,
if p and q are prime and p [ q,
then p
Under what conditions does a divisor of a product
=
be
-
q But since p is
.
±q.
necessarily divide
b
or c?
Theorem 1.4 gave one answer to this question. Here is another.
....
...
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tm
18
Chapter 1
Arithmetic in l Revisited
Theorem 1.5
Let p be an integer with p * 0, ±1. Then p is prime if and only if p has this
property:
whenever p I be, then p I b or p I c.
Proof"" Since this is an "if and only if" statement, there are two parts to the proof.
Step I Assume that p is prime andprove that p has the property stated in the theorem.
Proof of Step 1: If p is prime and divides be, consider the god of
p and b.
b) must be a positive divisor of the prime p. So the only possibilities
are (p, b) = 1 and (p, b) = ±p (whichever is positive). If (p, b) = ±p, then
Pih. If (p, b) = 1, sinc.e p l bc, we must have pie by Theorem 1.4. In every
case, therefore, p I b or p I c. Hence, p has the property stated in the theorem.
Now (p,
Step 2 Asswne that p
is an integer that has the property stated in the theorem and
prove that p iS prime.
Proof of Step 2: This proof is left to the reader (Exercise
14).
•
Corollary 1.6
If p is prime and p I a1a2
Proof"" If p I a1 (aia3 •
•
•
•
•
•
an, then p divides at least one of the a"
a,,), then p I a1 or p I a1a3
we are finished. If p I a2 (¥4
Theorem l.5 again.
•
•
•
•
•
•
If p I tii• we are finished; if
using Theorem l.5 repeatedly. After at most
that is divisible by p.
Choose
an
a,, by Theorem l. 5. If p lato
a,,), then p I ai or p I "3a4
•
•
• a,. b y
not, continue this process,
n steps, there must be an a1
•
integer other than 0, ± 1. If you factor it "as much as possible," you will
find that it is a product of one or more primes. For example,
12 =
4
60
=
12 5
113
=
113
•
3
•
=
2. 2. 3,
=
2
•
2
•
3
•
5,
(prime).
In this context, we allow thepossibility of a "product" withjuSt one factor in case the number
we
begin with is actually a prime. What was done in these examples can always be done:
Theorem 1.7
Every integer n except 0, ±1 is a product of primes.
Proof"" First note that if n is a product of primes, say n = PtP2 flk, then· -n =
(-p1)P2 ·Pk is also a product of primes. Consequently, we need prove
•
•
•
• •
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1.3
the theorem only when
n
Primes and Unique Factorization
19
> 1. The idea of the proof can be summarized
like this:
Let S be the set of all integers greater than 1 that are not a product of
primes. Show that S is the empty set. Then, since there are no integers
in S, it must be the case that every integer greater than 1 is a product of
primes (otherwise, it would be in S).
Proof that S is empty: The proof is by contradiction: We as.rume that Sis
not empty and use that assumption to reach a contradiction. So
assume
that
Sis not empty. Then S contains a smallest integer m by the Well-Ordering
Axiom. Since m E S, m is not itself prime. Hence m must have positive divi­
sors other than 1 or m, say m
=
ab with 1< a< m and 1< b< m. Since
both a and bare less than m (the smallest element of S), neither a nor bis in
S. By the definition of S, both a and bare the product of primes , say
a= P1P2 · • ·p ,
and
with r;;::; 1, s;;::; 1, and each p1, (/jprime. Therefore
is a product of primes, so that m it S. We have reached a contradiction:
m E S and m it S. Therefore, S must be empty.
•
Technology Tip: To find the prime factorization of integers
as
large
as
10-12 dig­
its on a TI graphing calculator, download the FACTOR program on our website
(www.CengageBrain.com). The program uses Theorem 1.10, which is proved on
page 21, to do the factorization. Maple and Mathematica can find the prime fac­
torization of these and much larger integers very quickly.
An integer other than 0, ± 1 that is not prime is called composite. Although a com­
posite integer may have several different prime factorizations, such
as
45= 3. 3. 5,
45= (-3). 5. (-3),
45 = 5. 3. 3,
45= (-5). (-3). 3,
these factorizations are essentially the same. The only differences are the order of the
factors and the insertion of minus signs. You can readily convince yourself that every
prime factorization of 45 has exactly three prime factors, say q1q2q3• Furthermore,
by rearranging and relabeling the q's, you will always have 3 = ±q., 3 = ±'12• and
5= ±q3• This is an example of the following theorem.
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20
Chapter 1
Arithmetic in 1. Revisited
Theorem 1.8
The Fundamental Theorem of Arithmetic
Every integer n except O,
±1 is a product of primes. This prime factorization
is unique in the following sense: If
and
n = P1P2 ···Pr
with each Pi. q1 prime, then r = s (that is, the number of factors is the same)
and after reordering and relabel ing the q's,
P3 = ±q3,
Proof.,. Every integer n except 0,
• • • 1
Pr
=
±q,.
±1 has at least one prime factorization by
Theorem 1. 7. Suppose that n has two prime factorizations, as listed in
the statement of the theorem. Then
P1(pz.P3
so that Pt I qtq2
·
•
•
p ,) = q1qzq3
• •
•
qa,
q8• By Corollary 1.6, Pt must divide one of the q1• By
q 's if necessary, we may assume that Pt I q1•
Since Pt and q1 are prime, we must have p1 = ±q1• Consequ ently,
•
• •
reordering and relabeling the
±q1PzP3 · · ·Pr = q1qz q3
·
·
·
%·
Dividing both sides by qt shows that
pi_± p3p4
so that Pi I q2q3
• •
•
·
·
•
p,) = q2q3q 4
·
•
•
qa,
q,. By Corollary 1.6,p2 must divide one of the q1; as
±qz and
before, we may assume P2 I q2• Henoe, P2 =
±qzP3p4
·
·
·Pr
=
q2q3q 4
·
•
·
qr
Dividing both sides by q2 shows that
p3(±p4
• . •
p,) = q3q-4
• • •
q8.
We continue in this manner, repeatedly using Corollary
1.6 and elimi­
nating one prime on each side at every step. If r = s, then this process
p1 = ±q" P2 = ±q2, • • • , p, = ±q,. So
to complete the proof of the theorem, we must show that r = s. The
leads to the desired conclusion:
proof that r = s is a proof by contradiction: We assume that r =fo s
(which means that r > s or that r < s), and show that this assumption
leads to a contradiction.
First, suppose that r > s. Then afters steps of the preceding process, all
the q's will have been eliminated and the equation will read
±Pi+tP1+2
· ·
·P,
=
L
This equ ation says (among other things) that p, 11. Since the only divi­
sors of 1 are ±1, we have p, =
±1 . However, since p, is prime, we know
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1.3
Primes and Unique Factorization
21
thatp, ¢ ±1 by the definition of "prime". We have reached a contradic­
tion (p,
±1 and p, * ± 1). So r > s cannot occur. A similar argument
=
shows that the assumption r < s also leads to a contraction and, hence,
cannot occur. Therefore, r = s is the only possibility, and the theorem is
proved.
•
Technology Tip: The FACTOR program for
(www.CengageBrain.com) factors an integer
quickly. For example, if n
=
94,017, then n = 3
H•?94017
TI
calculators on our website
as a product of primes relatively
n
·
7 · 112 • 37,
as
shown in F igure 1.
3
7
u
Done
FIGURE1
On Maple, the command ifactor(n ); will produce the prime factorization of
n.
If consideration is restricted to positive integers, then there is a stronger version of
unique factorization:
Corollary 1.9
Every integer
n =
p1p2p3
Pa :S
•
•
•
•
•
n
> 1 can be written in one and only one way in the form
·Pr• where the p1 are p ositive primes such that p1 s p2 s
S Pr·
Proof .. Exercise 12
•
Primality Testing
In theory it is easy to determine if a positive integer n is prime. Just divide
n
by every
integer betwee n 1 and n to see if n has a factor other than 1 or n. Actually, you need only
check prime divisors because any factor of n (except 1) is divisible by at least one prime.
The following primality test greatly reduces the number of divisions that are necessary.
Theorem 1.10
Let
n
> 1. If
n
has no positive prime factor less than or equal to
Vn,
then
n
is prime.
Before proving this theorem, it may be helpful to see how it is used.
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22
Chapter 1
Arithmetic In l Revisited
EXAMPLE 2
To prove that 137 is prime, the theorem says that we must verify that 137 has no
positive prime factors less than or equal to
VI37 ...
11. 7; that is, we need only
show that 2, 3, 5, 1, and 11 are not factors of 137. You can easily verify that
none of them divide 137. Hence, 137 is prime by Theorem 1.10.
The proof of Theorem 1.10 (like several earlier in this chapter) is somewhat more
detailed than is necessary. In particular, the underlined parts of the proof are normally
omitted .
Proof of Theorem 1.10 ... The proof
is by contradiction. Suppose that n is not
n has at least two positive prime factors, say Pt and P21
so that n = pJ P2k for some positive integer k. By hypothesis, n has no
positive mime divisors less than or equal to Vn. Hence, Pt > Vn and
P2 > Vn. Therefore,
prime. Then
n = P1P2k
which says that n
> n,
�
P1P2 > VnVn = n,
a contradiction. Since the assumption that
prime has led to a contradiction. we conclude that n is prime.
n is not
•
Theorem 1.10 is useful when working by hand with relatively small numbers.
Testing very large integers for primality, however, requires a computer and techniques
that are beyond the scope of this book.
• Exercises
A. 1. Express each number as a product of primes:
2.
(a)
5040
(b)
-2345
(c)
45,670
(d)
2,042,040
(a)
Verify that 25 - 1 and 27
-
1 are prime.
(b) Show that 211 - 1 is not prime.
3. Which of the following numbers are prime:
(a)
701
(b) 1009
(c)
1949
(d)
1951
4. Primes p and q are said to be twin primes if q
= p + 2. R>r example, 3 and 5 are
twin primes; so are 11 and 13. Find all pairs of positive twin primes less than 200.
5.
(a)
List all the positive integer divisors of 3"51, wheres, t E Zand s, t > 0.
(b) If
r,
s, t E Z are positive, how many positive divisors does 2'3151 have?
6. If p > 5 is prime and pis divided by 10, show that the remainder is
.......
1, 3,
7, or 9.
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it.
..
..
......
1.3
7.
8.
Primes and Unique Factorization
If a, b, c are integers and p is a prime that divides both a and
thatplbor ple.
+
a
23
be, prove
(a) Verify that x - l is a factor of X' - 1.
(b) If n is a positive integer , prove that the prime factorization of 2'bi 311 - 1
includes 11as one of the prime factors. [Hint:(2211 3") =(22 3r.J
·
•
•
9. Letpbe an integer other than 0, ± l. Prove that p is prime if and only if it
has this property: Whenever r and s are integers such that p = rs, then r =
±1 ors= ±1.
I 0. Letpbe an integer other than 0, ± I. Prove thatpis prime if and only if for
each a E Zeither(a,p) = l orp!a.
11. If a, b, c, dare integers and p is a prime factor of both a thatpis a prime fa ctor of (a+ c) - (b + d).
b and c - d , prove
12. Prove Corollary 1.9.
13. Prove that every integer n > l can be written in the formp['/121
p1 distinct positive primes and every r1 > 0.
• •
•
p�·, with the
14. Let p be an integer other than 0, ± 1with this property: Whenever band
are integers such thatpI be, then p I b or p I c. Prove thatpis prime.
[Hint: If dis a divisor of p, say p =dt, thenpId orpI t. Show that this
implies d = ±por d = ±l .]
c
15. Ifpis prime andpId', is it true that p" Id'? Justify your answer.
[Hint: Corollary 1.6.]
16. Prove that (a, b) =1 if and only if there is no primepsuch thatpI aand p I b.
17. If pis prime and (a,
18.
b) =p, then (a2, il) =?
Prove or disprove each of the following statements:
(a) If pis prime andpI (a2 + b2) andpI (c2 + Jl-), then p I (a2 - c2).
(b) If pis prime andpI(a2 + b 2) andpI (c2 + t:P), then p I (a2 + c2).
(c) If pis prime andpI aandpI (a2 + il), thenpI b.
p'f and b = P/.'l'!f
B. 19. Suppose that a = Pi.' P'i
[If, where p1, Pz, ... , Pk are
distinct positive primes and each r1, s1 2: 0. Prove that aI b if and only if
r1 s s1 for every i.
· •
·
·
· ·
p',; and b fl'i'P'i�
20. If a p�'P'iP?
pt, where P1> p2,
positive primes and each r1, s1 2: 0, then prove that
=
·
=
· •
(a) (a, b) = p rp; PJ"'
•
'
· ·
· · •
• • •
, Pk are distinct
·Pl!, where for each i, n1 =minimum of r1, s1•
(b) [a, b] =p�'�P� ···pi, where t1 =maximum of r1> s1• [See Exercise 31in
Section 1.2.]
21. If c2 =aband (a, b) = 1, prove that aand bare perfect squares.
22. Let n = p�'p'i
p'/, where Ph pz, ... , Pk are distinct primes and each r1 2: 0.
Prove that n is a perfect square if and only if each r1 is even.
· · ·
23.
Prove that aI b if and only if a2 i b2. [Hint: Exercise 19.]
....
..
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24
Chapter 1
Arithmetic
in
l.
Revisited
24. Prove that a I b if and only if d' I b".
25. Let p be prime and 1 s k <p. Prove that p divides the binomial coefficient
Recall that
[
(i)
(:).
p
==
k!(p
� k)r]
26. If n is a positive integer, prove that there exist n consecutive composite
integers.[Hint: Consider (n+ 1)! +2, (n + 1)1 + 3, (n + 1)! + 4, . .]
.
.
27. If p > 3 is prime, prove that# + 2 is composite. [Hint: Consider the possible
remainders when pis divided by 3.]
28. Prove or disprove: The sums
1+2+4,
1+2 + 4 + 8,
1+2 + 4 + 8+ 16, ...
are alternately prime and composite.
29. If n
E
Z and n of- 0, prove that n can be written uniquely in the form n
where k
=
i'm,
2!: 0 and m is odd.
30. (a) Prove that there are no nonzero integers a, b such that cl= 2b2.
[Hint: Use the Fundamental Theorem of Arithmetic.]
(b) Prove that
Vi. is irrational. [Hint: Use proof by contracfution (Appendix.A).
Assume that Vi."" a/b (with a, b
E
Z) and use part (a) to reach a contradiction.]
31. If p is a positive prime, prove that Vfi is irrational. [See Exercise 30.]
32. (Euclid ) Prove that there are infinitely many primes. [Hint: Use proof by
contradiction (Appendix A) . Assume there are only finitely many primes
p1, p2,
Pk• and reach a contradiction by showing that the number
, Pk·l
p1p2
Pk + 1 is not divisible by any of Pi. p2,
33. Let p > 1. If 2P - 1 is prime, prove that p is prime. [Hint: Prove the
contrapositive: If p is composite, so is 'lf - I.]
• • •
·
,
• • •
· •
Note: The converse is false by Exercise 2(b).
C. 34. Prove or disprove: If n is an integer and n
> 2, then there exists a prime p such
that n<p < n! .
35. (a) Let a be a positive integer. If Va is rational, prove that Va is an integer.
(b) Let r be a rational number and a an integer such that I'=
is
an
integer. [Part (a) is the case when n
36. Let p, q be primes with p 2::
=
5, q 2:: 5. Prove that 24 I (p2 -
...
........
Prove that
r
q2).
....
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a.
2.]
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2
CHAPTER
Congruence in "lL and Modular Arithmetic
Basic concepts of integer arithmetic are extended here to include the idea of
"congruence modulo n." Congruence leads to the construction of the set Z11 of all
congruence classes of integers modulo n. This construction will serve as a model
for many similar constructions in the rest of this book. It also provides our first
example of a system of arithmetic that shares many fundamental properties with
ordinary arithmetic and yet differs significantly from it
•
Congruence and Congruence Classes
T he concept of "congruence" may be thought of as a generalization of the equality
relation. Two integers a and b
are
equal if their difference is 0 or, equivalently, if their
difference is a multiple of 0. If n is a positive integer, we say that two integers are con­
gruent modulo n if their difference is a multiple of n. To say that a - b
=
nk for some
integer k means that n divides a - b. So we have this formal definition:
Definition
Let a, b, n be integers with n > 0. Then a is congruent to
[written "a = b (mod n)"], provided that n divides a - b.
b
modulo n
EXAMPLE 1
17 = 5 (mod 6) because 6 divides 17 - 5
because 7 divides 4 - 25
6-.(-4)
=
Remark
=
-2
12. Similarly, 4 = 25 (mod 7)
10.
In the notation
"
a
are really parts of a single symbol;
"
a
=
1 , and 6 = -4 (mod 5) because 5 divides
= b (mod n)," the symbols " = " and "(mod n)"
"a = b" by itself is meaningless. Some texts write
=n b" instead of "a= b (mod n)." Although this single-symbol notation is advanta­
geous,
we
shall stick with the traditional "(mod n)" notation here.
26
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26
Chapter 2
Congruence in Zand Modular Arithmetic
The symbol used to denote congruence looks very much like an equal sign. This is
no accident since the relation of congruence has many of the same properties as the
relation of equality. For example,
we know that equality is
reflexive: a =
a for every integer a;
symmetric: if a=b, then b =a;
transitive: if a= b and b = c, then a= c.
We now
see that congruence modulo n is also reflexive, symmetric, and transitive.
Theorem 2.1
Let n be a positive integer. For all a, b, cEZ,
(1) a= a (mod n);
(2) if a = b (mod n), then b =a(mod n);
(3) if a = b (mod n) and b = c (mod n), then a = c (mod n).
Proof ... (1) To prove that a= a(modn), we must show thatn I (a - a). But
a - a = 0 andn I 0 (see Example 2 on page 9). Hence, n I (a - a) and
a= a(modn).
(2) a= b (mod n) means that a - b = nk for some integer k. Therefore,
- a = -(a - b) = -nk = n(-k). The first and last parts of this
b
equation say that n I
(3)
If a= b (modn)
congruence, there
(b
�
a). Hence, b = a(modn).
and b = c (mod n), then by the definition of
are integers k and t such that a - b
b - c =nt. Therefore,
=
nk and
(a - b) + (b - c) = nk + nt
a - c =n(k + t).
Thus n I
(a - c) and, hence, a= c (mod n).
•
Several essential arithmetic and algebraic manipulations depend on this key fact:
If
a = b and c = d, then a + c = b +
d and
ac
=bd.
We now show that the same thing is true for congruence.
Theorem 2.2
Ifa= b (mod n) and c = d (mod n}, then
(1) a + c = b + d (mod n);
(2) ac = bd (mod n).
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2.1
Congruence and Congruence Classes
Proof• (1) To prove that a+ c = b + d (mod n), we must show that n
(a+ c) -(b+ti). Since a = b (mod n) and c =d (mod n),
n J (a-b) and n I (b- ti). Hence, there are i ntegers k and t
a - b = nk
and
c
divides
we know that
such that
- d = nt.
We use these facts to show that n divides (a+ c)-(h+
ti):
[Aritlunetic]
(a+ c)-(b+ti)= a+ c-b-d
= (a-b) + (c-ti)
=
27
[Rearrange terms.]
[a - b =
nk+ nt
nk a nd c
- d = nt.]
[Factor right side]
(a+ c)-(b+ d) = n(k+t)
The last equation says that n divides (a+ c) -(b+
ti). Hence, a+ c =
b+ d(modn).
(2) We must prove that n divides ac-bd *
ac-bd= ac+O-bd
= ac-bc+ be- bd
[-be+ be= O.]
= (a - b) c + b(c - d) [Factor first nro terms and last two tenns.]
= (nk)c+ b(nt)
ac-bd= n(kc
+
[a-b = nk
bt)
and c
-d= nt by(*) above.]
[Factor nfrom each term.]
The last equation says that n
I (ac
- bd). T herefore, ac
With the equality relation, it's easy to
see
= bd (mod 11).
•
what numbers are equal to a given
number a-just a itself. With congr uence, however, the story is different and leads to
some interesting consequences.
Definition
Let a and
(denoted
n
be integers with n > 0. The
congruence class of a modulo n
[a]) is the set of all those integers that are congruent to a modulo
n, thatlsi
[a]= {bJbEZ
and
b s a (mod n)}.
To say that b
that b
=a (mod n) means that b-a = kn
= a+ kn. Thus
[a]= {b I b =a
(mod
for some integer k or, equivalently,
n)} = {b I h =a+ kn
with kE.Z}
= {a +kn I kEl}.
*The first two lines of this proof
for a suitable expression
are a standard algebraic
technique: Rewrite
0
in the form
-X + X
X.
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28
Chapter 2
Congruence inland Modular Arithmetic
EXAMPLE 2
In congruence modulo 5, we have
[9]
=
=
{9+5k I kEZ}
{9, 9 ± s, 9 ± 10, 9 ± 15,
=
... }
{ .... -11, -6, -1, 4, 9, 14, 19, 24, .. . }.
EXAMPLE 3
The meaning of the symbol
"[ ]" depends on the context.
In congruence
modulo 3, for instance,
[2]
{2 + 3klkEZ}
=
=
{
. . .
, -7,
-
4, -1, 2, 5, 8,
. . . },
but in oongruence modulo 5 the congruence class [2] is the set
{2
+
5k I kEZ}
=
{ ... , -13,
-8, -3,
2,
7, 12,
.
. .}.
This ambiguity will not cause any difficulty when only one modulus is
under discussion. On the few occasions when several moduli are discussed
simultaneously, we avoid confusion by denoting the congruence class of
modulo n by
a
[a]n·
EXAMPLE 4
In congruence modulo 3, the congruence class
[2]
=
{.
.
.
, -7, -4, -1, 2, 5, 8,
.
.
}.
•
Notice, however, that [-1] is the same class because
[-1]
Furthermore,
=
{-1+3k I kEZ}
2= -1
=
{.
.
. ' -1,
-4, -1, 2, 5, . .
. }.
(mod 3). This is an example of the following theorem.
Theorem 2.3
a= c {mod
n) if and only if [a]
Since Theorem
2.3
=
[c].
is an "if and only if" statement, we must prove two different
things:
1.
If
a= c (mod n),
2.
If
[a]
=
then
[a]
=
[c].
[c], then a= c (mod n).
Neither of these proofs will
use
the definition of congruence. Instead, the proofs will
use only the fact that congruence is reflexive, symmetric, and transitive (Theorem 2.1 ).
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2.1
Congruence and Congruence Classes
29
Proof of Theorem 2.3. First, assume that a"" c (modn). To prove that [a]= [c], we first
show that [a] !::[c]. To do this, letbE[a]. Then by definitionb= a(modn). Since
a= c (modn), wehaveb = c(modn)bytransitivity. Therefore,bE [c] and
[a]!:: [c]. Reversing the roles of a andc in this argument and using the fact that
c=a by symmetry, show that [c]!:: [a]. Therefore, [a]= [c].
[a] = [c]. Since a=a (mod n) by reflexivity,
aE[c]. By the definition of [c] , we see that
Conversely, assume that
we have aE[a] and, hence,
a= c (modn).
•
A and Care two sets, there are usually three possibilities: Either A and Care dis­
A = C, or A n C is nonempty but A * C. With congruence classes, however,
there are only two possibilities:
If
joint, or
Corollary 2.4
Two congruence classes modulo
n are
either disjoint or identical.
Proof• If [a] and [c] are disjoint, there is nothing to prove. Suppose that
[a]
b with hE [a] and b E[c ].
b = c (mod n).
Therefore, by symmetry and transitivity, a= c (mod n). Hence, [a] = [c]
n
[c] is nonempty. Then there is
an integer
B y the definition o f congruence class, b = a (mod n) a n d
by Theorem 2.3.
•
Corollary 2.5
Let
n > 1 be an integer and
(1)
consider congruence modulo
n.
If a is any integer and r is the remainder when a is divided by n, then
[a] = [r].
(2) There are exactly n distinct congruences classes, namely, [OJ, [1],
[ 2],
[n - 1],
o
o
•
I
Proof•(l) Let aEZ By
Thus
the DivisionAlgorithm, a= nq + r, with Os r < n.
a - r= qn, so that a = r (mod n). By Theorem 2.3, [a] = [r].
(2) If [a] is any congruence class, then (1) shows
0 s r < n. Hence, [a] must be one of [ O], [l], [2],
that
...,
To complete the proof, we must show that these
To do this, we first show that no two of 0, 1, 2, . ..
n
,n
[a] = [r] with
[n - 1 ] .
classes are all distinct.
- 1 are congruent
modulo n. Suppose that s and t are distinct integers in the list 0, 1, 2, . ..
n
,
- 1. Then one is larger than the other, say t, so that 0s s < t < n.
Consequently, t - s is a positive integer that is less than n. Hence, n does
- s, which means that t ¢ s. Thus, no two of 0, 1, 2, .. . ,
n - 1 are congruent modulo n. Therefore, by Theorem 2.3, the classes [O],
[l], [2], .. . , [n - 1 ] are all distinct. •
not divide t
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30
Chapter 2
Definition
Congruence in 1. and Modular Arithmetic
The set of all congruence classes modulo n is denoted Zn (which is read
"Z mod n").
There are several points to be careful about here. The elements of
Zh are classes,
not single integers. So the statement [5] EZh is true, but the statement 5 E Zti is not.
F urthermore, every element of Zti can be denoted in many different ways. For example,
we know that
2 = 5 (mod 3)
2 =-I
Therefore, by Theorem 2.3, [2]
of
=
(5]
==
(mod
2 = 14 (mod 3).
3)
[-I] "" [14] in 71+ Even though each element
Zti (that is, each congruence class) has infinitely many different labels, there are only
finitely many distinct classes by Corollary 2.5, which says in effect that
The set Z� has exactly n elements.
For example, the set Z3 consists of the three elements [OJ, [l], [2).
• Exercises
A. I. Show that
aP-
I= 1 (modp) for the givenp and a:
(a) a=2,p=5
2.
(a)
If k
=
(c)
(b) a=4,p=7
a
= 3, p=11
1 (mod 4), then what is 6k + 5 congruent to modulo 4?
(b) If r = 3 (mod 10) ands= -7 (mod 10), then what is 2r + 3s congruent to
modulo 10?
3. Every published book has a ten-digit ISBN-10 number (on the back cover
or the copyright page) that is usually of the form x1-x2x3xrx5xl)X1XsXrX10
(where each xi is a single digit).* The first 9 digits identify the book. The last
digit x10 is a
check digit; it is chosen so that
10x1 + 9x2 + 8x3 + 7x4 + 6x5 + 5x6 + 4x1 + 3 x8 + 2x9 + x10 = 0 (mod 11).
If an error is made when scanning or keying an ISBN number into a computer,
the left side of the congruence w ill not be congruent to 0 modulo 11, and the
number will be rejected as invalid. t Which of the following are apparently valid
ISBN numbers?
(a) 3-540-90518--9
(b) 0--031-10559--5
"Sometimes the last digit of an ISBN number is the letter
number
10.
X.
(c) 0--385--49596--X
In such cases, treat X as if it were the
trhe procedures in Exercises 3 and 4 will detect every single digit substitution error (for instance,
3 is entered as 8 and no other error is made).They will detect about 90% of transposition errors (for
instance, 74 is entered as 47 and no other error is made). However, they may not detect muHiple errors.
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2.1
Congruence and Congruence Classes
31
4. Virtually every item sold in a store has a 12-digit UPC barcode which is scanned
at the checkout counter. The first
1 1 digits of a UPC number d1difl3•
• • •
d11d1
2
identify the manufacturer and product. The last digit d12 is a check digit which
is chosen so that
If the congruence does not hold, an error has been made and the item must
be scanned again, or the UPC code entered by hand. Which of the following
UPC numbers were scanned incorrectly?
(a) 037000356691
5.
(a)
(b)
Which of [O], [l], [2], [3] is equal to
Theorems 2.2 and 2.3.)
(b) Which of
(c) 040293673034
833732000625
[52000_! in Zt? [Hint: 5 = I (mod 4); use
[O], [1], [2], [3,] [4] is equal to [42001] in�?
6. If
a= b(mod n) and k In, is it true that a = b(mod k)? Justify your answer.
7. If
a E Z. prove that a2 is not congruent to 2modulo 4or to 3 modulo 4.
8. Prove that every odd integer is congruent to
1 modulo 4 or to 3 modulo 4.
9. Prove that
(a)
I 0.
(n
- a)2 = a2 (mod n)
(b) (2n
- a)2 = a2 (mod 4n)
If a is a nonnegative integer, prove that a is congruent to its last digit mod 10
[for example, 27= 7 (mod 10)].
B.11. If a, bare integers such that
that
a=
a= b(modp) for e very positive prime p, prove
b.
5 and p is prime, prove that [p] [l] or [p]
[Hint: Theorem 2.3 and Corollary 2.5.]
12. If p <'!:
=
13. Prove that
=
[5] in �·
a = b (mod n) if and only if a and b leave the same remainder when
divided by n.
14.
(a) Prove or disprove: If ab= 0 (mod n), then a= 0 (mod n) or b= 0 (mod n).
(b) Do part (a) when n is prime.
15. If (a, n)
16. If
[a ]
=
=
1, prove that there is an integer bsuch that ab = 1 (mod n).
[ 1 ] in Z,., prove that
(a, n)
=
1.
Show by example that the converse
may be false.
17. Prove that
10" = (-lf (mod 11) for every positive n.
18. Use congruences (not a calculator) to show that
(125698) (23797) ./:- 2891235306. [Hint: See Exercise 21.]
19. Prove or disprove: If [a]
20.
(a)
=
Prove or disprove: If
[b] in Z,., then (a, n) = (b, n).
a2 = b2 (mod n), then a = b(mod n) or
a = -b (mod n).
(b) Do part (a) when n is prime.
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32
Chap ter 2
Congruence inland Modular Arithmetic
21. (a) Show that 10"
=
1 (mod 9) for every positive n.
(b) Prove that every positive integer is congruent to the sum of its digits mod
9 [for example, 38 = 11 (mod 9)].
22.
(a) Give an example to show that the following statement is false: If ah
(mod n) and a "1= 0 (mod n), then b = c (mod n).
(b) Prove that the statement in part (a) is true whenever (a, n)
=
=
ac
L
EXCURSION: The Chinese Remainder Theorem (Section 14.1) may be
desired.
covered at this point if
Ill
Modular Arithmetic
The finite set "11,, is closely related to the infinite set Z. So it is natural to ask if it is
possible to define addition and multiplication in "1L,. and do some reasonable kind of
arithmetic there. To define addition in Z,,, we must have some way of taking two classes
in "1L,. and producing another class-their sum. Because addition of integers is defined,
the following tentative definition seems worth investigating:
The sum of the classes [a] and [c] is the class containing a+ c or, in symbols,
[a] Ee [c]
=
[a + c],
where addition of classes is denoted by Ef) to distinguish it from ordinary addition of
integers.
We can try a similar tentative definition for multiplication:
The product of [a] and [c] is the class containing ac:
[a] 0 [c]
=
[ac],
where 0 denotes multiplication of classes.
EXAMPLE 1
In� we have [3] Ee [4]
=
[3 + 4]
=
[7]
=
[2] and [3] 0 [2]
=
[3 2]
·
=
[6]
=
[l].
Everything seems to work so far, but there is a possible difficulty. Every element of
"1L,. can be written in many different ways. In�. for instance, [3] [13] and [ 4] [9]. In
the preceding example, we saw that [3] Ee [4]
[2 ] in�· Do we get the same answer if
we use [ 13] in place of [3] and [9] in place of [ 4]? In this case the answer is "yes" because
=
=
=
[13] ® [9]
=
[13 + 9]
=
[22]
=
[2] .
But how do we know that the answer will be the same no matter which way we write
the classes?
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2.2
Modular Arithmetic
33
To get some idea of the kind of thing that might go wrong, consider these five
classes of integers:
A
=
{.
. .
, -14, -8, -2, 0, 6, 12, 18,
B=
{... ,
c
{. . . ,-9,
D
E
=
-5, -1, 3, 7, 11, 15,
. . .
.
•
.
, -16, -10, -4, 2, 8, 14, 20,
=
{. .
=
{... ;
.
-ll, -7, -3, l, 5, 9, 13,
.. .
}
}
}
.
•
-18, -12, -6, 4, 10, 16, 22,
}
.
. ..}.
These classes, like the classes in �. have the following basic properties: Every integer
is in one of them, and any two of them are either disjoint or identical. Since 1 is in B
and 7 is in C, we could define B + C as the class containing 1 + 7
D.
=
8, that is,B + C =
But Bis also the class containing -3 and C the class containing 15, and so B + C
ought to be theclass containing-3 + 15 = 12. But 12 is in A, so thatB + C = A . Thus
you get different answers, depending on which "rep resentatives" you choose from the
classes B and
C.
Obviously you can't have any meaningful concept of addition if the
answer is one thing this time and something else another time.
In order to remove the word "tentative" from our definition of addition and mul­
tiplication in
Z,.,
we must first prove that these operations do not depend on the
choice of representatives from the various classes. Here is what's needed:
Theorem 2.6
If [a]
=
[b] and [c]
=
[d] in�. then
[a + c]
Proof"' Since [a]
[c]
=
=
and
[b + d]
[ac]
=
[bd].
[b], we know that a= b (modn) by Theorem 2.3. Similarly,
[d] implies that c = d (mod n). Therefore, by Theorem 2.2,
=
a + c = b + d (mod n)
Hence,
and
ac= bd(modn).
by Theorem 2.3 again,
[a + c]
=
[b + d]
and
[ac]
=
[bd].
•
Because of Theorem 2.6, we know that the following formal definition of addition
and multiplication of classes is independent of the choice of representatives from each
class:
Definition
Addition and multiplication in Zn are defined by
[a] EB [c]
=
[a+ c]
and
[a] 0 [c)
=
[ac].
CopJftglll.20t2C,...l.
.
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....... mJ"�--*-ad..-d.u;,"lflKl.b�a.mliag-.m---�l...Amiof;--•rilht1u_,,,.��-..,.1imllljf........_:Dgbl.!lllWtrktioas ...... it.
34
Chapter 2
Congruence in Zand Modular Arithmetic
EXAMPLE 2
Here are the complete addition and multiplication tables for Zs (verify that
these calculations are correct):*
(f)
[OJ
[lJ
[2J
[3J
[4J
8
[OJ
[lJ
[2J
[3J
[4J
[OJ
[OJ
[OJ
[I]
[2J
[3}
[4J
[OJ
[OJ
[OJ
[OJ
[OJ
[1]
(lJ
[21
[3J
[4J
[OJ
[1]
[OJ
[l]
[2J
[3J
[4]
[2J
[2]
[3J
[4J
[OJ
[l ]
[2J
[OJ
[2 }
[41
[l]
[3}
[3J
[31
[4]
[OJ
[l]
[2J
[3J
[OJ
[3J
[l J
[4J
[21
[4J
[41
[OJ
[IJ
[21
[3J
[4J
[OJ
[4J
[3J
[2J
[l}
And here are the tables for "14,:
(f)
[OJ
[lJ
[2J
[3J
[4J
[5J
[OJ
[OJ
[lJ
[3J
[4J
[5J
[lJ
[ ll
[2J
[2 1
[3}
[4J
[5J
[OJ
[2]
[2]
[3J
[4]
[5J
[OJ
[l J
[3J
[3]
[4J
[5J
[OJ
[lJ
[2J
[41
[4}
[5]
[OJ
[lJ
[2J
[3J
[SJ
[5J
[OJ
[lJ
[2J
[3J
[4J
8
[OJ
[lJ
[2J
[3J
[4J
[5J
[OJ
[OJ
[OJ
[OJ
[OJ
[OJ
[OJ
[1]
[OJ
[lJ
[2J
[3J
[4J
[5J
[2J
[OJ
[2J
[4J
[OJ
[2J
[4J
[3J
[OJ
[3J
[OJ
[3J
[OJ
[3J
[41
[OJ
[4J
[21
[OJ
[4J
[2J
[5J
[OJ
[5J
[41
[3J
[2]
[IJ
Properties of Modular Arithmetic
Now that addition and multiplication are defined in Z,.,we want to compare the properties
of these "miniature arithmetics" with the well-known properties of Z The key facts about
arithmetic in Z (and the usual titles for these properties) are as follows. For all a, b , cEZ:
1. If
a, bEZ, then a+bEZ
[Closurefor additionJ
(a+b)+ c.
[Associative addition}
2.
a+(b+c)
3.
a+b
4. a+0
""'
=
b +a.
=
a
=
0+a.
[Commutative additionI
[Additive ickntity}
*These tables are read like this: If [a] appears in the left-hand vertical column and [c] in the top
[i!] ffi [c] appears at the intersection
[a] and the vertical column containing [c].
horizontal row of the addition table, for example, then the sum
of the horizontal row containing
eap,rigm.20:12�1..umiq.A:l.lliala 11--4.....,-aathl t:IDJllilrd,. llC...t,, ardufticlMd.io.wmlllarls,_,. 0.1"�dpll.-mkd.��_,,._ ....... 8om.1M11Bam:.ndkir�.Bdbmbll_...._
...._._q-��._.fld.__...,.a11N:t... �a--.�c...,.� ......
... rir;bl1a-...,,,..��·..,..._w..._._..:dPLI�...-. ..
2.2
Modular Arithmetic
36
5. For each a E Z, the equation
a+
6. If
x
a,
= 0 has a solution in Z.
7.
a(_bc) =(ab)c.
8.
a(_h+ c) =ab+ ac and
9.
10.
[Closurefor multiplication]
bEZ, then abEZ.
[Associative multiplication]
(a + b)c =ac + be.
[ DistributiVe laws]
ab =ha
[Commutative multiplication]
[Multiplicative identity]
a 1 =a =1 a
•
11. If
·
ab =0, then a = 0 or b = 0.
By using the tables in the preceding
example,
you can verify that the first ten of
these properties hold in Zs and Z6 and that Property 11 holds in Zs and fails in
�·But using tables is not a very efficient method of proof (especially for verify­
ing associativity or distributivity). So the proof that Properties 1-10 hold for
any Z,. is based on the definition of the operations in Z,. and on the fact that
these properties
are
known to be valid in Z.
Theorem 2.7
For any classes [a], [b], [c] in Z,.,
1. If [a]E°Zn and [b]E Z,., then [ a]®[b]E Z,..
2.
[a]®([ b]®[c]) =([a]®[b]) ®[c].
3.
[a]® [b] =[b] Et3 [a].
4.
[a]®[O] =[a] =[O]®[ a].
5.
For each [a] in Z,., the equation [a] Et3 X =[O] has a solution in Zr,.
6. If [a]E°Zn and [b]E Zn, then [a] 0 [b]E Zn.
7.
[a] 0 {[b] 0 [c]) = ([a] 0 [b] ) 0 [c].
8.
[a] 0 ([b]®[c]) =[a] 0 [b) ®[a) 0 [c] and
([a]®[b]) 0 [c] =[ a] 0 [c] Et3 [b] 0 [c] .
9.
[a] 0 [b]=[b] 0 [ a].
10. [a] 0 [1]=[a]=[1] 0 [a].
Proof" Properties 1
and 6
are
an immediate consequence of the definition of Et3
and0inZ,..
To prove Property 2, note that by the definition of addition,
[a]®([b] EtJ [ c]) = [a]®[b + c] =[a + (h+ c)].
In Z we know that a+
(b+ c) =(a + b)+ c. So the classes of these
[a + (h + c)] =[(a+ b) + c). By
integers must be the same in Zn; that is,
the definition of addition in Zr,, we have
[(a+ h) + c ] =[a + b] EtJ [c] =([a]® [bD ® [c ].
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36
Chapter 2
Congruence inland Modular Arithmetic
2. The proofs of Properties 31 7, 8, and 9 are
10).
Properties 4 and 10 are proved by a direct calculation; for instance,
[a] 0 [l] [a· l] [a].
For Property 5, it is easy to see that X
[-a] is a solution of the
[a + (-a)] [O]. •
equation since [a] ffi [-a]
This proves Property
analogous (Exercise
=
=
=
=
=
Exponents and Equations
The same exponent notation used in ordinary arithmetic is also used in Zr,. If
and k is a positive integer, then
[a]k denotes the product
[a] 0 [a] 0 [a] 0
·
·
·
0 [a]
[a] EZn,
(k factors).
EXAMPLE 3
In Z5,
[3]2
[3] 0 [3]
=
=
[4]
and
[3]4
=
[3] 0 [3] 0 [3] 0 [3]
=
[l].
As noted on page 9, the set 7L11 has exactly n elements. Consequently, any equation
in 7L11 can be solved by substituting each of these
which ones
are
n
elements in the equation to see
solutions.
EXAMPLE 4
To solve x1 Ee
[5] 0 x
=
[O] in Zt,, substitute each of [O], [1], [2], [3], [4], and [5]
in the equation to see if it is a solution:
x
x2 Ee [5] 0 x
[OJ
[OJ0[0J ffi [5J0[0]
[l]
[1J0[1J ffi [5J0[1]
[2]
[2J0 [2J Ee [5]0[2]
[3]
[3J0[3J ffi [5J0[3]
=
[3J ffi [3J
=
[O]
Yes; solution
[4]
[4J0[4J ffi [5J0[4]
=
[4J ffi [2J
=
[O]
Yes; solution
[5]
[5]0[5] Ee [5]0[5]
=
[lJ Ee [lJ
=
[2]
No
Is
=
=
=
=
[OJ ffi [OJ
=
[O]
Yes; solution
[lJ ffi [5J
=
[O]
Yes; solution
[ 4] Ee [4]
So the equation has four solutions: [O],
Example
x2 ffi [5] 0 x
=
[2]
[O]?
No
[1], [3], and [4].
4 shows that solving equations in Z,, may be quite different from solving
equations in 7L. A quadratic equation in 7L has at most two solutions, whereas the
quadratic equation x1 ffi
[5]0x
=
[OJ has four solutions in Z6•
• Exercises
A. I. Write out the addition and multiplication tables for
(a)
(b) �
Z2
(c)
7L7
(d)
Z-12
In Exercises 2--8, solve the equation.
2.
x1 ffi x
=
[O] in �
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2.3
3. x2
4.
The Structure of ZP (p Prime) and Zn
37
=[lJ in.ls
x4 =[lJ in Zs
5. x2 EB [3J 0 x®[2J = [OJ in Zt,
6. x2 EB [SJ 0 x = [OJin £9
7. x3 EB x2® x®[lJ =[OJ in Zs
8. x3
9.
+
x2 =[2J inZ10
(a) Find an element [aJ in Z7 such that every nonzero element of Z7 is a power
of [aJ.
(c) Can you do part (a) in�?
(b) Do part (a) in :z'.s.
10. Prove parts 3, 7, 8, and 9 of Theorem 2.7.
11. Solve the following equations.
(a) x®x ®x =[OJinZ
3
(b) x®x ®x ®x =[OJ inZ4
(c) xEBx®x®x®x =[OJ in Zs
12. Prove or disprove: If
[aJ 0 [bJ = [OJin Z,,, then [a] = [OJ or [b] = (O].
13. Prove or disprove: If
[a] 0 [bJ =[a] 0 [cJand [a] :f: [ O ] in Z,,, then [bJ =[c].
B. 14. Solve the following equations.
(a) x1+x=[OJinZs
(b) x2 +x =[O] in�
(c) If pis prime, prove that the only solutions of x2+ x =[O] in � are [OJand
[p - lJ.
15. Compute the following products.
(a) ([aJ ®[b])2 inZ2
(b) ([aJ ®[b])3 inZ3
[Hint: Exercise 1 l(a) may be helpful.]
(c) ([aJ® [b])5 in.ls
[Hint: See Exercise l l(c).J
(d) Based on the results of parts (a)-(c), what do you think ([a]® [b])7 is
equal to inZ7?
16.
(a) Find all [a] in Zs for which the equation [a] 0 x =[IJhas a solution. Then
do the same thing for
(b) Zi
II
(d) �
The Structure of ZP (p Prime) and Zn
We now present some facts about the structure of Z,, (particularly when n is prime)
that will provide a model for our future work. First, however, we make a change of
notation.
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38
Chapter 2
Congruence in Zand Modular Arithmetic
New Notation
We have been very careful to distinguish integers in Z and classes in Z,, and have
even used different symbols for the operations in the two systems. By now, however,
you should be reasonably comfortable with the fundamental ideas and familiar with
arithmetic in Z,,. So we shall adopt a new notation that is widely used in mathemat­
ics, even though it has the flaw that the same symbol represents two totally different
entities.
W henever the context makes clear that we are dealing with
ate the class notation
"[a]"
and write simply
"a." In "14,,
Z,,, we shall abbrevi­
for instance, we might say
6 = 0, which is certainly true for classes in � even though it is nonsense if 6 and
0 are ordinary integers. We shall use an ordinary plus sign for addition in Z,, and
either a small dot or juxtaposition for multiplication. For example, in
Zs
we may
write things like
3. 4 = 2
or
4+1=0
or
4 + 4= 3.
On those few occasions where this usage might cause confusion, we will return to the
brackets notation for classes.
EXAMPLE 1
In this new notation, the addition and multiplication tables for Z3 are
+
0
0
0
2
2
CAUTION:
2
1
2
2
0
2
0
0
0
0
0
2
0
2
0
2
0
1
Exponents are ordinary integers-not elements of Z,,. In Z3,
for instance, 24 = 2 • 2 · 2 • 2 = 1and21=2, so that24 * 21
even though 4 = 1in Z3•
The Structure of Zp When p Is Prime
Some of the
�
do not share all the nice properties of Z. For instance, the product
of nonzero integers in Z is always nonzero, but in � we have2
·
3 = 0 even though
2 * 0 and 3 * 0. On the other hand, the multiplication table on page 34shows that the
product of nonzero elements in Z5 is always nonzero. Indeed, Zs has a much stronger
property than Z. W hen
a
a
* 0, the equation
ax
= 1 has a solution in Z if and only if
= ±1. But the multiplication table for Zs shows that, for any
ax
a
* 0, the equation
= 1 has a solution in Zs; for example,
x = 3 is a solution of 2x
x
=
4 is a solution of 4x
=
=
1
I.
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2.3
More generally, whenever
n
The Structure
of
ZP (p Prime) and Zn
39
is prime, Z,. has special properties:
Theorem 2.8
If p > 1 is an integer, then the following conditions are equivalent:*
(1)
p is prime.
(2)
For any a *
O in Zp,
the equation ax= 1 has a solution in
Zp.
(3) Whenever be=O in Zp, then b=O or c=0.
The proof of this theorem illustrates the two basic techniques for proving state­
ments that involve Z,,:
(i) Translate equations in Z,. into equivalent congruence statements in Z. Then
the properties of congruence and arithmetic in Z can be used. The brackets
notation fur elements of Z,, may be necessary to avoid confusion.
Z,. dim1ly, without involving arithmetic in Z.
(ii) Use the arithmetic properties of
In this case, the b rackets notation in
Proof ofTheorem 2.8 �
and [a) #-
use
the first technique. Supposepis prime
[O] in �- Then in Z, a '!/= 0
(modp) by Theorem 2.3. Hence,
a and pis a posi­
1. Since (a,p) also divides
a and p -r a, we must have (a,p)=1. By Theorem 1.2, au + pv=1 for
some integers u and v. Hence, au - 1=p(-v), so that au= 1 (modp).
Therefore [au]=[l] in� by Theorem 2.3. Thus [a][u]=[au]=[1], so
that x = [u] is a solution of [a]x = [l].
p -r
a by
(1 )::::? (2) We
Z,. isn't needed.
the definition of congruence. Now the gcd of
tive divisor of p and thus must be eitherpor
(2) => (3) We use the second technique. Suppose ab = 0 in �· If
a=0, there is nothing to prove. If a :/= 0, then by (2) there exists u E 'Ip
such that au=1 . Then
0=u
•
0=u(ab)= (ua)b=(au)b=1
In every case, therefore,
we
have
(3) => (1) Back to the first
a=
0 orb
=
•
b=b
0.
technique. Suppose that
b and e are any
integers and that pI be. Then be= 0 (modp). So by Theorem 2.3,
[b][e]=[be]=[O]
Hence, by (3),
we
have
in
Zp.
[b] = [O] or [e]=[OJ. Thus, b= 0 (mod p) or e = 0
b orp I c by the definition
(mod p) by Theorem 2.3, which means thatp I
of congruence. Therefore, p is prime by Theorem 1.5.
•
The Structure of�
= 1 need not have a solution in Z,.. For instance,
= 1 has no solution in �. as you can easily verify. The next result tells
when ax=1 does have a solution in Z,,. For clarity, we use brackets notation.
When n is not prime, the equation ax
the equation 2x
us exactly
•see page
508 in Appendix A for the meaning
of "the following conditions are equivalent" and what
must be done to prove such a statement.
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40
Chapter 2
Congruence inland Modular Arithmetic
Theorem 2.9
Let a and n be integers with
The equation
n>
1. Then
[a]x= [1] has a solution in Zn If and only if {a, n)= 1 in Z.
Proof" Since this is an "if and only if" statement, the proof has two parts.
we assume that the equation has a solution and show that (a, n) = 1.
[w] is a solution of [a]x= [ l ], then
First
If
[ a][w]= [1]
[a w]= [1]
aw= 1 (mod n) inZ
[Multiplication in z;J
[11ieorem 2.3]
aw - 1 = kn for some integer k
aw+ n(-k) = 1
[Definition of congruence]
[Rearrange terms]
(a, n) by d. Since dis a common divisor of a and n, there are inte­
r ands such that dr= a and ds= n. So we have
Denote
gers
aw+
drw +
n(-k) = 1
ds(-k) = 1
d(rw - sk) = 1.
So
d 11. Since dis positive by definition, we must have d= 1, that is, (a, n)= 1.
Now we assume that (a, n)= 1 and show that [a]x= fl]has a solu­
tion in Z,,. Actually,
we've already
done this. In the proof of (1) � (2)
(a,p)= 1.
(a, n)= 1, and shows
of Theorem 2.8, the primeness of pis used only to show that
From there on, the proof is valid in any Z,. when
that
[a]x= [l] has a solution in Z,,.
•
Units and Zero Divisors
Some special terminology is often used when dealing w ith certain equations. An ele­
a in Z,, is called a unit if the equation ax= 1 has a solution. In other words, a is
b in Z,. such that ab= 1. In this case, we say that b is the
inverse of a. Note that ab= 1 also says that bis a unit (with inverse a).
ment
a unit if there is an element
EXAMPLE 2
Both 2 and 8
are units
8 = 1. 8 is the inverse of 2 and 2 is the
L. because 3 3= 1. So 3 is its own inverse.
in Z15 because 2
inverse of 8. Similarly, 3 is a unit in
·
•
EXAMPLE 3
Part (2) of
is a unit.
Theorem 2.8 says that whenp is prime, every nonzero element of ZP
Here is a restatement of Theorem 2.9 in the terminology of units.
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it.
...
.......
......
2.3
The Structure of Z P (p Prime) and Zn
41
Theorem 2.10
Let a and n be integers with n >
1. Then
[a] is a unit in Z n if and only if (a, n)
=
1
in z.
A nonzero element a of Z,. is called a zero divisor if the eq uation
nonzero solution (that is, if there is a
nonzero element
c
ax
in Z,. such that ac
=
=
0 has a
0).
EXAMPLE 4
Both 3 and 5 are zero divisors in Z 15 because 3
in L. because 2 2
divisor
·
=
•
5
=
0. Similarly, 2 is a zero
0.
EXAMPLE 5
Part (3) of Theorem 2.8 says that when pis prime,
there
are
no zero divisors in z .
,
• Exercises
A. 1. Find all the units in
(a)
Z7
(b) Zs
in
(c) Zg
(d)
Z10•
(b) Zs
{c) Zg
(d)
Z1o·
2. Find all the zero divisors
3. Based on Exercises
1 and 2j
make a conjecture about units and zero divisors
in�.
4. How many solutions does the equation 6x
(b)
Z;,?
=
4 have in
(c)
Zg?
5. If a is a unit andbis a zero divisor in Z,., show that ab is a zero divisor.
6. If
n is composite, prove that
there is at least one zero divisor in�. (See
Exercise 2.)
7. Without using Theorem 2.8, prove that if pis prime and ab
a
8.
=
(a)
0 orb =
0. [Hint: Theorem 1.8.]
Give three examples of equations of the form ax
= bin
=
0 in Zp, then
Z12 that have no
nonzero solutions.
(b)
For each of the equations in part (a), does the equation ax
=
0 have a
nonzero solution?
B. 9.
(a)
If a is a unit in�. prove that a is not a zero divisor.
( b)
If a is a zero divisor in
Z,., prove that a is not
a unit.
[Hint: Think
contrapositive in part (a).]
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42
Chapter 2
Congruence inland Modular Arithmetic
10. Prove that every nonzero element of Z.. is either a unit or a zero divisor, but
not both. [Hint: Exercise 9 provides the proof of "not both".]
11. Without using Exercises 13 and 14, prove: If a,
b E Z.. and a is a unit, then the
equation ax = b has a unique solution in Z.,. [Note: You must find a solution
for the equation and show that this solution is the only one.]
12. Let a,
b, n be integers with n > 1 and letd =(a, 11). If the equation [a]x = [b]
[r] is a solution, then [ar]
[b] so that ar - b kn for some integer k.]
has a solution in Z,,, prove thatd I b. [Hint: If x
=
=
=
13. Let
a, b, n be integers with n > 1. Let d (a, n) and assumed I b. Prove that
[b] has a solution in Z.. as follows.
=
the equation [a]x
=
(a) Explain why there are integers u,
v, a"
bl> n1 such that au+ nv = d,
a= da1, b = dhto n= dn1•
(b)
Show that each of
[uh1], [uh1+ n1], [uh1+ 2ni] , [uh1o+ 3ni], .. . , [uh1+ (d - l)ni]
is a solution of [a]x = [b].
14. Let a,
b, n be integers with n > 1. Letd= (a, n) and assumed I b. Prove that
[b] has d distinct solutions in Z,, as follows.
the equation [a]x
=
(a) Show that the solutions listed in Exercise 13 (b) are all distinct.
[Hint: [r]
(b)
=
[s] if and only if n I (r - s).]
If x = [r] is any solution of [a]x= [b] , show that [r]
==
[uh1+ kn1] for some
integer k with 0 s ks d - 1. [Hint: [ar] - [auh i]= [O] (Why?), so that
n I (a(r - uh 1)). Show that n1 I (a1(r - uh1)) and use Theorem 1.4 to show
that n1 I (r - uh1).]
15. Use Exercise 13 to solve the following equations.s
(a) 1 5x = 9 in .l18
(b) 25x=
10 in "145•
a ""- 0 and bare elements of Z,, and ax== b has no solutions in Z,,, prove that
a is a zero divisor.
16. If
17. Prove that the product of two units in Z,, is also a unit.
18. The usual ordering of Z by< is transitive and behaves nicely with respect to
addition. Show that there is
(i) if a <
(ii) if
a
ordering of Z,, such that
no
b and b< .c, then a <
< b, then a+
c
<b+
c
c;
for every
c
in Z,,.
[Hint: If there is such an ordering with 0 < 1, then adding 1 repeatedly to both
< n - 1 by (ii). Thus 0 < n - 1 by (i). Add 1
to each side and get a contradiction. Make a similar argument when 1 < O.]
sides shows that 0 < 1 < 2 < ·
·
·
APPLICATION: Public Key Cryptography (Chapter 13) may be covered
at this point if desired.
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CHAPTER
3
Rings
ALTERNATE ROUTE: H you want to cover groups before studying rings,
you should read Chapters 7 and 8 now.
We have seen that many rules of ordinary arithmetic hold not only in Z but also in
the miniature arithmetics Zn. You know other mathematical systems, such as the
real numbers, in which many of these same rules hold. Your high-school algebra
courses dealt with the arithmetic of polynomials.
The fact that similar rules of arithmetic hold in different systems suggests
that it might be worthwhile to consider the common features of such systems.
In the long run, this might save a lot of work: If we can prove a theorem about one
system using only the properties that it has in common with a second system,
then the theorem is also valid in the second system. By " abstracting" the com­
mon core of essential features, we can develop a general theory that includes
as special cases Z, Zno and the other familiar systems. Results proved for this
general theory will apply simultaneously to all the systems covered by the theory.
This process of abstraction will allow us to discover the real reasons a particular
statement is true (or false, for that matter) without getting bogged down in non­
essential details. In this way a deeper understanding of all the systems involved
should result.
So we now begin the development of abstract algebra This chapter is just
the first step and consists primarily of definitions, examples, and terminology.
Systems that share a minimal number of fundamental properties with Z and Zn
are called rings. Other names are applied to rings that may have additional prop­
erties, as you will see in Section 3.1. The elementary facts about arithmetic and
algebra in arbitrary rings are developed in Section 3.2. In Section 3.3 we consider
rings that appear to be different from one another but actually are "essentially the
same" except for the labels on their elements.
43
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...
loll...-.:ilt.
44
Chapter 3
Ill
Rings
Definition and Examples of Rings
We begin the process of abstracting the common features of familiar systems with this
definition:
Definition
A ring is a nonempty setRequippedwith two operations* (usually written
as addition and multiplication) that satisfy the fo llowi ng axioms. For all a,
b,cER:
1. If a ERandbER, then a+ bER.
[Closure for addition]
2. a+ (b+ c) = (a+ b) + c.
[Associative addition]
3. 8+ b = b+ 8.
[Commutative addition]
4. There is an element On in R such
[Additive identity
that a+ OR = a = On + a for every
a ER.
or zero element]
5. For each a ER, the equation
a+ x =On has a solution in R.t
6. lfaERandbER,thenabER.
[Closure for multiplication]
7. a(bc)
[Associative multiplication]
=
(ab)c .
8. a(b+ c)
(a +
=
b)c =
ab + ac and
[Distributive laws]
ac + be.
These axioms are the bare minimum needed for a system to resemble Z and Zn. But
Z and Zn have several additional properties that are worth special mention:
Definition
A commutative ring is a rin g Rthat satisfies this axiom:
9. ab = bafor all a, b ER.
Definition
A ring with identity is a ring
axiom:
(Commutative multiplication]
R that contains an element 1n satisf ying this
[Multiplicative identity]
*"Operation" and "closure" are defined in Appendix
trhose who have already read Chapter
B.
7 should note that Axioms 1-5 simply say that a ring is an
abelian group under addition.
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3.1
Definition and Examples of Rings
45
In the following examples, the verification of most of the axioms is left to the
reader.
EXAMPLE 1
With the usual addition and multiplication,
Z (the integers)
R (the real numbers)
and
are commutative rings with identity.
EXAMPLE 2
The set Z11, with the usual addition and multiplication of classes, is a commuta­
tive ring with identity by Theorem
2.7.
EXAMPLE 3
Let Ebe the set of even integers with the usual addition and multiplication.
Since the sum or product of two even integers is also even, the closure
axioms (1 and
6) hold. Since 0 is an even integer, Ehas an additive identity
element (Axiom 4). If a is even, then the solution of
a+
x =
0 (namely-
also even, and so Axiom 5 holds. The remaining axioms (2, 3,
hold for
all integers and, therefore, are true whenever a, b, care even.
Consequently, Eis a commutative ring. Edoes not have
because no
integer
a) is
7, 8, and 9)
even integer e has the property
that
ae = a
an
=
identity, however,
ea for every even
a.
EXAMPLE 4
The set of odd integers with the usual addition and multiplication is not a
ring. Among other things, Axiom 1 fails: The sum of two odd integers is
not odd.
Although the definition of ring was constructed with 7L and 7L11 as models, there
are many rings that aren't at all like these models. In these rings, the elements may not
be numbers or classes of numbers, and their operations may have nothing to do with
"ordinary" addition and multiplication.
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46
Chapter 3
Rings
EXAMPLE S
The set T= {r, s, t, z} equipped with the addition and multiplication defined
by the following tables is a ring:
+
z
z
z
r
s
r
r
z
t
s
s
s
t
z
t
s
r
r
s
z
r
s
z
z
z
z
z
r
z
z
r
r
r
z
z
s
s
z
z
z
You may take our word for it that associativity and distributivity hold
(Axioms 2, 7, and 8). The remaining axioms can be easily verified from the
operation tables above. In particular, they show that Tis closed under both
addition and multiplication (Axioms 1 and
6) and that addition is commuta­
tive (Axiom 3).
The element z is the additive identity-the element denoted OR in Axiom 4. It be­
haves in the same way the number 0 does in Z (that's why the notation 0R is used in the
axiom), but z is not the integer 0-in fact, it's not any kind of number. Nevertheless,
we shall
call z the "z.ero element" of the ring T.
In order to verify Axiom 5, you must show that each of the equations
r+x=z
s+x=i
has a solution in T. This is easily seen
z+x=z
t +x =z
to be the case from the addition table; for
example, x= r is the solution of r + x = z because r + r = z.
Finally, note that Tis not a commutative ring; for instance, rs= rand
sr = z, so that rs -:¢: sr.
EXAMPLE 6
Let M(IR) be the set of all 2 X 2 matrices over the real numbers, that is, M(IR)
consists of all arrays
where a, b, c, dare real numbers.
Two matrices are equal provided that the entries in corresponding positions are equal;
that is,
(; �) (; �)
=
R>r example,
0
1
) (
=
if and only if
2 + 2
1
-
4
�)
b ut
a = r, b = s, c = t, d =
G
3
2
) (
*
3
5
1
2
)
u.
.
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3.1
Definition and Examples of Rings
47
Addition of matrices is defined by
( )
)
(a'
+
d
c'
a
b'
b
c
=
(a+a'
d'
c+c'
)
b + b'
d+d'.
R>r example,
) (
-2
1
+
4
7)
6
0
=
(
-2+ 7) = ( 7
3+4
l+O
5+6
5)
11
1 ·
Multiplication of matrices is defined by
(a
c
)
b)(w
d
x
y
R>r example,
=
-5)
7
(2.
=
0
(aw +by
cw+dy
z
•
)
ax+ bz .
ex+dz
1 +3. 6
2(-5)+3·7 )
1+(-4)6
0(-5)+( -4)7
(-� -�
}
Reversing the order of the factors in matrix multiplication may produce a different
answer, as is the case here:
)(
-5
7
2
()
3) (
=
-4
1.
.
2+( -5)0
1. 3+(-5)(-4))
6. 2 + 7. 0
6. 3 + 7(-4)
)
23
-10 .
So this multiplication is not commutative. With a bit of work, you can verify that
M(�) is a ring with identity. T he zero element is the 1..ero matrix
which is denotedOandX=
(
-a
-c
-b'\.
- d} is a so1ut10n of
.
Weclaim thatthemultiplicative identity e lement(Axiom l O)isthematrix/ =
To prove this claim, we first multiply a typical matrix in
(
a
c
b)( l
d
0
) (
0 =
1
a·
1+b·0
c·l+d·O
G )
0
I
.
M(R) on the right by I:
a·0+b·1)
c·O+d·l
=
( J
a
c
b ·
d
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48
Chapter 3
Rings
Since multiplication is not commutative here, we also need to check left multiplication
by las well;
b
) ( •a
=
d
1
+ 0 ·c
O•a+I·c
l·b+O· tf\
O·b + l ·d)
-(
a
c
This proves that I satisfies Axiom 10. * Consequently, I is called the identity matrix.
Note that the product of nonzero elements of
M(R) may be the :zero element; for
example,
6·2
-9)6 (4(-3)
2(-3) + 3·2
+
=
4(-9) + 6•6
2(-9) + 3·6
0
) (
=
0
EXAMPLE 7
If R is a commutative ring with identity, then
M(R) denotes the. set of
all
2 X 2 matrices with entries in R. With addition and multiplication defined as
in Example 6,
M(R) is a noncommutative ring with identity, as you can read­
ily verify. For instance, M(Z) is the ring of 2 X 2 matrices with integer entries,
M(O) the ring of 2 X 2 matrices with rational number entries, and M(lL,J the
ring of
2 X 2 matrices with entries from.l,,.
EXAMPLE 8
Let The the set of all functions from IR to R, where R is the set of real
numbers. As in calculus,/+
(f + g)(x)
=
g and fg are the functions defined by
f(x)
+
g(x)
and
(fg)(x)
=
f(x)g(x).
You can readily verify that Tis a commutative ring with identity. The zero ele­
ment is the function h given by h(x)
0 for all x E !R. The identity element is the
=
function
e
given by
e(x)
=
1 for all
x E IR. Once again the product of nonzero
elements of Tmay tum out to be the zero element; see Exercise
36.
We have seen that some rings do not have the property that the product of two
nonzero elements is always nonzero. But some of the rings that do have this property,
such as .l, occur frequently enough to merit a title.
Definition
An integral domain is a commutative ring R with identity 1R * OR that
satisfies this axiom:
11. Whenever a, bER and ab= OR, then a= OR orb= OR.
•checking a possible identity element under both right and left multiplication is essential. There
are rings in which an element acts like an identity when you multiply on the right, but not when you
multiply on the left. See Exercise
11.
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3.1
Definition and Examples of Rings
49
The condition lR if:. OR is needed to exclude the zero ring (that is, the single-element
ring {OR}) from the class of integral domains. Note that Axiom 11 is logically equiva­
lent to its contrapositive.*
Whenever a '# OR and b
*
OR, then ab * OR.
EXAMPLE 9
The ring z of integers is an integral domain. If pis prime, then zp is an integral
domain by Theorem 2.8. On the other hand, Z6 is not an integral domain because
4
·
3 = 0, even though 4 *
0 and 3 * 0.
You should be familiar with the set
fractions
a/b with a, b EZ
Q
of rational numbers, which consists of all
and b '¢: 0. Equality of fractions, addition, and mul tiplica­
tion are given by the usual rules:
a
r
b
s
if and only if
as= hr
a
-
b
·
c
-
d
�
ac
-
bd
Q is an integ ral domain. But Q has an additional property that
does not hold in Z: Every equation of the form ax = 1 (with a * 0) has a solution in
It is easy to verify that
Q. Therefore, Q is an example
Definition
A field
of the next definition.
is a commutative ring
R with
identity 1,q :f. OR that satisfies this
axiom:
12. For each a* 08 in
R, the equation ax= 1R has a solution in R.
Once ag ain the condition IR * OR is needed to exclude the zero ring. Note that
Axiom 11 is not mentioned explicitly in the definition of a field. However, Axiom 11
does hold in fields, as we shall see.in Theorem 3.8 below.
EXAMPLE 10
The set� of real numbers, with the usual addition and multiplication, is a field.
If p is a prime, then zp is a field by Theorem 2.8.
EXAMPLE 11
The set IC of complex numbers consists of all numbers of the form a + bi,
where a, b E� and i2 = -1. Equality in IC is defined by
a+ bi= r + si
if and only if
a= rand b
=
s.
*See Appendix A for a discussion of contrapositives.
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50
Chapter 3
Rings
The set C is a field with addition and multiplication given by
(a+ bt)+ (c +di) =(a+ c) +(b + d)i
(a+ b1)(c+ di) =(ac - bd) + (ad+ bc)i.
The field R of real numbers is contained in C because � consists of all complex
numbers of the form a+ Oi. If a+ bi *
(a + bl)x = 1 is x =c+ di, where
c =a/(a2+ /l-) ER
0 in C, then the solution of the equation
d = -b/(a2+ b'-)E IR (verify!).
and
EXAMPLE 12
Let K be the set of all 2 X 2 matrices of the form
)
b
a '
where a and b are real numbers. We claim thatK is a field. For any two matrices in K,
(
a
-b
(
a
-b
) (
( a+
-b
-d c}
d\ ( ac
b) (
=
b
d'\
c
+
a
c
=
-
- bd
c
a
-d
·
d
b +·d'\
a+ c }
ad
-ad - be
c}
+be)
ac - bd
·
In each case the matrix on the right is in K because the entries along the main
diagonal (upper left to lower right) are the same and the entries on the opposite
diagonal (upper right to lower left) are negatives of each other. Therefore, K is
closed under addition and mu ltiplication. K is commutative because
) (
b
a
-
ac-bd
d - be
-a
) (
ad+bc
ac - bd
-
a
-b
Clearly, the zero matrix and the identity matrix I are in K. If
A=
(
a
-b
)
b
a
is not the zero matrix, then verify that the solution of
X=
(aid
bid
-bid
aid
)
EK,
AX =I is
where
d = a2 +b1•
W henever the rings in the preceding examples are mentioned, you may assume
that addition and multi plication are the operations defined above, unless there is some
specific statement to the contrary. You should be aware, however, that a given set (such
as Z) may be made into a ring in many different ways by defining different addition
and multiplication operations on it. See Exercises 17 and 22-26 for examples.
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3.1
Definition and Examples of Rings
61
Now that we know a variety of different kinds of rings, we can use them to produce
new rings in the following way.
EXAMPLE 13
Let Tbe the Cartesian product Zr, X Z, as defined in Appendix B. Define
addition in Tbythe rule
(a, z) +(a', z') =(a+ a', z + z').
The plus sign is being used in time ways here: Jn the first coordinate on the right-band
side of the equal sign,+ denotes addition in Zr,; in the second coordinate+
, denotes
adlition in Z; the+ on the left of the equal sign is the addition in Tthat is being defined.
Si.tx:e Zr, is a ring and a, a' E Zr,, the first coordinate on the right, a+
a', is in Zr,. Similarly
z+ z' E Z. Therefore, addition in Tis closed. Multipli:ation is defined similarly:
(a, z)(a', z') = (aa', zz').
(3, 5) + (4, 9) (3-+ 4, 5 + 9) (1, 14) and (3, 5)(4, 9)
(3 4, 5 9) = (0, 45). You can readily verify that Tis a commutative ring with
identity. The zero element is (0, 0), and the multiplicative identity is (1, 1). What
For example,
•
=
=
=
•
was done here can be done for any two rings.
Theorem 3.1
Let R and S be rings. Define addition and multiplication on the Cartesian
product R x S by
(r, s) + (r', s') = (r + r', s + s')
and
(r,
s)(r', s')
=
(rr', ss').
Then Rx Sis a ring. If Rand Sare both commutative, then so is Rx S. If both
Rand S have an identity, then so does R x S.
Proof• Exercise 33.
•
Subrings
If R is a ring and S is a subset of R, then Smay or may not itself be a ring under the
operations in R. In the ring Z of integers, for example, the subset E of even integers is
a ring, but the subset 0 of odd integers is not,
as
we saw in Examples 3 and 4. When
a subset S of a ring R is i tself a ring under the addition and multiplication in R, then
we say that Sis a subring of R.
EXAMPLE 14
Z is a subring of the ring Q of rational numbers and Q is a subring of the field
R of all real numbers. Since Q is itself a field, we say that Q is a subfield of Ill.
Similarly, ll is a subfield of the field C of complex numbers.
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62
Chapter 3
Rings
EXAMPLE 15
The matrix rings M(Z) and M(O!) in Example 7 are subrings of
M(lll).
EXAMPLE 16
The ring Kin Example 12 is a subring of
M(lll).
EXAMPLE 17
Let Tbe the ring of all functions from Ill to R in Example 8. Then the subset S
consisting of all continuous functions from R to R is a subring of T. To prove
this, you need one fact proved in calculus: T he sum and product of continuous
functions are also continuous. So Sis closed under addition and multiplication
(Axioms l and 6). You can readily verify the other axioms.
Proving that a subset S of a ring Ris actually a subring is easier than proving directly
that Sis a ring. For instance, since a + b
=
b + a for all elements of
R, this fact is also true
when a, b happen to be in the subset S. Thus Axiom 3 (commutative addition) automati­
cally holds in any subset S of a ring. In fact, to prove that a subset of a ring is actually a
subring, you need only verify a few of the axioms for a ring, as the next theorem shows.
Theorem 3.2
Suppose that Ris a ring and that S is a subset of R such that
(i) Sis closed under addition (if a,
b ES, then a+ bES);
(ii) Sis closed under multiplication (if a, bES, then ab ES);
(iii} OR ES;
(iv) If a ES, then the solution of the equation
a+ x
= OR is in
S.
Then S is a subring of R.
Note condition (iv) carefully. To verify it, you need not show that the equation
a+ x = OR has a solution-we already know that it does because Ris a ring. You need
only show that this solution is an element of S (which implies that Axiom 5 holds for S).
Proof of Theorem 3.2 ... As noted before the theorem, Axioms 2, 3, 7, and 8 hold
for all elements of R, and so they necessarily hold for the elements of the
subset S. Axioms 1, 6, 4, and 5 hold by (i}-(iv).
•
EXAMPLE 18
The subset S = {O, 3} of� is closed under addition and multiplication
(0 + 0
= O; 0 + 3 = 3; 3 + 3 = O; similarly,
0 0
•
=0 =0
·
3; 3
•
3 =
3). By
the
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...... mJ"��-cbuGd.-.m.lly.n.cl.bl�lmmliog��l...Amiiog...- .. :dgbtm-__,_�OOllll!m·a;J'tlmlo1f..._...._:ligl:U�:MpiNit.
3.1
Definition and Examples of Rings
53
definition of S we have 0 ES. Finally, the equation 0 + x = 0 has solution
x = 0 ES, and the equation 3 + x = 0 has solution x = 3 ES. Therefore, Sis a
subring of z6 by Theorem 3.2.
EXAMPLE 19
Let S be the subset of M(lll) consisting of all matrices of the form
Then Sis closed under addition and multiplication because
r
(ab Oc) + (s' 0t) - (a+
b+s
0+0 - a+r
O
c+t
b+s c+t
)es
) (
(ab O )(' 0) (hr+cs ct)
ar
c
s
The identity matrix is in S (let
(: )
t
-
0
0
.
c
and
ES.
a = 1, b = 0, c = 1) and the solution of
0
-c
)
ES.
Hence S is a subring by Theorem 3.2.
EXAMPLE 20
The set
that
Z[Vl]
=
{a + bVl I a, b EZ} is a subring of
(a + bVl}(c + dv'2)
= ac
=
So
R You can easily verify
+adVl + bc'\/2+bdVl v'2
·
(ac + 2hd) + (ad+hc)\12) e Z[V2].
Z [\12] is closed under multiplication. See Exercise 13 for the rest of the proof.
• Exercises
A. 1. Th e following subsets of Z (with ordinary addition and multiplication) satisfy
all but one of the axioms for a ring. In each case, which axiom f ails?
(a) The set S of all odd integers and
0.
(b) The set of nonnegative integers.
2. Let R
=
{O, e,
,b c} with addition and multiplication defined by the tables on
page 54. Assume associativity and distributivity and show that R is a ring with
identity. Is R commutative? Is R a field?
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54
Chapter 3
Rings
3.
+
0
e
b
c
0
0
e
b
c
0
e
b
c
0
0
0
0
0
e
e
0
c
b
e
0
e
b
c
b
b
c
0
e
b
0
b
b
0
c
c
b
e
0
c
0
c
0
c
Let F= {0, e, a, b} with operations given by the following tables. Assume
associativity and distributivity and show that Fis a field.
+
0
0
0
e
a
b
e
e
0
b
a
a
a
b
0
b
b
a
e
e
0
e
0
0
0
0
0
e
0
e
a
b
e
a
0
a
b
e
0
b
0
b
e
a
b
a
a
b
4.
Find matrices A and C in M(ll) such that AC= 0, but CA =#- 0, where 0 is the
zero matrix. [Hint: Example 6.]
5.
Which of the following six sets are subrings of M(R)? Which ones have an identity?
(a) All matrices of the form
(b) All matrices of the form
(c) All matrices of the form
(d) All matrices of the form
(e) All matrices of the form
(f) All matrices of the form
6.
(�
(�
(:
(:
(�
(�
�) with rEQ.
!)
�)
�)
�)
�)
with a, b, c EZ.
with a, b, c ER
with a E II!.
with a ER
with a E ll.
(a) Show that the set R of all multiples of 3 is a subring of Z.
(b) Let k be a fixed integer. Show that the set of all multiples of k is a subring of Z.
7.
Let K be the set of all integer multiples of v2, that is, all real numbers of the
form nVl with n EZ. Show that K satisfies Axioms 1-5, but is not a ring.
8.
Is the subset {l,-1, i, -i} a subring of Cl
9.
Let R be a ring and consider the subset R* of RX R defined by R* = {(r, r) Ir ER}.
(a) If R
=
Z6, list the elements of R*.
(b) For any ring R, show that R* is a subring of R
x
R.
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3.1
10. Is S= {(a, b)
I a+ b =O}
11. Let Sbe the subset of
Prove that Sis a ring.
(b)
Show thatJ =
every A in S).
(c)
1
( )
0
O
Show thatJ is not a
JB 1" B.
55
a subringof Z X Z? Justify your answer.
M(R)
(a)
Definition and Examples of Rings
1
consistingof all matrices of the form
is a
(: :)
.
right identity in S(meaningthat AJ =A for
left identity in Sby
findinga matrix B in S such that
For more information about S , see Exercise 4 .
1
{a + bi la, bEZ}. Show thatZ[i] is a subringof C.
12. Let Z[1] denote the set
13. Let Z Vl denote the set
of
[Ii.
(
)
[See Example 20.]
{a+ bVl I a, bEZ}.
14. Let Tbe the ringin Example 8. Let S =
subringof T.
Show that Z \12 is a subring
(
{/E Tlf(2)
=
)
O}. Prove that Sis a
15. Write out the addition and multiplication tables for
(a) Z2
16. Let A
X Z3
(b) Z2
C !)
=
and 0
x Z2
(� �)
=
such that AB = 0.
(a)
(b)
(c) Z3
X Z3
in M(R). Let S be the set of all matrices B
List three matrices in S. [Many correct answers are possible.]
Prove that Sis a subring of
M(R).
[Hint: If B and Care in S, show that
B + C and BC are in Sby computingA(B + C) and A(BC).]
17. Define a new multiplication inZ by the rule: ab = 0for all
a, b,EZ
Show that
with ordinary addition and this new multiplication Z
,
is a commutative ring.
18. Define a new multiplication in Z by the rule: ab
=
I for all
a, b,EZ. With
ordinary addition and this new multiplication , isZ is a ring?
19. Let S:
{a, b, c} and let P(,S) be the set of
P(,S) as follows:
elements of
S={a, b,
A=
{a};
D = { a, b};
c};
B = {b};
C=
M
+
N= (M - N)
U
E={a, c};
{c};
Define addition and multiplication in
all subsets of S; denote the
0=
P(S) by
(N - M)
F= {b, c};
0.
these rules:
and
MN=MnN.
Write out the addition and multiplication tables for P(S). Also,
B. 20. Show that the subset R
an identity?
:
{O, 3, 6, 9, 12,
21. Show that the subset S= {O, 2,
identity?
see
Exercise 44.
15} of Zl8 is a subring. Does R have
4, 6, 8}of Zio is a subring.
Does Shave an
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56
Chapter 3
Rings
22. Define a new addition ® and multiplication 0 on Z by
a® b
:=:
a+ b
and
1
-
a0 b
=
a + b - ab,
where the operations on the right-hand side of the equal signs are ordinary
addition, subtraction, and multiplication. Prove that, with the new operations
® and 0, Z is an integral domain.
23. Let Ebe the set of even integers with ordinary addition. Define a new
multiplication *on Eby the rule "a*b
=
ab/2" (where the product on the
right is ordinary multiplication). Prove that with these operations Eis a
commutative ring with identity.
24. Define a new addition and multiplication on Z by
a® b
=
a+b
-
and
1
a0b
ab
=
-
(a + b) + 2
Prove that with these new operations Z is an integral domain.
25. Define a new addition and multiplication on Q by
r® s = r + s + 1
and
Prove that with these new operations
r 0 s = rs + r + s.
Q is a commutative ring with identity. Is
it an integral domain?
26. Let L be the set of positive real numbers. Define a new addition and
multiplication on L by
a® b = ab
and
a® b
=
J<>rP.
(a) Is La ring under these operations?
(b) Is La commutative ring?
(c) Is La field?
27. Let S be the set of rational numbers that can be written with an odd
denominator. Prove that Sis a subring of Q but is not a field.
28. Let p be a positive prime and let R be the set of all rational numbers that can
be written in the form r/p' with r, iEZ, and i � 0. Note that Z � R because
each n EZ can be written as n/JJ°. Show that Ris a subring of Q.
29. The addition ta ble and part of the multiplication table for a three-element ring
are given below. Use the distributive laws to complete the multiplication table.
+
r
s
r
r
s
s
s
r
r
s
r
r
r
r
s
r
s
t
r
r
30. Do Exercise 29 for this four-element ring:
+
w
x
y
z
w
x
y
z
w
w
x
y
z
w
w
w
w
w
x
x
y
z
w
x
w
y
y
y
z
w
x
y
w
w
z
z
w
x
y
z
w
w
y
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-...d.'lm:mJ"��._alll.....UO,.dlk1.'1119�..,..�Cmg.Qei...mos--a..:rigM1D__,_mdllllli:lml.romim•..-tilll9V....:DafUllWlrictims
...
-..n:11t.
3.1
Definition and Examples of Rings
31. A scalar matrix in M(IR) is a matrix of the form
numberk.
(� �)
57
for some real
(a) Prove that the set of scalar matrices is a subring of M(IR).
(b) If Kis a scalar matrix, show that KA= AKfor every A
in M(�).
(c) If Kis a matrix in M(R) such that KA = AK for every A in M(IR), show
that Kis a scalar matrix.
fact that KA
[Hint: If K
= AKto show that b
argument with A
=
(� �)
)
(: !).
0 and
t o show that
c=
a=
let A
=
G �).
Use. the
0. Then make a similar
d.]
{a E R I ar = ra for every r ER}. In other
R that commute with every other
element of R. Prove that Z(R) is a subring of R. Z(R) is called the center of
the ring R. [Exercise 31 shows that the center of M(IR ) is the subring of scalar
32. Let
R be a ring and let Z(R
=
=
=
words, Z(R) consists of all elements of
matrices.]
33. Prove Theorem
3.1.
34. Show that M(Z2)(all
2 2
X matr ices with entries in Z2) is a 16-element
noncommutative ring with identity.
35. Prove or disprove:
(a) If Rand Sare integral domains, then R
(b) If R and Sare fields, then R
X Sis an integral domain.
X S is a field.
36. Let T be the ring in Example 8 and let f, g be given b y
ifx :<;;2
ifx >
2
g(x)
{2 =
x
ifx:s;2
ifx > 2.
0
Show that/, gE Tand that/g =Or. Therefore T is not an integral domain.
37.
(a) If Ris a ring, show that the ring M(R ) of
Ris a ring.
all 2 X 2
matrices with entries in
(b) If R has an identity, show that M(R) also has an identity.
38. If R is a ring and a ER, let AR=
of
{r ER I ar= OR}. Prove that AR is a subring
R. AR is called the right annihilator of a. {For an example, see Exercise 16 in
which the ring Sis the right annihilator of the matrix A.]
Q(V2) = (r + sv'2 I r, s E Q}. Show that O(v'2) is a subfield of R.
[Hint: To show that the solution of (r + M
lx= 1 is actually in O(V2),
multiply 1/ ( r + sv'2) by (r
M)/(r
39. Let
-
- s\12).]
40. Let dbe an integer that is not a perfect square. Show that
{a +
bW I a, b E Q} is a subfield of
C.
O(\ld)
[Hint: See Exercise
..
39.]
=
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......
tm
58
Chapter 3
Rings
11.
41. Let S be the ring in Exercise
(a)
Verify that each of these matrices is a right identity in S:
(� k)
(b)
1
I
2
2
x+y=l.
x
+y
= 1, show that
.7
.3 'and
)
{x
\y
Prove that the matrix
(c) If
'
)
(·7.3
x
y
-�}
(-12
is a right identity in S if and only if
G ;)
is not a left identity in S.
42. A division ring is a (not necessarily commutative) ring R with identity
I R¢ OR that satisfies Axioms
11 and 12 (pages 48 and 49). Thus a field is a
commutative division ring. See Exercise 43 for a noncommutative example.
b are nonzero elements of R.
(a) If bb = b, prove that b = IR. [Hn
i t: Let be the solution of bx = lR and
Suppose R is a division ring and
a,
u
no te that bu
(b)
= b2u.]
If u is the solution of the equation ax= IR, prove that u is also a solution
of the equation xa
=
IR. (Remember that R may not be commutative.)
[Hn
i t: Use part (a) with
b=
ua.
]
43. In the ring M(C), let
1=
G �)
i
= (i0
�)
J
•
-1
0
= (-1
�)
k= e
�)
The product of a real number and a matrix is the matrix given by this rule:
r(�
;) (: :)
=
The set Hof real quaternions consists of all matrices of the form
al
+ bj + cj +dk
=
( ) +b(;
a
1
0
o
O\
1) +d(o.1• 0;)
(
+
0 -i}
1
0
0)
+ (-c0 0 +(di0 di0
-bi)
o
1
c
_
)
c
-
where
(a)
(b)
a,
b,
c,
bi
( a+
+ di
-c
c
a
)
+di)
- bi '
and dare real numbers.
Prove that
jl = j2 = kl = -1
ij =-ji = k
jk =-kj =
ki =-ik = j.
i
Show that His a noncommutative ring with identity.
eap,ngm.20:12�1..Mmiq.A:l.llialaa--&.....,-aatn.t:IDJllilrd,. llC...t,,ar�io.wtdliarl:a,_,. 0..1"�dpll.-mkd.�1r1C11Hm.�M
.ftom.1M•Bam:.ndkir�.Bdbmbll_...._
........ q-��
fld.�dlN:t Cl'Na!S---.�C...�
rigbtlD...,,,..��-...,.--il......_,.:dPLI� ........
...
...
.......
......
3.2
Basic Properties of Rings
59
[Hint: If M al +
bi+cj + dk, then verify that the solution of the equation Mx = 1 is the
matrix tal - tbi - tcj - tdk, where t
1f(a2 + I} + c2 + d2).]
(c) Show that His a division ring (defined in Exercise 42).
=
=
(d) Show that the equation
x2 -1 has infinitely many solutions in H.
[Hint: Consider quaternions of the form 01 +bi+cj - dk, where
b2 + c2 + d2 = l.]
=
44. Let S be a set and let P<..S) be the set of all subsets of S. Define addition and
multiplication in P(S) by the rules
M + N = (M - N) U (N - M)
(a)
and
MN= Mn N.
Prove that P(S) is a commutative ring with identity. [The verification of
additive associativity and distributivity is a bit messy, but an informal
discussion using Venn diagrams is adequate for appreciating this example.
See Exercise
19 for a special case.]
(b) Show that every element of P(S) satisfies the equations
x
+x = Or<SJ·
x2 = x and
C. 45. Let C be the set Iii X R with the usual coordinatewise addition (as in
Theorem
3.1) and a
new multiplication given by
(a, b)(c, d) = (ac - bd, ad+be)
Show that with these operations C is a field.
46. Let
I
r
ks + I for some k with
, (s - l)r} of .Z,. is a ring with
and s be positive integers such that r divides
s k 181
r.
Prove that the subset
{O, r, 2r, 3r,
. •
.
identity ks+ l under the usual addition and multiplication in Z,4• Exercise 21
is a special case of this result.
APPLICATION: Applications of the Chinese Remainder Theorem
(Section 14.2) may be covered at this point if desired.
Ill
Basic Properties of Rings
When you do arithmetic in Z, you often use far more than the axioms for an integral
domain. For instance, subtraction appears regularly,
as
do cancelation and the various
rules for multiplying negative numbers. We begin by showing that many of these same
properties hold in every ring.
Arithmetic in Rings
Subtraction is not mentioned in the axioms for a ring, and we cannot just assume
that such an operation exists in
an
arbitrary ring. If we want to define a subtraction
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l(lllN .
60
Chapter 3
Rings
operation in a ring, we must do so in terms of addition, multiplication, and the ring
axioms. The first step is
Theorem 3.3
For any element a in a ring R, the equation a + x =OR has a unique solution.
Proof ... We know that a+ x = 0 R has at least one solution u by Axiom 5. If
vis
also a solution, then a+u=OR and a+v=OR, so that
v=�+v=�+aj+v=�+aj+v=u+�+�=u+�=L
Therefore, u is the only solution.
•
We can now define negatives and subtraction in any ring by copying what happens
in familiar rings such as Z. Let R be a ring and aE R. By Theorem 3.3 the equa­
tion a+ x =OR has a unique solution. Using notation adapted from Z, we denote this
unique solution by the symbol "-a." Since addition is commutative,
-a is the unique element of R such that
a +(-a) = OR = (-a) + a.
In familiar rings, this definition coincides with the known concept of the negative of
an element. More importantly, it provides a meaning for "negative" in any ring.
EXAMPLE 1
In the ring z6, the solution of the equation 2 + x = 0 is 4, and so in this ring
-2=4. Similarly, -9 = 5 in Z14 because 5 is the solution of 9 + x = 0.
Subtraction in a r ing is now defined by the rule
b
- a means b +(-a).
In Z and other familiar rings, this is just ordinary subtraction. In other rings we have
a new operation.
EXAMPLE 2
In�wehavel- 2= 1+(-2)= 1+4=5.
In junior high school you learned many computational and algebraic rules for deal­
ing with negatives and subtraction. The next two theorems show that these rules are
valid in any ring. Although these facts are not particularly interesting in themselves, it
is essential to establish their validity so that we may do arithmetic in arbitrary rings.
Theorem 3.4
If a + b = a
+c
in a ring R, then
b = c.
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B as ic Properties of Ri ngs
3.2
61
Proof"' Adding - a to both sides of a + b=a + c and then using associativity
and negatives show that
-a + (a + b) = -a + (a + c)
(-a
+
a) + b= (-a + a) +
OR + b =OR+
b = c.
c
c
•
Theorem 3.5
For any elements a and b of a ring R,
(1) a ·OR= OR=OR· a. In particular, OR· OR= OR.
(2) a(-b) = -ab
and
(-a)b = -ab.
(3) -(-a)=a.
(4) -(a + b} =(-a) + {-b).
(5) - {a - b} = -a + b.
(6) (-a)(-b) =ab.
If R has an identity, then
Proof"' (1) Since OR + OR= OR, the distributive law shows that
a· OR + a ·OR = a(OR + ORl
=
a OR =a· OR + OR.
·
Applying Theorem 3.4 to the first and last parts of this equation shows
that a·OR= OR. The proof that
OR· a= OR is similar.
(2) By definition, -ab is the unique solution of the equation
ab +
=OR,
x
and so any other solution of this equation must be equal
to -ab. But x =a(-b) is a solution because, by the distribution law
and (1),
ab + a(_-b) = a[b + (-b)] = a[OR ] =OR.
Therefore,
a(_-b) = -ab. The other part is proved similarly.
(3) By definition,
a is a
-(-a) is the unique solution of (-a) + x =0R· But
solution of this equation since
(-a) + a =OR. Hence, -(-a)=a
by uniqueness.
(4) By definition, -(a+ b) is the unique solution of (a + b) + x =
OR, but (-a) + (-b) is also a solution, because addition is commutative,
so that
(a+ b) + [(-a) + (-b)] =a + (-a} + b + (-b)
=
OR
+
OR
=OR.
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62
Chapter 3
Rings
Therefore, -(a+ b) =(-a)+ (-b) by uniqueness.
(5) By the definition of subtraction and (4) and (3),
-(a - b) =-(a+(-b)) =(-a)+(-(-b)) =-a+ b.
(6) ( -a)(-b) = -(a (-b)) [By th£ second equation in (2), with -h in
place of b]
-(-ab)
=ab
[By th£first equation in (2)]
=
[By (3), with ab in place of a]
(7) By (2),
(-1.R)a
=
-(lRa)= -(a) = -a.
•
When doing ordinary arithmetic, exponent notation is a definite convenience, as is
its additive analogue (for instance, a+ a +a =3a). We now carry these concepts over
to arbitrary rings. If R is a ring, a ER, and n is a positive integer, then we define
a•= aaa ••·a
(n factors).
It is easy to verify that for any a E R and positive integers
m
and n,
and
If R has an identity and a * OR, then we define r./l to be the element lR. In this case, the
exponent rules are valid for all m, n 2!: 0.
If R is a ring, a ER, and n is a positive integer, then we define
= a + a + a + • ·+a. (n summands)
=(-a)+(-a)+(-a)+···+(-a). (nsummands)
na
-na
•
Finally, we define Oa = 0R· In familiar rings this is nothing new, but in other rings it
gives a meaning to the "product" of an integer n and a ring element a.
EXAMPLE 3
Let R be a ring and a, b ER. Then
(a + b)2 =(a+b)(a +b) =a(a+ b)+b(a+b)
=aa+ab+ha+bh =ti- +ab+ ha+b1.
Be careful here. If ab * ha, then you cant combine the middle terms. If R is a com­
mutative ring, howevei; then ab ha and we have the familiar pattern
=
(a+b)2=rl-+ab+ba+�=ri-+ab+ ab+�=ri-+�+�
For a calculation of (a +b)" in a commutative ring, with n > 2, see the Binomial
Theorem in Appendix E.
It's worth noting that subtraction provides a faster method than Theorem 3.2 for
showing that a subset of a ring is actually a subring.
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..
..
3.2
Basic Properties of Rings
63
Theorem 3.6
Let S be a nonempty subset of a ring R such that
(1)Sis closed under subtraction (if a, bES, then a - bES};
(2)S is closed under multiplication (ifa, bES, then abES).
Then Sis a subring of R.
Proof"' We show that S satisfies conditions (i)-(iv) of Theorem 3.2 and hence
is a subring. The conditions will be proved in this order: (ii), (iii), (iv),
and (i).
(ii) Hypothesis (2) here is identical with condition (ii) of Theorem 3.2 .
Hence, S satisfies condition (ii).
(iii) Since Sis nonempty, there is some element c with
(with a= c and b = c), we
see that c
cES.
- c =OR is in S.
Applying (1)
Therefore, S
satisfies condition (iii) of Theorem 3.2.
(iv) If a is any element of S, then by (1), OR
-a is
the solution of a +
x = OR,
- a=
-ais also in S. Since
condition (iv) of Theorem 3.2 is
satisfied.
(i) If a, bE S, then -b is in S by the proof of (iv). By (1), a
a + b is in S. So S satisfies condition
Therefore, Sis a subring of R by Theorem 3.2.
(i) of Theorem
-
( -b) =
3.2.
•
Units and Zero Divisors
Units and
zero
divisors in Z,. were introduced in Section 2.3. We now carry these con­
cepts over to arbitrary rings.
Definition
An element a in a ring R with identity is called a unit if there exists u ER
such that au= 1n= ua. l n thiscase theelemerit u iscalled the(multiplica­
tive) inverse of a and is denoted
.,-1•
EXAMPLE 4
The only units in Z
are
1 and -1.
EXAMPLE 5
By Theorem 2.10, the units in Z15 are 1, 2, 4, 7, 8, 11, 13, and 14. For instance,
8 = 1, so 2-1 = 8 and s-1 = 2.
2
·
..........
..
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....:Dgbl.!lllWtrktkJas
...
.......it.
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tinl.p:dJCCIGl mAJM__....fmn.
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64
Chapter 3
Rings
EXAMPLE 6
Every nonzero element of the field �is a unit: If
a
-;; = 1. The same
1
:f. 0, then a
•
thing is true for every field F. By definition, F satisfies Axiom 12: If a # Op, then
the equation ax = 1, has a solution in F. Hence,
Every nonzero element of a field is a unit.
EXAMPLE 7
. (a �)
Amatr1x
c
in M(�) such that ad - be -:/= 0 is a unit b ecause, as you can
(
easily verify,
0
1
)
and
d
ad-be
-c
ad-be
In particulat; each of these matrices is a unit:
A
=
G �).
B
=
-b
ad-be
a
ad-be
c=
C-� �).
)(
a
=
!) G �).
c
(1;3 �).
Units in a matrix ring are called inYertible matrices.
EXAMPLE 8
Let Fbe a field andM(F) the ring of 2 X 2 inatrm> with entries in F. If
A
=
(: !)
EM(F) and ad-be
'¢ OF> then ad-be is a unit in Fby Example 6.
Thecom]:Utations inF.xample7,with d 1
a
aninvertiblematrix[unit inM{F)]with
inverse
Definition
rqiacedb y(ad-bet1,showthatAis
(-cd(ad
- bc)-1
(ad bc -1
- bc
_
-
)
-b(ad bc)-1
a(ad- bc)-1 ·
)
An element a in a ring Risa zero divisor provided that
(1)
a ;e- OR.
{2} There exists a nonzero element c in R such that ac = OR or ca = OR.
Note that in requirement (2), the element c is not unique: Many elements in the ring
may satisfy the equation
ax=
OR or the equation
xa
=OR (Exei:cise
.......8om.1M•Bam:.ndkir�.Bdbmbll_...._
...--if�:dpa.R!Jlltril:tlima......iL
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....... q-��
fld.__...,.dlN:t �a--.�c.a.�
rigbtlD---��-
...
...
........
6). Furthermore,
3.2
Basic Properties of Rings
ac
in a noncommutative ring, it is possible to have
Section 3.1).
66
= OR and ca :!- OR (Exercise 4 in
EXAMPLE9
Both 2 and 3 are zero divisors in Z6 because 2 • 3
zero divisors in Z12 because 4
9
•
=
=
0. Similarly, 4 and 9 are
O.
For a zero divisor A in a matrix ring, it is possible to find a matrixC such that
AC=OandCA=O.
EXAMPLE 10
Let F be a field. A nonzero matrix
zero divisor because,
as
(: �)
you can easily
inM(.F) such that ad
verify,
-
be=OF is a
In particular, each of the.ie matrices is a zero divisor in the given ring:
A= G �)
inM(R),
B=
4/3
(
-2
��) inM(Q), and C =
(� !)
inM(Z6).
EXAMPLE 11
1 1: If
Every integral R domain satisfies Axiom
ab=OR, then a=OR orb =OR.
In other words, the product of two nonzero elements cannot be 0. Therefore,
An integraJ domain contains no zero divisors.
F inally, we present some useful facts about integra l domains and fields.
Theorem 3.7
Cancelation is valid in any integral domain R:
If a
#OR and ab= ac in R, then
b
=c.
Cancelation may fail in rings that are not integral domains. In Z12, for instance,
2
•
4= 2
·
10, but 4 :!-
Proof ofTheorem 3.7
10.
...
If ah = be, then ah
a :!- OR, we must haveb
dicting Axiom
... .......
-
c
=OR
-
be = OR, so that a(b
(if not,
1 1). Therefore,b =c.
-
c)=OR.
Since
then a is a zero divisor, contra­
•
...
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66
Chapter 3
Rings
Theorem 3.8
Every field
F is
an integral domain.
Proof • Since a field is a commutative ring with identity by definition , we need
only show that Fsatisfies Axiom 11: If ah = 0p, then a = 0For h = Op.
So suppose that ah = Op If h = Op, there is nothing to prove. If b #: Op,
then b is a unit (Example 6). Consequently, by the definition of unit and
part (1) of Theorem 3.5,
a= alp= abh-1 = Opb-1 = Op
So in every case, a =Op orb= Op. Hence, Axiom 11 holds and Fis an
integral domain . •
The converse of Theorem 3.8 is false in general (Z is an integral domain that is not
a field), but true in the finite case.
Theorem 3.9
Every finite integral domain Risa field.
Proof• Since Ris a commutative ring with identity, we need only show that for
each a#: OR, the equation ax= lR has a solution. Let ah a,;,
, a,, be
the distinct elements of Rand suppose a,#: OR. To show that a,x= 1R
has a solution, consider the products a1ai. a,a,;, a,a3,
, a,a,.. If a1#: a1,
then we must have a,a1 #: a,a1 (because ata1= l4a1 would imply that 0i = a1
by cancelation). Therefore, a1a1, a1a'b . . . , a,a,. are n distinct elements of
R. However, Rhas exactly n elements all together, and so these must be
all the elements of Rin some order. In particular, for some}, C4a1= 1R.
Therefore, the equation arJC = 1R has a solution and Ris a field. •
• • •
• • •
• Exercises
A. 1. Let R be a
(a)
ring and a,b ER.
(a+ h)(a-b) =?
(b) (a+ b)3 =?
(c) What are the answers in parts (a) and (b) if R is commutative?
2.
Find the inverse of matrices A, B, and C in Example 7.
3.
An element e of a ring R is said to be idempotent if e1= e.
(a) Find four idempotent elements in the ring M(R).
(b) Find all idempotents in Z12•
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3.2
Basic Properties of Rings
4. For each matrix A find a matrix Csuch that AC = 0 or CA
A=
5.
(
= 0:
-10)·
5
-2
67
1/4
4 '
)
3/2 .
(a) Show that a ring has only one zero element. [Hint: If there were more
than one, how many solutions would the equation OR+ x =OR have?]
(b) Show that a ring R with identity has only one identity element.
(c) Can a unit in a ring R with identity have more than one inverse? Why?
6. (a) Suppose A and Care nonzero matrices in M(IR) such that AC= 0. If k
is any real number, show that A(kC) = 0, where kC is the matrix C with
every entry multiplied by k. Hence the equation AX= 0 has infinitely
many solutions.
(b) If A =
7. Let
G �)
R be a ring
.find four solutions of the equation AX= 0.
with identity and let S
= {nlR I nE Z}. Prove that Sis a
a ER is on page 62. Also see
subring of R. [The definition of na with n E Z,
Exercise
27.]
8. Let R be a ring and
b a fixed element of R. Let
is a subring of R.
9. Show that the set S of matrices of the form
numbers is a subring of
I 0. Let
M(IR).
T = {rb
(: �)
I rER}.
Prove that T
.with a and b real
R and S be rings and consider these subsets of RX S:
R=
{(r,Os) I rER}
(a) If R = Z3 and S
""
and
S ={(OR,
s) J SES}.
Z5• What are the sets Rand S?
(b) For any rings R and S, show that Ris a subring of RX S.
(c) For any rings Rand S, show that Sis a subring of RX S.
11.
Let
R be a ring and ma fixed integer. Let S = {r ER J mr =OR}· Prove that S
is a subring of R.
12.
Let
a
and b be elements of a ring
R.
(a) Prove that the equation a +
x
= b has a
unique solution in
R.
(You
must prove that there is a solution and that this solution is the only
one.)
(b) If Ris a ring with identity and a is a unit, prove that the equation ax = b
has a unique solution in R.
13.
Let Sand The subrings of a ring
R. In (a) and (b), if the answer is "yes,"
prove it. If the answer is "no," give a counterexample.
(a) Is Sn Ta subring of R?
(b) Is SU Ta subring of R?
..
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...... tm
68
Chapter 3
Rings
14. Prove that the only idempotents in an integral domain Rare OR and
Exercise
15.
lR. (See
3.)
(a) If a and b are units in a ring Rwith identity, prove that ab is a unit whose
inverse is (ab)-1 b -1a- 1•
(b) Give an example to show that if a andb are units , then a-1b-1 need not be
=
the multiplicative inverse of ab.
16. Prove or disprove: The set of units in a ringRwith identity is a subring ofR.
17. If u is a unit in a ring Rwith identity, p rove that u is not a zero divisor.
18. Let
a be a nonzero element of a ring Rwith identity. If the equation ax=lR
u and the equ ation ya=lR has a solution u, prove that u=u.
has a solution
19. LetR and Sbe rings with identity. What are the units in the ring R X S?
20. LetR and Sbe nonzero rings (meaning that each of them contains at least
one nonzero element). Show thatR X S contains zero divisors.
21. Let Rbe a ring and let
a be a nonzero element ofRthat is not a zero divisor.
Prove that cancelation holds for a; that is, prove that
(a) If ab= acinR, thenb=c.
(b) If ha = ca inR, thenb =
22.
c.
(a) If ab is a zero divisor in a ring R, prove that a orb is a zero divisor.
(b) If a orb is a zero divisor in a commutative ring Rand ab# OR, prove that
ab is a zero divisor.
23. (a) LetR be a ring and
a, b ER. Let m and n be nonnegative integers and
prove that
(i)
(m+ n)a=ma+ na.
(ii)
m(a+ b)=ma+ mb.
(iii)
m(ab) = (ma)b = a(mb).
(iv)
(ma)(nb) == mn(ab).
(b) Do p art (a)when m and n are any integers.
m and n be p ositive integers.
"<+"
"
(a) Show that a"'a =a
and (aj" =a-.
24. LetR be a ring and a, b ER. Let
(b) Under what conditions is it true that (abf=a"b"?
25. Let Sbe a su bring of a ringRwith identity.
(a) If Shas an identity, show by example that ls may not be the same as lR.
(b) If both Rand Sare integral domains , prove that ls= lR.
B. 26. Let She a subring of a ring R. Prove that
the equation a +
Os=OR. [Hint: For a E S, consider
x=a.]
27. LetRbe a ring with identity and h a fixed element ofRand let S=
Is Snecessarily a subring of Kl [Exercise 7 is the case when b
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....._._q-��._.fld.__...,.dkt._owadl._.....--..c...� ........ rir;bl1a-...,,,..��·...,.
=
{nb [ nEZ}.
lR.]
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... w......_..:dPLI�...-...
3.2
28. Assume that R=
Basic Properties of Rings
69
{OR, lR, a, b} is a ring and that a and bare units. Write out
the multiplication table of R.
29. Let Rbe a commutative ring with identity. Prove that Ris an integral domain
if and only if cancelation h olds in R (that is ,
b=
a if: OR and ab = ac in Rimply
c).
30. Let Rbe a commutative ring with identity and b ER. Let Tbe the subringof all
multiples of b (as in Exercise 8).
If u is a unit in Rand u ET, prove that T
31. A Boolean ring is a ring Rwith identity in which x2 = x for every
examples,
(a)
see
x ER.
=
R.
For
Exercises 19 and 44 in Section 3.1. If Ris a Boolean ring, prove that
a+ a= OR for every a ER, which means that a=
(a + a)2.]
(b) Ris commu tative.
-a.
[Hint: Expand
[Hint: Expand (a+ b)2.]
32. Let Rbe a ring without identity. Let Tbe the set R X Z. Define addition and
multiplication in Tby these rules:
(r, m) + (s, n) = (r + s, m + n).
(r, m)(s, n) = (rs
+
ms
+ nr, mn).
(a) Prove that Tis a ring with identity.
(b) Let R consist of all elements of the form (r, 0) in T. Prove that Ris a
subring of T.
33. Let R be a ring with identity.
34. Let Fbe a field and A
=
If ab and a are units in R, prove that b is a unit.
(: !)
a matrix in M(F).
(a) Prove that A is invertible if and only if
8, and 10 and Exercise I7].
(b) Prove that A is a zero divisor
35. Let A
(a) If
=
( )
a
b
e
d
ad
-
be if;
O_,... [Hint: Examples 7,
if and only if ad - be= o_,...
.
.
.
.
b e a matrix with mteger entries.
ad - be= ±I, show that A is invertible in M(T). [Hint: Example 7.]
(b) If ad - be * 0,
M(Z).
1, or
-1, show that A is neither a unit nor a zero divisor in
[Hint: Show that A has an inverse in M(lli) that is not in M(Z ); see
Exercise S(c ). For zero divisors , see Exercise 34(b ) and Example 10.]
36. Let R be a commutative ring with identity. Then the set M(R) of 2 X 2
(: �)
matrices with entries in R) is a ring with identity by Exercise 37 of Section
If A
=
M(R).
�
3.1.
E M(R) a d ad - be is a unit in R, show that A is invertible in
[Hint: Replace ad_ be by (ad - be)-1 in Example 7.]
37. Let Rbe a ring with identity and a, bER. Assume that
a is not a zero divisor.
ab = IR, if and only if ha= IR. [Hint: Note that both ab = IR and
ha= IR imply aha= a (why?); use Exercise 21.]
Prove that
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70
Chapter 3
Rings
38. Let R be a ring with identity and a, b ER. Assume that neither a nor b is a
zero divisor. If ab is a unit, prove that a and bare units. [Hint: Exercise 21.]
39. (a) If Ris a finite commutative ring with identity and a ER, prove that a is
either a zero divisor or a unit. [Hint: If a is not a
zero
divisor, adapt the
proof of Theorem 3.8, using Exercise 21.]
(b) Is part (a) true if
Ris infinite? Justify your answer.
40. An element a of a ring is nilpotent if a" = OR for some positive integer n.
Prove that R has no nonzero nilpotent elements if and only if OR is the unique
solution of the equation
x'- =OR.
TIU! following ckfinition is needed for Exercises 41-43. Let R be a ring with ickntity.
OR, then R is said to have
R is said to have characteristic zero.
If there is a smallest positive integer n such that nl R
characteristic n. If no such n exists,
7L,, has characteristic
n.
Prove that a finite ring with identity has characteristic n for some
n
41. (a) Show that Z has characteristic zero and
(b)
42.
=
What is the characteristic of
Zi X 7L6?
> 0.
43. Let R be a ring with identity of characteristic n > 0.
(a) Prove that
(b)
na
=OR for every a ER.
If R is an integral domain, prove that n is prime.
C. 44. (a) Let a and b be nilpotent elements in a commutative ring R (see
Exercise 40). Prove that a + band ab are also nilpotent. [You will need the
Binomial Theorem from Appendix E.]
(b)
Let N be the set of all nilpotent elements of R. Show that N is a subring
of R.
3
45. Let R be a ring such that x = x for every x ER. Prove that
R is commutative.
46. Let R be a nonzero finite commutative ring with no zero divisors. Prove that
Ris a field.
II
Isomorphisms and Homomorphisms
If you were unfamiliar with roman numerals and came
across a discussion of
integer
arithmetic written solely with roman numerals, it might take you some time to realize
that this arithmetic was essentially the same as the familiar arithmetic in Z except for
the labels on the elements. Here is a less trivial example.
EXAMPLE 1
Consider the subset S = {O, 2, 4, 6, 8} of Z10• With the addition and multiplica­
tion of Z10, Sis actually a commutative ring, as can be seen from these tables:*
•The reason the elements of Sare listed in this order will become clear in a moment
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3.3
Isomorphisms and Homomorphisms
71
+
0
6
2
8
4
0
6
2
8
0
0
6
2
8
4
0
0
0
0
0
0
6
6
2
8
4
0
6
0
6
2
8
4
2
2
8
4
0
6
2
0
2
4
6
8
8
8
4
0
6
2
8
0
8
6
4
2
4
4
0
6
2
8
4
0
4
8
2
6
4
A careful examination of the tables shows that Sis a field with five elements and that
the multiplicative identity of this field is the element 6.
We claim that Sis "essentially the same" as the field Zs except for the labels on the
elements. You can
see
this as follows. Write out addition and multiplication tables
for Zs.* To avoid any possible confusion with elements of S, denote the elements of
Zs by 0, T,
2, 3, 4. Then relabel the entries in the Z5 tables according to this scheme:
Relabel 0
as
relabel
0,
relabel 3 as 8,
T as 6,
relabel
2 as 2,
relabel 4 as 4.
Look what happens to the addition and multiplication tables for Zs:
0
+
j
1
-,.
-
0
6
2
8
J
�
4
6
1
jJ
0
j
2
2
z
8
J
�
l
j
8
:z
4
Ji
4
JJ
4
0
j
0
1
6
r
6
2
j
g
0
j
6
4
4
A
0
j
.
�
j
Ji
4
�
8
j
8
2
6
1
j
2
6
j
8
2
l
2
z
2
1
-,.
-
0
6
2
8
J
8
�
4
jJ
j
g
;a
j
j
0
0
0
0
0
0
j
j
1
"J.
1
�
6
0
6
2
8
4
l
j
l
Ji
j
1
2
0
2
4
6
8
:z
j
:z
r
Ji
j
8
0
8
6
4
ff
j
A
j
-
4
0
4
8
2
z
2
j
6
By relabeling the elements of Zs, you obtain the addition and multiplication
tables
for S. Thus the operations in Zs and S work in exactly the same way-the
only difference is the way the elements are labeled. As far as ring structure goes,
Sis just the ring Z5 with new labels on the elements. In more technical terms, Zs
and S are said to be isomorphic.
In general, isomorphic rings are rings that have the same structure, in the sense that
the addition and multiplication tables of one are the tables of the other with the ele­
ments suitably relabeled, as in Example 1. Although this intuitive idea is adequate for
small finite systems, we need a rigorous mathematical definition of isomorphism that
agrees with this intuitive idea and is readily applicable to large rings as well.
There are two aspects to the intuitive idea that rings R and S are isomorphic:
relabeling the elements of R and comparing the resulting tables with those of S to
verify that they are the same. Relabeling means that every element of R is paired with
a unique element of S (its new label). In other words, there is a function f.R-+ S that
*The "11..5 tables (in congruence class notation) are shown in Example 2 of Section 2.2.
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72
Chapter 3
Rings
assigns to each
r ER
its new label/(r) ES. In the preceding example, we used the rela­
beling function f: Z5 -+ S, given by
/(0) = 0
/(l)= 6
/(2)= 2
/(3)= 8
/(4)= 4.
Such a function must have these additional properties:
(i) Distinct elements of R must get distinct new labels:
If r '# r' in R, then/(r) '# /(r') in S.
(ii) Every element of S must be the label of some element in R:*
For each
Statements (i) and
s ES,
there is an rE R such that/(r) =
(ii) simply say that the functionf must
tive, that is,/must be a
bijection. t
In order for a bijection (relabeling scheme)/to be
an
s.
be both injective and surjec­
isomorphism, applying/to
the addition and multiplication tables of R must produce the addition and multiplica­
tion tables of S. So if
a+ b =
c
in the R-table, we must have/(a)+ f(b) = f(c) in the
S-table, as indicated in the diagram:
�ft))
R�
I �
S +
/(a)
a,
�
However, since
f{c)
a+ b = c, we must also have/(a + b) = f(c). Combining this
fact that/(a) + f(b) = f(c),
we
with the
see that
f(a + b)
=
f(a) + f(b).
This is the condition that f must satisfy in order for f to change the addition tables
of R into those of S. The analogous condition on f for the multiplication tables is
/(ab)= f(a)f(b). We now can state a formal definition of isomorphism:
Definition
A
ring R is isomorphic to a ring S (in symbols, R = S) if there is a function
f:R-+ S such that
(I) f is injective;
(ii) f is surjective;
{iii) f{a + b) = f(a}+ f(b)
and
In this case the function f is called an
f{ab)= f(a) f(b) for all a, b ER.
isomorphism.
*otherwise, we couldn't possibly get the complete tables of S from those of R.
f1njective, surjective, and
bijective functions are discussed in Appendix
B.
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3.3
CAUTION:
Isomorphisms and Homomorphisms
73
In order to be an isomorphism, a function must satisfy all
three of the conditions in the definition. It is quite possible
for a function to satisfy any two of these conditions but not
the third; see Exercises 4, 25, and 32.
EXAMPLE 2
In Example 12 on page 50, we considered the field K of all 2
the form
x
2 matrices of
b
where a and are real numbers. We claim that K is isomorphic to the field
C of complex numbers. To prove this, define a functionfK-+ C by the
rule
I(-� !) a+ bi.
=
To show that/is injective, suppose
( b) = i( s).
-b a -s
b =as. bi = si
f
Then by the definition off,
we must have
rand
a=
r
a
r
+
r
in C. By the rules of equality in C,
+
Hence, in K
so that/ is injective. T he function/is surjective because any complex number
f
+
is the image under of the matrix
a
bi
in K. Finally, for any matrices A and Bin K, we must show that/(A + B)
f(A) + f(B) andf(AB) f(A)f(B). We have
=
=
I
b +cf\
[( a b) + ( cf\ ] (
-b a -d c} -b - d a+ c}
= (a+ c) + (b + d)i
(a + bi) + ( + di)
( b) +i( d)
-b a -d c
c
_
i
a+
c
=
=
c
i
a
c
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74
Chapter 3
Rings
and
i[(
a
-b
b
a
)(
c
-d
d\] i (
c}
_
=
=
=
)
ac - bd
-ad - be
ad+be
ac - bd
(ac - bd) + (ad+bc)i
(a+bi)(c + di)
1( : !)1 ( ; J·
_
_
Therefore,/is an isomorphism.
It is quite possible to relabel the elements of a single ring in such a way that the ring
is isomorphic to itself.
EXAMPLE 3
Letf"C-+ C be the complex conjugation map given byf(a +
function fsatisfies
l[(a t bi)+ (c +di)]
=
=
=
bi)
= a-bi.* The
f[(a+c) + (b+d)i]
(a+ c) - (b+d)i (a - bi)+ (c - di)
f(a+ bi)+f (e+di)
=
and
/[(a+ bi)(c +di)]= f[(ac - bd) + (ad+ bc)i]
(ac - bd) - (ad+bc)i (a - bi)(c - di)
f(a + bi)f(c+di).
=
=
=
You can readily verify that/is both injective and surjective (Exercise 17).
Therefore/ is an isomorphism.
EXAMPLE 4
If R is any ring and t.R:R-+ R is the identity map given by t.R(r)
=
r,
then for
any a, bER
"R (a+ b)
Since
=a+b
=
t.R(a) + 1.Jb)
and
t.R(ab)
=
ab
= 1.Ja)1.Jb).
'R is obviously bijective, it is an isomorphism.
Our intuitive notion of isomorphism is symmetric: "R is isomorphic to S" means
the same thing as "Sis isomorphic to R". The formal definition of isomorphism is not
"The function fhas a geometric interpretation in the complex plane, where a+ bi is identified with
the point (a, b): It reflects the plane in the x-aKis.
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...._._q-��._.fld.__...,.a11N:t... �a--.�c.a.� ........ rir;bl1a-...,,,..��·...,. ... w....._.,....� ........
3.3
Isomorphisms and Homomorphisms
75
symmetric, however; since it requires a function from R onto S but no function from
S onto R. This apparent asymmetry is easily remedied. If f.R -+S is an isomorphism,
then/is a bijective function of sets. Therefore,fhas an inverse function g:S-+ R such
that go f = t.R (the identity function on R) and/• g = Ls-* It is not hard to verify that
the function g is actually an isomorphism (Exercise 29). Thus R = S implies that
S = R, and symmetry is restored.
Homomorphisms
Many functions that are not injective or surjective satisfy condition (iii) of the definition
of isomorphism. Such functions are given a special name.
Definition
LetR and S be rings. A function f:R-+S is said to be a homomorphism
f(a + b)
=
f(a) + f{b) and
f(ab)
=
if
f(a)f(b) for all a, b ER.
Thus every isomorphism is a homomorphism, but as the following examples show,
a homomorphism need not be an isomorphism because a homomorphism may fail to
be injective or surjective.
EXAMPLE 5
For any rings R and S the zero map z:R-+ S given by z(r) = Os for every r ER is
a homomorphism b ecause for any a, b ER
z(a + b) = Os = Os+ Os = z(a) + z(b)
and
z(ab) = Os= Os· Os= z(a)z(b�
When both R and S contain nonzero elements, then the zero map is neither
injective nor surjective.
EXAMPLE 6
The function/:Z-+ "Z<i given byf(a) = [a] is a homomorphism because of the
way that addition and subtraction are defined in Z6: for any a, b E Z
f(a + b) = [a + b] = [a] + [h]
::=
f(a) + f(b)
and
f(ab) = [ab]
=
[a][b] ""'f(alf(b ).
The homomorphism/is surjective, but not injective (Why?).
*See Appendix B for details.
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76
Chapter 3
Rings
EXAMPLE 7
The map g:R--+. M(R) given by
g(r) ( 0 0\r}
r, s
g(r) g(s) (-� �) (-� �) ( s r � s)
) g( )
(
- -(r s) r
=
-r
is a homomorphism because for any
+
+
=
_7o_
=
0
-
and
ER
0
+
-
+s
r+s
g(r)g(s) (-� �)(-� �) (-�s �) g(rs).
=
=
=
The homomorphism g is injective but not surjective (Exercise 26).
CAUTION:
Not all functions are homomorphisms. Th e properties
f(a + b)
=
f(a) + f(b)
f(ab)
and
=
fail for many functions. For example, if fR --+.
f(x)
x + 2, then
f(a)f(b)
� given by
=
/(3 + 4)
=
/(7)
=
9
but
/(3)
+ f(4)
so that/(3 + 4) * /(3) + /(4). Simil arly,/(3
because
/(3
•
4)
=
/(12)
=
14,
but
/(3)/(4)
=
·
5 +6
=
11
4) o:fo /(3) /(4)
=
5
•
6
=
30.
Theorem 3.10
Let f:R--+.S be a homomorphism of rings. Then
(1) {{OR)
=
(2) f(-a)
Os.
=
(3) f(a - b}
-f(a) for every a ER.
=
f{a} - f(b) for all a, b ER.
If R is a ring with identity and f is surjective, then
(4) S is a ring with
identity f(1R)·
{5) Whenever u is a unit In R, then f(u) is a unit in Sand f(ur1
..
..
=
f(u-1).
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dlremad.'lmm,-��.,.,,
�dllel.'lbe�lmmliog��l...Amiiog...:dgbtm-__,_�OOllll!m·a;J'tlmlo1f..._...._:ligl:U�:MpiNit.
3.3
Isomorphisms and Homomorphisms
Proof .. (t) f(oiJ+ f(oiJ= f(oR + oR)
77
[/is a homomorphism.]
/O
( R) +/(OR) = f(OR)
O
( R+OR= ORinR]
f(OiJ + f(OiJ= f(OiJ +Os
( /O
( R) +Os= f(OiJ in SJ
[Suhtract f(OiJfrom both sides.].
/O
( R)= Os
(2) First, note that
f(a) +/(-a) = f(a
+
(-a))
[/is a homomorphism.]
[a+ (-a)= OR)
= f(OR)
= Os
[Part (I)].
Therefore,/(-a) is a solution of the equation/(a) + x =Os. But the
unique solution of this equation is-/a
( ) by Theorem 3.3. Hence
f(-a) = -f(a) by uniqueness..
( ) f(a - b) = f(a + (-b))
3
= f(a) + f(-b) )
[Definition of subtraction]
(f iSahomomorphism.]
= f(a) + (-f(b))
[Part (2)]
= f(a) - f(b)
[Definition of subtraction].
(4) We shall show that/l
( iJE S is the identity element of S. Lets
be any element of S. Then since/is surjective, s = f(r) for some rER.
Hence,
s
·/(IR) = f(r')f(IR) = f(r lR) = f(r) = s
•
and, similarly,fl
( _,J • s = s. Therefore, Shas/1
( _,J as its identity element.
( 5) Sinceu is a unit in R, there is an element v in R such that
uv = 1R =vu. Hence, by (4)
f(u)f(v)
=
f(uv)= f(liJ=
ls-
Similarly ,vu= IR implies thatf(v)f(u) 15• Therefore,/(u) is a unit in
S, with inverse/(v). In otherwords,f( u)-1 = f(v). Sincev = u-1, we see
that/(u)-1 = f(v) = /(u-1). •
==
Iff.R 4 S is a function, then the image of /is this subset of S:
Im/= {SES Is= f(r) for some rER} = if(r) I rER}.
If f is surjective, then Im f = S by the definition of surjective. In any case we have:
Corollary 3.11
Iff:R 4 S is a homomorphism of rings, then the image off is a subring of S.
Proof .. Denote Im/byL Iis nonemptybecause05 =f(OiJEibyl
( )ofTheorem31
. 0.
The definition of homomorphism shows that I is closed under multiplica­
tion: Iff(a),f(h)El, tltenf(a)f(b) = f(ab)El. Similarly, [�closed under
subtraction because/a
( ) - f(b)= f(a - b)E/ by Theorem 3. 10. Therefore, I
is a subring of S by Theorem 3.6. •
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.... ._
78
Chapter 3
Rings
Existence of Isomorphisms
If you suspect that two rings are isomorphic, there are no hard and fast rules for
finding a function that is an isomorphism between them. However the properties of
homomorphisms in T heorem 3.10 can sometimes be helpful.
EXAMPLE 8
If there is an isomorphism/from Z12 to the ring Z3 X Z4, thenf(l) = (1, 1) by
part (4) of Theorem 3.10. Since/is a homomorphism, it has to satisfy
/(2) =/(1+ 1)
=
/(1)+ /(1)
=
(1, 1) + (1, 1)
=
(2, 2)
/(3) =/(2+ 1) =/(2) + /(1) = (2, 2) + (1, 1) = (0, 3)
/(4)
==
/(3+ 1) = /(3)+ /(1) = (0, 3) + (1, 1) = (1, 0).
Continuing in this fashion shows that if/is an isomorphism, then it must be
this bijective function:
/(1) = (1, 1)
/(4) = (1, 0)
/(7) = (1, 3)
/(10) = (1, 2)
/(2) = (2, 2)
/(5) = (2, 1)
/(8) = (2, 0)
/(11) = (2, 3)
/(3)
/(6) ""'(0, 2)
/(9)
=
(0, 3)
=
(0, 1)
f(O)
=
(0, 0).
All we have shown up to here is that this bijective function/is the only possible
isomorphism. To show that this/actually is an isomorphism, we must verify
that it is a homomorphism. This can be done either by writing out the tables
(tedious) or by observing that the rule off can be described this way:
f([a]1 2) = ([a]3, [a]4) ,
where [afo denotes the congruence class of the integer a in Z12, [a]3 denotes the
class of a in Z3, and [a]4 the class of a in z_.. (Verify that this last statement is
correct.) Then
f([a]i2+ [b]i:z) =/([a+ b]ii)
[Definition of addition in Zn]
= ([a+ bh, [ a+ b14)
[Definition offl
= ([a]3 + [b]3, [a]4 + [b]4)
[Definition of addition in Z1 and Z4]
= ([a]3, [a]4)+ ([bh, [b]4)
[Definition of addition in Z3 X Z4]
= f([a]ii) + f([b]1i)
[Definition offl.
identical argument using multiplication in place of addition shows that
f([a]t2[bfo) = /([a]tz)f( [b]ll). T herefore,fis an isomorphism and Z12 = Z3 X Z,..
An
Up to now we have concentrated on showing that various rings are isomorphic,
but sometimes it is equally important to demonstrate that two rings are not isomorphic.
To do this, you must show that there is no possible function from one to the other
satisfying the three conditions of the definition.
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...__.___,.��._.fld.__...,.a11N:t... Cl'Na!Sa--.�c.a.� ......
.. rir;bl1a-...,,,..��·...,. ... w......_..:dPLI�...-. ..
3.3
Isomorphisms and Homomorphisms
79
EXAMPLE 9
� is not isomorphic to Z12 or to Z because it is not possible to have a surjective func­
tion from a six-element set to a larger set (or an injective one from a larger set to "4,).
To show that two infinite rings or two finite rings w ith the same number of elements
are
not isomorphic, it is usually best to proceed indirectly.
EXAMPLE 10
The rings Z4 and Z2 X Z2 are not isomorphic. To show this, suppose on the
contrary that/:,l4�Z2 X Z2is an isomorphism. ThenJlO)
( 0, 0) and
=
/(1) = (1, 1) by T heorem 3.10 . Consequently,
/(2)
=/(1
+
1) = f(l)
Since f is injective and /(0)
�
+ f( l)
= (1, l)
+
(1, l) = (0, O}
/(2), we have a contradiction. Therefore, no
isomorphism is possible.
Suppose that/:R�Sis an isomorphism and the elements a,
b, c, . .. of Rhave a par­
ticular property. If the elements/(a),f(b),f(c), ... of Shave the same property, then
say that the property is preserved by isomorphism. According to parts (1),
we
(4), and (5) of
Theorem 3.10, for example, the property of being the zero element or the identity element
or a unit is preserved by isomorphism. A property that is preserved by isomorphism can
sometimes be used to prove that two rings are IWt isomorphic, as in the following examples.
EXAMPLE 11
In the ring Zs the elements l, 3, 5, and 7 are units by Theorem 2.10. Since
being a unit is preserved by isomorphism, any isomorphism from Z8 to another
ring with identity will map these four units to four units in the other ring.
Consequently, Z8 is not isomorphic to any ring with less than four units. In
particular, ls is not isomorphic to l4 X Z2 because there are only two units in
this latter ring, namely
(1, 1) and (3, l) as you can readily verify.
EXAMPLE 12
None of
fields
0, R, or C is isomorphic to Z because every nonzero element in the
0, R, and C is a unit, whereas Z has only two units (1 and -1).
EXAMPLE 13
Suppose Ris a commutative ring andfR�Sis an isomorphism. Then for any
a, b ER, we have ab = ba in R. Therefore, in S
f(a)f(b) =/(ab) = f(ba) =f(b)f(a).
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80
Chapter 3
Rings
Hence, Sis also commutative because any t wo elements of Sare of the for
f(b ) (since/is surjective). In other
mf a
( ),
words, the property of being a commutative
ring is preserved by isomorphism. Therefore, no commutative ring can be iso­
morphic to a noncommutative ring.
• Exercises
A. 1. Let/:Z�-+Z2 X Z3 be the bijection given by
0-+(0,0),
2-+(0,2),
1-+(l,1),
5-+ (1, 2).
4-+ (0, 1),
3-+(1,0),
Use the addition and multiplication tables of "14 and Z2
X Z3 to show that/is
an isomorphism.
2. Use tables to show that ll.2 X Z2 is isomorphic to the ring Rof Exercise 2 in
Section 3.1.
3. Let Rbe a ring and let R* be the subring of R X Rconsisting of all elements
of the form (a,
isomorphism.
a). Show that the functionf:R-+ R* given byf(a)
4. Let Sbe the subring {O,
.
2,4, 6, &} of ll.10 and let ll.5
=
=
a,
( a) is an
{O, T, 2, 3, 4,} (notation
as in Example 1) Show that the following bijection from 71..5 to Sis not an
isomorphism:
o
____,,.
____,,.
T
o
2
2
____,,.
4
3
_____,,.
6
4
____,,.
s.
5. Prove that the field R of real numbers is isomorphic to the ring of all
matrices of the form
by f(a)
=
G �)
.with a E Iii.
2X2
[Hint: Consider the function/given
(� �).]
6. Let Rand S be rings and let
R be the subring of R X Sconsisting of all
elements of the for m (a, Os)· Show that the functionf:R-+ R given by
f(a)
(a, Os) is an isomorphism.
=
7. Prove that !fl is isomorphic to the ring S of all 2
(� �)
X 2 matrices of the form
.whereaER.
O(v2) be as in Exercise 39 of Section 3.1. Prove that the function
/:Q(Vl)-+ O(Vl) given by /(a + bVl) = a bVl is an isomorphism.
8. Let
-
9. If /:Z -+ 7l. is an isomorphism, prove that/is the identity map.
are/(1 ) ,/( 1 + 1),
..
[Hint: What
. ?]
10. If Ris a ring with identity and/:R-+ Sis a homomorphism from Rto a
ring S, prove thatf(IR) is an idempotent in S. [Idempotents were defined in
Exercise 3 of Section
......
..
....
3.2.]
......
..
.......
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......... ��
�.dkl. OMadl
�c.g..;ge�
-rlgbtlD....W��--il
:dgbb� ........
3.3
11.
Isomorphisms and Homomorphisms
State at least one reason why the given function is not a homomorphism.
(a) /:Ill-+ R and/(x)
==
Vx.
(b) g:E-+ E, where Eis the ring of even integers and/(x)
(c) h:R-+ R and/(x)
=
2'".
(d) k:Q 4 0, where k(O )
12.
81
=
0
(�) �
and k
=
=
3x.
if a'/: 0.
Which of the following functions are homomorphisms?
(a) f Z -+ .l, defined byf(x)
= -x.
(b) fZ2-+Z2, defined byf(x) = -x.
(c) g:CI! -+ 0, defined by g(x)
=
(d) h:R-+ M(R), defined by h(a)
x2
=
I
+
1
°
( : �}
-
(e) /:Z12 -+.l.i, defined by/([x]12) = [x1, where [u],. denotes the class of the
integer u in Z,..
13.
Let R and S be rings.
(a) Prove that/:R X S-+Rgiven byf((r , s)) = r is a surjective homomorphism.
(b) Prove thatg:R X
S-+.S given by g((r, s))
=sis a surjective homomorphism.
(c) If both Rand Sare nonzero rings, prove that the homomorphisms/ and g
are not injective.
14.
Letf:Z -+.Z6 be the homomorphism in Example 6. Let K = {aEZ lf(a) = [O]}.
Prove that K is a subring of Z.
15.
Let/:R-+ S be a homomorphism of rings. If r is a zero divisor in R , isf(r) a
zero divisor in S?
B. 16. Let T, R, and Fbe the four-element rings whose tables are given in Example 5
of Section 3.1 and in Exercises 2 and 3 of Section 3.1. Show that no two of
these rings are isomorphic.
17.
Show that the complex conjugation function/:C -+ C (whose rule is
f(a +bi)= a- bi) is a bijection.
18.
Show that the isomorphism of Zs and Sin Example 1 is given by the function
whose rule is/([x]s) [6x]10 (notation as in Exercise 12(e)). Give a direct
proof (without using tables) that this map is a homomorphism.
=
19.
Show that S {O, 4, 8, 12, 16, 20, 24} is a subring of Z28• Then prove that the
map/:Z7-+ S given by/([x ],) = [8xh8 is an isomorphism.
20.
Let E be the ring of even integers with the • multiplication defined in
Exercise 23 of Section 3.1. Show that the map f:E-+ Z given byf(x) = x/2 is
an isomorphism.
21.
Let Z* denote the ring of integers with the EB and 0 operations defined in
Exercise 22 of Section 3.1. Prove that L is isomorphic to L*.
=
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82
Chapter 3
Rings
22.
Let 71_ denote the ring of integers with the ffi and 0 operations defined in
Exercise 24 of Section 3.1. Prove that 71_ is isomorphic to Z.
23.
Let C be the field of Exercise 45 of Section 3.1. Show that C is isomorphic to
the field C of complex numbers.
24.
(a) Let R be the set� X �with the usual coordinatewise addition, as in
Theorem 3.1. Define a new multiplication by the rule (a, b)(c, d)
(ac, be). Show that R is a ring.
=
(b) Show that the ring of part (a) is isomorphic to the ring of all matrices in
M(R) of the form
25.
(: �}
Let L be the ring of all matrices in M(Z) of the form
A: 0)
functio�/L
: -+ � given by
1\
not an 1somorph1sm.
c
= a
(: �)
.Show that the
is a surjective homomorphism but
26.
Show that the homomorphism gin Example 7 is injective but not surjective.
27.
(a) If gR
: -+ SandfS-+ Tare homomorphisms, show that/0g:R-+ Tis a
homomorphism.
(b) If fandgare isomorphisms, show that fogis also an isomorphism.
28.
(a) Give an example of a homomorphismfR -+ S such that R has an identity
but S does not. Does this contradict part (4) of Theorem 3.10?
(b) Give an example of a homomorphismf:R-+ S such that Shas an identity
but Rdoes not.
29.
Let/:R-+ S be an isomorphism of rings and let gS
: -+ R be the inverse
function of / (as defined in Appendix B). Show that gis also an isomorphism.
[Hint: To show g(a + b) g(a) + g(b), consider the images of the left- and
right-hand side under/and use the facts that/is a homomorphism and/ogis
the identity map.]
=
30.
Let/:R-+ S be a homomorphism of rings and let K = {rER lf(r) =Os }·
Prove that K is a subring of R.
31.
Let/:R-+ S be a homomorphism of rings and Ta subring of S.
Let P = {rER lf(r)ET}. Prove that Pis a subring of R.
32.
Assume n == l (mod m). Show that the function f: Z,,, -+ lm11 given by
f([x],,,) = [nx]""' is an injective homomorphism but not an isomorphism when
n <2: 2 (notation as in Exercise 12(e)).
33.
(a) Let The the ring of functions from IR to� as in Example 8 of Section 3.l.
Let (}T
: -+ R be the function defined by fJ(/) =/(5). Prove that(}is a
surjective homomorphism . Is(}an isomorphism?
(b) Is part (a) true if 5 is replaced by any constant c ER?
34.
If f:R-+ Sis an isomomorphism of rings, which of the following properties
are preserved by this isomorphism? Justify your answers.
(a) aERis azero divisor.
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3.3
Isomorphisms and Homomorphisms
83
(b) a ERis idempotent.*
(c) Ris an integral domain .
35. Show that the first ring is not isomorphic to the second.
(a) EandZ
(b) � X �
(c) �
(d) 0 and�
(e) Z
36.
X Z 14 and li6
X R X �andM(R)
(t) Z.. X Z4 and Z16
X Z2andZ
(a) Iff:R--+ Sis a homomorphism of rings, show that for any
n EZ,f(nr) = nf(r).
r
ER and
(b) Prove that isomorphic rings with identity have the same characteristic.
[See Exercises 41-43 of Section 3.2.]
(c) If/:R--+Sis a homomorphism of rings with identity, is it true that Rand
S have the same characteristic?
37.
(a) Assume that e is a nonzero idempotent in a ring Rand that e is not a zero
divisor.* Prove that e is the identity element of R. [Hint: e2 = e (Why?). If
a ER, multiply both sides of e2 = e by a.]
(b) Let
S be a ring with identity and Ta ring with no zero divisors. Assume
that/:S...+Tis a nonzero homomorphism of rings (meaning that at least
one element of Sis not mapped to OT)· Prove that/(18) is the identity
element of T. [Hint: Show that/(18) satisfies the hypotheses of part (a).]
38. Let Fbe a field andf:F--+ R a homomorphism of rings.
(a) If there is a nonzero element c of Fsuch that/(c) =OR, prove that/is
the zero homorphism (that is,/(x) =OR for every xEF). [Hint: c-1 exists
(Why?). If xEF, consider/(xcc-1).]
(b) Prove that/is either injective or the zero homomorphism. (Hint: If/is not
the zero homomorphism and/(a) = f(b), then/(a - b) OR.]
=
39. Let Rbe a ring without identity. Let Tbe the ring with identity of Exercise 32
in Section 3.2. Show that Ris isomorphic to the subring R of T. Thus, if Ris
identified with R, then R is a subring of a ring with identity.
C. 40. For each positive integer k, let kZ denote the ring of all integer multiples of k (see
Exercise 6 of Section 3.1). Prove that if m # n, then mZ is not isomorphic to nZ.
41. Let m, n EZ with (m, n) = 1 and let/:Z_--+ Zm X Zn be the function given
by /([aJ-) = ([a]m, [a]n). (Notation as in Exercise 12 (e). Example 8 is the case
m
3,n
4.)
=
=
(a) Show that the map/is well defined, that is, show that if [a],,,,,= [bJmn in
Z...,., then [a]m
[bJm in Zm and [a]n= [b]n inZ,,.
=
(b) Prove that/is an isomorphism. [Hint: Adapt the proof in Example 8: the
difference is that proving/is a bijection takes more work here.]
42. If (m, n) # 1, prove thatZmn is not isomorphic to Zm
x Z,,.
•idempotents are defined in Exercise 3 of Section 3.2.
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CHAPTER
4
Arithmetic in f[x]
In Chapter 1 we examined grade-school arithmetic from an advanced standpoint
and developed some important properties of the ring Z of integers. In this chapter
we follow a parallel path, but the starting point here is high-school algebra-­
specifically, polynomials with coefficients in the field IR of real numbers, such as
x2 - 3x - 5,
6x3
-
3x2 + 7x + 4,
x12 -1.
Dealing with polynomials means dealing with the mysterious symbol "x", which
is used in three different ways in high-school algebra First, xoften "stands for'' a
number, as in the equation 12x - 8
O, where xis the number
�
· Second, xsome­
times doesn't seem to stand for any particular number but is treated as if it were a
=
number in simplification exercises such as this one:
x3+ x
x(x2 + 1)
x
- x2+1 - ·
x2+1-
--
Third, xis also used as the variable in the rules of functions such as f(x)
3x + 5.
Now that you know what rings and fields are, we shall consider polynomials
=
with coefficients in any ring and attempt to clear up some of the mystery about
the nature of x. In Sections 4.1-4.3, we shall see that when xis given a meaning
similar to the second way it is used in high school, then the polynomials with coef­
ficients in a field Fform a ring (denoted F[x]) whose structure is remarkably similar
to that of the ring Z of integers. In many cases the proofs for Z given in Chapter 1
carry over almost verbatim to F[x].
In Sections4.4-4.6 we consider tests to determine whether a polynomial is irre­
ducible (the analogue of testing an integer for primality). Here the development is
not an exact copy of what was done in the integers. The reason is that the polyno­
mial ring Ff x] has features that have no analogues in the ring of integers, namely,
the concepts of the root of a polynomial and of a polynomial function (which cor­
respond to the first and third uses of xin high school).
86
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_ .... ..,_.... __ ... _..,. ................. ..,.....Co...
og l.olmlo&---� ..---·..,-11..-.-�-·...-.11.
86
Chapter 4
•
Arithmetic in
F[x]
Polynomial Arithmetic and the Division Algorithm
The underlying idea here is to define "polynomial" in a way that is the obvious exten­
sion of polynomials with real-number coefficients. Let R be any ring. A polynomial
with coefficients in R is an expression of the form
ao +
a1x
+ a,.i'- +
·
·
·
+ a,,x",
where n is a nonnegati ve integer and ai ER.
This informal definition raises several questions: What is XI Is it an element of R1
If not, what does it mean to multiply
x
by a ring element? In order to answer these
questions, note that an expression of the form ao
+
a1x
+ ¥1 +
·
·
•
+ a,.X' makes
sense, provided that the a1 and x are all elements of some larger ring. An analogy might
be helpful here. The number
3
-
41T +
l27T2 +
1T� and 8
1T is not in the ring Z of integen;, but expressions such as
1T2. + fru5 make sense in the real numbers. Furthermon;
-
it is not difficult to verify that the set of all numbers of the form
ao
+ a11T + a2TT2 +
·
·
·
+ a,.11',
is a subring of � that contains both Z and
with n 2!: 0 and a1EZ
1T (Exercise 2).
For the present we shall think of polynomials with coefficients in a ring R in much
the same way, as elements of a larger ring that contains both R and a special element
x
that is not in R. This is analogous to the situation in the preceding paragraph with
R in place of Z and
the element
answer,
x
x
in place of 1T, except that here we don't know anything about
or even if such a larger ring exists. The following theorem provides the
as well
as
a definition of "polynomial".
Theorem 4.1
If R is a ring, then there exists
a
ring T containing an element
x that is not in
R and has these properties:
(i) R Is a subring
(ii) xa = ax for
of T .
every a ER.
(iii} The set R[x] of all elements of T of the form
(where n 2!: O and
is a subring
a1ER)
of T that contains R.
(iv) The representation of elements of R[x] is unique: If n ::!'> m and
l1o
then
{v} aa
+ a1x + a.,x2 +
·
•
·
+ a/' =b0 + b1x + b.,x2 +
·
·
·
+ b,.,x"',
a1=b1fori=1, 2, .. . , n and b1=OR for each i> n.
+ a1x + a�2 +
·
·
·
+ anJ(I =OR if and only If a1 =OR for every I.
Proof" See Appendix G. We shall assume Theorem 4.1 here.
The elements of the ring
R[x] in Theorem 4.1 (iii)
are
•
called polynomials with
coefficients in R and the elements a1 are called coefficients. The special element
x
is
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4.1
Polynomial Arithmetic
and
the Division Algorithm
sometimes called an indeterminate.* To avoid any misunderstandings in Theorem
87
4.1,
please note the following facts.
I. Property (ii) of T heorem 4.1 does not imply that the ring Tis commutative, but
only that the special element
x commutes with each element of the subring R
(whose elements may not necessarily commute with each other).
2, Property (v) is the special case of property (iv) when each b,
=
OR.
3. The first expression in property (v) is not an equation to be solved for x. In this
context, asking what value of x makes ao + a1x + a-iX2 +
+ a,;<'
OR is as
meaningless as asking what value of 1T makes 3 + 51T - 77T2
0 because x (like
1T) is a specific element of a ring, not a variable that can be assigned values.t
·
·
·
=
=
EXAMPLE 1
The rings Z[x],
Cl![x], and n[x] are the rings you are familiar with from high
3 + 5x - 7x'- is in all three of these rings, but 3 + 7 .5x'- is
only in Q[x] and �x] because the coefficient 7.5 is not an integer. Similarly,
4.2 + 3x + V5x4 is in R[x] but not in the other two rings since V5 is not a
school . For instance,
rational number. Terms with zero coefficents are usually omitted , as they were
in the preceding sentence.
EXAMPLE 2
4 - 6x + 4x3 E E[x]. However, the
Eix], because it cannot be written with even coefficients.
Let Ebe the ring of even integers. Then
polynomial xis not in
Polynomial Arithmetic
The rules for adding and multiplying polynomials follow directly from the fact that
R[x] is a ring.
EXAMPLE 3
If f(x)
=
1
+
5x - x1 + 4x3 + 2x4 and g(x)
=
4
+
2x + 3r + x3 in Z7[x], then
the commutative, associative, and distributive laws show that
f(x) + g(x)
=
=
=
(I
+
5x - x'- + 4x3 + 2x4) + (4 + 2x + 3x2 + x3 + Ox4)
(1 + 4) + (5 + 2)x + ( 1 + 3)x2 + (4 + l )x3 + (2 + O)x4
5 + Ox + 2x2 + 5x3 + 2x4 5 + 2x2 + 5x3 + 2x4•
-
=
G, there
T and is
*AHhough in common use, the term "indeterminate" is misleading. As shown in Appendix
is nothing undetermined or ambiguous about x. It is a specific element of the larger ring
not an element of R.
tvariables and equations will
be dealt with
in Section
4.4.
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88
Chapter 4
Arithmetic in
F[x]
EXAMPLE 4
The product of
1 - 1x + x2 and 2 + 3x in Q[x] is found
tive law repeatedly:
(1 - 1x + x2)(2
+
3x)
by using the distribu­
=
1(2 + 3x) - 1x(2 + 3x) + x2(2 + 3x)
1(2) + 1(3x) - 7x(2) - 7x(3x) + x2(i) + x2(3x)
2 + 3x - 14x - 2lx2 + 2x2 + 3x3
=
2 - llx - 19x2 + 3xl,
=
=
The preceding examples are typical of the general case. You add polynomials by
adding the corresponding coefficients, and you multiply polynomials by using the
distributivelaws and collecting like powers of x. Thus polynomial addition is g iven by
the rule:*
(ao + a1x + a-i.:C +
+ a;,X") + (ho + h1x + b2x2 +
+b;,X")
+ (a,, + b1.)::t!'
(Oo + ho) + (a1 + bi)x + (a + bi)xi- +
2
·
·
·
·
=
·
·
·
·
·
and polynomial multiplication is given by the rule:
+ bmx!")
(ao + a1x + a,;x2 +
+ a,,x")(b0 + h1 x + h x2 +
2
2
+
+
+
(aJJ
+
1 a1bo)x (aob
aobo
a1h1 + ¥o)X +
+ a,,bmxi+"'.
2
·
·
·
•
·
=
·
·
·
·
For each k � 0, the coefficient of ;/<in the product is
aohk + a1bk-1 + ¥k-2. +
·
·
·
+ ak-2.b,, + ak-1h1 + a,)Jo
k
=
�aA-e.
t=O
whereai =OR if i> n andb1
=
ORifj>
m.
R[x] that if R is com­
R has a multiplicative identity
also the multiplicative identit y of R[x] (Exercise 8).
It follows readily from this descrip tion of multiplication in
mutative, then so is
lR, then lR is
Definition
R[x] (Exercise 7).
F urthermore, if
Let f(x)
a0 + a1x + l!i!X2 +
+ ilnx" be a polynomial in R[x] With an oft OR.
Then a11 is called the leading coefficient of f(x). The degree of f(x) is the
integer n; it is denoted "deg f(x)". In other words, deg f(x} is the largest
exponent of x that appears with a nonzero coefficient, and this coefficient
is the leading coefficient.
=
·
·
·
EXAMPLE 5
3 - x + 4x2 - 7x3 ER[x] is 3, and its leading coefficient is -7.
1
(3 + 5x) 1 and deg (x 2) 12. The degree of 2 + x + 4x2 ox3 + Ox5 is 2 (the largest exponent of x with a nonzero coefficient); its leading
coefficient is 4.
The degree of
Similarly, deg
=
=
*We may assume that the same powers of x appear by inserting zero coefficients where necessary.
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...._._q-��._.fld.__...,.a11N:t... �a--.�c.a.� ......
.. ftebtb....,.. ....... � • ...,. ... w......_..:dPLI�...-. ..
4.1
Polynomial Arithmetic and the Division Algorithm
89
The ring R that we start with is a subring of the polynomial ring R(x]. The elements
of R, considered as polynomials in R[x], are called constant polynomials. The polyno­
mials of degree 0 in R[x] are precisely the nonzero constant polynomials. Note that
the constant polynomial OR does not have a degree
(because no power of x appears with nonzero coefficient).
Theorem 4.2
If R is an Integral domain and f(x), g(x) are nonzero polynomials in R[x], then
deg[f(x)g(x) ]
=
de g f (x)
+
deg g(x).
Proof,. Supposef(x) = ao + aix + a2X'- +
+ a,.x' and g(x) = b0 + b1x +
�+
+ bmx!" with a,,* OR and bm :FOR, so that degf(x) = n and
· · ·
· · ·
deg g(x) = m. Then
f(x)g(x)
=
aJJo + (aoh1
+
a1bn)x + (a.jl0 + a1b1 + � +
·
· ·
+
aHb,,.X'+"'.
The largest exponent of x that can possibly have a nonzero coefficient is
But a,.b,,. :F OR because R is an integral domain and aH :F OR and
b,,. :F OR. Therefore,f(x)g(x) is nonzero and deg[/(x)g(x)] = n + m =
deg/(x) + deg g(x). •
n + m.
Corollary 4.3
If R is an integral domain, then so is R[x].
Proof"' Since R is a commutative ring with identity, so is R[x] (Exercises 7 and 8).
The proof of Theorem 4.2 shows that the product of nonzero polynomials
is nonzero. Therefore, R[x] is an integral domain. •
in R[x]
The first five lines of the proof of Theorem 4.2
this conclusion.
are
valid in any ring and lead to
Corollary 4.4
Let R be a ring. If f(x), g(x), and f(x)g(x) are nonzero in R[x], then
deg [f(x)g(x)] � de g f(x)
+
deg g(x).
EXAMPLE 6
In �[x], let/(x) = 2x4 and g(x) = 5 x. Thenf(x)g(x) = (2x4)(5x) = 4x5,
so deg [f(x)g(x)] deg/(x) + deg g(x). However, if g(x) = 1 + Ji'-, then
=
f(x)g(x) = 2x4(1 + 3x3) = 2x4 + 2 3X' = 2x4 + OX'= 2x4,
•
which has degree 4. But degf(x) + deg g(x) = 6. So deg [f(x)g(x)] < deg/(x) +
degg(x).
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90
Chapter 4
Arithmetic in F[x]
For information on the degree of the sum of polynomials,
see
Exercises 4 and 12.
Corollary 4.5
LetR be an integral domain and f(x)ER[x]. Then
f(x) is a unit inR[x] if and only if f(x) is a constant polynomial that is a unit inR.
In particular, if Fis a field, the units in F[x] are the nonzero constants inf.
Remember that the proof of an "if and only if" statement requires two separate proofs.
Proof of Corollary 4.5 ... First, assume thatf(x) is a unit in R[x]. Thenf(x)g(x) =lR
for some g(x) in R[x]. By Theorem 4.2,
deg/(x) + degg(x) =deg [f(x)g(x)] =deg lR =0.
Since the degrees of polynomials are nonnegative, we must have
deg/(x) =0 and degg(x) =0. Therefore,/(x) and g(x) are constant poly­
nomials, that is, constants in R. Sincef(x)g(x) =lR,f(x) is a unit in R.
Conversely, assume that.f(x) is a constant polynomial that is a unit in R,
say f(x) =b, with b a unit in R. Let h(x) =b-1• Thenf(x)h(x) =bb-1 =lR.
Therefore,f(x) is a unit in R[x].
The last statement of the corollary follows immediately since
every nonzero element of a field is a unit in the field (see Example 6 in
Section 3.2).
•
EXAMPLE 7
The only units in Z[x] are 1 and -1, since these are the only units in Z. The units in
R[x] (or in Q[x] or in C[xD are all nonzero constants, since
R, 0, and C are fields.
Corollary 4.5 may be false if R is not an integral domain (Exercise 11).
EXAMPLE 8
5x + 1 is a unit in Z25[x] that is not a constant because (as you should verify)
(Sx + l)(20x + l) =l.
The Division Algorithm in F[x]
Our principal interest in the rest of this chapter will be polynomials with coefficients in
a field F (such as Q or IR or Z5). As noted in the chapter introduction , the domain F[x]
has many of the same properties as the domain Z of integers, including the Division
Algorithm (Theorem 1.1), which states that for any integers a and b with b positive,
there exist unique integers q and r such that
a=bq+r
and
0�
r
< b.
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Polynomial Arithmetic and the Division Algorithm
4.1
For polynomials, the only changes
are
to
require the divisor to
be
91
nonzero and
to
replace the statement "O $ r < b'' by a statement involving degrees. Here is the formal state­
ment (with/(x) in place of a,g(x) in plaoe of b, and q(x), r(x) in plaoe of
q,
r
respectively).
Theorem 4.6 The Division Algorithm in f[x]
Let F be a field and f(x}, g(x) EF{x] with g(x) # Op Then there exist unique
polynomials q(x) and r(x) such that
f(x)
=
g(x)q(x)
+
r(x)
and either r(x)
=
or
OF
deg r(x) < deg g(x).
Example 9 shows how polynomial division works and why the Division Algorithm
is valid in one particular case.
EXAMPLE 9
dividef(x)
We shall
3x5 + 2x4 + 2x3 + 4x2+ x - 2 by g(x) 2x3
italic column on the right keeps track of what happens at each step.*
=
=
+
1.
The
dhlisorg(_x)
! 113�ir+ +
+-quotient q(x)
2i' 2x1 4x2
2 +- diVidendf(x)
3� +
�x2
+-(ir)g(x)
2
2x"'+2x'!+ �+x-2 +-f(x)-(�x2)g(x)
2
+x
2x4
+-xg(x)
2x3+�x2
+-f(x) - (�x2)g(x) xg(x)
2x3
+ 1 +-lg(x)
5
- 3 +-f(x)-(iryx)-xg(x) - lg(x)
der r(x)
2x2
f(x) -g(x) (ir x
1)
x
2J!-+
+
1
+
+
+x
2
remain
-
-2
-
-----+
+
+
=
=
f(x) -g(x)q(x)
The last line on the left side and the last three lines on the right side show that
f(x)
g(x)q(x) + r(x).
f(x) - g(x)q(x) r(x) or equivalently,
So the Division Algorithm holds for the polynomialsf(x) andg(x).
=
•Division Re(esher:Thefir&tterm of the quotient
dividend
divisor
j-r
is obtained by dividing the leading term ofthe
(at") by the leading term of the divisor (2x"):at"/2x3
( (irpCx))
=
=
ixt.
The product of this term and the
is then subtracted from the dividend resuHing in 2x' +
2x"
+
j-r x
+
-
2, as
shown. The process is repeated, using this last expression as the dividend and the same divisor, and
continues until you reach a polynomial with degree smaller than the degree of the divisor.
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92
Chapter 4
Arithmetic in
F[x]
Of course, an example is not a proof, even though you can readily convince your­
5).
self that the same procedure works with other divisors and dividends (Exercise
Consequently, skipping the proof until you are familiar with mathematical induc­
tion, would be quite reasonable. Th at's why the proof of Theorem
4.6 is marked
optional.
Proof of Theorem 4.6 The Division Algorithm
(Optional)"
q(_x) and r(x).
degf(x) <degg(x), then the theorem is true
with q(x) =Op and r(x) =f(x) becausef(x) =g(x)Op + f(x).
Case 2: If f(x) of:. Op and degg(x) $ degf(x), then the proof of exis­
tence is by induction on the degree of the dividend f (x) .* If degf(x)
0,
then degg(x) =0 also. Hence,.f(x) =a and g(x) =b for some nonzero
a, b E F. Since Fis a field, b is a unit and a =b(b-1a) + Op. Thus the
theorem is true with q(_x) =b-1a and r(x) =OF'
We first prove the existence of the polynomials
Case 1: If f(x) =Op or if
=
Assume inductively that the theorem is true whenever the dividend
has degree less than n. This part of the proof is presented in two columns.
The left-hand column is the formal proof, while the right-hand column
refers to Example 9. The example will help you understand what's being
done in the proof.
PROOF
EXAMPLE9
We must show that the theorem is true whenever
the dividendf(x) has degree n, say
a x + ao
1
Op. The divisor g(x) must have the
f(x)
with a,, *
form
=
a,,:x!' +
· •
g(x) =b;,,X"' +
with h,.. * Op and m $
· ·
n.
+
·
·
+
hx
1
+
h0
n=5
f(x)
3x5
=
,.......,-.,
t
2x3 +
4x2 + x
-
2
a,.x"
m=3
g(x) =2x3
+ l
,.......,-.,
b,.x1"
We begin as we would
in the long division of g(x) intof(x). Since Fis a
field and b,,. * Op, bm is a unit. Multiply the divi­
sor g(x) by
+ 2x4
a,,b,,.-1:x!'-m to obtain
a,,bm-l:x!'-"'g(x)
= a,,b,,.-1x"-"'(h;,,X"' +
· ·
·
+
h1x
+
h0)
3
2x2g(x)
3
=
=
2x2{2x1
first term of
the quotient
+ l)
3
3x5 +-x2
2
*We use the Principle of Complete Induction; see Appendix C.
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4.1
Polynomial Arithmetic and the Division Algorithm
93
Since a,)J,,.-1X'-"'g(_x) and f(x) have the same
degree and the same leading coefficient, the
difference
f(x) - a,.bm -tx"�,.g(x)
is a polynomial of degree less than n (or possibly
the zero polynomial). Now apply the induction
hypothesis with g(x) as divisor and the poly­
1
nomial .f(x) - a,;,,,.- x!'-"'g(x) as dividend (or
use Case 1 if this dividend is zero). By induction
there exist polynomials q1(x) and r(x) such that
fourth line of long dMsion
f(x) -a,.b,,.-Jxr-"'g(x) = g(x)q1(x)+r(x)
and
r(x) = Op
or
deg r(x) < deg g(_x).
q1(x)
=
x
+ 1
r(x)
last part of
5
=
-r - 3
2
remainder
the quotient
Therefore,
f(x)
=
r(x)
=Op
g(x)[a,.b,,. -tr--m+ q1(x)]
or
deg r(x)
<
+
r(x)
and
deg g(x).
Thus the theoremistruewith q(x) = a,.bm-1x"-"'+q1(x)whendegf(x) = n. This completes
the induction and shows that q(x) and r(x) always exist for any divisor and dividend.
To prove that q(x) and r(x) are unique, suppose that q2(x) and r.J.x) are polynomials
such that
f(x)
= g(x)q1(x)+ r2(x)
and
Then
g(x)q(x)+r(x)
=
r2(x)
= Op
or deg r2(x) < deg g(x).
f(x) = g(x)q2(x) + r1(x),
so that
g(x)[q(x) - q2(x)]
= r2(x)
-r(x).
q(x) - q2(x) is nom.ero, then by Theorem 4.2 the degree of the left side is deg g(x)+
deg[q(x) - qz(x)], a number greater than or equal to deg g(x). However, bothrz(x) and r(x)
have degree strictly less than deg g(x), and so the right-hand side of the equation must also
have degree strictly less than deg g(x) (Exercise 12). This is a contradiction. Therefore
q(x) - q2(x) = Op, 01; equivalently, q(,x) = q.J..x). Since the left side is zero, we must have
r1(x) - r(x) = O.F> so that r1(x) = r(x). Thus the polynomials q(,x) and r(x) are unique. •
If
• Exercises
NOTE: R denotes a ring and F afield
A. 1.
Perform the indicated operation and simplify your answer:
2
(a) (3x4+ 2x3 -4r+ x+ 4)+ (4x3+ x + 4x+ 3) inZ5[x]
(b) (x + 1)3 inZ3[x]
(c) (x - 1)5 inZ5[x]
2
(d) (x - 3x +2)(2x3 - 4x+ 1) inZ7[x]
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94
Chapter 4
Arithmetic in F[x]
2.
Show that the set of all real numbers of the form
a0
2
+ aJ'lT + a27r +
·
·
·
+ a,,'TT",
with n � 0 and a1 E Z
is a subring of R that contains both Z and 7r.
3.
(a) List all polynomials of degree 3 in Z2[x].
(b) List all polynomials of degree less than 3 in Z3[x].
4.
In each part, give an example of polynomials/(x), g(x) E Q[x] that satisfy the
given condition:
(a) The deg of f(x) + g(x) is less than the maximum of deg/(x) and deg g(x).
(b) Deg [f(x) + g(x)] =max {deg/(x), degg(x)}.
polynomials q(x) and r(x) such that/(x) =g(x)q(x) + r(x), and r (x) =0
or deg r(x) < deg g(x):
5. Find
(a) f(x) = 3x4 - 2x3 + 6x2 - x + 2 and g(x)
(b) f(x) = x4 - 7x + l and g(x) =2x2 + l
in
==
x2 + x +
1
in Q[x].
Q[x].
(c) f(x) =2x4 + x?- - x + l andg(x) =2x - 1 in Zs[x].
(d) f(x) =4x4 + 2x3 + 6x2 + 4x + 5 andg(x) =3x?- + 2 in .l7[x].
6.
Which of the following subsets of R(x] are subrings of R[x]? Justify your answer:
(a) All polynomials with constant term OR.
(b) All polynomials of degree 2.
(c) All polynomials of degrees
k,
where k is a fixed positive integer.
(d) All polynomials in which the odd powers of x have zero coefficients.
(e) All polynomials in which the even powers of x have zero coefficients.
7.
If R is commutative, show that R[x] is also commutative.
8. If R has multiplicative identity lR, show that lR is also the multiplicative
identity of R[x].
9.
If c E R is a zero divisor in a commutative ring R, then is c also a zero divisor
in R[x]?
I 0. If F is a field, show that F[x] is not a field. [Hint: Is x a unit in F[ x]?]
B. 11.
12.
Show that 1 + 3x is a unit in .l.i[x]. Hence, Corollary 4.5 may be false if R is
not an integral domain.
If f(x),g(x) E R[x] and/(x) + g(x) *- OR, show that
deg[f(x) + g(x)] s max {deg/(x), degg(x)}.
13.
Let R be a commutative ring. If a,, *- OR and f(x)
0() + a1x + a2x?- +
+
a,,x!' (with a"*- O� is a zero divisor in R[x], prove that a,, is a zero divisor in R.
14.
(a) Let R be an integral domain and/(x),g(x)
=
· ·
E R[x]. Assume that the
leading coefficient of g(x) is a unit in R. Verify that the Division Algorithm
holds forf(x) as dividend andg(x) as di'\isor. [Hint: Adapt the proof of
Theorem 4.6. Where is the hypothesis that F is a field used there?]
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........
4.2
(b)
15. Let
Divisibility In
F[x]
95
Give an example in Z[x] to show that part (a) may be false if the leading
coefficient of
R
g(x) is not a unit. [Hint: Exercise 5(b)
be a commutative ring with identity and
with Zin place of
Q.]
a E R.
(a) If d OR, show that lR+ax is a unit in R[x]. [Hint: Consider
a2x2.]
(b) If a4 OR, show that IR+ax is a unit in R[x].
=
l
-
ax+
=
16. Let
R be a commutative ring with identity and a E R . If 1R+ ax is a unit in
R[ x], show that d' OR for some integer n > 0. [Hint: Suppose that the inverse
of lR+ax is b0+b1x+b2i1'+
+bd'. Since their product is lR, b0 lR
(Why?) and the other coefficients are all OR.]
=
· · ·
17. Let
R
==
be an integral domain. Assume that the Division Algorithm always
holds in R[x]. Prove that
R
is a field.
18. Let 1p:R[x]-+ R be the function that maps each polynomial in
constant term (an element of
R[x] onto its
R). Show that 'Pis a surjective homomorphism
of rings.
19. Let 1p:Z[x]-+ Zn[x] be the function that maps the polynomial ao+a1x+ ·
+
ad' in Z[x] onto the polynomial [ao]+[a1]x+ +[a.k]x", where [a] denotes
the class of the integer a in Z,.. Show that <p is a surjective homomorphism of
· ·
· ·
·
rings.
20. Let
D:R[x]-+ R[x] be the derivative map defined b y
D(ao+a1x+a,.x"+
·
·
·
+"i,x")
=
a1+2a2x+3a3x2+
·
· ·
+na,.xi-1•
Is D a homomorphism of rings? An isomorphism?
C.21. Let
h:R-+ Sbe a homomorphism of
rings and define a function h:R[x]-+ 5lx]
by the rule
h(ao+a1x+
·
· ·
+anx") = h(ao)+h(a1)x+h(a,.)x2 +
·
· ·
+h(a,Jx".
Prove that
(a)
his a homomorphism of rings.
(b)
his injective i f and only if his injective.
(c)
his surjective if and only if his surjective.
(d) If R
22. Let
R
=
S, then
R[x] = S[x].
be a commutative ring and let k(x) be a fixed polynomial in
that there exists a unique homomorphism <p:R[x]-+
<p(r) = r for all r E R
II
and
R[x]. Prove
R[x] such that
<p(x)
=
k(x).
Divisibility in F[x]
All the results of Section 1.2 on divisibility and greatest common divisors in Z now
carry over, with only minor modifications, to the ring of polynomials over a field.
Throughout this section,
F always denotes afield
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96
Chapter 4
Definition
Arithmetic in
F[x]
Let F be a field and a(x), b(x) E f[x ] with b{x) nonzero. We say that b(x)
divides a(x) [or that b(x) is a factor of a{x)], and write b(x) I a(x) if a(x)
b(x)h(x) for some h(x) E f[x].
=
EXAMPLE 1
(2x + l) I (6x2 - x - 2) in O[x]
6x2 - x - 2 (2x + l)(3x - 2).
+ 1 also divides 6x2 - x - 2. For
lOx + 5 divides 6x2 - x - 2 because 6x2 - x - 2
because
=
Furthermore, every constant multiple of 2x
5(2x + 1)
instance,
=
H
5(2x + l)
(lx, - 2)
Example
1
=
l
illustrates the first part of the following result.
Theorem 4.7
Let F be a field and a(x}, b(x)
E
f[x] with b(x)
nonzero.
(1) If b(x) divides a(x), then cb(x) divides a(x) for each nonzero c E F.
(2) Every divisor of a(x) has degree less than or equal to deg a(x).
Proof� (1) If b(x) I a(x), then a(x)
a(x)
Therefore,
(2)
=
·
=
b(x)h(x)
for
cc-1b(x)h(x)
some
=
h(x)
E
F[x]. Hence,
cb(x)[c-1h(x)].
cb(x) I a(x).
Suppose
deg
lp b(x')h(x)
=
a(x)
b(x) I a(x), say a(x) b(x')h(x). By Theorem 4.2,
deg b(x) + deg h(x).
=
=
Since degrees are nonnegative, we must have 0 s deg b(x) s deg a(x).
•
As we learned earlier, the greatest common divisor of two integers is the largest
integer that divides both of them. By analogy, the greatest common divisor of two
polynomials a(x), b(x) E F[x] ought to be the polynomial of highest degree that divides
both of them. But such a greatest common divisor would not be unique because each
constant multiple of it would have the same degree and would also divide both
and
b(x).
In order to guarantee a unique gc d ,
introducing a new concept. A polynomial in
coefficient is lp For instance, x3 +
Definition
x
+
we
a(x)
modify this definition slightly b y
F[x] i s
said t o be monic i f its leading
2 is monic in Q[x] , but 2x + 1 is not.
Let F be a field and a(x), b(x) E f[x ] , not both zero. The greatest common
divisor (gcd) of a(x) and b(x) is the monic polynomial of highest degree
that divides both a(x) and b{x).
In other words, d(x) is the gcd of a(x) and b(x) provided that d(x) is monic and
(1) d(x) la(x) and d(x) lb(x);
(2) lfc(x) la(x) and c(x) lb(x), then deg c(x) s deg d(x).
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4.2
Divisibility in F[x]
Polynomials a(x) and b(_x) have at least one monic common divisor (namely
97
lp). Since
the degree of a common divisor of a(x) and b(x) cannot exceed either deg a(x) or deg b(x)
by Theorem 4.7, there must be at least one monic common divisor of highest degree. In
Theorem 4.8 below we shall show that there is only one monic common divisor of highest
degree, thus justifying the definition's reference to tJw greatest common divisor.
EXAMPLE 2
To find the gcd of
of highest degree
3x1 + x +
are just
6 and 0 in O[x], we note that the common divisors
3x2 + + 6 of degree 2. These include
the divisors of
x
3x2 + x + 6 itself and all nonzero constantmultiples of this polynomial-in
particular, the monic polynomial
1
t<3x2
+ x + 6)
x2 + 31 x + 2.
=
Hence, i1-
+ jx + 2 is a gcd of 3x1 + x + 6 and 0.
EXAMPLE 3
You can easily verify these factorizations in CJ![x]:
a(x)
=
2x4 +
b(x)
- 5x - 2 (2x + l)(x + 2)(x + l)(x 2x3 - 3x1 - 2x (2x + l)(x - 2)x.
5x3
=
=
=
It appears that 2x + 1 is a common divisor of highest degree of
In this case, the constant multiple
�
(2x + 1)
sor of highest degree. For a proof that x +
divisor,
see
1),
Exercise 5(g).
=
k
x+
�
is a
a(x) and b(x).
monic common
divi­
actually is the greatest common
The remainder of this section , which is referred to only a few times in the rest of
the book, may be skimmed if time is short-read the theorems and corollaries, but
skip the proofs.
Theorem 4.8
Let F be a field and a(x),
b(x} E F[x], not both zero. Then there is a unique great­
d(x) of a(x) and b(x). Furthermore, there are (not neces­
polynomials u(x) and v{x} such that d{x)
a(x)u{x) + b(x)v(x).
est common divisor
sarily unique)
=
Steps 1 and 2 of the proof
are patterned after the proof of Theorem
Proof ofTheorem 4.8 ... Let s be the set of
all linear combinations of
1.2.
a(x)
and
b(x), that is,
S
Step I
=
{a(x)m(x)
+
b(x)n(x) lm(x), n(x) EF[x]}.
Find a monic polynomial ofsmallest degree in S.
Proof of Step 1: S contains nonzero polynomials (for instance, at least
one of a(x) lp + b(x) Op or a(x) O� + b(x) lp). So the set of all
•
·
•
•
..
..
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...
....... it.
98
Chapter 4
Arithmetic in F[x]
degrees of polynomials in S is a nonempty set of nonnegative integers,
which has a smallest element by the Wel l-Ordering Axiom. Hence, there
is a polynomial w(x) of smallest degree in S. If d is the leading coef­
ficient of w(x), then t(x) = d -'w(x) is a monic polynomial of smallest
degree in S. By the definition of S,
t(x)
Step 2
a(x)u(x) + b(x)t(x) for
=
some
u(x), v(x)EF[x].
Prove that t(x) is a gcd of a(x) and b(x).
Proof of Step 2: We m us t prove
gcd:
that t satisfies the two conditions in the
definition of the
(1) t(x) I a(x) and t(x) lb(x);
(2) If c(x) I a(x) and c(x) I b(x), then deg c(x) s deg t(x).
Proof of (1): In the proof of Step 2 of Theorem 1.2, replace a, b,
c, t, q, r, u, v, k, ands with a(x), b(x), c(x), t(x), q(x), r(x), u (x),
v(x), k(x), and s(x), respectively, to show that t(x) is a common
divisor of a(x) and b(x).
Proof of (2): With the same r epl acements as in the proof of (1),
repeat the proof of Step 2 of Theorem 1.2, u ntil you reach this
statement:
t(x)
a(x)u(x) + b(x)v(x)
=
=
=
[c(x)k(x)]u(x) + [c(x)s(x) ]v(x)
c(x)[k(x)u(x) + s(x)v(x)].
The first and last terms of this equation show that c(x) I t(x). By
Theorem 4.7, deg c(x) s deg t(x).
This s h ows
Step 3
that t(x) is a greatest common divisor
of f(x) and g(x).
Prove that t(x) is the unique gcd of a(x) and b(x).
Proof of Step 3: Suppose that d(x) is any gcd of a(x) and b(x). To prove
uniqueness, we must show that d(x) = t(x). Since d(x) is a common divi­
sor, we have a(x) = d(x'Jf(x) and b(x) = c(x)g(x) for some.f(x), g(x) E F[x].
Therefore,
t(x)
=
a(x)u(x) + b(x)v (x)
=
=
[c(x'Jf(x)]u(x) + [d(x)g(x)]v(x)
d(x)[f(x)u(x) + g(x)v(x)].
By Theorem 4.2,
deg t(x)
Since they are
=
deg d(x) + deg [f(x)u(x)
+
g(x)v(x)].
gcd's, t(x) and d(x) have the same degree. Hence,
deg [f(x)u(x) + g(x)v(x)]
=
0,
thatf(x)u(x) + g(x)t(x) c for some constant cEF. Therefore,
t(x) c(x)c. Since both t(x) and d(x) are monic, the leading coefficient
on the left side is lFand the leading coefficient on the right side is c. So
we must have c = lp Therefore, c(x) = t(x)
a(x)u(x) + b(x)t(x) is the
unique gcd of a(x) and b(x). •
so
=
=
=
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4.2
99
Divisibility in F[x]
Corollary 4.9
Let F be a field and a(x), b(x) Ef[x], not both zero. A monic polynomial
the greatest common divisor of a{x) and b(x) if and only if d(x)
d(x}EF[x] is
satisfies these conditions.
d{x) la(x) and d(x) lb{x).
(ii) if c(x) la(x) and c(x) lb(x), then c(x) ld(x).
(i)
Proof• Adapt the proof of
Polynomialsf(x) and g(x)
Corollary 1.3 to F[x].
are
said to be
•
relatively prime if their
greatest common
divisor is lp
Theorem 4.10
Let F be a field and a(x), b(x), c(x)EF[x]. If a(x)
relatively prime, then a(x) I c(x).
Proof• Adapt the proof of
Theorem l.4 to F[x].
I b{x)c(x) and a(x) and b(x) are
•
• Exercises
NOTE: F denotes afield.
A. I. Iff(x) EF[x], show that every nonzero constant polynomial divides.f(x).
2. Iff(x)
3. If
4.
=
c,;t' +
·
·
·
+co with c11 :/=
a, b EF and a :/= b,
show that
OF, what is the gcd of f(x) and Op?
x + a and x + b are relatively prime in F[x].
(a) Letf(x), g(x)EF[x]. If/(x) lg(x) andg(x) lf(x) , showthat/(x)
some nonzero c E F.
(b) If f(x)
and g(x) in part (a) are monic, show
thatf(x)
=
=
cg(x) for
g(x).
5. The Euclidean Algorithm for finding gcd's is described for integers in Exercise
15
of Section 1.2. The process given there also works for polynomials over a
field, with one minor adjustment. For integers, the last nonzero remainder is
the
gcd. For polynomials the last nonzero remainder is a common divisor of
highest degree, but it may not be monic. In that case, multiply it by the inverse
of its leading coefficient to obtain the gcd. Use the Euclidean Algorithm to
find the gcd of the given polynomials:
(a) x4 - xi - x2 + 1 and x3 - 1 in O[x]
(b) x! + x4 + 2x3 - x2 - x - 2andx4 + 2x3 + 5x2+ 4x + 4in O![x]
(c) x4 + 3x3 + 2x + 4 and x2 -
1 in Z5[x]
(d) 4x4 + 2x3 + 6x2 + 4x + 5 and 3x3 + 5x2 + 6x in .Z7[x]
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...
�...-.:lit.
100
Chapter
Arithmetic in F[x]
4
(e) x3 - ix'-+ 4x - 4i and x2 + 1 in C[x]
(f) x4 + x + 1 and x2 + x + 1 in Z2[x]
(g) 2x4 + 5x3 - 5x - 2 and 2x3 - 3x 2 - 2x in Q[x].
6. Express each of the gcd's in Exercise 5 as a linear combination of the two
polynomials.
B. 7. Letf(x)EF[x] and assume thatf(x) lg(x) for every nonconstant g(x)EF[x]. Show
that/(x) is a constant polynomial. [Hint:f(x) must divide both x + 1 and x.]
8.
Let/(x), g(x)EF[x], not both zero, and let d(x) be their gcd . If h(x) is a
common divisor of f(x) and g(x) of highest possible degree, then prove that
h(x)
=:
cd(.x) for some nonzero c EF.
9. Iff(x) =F OFandf(x) is relatively prime to OF, what can be said aboutf(x)?
IO. Find the gcd of
x +a + band x 3 - 3abx + a3 + b3 in Q[x].
11. Fill in the details of the proof of Theorem 4.8.
12. Prove Corollary 4.9.
13. Prove Theorem 4.10.
14.
Let/(x), g(x), h(x)E F[x], with/(x) and g(x) relatively prime. If f(x) jh(x) and
g(x) jh(x), prove thatf(x)g(x) I h(x).
15.
Let/(x), g(x), h(x)EF[x], with/(x) and g(x) relatively prime.
prove that h(x) and g(x) are relatively prime.
16.
If h(x) lf(x),
Let/(x), g(x), h(x)EF[x], with /(x) and g(x) relatively prime. Prove that the
h(x) and g(x).
gcd of f(x)h(x) and g(x) is the same as the gcd of
11
lrreducibles and Unique Factorization
F always denotes a field. Before carrying over the results of
1.3 on unique factorization in Z to the ring F[x], we must first examine an area
in which Z differs significantly from F[x]. In Z there are only two units,* namely ±1,
but a polynomial ring may have many more units (see Corollai:y 4.5).
An element a in a commutative ring with identity R is said to be an associate of an
element bof R if a
bu for some unit u. In this case h is also an associate of a because
u-• is a unit and h = au-1• In the ring Z, the only associates of an integer n are n and
-n because ±1 are the only units. If F is afield, then byCorollary4.5, the units inF[x]
Throughout this section
Section
=
are the nonzero constants. Therefore,
ft.x) is an as.wciate of g(x) in Flxl if and only ifft.x)
=
cg(x) for some nonzero c
E F.
Recall that a nonzero integer p is prime in Z if it is not ± 1 (that is, p is not a unit
in Z) and its only divisors are ±I (the units) and ±p (the associates of p). In F[x] the
units are the nonzero constants, which suggests the following definition.
•"Unit" is defined just before Example 4 in Section 3.2.
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4.3
Definition
lrreducibles and Unique Factorization
101
Let f be a field. A nonconstant polynomial p(x} <= f[x] is said to be
irreducible* if its only divisors are its associates and the nonzero constant
polynomials (units). A nonconstant polynomial that is not irreducible is
said to be reducible.
EXAMPLE 1
The polynomial
x + 2 is irreducible in O[x] because,
by Theorem 4.2, all its
are nonzero constants.
+ 2), say x + 2 = f(x)g(x), and if degf(x) = 1, then g(x) has degree
0, so that g(x) = c. Thus c1(x + 2) f(x), and/(x) is an associate of x + 2. A
similar argument in the general case shows that
divisors must have degree 0 or 1. Divisors of degree 0
Iff(x) I (x
=
every polynomial of degree
I in
F(xl is irreducible in Flxl.
The definition of irreducibility is a natural generalization of the concept of primal­
ity in
Z. In most high-school texts, however, a polynomial is defined
to be irreducible
if it is not the product of polynomials of lower degree. The next theorem shows that
these two definitions
are equivalent.
Theo rem 4.11
Let f be a field. A nonzero polynomial f{x) is reducible in f[x] if and only if f(x)
can be written as the product of two polynomials of lower degree.
Proof• First, assume thatf(x) is reducible. Then it must have a divisor g(x) that
is neither
an associate
nor a nonzero constant, sayf(x)
either g(x) or h(x) has the same degree
= g(x)h(x). If
asf(x), then the other must have
degree 0 by Theorem 4.2. Since a polynomial of degree 0 is a nonzero
constant in F, this means that either g(x) is a constant or an associate
of flx), contrary to hypothesis. Therefore, both
g(x) and h(x) have lower
degree than/(x).
Now assume that/(x) can be written as the product of two polyno­
mials of lower degree, and see Exercise 8.
•
are presented in Sections 4.4 to 4.6. For now,
an absolute one. For instance, x'- + 1
is reducible in C[x] because x2 + 1
(x + i)(x - i) and neither factor is a constant or
an associate of x1 + 1. But x'- + 1 is irreducible in O[x] (Exercise 6).
The following theorem shows that irreducibles in F[x] have essentially the same
divisibility properties as do primes in Z. Condition (3) in the theorem is often used to
Various other tests for irreducibility
we note that the concept of irreducibility is not
=
prove that a polynomial is irreducible; in many books, (3) is given as the definition of
"irreducible".
•vou could just as well call such a polynomial "prime'', but "irreducible" is the customary term with
polynomials.
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102
Chapter 4
Arithmetic in
F[x]
Theorem 4.12
Let F be a field and
p(x) a nonconstant polynomial in F[x]. Then the following
conditions are equivalent:*
(1) p(x) is irreducible.
b(x) and c(x) are any polynomials such that p(x) I b(x)c(x), then
p(x) I b(x) or p(x) I c(x}.
(3) If r(x) and s(x} are any polynomials such that p(x) = r(x)s(x), then r(x}
or s(x) is a nonzero constant polynomial.
(2) If
Proof• (1) => (2) Adapt the proof of Theorem 1.5 to F[x]. Replace statements
-:±p by statements about the associates of p(x); replace statements
-:±1 by statements about units (nonzero constant polynomials) in
F[x]; use Theorem 4.10 in place of Theorem 1.4.
about
about
(2) => (3) If p(x) = r(x)s(x), then p(x) I r(x) or p(x) ls(x), by (2). If
p(x) I r(x), say r(x) p(x)v(x), then p(x) r(x)s(x) p(x)v(x)s(x). Since
F[x] is an integral domain we can cancel p(x) by Theorem 3.7 and con­
c lude that IF= v(x)s(x). Thus s(x) is a unit, and hence by Corollary 4.5,
s(x) is a nonzero constant. A similar argument shows that if p(x) ls(x),
then r(x) is a nonzero constant.
=
=
=
,
(3) => (1) Let l'(x) be any divisor of J(x), say p(x) = l'(x)t(x). Then
l'(x) is a nonzero constant or d(_x) is a nonzero constant. If
d(x) = d 'i= OF> then multiplying both sides of p(x) = c(x)d(_x) = dc(x) by
d-1 shows that l'(x) = a1J(x). Thus in every case, c(x) is a nonzero con­
stant or an associate of p(x). Therefore,p(x) is irreducible.
•
by (3), either
Corollary 4.13
Let F be a field and p(x) an irreducible polynomial in F[x]. If p(x} la1(x)a�x)
then p(x) divides at least one of the a1(x).
Proof• Adapt the proof of Corollary 1.6 to F[x].
•
· ·
an(X),
•
Theorem 4.14
Let F be
a
field. Every nonconstant polynomial f(x) in F[x] is a product of
irreducible polynomials in
sense : If
F[x].t This factorization is unique in the following
and
"For the meaning
of
''the following conditions are equivalent" and what must be done to prove
Theorem 4.12, see page 508 of Appendix A. Example 2there is the integer analogue ofTheorem 4.12.
fWe allow the possibility of a product with just
one factor in case f(x) is itself irreducible.
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4.3
with each
pt.,x)
q1(x)
and
lrreducibles and Unique Factorization
103
irreducible, then r = s {that is, the number of irre­
ducible factors is the same). After the
qtx)
are reordered and relabeled, if
necessary,
pt.,x) is an associate of qt.,x)
Proof� To show that/(x) is a product of
Theorem 1.7 to
(i
= 1, 2, 3, ... ,
r).
irreducibles, adapt the proof of
F[x]: Let Sbe the set of
all nonconstant polynomials
that are not the product of irreducibles, and use a proof by contradiction
to show that Sis empty. To prove that this factorization is unique up to
q1(x)112(x)
qJ.x)
p,(x)] =
q1(x)q2(x)
qJ..x), so that p1(x) divides q1(x)q2(x)
qJ..x). Corollary
4.13 shows thatp1(x) I qj(x) for someJ. After rearranging and relabeling the q(x)'s if necessary, we may assume that p1(x) I q1(x). Since q1(x)
is irreducible, p1(x) must be either a constant or an associate of q1(x).
However, p1(x) is irreducible, and so it is not a constant. T herefore, p1(x)
is an associate of q1(x), withp1(x) = c1q1(x) for some constant c1• Thus
associates, suppose/(x)
=
p1(x)J11.(x)
·
·
•
p1(x)
=
with each pi(x) and qj.x) irreducible. Then p1(x)W2(x)
·
•
·
•
q1(x)[ctP2(x)p3(x)
Canceling
·
· ·
p,(x)]
= p1(x)p.J..x)
•
·
• p,(x)
=
• • •
•
·
•
• •
q1(x)q2(x)
· ·
·
qjx).
q1(x) on each end, we have
p2(x)[ctP3(x)
·
·
•
p,(x)]
=
q.f...x)q3(x)
•
•
•
qJ..x).
Complete the argument by adapting the proof of Theorem 1.8 to F[x],
replacing statements about ±fJJwith statements about associates of
q i(x).
•
• Exercises
NOTE:
F denotes afield and pa poSitiVe prime integer.
A. 1. Find a monic associate of
2
(a) 3x3 + 2x
+
x + 5 in Q[x]
5
2
(b) 3x - 4x
+ 1 in Z5[x]
(c) ix3 + x - 1 in C[x]
2. Prove that every nonzero f(x) E
F[x] has a
unique monic associate in
F[x].
3. List all associates of
(b) 3x + 2 in Z7[x]
(a) x2 + x + 1 in Z5[x]
4. Show that a nonzero polynomial in ZJx] has exactly p
5. Prove thatf(x) and g(x)
g(x) lf(x).
are
associates in
F[x] if
-
1 associates.
and only if f(x) jg(x) and
x2 + 1 is irreducible in Q[x]. [Hint: If not, it must factor as
(ax+ b)(cx + d) with a, b, c, d E Q; show that this is impossible.]
6. Show that
7. Prove that/(x) is irreducible in F[x] if and only if each of its associates is
irreducible.
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104
Chapter 4
Arithmetic in F[x]
8. If f(x) E
F[x] can be written as the product of
two polynomials of lower
degree, prove that/(x) is reducible in F[x]. (This is the second part of the
proof of Theorem 4.11.)
9. Find all irreducible polynomials of
(a)
degree 2 in .l2[x]
(c)
degree 2 in .l3[x]
(b) degree 3 in .l2[x]
10. Is the given polynomial irreducible:
(a) x2 - 3 in Q[x]? In IR[x]?
(b) x2 + x
x3
11. Show that
12. Ex press
-
x4
-
2 in .l3[x]? In
-
.Z7[x]?
3 is irreducible
in
.l7[x].
4 as a product of irreducibles in O[x], in Bl[x], and in C[x].
13. Use unique factorization to find the gcd in C[x] of
and {x - l)(x - 3)(x 14. Show that x2 +
4)3•
(x - 3)3(x - 4)4(x
x can be factored in two ways in .l6[x]
fl
as the product of non­
constant polynomials that are not units and not associates of
B. 15.
-
x or x
+ 1.
(a)
(x + a)(x + b), show that there are
exactly (# + p)/2 monic polynomials of degree 2 that are not irreducible in
.l,,[x].
(b)
Show that there are ex actly
By counting products of the form
(#
-
degree 2 in Zp[x].
16. Prove that p(x) is irreducible in
p(x) ig(x) or p(x) is
p)/2 monic irreducible polynomials of
F[x] if and only if for every g(x)
E
relatively prime to g(x).
F[x], either
17. Prove (1) => (2) in Theorem 4.1 2 .
18. Without using statement (2), prove directly that statement ( l ) i s equivalent t o
statement
(3) i n Theorem
4.12.
19. Prove Corollary 4.13.
20. If p(x) and q(x) are nonassociate irreducibles in F[x], prove that p(x) and q(x)
are relatively prime.
21.
(a)
(b)
F ind a polynomial of positive degree in .l9[x] that is a unit .
Show that every polynomial (except the constant polynomials
3
and
6)
in �[x] can be written as the product of two polynomials of positive
degree.
22.
23.
(a)
Show that x3 +
a
is reducible in Z3[x] for each a E
Z3•
(b) Show that x5 +
a
is reducible in .Z5[x] for each a E
Z5•
(a)
Show that x2 + 2 is irreducible in
(b)
Factor x4
-
Z5[x].
4 as a product of irreducibles in .Z5[x].
24. Prove Theorem 4.14.
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4.4
25.
Polynomial Functions, Roots, and Reducibility
Prove that every nonconstant/(x) E F[x] can be written in the form
cp1(x)Pi(x)
p,.(x), with c E F and each pi(x) monic irreducible in F[x].
Show further that if f(x)
dq1(x)q (x)
qm(x) with d E F and each iit(x)
2
monic irreducible in F[x], then m = n, c = d, and after reordering and
relabeling if necessary, p1(x) = q1(x) for each i.
· ·
·
=
II
105
· ·
·
Polynomial Functions, Roots, and Reducibility
In the parallel development of F[x] and Z, the next step is to consider criteria for
irreducibility of polynomials (the analogue of primality testing for integers). Unlike
the situation in the integers, there are a number of such criteria for polynomials whose
implementation does not depend on a computer. Most of them are based on the fact
that every polynomial in F[x] induces a function from F to F. The properties of this
function (in particular, the places where it is zero) are closely related to the reducibility
or irreducibility of the polynomial.
Throughout this section, R is a commutative ring. Associated with each polynomial
a,.x' +
+ a :x?- + a1x + Oo in R[x] is a functionfR � R whose rule is
2
for each r E R,
f(r) ant" +
+ ar + a1r + Qo.
·
·
·
=
·
·
·
The function/ induced by a polynomial in this way is called a polynomial function.
EXAMPLE 1
The polynomial :x?- + 5x + 3 E R[x] induces the functionfR ��whose rule
isf(r) = r2+ 5r + 3 for each r E �.
EXAMPLE 2
The polynomial x4 + x + 1 E Z3[x] induces the functionfZ3 � Z3 whose rule
isf(r) = r4 + r + 1. Thus
f(O)
=
D4 + 0
+ 1
=
l,
/(2)
""
24 +
ft))
2 + 1
=
=
14 + 1 + 1
=
0,
1.
The polynomial x3 + x2 + 1 E Z3[x] induces the function g:Z3 �Z3 given by
3
g( O) = a3 + 02 + 1 = 1,
g(l) = 1 + 12 + 1 = 0,
3
g(2) 2 + 22 + 1 = 1.
=
Thusf and g are the same function on Z3, even though they are induced by
different polynomials in Z3[x]. *
Although the distinction between a polynomial and the polynomial function it
induces is clear, the customary notation is quite ambiguous. For example, you will see a
*Remember that functions (and g are equal if f(r) = g(r) for every r in the domain.
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106
Chapter 4
Arithmetic in F[x]
statement such asf(x)
polynomialx2 - 3x + 2
=
x2- 3x + 2. Depending on the context,f(x) might denote the
E
�x] or the rule of its induced function/ Ill -+ill. The sym­
bol x is being used in two different ways here. In the polynomial x2- 3x + 2, x is an
indeterminate (transcendental element) of the ring Rx
[ ]. * But in the polynomial func­
tionf:lll-+ Ill, the symbolx is used as a variable to describe the rule of the function. It
might be better to use one symbol for an indeterminate and another for a variable, but
the practice of using x for both is so widespread you may as well get used to it.
The use of the same notation for both the polynomial and its induced function also
affects the language that is used. For instance, one says "evaluate the polynomial
3.x2- 5x + 4 at x = 2" or "substitutex
=
2 in 3x2- 5x + 4" when what is really meant
is "find/(2) when/ is the function induced by the polynomial 3x2 - 5x + 4".
The truth or falsity of certain statements depends on whether x is treated as
an
indeterminate or a variable. For instance, in the ring lllfx], where x is an indetermi­
nate ( special element of the ring), the statement x2 - 3x + 2 = 0 is false because, by
Theorem 4.1, a polynomial is zero if and only if all its coefficients are zero. When x is a
variable, however, as in the rule of the polynomial functionf(x )
x2- 3x + 2, things
=
are different. Here it is perfectly reasonable to ask which elements of Ill are mapped to 0
by the function/, that is, for which values of the variablex is it truethatx2- 3x + 2
0. It
may help to remember that statements about the variable x occur in the ring R, whereas
=
statements about the indeterminate x occur in the polynomial ring R[x].
Roots of Polynomials
Questions about the reducibility of a polynomial can sometimes be answered by
considering its induced polynomial function. The key to this analysis is the concept
of a root.
Definition
Let R be a commutative ring and f(x) E R[x]. An element a ot R is said to
be a root {or zero) of the polynomiaff{i} if ((a}= OR, that is, if the induced
function f:R-+ R maps a to O�
EXAMPLE 3
The roots of the polynomialflx)
=
x2
-
3x + 2 E lll[x] are the values of the
variablex for whichfx
( ) = 0, that is, the solutions of the equation x2- 3x + 2
It is easy to see that the roots are l and 2.
=
0.
EXAMPLE 4
The polynomial
x2 + 1
E lll[x ] has no roots in R because there are no real·
number solutions of the equation x2 + 1
=
0. However, if x'- + 1 is considered
as a polynomial in C[x], then it has i and-; as roots because these are the
solutions in C of x2 + 1
=
0.
*See page 550 in Appendix G for more information.
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4.4
Theorem 4.15
Polynomial Functions, Roots, and Reducibility
107
The Remainder Theorem
Let F be a field, f(x) E F[x], and a E f. The remainder when f(x) is divided by
the polynomial x - a is f(a).
EXAMPLE 5
To find the remainder whenftx) = x19 + 3x24 + 5 is divided by x
the Remainder Theorem with a = 1. The remainder is
-
1, we apply
/(1) = 179 + 3. 124 + 5= 1+3+5=9.
EXAMPLE 6
To find the remainder whenftx) = 3x4 - 8x2 + 1 lx + 1 is divided by x + 2, we
apply the Remainder Theorem carefully. The divisor in the theorem is x - a,
not x + a. So we rewrite x + 2 as x - (-2) and apply the Remainder T heorem
with a = -2 .. The remainder is
f(-2) = 3(-2)4 - 8(-2)2
+
11(-2) + 1 = 48 - 32 - 22 + 1 = -5.
Proof of Theorem 4,15 ... By the Division Algorithm,f(x) = (x - a)q(x)
+ r(x),
where the remainder r(x) either is OF or has smaller degree than the
divisor x - a. T hus deg r(x) = 0 or r(x) = Op In either case, r(x)= c for
some c E F. Hence,f(x) = (x - a)q(x) + c, so that f( a) = (a - a)q(a) +
c =Op+ c = c. •
Theorem 4.16
The Factor Theorem
Let F be a field, f(x) E F[x], and a E F. Then a is a root of the polynomial f(x)
if and only if x - a is a factor of f(x) in F[x].
Proof ... First assume that a is a root off(x). Then we have
f(x) = (x - a)q(x)
f(x) = (x - a)q(x)
f(x) = (x - a)q(x)
+
r(x)
+ f(a)
[DMsion Algorithm]
[Remainder 11ieorem]
[a is a root of f(x), so f(a) = OF.]
Therefore, x - a is a factor of f(x).
Conversely, assume that x - a is a factor off(x), say ft.?c) (x - a)g(x).
Then a is a root off(x) becausefta) =(a - a)g(a) =O,g(a) = Op •
=
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....... my�mmal._oot...uu:rlflKl.b�a.mliag-.m---�l...Amiof;--•rilht1u_,,,.��-..,.1imllljf....._._._:Dgtu�...-.it.
108
Chapter 4
Arithmetic in F[x]
EXAMPLE 7
To show that
x7 - -x:S + 2x" - 3x2 - x + 2 is reducible in O[x], note that
x - 1 is a factor.
1 is a
root of this polynomial. Therefore,
Corollary 4.17
Let F be a field and f(x) a nonzero polynomial of degree n in F[x]. Then f(x) has
at most n roots in F.
Proor ... Iff(x) has a root a1 inF, then by the Factor Theorem,f(x) = (x - a1)h1(x)
for some h1(x)
E F[x]. If h1(x) has a root a2 in F, then by the Factor
Theorem
f(x) =
(x
- aJ(x - a'1Jhi.(x)
for some
h2(x)
E
F[x].
If hi(x) has a root a3 inF, repeat this procedure and continue doing so
until you reach one of these situations:
(1) f(x) (x - a1)(x - tll) · · · (x - a,.)h,.(x)
(x - ak) hti.._x ) and hti.._x) has no
(2) f(x) = (x - a1)(x - aj
=
· ·
·
root inF.
In Case
(1), by T heorem 4.2, we have
= deg(x - a1) + deg(x - al) +
n=1 + 1+
+ 1 + deg h,.(x)
n = n + deg h,.(x)
degf(x)
·
Thus,
·
·
·
·
+
deg(x -
a,,) + deg
h,.(x)
·
deg h,.(x) = 0, so h,.(x) = c for some constant c E F andf(x)
factors as
f(x) = l'(x - a 1)(x - ai)
Clearly, the
n
numbers ab a2,
• • •
•
·
•
(x
- a,J .
, a,, are the only roots of f(x).
The arg ument in Case (2) is essentially the same Gust replace n by k)
and leads to this conclusion: n = deg/(x) = k + deghti.._x ). So the num­
ber of roots is k and k
s n.
•
Corollary 4.18
Let F be a field and f(x} E F[x], with deg f(x) � 2. If f{x} is irreducible in F[x],
then f(x} has no roots in F.
Proof... If f(x) is irreducible, then it has no factor of
Therefore,/(x) has no roots
the fo rm
x-
inFby the Factor Theorem.
a inF[x].
•
"If you prefer a proof by induction, see Exercise 29.
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4.4
Pol ynomi al Functions, Roots, and Reduci bi lity
109
The converse of Corollary 4.18 is false in general. For example, x4 + 2x2 + 1
(x2 + l)(x2 + 1) has no roots in Q but is reducible in Q[x]. However, the converse is
true for degrees 2 and 3.
=
Corollary 4.19
Let F be a field and let f(x} E
ir reducible in
F[x] be a polynomial of degree 2 or 3. Then f(x) is
F[x] if and only if f{x) has no roots in F.
Proof • Suppose f(x) is irreducible. Thenf(x) has no roots in Fby Corollary 4.18.
Conversely, suppose thatf(x) has no roots inF. Then/(x) has no :lirst­
degree factor in F[x] because every first-degree polynomial ex + din F[x]
has a root in F, namely -c-1d. Therefore, if f(x) r(x)s(x), neither r(x)
nor s(x) has degree 1. By Theorem 4.2, degf(x) deg r(x) + deg s(x).
Since/(x) has degree 2 or 3, the only possibilities for (deg r(x), deg s(x))
are (2, 0) or (0, 2) and (3, 0) or (0, 3). So either r(x) or s(x) must have
degree 0, that is, either r(x) or s(x) is a nonzero constant. Hence,/(x) is
irreducible by Theorem 4.12. •
=
=
EXAMPLE 7
To show that x1 + x + 1 is irreducible in Zs [x], you need only verify that none
of 0, 1, 2, 3t 4 E Zs is a root.
We close this section by returning to its starting point, polynomial functions.
Example 2 shows that two different polynomials in F[x] may induce the same function
from Fto F. We now see that this cannot oocur if Fis infinite.
Corollary 4.20
Let F be an infinite field and
f{x), g(x) E F[x]. Then f{x) and g(x) induce the
f(x) g(x) in F[x].
same function from Fto F if and only if
=
Proof • Suppose that/(x) and g(x) induce the same function from Fto F. Then
f(a) g(a), so thatf(a) - g(a) OF> for every a E F. This means that
=
=
every element of Fis a root of the polynomial/(x) - g(x). Since Fis
infinite, this is impossible by Corollary 4.17 unless f(x) - g(x) is the
zero polynomial, that is,f(x) g(x). The converse is obvious. •
=
• Exercises
NOTE: Fdenotesaffeld
A. I.
(a) Find a nonzero polynomial in Zz[x] that induces the zero function on Z2•
(b) Do the same in Z3[x].
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110
Chapter 4 Arithmetic in F[x]
2.
Find the remainder when.fix) is divided by g(x):
1
(a) f(x) x 0 + x1 and g(x) x - I in Cl![x]
=
=
(b) f(x)
(c) f(x)
(d) f(x)
3.
=
=
=
2x5 - 3x4+ x3 + 2x + 3 and g(x)
=
x - 3 inZ [x]
5
=
Determine if h(x) is a factor of fix):
(a) h(x)
=
x+ 2 and/(x)
(b) h{x)
=
x-
(c) h(x)
=
.!_ and/(x)
=
x - 3 and/(x)
=
x3 - 3x2 - 4x - 12 in lll[ x]
= 2x4+
2
x + 2 and/(x)
=
(d) h(x)
4.
2x5 - 3x4+ x3 - 2x2+ x - 8 and g(x) x - 10 in Cl![x]
1
l0x75 - 8x65+ 6.05 + 4x37 - 2x 5+ 5 and g(x) = x+ 1 in O[x]
=
x3 + x - �in O[x]
4
3x5 + 4x4 + 2x3 - x2+ 2x +
I
x6- � + x - 5 inZ7[x]
inZ [x]
5
(a) For what value of k is x - 2 a factor of x4 - 5x3 + 5x2+ 3x+ k in Q[x]?
(b) For what value of k is x
5.
+ 1 a factor of x4+ 2x3 - 3x2 + kx + 1 in Z [x]?
5
Show that x - lFdividesanx" + ... + azx2 +a1X +ao in F[x] if and only if
ao +a1 + az + ... + an
o,..
=
6.
(a) Verify that every element of Z3 is a root of x3 - x
E Z3[x].
(b) Verify that every element of Zs is a root of x5 - x
E Zs[x].
(c) Make a conjecture about the roots of x!' - x
E Zp[x]
(p prime).
7.
Use the Factor Theorem to show that x7 - x factors inZ [x] as
7
x(x - lXx - 2)(x - 3)(x - 4)(x - S)(x - 6), without doing any polynomial
multiplication.
8.
Determine if the given polynomial is irreducible:
(a) x2 -
7 in R[x]
(c) x2 +
7 in
(e) x3 -
9
C[x]
inZ11 [x]
(b)
x2
- 7 in O[x]
(d) 2x 3 + x2 + 2x + 2 in Z [x]
(t) x4
+
5
x2 + 1 inZ 3[x]
List all monic irreducible polynomials of degree 2 inZ3[x]. Do the same inZ [x].
5
10. Find a prime p > 5 such that x2+ 1 is reducible inZp[x].
9.
11.
B.12.
Find an odd prime p for which x - 2 is a divisor of x4 + x3 + 3x2 + x + 1 in
Zp[x].
Ifa E Fis a nonzero root of c,;X'+ Cn_,x"-1 +
1
+ c,.__1x +
that a-1 is a root of c� + c1x"- +
·
·
13.
·
·
·
·
+ c1x+
c,,.
c0
E
F[x], show
(a) If f(x) and g(x) are associates in F[x], show that they have the same roots
in F.
14.
(b) If fix), g(x)
E
(a) Supposer, s
E
F[x] have the same roots in F, are they associates in F[x]?
Fare roots of ax1 +bx+ c E F[x] (witha* Op). Use the
Factor Theorem to show that r + s = -a-1b and rs = a-1c.
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�.dkl. OMadl
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......
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4.4
Polynomial Functions, Roots, and Reducibility
111
(b) Suppose r, s, t E Fare roots of ax3+ hx2 + ex+ d E F[x] (with a *OF)·
Show that r + s + t = -a-1b and rs + st + rt = a-1e and rst "" -a-1d.
15. Prove that x'-+ I is reducible in Zp[x] if and only if there exist integers a and h
such thatp =a+ h and ah= l (modp).
16. Letf(x), g(x) E F[x] have degree 5; n and let
of F. Iff(e1)
=
g(cJ for i
=
0, 1,
. . .
, n,
c0, e1,
•
•
• ,
prove thatf(x)
e11 be distinct elements
g(x) in F[xJ.
=
17. Find a polynomial of degree 2 in �[x] that has four roots in Z6• Does this
contradict Corollary 4.17?
1p:C-+. C be an isomorphism of rings such that 1p(a) = a for each
a E Q. Supposer E C is a root of f(x) E Q[x] . Prove that cp(r) is also a
root of /(x).
18. Let
19. We say that a E F is a multiplerootof/(x) E F[xJ if (x
f(x) for some k � 2.
-
a)" is a factor of
(a) Prove that a E Ris a multiple root of f(x) E R[x] if and only if a is a
root of bothf(x) andf'(x), wheref'(x) is the derivative of f(x).
(b) Iff(x) E R[x] and i f f(x) is relatively prime tof'(x). prove thatf(x) has
no multiple root in R.
20. Let R be an integral domain. Then the Division Algorithm holds in R[x]
whenever the divisor is monic, by Exercise 14 in Section 4.1. Use this fact to
show that the Remainder and Fact or Theorems hold in R[x].
21. If R is an integral domain andf(x) is a nonzero polynomi al of degree n in
R[x], prove thatf(x) has at most n roots in R. [Hint: Exercise 20.]
22. Show that Corollary 4.20 holds if F is an infinite integral domain. [Hint: See
Exercise 21.]
23. Let/(x), g(x), h(x) E F[x] and r E F.
(a) Iff(x)
=
g(x) + h(x) in F[x], show that.f(r) =g(r)+ h(r) in F.
(b) Iff(x) =g(x)h(x) in F[x], show thatf(r) =g(r)h(r) in F.
Where were these facts used in this section?
24. Let
a be a fixed element of F and define a map 'Pa:F[x]-+. Fby 'Pa[f(x)]
=
f(a)
.
Prove that
'Pa is a surjective homomorphism of rings. The map 'Pa is called an
e\•aluation homomorphism; there is one for each a E F.
25. Let Ol[1T] be the set of all real numbers of the form
r0 + r11T + r21T2 +
·
·
·
+ a,.1T",
with n � 0 and r1 E
Q.
(a) Show that 0[1T] is a subring of R.
(b) Show that the function 8:0[x]-+. 0[1T] defined by 8(f(x))
=
f(1T) is an
isomorphism. You may assume the following nontrivial fact:
1T
is not
the root of any nonzero polynomial with rational coefficients. Therefore,
Theorem 4.1 is true with R = 0 and 1T in place of
Exercise 26.
x. However, see
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112
Chapter 4
Arithmetic in F[x]
26. Let O![v'2] be the set of all real numbers of the form
ro +
r1
v'2 + r2(v'2)2 +
· · ·
+
r,,(v'2)", with n ;?: 0 and r1
E
Q.
(a) Show that 0![\12] is a subring of R.
(b) Show that the function 8:0[x]-+ 0['\/2] defined by 8(f(x)) /(v'2) is a
surjective homomorphism , but not an isomorphism. Thus Theorem 4.1 is
not true with R CJ! and v'.2 in place of x. Compare this with Exercise 25.
=
=
27. Let The the set of all polynomial functions from Fto F. Show that Tis a
commutative ring with identity, with operations defined as in calculus: For
eachr E F,
(f + g)(r) = j(r) + g(r)
and
(fg)(r)
=
f(r)g(r).
[Hint: To show that Tis closed under addition and multiplication, use
Exercise 23 to verify that/+ g andfg are the polynomial functions induced
by the sum and product polynomials/(x) + g(x)andf(x)g(x), respectively.)
28. Let The the ring of all polynomial functions from Z3 to Z3 (see Exercise 27).
(a) Show that Tis a finite ring with zero divisors. [Hint: Consider/(x)
and g(x) x2 + 2x.J
= x
+ 1
=
(b) Show that Tcannot possibly be isomorphic to Z3[x]. Then see Exercise 30.
29. Use mathematical induction to prove Corollary 4.17.
C. 30.
If Fis an infinite field, prove that the polynomial ring F[x] is isomorphic to
the ring Tof all polynomial functions from Fto F(Exercise 27). [Hint: Define
a map ip:F[x]-+ Thy assigning to each polynomial/(x) E F[x] its induced
function in T; ip is injective by Corollary 4.20.]
31. Let cp:F[x]-+ F[x] be an isomorphism such that cp(a) =a for every a E F.
Prove that/(x) is irreducible in F[x] if and only if rp(/(x))is.
32. (a) Show that the map cp:F[x] -+ F[x] given by cp(f(x))
isomorphism such that i:p(a) = a for every a E F.
==
f(x + IF) is an
(b) Use Exercise 31 to show that/(x) is irreducible in F[x] if and only if
f(x+ lF)is.
II
irreducibility in Q[x]*
The central theme of this section is that factoring in Q [x] can be reduced to factoring
in :Z[x]. Then elementary number theory can be used to check polynomials with inte­
ger coefficients for irreducibility. We begin by noting a fact that will be used frequently:
If f(x) E Cl!lxl, then c/(x) bas integer
coefficients for some nonzero integer c.
*This section is used only in Chapters 11, 12, and
15. It may be omitted until then, if desired. Section 4.6
is independent of this section.
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-.n-:11t.
4.5
Irreducibility in
Q[x]
113
For example, consider
f(x)
=
XS
+
2
3
1
3x4 + 4x3 - 6'
The least common denominator of the coefficients of f(x) is 12, and 12/{x) has integer
coefficients:
12/(x) =
]
r
2
3
1
12lx5 + 3x4 + 4x3 - '6
9x3 - 2.
= 12x5 + 8x4 +
According to the Factor Theorem, finding first-degree factors of a polynomial
g(x) E O![x] is equivalent to finding the roots of g(x) in O. Now, g(x) has the same
roots as cg(x) for any nonzero constant c. When c is chosen so that cg(x) has integer
coefficients, we can find the roots of g(x) by using
Theorem 4.21
Rational Root Test
Let f(x) = a/ + a11_t�--'1 + · · · + a1x + a0 be a polynomial with integer coef­
ficients. If r :/: O and the rational number r/s (in lowest terms) is a root of f(x),
then rlao and slan.
Proof.. First consider the case whens
1, that is, the case when the integer r
is a root of f(x), which means that a,.r" + 4.-tr"-1 +
+ a1r + ao = 0.
Hence,
=
·
"o
=
"o =
- a,/' - tlit-1r--1 r(--o,.t"'-t a,._1,,.-2
. • •
_
_
•
· ·
- air
•
•
_
ai),
which says that r divides "o·
In the general case, we use essentially the same strategy. Since r/s is a
root of f(x), we have
a,.(�) a.-{�=:)
+
+
· · ·
+
a
{0
+ "o
=
0.
We need an equation involving only integers (as in the case whens
So multiply both sides bys", rearrange, and factor as before:
a,.r" + a,._.sr"-1 +
·
·
·
+
a1s"-1r + ar/'
at/' = -a,/'
(*)
Of/'
=
-
=
=
1).
0
a,,_1.rr"-1
-
•
•
•
- a1s"-1r
r[-a,l'-t - a...-1.rr"-2 - ... - a1s"-1].
This last equation says that r divides¥, which is not quite what we
want. However, since r/s is in lowest terms, we have (r, s)
1. It follows
that (r, s")
1 (a prime that divides s" also divides s, by Corollary 1.6).
Since r I� and (r, s") = 1, Theorem 1.4 shows that r I ao· A similar argu­
ment proves that s I a,, Gust rearrange Equation (*) so that a,,r' is on one
side and everything else is on the other side). •
=
=
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.
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oot...un;,.dkt.bi�
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114
Chapter 4
Arithmetic in F[x]
EXAMPLE 1
The possible roots in Q of f(x)
2x4 + x3 2lx2 14x + 12 are of the form
r/s, where r is one of ±1, ±2, ±3, ±4, ±6, or ±12 (the divisors of the constant
term, 12) ands is ±1 or ±2 (the divisors of the leading coefficient, 2). Hence,
the Rational Root Test reduces the search for roots of f(x) to this finite list of
possibilities:
- -
=
1
1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12, 2'
1 3
3
-2· 2' -2'
It is tedious but straightforward to substitute each of these inf(x) to find that -3
and
�
are the only roots of f(x) in O.* By the Factor Theorem, both x
x + 3 and x -
�
- (-3) =
are factors of f(_x). Division shows that
f(x)
=
( - �)(2x1 - - 8).
- -
(x + 3) x
4x
The quadratic formula shows that the roots of 2x2 4x 8 are 1 ± VS,
neither of which is in O. Therefore, 2x2 4x - 8 is irreducible in Q[x] by
Corollary 4.19. Hence, we have factoredf(_x) as a product of irreducible poly­
nomials in O[x].
EXAMPLE 2
The only possible roots of g(x) x3· + 4x2 + x - 1 in Q
1 and -1 (Why?).
Verify that neither 1 nor -1 is a root of g(x). Hence g(x) is irreducible in Q[x]
by Corollary 4.19.
are
=
If f(x) E O![x], then cf(x) has integer coefficients for some nonzero integer c. Any
factorization of cf(x) in Z[x] leads to factorization of f(x) in Q[x]. So it appears that
tests for irreducibility in O![x] can be restricted to polynomials with integer coefficients.
However, we must first rule out the possibility that a polynomial with integer coeffi­
cients could factor in O![x] but not in Z[x]. In order to do this, we need
Lemma 4.22
Let f(x}, g(x), h(x} E Z[x] with f(x) g(x)h(x). If pis a prime that divides every
coefficient of f(x), then either p divides every coefficient of g(x) or p divides
every coefficient of h(4
=
*A
of
of y = l{x) are the roots of ll:X), you can eliminate any numbers from the list that aren't near
1
3
intercept. In this case, the griiph indicates thiit you need only check -3, 2' iind -2'
graphing calculator will reduce the iimount of computation significiintly. Since the x-intercepts
the graph
iin
.......
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k
...
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4.5
Proof• Letflx) = ao + a1x +
h(x) = c0 + c1x +
·
·
··· + a�, g(x)
·+
Irreducibility in Q[x]
115
b0+ b1x +
+ b,,.x", and
c,,x!'. We use a proof by contradiction. If the
=
·
· ·
lemma is false, thenp does not divide some coefficient of g(x) and some
coefficient of J(x). Let b, be the first coefficient of g(x) that is not divis­
ible by p, and let
c, be thefirst coefficient of h(x) that is not divisible by
r and p I c1 for j < t. Consider the coefficient ar+t of
p. Then p I b1 for i <
f(x). Sinceflx) = g(x)h(x),
ar+t
=
hocrH + . . .
+
br-1C1+1
+
b,c, + h.+1Ct-I + ... + br+.CD·
Consequently,
b,c, = ar+t - [boCr+t +
•
•
•
+ b r-1C1+1J
- (hr+JCr-1 +
• • •
+
b,+,cnJ.
Now, p I a.+i by hypothesis. Also, p divides each term in the first pair of
brackets because r was chosen so thatp lb, for each i < r. Similarly,p
divides each term in the second pair of brackets because p I c1 for each
j < t. Since p divides every term on the right side, we see that p I b,c.,.
Therefore, p I b, or p I Ct by Theorem 1.5. This contradicts the fact that
neither b, nor c, is divisible by p. •
Theorem 4.23
Let f(x) be a polynomial with integer coefficients. Then f(x) factors as a prod­
uct of polynomials of degrees m and n in O[x] if and only if f{x) factors as a
product of polynomials of degrees m and n in Z[x].
Proof •Obviously, if flx) factors in Z[x], it factors in O[x]. Conversely, suppose
flx)
=
g(x)h(x) in O[x]. Let c and dbe nonzero integers such that cg(x)
and dh(x) have integer coefficients. Then cdflx)
[cg(x)][dh(x)] in Z[x]
with deg cg(x)
deg g(x) and deg dh(x)
deg h(x). Let p be any prime
=
=
=
divisor of cd, say cd = pt. Thenp divides every coefficient of the polyno­
mial cdflx). By Lemma 4.22,p divides either every coefficient of cg(x)
or every coefficient of dh(x), say the former. Then cg(x) = pk(x) with
k(x) E Z[x] and degk(x)
degg(x). Therefore,pif(x) = cdf(x)
[cg(x)][dh(x)] = [pk(x)][dh(x)]. Cancelingp on each end, we have
=
k(x)[dh(x)] in Z[x].
tf(x)
Now repeat the same argument with any prime divisor of
=
=
t and cancel
that prime from both sides of the equation. Continue until every prime
factor of cd has been canceled. Then the left side of the equation will be
±flx), and the right side will be a product of two polynomials in Z[x],
one with the same degree as g(x) and one with the same degree as h(x).
•
EXAMPLE 3
We claim thatflx)
x4 - Sx2 + 1 is irreducible in O[x]. The proof is by con­
tradiction. Ifflx) is reducible, it can be factored as the product of two noncon­
=
stant polynomials in O[x]. If either of these factors has degree 1, then/(x) has
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116
Chapter 4
F[x]
Arithmetic in
a root in Q. But the Rational Root Test shows that.f(x) has no roots in Q. (The
only possibilities are ±1, and neither is a root.) Thus if f(x) is reducible, the
only possible factorization is as a product of two quadratics, by Theorem 4.2.
In this case Theorem 4.23 shows that there is such a factorization in .l[x].
Furthermore, there is a factorization as a product of monie quadratics in .l[x]
by Exercise 10, say
(x1 + ax +
b)(x2 +
ex + d) = x4
-
sx2
+1
with a, b, e, d E .l. Multiplying out the left-hand side,, we have
x4 +
(a
+ c)r +
(ac + b + d)x2 + (be + ad)x
Or - Sx2 + Ox + 1.
= x4 +
+ bd
Equal polynomials have equal coefficients; hence,
ac + b + d= -S
a + c=O
Since a + c = 0, we have a =
bd=1.
bc+ad=O
-c, so that
�s = ac +
+ d = -c2 + b + d,
b
or, equivalently,
5=c2-b-d.
However, bd= 1 in .l implies that b = d=1orb = d = -1, and so there are only
these two possibilities:
or
S=e-1-1
7 = c2
s=c2+1 + 1
3 = c2.
There is no integer whose square is 3 or 7, and so a factorization of f(x) as a
product of quadratics in Z[x], and, hence in Q[x], is impossible. Therefore,f(x)
is irreducible in Q[x].
The brute-force methods of the preceding example are less effective for polynomi­
als of high degree because the system of equations that must be solved is complicated
and difficult to handle in a systematic way. However, the irreducibility of certain poly­
nomials of high degree is easily established by
Theorem 4.24
Eisenstein's Criterion
Let f(x}
atpX11 +
+ a1x + a0 be a nonconstant polynomial with Integer
coefficients. If there is a prime p such that p divides each of a0, a1,
, liln-1
but p does not divide an and p2 does not divide aa, then f(x) is irreducible in Q[x].
=
·
·
·
•
Proof• The proof is by contradiction. If f(x)
it can be factored in .l[x], say
f(x)
= (b0
+ b1x +
·
·
·
•
•
is reducible, then by Theorem 4.23
+ b,x')( c0 + c1x
+
·
·
·
+
c1x'),
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4.5
where each h,,
Irreducibility in
Q[x]
117
b0c0• By hypothe­
by Theorem 1.5, say p I b0• Since p2 does
not divide ao, we see that c-0 is not divisible byp. We also have an = b,c..,
Consequently, p does not divide hr (otherwise a,, would be divisible by p,
contrary to hypothesis). There may be other b1 not divisible by p as well.
Let bk be the first of the b1 not divisible by p; then 0 < k s r < n and
c1
E Z, r <:?: 1, and s <:?: 1. Note that ao =
sis, p I ao and, hence, p I b0 or p I c0
and
By the rules of polynomial multiplication,
ak =hock+ h1ck-1 +
·
·
·
+ bk-1C1 + h,.co,
so that
hfA> = ak - hoct. - h1ct.-1 -
•
•
.
-bk-1C1•
Since p I ak and p Ih, for i < k, we see that p divides every term on the
right-hand side of this equation. Hence, p I hk Co· By Theorem 1.5, p must
divide bk or c0• This contradicts the fact that neither bk nor
by p. Therefore,f(x) is irreducible in Q[x]. •
c0 is divisible
EXAMPLE 4
x17 + 6x13 - 15x" + 3x2 - 9x + 12 is irreducible in O[x] by
Eisenstein's Criterion withp = 3.
The polynomial
EXAMPLE 5
The polynomial x? +S is irreducible in Q[x] by Eisenstein's Criterion with
p = S. Similarly, � + 5 is irreducible in Q[ x] for each n <:?: 1. Thus
there are irreducible polynomials of every degree in Qlxt.
Although Eisenstein's Criterion is very efficient, there are many polynomials to
which it cannot be applied. In such cases other techniques are necessary. One such
method involves reducing a polynomial mod p, in the following sense. Let p be a posi­
tive prime. For each integer a, let
a,.XC +
·
·
·
polynomial
+ a1x +
[a¥ +
[a] denote the congruence class of a in Zp. If f(x) =
](x) denote the
[ai] x + [a.ii in Zp[x]. For instance, if f(x) = 2x4 - 3x2 +
ao is a polynomial with integer coefficients, let
·
·
·
+
Sx + 7 in .l[x], then in Z3[x],
](x) = [2]x" - [3]x2 + [5]x + [7]
= [2]x"
-
[O]x2 + [2]x+ [l] = [2]x4 + [2]x+ [ l].
Notice that f(x) and
](x) have the same degree. This will always be the case
/(x) is not divisible by p (so that the leading
coefficient of ](x) will not be the zero class inZ,).
when the leading coefficient of
CopJftglll.20t2C...�
.,.
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118
Chapter 4
Arithmetic in F[x]
Theorem 4.25
Let f(x)
=awl+· · · +a,x + a0 be a polynomial with integer coefficients, and
let p be a positive prime that does not divide ak. If f(x) is irreduc ible in
then
f(x) is irreducible in Q[ x] .
Zp[x],
Proof.. Suppose, on the contrary, thatf(x) is reducible in Q[x]. Then by
Theorem 4.23,f(x) g(x)h(x ) withg(x), h(x) nonconstant polynomials
in Z[x]. Sinoe p does not divide ak> the leading coefficient of f(x), it
cannot divide the leading coefficients of g(x) or h(x) (whose product is
a1c). Consequently, deg g(x) = deg g(x) and deg h(x) = deg h(x). In par­
ticular, neither g(x) nor h(x) is a constant polynomial in �[x ].
Verify that/(x) g(x)h(x) in Z[x] implies that f(x) = g(x)h(x) in
Z,[x] (Exercise 20). This contradicts the irreducibility of f(x) in Z,[x].
Therefore,/(x) must be irreducible in Q[x]. •
=
=
The usefulness of Theorem 4.25 depends on this fact: For each nonnegative in­
teger k, there are only finitely many polynomials of degree k in Z, [x] (Exercise 17).
Therefore, it is always possible, in theory, to determine whether a given polynomial in
Z,[x] is irreducible by checking the finite number of possible factors. Depending on
the size of p and on the degree of f(x), this can often be done in a reasonable amount
of time.
EXAMPLE 6
To show that/(x) x5 + 8x4 + 3x2 + 4x + 7 is irreducible in Q[x], we reduce
mod 2. In Zl[x], ftx) = � + x2 + 1. * It is easy to see that f(x) has no roots in
Z2 and hence no first-degree factors in Z2[x]. The only quadratic polynomials in
2
Z2[x] are x2, x2 + x, x2 + 1, and x2 + x + 1. Howevei:; if x , x2 + x = x(x + 1),
or x2 + 1 (x + l)(x + 1) were a factor, then /(x) would have a first-degree
factor, which it doesn't. You can use division to show that the remaining qua­
dratic, x2 + x + 1, is not a factor of f(x). Finally, ](x) cannot have a factor
of degree 3 or 4 (if it did, the other factor would have degree 2 or 1, which is
impossible). Therefore, ](x) is irreducible in Z2[x]. Henoe,f(x) is irreducible
in O[x].
=
=
CAUTION:
If a polynomial in Z[x] reduces mod p to a polynomial that
is reducible in Z,[x], then no conclusion can be drawn from
Theorem 4.25. Unfortunately, there may be many p for
which the reduction of f(x) is reducible in Z,[x], even when
f(x) is actually irreducible in Q[x]. Consequently, it may
take more time to apply Theorem 4.25 than is first apparent.
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l
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4.5
Irreducibility in
Q[x]
119
• Exercises
A. 1. Use the Rational Root Test to write each polynomial as a product of irreduc­
ible po lynomials in Q[x]:
(a) -x4 +x3 + x2 + x
(c)
3x 5 + 2x" - 7x3 +
+
2
(b)
X5 + 4x4
+ x3 - X2
(d) 2x4 - 5x3 + 3x2 + 4x -
2x2
6
(f) 6x4 - 3lx3 + 25x2 + 33x + 7
(e) 2x" + 7x3 + 5x2 + 7 x + 3
2. Show that Vpis irrational for every positive prime integer p. [Hint: What are
the roots of x2 - p? Do you prefer this proof to the one in Exercises 30 and 31
of Section 1.3?)
3. If a monic polynomial with integer coefficients has a root in 0, show that this
root must be an integer.
4. Show that each polynomial is irreducible in O[x], as in Example
(a) x4 + 2x3 + x + 1
(b) x4 -
3.
2x2 + 8x + 1
5. Use Eisenstein's Criterion to show that each polynomial is irreducible in Q[x]:
(a) x5 - 4x + 22
(b) 10 - l5x + 25x2 - 7x4
(c) 5x11 - 6x4 + 12x3 + 36x - 6
6. Show that there are infinitely many integers k such that :i' + 12x5 - 2lx + k
is irreducible in O[x].
7. Show that each polynomial/(x) is irreducible in Q[x] by finding a primep
such thatf(x) is irreducible in Z,[x]
(a) 7x3 + 6x2 + 4x + 6
(b) 9x4 + 4x3 - 3x + 7
8. Give an example of a polynomialf(x) E Z[x] and a prime p such thatf(x)
is reducible in Q[x] but J(x) is irreducible in Z,[x]. Does this contradict
Theorem 4.25?
9. Give an example of a polynomial in Z[x] that is irreducible in Q[x] but factors
when reduced mod 2, 3, 4, and 5.
IO. If a monic polynomial with integer coefficients factors in Z[x] as a product of
polynomials of degrees m and n, prove that it can be factored as a product of
monic polynomials of degrees m and n in Z[x].
B. 11. Prove that 30X' - 91 (where n E Z, n > 1) has no roots in O.
12. Let Fbe a field andf(x) E F[x]. If c E F andf(x + c) is irreducible in F[x],
prove thatf(x) is irreducible in F[x]. [Hint: Prove the contrapositive.]
13. Prove thatf(x)
x4 + 4x + 1 is irreducible in O[x] by using Eisenstein's
Criterion to show thatf(x + 1) is irreducible and applying Exercise 12.
=
14. Prove thatf(x)
=
x4
+ x3 +
hint for Exercise 21 withp
x2 + x
=
+ l is irreducible in O[x]. [Hint: Use the
5]
.
15. Let/(x)
� + a,._1xi-1 +
+ a1 x + '1(1 be a polynomial with integer
coefficients. If p is a prime such thatp I at> p I �
,p I a,, but p .I-' ao and
=
· ·
·
•
.
.
.
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120
Chapter 4
Arithmetic in F[x]
p2 .I' a,., prove
that/(x) is ir reducible in
Cl![x]. [Hint: Let y = 1/x inf(x)/X'; the
resulting polynomial is irreducible, by Theorem 4.24.]
16. Show by example that this statement is false: Iff(x) E
prime p satisfying the hypotheses of Theorem 4.24,
17. Show that there are ,/+t
-
ti' polynomials of
degree kin
18. Which of these polynomials are irreducible in
(a) x4 - x2 + 1
(c) x5
4
+ 4x
(a) r
+
a
+1
(d) r + sx2 +
+ 2x3 + 3x2 - x + 5
Z,,[x].
Q[x]:
(b) x4 + x
19. Write each polynomial as
Z[x] and there is no
thenf(x) is reducible in O[x].
4x
+
7
product of irreducible polynomials in
2x4 - 6x2 - 16x - 8
(b) x1
-
Q[x].
2x6 - 6x4 - 15x2 - 33x - 9
a,,X' +
+ a1x + ao, g(x) b.,X +
+ b1x + b0, and h(x)
+ c1x + Co are polynomials in Z[x] such that/(x) g(x)h(x), show
that in Z,,[x], f(x) = -g(_x'jh(x). Also, see Exercise 19 in Section 4.1.
20. If f(x)
c;r +
=
·
·
·
·
·
=
·
·
·
=
·
=
C.21. Prove that for p prime,f(x) = ;r1
in O[x]. [Hint: (x f(x + 1) = [(x + l}"
(Appendix
l)f(x)
-1]/x.
+ xJ'-1 +
·
·
·
+
x2 + x + 1 is i rreducible
;I' - 1 , so that/(x) = (:i' - 1)/(x - 1) and
Expand (x + 1)P by the Binomial Theorem
=
E) and note that p divides
(:)
when k > 0. Use Eisenstein's
Criterion to show that/(x + 1) is ir reducible; apply Exercise 12.]
EXCURSION: Geometric Constructions (Ch.apter 15) may be covered
this point if de$ired.
II
at
Irreducibility in R[x] and C[x]*
Unlike the situation in
ducible polynomials in
nomial in
O[x], it is possible to give an explicit description of all the irre­
ll[x] and qx]. Consequently, you can im mediately tell i f a poly­
lll[x] or qx] is irreducible
without any elaborate tests or criteria. These facts
are a consequence of the following theorem, which was first proved by Gauss in 1799:
Theorem 4.26
The Fundamental Theorem of Algebra
Every nonconstant polynomial in C[x] has a root in IC.
This theorem is sometimes expressed in other terminology by saying that the field
IC is
aJgebraically closed. Every known proof
of the theorem depends significantly on
facts from analysis and/or the theory of functions of a complex variable. For this rea­
son, we shall consider only some of the implications of the F undamental Theorem on
irreducibility in
C[x] and ll[x]. For a proof, see Hungerford [5].
*This section is used only in Chapters 11and12. It may be omitted until then, if desired.
...
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it.
Irreducibility in �[x] and C[x]
4.6
121
Corollary 4.27
A
polynomial is irreducible in C[x] if and only if It has degree 1.
Proof ..A polynomial.ftx) of
degree � 2 in C[x] has a root in C by Theorem 4.26
and hence a first-degree factor by the Factor Theorem. Therefore f(x) i s
reducible i n C[x], an d every irreducible polynomial i n C[x] must have
degree 1. Conversely, every first-degree polynomial is irreducible
(Example
1 in Section 4.3).
•
Corollary 4.28
Every nonconstant polynomial f(x) of degree n in IC[x] can be written in the
form c(x - a1}(x - a2) • • • (x - an) for some c, a1, a2,
, an E C. This factor­
ization is unique except for the order of the factors.
•
Proof "'By Theorem 4.14,/(x) is a product of
•
•
irreducible polynomials in
C[x].
n of
Each of them has degree 1 by Corollary 4.27, and there are exactly
them by Theorem 4.2. Therefore,
f(x)
= (r1 x + s1)(riX + s2J • · · (r,.x + s,.)
1
= r1(x - (-r1-1sJ)r2(x - (-r2- sv) · · • rJ...x - (-r,.-1s,.))
=
- a1)(x - a-i) • • • (x - a,.),
1
where c = r1r2 • • r,. and a1 = r1- s1• Uniqueness follows from Theorem 4.14;
c(x
•
see Exercise 25 in Section 4.3.
•
To obtain a description of all the irreducible polynomials in IR[x], we need
Lemma 4.29
If f(x)
a
is a polynomial in R[x] and a+ bi is a root of f(x) in C, then a - bi is also
root of f(x).
Proof "' If c = a + bi
E
any c, d E C,
(c
Also note that
·
·
•
+
a1x
C (with a, b
+ d)
=
c+d
and
cd
=
ed
.
c = c if and only if c is a real nwnber. Now, if.ftx)
c is a root of f(x), then/(c) = 0, so that
=
a,.:X' +
+ ao and
0
=
0
=
f(c)
a,.c" +
=
Q,.c"
=
=
=
Therefore
R), let c denote a - bi. Verify that for
E
+
a,.c" +
· ·
·
· ·
+ a1c + tlo
+ a,c +
. . .
·
+
a1c
ao
+ ao
[Becawe each a,E R.]
f(c).
c = a - bi is also a root of Jtx).
•
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.
ICDl
dllklDlil.
llllnl•_..,.lillll��:Dgbb�...-.:lit.
122
Chapter 4
Arithmetic
in
F[x]
Theorem 4.30
polynomial f(x) is irreducible in R(x] if and only if f(x) is a first-degree poly­
nomial or
A
f(x}
=
ax2 + bx + c
with b2
-
4ac
< 0.
Proof.,. The proof that the two kinds of polynomials mentioned in the theo­
rem are in fact irreducible is left to the reader (Exercise 7). Conversely,
supposef(x) has degree� 2 and is irreducible in R[x]. Then/(x) has a
root w in C by Theorem 4.26. Lemma 4.29 shows that w is also a root
of f(x). Furthermore, w :# w (otherwise w would be a real root of f(x),
contradicting the irreducibility of f(x)). Consequently, b y the Factor
Theorem, x - w and x - ware factors of f(x) in C[x]; that is,f(x) =
(x - w)(x - W)h(x) for some h(x) in C[x]. Let g(x) = (x - w)(x - W);
thenf(x) = g(x)h(x) in C[x]. Furthermore, if w = r + si (with r, s E R),
then
g(x)
=
=
(x - wXx - W)
=
(x - (r + si))(x
-
(r
-
si))
x1 - 2rx + (r1 + ?).
Hence, the coefficients of g(x) are real numbers.
We now show that h(x) also has real coefficients. The Division
Algorithm in R[x] shows that there are polynomials q(x), r(x) in ll.[x]
such thatf(x) g(x)q(x) + r(x), with r(x) 0 or deg r(x) <deg g(x). In
C[x], however, we havef(x) g(x)h(x) + 0. Since q(x) and r(x) can be
considered as polynomials in C[x], the uniqueness part of the Division
Algorithm in C[x] shows that q(x) = h(x) and r(x) = 0. Thus h(x) =
q(x) E ll.[x]. Sincef(x) g(x')li(x) andf(x) is irreducible in R[x] and
deg g(x) 2, h(x) must be a constant d E n.. Consequently,f(x) dg(x)
is a quadratic polynomial in R[x] and hence has the form ax2 + bx+ c
for some a, b, c E Ill. Sincef(x) has no roots in R, the quadratic formula
(Exercise 6) shows that b2 - 4ac < 0. •
=
=
=
=
=
=
Corollary 4.31
Every polynomial f(x) of odd degree in R[x] has a root inn..
Proof... By Theorem 4.14,f(x) p1(x)p (x)
pJ..x) with eachp1(x) irreduc­
2
ible in R[x]. Eachp1(x) has degree 1 or 2 by Theorem 4.30. Theorem 4.2
shows that
=
degf(x)
=
•
•
•
degp1(x) + degP2(x) +
·
·
·
+ degpk(x).
Sincef(x) has odd degree, at least one of thep1(x) must have degree L
Therefore,f(x) has a first-degree factor in R[x] and, hence, a root in ill. •
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4.6
Irreducibility i n
llll:[x] and C[x]
123
It may seem that the Fundamental Theorem and its corollaries settle all the basic
questions about polynomial equations. Unfortunately, things aren't quite that simple.
None of the known proofs of the Fundamental Theorem provides a constructive way
to find the roots of a specific polynomial.* Therefore, even though we know that every
polynomial equation has a solution in C, we may not be able to solve
a
particular
equation.
Polynomial equations of degree less than 5 are no problem. The quadratic formula
shows that the solutions of any second-degree polynomial equation can be obtained
from the coefficients of the polynomials by taking sums, differences, products, quotients,
and square roots. There are analogous, but more complicated, formulas involving cube
and fourth roots for third- and fourth-degree polynomial equations (see page 423 for one
version of the cubic formula). However, there are no such formulas for finding the roots
of all fifth-degree or higher-degree polynomials. This remarkable fact, which was proved
nearly two centuries ago, is discussed in Section 12.3.
• Exercises
A. I. Find all the roots in C of each polynomial (one root is already given):
(a) x4 - 3x3 + -x'- + 1x - 30; root 1 - 2i
(b) x4 - 2x3 - -x'- + 6x - 6; root l + i
(c)
x4
-
4x3 + 3-x'- + 14x + 26; root 3 + 2i
2. Find a polynomial in IR[x] that satisfies the given conditions:
(a) Monie of degree 3 with 2 and 3 + i as roots
(b) Monie of least possible degree with 1 - i and 2i as roots
(c) Monie of least possible degree with 3 and 4i - 1 as roots
3. Factor each polynomial as a product of irreducible polynomials in Q[x], in
lli(x], and in C[x]:
(a) x4
4. Factor
-
2
(b) x3 + 1
(c) x1 - x2 - 5x + 5
x'- + x + 1 + i in C[x].
B. 5. Show that a polynomial of odd degree in IR[x] with no multiple roots must
have an odd number of real roots.
*It may seem strange that it is possible to prove that a root exists without actually exhibiting one,
but such "existence theorems" are quite common in mathematics.
A very rough
analogy is the
situation that occurs when a person is ki lied by a sniper's bullet The police know that there
is a
killer, but actually finding the killer may be difficult or impossible.
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124
Chapter 4
Arithmetic in F[x]
6. Letf(x) =
are
ax2 +bx +c
-b + v'b2
E lll[x] with a::/:: 0. Prove that the roots of f(x) in C
-
4ac
2a
and
-b - v'b2 2a
[Hint: Show that ar +bx+
c = 0 is equivalent to x2
+bx +
[Hint: See Exercise 6].
c E
complete the square to find x.]
7. Prove that every ax2
8. If a+
bi is a root of :x1
is also a root?
-
4ac
+ (b/a)x =
lll[x] with b2 - 4ac < 0
3x2 +2ix + i -
-c a;
/
is irreducible in
1 E C[x], then is it true that
then
R[x].
a -
bi
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CHAPTER
5
Congruence in f[x] and Congruence-Class Arithmetic
In this chapter we continue to explore the analogy between the ring Z of integers
and the ring F[x] of polynomials with coefficients in a field F. We shall see that the
concepts of congruence and congruence-class arithmetic carry over from Z to
F[x] with practically no changes. Because of the additional features of the polyno­
mial ring F[x] (polynomial functions and roots), these new congruence-class rings
have a much richer structure than do the rings Zn. This additional structure leads
to a striking result: Given any polynomial over any field, we can find a root of that
polynomial in some larger field.
•
Congruence in F[x] and Congruence Classes
The conoept of congruence of integers depends only on some basic facts about divisibility
in Z. If Fis a field, then the polynomial ring F[x] has essentially the same divisibility
properties as does Z. So it is not sur prising that the concept of congruence in Z and its
basic properties (Section 2.1) can be carried over to F[x] almost verbatim.
Definition
Let F be a field and f(x), g(x), p{x)Ef[X] with p(x) nonzero. Then f(x) is
congruent to g(x) modulo p(x}--written f(x) = g(x) (mod p(x)}--provided
that p(x) divides f(x) - g(x).
EXAMPLE 1
In
O[x], x'- + x + 1 = x + 2 (mod x + 1) because
(x1 +
x
+ 1) - (x + 2)
=
:x?- - 1
=
(x + l)(x - 1).
125
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_ .... ..,_... ,,,__ ... _..,. _ ... _.....,...,_..c.g,..1.Nmlo&---riP<"'____ _,_11..-.-liajoll-. ....... ll
126
Chapter 5
Congruence in F[x] and Congruence-Class Arithmetic
EXAMPLE 2
In lll[x], 3x4 + 4x2 + 2x + 2
division shows that
(3x4 + 4x2 + 2x
+
=
x3 + 3x2 + 3x + 4 (mod x1 + 1) because
2) - (x3
+
3x1 + 3x + 4)
=
=
3x4
-
x3
+
x2 - x - 2
(x2 + l)(3x1 - x - 2).
Theorem 5.1
Let F be a field and p(x) a nonzero polynomial In f[x]. Then the relation of
congruence modulo p(x) is
(1) reflexive: f(x}
=
f(x) (mod p(x)} for all f(x) E F[x];
(2) symmetric: if f(x)
=
g(x) (mod p(x)}, then g(x)
(3) transitive: if f(x} = g(x) {mod p(x)} and g(x)
f(x) = h(x) (mod p(x}).
=
=
f(x) (mod p(x));
h(x) {mod p(x)), then
Proof• Adapt the proof of Theorem 2.1 withp(x),Jt:x),g(x), h(x) in place of
n, a,
b,c.
•
Theorem 5.2
Let F be a field and p(x) a nonzero polynomial in F[xJ. If f(x)
and h(x) = k(x) (mod p(x)), then
{1} f(x)
+
h(x)
(2) f(x}h(x)
=
=
g(x)
+
=
g(x) (mod p(x))
k(x) (mod p(x}},
g(x)k(x) (mod p(x)).
Proof• Adapt the proof of Theorem 2.2 with p(x),ft.x), g(x), h(x), k(x) in place
of n, a, h, c, d.
Definition
•
Let F be a field and f(x), p(x) EF[x] with p(x) nonzero. The congruence class
(or residue class) of f(x) modulo p(x) is d enoted [f(x)] and consists of all
polynomials in f[x] that are congruent to f(x) modulop(x), that is,
[f{x)] = {g(x) I g(x) E f[ x] and g(x) ""' f(x) (mod p(x))}.
Sinceg(x) = f(x) (modp(x)) means thatg(x) -f(x) k(x)p(x) for some k(x) E F[x]
equivalently, that g(x) = f(x) + k(x)p(x), we see that
=
or,
[/(x)]
=
=
{g(x)lg(x)
{f(x)
+
f(x) (mod p(x))}
k(x)p(x)lk(x)EF[x]}.
=
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Congruence in F[x] and Congruence Classes
5.1
127
EXAMPLE 3
Consider congruence modulo r +1in R[x]. The congruence class of 2x +1is
the set
{(2x
+
1) + k(x)(x2 + 1) lk(x) E R[x]}.
The Division Algorithm shows that the elements of this set are the polynomials in R[x]
that leave remainder 2x +1when divided by x2 + l.
EXAMPLE 4
Consider congruence modulo r + x+1inZl[x]. To find the congruence
class of x'-, we note thatx2=x+1(mod x1+x+1) becausex2- (x + 1) =
x2- x- 1=(x2+x+1)1 (remember that 1+1= 0 inZ2, so that 1= -1) .
Therefore, x +1is a member of the congruence class [x2]. In fact, the next
theorem shows that [x + l] = [�.
Theorem 5.3
f(x) = g(x) {mod p(x))if and only if [ f(x) ] = [g(x)].
Proof"" Adapt the proof of Theorem 2.3 with/(x), g(x), p(x), and Theorem 5.1
in place
of
a, c,
n, and Theorem 2.1.
•
Corollary 5.4
Two congruence classes modulo p(x) are either disjoint or identical.
Proof ... Adapt the proof of Corollary 2.4.
•
Under congruence modulo n in Z, there are exactly n distinct congruence classes
(Corollary 2.5). These classes are [OJ, [1], . . . , [n - 1.] Note that there is a class for each
possible remainder under division by n. In F[x] the possible remainders under divi­
sion by a polynomial of degree n are all the polynomials of degree less than n (and, of
course, 0). So the analogue of Corollary 2.5is
Corollary 5.5
Let F be a field and p(x) a polynomial of degree n in F[x], and consider congru­
ence modulo p(x).
(1)
tf f(x) E F[x] and r{x)is the remainder when f(x)is divided
by p(x), then
[f(x)] = [r{x}].
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128
Chapter
5
Con gruence in F[x] and Congruence-Class Arithmetic
(2) Let S be the set consisting of the zero polynomial and all the poly­
nomials of degree less than
n
in F{x]. Then every congruence class
modulo p(x) is the class of some polynomial in S, and the congru­
ence classes of different polynomials in S are distinct
Proof•(l) By the Division Algorithm,.ftx) =p(x)q(x)+ r(x),
with r(x) =Op or
deg r(x) < n. Thus,f(x) - r(x) = p(x)q(x), so thatf(x) == r(x ) (mod p(x)).
By Theorem 5.3, [ /(x)] = [r(x)].
(2) Sinc.e r(x) = Op or deg r(x) < n, we see that r(x) ES. Hence, every
congruenc.e class is equal to the congruenc.e class of a polynomial in S.
Two different polynomials in S cannot be congruent modulo p(x ) because
their difference has degree less than n, and henc.e is not divisible byp(x).
Therefore, different polynomials in S must be in distinct congruence
classes by Theorem 5.3. •
The set of all congruence classes modulo p(x) is denoted
F[x]/(p(x)),
which is the notational analogue of Z,..
EXAMPLE 5
+ 1 in R[x]. There is a congruence class for
+ 1. Now, the possible remainders
are polynomials of the form rx + s (with r, sER; one or both of r, smay
possibly be 0). Therefore, R[x]/(x2+ 1) consists of infinitely many distinct
Consider congruence modulo r
each possible remainder on division by x2
congruence classes, including
[O], [ x ], [x
+
1],
[5x +
3],
[� ]
x
+
2 . [x
-
7], . ...
Corollary 5.5 states that [rx + s] = [ex+ d] if and only if rx +sis equal (not
just congruent) to ex + d. By the definition of polynomial equality, rx + s =
ex + d if and only if
r
= c and s = d. Therefore, every element of R[x]/(x2 + 1)
s].
can be written uniquely in the form [rx+
EXAMPLE 6
Consider congruence modulo r + x+ 1 in Zi[x]. The possible remainders on
division by x1 + x+ 1 are the polynomials of the for m ax+ b with a, b E Z1•
0, 1, x, and x + 1. Therefore,
Z2[x]/(r + x + 1) consists of four congruenc.e classes: [OJ, [ l], [x], and [x + l].
Thus there are only four possible remainders:
EXAMPLE 7
The pattern in Example 6 works in the general case. Let n be a prime integer,
so that Z,. is a field and the Division Algorithm holds in Z,.[x]. If p(x ) E Z,.[x]
has degree k, then the possible remainders on division by p(x) are of the form
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5.1
Congruence in F[x] and Congruence Classes
129
+ 4t-1xl"-1, with a1EZ,,. There are n possibilities for each of
ao + a1x +
the k coefficients ao, .. . , ak-1' and so there are r/' different polynomials of this
form. Consequently, by Corollary 5.5, there are exactly ti' distinct congruence
classes modulop(x) in .l,,[xJ/(p(x)).
·
·
·
• Exercises
NOTE: F denotes a field andp(x) a non=ero polynomial in F[x].
A. 1.
Let/(x), g(_x),p(x) EF[x], withp(x) nonzero. Determine whetherf(x) = g(_x)
(modp(x)).Show your work.
(a) f(x) = � - 2x4 + 4x3 + x +
p(x) = x2 + I; F = Q
1; g(x)
= 3x" + 2x1 - sx2
(b) f(x) = x4 + x1 + x + 1; g(x) = x4 + x1 + x1 +
p(x) = x2 + x; F = Z 2
-
9;
1;
(c) f(x) 3� + 4x4 + 5x1 - 6x2 + Sx - 7;
g(_x) = 2r + 6x4 + x3 + 2x2 + 2x - S;p(x) = x3 - x2 + x - 1; F = IR
=
2.
If p(x) is a nonzero constant polynomial in F[x], show that any two
polynomials in F[x] are congruent modulo p(x).
3.
How many distinct congruence classes are there modulo x1 + x + 1 in Z2[x]?
List them.
4.
Show that, under congruence modulo x3 + 2x + 1 in Z3 [x], there are exactly
27 distinct congruence classes.
5.
Show that there are infinitely many distinct congruence classes modulo x1 - 2
in Q[x). Describe them.
6. Let a E F. Describe the congruence classes in F [x] modulo the polynomial x - a.
7.
B. 8.
Describe the congruence classes in F[x] modulo the polynomial x.
Prove or disprove: If p(x) is relatively prime to k(x) andflx)k(x) = g(_x)k(x)
(modp(x)), thenfix) = g(x) (modp(x)).
9.
Prove thatf(x) = g(x) (modp(x)) if and only if f(x) and g(x) leave the same
remainder when divided byp(x).
IO.
Prove or disprove: Ifp(x) is irreducible in F[x] andf(x)g(_x) =Op (modp(x)),
then fix) =OF (modp(x)) or g(_x) =OF (modp(x)).
1 1. Ifp(x) is
reducible in F[x], prove that there existf(x), g(x) EF[x] such that
fix) '¢OF (mod p(x)) and g(x) '=I= Op (modp(x)) butflx)g(_x) = Op (mod p(x)).
12.
If fix) is relatively prime to p(x), prove that there is a polynomial g(x) E F[x]
such that/(x)g(x) = lp(modp(x)).
13.
Supposef(x), g(_x) E R[x] andf(x) = g(x) (mod x).What can be said about the
graphs of y =fix) andy = g(x)?
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130
Chapter 5
m
Congruence in F[x] and Congruence-Class Arithmetic
Congruence-Class Arithmetic
Congruence in the integers led to the rings Zn. Similarly, congruence in
F[ x] also pro­
duces new rings and fields. These tum ou t to be much richer in structure than the rings
Zn· The development here closely parallels Section 2.2.
Theorem 5.6
Let f be a field and p(x) a nonconstant polynomial in f[x].
[ h{x)] = [ k(x) ] in f[x]/(p(x}}, then,
[f{x) + h(x)] = [g(x) + k(x) ]
[f(x)h(x)]
and
If
=
[f(x)] = [g(x)] and
[g(x)k(x} ].
Proof.,. Copy the proof of Theorem 2.6, w ith Theorems 5.2 and 5.3 in place of
Theorems 2.2 and 2.3 .
Because of Theorem
5.6
•
we can now define addition and multiplication of c on­
gruence classes just as we did in the integers and be certain that these operations are
independent of the ch oice of represen tatives in each cong ruence class.
Definition
Let F be a field and p(x} a nonconstant pofynomial in f[x]. Addition and
multiplication in F[x]/(p( x)) are defined by
[f(x}] + [g(x)]
=
[f(x) + g(x}),
[f(x)][g(x)] = [f(x}g (x)].
EXAMPLE 1
Consider congruence modulo x?and
[3x +
+ 1 in R[x]. The sum of the classes [2x + 1]
5] is the class
[(2x + 1) + (3x + 5)]
=
[5x + 6].
The product is
[2x + 1][3x + 5] = [(2x + 1)(3x + 5)] = [6x 2+ 13x +
5].
As noted in Example 5 of Section 5.1, every c ongruence class in
R[x]/(X'- + 1)
[ax + b]. To express the class [6x?- + 13x + 5] in this
form, we divide 6x2 + l3x + 5 by x2 + 1 and find that
can be written in the form
6x?- + l3x + 5
It follov.is that 6r +
[13x - l].
=
Bx + 5 = l3x -
6(x?- + 1) + (13x - 1).
1 (mod x?-+ 1), and hence [6x2 + 13x +
5] =
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5.2
Congruence-Class Arithmetic
131
EXAMPLE 2
5.1, we saw that Z:z[x]/(x2 + x + 1) consists of four
[x], and [x + 1]. Using the definition of addition of classes,
that [x + l] + [l]
[x + 1 + l] [x] (remember that 1 + 1 0
In Example 6 of Section
classes: [O], [l],
we see
=
=
=
Z-2). Similar calculations
Z2[x]/{x2 + x + 1):
in
produce the following addition table for
+
[O]
[l]
[x]
[x
+
l]
[O]
[O]
[1]
[x]
[x
+
l]
[l]
---------J!L
_
[x]
[x
_______
+
+
[x
1]
J�L-----" [x
[x
[x]
1]
+
1]
[x]
[l]
[O]
[1]
[x]
Zi[x]/(x2 + x
Most of the multiplication table for
+
[OJ
l]
+
1) is easily obtained from
the definition:
[O]
_[tJ
_
__ _____
_
[l]
[O]
JgL_
___
+
l]
[x
[x
j rx1
[x
I
J!L
[O]
! [x]
: [O]
______
[x]
[O]
[x)
[x
[O]
[O]
+
+
l]
+
l]
[O]
l]
To fill in the rest of the table, note, for example, that
[x] [x
•
+
1)
[x(x
=
+
l)]
=
[x2 + x].
[x] shows that x2 + x (x2 + x + 1) + 1.
2
1 (mod x2 + x + 1), so that [x2 + x] [l]. A similar calcu­
lation shows that [x] [x]
[x2] [x + l] (because x2 (x2 + x + 1) + (x + 1)
in Z2[x]. Verify that [x + l][x + 1]
[x ].
Now division or simple addition in Z
Therefore,
x2 + x
=
=
=
•
=
=
=
=
If you examine
Z [x]/(x2
2
+
x
+
the
tables
in
the preceding example, you will see that
1) is a commutative ring with identity (in fact, a field). In view
of our experience with Z and Z,., this is not too surprising. What is unexpected is the
upper left-hand comers of the two tables (the sums and products of [O] and [lD. It is
easy to see that the subset F*
{[O], [l]} is actually a subring of Zi[x]/(x1 + x + 1)
and that F* is isomorphic to Z2 (the tables for the two systems are identical except for
the brackets in F*). These facts illustrate the next theorem.
=
Theorem 5.7
Let F be a field and
F[x]/(p(x))
p(x)
a nonconstant polynomial in
of congruence classes modulo
identity. Furthermore,
...
.......
p(x)
f[x].
Then the set
is a commutative ring with
F[x]/(p(x)) contains a subring F* that is isomorphic to F.
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132
Chapter 5
Congruence in F[x] and Congruence-Class Arithmetic
Proof"" To prove that F[x)/(p(x)) is a commutative ring with identity, adapt the
proof of Theorem 2.7 to the present case. Let F* be the subset of
F[x]/(p(x)) consisting of the congruence classes of all the constant
polynomials; that is, F*
{[a) I a E F}. Verify that F* is a subring of
F[x]/(p(x)) (Exercise 10). Define a map <p:F � F* by 'P(a) [a). This
=
=
definition shows that 'P is surjective. The definitions of addition and
multiplication in
rp(a+ b)
=
F[x]/(p(x)) show that
[ a+ b]
'P(ab)
=
=
[ab]
[a] + [b]
=
[a ] [b]
•
rp(a) + rp(b)
=
=
and
'P(a) rp(b).
•
Therefore, rp is a homomorphism.
To see that rp is injective, suppose 'P(a)
rp(b). Then [a ]
[b], so that
a= b (modp(x)). Hence,p(x) divides a - b. However,p(x) has degree<=:: 1,
and a - b E F. This is impossible unless a - b
0. Therefore, a
b and
'P is injective. Thus 'P:F � F* is an isomorphism, •
=
=
=
=
F and a polynomial p(x) in F[x]. We have now constructed a
copy of F. What we would really like is a
ring that contains the field F itself. There are two possible ways to accomplish this, as
We began with a field
ring F[x] /(p(x)) that contains an isomorphic
illustrated in the following example.
EXAMPLE 3
In Example 2, we used the polynomial :x?-+ x + 1 in Z2[x] to construct the ring
2
Z2':x]/(x + x+ 1), which contains a subset F* {[O], [l]} that is isomorphic to
Z2• Suppose we identify Z2 with its isomorphic copy F* inside Z2':x]/(x2+ x+ 1)
=
and write the elements of F*
as
if they were in Z,.. Then the tables in Example 2
become
1
[x)
[x+ l]
0
1
[x)
[x + l]
1
0
[x+ 1]
[x]
[x]
[x+ 1)
0
[x]
1
[x +1]
[x+1]
[x]
1
0
0
1
[x]
[x+ l]
0
0
0
0
0
1
0
1
[x]
[x+ l]
[x]
[x+ l]
0
[x]
0
[x+ 1]
[x+ l]
1
[x]
+
0
0
1
1
We now have a ring that has Z2 as a subset. If this procedure makes you a bit
uneasy (is Z2 really a subset?), you can use the following alternate route to the
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5.2
Congruence-Class Arithmetic
133
same end. Let Ebe any four-element set that actually contains Z2 as a subset,
say E = {O, 1, r, s}. Define addition and multiplication in Eby
+
r
s
0
0
1
0
0
1
r
s
0
1
1
0
s
r
1
r
r
s
0
r
s
s
r
1
s
0
1
r
s
0
0
0
0
0
1
r
s
0
r
s
1
0
s
1
r
A comparison of the tables for Z2 [x]/ (x2+ x+ 1 ) and those for E shows that
these two rings are isomorphic (replacing [x] by rand [x+ l] bys changes
one set of tables into the other ). Therefore, Eis essentially the same ring we
obtained before. However, E does contain Z2 as an honest-to-goodness subset,
without any identification.
What was done in the prec.eding example can be done in the general case. Given
a field F and a polynomial p(x) in F[x], we can construct a ring that contains Fas
a subset. The customa ry way to do this is to identify Fwith its isomorphic copy F*
inside F[x]/(p(x)) a nd to consider Fto be a subset of F[x]/(p(x)). If doing this
makes you uncomfortable, keep in mind that you can always build a ring isomorphic
to F[x]/(p(x)) that genuinely contains Fas a subset, as in the preceding example.
Because this latter approach tends to get cumbersome, we shall follow the usual
custom and identify Fwith F* hereafter. Consequently, when a, b E F, we shall write
b[x] instead of [b] [x] and a+ b[x]instead of [a ]+ [b][x] = [a+ bx]. Then Theorem 5.7
can be reworded:
Theorem 5.8
Let F be a field and p(x) a nonconstant polynomial in F[x]. Then F[x]/(p(x))is a
commutative ring with identity that contains f.
If a andn are integers such that (a, n) = 1, then byTheorem2.10, [a ]is a unit inZ,,.
Here is the a nalogue for polynomials. •
Theorem 5.9
Let F be a field and p(x) a nonconstant polynomial in f[x ] . If f(x) Ef[x] and f(x)
is relatively prime to p(x), then [f(x)] is a unit in F[x]/(p(x)).
Proof• By Theorem 4.8 there are polynomials u(x)and v(x)such that/(x)u(x)+
p(x)v(x) 1. Hence,f(x)u(x) - 1 -p(x)v(x) p(x)(-v(x)), which
[l] by Theorem 5.3. Therefore, [f(x)] [u(x)]
implies that [f(x)u(x)]
[l], so that [f(x)] is a unit in F[x]/(p(x)). •
[f(x)u(x)]
=
=
=
=
=
=
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134
Chapter 5
Congruence in F[x] and Congruence-Class Arithmetic
EXAMPLE 4
Since x2
-
2
is irreducible in Q[x], 2x + S and x2
shows that its inverse is [u(x)], where (2x +
S)u(x)
2
--
(Why?) Hence, [2x + SJ is a unit in the ring 0[x]/(x2
+
are relatively prime in O[x].
2). The proof of Theorem S.9
(x2
-
2)t.(x)
=
1. Using the
Euclidean Algorithm as in Exercise 1 S of Section 1.2, we find that
(2x
Therefore,
x
+
+
s
7
s)(-1�x )
+
[-� :1]
1
+
(x2
2)
- (�)
is the inverse of [2x +SJ in
=
1.
Q[x]/(x2
-
2).
• Exercises
A. In Exercis es 1-4,
write out the addition and multiplication tables for the congruence­
class ring F[x)/(p(x)). In e ach case, is F[x]/(p(x)) afield?
1. F = .l ;p(x) = x 3 + x + 1
2. F= .l3;p(x)= x2 + 1
2
B. In
Exercises 5-8, each e lement of the g iven congruence-class ring can be written
in the form [ ax+ b] (Why?). Determin e the rules for addition and multiplication
of congruence class es. (In other words, if the product [ax + b][cx + d ] is the
class [rx + s], describe how to.fin d rand s from a, b, c,d,and simi larlyfor
a ddition.)
5.
R[x]/(x2 +
6.
O[x]/(x2
1) [Hint: See Example l.]
-
2)
7.
O[x]/(x2
-
3)
8. O[x]/(x2)
9. Show that IR[x]/(x2 + 1) is a field by verifying that every nonzero congruence
class [ax+ b] is a unit.
where c
=
[Hint: Show that the inverse
-a/(tl- + ll-) and d = b/(d- + b2).]
10. Let Fbe a field and p(x) EF[x]. Prove that F* =
F[x]/(p(x)).
of
[ax + b] is [ex+ dJ,
{[ a ] I a E.F}
is a subring of
11. Show that the ring in Exercise 8 is n ot a field.
12. Write out a complete proof of Theorem S.6 (that is, carry over to F[x] the
proof of the analogous facts for Z).
13. Prove the first statement of Theorem S.7.
14. In each part explain why [.flx)] is a unit in F[x]/(p(x)) and find its inverse.
[Hint: To find the inverse, let u(x) and v(x) be as in the proof of Theorem S.9.
u(x) = ax + band v(x) = ex + d. Expandingf(x)u(x) +
p(x)u(x) leads to a system of linear equations in a, b,c,d. Solve it.]
You may assume that
(a) [f(x)]
=
(b) (f(x)]
=
[2x - 3] E Q[x]/(x2 - 2)
[x 2 +x +
l] EZ3[x]/(x2 + 1)
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........ q-��
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...
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5.3
C. 15.
16.
Ill
The Structure of
F[x]/(p(x)) When p(x) Is
Irreducible
135
Find a fourth-degree polynomial in Z2[x] whose roots are the four elements of
the field.Z:z[x]/(x2 + x + 1), whose tables are given in Example 3. [Hint: The
Factor Theorem may be helpful.]
Show that O[x]/(x2 - 2) is a field.
The Structure of
F[x]/(p(x)) When p(x) Is Irreducible
Whenp is a prime integer, then Theorem 2.8 states, in effect, that ZP is a field (and, of
course, an integral domain). Here is the analogous result for F[x] and an irreducible
polynomialp(x).
Theorem 5.10
Let F be a field and p(x) a nonconstant polynomial in F[x}. Then the following
statements are equivalent:
(1) p(x) is irreducible in F[x].
(2) F[x]/(p{x)) is a field.
(3) F[x] /(p(x )) is an integral domain.
Theorem 5.10 and most of its proof are a copy of Theorem 2.8 and its proof , with
Z replaced by F{x] and � by F(x) /(p(x) ), and the necessary adjustments made for the
differences between prime integers and irreducible polynomials.
Proof ofTheorem 6.10 ... (1) � (2) By Theorem 5.7, F(x) /(p(x)) is a commutative
ring with identity, and thus satisfies Axioms 1-10. To prove that
F(x) /(p(x)) is a field, we must verify that every nonzero element in
F(x)/(p(x) ) is a unit (Axiom 12, page 49). Suppose that [a(x)] ::F [O] in
F(x)/(p(x)). We must find [u(x)] such that [a(x)] [u(x)] [lF]· Since
[ a(x) ] :#: [O], we know that a(x) ¢ 0 (modp(x)) by Theorem 5.3. Hence,
p(x) ,t a(x) by the definition of congruence. Now the gcd of a(x) and
p(x) is a monic polynomial that divides both a(x) and p(x). Since p(x)
is irreducible, the gcd is either 1For a monic associate of p(x) (the only
monic divisors of p(x) ). As explained on page 100, an associate of p(x)
is a polynomial of the form cp(x) , with Op:#: c EF. Consequently, a(x)
is not divisible by any associate of p(x) (because a(x) is not divisible by
p(x) ) . Since the gcd also divides a(x) andp(x) .r a(x) , the gcd of a(x) and
p(x) must be lp. By Theorem 4.8, there are polynomials u(x) and v(x) so
that a(x)u(x) + p(x)v(x) 1F- Hence, a(x)u(x) - lp p(x)(-v(x)), so
that a(x)u(x) = lp(modp(x) ) . Therefore, [a(x)u(x) ] [l� in F(x)/(p(x))
by Theorem 5.3. Thus, [a(x)][u(x) ] [a(x)u(x) ] [lm, so that [a(x)] is a
unit. Hence, F(x)/(p(x)) satisfies Axiom 12 and F(x) /(p(x) ) is a field.
=
=
=
=
=
=
(2) � (3) This is an immediate consequence of Theorem 3.8.
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136
Chapter 5
Congruence in F[x] and Congruence-Class Arithmetic
(3) => (1)
We shall verify statement (2) of Theorem
4.12 to show
that
p(x) is irreducible. Suppose that b(x) and c(x) are any polynomials in F[x]
and p(x) I b(x)c(x). Then b(x)c(x) = Op ( mod p(x)). So by Theorem 5.3,
[b(x)][c(x)]
=
Because F(x)/(p(x)) is
an
= (OF] in F(x)/(p(x)).
integral domain by
(3), we have (a(x)]
(Op]. Thus, b(x) = Op (mod p(x)) or c(x)
Theorem 5.3, which means that p(x) I b(x) orp(x) I
or
[b(x)]
[b(x)c(x)]
=
=[Op]
OF (mod p(x)) by
c(x) by the definition
=
of congruence. Therefore,p(x) is irreducible by Theorem 4.12.
•
Theorem 5.10 can be used to construct finite fields. Ifpis prime and.f(x) is irreduc­
ible in
L;,[x]
of degree k, then Z,,[x] /(f(x)) is a field by Theorem 5.10. Example 7 in
5.1 shows that this field has P' elements . F inite fields are discussed further in
Section 11.6, where it is shown that there are irreducible polynomials of every positive
degree in zp [ x] and, hence, finite fields of all possible prime power orders. See Exercise 9
Section
for an example.
Let Fbe a field and p(x) an irreducible polynomial in F[x]. Let K denote the field of
congruence classes
F[x]/(p(x)). By Theorems 5.8 and 5.10, Fis a subfield of the field
F[x] can be consid­
K. One also says that K is an extension field of F. Polynomials in
ered to have coefficients in the larger field K, and
we
can ask about the roots of such
polynomials in K. In particular, what can be said about the roots of the polynomial
p(x) that we started with? Even thoughp(x) is irreducible in F[x], it may have roots in
the extension field K.
EXAMPLE 1
The polynomial p(x) = r+
x + 1 has no roots in Z2 and is, therefore, irreducible
Z2[x]/(x2 + x + 1) is an extension
in Zix] by Corollary 4.19. Consequently, K =
field of Z2 by Theorem 5.10. Using the tables for Kin Example 3 of Section
we see
5.2,
that
[ x]2 + (x] + 1 = [x + 1] + [x] + 1
= 1 + 1 = 0.
This result may be a little easier to absorb if we use a different notation. Let
[x]. Then the calculation above says that a2 +a+ 1 O; that is, a is a root
r+ x + 1. It's important to note here that you don't really
need the tables for Kto prove that a is a root of p (x) because we know that
r+ x + 1=0 (mod r+ x + 1). Consequently, [x1 + x + l] = 0 in K, and
a=
=
in K of p(x)
=
by the definition of congruence-class arithmetic ,
a2 +a+ 1
=
[x]2 + [x]
+
1
=
(x2 + x + l]
=
0.
For the general case we have
Theorem 5.11
Let F be a field and p(x) an irreducible polynomial in F [x ] . Then F[x]/(p(x)) is an
extension field of F that contains a root of p(x).
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..
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5.3
The Structure of
F[x]/(p(x)) When p(x) Is
Irreducible
137
Proof" Let K= F[x]/(p(x)). Then Kis an extension field of F by Theorems 5.8
and 5.10. Let p(x)=
hence, in K. Let
a,,x!'+
+ a1x + ao, where each a1 is in F and,
a= [x] in K. We shall show that a is a root of p(x). By
·
·
·
the definition of congruence-class arithmetic in K,
"
a,,a
+ ...+ a,a + ao= a,,[x]" + ...+ a,[x] + ao
= [a,,x" +
+ a1x+ ao]
·
= [p(x)]=
Therefore,
a
EKis a root of p(x).
·
·
OF
[Because p(x)
=
Op (modp(x)).].
•
Corollary 5.12
Let F be a field and f(x) a nonconstant polynomial in f[x]. Then there is an
extension field K of F that contains a root of f(x).
Proof" By Theorem 4.14,/(x) has an irreducible factor p(x) in F[x]. By Theorem
5.11, K =
F[x]/(p(x)) is an extension field of
Fthat contains a root of p(x).
Since every root of p(x) is a root of/(x), Kcontains a root of f(x).
The implications of Theorem 5.11
run
•
much deeper than might first appear.
Throughout the history of mathematics, the passage from a known number system to a
new, larger system has often been greeted with doubt and distrust. In the Middle Ages,
some mathematicians refused to acknowledge the existence of negative numbers. When
complex numbers were introduced in the seventeenth century, there was uneasiness-­
which extended for nearly a century-because some mathematicians would not accept
the idea that there could be a number whose square is -1, that is, a root of
cause for these difficulties
was
x2+
1. One
the lack of a suitable framework in which to view the
situation. Abstract algebra provides such a framework. Theorem 5.11 and its corollary ,
then, take care of the doubt and uncertainty.
It is instructive to consider the complex numbers from this point of view. Instead
of asking about a number whose square is -1, we ask, "Is there a field containing
R
in which the polynomial
x2+
x2 + 1 is irreducible in R[x],
R[x]/(x2+ 1)is an extension field of
namely a = [x].In the field K, a is an element whose
1 has a root?" Since
Theorem 5.11 tells us that the answer is yes: K =
R that contains a root of x2+ 1,
square is -1. But how is the field Krelated to the field of complex numbers introduced
earlier in the book?
As is noted in Example 5 of Section 5.1, every element of K=
R[x]/(x2+ 1) can
[ax + b] with a, b E R. Since we are identifying each
the element [r] in K, we see that every element of Kcan be written
be written uniquely in the form
element r E IR with
uniquely in the form
[a + bx] = [a] + [b][x]= a+ ba.
Addition in Kis given by the rule
(a+ ba)+ (c+ da) = [a+ bx]+ (c+ dx] = [(a+ bx)+ (c+ dx}]
= [(a + c)+ (b + d)x]= [a + c]+ [b+ dJ[x].
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138
Chapter 5 Congruence in F[x] and Congruence-Class Arithmetic
so that
(a+ ha)+ (c+ da)= (a+ c)+ (b+ d )a.
Multiplication inKis given by the rule
(a
+
ba)(c+da)= [a
+
bx][c+dx]= [(a
bx)(c+ dx)]
+
= [ac+ (ad+ bc)x+bdx2]
= ac+ (ad+ bc)a+
a is a root of
inKbecomes
However,
x2 +
1, and so a2
=
(a+ ba)(c+ da) = (ac
If the symbol
a
bda2•
-1. Therefore, the rule for multiplication
-
btf)+(ad+ bc)a.
is replaced by the symbol i, then these rules become the usual rules for
adding and multiplying complex numbers. In formal language, the field K is isomor­
phic to the field C, with the isomorphism/being given by f{a+ha)=
a+ bi.
Up to now we have taken the position that the field C of complex numbers was
already known. The fieldKconstructed above then turns out to be isomorphic to the
known field C. A good case can be made, however, for not assuming any previous
definition
1). Such a definition
knowledge of the complex numbers and using the preceding example as a
instead. In other words, we can define C to be the field R [x]/ (x2+
is obviously too sophisticated to use on high-school students, but for mature students
it has the definite advantage of removing any lingering doubts about the validity of
the complex numbers and their arithmetic.* Had this definition been available several
centuries ago, the introduction of the complex numbers might have caused no stir
whatsoever.
• Exercises
NOTE: F always denotes a field.
A. 1. Determine whether the given congruence-class ring is a field. Justify your
answer.
(a)
Z3[x]/(x3+2x1+ x + 1)
(b) Z5[x]/(2x3 - 4x2+2x + 1)
(c) Z2[x]/ (x 4+x2+ l)
B. 2. (a) Verify that 0(v'2)=
(b)
Show that
{r+sv'2 Ir, s E Q}
is a subfield of R.
O(v'f) is isomorphic to Q[x]/(x2 -
2). [Hint: Exercise 6 in
Section 5.2 may be helpful.]
•o
nly a minor rearrangement of this book is needed to accommodate such a definition. A few
examples in Chapter 3 would have to be omitted, and the discussion
and
IR[x] (Section 4.6) would
of irreducibility in C[x]
S is
have to be postponed. All the intervening material in Chapter
independent of any formal knowledge of the complex numbers.
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5.3
The Structure
of
F[x]/(p(x)) When p(x) Is Irreducible
3. If a E: F, describe the field F[x]/(x
4.
- a).
Let p(x) be irreducible in F[x]. Without using Theorem 5.10, prove that if
[f(x)][g(x)] [Op] in F[x]/(p(x)), then [f(x)] [Op ] or [g(x)] [Op]. [Hint:
Exercise 10 in Section 5 . 1 .]
=
5. (a) Verify that
(b) Show that
6.
139
=
0('\13)
=
{r + s'\131 r, sE:O}
=
is a subfield of�.
O(v'J) is isomorphic to Q[x]/(x2
- 3).
Let p(x) be irreducible in F[x]. If [f(x)] #:-[Oil in F[x]/(p(x)) and h(x) E:
F[x], prove that there exists g(x) E: F[x] such that [f(x)][g(x)] [h(x)] in
F[x]/(p(x)). [Hint: Theorem 5.10 and Exercise l2(b) in Section 3 .2.]
=
7.
Iff(x) E: F(x] has degree n, prove that there exists an extension field E of
c0(x - c1)(x - ei) · · (x - c,,) for some (not necessarily
distinct) c; E: E. In other words, E contains all the roots of f(x).
Fsuch thatf(x)
=
·
8. If p(x) is an irreducible quadratic polynomial in
F [x], show that F [xlf(p(x))
contains all the roots of p(x).
9. (a) Show that Z2[x]/(x3 +
x
+ 1) is a field.
(b) Show that the field Z:z[x]/(x3 +
10. Show that
x + 1) contains all three roots of xl + x + 1.
Q[x]/(x2 - 2) is not isomorphic to 0[x]/(x2 - 3). [Hint:
Exercises 2
and 5 may be helpful.]
11. Let K be a ring that contains� as a subring. Show that p(x)
=
3x2 +
no roots in K. Thus, Corollary 5.12 may be false if Fis not a field.
were a root, then 0
=
2
·
3
and
3u
2
+1
=
1 E: ZJ:x] has
[Hint: If u
0. Derive a contradiction.]
2x3 + 4x2 + Sx + 3 E: Z16[x] has no roots in any ring K that contains
Z16 as a subring. [See Exercise 11 .]
12. Show that
C.13. Show that every polynomial of degree 1 , 2, or 4 in Z2[x] has a root in
Zi[x]/(x4 + x + 1).
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CHAPTER
6
Ideals and Quotient Rings
Congruence in the integers led us to the finite arithmetics Zn and helped moti­
vate the definition ofa ring. Congruence in the polynomial ring f[x] resulted in a
new class of rings consisting of the various F[x]/(p{x)). These rings enabled us to
construct extension fields ofF that contained roots of the polynomial p(x). In this
chapter the concept of congruence is extended to arbitrary rings, producing
additional rings and a deeper understanding ofalgebraic structure.
You will see that much ofthe discussion is an exact parallel ofthe development
of congruence in Z (Chapter 2) and in f[x] (Chapter 5). Nevertheless, the results
here are considerably broader than the earlier ones.
•
Ideals and Congruence
Our goal is to develop a notion of congruence in arbitrary rings that includes
as
spe­
cial cases congruence modulo n in Zand congruence modulo p(x) in F[x]. We begin by
taking a second look at some examples of congruence in Zand .l'lx] from a somewhat
different viewpoint than before.
EXAMPLE 1
In the ring Z,
a=
b(mod3) means that
a -
bis a multiple of 3. Let /be the set
of all multiples of3, so that
I=
{O,
±3, ±6, ... }.
Then congruence modulo3may be characterized like this:
a=
b(mod3)
means
a -
bEI.
141
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142
Chapter 6
Ideals and Quotient Rings
Observe that the subset !is actually a suhring of 7L (sums and products of mul­
tiples of 3 are also multiples of 3 ). Furthermore, the product of any integer and
a multiple of 3 is itself a multiple of 3. Thus the subring I has this property:
Whenever
k E7L and i EI, then ki E /.
EXAMPLE 2*
The notation f(x) "" g(x) (mod x2 - 2) in the polynomial ring Q[x] means that
f(x) - g(x) is a multiple of X1- - 2. Let I be the set of all multiples of X1- - 2 in Q[x],
that is, I= {h(xXx2- 2) !h(x)EO[x]}. Once again, it is not difficult to check that /is
a subring of O[x] with this property:
Whenever
k(x)E Q[x]
and
t(x)El,
then
k(x)t(x)El
(the product of any polynomial with a multiple of r - 2 is itself a multiple of:? Congruence modulo
2).
x2 - 2 may be described in terms of I:
f(x) = g(x) (mod X1- - 2)
means
fi..x) - g(x)EI.
These examples suggest that congruence in a ring R might be defined in terms
of certain subrings. If
mean a
-
I were such a subring, we might define a = b (mod I) to
I might consist of all multiples of a fixed element, as in
b EI. The subring
the preceding examples, but there is no reason for restricting to this situation. The
examples indicate that the key property for such a subring I is that it "absorbs prod­
ucts": Whenever you multiply an element of I by any element of the ring (either inside
or outside I), the resulting product is an element of I. The set of all multiples of a fixed
element has this absorption property. We shall see that many o ther subrings have it as
well. Because such subrings play a crucial role in what follows, we pause to give them
a name and to consider their basic properties.
Definition
A subring I of a ring R is an Ideal provided:
Whenever n� R and a E /, then ra E 1 and ar E /.
The do uble absorption condition that ra EI and ar El is necessary for noncommutative rings.
When R is commutative,
as
in the preceding examples, this condition reduces to
ra
EI.
EXAMPLE 3
The zero ideal in a ring R consists of the single element OR. This is a subring that absorbs all
products since rOR = OR = ORr for every r ER. The entire ring R is also an ideal.
*Skip this example if you have not read Chapter 5.
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6.1
I deals and Congruence
143
EXAMPLE 4
In the ring Z[x] of all polynomials with integer coefficients, let /be the set of
polynomials whose constant terms
are
even integers. Thus X' +
but 4x2 + 3 is not. Verify that I is an ideal in Z[x] (Exercise
2).
x
+ 6 is in /,
EXAMPLE 5
Let The the ring of all functions from IR to R, as described in Example S
of Section 3.L Let /be the subset consisting of those functions g such that
g(2)
0. Then / is a subring of T(Exercise
function in T and if g E /, then
=
(fg)(2)
=
f(2)g(2)
14 of
=
/(2)
Section
.
0
=
3.1). If/is any
o.
Therefo re,fg EI. Similarly, gfEl, so that I is an ideal in T.
EXAMPLE 6
The subring Z of the r ational numbers is not an ideal in
have the absorption property. For instance,
.
2'
5
is not
k
Q because Z fails to
E Q and 5 E Z ,but their product,
.
71
m 1L.
EXAMPLE 7
Verify that the set I of all matrices of the form
subring of the ring M(R) of
all
2
X
(: �)
with a, b E IR forms a
2 matrices over the reals. It is easy to see
that I absorbs products on the left:
But I is not an ideal in M(R) because it may not absorb products
instance,
on
the right-fur
One sometimes says that I is a left ideal, but not a two-sided ideal, in M(!R).
The following generalization of Theorem 3.6 often simplifies the verification that a
particular subset of a ring is an ideal .
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144
Chapter
6
Ideals and Quotient Rings
Theorem 6.1
A nonempty subset I of a ring Ris an ideal if and only if it has these properties:
(i)
ifa, b E /,then a-b E /;
(ii) if r E Rand a E /, then ra E I and ar E
Proof ... Every
/.
ideal certainly has these two properties. Convenely, suppose I
has properties (i) and (ii). Then I absorbs products by (ii), so we need
only verify that I is a subring. Property (i) states that I is closed under
subtraction. Since/is a subset of R, the product of any two elements
of I must be in I by (ii). In other words, I is closed under multiplication.
Therefore, I is a subring of R by Theorem 3 .6. •
Finitely Generated Ideals
In the first example of this section
we saw
that the set I of all multiples of 3 is an ideal
in Z . This fact is a special case of
Theorem 6.2
Let R be a commutative ring with identity,c ER, and I the set of all multiples
of c in R, that is, I
=
{re Ir ER}.
Then I is an ideal.
Proof.. If ri, r2, rER and r1c, rie El, then
and
because r1
tive,
(r1c)r
- r2 and rr1 are elements of R. Similarly, since R is commuta­
(rr1)c E I. Therefore, I is an ideal by T heorem 6.1. •
=
The ideal I in Theorem 6 .2 is called the principal ideal generated by
c
and hereafter
w ill be denoted by (c). In the ring Z, for example, (3) indicates the ideal of all multiples
of 3. In any commutative ring R with identity, the principal ideal (1R) is the entire ring
R because r = rlR for every r ER. It can be shown that every ideal in Z is a principal
ideal (Exercise 40). However, there are ideals in other rings that are not principal , that
is, ideals that do not consist of all the multiples of a particular element of the ring.
EXAMPLE 8
We have seen that the set I of all polynomials with even constant terms is an
ideal in the ring Z[x]. We claim that/is not a principal ideal . To prove
this,
suppose, on the contrary, that I consists of all multiples of some polynomial
p(x). Since the constant polynomial 2 is in/, 2 must be a multiple of p(x).
By Theorem 4.2, this is possible only if p(x) has degree 0, that is, if p(x) is a
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6.1
constant, say p(x) =
c.
I deals and Congruence
146
Sincep(x) EJ, the constant c must be an even integer. Since
2 is a multiple of p(x) = c, the only possibility is c = ±2. On the other hand, x EI
because it has even constant term 0. Therefore, x must be a multiple of p(x) = ±2.
However, if ±2g(x) = x, then g(x) has degree 1byTheorem4.2, say g(x) = ax
But ±2(ax
+ b) =
+ b.
x implies that ±2a = 1 because the coefficient of x must be the
same on both sides. This is impossible because a is
an
integer. Therefore, I does not
consist of all multiples of p(x) and is not a principal ideal.
In a commutative ring with identity, a principal ideal consists of all multiples of a
fixed element. Here is a generalization of that idea.
Theorem 6.3
Let R be a commutative ring with identity and c1, c2,
I=
{r1c1 + r2c2 +
·
·
Proof• Exercise 14.
·
+ fnCn I r1, r2,
•
•
•
•
•
•
, Cn ER. Then the set
, fn ER} is an Ideal in R.
•
The ideal I in Theorem 6.3 is called the ideal generated by
sometimes denoted by (c1,
c2,
•
•
•
, c,,). Such an ideal is said to
principal ideal is the special case n = 1, that is,
an
ci. c2,
•
•
,
•
C8
and is
be finitely generated. A
ideal generated by a single element.*
The generators of a finitely generated ideal need not be unique, that is, the ideal gener­
ated by cu
C:z,
though no
c1
•
•
•
, en might be the same set as the ideal generated by d1o ti,;, ... , d,to even
is equal to any �(Exercise 16).
EXAMPLE 9
In the ring Z[x], the ideal generated by the polynomial x and the constant poly­
nomial 2 consists of all polynomials of the form
f(x)x + g(x)2,
with f(x), g(x)EZ[x].
It can be shown that this ideal is the ideal I of all polynomials with even
constant term, which was discussed in Example 8 (Exercise 15).
Congruence
Now that you are familiar with ideals, we can define congruence in an arbitrary ring:
Definition
Let/ be an ideal in a ring Rand leta,bER.Then a iscongruenttob modulo
I [written a= b (mod /)] provided that a
- b El.
*When a commutative ring does not have an identity, the ideal generated by c1, c2,
•
•
•
, en is defined
somewhat differently (see Exercise 3.1).
...
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146
Chapter 6
Ideals and Quotient Rings
Example 1 shows that congruence modulo
3
in the integers is the same thing as
(3) of all multiples of 3.
2 shows that congruence modulo x2 - 2 in 0[x] is the same as con­
congruence modulo the ideal /, where I is the principal ideal
Similarly, Example
gruence modulo the principal ideal (x2 - 2). Thus congruence modulo an ideal includes
as
a special case the concepts of congruence in Zand F[x] used earlier in this book.
EXAMPLE 10
Let Tbe the ring of all functions from� to Rand let /be the ideal of all func­
tions g such that g(2 ) = 0. If f(x) = r + 6 and h(x) = Sx, then the f unction
f- his in /because
(f- h)(2)
Therefore, f = h (mod
=
/(2) - h(2) = (22 + 6)
-
(5 2)
•
=
0.
I).
Theorem 6.4
Let l be an ideal in a ring R. Then the relation of congruence modulo I is
(1)
reflexive: a= a (mod /) for every a E R;
(2)
symmetric: If a: b (mod/},then b =a (mod/);
(3) transitive: if a= b (mod/) and b =
c
(mod /),then a= c (mod/).
This theorem generalizes Theorems 2.1 and 5.1. Observe that the proof is virtually
identical to that of Theorem 2.1-just replace statements like
"n lk" or
"k = nt"
with the statement
"kEI''.
"k is divisible by n" or
Proof of Theorem 6.4 � (1) a - a = ORE/; hence, a,.. a (mod I).
(2) a= b (mod I) means that a - b = i for some i EL Therefore, b - a =
- (a - b) =
-i. Since I is an ideal , the negative of an element of I is also
in I, and so b - a= -iEL Hence, b
(3)
=a
(mod I).
If a= b (mod I) and b = c (mod I), then by the definition of con­
gruence, there are elements i and j in I such that a - b
Therefore, a-
c=
=
i and b -
c = j.
(a - b) + (b - c) = i + j. Since the ideal / is closed under
addition, i +}El and, hence, a= c (mod I).
•
Theorem 6.5
Let I be an ideal in a ring R. If a = b (mod/) and c = d (mod/},then
(1) a+ c = b + d (mod
/);
(2) ac = bd (mod/).
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6.1
I deals and Congruence
14 7
This theorem generalizes Theorems 2.2 and 5.2. Its proof is quite similar to theirs
once you make the change to the language of ideals.
Proof ofTheorem 6.5 ... (1) By the definition of congruence, there are i,JEI such
that a - b = i and c - d= j. Therefore, (a+e) - (b+d) = (a - b)+
(e - d)=i+}El. Hence, a+c=b+d(mod I).
ae - bd=ae - be+be - bd=(a - b)e+b(e - d)=ie+bj. Since
I absorbs products on both left and right, ieEI and bjEI. Hence,
ae - bd= ie+bjEI. Therefore, ae = bd (mod I). •
(2)
the ideal
If
I is an ideal
in a ring
R and aER , then the congruence cla&'i of a modulo I is the
a modulo I, that is, the set
set of all elements of R that are congruent to
{bERlb=a (modI)} = {bERlb-aE/}
= {bERlb - a=
i, with
iEI}
={bERlb =a+i, with iEI}
={a+iliEJ}.
Consequently, we shall denote the congruence class of
rather than the symbol
a modulo I by the symbol a+I
[a] that was used in Z and F(x]. The plus sign in a+I is just a
formal symbol; we have not defined the sum of an element and an ideal. In this con­
text, the congruence class
a+I is usually
called a (left) coset of
I in R.
Theorem 6.6
Let
I be an ideal in a ring Rand let a,
I=c+I.
c
E R. Then a=c (mod /) if and only
if a+
Proof " With only minor notational changes, the proof of Theorem 2.3 carries
over almost verbatim to the present case. Simply replace "mod n" by "mod
I'' and
"[a]" by "a
+ f'; use Theorem 6.4 in place of Theorem 2.1.
•
Corollary 6.7
Let I be an ideal in a ring R. Then two cosets of I are either disjoint or identical.
Proof" Copy theproof of Corollary 2.4 with the obvious notational changes.
•
If I is an ideal in a ring R, then the set of all cosets of I (congruence classes modulo
I)
is denoted R/I.
EXAMPLE 11
Let I be the principal ideal (3) in the ringZ. Then the cosets of I are just the
congruence classes modulo 3, and so there are three distinct cosets: 0+I=[O],
1+I= [1], and 2+I= [2]. The set.Z//of all cosets isprecisely
the setZ3in
our previous notation.
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148
Ideals and Quotient Rings
Chapter 6
EXAMPLE 12
Let I be the ideal in Z[x] consisting of all polynomials with even constant
terms. We claim that Z[x]/I consists of exactly two distinct cosets, namely,
0 +I and l + I. To see this, consider any coset/(x) + I. The constant term of
f(x) is either even or odd. If it is even, then/(x) E/, so thatj(x) ""'0 (mod l).
Therefore,/(x) +I= 0 +/by Theorem 6.6. Iff(x) has odd constant term,
then/(x) - I has even constant term, so that/(x)""' I (modi). Thus/(x) +I=
I +/by Theorem 6.6.
EXAMPLE 13
Let T be the ring of functions from Rto Rand let /be the ideal of all functions
that g(2)= 0. Note that for each real number r, the constant function/,
(whose rule is/,(x)= r) is an element of T. Let h(x) be any element of T. Then
h(2) is some real number, say h(2) c, and
g such
=
(h - /.)(2) = h(2) - /.(2)
= c
- c=
0.
Thus h - f.EI, so that h ""'.fc (mod I) and, hence, h + I= f. + I. Consequently,
every coset of I can be written in the formf,. +I for some real number r.
Furthermore, if c '# d, thenf.(2) * fd(2), so that [/.- /,d(2) oF 0 and/. - f6¢. I.
Hence,/. �/,(mod I) andJ., +I* fa+ I. Therefore, there are infinitely many dis­
tinct cosets of I, one for each real number
r.
• Exercises
NOTE: R denotes a ring.
A. I.
Show that the set K of all constant polynomials in Z[x] is a subring but not an
ideal in Z[x].
2.
Show that the set I of all polynomials with even constant terms is an ideal in
.l[x].
3.
(a) Show that the set!= {(k, 0) lkE.l} is an ideal in the ringZ
X
Z.
(b) Show that the set T = {(k, k) lkE.l} is not an ideal in Z X Z.
4.
Is the set J
over Ill?
5.
Show that the set K
=
an ideal in the ring M(�) of 2 X 2 matrices
{(� �)Ire IR}
{(� �)I EH}
=
a,
b
is a subring of M(IR) that absorbs
products on the right. Show that K is not an ideal because it may fail to
absorb products on the left. Such a set K is sometimes called a right ideal.
6.
(a) Show that the set of nonunits in Z8 is an ideal.
(b) Do part (a) for Z9• [Also, see Exercise 24.]
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6.1
I deals and Congruence
149
7. LetcERand letJ= {rcjrER}.
(a) If Ris commutative , prove thatJis an ideal (that is , Theorem 6.2 is true
even when Rdoes not have an identity).
(b) If Ris commutative but has no identity , is can element of the ideal
I?
(Hint: Consider the ideal {2k I k EE} in the ring E of even integers. Also see
Exercise
33.]
(c) Give an example to show that if Ris not commutative, then Ineed not be
an ideal.
8. If Iis an ideal in Rand Jis an ideal in the ring S, prove that
IX J is an ideal in
the ringR X S.
9. Let R be a ring with identity and let /be an ideal in R.
(a) If IRE/, prove that/= R.
(b) If I contains a unit , prove that
I= R.
10. If Iis an ideal in a fei ld F, prove that I= (Op) or I= F. [Hint: Exercise 9 .]
11. List the distinct principal ideals in each ring:
(a) Zs
(b) ll..9
(c) ll..12
12. List the distinct principal ideals in Z2 X Z3•
13. If Ris a commutative ring with identity and (a) and (b) are principal ideals
such that (a) = (b), is it true that a = b? Justify your answer .
14. Prove Theorem 6.3.
15. Show that the ideal generated by x and 2 in the ring Z[x] is the ideal Iof all
polynomials with even constant terms (see Example 9).
16. (a) Show that (4, 6) = (2) in Z, where (4, 6) is the ideal generated by 4 and 6
and (2) is the principal ideal generated by 2.
(b) Show that (6, 9, 15)= (3) in Z.
17. (a) If Iand J are ideals in R, prove that In Jis an ideal.
(b) If [h] is a (possibly infinite) family of ideals in R, prove that the
intersection of all the Ik is an ideal.
18. Give an example in Z to show that the set theoretic union of two ideals may
not be an ideal (in fact, it may not even be a subring).
19. If I is an ideal in Rand S is a subring of R, prove that In S is an ideal in S.
20. Let Iand J be ideals in R. Prove that the set K = {a + b I a EI, b E J} is an
ideal in Rthat contains both Iand J. K is called the sum of Iand Jand is
denoted I+ J.
21. If dis the greatest common divisor of a and b in Z, show that (a) + (b) = (d).
(The sum of ideals is defined in Exercise 20.)
22. Let Iand Jbe ideals in R. Is the set K = {ab I a EI, b E J} an ideal in R?
Compare Exercise 20.
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150
Chapter 6
Ideals and Quotient Rings
23. (a) Verify that I= {O, 3} is an ideal in Z6 and list all its distinct cosets.
(b) Verify that I= {O, 3, 6, 9, 12} is an ideal in Z15 and list all its distinct cosets.
B. 24. Let R be a commutative ring with identity, and let N be the set of nonunits in
R. Give an example to show that N need notbe an ideal.
25. Let J be an ideal in R. Prove that Iis an ideal, where
I= {rER lrt =OR for every tEJ}.
26. Let Ibe an ideal in R. Prove that Kis an ideal, where
K = {aE RlraElfor every rER}.
27. Let fR � S be a homomorphism of r ings and let
K= {rERlf(r)
==
08}.
Prove that K is an ideal in R.
28. If I is an ideal in R, prove that l[x] (polynomials with coefficients in/) is an
ideal in the polynomial ring R[x].
29. If (m, n) = 1 in z, prove that (m) n (n) is the ideal (mn).
30. Prove that the set of nilpotent elements in a commutative ring R is an ideal.
[Hint: See Exercise 44 in Section 3.2.]
31. Let R be an integral domain and a, b E R. Show that (a)
=
(b) if and only if
a = bufor some unit uER.
32. (a) Prove that the set J of all polynomials in Z[x] whose constant terms are
divisible by 3 is an ideal.
(b) Show that J is not a principal ideal.
33. Let R be a commutative ring without identity and let aER. Show that
A = {ra + na IrER, nEZ} is an ideal containing a and that every ideal
containing a also contains A. A is called the principal ideal generated by a.
34. If Mis an ideal in a commutative ring R with identity and if a ER with a Ii!'. M,
prove that the set
J= {m + ralrER andmEM}
is an ideal such that M � J.
35. Let /be an ideal in Z such that (3) �/r;;;,_ Z. Prove that either I= (3) or I= Z.
36. Let Iand Jbe ideals in R. Let IJ denote the set of
all possible finite sums of
elements of the form ab (with aE/, bEJ), that is,
IJ= {a1b1 + a-ib2 +
·
·
·
+ a,,bn
I n :.?: 1, akE/, bk EI}.
Prove that IJ is an ideal, IJ is called the product of I and J.
37. Let R be a commutative ring with identity
lR ::/: OR whose only ideals are
(OR} and R. Prove that R is a field. [Hint: If a::/: OR, use the ideal (a) to find a
multiplicative inverse for a.]
38. Let Ibe an ideal in a commutative ring R and let
J = {rER [r" Elfor some positive integer n}.
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6.1
Prove that J is an ideal that contains I.
Theorem from Appemiix E. Exercise
I deals and Congruence
[Kmt:
30
151
You will need the Binomial
is the case when I= (OR).]
39. (a) Show that the ring M(IR) is not a division ring by exhibiting a matrix that
has no multiplicative inverse. (Division rings are defined in Exercise 42 of
Section
(b)
3.1.)
Show that M(R) has no ideals except the zero ideal and
M(lli) itself.
If J is a nonzero ideal, show that J contains a matrix A with a
[Hint:
nonzero entry c in the upper left-hand comer. Verify that
o
•
'r
o
(01 ) ( ) (01 0 )
(� �)
o
.A
•
show that
o
o
--
o
and that this matrix is in J. Similarly,
is in J. What is their sum? See Exercise 9.]
40. Prove that every ideal in Z is principal.
[Hint: IfI is a nonzero ideal, show that
I must contain positive elements and, hence, must contain a smallest positive
element
c (Why?).
Since
cE/, every multiple of c is also in /; hence, (c) . � L
To show thatI!;;;;; (c), let a be any element of L Then a = cq +
(Why?). Show thatr
=
0
so that a=
r with
0
sr<c
cqE(c).J
41. (a) Prove that the set Sof rational numbers (in lowest terms) with odd
denominators is a subring of Q.
(b) Let I be the set of elements of S with even numerators. Prove that Iis an
ideal in S.
(c)
Show that S/Iconsists of exactly two distinct cosets.
42. (a) Let p be a prime integer and let Tbe the set of rational numbers
(in lowest
terms) whose denominators are not divisible by p. Prove that Tis a ring.
(b)
Let I be the set of elements of Twhose numerators are divisible by p.
Prove thatI is an ideal in T.
(c)
Show that T/Iconsists of exactly p distinct cosets.
43. Let Jbe the set of all polynomials with zero constant term in Z[x].
(a) Show that J is the principal ideal
(b)
(x) in Z[x].
Show that Z[x]/J consists of an infinite number of distinct cosets, one for
eachnEZ.
44. (a) Prove that the set Tof matrices of the form
subring of
(b)
M(R).
Prove that the set Iof matrices of the form
in the ring T.
(c)
a
b
( )
(� �)
0
a
with a, b E IR is a
with b
Show that everycoset in T/ Ican be written in the form
E IR is an ideal
(� �)
+I.
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152
Chapter 6
Ideals and Quotient Rings
45.
(a)
Prove that the set S of matrioes of the form
subring of M(R).
(b) Prove that the set I of matrices of the form
in the ring
S.
( !)
( )
a
0
0
ho
0
(c) Show that there are infinitely many distinct cosets in
with
a, b, cE Ris a
with b E IR is an ideal
SfI, one for each
pair
in!R x R.
C. 46. Let F be a field. Prove that every ideal in
F[x] is principal. [Hint: Use the
Division Algorithm to show that the nonzero ideal I in F[x] is (p(x)), where
p(x) is a polynomial of smallest possible degree in I.]
Z,. has an identity if and
= u and Sis the ideal (u).
47. Prove that a subring S of
in Ssuch that
m
u2
only if there is an element u
Quotient Rings and Homomorphisms
We now show that the set of congruence classes modulo an ideal is itself a ring. As you
might expect, this is a straightforward generalization of what
classes in Zand
F[x]. However,
we
did with congruence
you may not have expected these rings of congruence
classes to have close connections with some topics studied in Chapter 3, isomorphisms
and homomorphisms. These connections are explored in detail and provide new insight
into the structure of rings.
Let I be an ideal in a ring
R.
RfI are the co sets of I (con­
+I = {a + i I i E I}. In order
The elements of the set
gruence classes modulo I ) , that is, all sets of the form
a
to define addition and multiplication of cosets as we did with congruence classes in Z
and
F[x],
we need
Theorem 6.8
Let/ be an ideal in a ring R. If a+I= b +I and c +I= d+I in RfI, then
(a+ c) +I= (b + d) +I and
ac +I=
bd+I.
Proof ... This is a generalization of Theorem 2.6 , in slightly different notation.
Replace
"[a]" by "a + f' and copy the proof
of Theorem 2.6, using
Theorems 6.5 and 6.6 in place of Theorems 2.2 and 2.3.
•
We can now define addition and multiplication in
F[x]f(p(x)):
RfI just as we did in Z,, and
+ /(congruence class of a) and the coset c +I
c) is the coset (a+ c)+I (congruence class of a+ c). In symbols,
The sum of the coset
(congruence class of
a
(a+ I) + (c +I)= (a+ c) +I.
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6.2
Quotient Rings
and
Homomorphisms
153
This statement may be a bit confusing because the plus sign is used with three entirely
different meanings:
as
a formal symbol to denote a coset:
as an operation on elements of R:
a +I;
a + c;
as the addition operation on cosets that is being defined.*
The important thing is that, because of Theorem
6.8, coset addition is independent
of the choice of representative elements in each coset. Even if we replace a+ /by an
equal coset b + I and replace c + I by an equal coset d +I, the resulting coset sum,
namely (b + d) +I, is the same as (a + c) + L
Multiplication of cosets is defined similarly and is independent of the choice of
representatives by Theorem
6.8:
(a + l)(c + I)
= ac
+I.
EXAMPLE 1
If I is the principal ideal (3) in Z, then addition and multiplication of cosets is
the same as addition and multiplication of congruence classes in Section 2.2.
Thus Z/I is just the ring Z3•
EXAMPLE 2+
If Fis a field, p(_x) is a polynomial in F[x], and I is the principal ideal (p(x)),
then cosets of I are precisely congruence classes modulo p(x), so that addition
and multiplication of cosets are done exactly as they were in Section 5.2. Thus
F[x]/I is the congruence-class ring F[x]/(p(x)).
EXAMPLE3
Let I be the ideal of polynomials with even constant terms in .l[x]. As we saw
in Example 12 of Section 6.1, Z[x]/I consists of just two distinct cosets, 0 +I
1 + I. We have (1 +I) +(1 +I)=: (I + 1) +I= 2 +I, but 2 E J, so that
+ I= 0 + I. Similar calculations produce the
following tables for .l[x]/I. It is easy to see that Z[x] /I is a ring (in fact, a field)
and
2
=
0 (mod I) and, hence, 2
isomorphic to Z2:
O+I
l+I
O+I
l+I
O+I
O+I
l +I
O+I
O+I
O+I
l+J
l+J
O+I
l+J
O+I
l+J
+
"This ambiguity can be avoided by using a different notation for cosets, such as [a], and a different
symbol for coset addition, such as Ef). The notation above is customary, however, and once you're
used to it, there should be no confusion.
+skip this example if you have not read Chapter
5.
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154
Chapter 6
Ideals and Quotient Rings
These examples illustrate the following theorem, which should not be very surpris­
ing in view of your previous experience with Z and F[x].
Theorem 6.9
Let I be an ideal in a ring R. Then
(1) R/ I
is a ring, with addition and multipl icatlon of cosets as defined
previously.
(2)
If R is commutative, then R/I is a commutative ring.
{3) If R has an Identity, then so does the ring
R/ I.
Proof• (1) With the usual change of notation ("a+I" instead of "[a]"), the
proof of Theorem 2.7 carries over to the present situation since that
proof depends only on the fact that Z is a ring. Don't take our word for
it, though; write out the proof in detail for yourself.
(2) If R is commutative and a, c ER, then ac = ca. Consequently, in
R//we have (a+I)(c +I) = ac+I= ca+ I= (c+ I)(a+I). Hence,
RfIis commutative.
(3) The identity in RfIis the coset lR +I because (a+ l)(lR + 1) =
alR +I= a+ I and similarly ClR + I)(a+I)= a+ I. •
The ring RfI is called the
quotient ring (or factor ring)
of
R by
I. One sometimes
speaks of factoring out the ideal I to obtain the quotient ring RfI.
Homomorphisms
Quotient rings are the natural generalization of congr uence-class arithmetic in Z and
F[x]. As is often the case in mathematics, however, a concept developed with one idea
in mind may have unexpected linkages with other important mathematical concepts.
That is precisely the situation here. We shall now
see
that the concept of homomor­
phism that arose in our study of isomorphism of rings in Chapter 3 is closely related
to ideals and quotient rings.
Definition
Let f:R __.. S
be a
homomorphism of rings. Then the kernel off is the set
K = {rERlf{r) =Os}·
Thus, the kernel of f is the subset of
f maps to
Os in
S. Note that
OR
R consisting
of those elements of
is in the kernel since f(Oi) =
Os by
R
that
Theorem 3.10.
However, the kernel may also contain nonzero elements.
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6.2
Quotient Rings
and
Homomorphisms
156
EXAMPLE 4
In Example 6 of Section 3.3 we saw that the function/:Z--+� defined by
f(r) = [r] EZ6 is a homomorphism of rings. Its kernel K contains many nonzero
integers. For instance, 12 EK because/(12) = [12] = [O] in�· In fact every
multiple of 6 is in the kernel because
K= {rEZl /(r) = [O]} = {rEZl[r] = [O]}
= {rE Z Ir= 0 (mod 6 )}
=
{rEZ 161 r}
= {all multiples of 6}
[Defmitiono f f]
[Theorem 2.3]
[Deflllitionof congruence mod 6]
[6 I r means r is a multiple of 6].
So the kernel K is the principal ideal (6) in Z.
EXAMPLE 5
The function O:R[x]-+ R that sends each polynomial in R[x] to its constant
term in Ris a ring homomorphism (Exercise 1). Its kernel consists of all
polynomials with constant term 0. But every polynomial with 0 constant term
is divisible by x. So the kernel is the principal ideal (x) in R[x].
Examples 4 and 5 provide examples of the following theorem.
Theorem 6.10
Let f:R--+ S be a homomorphism of rings. Then the kernel Koff is an ideal in
the ring R.
Proof..we shall use Theorem 6.1 to show that K = {rERl/(r) =Os} is an ideal.
We must
verify that is a nonempty subset of R that is closed under sub­
traction and absorbs products. First, K is nonempty because 0REK as
noted before Example 4. To prove that K is closed under subtraction, we
must show that for a, b EK, the element a - b is also in K. To show
a - b EK, we must show that/(a - b) = Os. This follows from the fact
that/ is a homomorphism and that/(a) = Os andf(b) = Os (because a,
bEK):
f(a - b)
=
f(a) - f(b) = Os - Os= Os.
To prove that K absorbs products we must first verify that raEK for any
rER and aEK, that is, thatf(ra) =Os; here's the proof:
f(ra) = f(r)f(a) = f(r) Os= Os.
similar argument shows that arEK. T herefore K is an ideal by
Theorem 6.1. •
A
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156
Chapter 6
Ideals and Quotient Rings
In Examples 4 and 5, the kernel of the homomorphism contained
many nonzero
OR, in
elements. Sometimes, however, the kernel of a homomorphism contains only
which case we have an interesting result.
Theorem 6.11
Let f:R � S be a homomorphism of rings with kernel K. Then K = (OR) if and
only if f is injective.
Proof.. Suppose that K = (O.R). We must show that/is injective, so assume
that a, bER and/(a) =fib). Because/is a homomorphism,
f(a - b) f(a) -f(b) = Os. Hence, a - bis in the kernel K = (O.R),
=
which means that a
- b = OR and a= b. Therefore/is injective.
Conversely, suppose/is injective. If cEK, we must show that c =OR.
By the definition of the kernel,/(c) = Os. By Theorem 3.10,/(0.R) =Os=
f(c). Therefore, c =OR because/is injective. Hence, the kernel consists
of the single element OR, that is, K
= (O.R).
•
EXAMPLE 6
In Example 7 of Section 3.3 we saw that the function g:lll � M(R) given by
O 0
g(r)
is a ring homomorphism. Its kernel of g consists of all real
-r
r
o
0
0
0
numbers r such that g(r) =
, that is, such that
-r
r
0
0
0
0
This can only occur when r = 0. So the kernel is the zero ideal (0). Hence, g is
=
(
)
( o) ( )
( o)
=
.
injective by Theorem 6.11.
Theorem 6.10 states that every kernel is an ideal. Conversely, every ideal is the
kernel of a homomorphism:
Theorem 6.12
Let I be an ideal in a ring R. Then the map 7T:R �Rf I given by '1T(r) = r + I is
a surjective homomorphism with ke rnel /.
The map '1T is called the natural homomorphism from R to RfI.
Proof of Theorem 6.12 ... The map '1T is surjective because given any coset r + I in
RfI, '1T(r)
=
r +I. The definition of addition and multiplication in RfI
shows that '1T is a homomorphism:
'1T(r
+ s) = (r + s) + I= (r + I) + (s + I) = '1T(r) + '1T(s);
'1T(rs) = rs + I= (r + I)(s
+ I) = '1T(r) '1T(S).
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Quotient Rings and Homomorphisms
6.2
The kernel of
Tr is the set of
(the zero element in
R/ I).
elements rERsuch that
157
Tr(r)=OR+ I
However, Tr(r)=OR + I if and only if
r
+ I=
OR+ I, which occurs if and only if r =OR (mod I), that is, if and only if
r E I. Therefore, I is the kernel of Tr.
•
The natural homomorphism Tr in Theorem 6.12 is a special case of a more general
situation. If
S is a surjective homomorphism of rings,
f:R -+
homomorphic image of
image of
R),
R. If /is
we
say that S is a
actually an isomorphism (so that Sis an isomorphic
then we know that
R
and S have identical structure. Whenever one
of them has a particular algebraic property, the other one has it too . If f is not an
isomorphism, then properties of one ring may not hold in the other. However, the
properties of S and the homomorphismf often give us some useful information
about
R.
An analogy with sculpture and photography may be helpful: If
is an isomorphism, then S is
an
f:R -+ S
R. If f is only a
two-dimensional photographic image of R in
exact, three-dimensional replica of
surjective homomorphism, then S is a
which some features of Rare accurately reflected but others are distorted or missing.
The next theorem tells us precisely how
R, S, and
the kernel off are related in these
circumstances.
Theorem 6.13
Let
f:R-+ S
First Isomorphism Theorem
be a surjective homomorphism of rings with kernel
quotient ring R/K
K. Then the
is isomorphic to S.
The theorem states that every homomorphic image of a ring R is isomorphic to a
quotient ring R/ K for some ideal K. Thus if you k now all the quotient rings of R, then
you know all the possible homomorphic images of
R. The ideal K measures how much
information is lost in passing from the ring R to the homomorphic image R/ K. When
K= (Oa),
then/is an isomorphism by Theorem 6.11, and no information is lost. But
when K is large, quite a bit may be lost.
Proof of Theorem 6.13 ... We shall define a function <p. from R/K to sand then
show that it is an isomorphism. To define <p,
each coset
r + K of R/K an element of
we
must associate with
S. A natural choice for such an
element would bef(r) ES; in other words, we would like to define
cp:R/K-+ Sby the rule <p(r + K) = f(r). The only possible problem is that
a coset can be labeled by many different elements of R. So
that the value of
cp depends only on the
we
must show
coset and not on the particular
representative r chosen to name it. If r +
K) by Theorem 6.6, which means that
K= t + K, then r = t (mod
r - t EK by the definition of
congruence. Consequently, since/is a homomorphism,/(r) - f(t)=
f(r - t) = 05• Therefore, r + K = t + K implies that/(r) = f(t). It
follows that the map cp:R/K-+ S given by the rule cp(r + K) = f(r) is a
well-defined function, independent of how the coset is written.
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158
Chapter 6
Ideals and Quotient Rings
If SES, thens=f(r) for some rER because/is surjective. Thus
s=f(r) = q:{_r +K), and cp is surjective. To show that <p is injective, we
assume that <p(r + K ) = q:{_c + K) and show that r + K= c + K, as follows:
cp(r+ K) = cp(c + K)
f(r) = f(c)
f(r ) - f(c) =Os
f (r - c) =Os.
[Definition of <p]
[f is a homomorphism.]
Thus, r - cEKand hence, r = c (mod .K). So r+ K= c + Kby
Theorem 6.6. Therefore, cp is injective.
Finally, cp is a homomorphism because/is
cp[(c + K)(d+ K)] = cp(cd+ K) =f(cd') =f(c}f(d)
= cp(c + K)<p(d+ K)
and
cp[(c + K) + (d+ .K)] = <p[( c + d) + .K] =f(c + d) =f(c) + f(d)
= cp(c + K) + <p(d + K).
Therefore, cp:R/K-4 Sis
an
isomorphism.
•
The First Isomorphism Theorem is a useful tool for determining the structure of
quotient rings, as illustrated in the following examples.
EXAMPLE 7
In the ring Z[x], the principal ideal (x) consists of all multiples of x, that is,
all polynomials with constant term 0. W hat does the quotient ring Z[x]/(x)
look like? We can answer the question by using the function 8:Z[x] -4 Z,
which maps each polynomial to its constant term. The function 8 is certainly
surjective because each kEZ is the image of the polynomial x + kin Z[x ].
Furthermore, (J is a homomorphism of rings (Exercise 1). The kernel of 8
consists of all those polynomials that are mapped to 0, that is, all polynomials
with constant term 0. Thus the kernel of (J is the ideal (x). By Theorem 6.13 the
quotient ring Z[x]/(x) is isomorphic to Z.
EXAMPLE 8
Let T he the ring of functions from Rto Rand /the ideal of all functions
g such that g(2)
0. In Example 13 of Section 6.1 we saw that T/I con­
sists of the cosets.f.. +I, one for each real number r, wheref,:R-4 IR is the
constant function given by f,(x) = r for every x. This suggests the possibility
that the quotient ring T/I might be isomorphic to the field R. We shall use
=
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� ...... k
Quotient Rings
6.2
Homomorphisms
and
159
Theorem 6.13 to show that this is indeed the case by constructing a surjective
rp:T-+ R be the
= /(2). Then rp is surjective because for every real
number r, r = /,(2) = rp(f,). Furthermore, rp is a homomorphism of rings:
homomorphism from Tto �whose kernel is the ideal/. Let
function defined by rp(f)
rp(f +
h ) = (f
rp(fh)
=
+
h)(2) = /(2)
(fh)(2)
By definition, the kernel of
=
+
h(2) = rp(/)
+
rp(h)
f(2)h(2) = rp(f)rp(h).
rp is the set
{gETlrp(g) =
0}
= {gETlg(2) =
O}.
Thus the kernel is precisely the ideal I. By Theorem 6.13, T//is isomorphic to
R.
EXAMPLE 9
What do the homomorphic images of the ring Z look like? To answer this
question, suppose thatf:Z--+ Sis a surjective homomorphism. If /is actually
an isomorphism, then S looks exactly like Z, of course (in terms of algebraic
structure). If f is surjective, but not an isomorphism (that is, not injective), then
the kernel Kof f is a nonzero ideal in Z by Theorem 6.11. Since K is
in Z, Kmust be a principal ideal, say K =
(n) for some n *
an
ideal
0, by Exercise 40
in Section 6.1. By Theorem 6.13, Sis isomorphic to ZK
/
=Z
/(n)
""Zn. Thus
every homomorphic image of .Z is isomorphic either to Z or to Zn for some
n.
• Exercises
A. 1. Show that the map
O:R[x]-+ R that sends each polynomial/(x) to its constant
term is a surjective homomorphism.
2. Show that every homomorphic image of a field Fis isomorphic either to F
itself or to the zero ring.
[Hint: See
Exercise 10 in Section 6.1 and Exercise 7
below.]
3. If Fis a field, Ra nonzero ring, andf:F-+ Ra surjective homomorphism,
prove that f is an isomorphism.
4. Let
(a)
[aJn denote the congruence class of
the integer
a modulo n.
Show that the mapf:.Z12 -+Z4 that sends [a]i2 to
[a]4 is a well-defined,
surjective homomorphism.
(b) Find the kernel off.
5. Let I be an ideal in an integral domain R. Is it true that R/ I is also an integral
domain?
6. The function rp:R[x]-+ R given by rp(f(x))
rings by Exercise 24 of Section 4.4 (with a
= /(2) is a homomorphism of
= 2). Find the kernel of rp. [Hint:
Theorem 4.16.]
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160
Chapter 6
Ideals and Quotient Rings
7.
If R is a ring, show that Rf(ORJ = R.
8.
Let R and S be rings. Show that 7T:R X S-+ R given by 7T(r, s) = r is a
surjective homomorphism whose kernel is isomorphic to S.
9.
R
=
{ G �)I
a, b, cEZ
}
is a ring with identity by Example 19
in Section 3.1.
(a) Show that th� mapf:R-+ Z given by
homomorphism.
t{: �)
=
a is a surjective
(b) What is the kernel off?
10.
(a) Let/:R-+ S be a surjective homomorphism of rings and let /be an ideal
in R. Prove that/(/) is an ideal in S, where/(/)= {sESls =/(a) for
some a El}.
(b) Show by example that part (a) may be false if /is not surjective.
11.
Z[Vl] is a ring by Exercise 13 of Section 3.1. Let/:Z(Vl]-+ Z(v'2] be the
function defined by f(a + bv'l) = a - bv'l.
(a) Show that/is a surjective homomorphism of rings.
(b) Use Theorem 6.11 to show that/is also injective and hence is an
isomorphism. [You may assume that V2 is irrational.]
12.
13.
Let Ibe an ideal in a noncommutative ring R such that ab - ba EI for all
a, b ER. Prove that Rf I is commutative.
Let Ibe an ideal in a ring R. Prove that every element in Rf Ihas a square root
only if for every a ER, there exists b ER such that a - b1EI.
if and
14.
Let Ibe an ideal in a ring R. Prove that every element in Rf Iis a solution of
1
= x if and only if for every a ER, a1 - a EI.
x
15.
Let !be an ideal in a commutative ring R. Prove that Rf !has an identity if
and only if there exists e ER such that ea - a E Ifor every a ER.
16.
Let I "i: R be an ideal in a commutative ring R with identity. Prove that Rf I is
an integral domain if and only if whenever abE I, either a EI or bEI.
17.
Suppose/ andJare ideals in a ringR and Ietf:R-+Rf IX RfJbe the
function defined by f (a) (a + I, a + J).
=
(a) Prove that/is a homomorphism of rings.
(b) Is/surjective? [Hint: Consider the case when R
=
Z, I= (2),J = (4).)
(c) What is the kernel off?
18.
Let R be a commutative ring with identity with the property that every ideal
in R is principal. Prove that every homomorphic image of R has the same
property.
19.
Letland Kbe ideals in a ring R, with K!;;; I. Prove thatlfK ={a+ Kia EI} is
an ideal in the quotient ring RfK.
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6.2
20.
Quotient Rings
and
Homomorphisms
161
Let/ :R -4Sbe a homomorphism of rings with kernel K. Let /be an ideal
in R such that I!;;: K. Show that f:R/I-4 S given by f(r + I) f(r)is a well­
defined homomorphism.
=
21.
Use the First Isomorphism Theorem to show that Z20/(5) = Z5•
22.
Let f :R -4Sbe a homomorphism of rings. If Jis an ideal in S and I =
{r ERlf(r ) EJ], prove that Iis an ideal in Rthat contains the kernel off
23.
(a)
Let R be a ring with identity. Show that the mapf:Z-4 Rgiven by
f(k) = k lR is a homomorphism.
(b) Show that the kernel of/is the ideal (n), where n is the characteristic of
R. [Hint: "Characteristic" is defined immediately before Exercise 41 of
Section 3.2. Also see Exercise 40 in Section 6.L]
24.
Find at least three idempotents in the quotient ring O[x]/(x4 +
[See Exercise 3 in Section 32
. .]
25.
Let R be a commutative ring and J the ideal of all nilpotent elements of R
(as in Exercise 30 of Section 6.1). Prove that the quotient ring R/Jhas no
nonzero nilpotent elements.
26.
Let Sand /be as in Exercise 41 of Section 6.1. Prove that S/I= Z2•
27.
Let T and I be as in Exercise 42 of Section 6.1. Prove that T /I = Zp.
28.
Let T and I be as in Exercise 44 of Section 6.1. Prove that T /I = R.
29.
Let Sand Ibe as in Exercise 45 of Section 6 1 Prove that S /I= IR X IR.
C. 30.
.
x2).
.
(The Second Isomorphism Theorem) Let I and Jbe ideals in a ring R. Then
In Jis an ideal in I, and J is an ideal in I + Jby Exercises 19 and 20 of
I J
=
Section 6. L Prove that /
. [Hint: Show that/:/-4 (I+ J)/Jgiven
J
by f(a) = a + Jis a surjective homomorphism with kernel I() J.]
�
;
31.
(The Third Isomorphism Theorem) Let I and Kbe ideals in a ring R such that
Kr;;;.!. Then I/Kis an ideal in R/Kby Exercise 19. Prove that (R/K)/(I/K)=
R/I. [Hint: Show that the mapf:R/K-4R//given by/(r + K) = r + Iis a well­
defined surjective homomorphism with kernel I/K.]
32.
(a) Let K be an ideal in a ring R. Prove that every ideal in the quotient
ring R/K is of the form I/K for some ideal Jin R. [Hint: Exercises 19
and22.]
(b) Iff!R-4 Sis a surjective homomorphism of rings with kernel K, prove
that there is a bijective function from the set of all ideals of S to the set of
all ideals of R that contain K. [Hint: Part (a)and Exercise 10.]
EXCURSION: The Chinese Remainder Theorem for R ings
(Section 14.3) may be covered at this point if desired.
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162
Chapter
m
Ideals and Quotient Rings
6
The Structure of R//When /Is Prime or Maximal*
Quotient rings were developed as a natural generalization of the rings z, and F[x]/(p(x)).
When pis prime and p(x)irreducible, then z, and F[x]/(p(x)) are fields. In this section
we explore the analogue of this situation for quotient rings of commutative rings. We
shall determine the conditions necessary for a quotient ring to be either an integral
domain or a field.
Primes in "lL and irreducibles in F[x] play essentially the same role in the structure
of the congruence class rings. Our first task in arbitrary commutative rings is to find
some reasonable way of describing this role in terms of ideals. According to Theorem 1.5,
a nonzero integer p (other than ±1) is prime if and only if p has this property:
Whenever p I he, then p I b or p I e . To say that p I a means that a is a multiple of p, that
is, a is an element of the principal ideal (p) of all multiples of p. Thus this property of
primes can be rephrased in terms of ideals:
If p 1' 0, ± 1, then pis prime if and only if
whenever be E (p), then b E (p) or c E (p).
The condition p 1' ±1 guarantees that 1 is not a multiple of p and, hence, that the ideal
(p) is not all of Z. Using this situation as a model, we have this
Definition
An ideal P in a commutative ring R is said to be prime if P 1' Rand whenever
bcEP, then b EPorcEP.
EXAMPLE 1
As shown above, the principal ideal (p) is prime in "lL whenever pis a prime
integer. On the other hand, the ideal P= (6) is not prime in Z because
2 3EPbut 2itPand 3itP.
•
EXAMPLE 2
The z ero ideal in any integral domain R is prime because ab= OR implies
a= OR orb= OR.
EXAMPLE 3
The implication (1 ) � (2) of Theorem4.12shows that if Fis a field and p(x) is
irreducible in F[x,] then the principal ideal (p(x)) is prime in F[x].
*This section is not used in the sequel and may be omitted if desired.
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6.3
The Structure of
R/I When
/Is Prime or Maximal
163
EXAMPLE 4
Let I be the ideal of polynomials with even constant terms inZ[x]. Then I is not
principal(Example 8 of Section 6.1) and clearly I'#; Z[x]. Letf(x) = a,,r' +
+ "o
andg(x) =bmX" +···+ho bepolynomials inZ[x] such tbatf(x)g(x)EI. Then the
constant term of f(x)g(x ), namely a;Po, must be even. Since the product of two odd
integers is odd, we conclude that either "o is even(that is,f(x) EI) or ho is even(that
is, g(x) E !). Therefore, /is a prime ideal.
· · ·
The ideal I in Example 4 is prime, and the quotient ring Z[x]f I is a field (see
Example 3 of Section 6.2). Similarly, Zf(p) = z, is a field whenp is prime. However,
the next example shows thatRf Pmay not al ways be a field whenPis prime.
EXAMPLE 5
The principal ideal (x) in the ring Z[x] consists of polynomials that are mul­
tiples of x, that is, polynomials with zero constant terms. Hence, (x) '#; Z[x]. If
f(x) = a,.x" + · + llo and g(x) = b,,;xm + · + b0 and/(x)g(x) E /,thenthe
constant term of f(x)g(x), namely ao/Jo, must be 0. This can happen only if
"o = 0 orb0
0, that is, only if f(x) E(x) or g(x) E(x). Therefore, (x) is a prime
ideal . However, Example 7 of Section 6.2 shows that the quotient ring .Z[x]/(x)
is isomorphic to .Z. Therefore, .Z[x]f(x) is an integral domain but not a field.
·
·
· ·
=
In light of Example 5, the next theorem is the best we can do with prime ideals.
Theorem 6.14
Let P be an ideal in a commutative ring R with identity. Then P Is a prime ideal
if and only if the quotient ring Rf Pis an integral domain.
Proof .. IfPis any ideal inR, then by Theorem 6.6, a+P=OR+PinRfPif
and only if a= OR(modP). Furthermore, a= OR(mod P) if and only if
aEP. So we have this useful fact:
(*)
a+P=O R+PinRfP
if and only if
aEP.
SupposePis prime. By Theorem 6 .9,Rf Pis a commutative ring
identity. In order to prove thatRfPis an integral domain, we must
show that its identity is not the zero element and that it has no zero
divisors. SincePis prime, P '#;R. Consequently, 1 R�Pbecause any ideal
containing 1 R must be the whole ring. However, 1 R�P implies that
lR+P:;:. OR+PinRfPby(•). Now we show thatRf Phas no zero
divisors. If (b +P)(c + P) = OR+P, then be +P = OR +P and be EP
by(*). HencebEPorcEP Thusb +P =OR+Pore+P =OR+P, so
thatRfPhas no zero divisors. ThereforeRfP is an integral domain.
with
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164
Chapter 6
Ideals and Quotient Rings
Now assume thatRfPis an integral domain. Then by definition
1 R +P * OR +Pand hence 1R �Pby(
* ). ThereforeP * R.To complete
the proof thatPis prime we assume that beE Pand show thatb EP or
cEP. N ow if bcEP, theninR/Pwehave(b + P)(c+ P) =be+P=
OR+Pby(•).Thusb +P=OR+Pore+P=OR+PbecauseR/Phas
no zero divisors. Henceb EPorcEPby(•).ThereforePis prime . •
Since the quotient ring modulo a prime ideal is not necessarily a field, it is natural
to ask what conditions
an
ideal must satisfy in order for the quotient ring to be a field.
EXAMPLE 6
Consider the ideal(3)in Z. We know that Z/(3) = Z3 is a field. Now consider
the ideal (3). Suppose J is an ideal such that(3)!;;;; J !;;;; Z. If J #; (3), then there
exists a E J with a�(3). In particular, 3 K a, so that 3 and a are relatively prime.
Hence, there are integers u and v such that3u + av = 1. Since3 and a are in
the ideal J, it follows that 1 E J. Therefore J = ZbyExercise 9 of Section 6.1,
and so there are no
ideals strictly between (3) and Z.
EXAMPLE 7
The quotient ring Z[x]/(x)is not a field (Example 5). Furthermore, the ideal
I
of poly nomials with even constant terms lies strictly between (x)and Z[x], that
is,(x)£ I£ Z[x].
Here is a formal definition of the proper ty suggested by these examples:
Definition
An ideal Min a ring R is said to be maximal if M * Rand whenever J is an
ideal such that
Mi= Ji= R, then
M = J or J = R.
Example 6 shows that the ideal (3)is maximal in Zand Example 7 shows that the
ideal (x)is not maximal in Z[x]. Note that a ring may have more than one maximal
ideal.The ideal {O, 2, 4} is maximal in Z6, and so is the ideal {O, 3}. There are infinitely
many maximal ideals in Z(Exercise 3). Maximal ideals provide the following answer
to the question posed above:
Theorem 6.15
Let M be an ideal in a commutative ring R with identity. Then M is a maximal
ideal if and only if the quotient ring R / M is a field.
Proof... We shall use the same fact that was used in the proof
(*)
a+ M= OR+ MinR/M
of Theorem 6.14:
if and only i f
aEM.
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6.3
The Structure of
R/I When /Is Prime or Maximal
166
Suppose R fMis a field. Then by definition lR+ M '#OR+ Mand
hence 1 R
ft.Mby ( *) . ThereforeM '# R . To show thatMis maximal, we
assume that J is an ideal with Mi;;;Ji;;;
;
; R and show that M= Jor
J= R. IfM= J, there is nothing to prove. If M '# J, then there exists
aEJwitha¢M. Hencea+ M-1: OR+ Min thefield R fM,anda+ M
has an inverse b+ M such that (a+ M)(b+ M) =ab+ M = lR+ M.
Then ab
=
1 R (mod M) by Theorem 6.6,so thatab - lR = m for some
mEM. Thus IR = ab -m. Sincea and m are in the ideal J, it follows
that IREJ andJ= R . ThereforeMis a maximal ideal.
Now assumeMis a maximal ideal in R . By Theorem 6.9, R fMis a com­
mutative ring with identity. In order to prove that R fMis a field,we first
show that its identity is not the zero element . SinceMis maximal,M 'I: R
Consequently, 1 R
ft.Mbecause any ideal containing 1 R must be the whole
ring. Howevei; 1 R¢Mimpliesthat lR+ M'# OR+ Min R M
f
by (t).
Next we show that every nonz.ero element of R M
f
has a multiplicative
inverse. Ifa + Mis a nonzero element of R fM, thena
ft.M(otherwise a + M
would be the zero coset). The set
J= {m + ralrE R andmEM}
is an ideal in R that containsMby Exercise 34 of Section 6.1 . Furthermore ,
a = OR+ lRa is in J, so that M 'I: J. By maximality we must have J = R .
Hence lREJ, which implies that lR = m +ca for somemEMand cE R .
Note that ca - 1R =
-
mEM,
so that ca
=
lR ( mod M), and hence
ca+ M= lR+ Mby Theorem 6.6. Consequently, the coset
c+ Mis the inverse of a+ Min RjM:
(c+ M)(a + M) = ca+ M= l R+ M.
So every nonzero element of R M
f
is a unit (Axiom 12 is satisfied).
Therefore, R M
f
is a field.
•
Corollary 6.16
In a commutative ring R with identity, every maximal ideal is prime.
Proof" IfMis a maximal ideal, then R/Mis a field by Theorem 6 .15. Hence,
RM
f
is an integral domain by Theorem 3.8. Therefore, Mis prime by
Theorem 6.14.
•
Theorem 6.15 can be used to show that several familiar ideals are maximal.
EXAMPLE 8
The ideal I of polynomials with even constant terms in Z[x]is maximal because
Z[x]f I is a field
(see Example 3 of Section 6.2).
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166
Chapter
6
Ideals and Quotient Rings
EXAMPLE 9
Let The the ring of functions from R to R and let !be the ideal of all functions
g such that g(2)
= 0. In Example 8 of
Section 6.2 we saw that T/I is a field
isomorphic to R. Therefore, I is a maximal ideal in T.
• Exercises
A. 1. If
n
is a composite integer, prove that (n) is not a prime ideal in Z.
2. If R is a finite commutative ring with identity, prove that every prime ideal in
R is maximal. [Hint: Theorem 3.9.)
3. (a) Prove that a nonzero integerp is prime if and only if the ideal (p) is
maximal in Z.
(b) Let Fbe a field andp(x) EFx
[ l. Prove thatp(x) is irreducible if and only if
the ideal (p(x)) is maximal in F[x).
4. Let R be a commutative r ing with identity. Prove that R is an integral domain
if and only if (O.R) is a prime ideal.
5. List all maximal ideals in Z6• Do the same in Z12•
6. (a) Show that there is exactly one maximal ideal in Z8• Do the same for "4·
[Hint: Exercise 6
in Section 6.1.]
(b) Show that Zio and Z15 have more than one maximal ideal.
7. Let R be a commutative ring with identity. Prove that R is a field if and only if
(OR) is a maximal ideal.
8. Give an example to show that the intersection of two prime ideals need not be
prime. [Hint: Consider (2) and (3) in Z:.]
9. Let R be an integral domain in which every ideal is principal. If (p) is a
nonzero prime ideal in R, prove that p has this property: Whenever p factors,
p
:=
cd, then c or dis a unit in R.
B. 10. Letp be a fixed prime and let J be the set of polynomials in .lx]
[
whose
constant terms are divisible by p. Prove that J is a maximal ideal in Z[x].
11. Show that the principal ideal (x - 1) in Z:[x] is prime but not maximal.
12. If p is a prime integer, prove that Mis a maximal ideal in ZX Z, where M
{(pa, b) I a, b EZ}.
=:
13. If I is an ideal in a ring R, then IXI is an ideal in RX R by Exercise 8 of
Section 6.1. Prove that (RXR)/(I XI) is isomorphic to R/IX R/I.
H
[ int: Show that the functionf:RX R�R/I X R/Igiven byf((a,
(a + I, b + I ) is a surjective homomorphism of
b))
=:
rings with kernel IX I.]
14. If P is a prime ideal in a commutative ring R, is the ideal PX P a prime ideal
in RX R? [Hint: Exercise 13.)
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6.3
15.
The Structure of
Rf/ When/ Is Prime or Maximal
167
(a) LetR be the set of integen equipped with the usual addition and
multiplication given by ab= 0 for all a, b ER. Show thatRis a
commutative ring .
(b) Show that M
=
{O, ±2, ±4, ±6, . .. } is a maximal ideal inRthat is not
prime. Explain why this result does not contradict Corollary 6.16.
M = {O, ±4, ±8, . . . } is a maximal ideal in the ring E of even
E/ Mis not a field. Explain why this result does not contradict
Theorem 6.lS.
16. Show that
integen but
17. Let /:R-+ S be a surjective homomorphism of commutative rings. If J is a
prime ideal in S, and I=
{r ERlf(r)EJ}, prove that /is a prime ideal inR.
18. Let P be an ideal in a commutative ringRwith P :¢: R. Prove that Pis prime
if and only if it has this
property: Whenever A and B are ideals inRsuch that
36 of Section 6.1. This
AB c;; P, then A c;;.: P or B c;; P. (AB is defined in Exercise
property is used as a definition of prime ideal in noncommutative rings.]
19. Assume that whenRis a nonzero ring with identity, then every ideal of
Rexcept R itself is contained in a maximal ideal (the proof of this fact is
beyond the scope of this book). Prove that a commutative ringRwith identity
has a unique maximal ideal if and only if the set of nonunits in Ris an ideal.
Such a ring is called a
local ring. (See Exercise 6 of Section 6.1 for examples of
local rings.)
20. Find an ideal in
C. 21.
Z X Z that is prime but not maximal .
(a) Prove thatR = {a+bila, b EZ} is a subring of C and that
M= {a+bil3laand3lb}
r +sifl.M, then 3 A'ror 3 Ks. Show
3 does not divide r2+s2 = (r +s1)(r - s1). Then show that any ideal
containing r +si and M also contains l.]
is a maximal ideal inR. [Hint: If
that
(b) Show that R/M is a field with nine elements.
22. Let R be as in Exercise
21. Show that J is not a maximal ideal in R,where J =
{a+bi IS I a and S lb}.[Hint: Consider the principal ideal K= (2+ i) inR.]
23. IfR and J are as in Exercise
22, show thatRfJ = Zs
24. IfR and Kare as in Exercise
22, show thatRK
/ = Z5•
X
Zs.
T= {a+bVlla,bEZ} is a subring of �and M=
{a+ hv'21SlaandSib} is a maximal ideal in T.
25. Prove that
ALTERNATIVE ROUTES: At this point there are three possibilities.
You may explore a new algebraic concept, groups (Chapter7)-if you
have not already done so-or continue further with either integral
domains (Chapter
10) or fields (Chapter 11).
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CHAPTER
7
Groups
The algebraic systems with which you are familiar, such as Z, Zn, the rational
numbers, the real numbers, and other rings all have two operations: addition and
multiplication. In this chapter, we introduce a different kind of algebraic structure-­
called a group-that uses a single operation. Groups arise naturally in the study of
symmetry, geometric transformations, algebraic coding theory, and in the analysis
of the solutions of polynomial equations.
ALTERNATE ROUTE: If you have not read Chapter 3 (Rings), you
should replace Section 7 1
. with Section 7.1A
. , which begins on page 183.
•
Definition and Examples of Groups
A group is an algebraic system with one operation. Some groups arise from rings by
ignoring one of their operations and concentrating on the other. As we shall
example,
see,
for
the integers form a group under addition (but not multiplication) and the
nonz.ero rational numbers form a group under multiplication (but not addition). But
many groups do not arise from a system with two operations. The most important of
these latter groups (the ones that were the historical starting point of group theory)
developed from the study of permutations.* Consequently, we begin with a consider­
ation of p ermutations.
Informally, a p ermutation of a set Tis just an ordering of its elements. For example,
there
are
six possible permutations of T = {1,2, 3}:
123
132
213
231
312
321.
•in the early nineteenth century, permutations played a key role in the attempt to find formulas for
solving higher-degree polynomial equations similar to the quadratic formula. For more information,
see Chapter
12.
169
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_ .... ..,_.... __ ... _..,. ..... ....... ....... ..,..... Coog... l.olmlll& __ ... .... ...... - ........ ..,-11..-.-�-·...-.11.
170
Chapter 7
Groups
Each such ordering determines a bijective function from. T to I'. m.ap I to the first
element of the ordering, 2 to the second, and 3 to the third.* For instance, 2 3 1
determines the function.f:T � Twhose rule isf(l) = 2;/(2) = 3;/(3) = 1. Conversely,
every bijective function from. T to Tdefines
an
ordering of the elements, namely,f(l),
f (2), /(3). Consequently, we define a permutation of a set T to be a bijective function
from Tto T. This definition preserves the informal idea of ordering and has the advan­
tage of being applicable to infinite sets. For now, however, we shall concentrate
sets and develop a convenient notation for dealing with their permutations.
on
finite
EXAMPLE 1
Let T:=
{l, 2, 3}. The permutation/whose rule isf(l)
may be represented by the array
G � �).
=
2,/(2)
=
3,/(3)
=
1
in which the image under f of an
element in the first row is listed immediately below it in the second row. Using
this notation, the six permutations of Tare
G
G
�) G � �) G
2
2
�) G � �) G
2
3
2
1
2
2
Since the composition of t wo bijective f unctions is itself bijective, the composi­
tion of any two of these permutations is one of the six permutations on the list
�ov� For instance, if f
tion given by
=
G�D
(f g)(l)
o
(Jo g)(2)
Thus f 0 g
( )
and g
=
=
=
G � �).
then/o g is the func­
2
f(g(l))
=
/(2)
f(g(2))
=
/(1) = 3
=
(f 0 g)(3) = f(g(3)) = /(3) = I.
1 2 3
. It is usually easier to make computations like this
2 3 1
by visually tracing an element's progress as we first apply g and then/; for
=
example,
•Bijective functions are discussed in Appendix B.
....
..
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7.1
)
Definition and Examples of Groups
171
If we denote the set of permutations of T by S3, then composition of functions
(
0
is an operation on the set S3 with this property:
Since composition of functions is associative,* we
)
(J o g
. .
o
h
=
fo (g o h)
. = (1 2 3)
123
and
that
for all/, g,
rif
Ve y that the identity permutation I
1 °/ = f
see
f• l =f
.
h ES3•
has this property:
forevery/ES3•
Every bijection has an inverse function;* consequently,
if/ES3, then there existsgES3 such that
Jo g = /
For instance, if f =
and
G � �)
and
,then g =
g of = I.
G � �)
because
2
2
2 3) (1 2 3) = (1 2
G 3 1 3 1 2 1 2
0
You should determine the inverses of the other permutations in S3 (Exercise
Finally, note that/og may not be equal tog 0f; for instance,
1).
2
2
2
G 2 D·G 1 D=G 3 D
but
G
2
�)o(!
2 = 2
D G �}
2
•see Appendix B.
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172
Chapter 7
Groups
By abstracting the key properties of S3 under the operation
Definition
o,
we obtain this
A group is a nonempty set G equipped with a binary operation* that satis­
fies the following axiomst;
Closure: lfaEG and bEG, then a*bEG.
2. Associativity: a* (b * c} =(a* b) * c for all a, b, c E G.
3. There is an element eEG (called the identity element) such that
a *e = a = e *a for every aEG.
4. For each aEG, there is an element dE G (called the inverse of a}
such that a *d = e and d *8 = e.
1.
A group is said to be abelian+ if it also satisfies this axiom:
5. Commutativity: a * b =
A group
b *a for all a, b E G.
G is said to be finite (or of finite order) if it has a finite number of elements
.
G is called the order of G and is denoted [ G[. A
In this case, the number of elements in
group with infinitely many elements is said to have
infinite order.
EXAMPLE 2
The discussion preceding the definition shows that S3 is a nonabelian group of
order 6, with the operation
*
being composition of functions.
EXAMPLE 3
The permutation group S] is just a special case of a more general situation. Let
n
be a fixed positive integer and let T be the set { 1, 2, 3,
.
.
•
, n}. Let S,. be the set
of all permutations of T(that is, all bijections T-+ T). We shall use the same
.
notation
sueh functlons
as we d'd
1
m
1or
c
. s3• In '"'6•
(."� c
1or mstance,
.
.
(1
2 3 4 5 6
4 6 2 3 5
)
1
denotes the permutation that takes 1to4, 2 to 6, 3 to 2, 4 to 3, 5 to 5, and 6 to
1.
Since the composite of two bijective functions is bijective,IS,. is dosed under
the operation of composition. For example, in S6
(Remember that in composition of functions, we apply the right-hand function
first and then the left-hand one. In this case, for instance, 4--+ 3 --+ 2, as shown
tsinary operations are defined i n Appendix
B.
of the Norwegian mathematician
N.
*In honor
•see Appendix
H. Abel (1802-1829).
B.
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:DafUllWlrictims-..n:11t.
7.1
Definition and Examples of Groups
173
by the arrows.) We claim that Sn is a group under this operation. Composition
of functions is known to be associative, and every bijection has an inveNe func­
tion under composition.t It is easy to verify that the identity permutation
(
1 2 3
·
·
·
I 2 3
n). . .
IS the identity element of
n
group on n symbols. The order of Sn is n!
=
.
.
S,.. Sn IS called the symmetric
n(n
l)(n
- -
2) ...2.1(Exercise20).
EXAMPLE 4
The preceding example is easily generalized. Let T be any nonempty set, possi­
bly infinite. Let A(T) be the set of all permutations of T (all bijective functions
T-+ T). The arguments given above for Sn carry over to A(T) and show that
A(T) is a group under the operation of composition of functions (Exercise
12).
EXAMPLE 5
Think of the plane as a sheet of thin, rigid plastic. Suppose you cut out a square,
pick it up, and move it around, i then replac.e it so that it fits exactly in the cut-out
space. Eight ways of doing this are shown below (where the square is centered at
the origin and its comers numbered for easy reference). We claim that any mo­
tion of the square that ends with the square fitting exactly in the cut-out space
has the same result as one of these eight motions (Exercise 14).
All Rotations Are Taken Counterclo<kwise Around lh£ Center.
r0
===
rotation of 0°
4
4
3
r1
=
3
rotation of 900
3
4
2
2
tsee Appendix
B for details.
*Flip it, rotate it, turn it over, spin it, do whatever you want, as long as you don't bend, break, or distort it
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tm
174
Chapter 7
r2
Groups
=
rotation of 180°
2
4
,,
.....----....
3
3
2
r3
4
= rotation of 270°
4
�
3
2
2
4
3
d = reflection in the x-axis
2
4
d
.....----....
3
3
2
4
t = reflection in the y-axis
4
4
t
�
3
3
2
2
......
.......
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8om.1M11Bam:.ndkir�.Bdbmbll_...._
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7.1
Definition and Examples of Groups
175
h = reflection in line y = x
4
3
3
2
4
4
2
2
v ""
reflection in line y =
-x
4
v
3
3
2
If you perform one of these motions and follow it by another, the result will be
one of the eight listed above; for example,
4
4
3
2
2
t
If you think of a motion as a function from the square to itself, then the idea of follow­
ing one motion by another is just composition of functions. In the illustration ab<We
(h followed by r1 is t), we can write r1 oh = t (remember r1 oh means first apply h, then
apply ri). Verify that the set
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176
Chapter 7
Groups
equipped with the composition operation has this table:
ro
r1
r2
rJ
ro
ro
r1
r2
rJ
r1
r,
'2
r3
ro
r1
r2
r3
ro
r1
rJ
r3
ro
d
h
d
h
v
t
'•
v
d
h
t
h
v
d
h
v
h
h
v
d
ro
r3
'2
r1
ro
r3
r1
1)
'1
ro
r3
r1
r1
ro
v
d
h
v
r2
v
'2
d
t
d
d
h
v
v
d
h
'1
Clearly D4 is closed under 0, and composition of functions is known to be associative.
The table shows that r0 is the identity element and that every element of D4 has an
inverse. For instance,
r3 o r1
=
r0
h. D4 is called the dihedral group of degree 4 or
=
r1 o r3
.
Therefore, D4 is a group. It is not abelian
the group of symmetries of the square.
because, for example, hod:# do
EXAMPLE 6
The group of symmetries of the square is just one of
An analogous procedure can be carried out with any regular polygon of
many symmetry groups.
The resulting group Dn is called the dihedral group of degree n. The group D3, for
n
sides.
rotations about the center of O", 120°, and 240°; and the three reflections shown
example,
consists of the six symmetries of an equilateral triangle (counterclockwise
here), with composition of functions
as
the operation:
2214752
t
/
u
3� ��
�
l
Symmetry
/ ....---....
l
groups
l
arise
frequently
I
in
,/
art,
2
I
2
architecture,
and
science.
Crystallography and crystal physics use groups of symmetries of various
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7.1
Definition and Examples of Groups
177
three-dimensional shapes. The first accurate model of DNA (which led to the
Nobel Prize for its creators) could not have been constructed without a recogni­
tion of the symmetry of the DNA molecule. Symmetry groups have been used by
physicists to predict the existence of certain elementary particles that were later
found experimentally.
Groups and Rings
A ring R has two associative operations, and it is natural to ask if R is a group under
either one. For addition the answer is yes:
Theorem 7.1
Every ring is an abelian group under addition.
Proof.,. An examination of the first five axioms for a ring (in Section 3.1) shows that
they are identical to the five axioms for an abelian group, with the operation *
being +,the identity element e being 0R• and the inverse of
a being -a.
•
EXAMPLE 7
By Theorem
7.1, each of
the following familiar rings is an abelian group under
addition:
Z,
Z,,,
Q,
Matrix rings, such as
IR,
C;
M(IR) and M(Z-i);
Polynomial rings such as Z[x], R[x], and Z,,[x].
Hereafter, when we use the word "group" without any qualification in referring
to these or other rings, it is understood that the operation is addition.
Multiplication, however, is a different story:
A nonzero ring R is
nel'er a
group under multiplication.
If R has no identity, Axiom 3 fails. If R has an identity, then OR has no inverse and
Axiom 4 fails. Nevertheless, certain subsets of a ring with identity may be groups
under multiplication.
Theorem 7.2
The nonzero elements of a field F form an abelian group under multiplication.
Hereafter we shall denote the set of nonzero elements in a field F by F*.
Proof of Theorem 7.2 ... Multiplication in F* satisfies the following ring axioms:
6 and 11 (closure), 7 (associativity), 10 (identity), 12 (inverses), and 9
(commutativity)--see pages 44, 48, and 49. So F* satisfies group axioms
1-5 and, therefore, is an abelian group under multiplication .
•
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.
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178
Chapter 7
Groups
EXAMPLE 8
Theorem 7.2 shows that each of the following is an abelian group under
multiplication:
0*
the nonzero rational numbers;
R*
the nonzero real numbers;
C* the nonzero complex numbers.
EXAMPLE 9
If p is prime, then z, is a field by Theorems 2. 7 and 2.8. Therefore, z,* is a
group under multiplication by Theorem 7.2.
EXAMPLE 10
The positive rational numbers
O** form an infinite
abelian group under multi­
plication, because the product of positive numbers is positive, 1 is the identity
element, and the inverse of
a
is lja. Similarly, the positive reals
IR** form an
abelian group under multiplication.
EXAMPLE 11
The subset
{ l, -1, i, -j}
of the complex numbers forms an abelian group of
order 4 under multiplication. You can easily verify closure, and l is the identity
element. Since i(- i)
inverse since (-1)(-1)
=
=
1, i and -i are inverses of each other; -1 is its own
I. Hence, Axiom 4 holds.
EXAMPLE 12
Neither the nonzero integers nor the positive integers form a group under mul­
tiplication. Although 1 is the multiplicative identity for each system, no integers
except for ± 1 have a multiplicative inverse, so Axiom 4 fails. For example, the
equation 2x
=
1 has no integer solution, so 2 has no inverse under multiplica­
tion in the integers.
EXAMPLE 13
When n is composite, the nonzero elements of Z,, do not form a group under
multiplication because (among other things) closure fails. In Zc,, for instance,,
2
·
3
=
O and in Z'}J}, 4
·
5
==
0. Similarly if n = ts, then in Z,,, ts = 0.
A ring R with identity always has at least one subset that is a group under
multiplication. Recall that a unit in R is an element a that has a multiplicative inverse,
that is, an element u such that au = IR
""' v.a.
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...._._q-��.,._fld.__...,.dlN:t... �._.......--=-.c.a..� ...... dllllrigbtlD...,,,..��- .. --W......_,.:dPLl� ...... iL
7.1
Definition and Examples of Groups
179
Theorem 7.3
If R is a ring with identity, then the set
U of all
units in R is a group under
multiplication.*
Proof• The product of units is a unit (Exen:ise 15 in Section 3.2). so u is closed under
multiplication (Axiom 1). Multiplication in R is associative, so Axiom 2 holds.
Since 1R is obviously a unit, Uhas an identity element (Axiom 3).Axiom 4
holds in U by the definition of unit. Therefore, Uis a group.
•
EXAMPLE 14
Denote the multiplicative group of units in Z,. by U,.. Ac.cording to Theorem 2.10,
U,. consists of alla E Z,. such that (a, n)
= 1 (when a is considered as an ordinary
= {l, 3, 5, 7}, and the group of units
{1, 2,4, 7, 8, 11, 13, 14}. Here is the operation table for Ug:
integer). Thus the group of units in Z 8 is U8
inZ15 is U15
=
1
3
5
7
1
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
7
5
3
1
EXAMPLE 15
Examples 7 and
10 of Section 3.2,
and Exercise
17
of Section
3.2 show that
the
group of units in M( R ) is
GL(2,R)
=
{ (: �)I
wherea,b, e, d E Randad
-
be'#= 0
}•
which is called the general linear group of degree 2 over R. It is an infinite
nonabelian group (Exercise 7).
EXAMPLE 16
Examples 8 and
10 of Section 3.2, andExercise 17
of Section
3.2 show that
the
group of units in M(Z0 is
GL(2,Z2)
=
{ (: �)I
wherea,b, e, d E
the general linear group of degree
2
Z2 andad -
be-:/:: 0
}•
over Z2• It is a nonabelian finite group of
order 6 (Exercise 7).
*Theorem 7.2 is a special case ofTheorem 7.3 because the units in 11 field are the nonzero elements.
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180
Chapter 7
Groups
New Groups from Old
The Cartesian product, with operations defined coordinatewise, allowed us
str uct new rings from known ones. The same is true for groups.
to
con­
Theorem 7.4
LetG {with operatoi n*) and H {with operation o) be groups. Define an operation•
onGXHby
(g, h) • (g', h') = (g * g', h 0 h').
Then G x H is a group. If G and Hare abelian, then so si G x H. If G and H
are finite, thenso s
i G x Hand I G x H I = I G llH �
Proof ... Exercise 26.
•
EXAMPLE 17
Both Z and� are groups under addition. In Z X � we have (3, 5) • (7, 4)
(3+7, 5 + 4) = (10, 3). The identity is (0, 0), and the inverse of (7, 4) is ( -7, 2).
=
EXAMPLE 18
Consider Ill* X D,., where R* is the multiplicative group of nonzero real num­
bers. The table in Example 5 shows that
(2, r1) (9, v) = (2 9, r1 o v) = (18, d).
•
•
The identity element is (1, ro), and the inverse of (8, r3) is (1/8, r1).
• Exercises
A. I.
2.
Find the inverse of each permutation in S 3•
Find the multiplicative inverse of each nonzero element in
(a) Z3
3.
(a) Z18
4.
(b) Zs
(c)Z7
What is the order of each group:
(b)
D4
(c) S4
(d) Ss
(e) Uu
Determine whether the set G is a group under the operation*·
(a) G= {2, 4, 6, 8} in Z10; a* b = ab
(b) G=Z; a* b =a - b
(c) G={nEZlnisodd};a*b=a+b
(d) G= {2xlxE0}; a. b =ah
CapJriliM 2012c.upe.i...m.g.A:a� llMlnrld. �11Dtbea:ip.rd.11Cumd,,-ar�mwt1aMar1:apn.. o.11)��-mim.JIDl11t1D111Hm.mAJH�finm:l.m.111eom:.udkir�).Bdlorilf..._.._.
...... ..,.��dou.ad.........,-dild... -Cl'l'Mdl1-'Dliag�c.-g..p�----rlgbt1D....,,,.�Oldlllll:-..,.tia:MllE.-.....-i.._.� ...... it.
7.1
Definition and Examples of Groups
5. Find the inverse of the given group element.
Example
or
16 in Section
(b)
[Hint:
Example 8 in Section
181
3.2-
7.1.A-and Exercise 2.]
(! �)inzs
6. Give an example of an abelian group of order 4 in which every nonidentity
element a satisfies a*
7.
a= e.
[Hint: Consider Theorem 7.4.]
(a)
Show that the group
(b)
Show by example that the groups
GL(2, Z2) has
order
6 by listing all its elements.
GL(2, IR) and GL(2, Z2) are nonabelian.
8. Use Theorem 2.10 to list the elements of each of these groups: U4, U6, U1o.
U'}J.}, U30.
9. Write out the operation table for the group D3 described in Example
10. Show that G
=
{ (_: !) I
a,
b E IR, not both
0}
6.
is an abelian group under
matrix multiplication.
11. Consider the additive group Z2 and the multiplicative group
L
=
{±1, ± i} of
complex numbers. Write out the operation table for the group Z2 X
L.
T be a nonempty set and A(T) the set of all p ermutations of T. Show that
A(T) is a group under the operation of composition of functions.
12. Let
13. Give examples of nonabelian groups of orders 12,
16, 30,
and 48.
[Hint: Theorem 7.4 may be helpful.]
B. 14. Show that every rigid motion of the square (as described in the footnote at the
beginning of Example 5) has the same result as an element of
D4• [Hint: The
position of the square after any motion is completely determined by the location
of corner 1 and by the orientation of the square-face up or face down.)
15. Write out the operation table for the symmetry groups of the following figures:
<·>o \�\
(b)
16. Let 1, �
•=
(a)
j, k be the following matrices with complex entries:
G �).
i
=
G -�).
j
=
(- � �).
k
=
e �).
Prove that
i2 = j2 = k2 =
jk
(b)
(c)C><J
Show that set Q
=
=
-kj
=
{l, � -1,
i
-1
ij = -ji = k
ki
=
-ik
=
j.
-i, j, k, -j, -k} is a group under matrix
multiplication by writing out its multiplication table. Q is called the
quaternion group.
CnpJIWll2012.C.....,LAmag.AIRqliba-wd.lbJ"mtbll� �Ol'�:iawldm«ia:PKL 0.10�dalD,.-tinl��_,-119�fa:m:J.1119eBOOll:.ndiloc�:Mlmilil......- ...
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182
Chapter 7
Groups
17. If Gis a group under the stated operation, prove it; if not, give a counter
ple:
exam
(a) G
=
(b) G =
18. Let K
=
Q; a* b
=
a+
b+3
{rEOlr #= O};a * b
=
ab/3
{rE �Ir #= 0, r #= 1}. Let Gconsist of these six functions from K to K:
f(x)
i(x)
1
= 1-x
--
=
x
= x x- 1
g(x)
= x1
x
k(x) -x- 1
h(x)
--
j(x)
=
1- x
-
=
Is Ga group under the operation of function composition?
19. Do the nonzero real numbers form a group under the operation given by a* b
Ia I b, where I a I is the absolute value of a?
=
20. Prove that Sn has order nl. [Hint: There are n possible images for 1; after one
has been chosen, there are n
1 possible images for 2; etc.]
-
21. Suppose Gis a group with operation *· Define a new operation# on Gby
a# b
=
b *a. Prove that Gis a group under#.
22. List the elements of the group D5 (the symmetries of a regular pentagon ).
[Hint: The group has order 10.]
23. Let SIJ...2, R) be the set of all 2 X 2 matrices
andad - be
=
(: !)
such thata, b,
c,
dE �
1. Prove that SL(2, R) is a group under matrix multiplication.
It is called the special
linear group.
24. Prove that the set of nonzero real numbers is a group under the operation *
defined by
a*b
=
ifa > 0
ab
a/b
{
ifa <
0.
25. Prove that Ill* X Ris a group under the operation* defined by (a, b) * (c, ti)
(ac, be+ d).
=
26. Prove Theorem 7.4.
27. If ab
=
ac in a group G, prove that b
=
c.
28. Prove that each element of a finite group Gappears exactly once in each row
and exactly once in each column of the operation table. [Hint: Exercise 27 .]
29. Here is part of the operation table for a group G whose elements are a, b, c, d.
Fill in the rest of the table.
[Hint: Exercises 27 and 28.]
a
b
c
d
a
a
b
c
d
b
b
a
c
c
d
d
a
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7.1.A
30.
Definition and Examples of Groups
183
A partial operation table for a group G {e, a, b, c1 d, f} is shown below.
Complete the table. [Hn
i t: Exercises 27 and 28.]
=
e
a
b
e
e
a
a
a
b
b
e
b
b
c
c
d
d
f
f
c
d
f
c
d
f
d
f
a
31.
Let Tbe a set with at least three elements. Show that the permutation group
A(T) (Exercise 12) is nonabelian.
32.
Let T be an infinite set and let A(T) be the group of permutations of T
(Exercise 12). Let M {/E A(T)I f(t) 'I> t for only a finite number of tE 1}.
Prove that Mis a group.
=
33. If a, b ER
with a i= 0, let Ta,b:R-4 �be the function given by T,,p(x) = ax + b.
Prove that the set G { r,,,,, I a, b ER with a if= O} forms a nonabelian group
under composition of functions.
=
34.
C.35.
36.
Let H { T1,,, I b E �} (notation as in Exercise 33). Prove that His an abelian
group under composition of functions.
=
If/ES,,, prove that/k=/for some positiveintegerk, where/k means
fo f of o • • • of (k times) and I is the identity permutation.
Let G {O, 1, 2, 3, 4, 5, 6, 7} and assume G is a group under an operation*
with these properties:
=
(i)
*bs
a
(ii) a* a
=
a
+ b for all a,
b E G;
0 for all a E G.
Write out the operation table for G. [Hint: Exercises 27 and 28 may help.]
Ill
Def in it ion and Examples of Groups
NOTE: If you have Natl Section 7.1, omitthis section and begin Section 7.2.
A group is an algebraic system with one operation. Some groups arise from familiar
systems, such as Z, Z,,, the rational numbers, and the real numbers, by ignoring one
of their operations and concentrating on the other. As we shall see, for example, the
integers form a group under addition (but not multiplication) and the nonzero ratio­
nal numbers form a group under multiplication (but not addition). But many groups
do not arise from a system with t wo operations. The most important of these latter
� 2012C......,.l...amlq..A.-.JUeilll; a--1. M&Jaott» .:opi.d.-....t.or�iinwidmorlapmt. n.. ra �..-• .-titdpa11c�maJM....,._.-16xim.IM<1Bol*...slot.ai.p.t(1). F.dlari.ll_
.t.m.d._my...,,._...t�1*-aot�1111Kt--�i--.�ac.gq11Le..aag�1Mrigbt1D�mklll:ioma.�•..,-tim9if�dgbl.t�NqWel.t.
184
Chapter 7
Groups
groups (the ones that were the historical starting point of group theory ) developed
from the study of permutations.* Consequently, we begin with a consideration of
permutations.
Informally, a permutation of a set Tis just an ordering of its elements. For example,
there are six possible permutations of T { 1, 2, 3}:
=
123
132
21 3
31 2
231
321.
Each such ordering determines a bijective function from T to T: map 1 to the first
element of the ordering, 2to the second, and 3 to the third.t For instance, 23 1 de­
termines the function/:T�Twhose rule is/(1) 2;/(2) 3;/(3) 1. Conversely,
every bijective function from T to T defines an ordering of the elements , namely ,
f (1), /(2), /(3), Consequently, we define a permutation of a set T to be a bijective
function from T to T. This definition preserves the informal idea of ordering and
has the advantage of being applicable to infinite sets. For now, however , we shall
concentrate on finite sets and develop a convenient notation for dealing with their
permutations.
=
=
=
EXAMPLE 1
Let T
=
{ 1, 2, 3}, The permutation/ whose rule is/(1)
may be represented by the array
(� � �)
=
2, f 2
( )
=
3,/(3)
=
I
.in which the image under fof an
element in the first row is listed immediately below it in the second row. Using
this notation, the six permutations of Tare
G
G
2
2
2
3
2
2
3
�) G
�) G
2
2
2
Since the composition of two bijective functions is itself bijective, the composi­
tion of any two of these permutations is one of the six permutations on the list
�ove: For instance, if f
tion given by
=
G�D
(f 0 g)(l )
and g
=
/(g( l ))
=
G � !).
=
/(2)
=
(Jo g)(2) /(g(2)) /(1)
=
(f g)(3)
0
=
=
/(g(3))
=
/(3)
then/<> g is the func­
2
=
3
=
l.
*In the early nineteenth century, permutations played a key role in the attempt to find formulas for
solving higher-degree polynomial equations similar to the quadratic formula. For more information,
see Chapter
12.
tsijective functions are discussed in Appendix
..
B.
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7.1.A
Thus fog
=
(� � �)
Definition and Examples of Groups
185
.It is usually easier to make computations like this
by visually tracing an element's progress as we first apply g and then/;
( (-;---3-).- ,\o (
321
'-
1
r
\21
1
)
23
1
23
=
3
""-----
-...
) (
3
for example,
_,-'
If we denote the set of permutations of Thy S3, then composition of functions
( o ) is an operation on the set S3 with this property:
If /ES3
andgES3, then/0 gES3•
Since composition of functions is associative,• we see that
forallf,g,hES1•
(fog) oh =Jo(goh)
Verify that the identity permutation I =
Jof=f
G � :)
/o I=f
and
has this property:
for every /ES3•
Every bijection has an inverse function;* consequently,
iffE S3, then there exists gES3 such that
fog=/
For instance, if f
and
=
G � �)
.then g
(1
2
2
3
1
)(
3
1
0
goj=I.
and
3
=
2
l
G�D
because
1
) (
3
2
=
I
2
2
)
3 ·
3
You should determine the inverses of the o ther permutations in S3 (Exercise l).
Finally, note thatjog may not be equal to go f;
for instance,
but
•see Appendix B.
....
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186
Chapter 7
Groups
By abstracting the key properties of S3 under the operation
Definition
o,
we obtain this
A group is a nonempty set G equipped with a binary operation * that
satisfies the following axiomst:
1. Closure : lfaEGandbEG,thena*bEG.
2. Associativity: a* (b* c)
= (a* b) * c for all a, b,
CE G,
3. There 1s an element eEG (called the id ent ity element) such that
a ,. e = a = e *a tor every a E G.
4. For each a EG, there is an element d E G (called
such that a *
d=
e and
d *a = e.
A group is said to be abel ian"
the inverse ofa)
if it also satisfies this axiom:
5. Commutativity: a *b = b *afor al I a, b EG.
A group G is said to be finite (or of finite order) if it has a finite number of elements.
In this case, the number of elements in G is called the order of G and is denoted IG� A
group with infinitely many elements is said to have infinite order.
EXAMPLE 2
The discussion preceding the definition shows that S3 is a nonabelian group of
order 6, with the operation * being composition of functions.
EXAMPLE 3
The permutation group S3 is just a special case of a more general situation. Let
n be a fixed positive integer and let Tbe the set {1, 2, 3, . . . , n}. Let Sn be the set
of all permutations of T(that is, all bijections T ...+ T). We shall use the same
1or
sueh funct10ns
as we d"d
1
notation
•
J!.
.
m
.
1or mstance,
•
s3• In s.6, J!.
(1 2
3 4 5 6)
4 6 2 3 5 1
denotes the permutation that takes 1 to 4, 2 to 6, 3 to 2, 4 to§ 3, 5 to 5, and 6 to
I. Since the composite of two bijective functions is bijective, S,. is closed under
the operation of composition. For example, in S6
(I
2
3
5
;>-�---�--�- \'
) -p
�_
2
__ _
....1
�
(I
2
3
t
5
\,_�---��-.a�- 3
5
6)
1
=
__
_
�
.. -..___.... __ "'_"'_'"' - - - -- ---- � - ---
B.
*In honor of the Norwegian mathematician N.
"tsinary operations are defined inAppendix
ISeeAppendix
B for details.
(1
�--
2
4
3
4
-�-"'2
5
6)
3
H. Abel (1802-1829).
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�--mySllpllNSMdl�4'M8fld:�dll&t:tbl���c.g·Le..q�tM:rigbttl>__,..�eoime.•..,-tio»��·d,al&t�:M1_111nk.
7.1.A
Definition and Examples of Groups
187
(Remember that in composition of functions, we apply the right-hand function
first and then the left-hand one. In this case, for instance,
4 -+ 3 -+ 2, as shown
by the arrows.) We claim that S11 is a group under this operation. Composition
of functions is known to be associative, and every bijection has an inverse func­
tion under composition.t It is easy to verify that the identity permutation
G��
·
·
·
:)
is the identity element of S11• S,. is called the symmetric
group on n symbols. The order of S11 is n!
=
n(n - l)(n - 2)
•
.
.
2.1(Exercise 20).
EXAMPLE 4
The preceding example is easily generalized. Let The any nonempty set,
possibly infinite. Let A(T) be the set of all permutations of T(all bijective
functions T-+ T). The arguments given above for S,. carry over to A(T) and
show that A(T) is a group under the operation of composition of functions
(Exercise 12).
EXAMPLE S
Think of the plane as a sheet of thin, rigid plastic. Suppose you cut out a
square, pick it up, and move it around,f then replace it so that it fits exactly in
the cut-out space. Eight ways of doing this are shown below (where the square
is centered at the origin and its corners numbered for easy reference). We claim
that any motion of the square that ends with the square fitting exactly
in the
14).
cut-out space has the same result as one of these eight motions (Exercise
AIL Rotations Are Taken
r0
=
Counterclodcwise Around the Center.
rotation of 0°
4
4
'o
3
3
2
tsee Appendix
2
B for details.
*Flip it, rotate it, turn it over, spin it, do whatever you want, as long as you don't bend, break, or
distort it.
..
.
.....
...
� 2012�l....Mmdilllg.A.-.llieilll; a-Nd..Mqm11.t».:opl9d.-=umd.arda(lliclilmd.inwilal9ariapmt. DmtotllKlnmlc
.--.mff1M7c�m..,- a. --......- Mta:m-.e8odt..slar"�•). Edlmlll. •
dllllimd._mynw--1-1*-mll___.,.dKt
�'-uial;�c.glPLNmiiag...._ IMrigbttD�mldkiamf�•..,-tm.if--..-Dpu�-.Weit.
188
Chapter 7
r1
Groups
=
rotation of 90°
4
'1
�
4
2
2
r2
= rotation of 180°
4
2
,
,
�
3
3
4
2
r3
= rotation of 270°
4
�
3
2
2
4
3
d = reflection in the x-axis
4
2
d
�
3
2
3
4
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7.1.A
Definition and Examples of Groups
189
t =reflection in the y-axis
4
4
t
.
�
3
3
2
2
h = reflection in line y
= x
3
4
h
�
3
2
4
4
2
2
v
=reflection in line y = -x
4
v
3
2
3
If you perform one of these motions and follow it by another, the result will be
one of the eight listed above; for example,
4
3
2
4
2
'
Dm
....
� :!Ol2Gapf;ei....-lag.AiltitlD ll--1MaJ'ODtb9cupimd. � ar�:iill wtdm«laJIKl.
lo�opbl.-tinlpmlJ'e<D•m.J".,.•ppmmm fiom.11119e8om: ndloc".a.pm(1}.:EcalarUl. R1
�--...,...,...-�dl-.ad.......un, dlaclllM�i.mia&�o.g.. u..iag-- .. rigbl:ID__,.,.�c:romHlllllUIJlimllil:�dgtiu�:raq111N:k.
190
Chapter 7
Groups
If )'OU think of a motion as a function from the square to itself, then the idea of fol­
lowing one motion by another is just composition of functions. In the illustration
above (h followed by r1 is t), we can write r1 o h = t (remember r1 oh means first appl y
h, then apply ri). Verify that the set
D4
h,
"" {ro, ri, rz, r3,
v,
d,
t}
equipped with the composition operation has this table:
h
h
0
ro
'1
'2
r3
ro
'o
'1
'2
r3
d
d
'1
'1
'2
f3
'o
h
'2
r2
'3
ro
r1
t
f3
f3
ro
'1
'2
v
d
d
h
d
h
d
h
v
t
h
ro
r3
'2
v
t
'1
ro
r3
r2
d
h
v
r2
'1
ro
f3
d
'3
r2
r1
ro
v
t
d
h
v
t
v
v
t
v
v
d
h
t
r1
Clearly D4 is closed under o, and composition of functions is known to be associarive. The table shows that r0 is the identity element and that every element of D4 has
an inverse. For instance, r3 ° r1
because, for example, hod :I=
o:::
r0 = r1
°
r3
.Therefore, D4 is a group. It is not abelian
doh. D4 is called the dihedral group of degree 4 or the
group of symmetries of the square.
EXAMPLE 6
The group of symmetries of the square is just one of many symmetry groups. An
analogous procedure can be carried out with any regular polygon of
resulting group Dn is called the dihedral group of degree
n.
n
sides. The
The group D1, for ex­
ample, consists of the six symmetries of an equilateral triangle (counterclockwise
rotations about the center of O", 120", and 240"; and the three reflections shown
here and on the next page), with composition of functions as the operation:
����
r1�1 .4f· s
�
2
2
2
l
3
�
�
2
2
I
3
I
2
3
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�--__,.�-�._lld........,.dlcl.bl��..,..._..c-s.��-rigbtto__,..�l;lDlllslltlllll:aqtio:wil....... �� ....... k.
7.1.A
Definition and Examples of Groups
191
3� ��
,
t
3
1
2
u
/�
,,:'
2
�
2
2
I
Symmetry groups arise frequently in art, architecture, and science. Crystallography
and crystal physics use groups of symmetries of various three-dimensional shapes.
The first accurate model of DNA (which led to the Nobel Prize for its creators) could
not have been constructed without a recognition of the symmetry of the DNA mol­
ecule. Symmetry groups have been used by physicists to predict the existence of certain
elementary particles that were later found experimentally.
Systems with Two Operations
We now examine some familiar systems with two operations to
see
what groups arise
when only one of the operations is considered.
EXAMPLE 7
We now show that each of the following is an abelian group under addition,
that is, with the operation* in the definition of a group being+:
Z
0
the integers;
the rational numbers;
Zn
the integers mod n;
the real numbers;
R
C
the complex numbers.
That each system is closed under addition is a fact from basic arithmetic
(Axiom
I). Likewise, addition in each of these systems is associative: For any
three numbers a, b,
c,
a+ (b + c) = (a +b) +
[Additive form of Axiom 2]
c
In each system, the identity element is 0 because
a+O=a=O+a
Similarly, the inverse of a is
a+ (-a)= 0
and
-a
[Additive fonn of Axiom 3]
because
-a+ a= 0
[Additive form of Axiom 4]
Finally, each group is abelian because for any two numbers a and b,
a+b=b+a
[Additive form of Axiom 5]
Hereafter, when we use the word "group" without any qualification in refer­
ring to Z, Zn,
0, R, or C, it is understood that the operation is addition. When
it comes to multiplication, we have this basic fact:
None of Z, Zn,
0, IR, or C is a group under multiplication.
....
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192
Chapter 7
Groups
To be sure, each has 1 as its multiplicative identity element, but unfortunately
0 has no inverse--the equation Ox
=
l has no solutions---o
-s Axiom 4 fails.
Nevertheless, certain subsets of these systems may be groups under multiplication.
EXAMPLE 8
Each of the following is an abelian group under multiplication:
0*
the nonzero rational numbers; JR* the nonzero real numbers;
C* the nonzero complex numbers.
Each system is closed under multiplication because the product of nonzero num­
bers is nonzero (Axiom 1). Basic arithmetic tells us that multiplication is associa­
tive and commutative (Axioms 2 and 5). The identity element in each system is 1
because a· 1
=a =
1
·a (Axiom 3).
The inverse of a is
1/a (Axiom 4).
EXAMPLE 9
Letp be a prime, and consider the nonzero elements of Zp under multiplica­
tion. If a :f:. 0 and b :f:. 0, then ab :f:. 0 by condition (3) of Theorem 2.8, so
closure holds (Axiom 1). The identity element is 1(Axiom3) and inverses exist
by condition
(2) of Theorem 2.8(Axiom 4) . Multiplication is associative and
2. 7 (Axioms 2 and 5). So the nonzero elements of ZP
commutative by Theorem
form an abelian group under multiplication.
EXAMPLE 10
Each of
0...
the positive rational numbers
and
R•• the positive real numbers
is an abelian group under multiplication. Both systems are closed under multi­
plication since the product of positive numbers is positive. The identity element
is 1 and the inverse of a is 1/a.
EXAMPLE 11
The subset L
=
{ 1, -1, i, -() of the complex numbers forms an abelian group
under multiplication. You can easily verify that closure holds and that 1 is the
identity element. Since i(-i)
=
-P
=
-(-1)
=
1, we see that i and -i are inverses
of each other;-1 is its own inverse since (-1)(-1)
=
1. Hence, Axiom 4 holds.
EXAMPLE 12
Neither the nonzero integers nor the positive integers form a group under multiplica­
tion.Although 1 is the multiplicative identity for each system, no integers except tor
± 1 have a multiplicative inverse, soAxiom 4 fails. For example, the equation 2x
=
1
has no integer solution, so 2 has no inverse under multiplication in the integers.
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7.1.A
Definition and Examples of Groups
193
EXAMPLE 13
When
n
is composite, the nonzero elements of Z,. do not form a group under
multiplication because (among other things) closure fails. In z6, for instance,
2
·
3
=
0 and in Zw, 4 5
·
=
0. Similarly if
n =
rs, then in Z11,
rs
=
0.
EXAMPLE 14
Let U11 be the set of units in Z.,. * By Exercise
17 of Section 2.3, the product of
two units is a unit, so U11 is closed under multiplication (which is known to be
associative and commutative). The identity
1 is a unit since l
•
1
=
1. So U11
is an abelian group under multiplication. By Theorem 2.10, U,. consists of all
a EZ11 such that (a, n) = 1 (when a is considered as an ordinary integer). Thus,
the group of units in Zs is U8
U15
=
{ l , 2,
{1, 3, 5, 7}, and the group of units in Z15 is
4, 7, 8, 11, 13, 14}. Here is the multiplication table for U8:
=
3
5
3
5
7
1
7
5
7
3
3
5
5
7
1
3
7
7
5
3
1
The next example involves matrices.t A 2 X 2 matrix over the real numbers, is an
array of the form
where a, b,
c,
dare real numbers.
Two matrices are equal provided that the entries in corresponding positions are
equals, that is,
if and only if
a = r, b = s, c = t, d = u.
For example,
0)=(2+2
0)
1
1 '
1- 4
but
G
Matrix multiplication is defined by
x)=(aw+by ax+bz)
.
z
cw+dy cx+dz
·
•Recall that an element a in Z,. is a unit if the equation ax=
1
has a solution (that is, if a has an inverse
under multiplication).
tit you
have taken a course in linear algebra, you can skip this paragraph.
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194
Chapter 7
Groups
-
For example,
(� -!)(!
)
....,5
1
(
21 + 3
•
o
t+
•
•
6
(-4)6
)
2( -5) + 3 7
o( �s) + (-4)7 •
(
Reversing the order of the factors in this product produces
) (l
3
.
-5)(2
7
0
-4
=
•
2
+ (-5)0 1
6. 2 + 7.
3 +
•
6 3+
0
)
( -5)( -4 )
•
7(-4)
=
- 20
24
(2
_!!).
.-
2
12
)
3
-10.
So matrix multiplication is not commutative. A straightforward (but tedious) compu­
tation shows that matrix multiplication is associative. It's easy to verify that
G �)(: !) e �) C �)G �).
=
==
Hence,
(
1 0).
0
1
is the .1dent1ty
. el ement.
EXAMPLE 15
We shall show that the set of matrices
{(: ! ) I
where
a, b, e, dER and ad - be :,t: O}
is a group under multiplication, called the general linear group of degree 2 over IR
and denoted GL(2, Ill). T he discussion before the example shows that GL(2, IR)
(
has associative multiplication and an identity element
)(
-be #: 0,
readily verify that when ad
(ac b) ad�be ad-_bbe
d -c
a
ad-be ad-be
=
1
So every matrix in GL(2,
-
0)
and
01
(
(Axioms 2 and 3). You can
)
ad�be ad--.}!be (a bJ
a
cd
ad-be ad-be
-c
=
(1
0)
01·
Ill) has an inverse (Axiom 4).
-
To finish the proof, we need only show that GL(2, ll) is closed under multiplication
(; :) in GL(2, Ill), that ad-be * 0
and wz
xy #: 0, and hence, (ad- be)(wz xy) ::/: 0. To prove that
(a b )(w x) = (aw+ by ax+ bz)
e d y z cw+dy cx+dz
(Axiom
1). Suppose that
(:
!
)
and
are
so
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7.1.A
Definition and Examples of Groups
is in GL(2, R), we must provethat(aw+by)(cx+dz)-
195
(ax+bz)(cw+dy)-:# 0.
Verify that
(aw + by)(cx + dz) - (ax +bz)(cw+ dy) = (ad - bc)(wz - xy) -:/= 0.
So the product matrix is inGL(2,
R). Therefore, GL(2, �)
plication and is a group, which is nonabelian (Exercise 7).
is closed under multi­
The discussion preceding Example 15 carries over to matrices whose entries are in
systems other than the real numbers, such
as
0, C, and 7L1(withp prime).
EXAMPLE 16
We shall show that
GL(2,
lL-}) =
{(: �)I
where
a, b, c, dEZ2 and ad - be-:/=
o}.
the general linear group of degree 2 over Z2, is a group under multiplication.
Matrix multiplication is associative, and the identity matrix is ob viously in
GL(2, Z2). The proof that GL(2, 7L2) is closed under multiplication is identical
to the oneforGL(2, �) in Example 15. If A=
in
Z2,
(
d(ad -
ad -
so
-c(ad
be
bc)-1
_
bc)-1
has an inverse
by
- bcJ""'L)
a(ad bc -l
-b(ad
)
_
(: !)
EGL(2,Z2) , thenad-bc-:/=O
Example 9. Verify that the inverse of
A
is
" .
.
.
.
.
m
i gi
, whJch JS th e same mverse matr x ven
Example 15 , with a change of notation:
(ad- bcJ"J
in place of ad
� be. Hence,
GL(2, Zll is a group. It is a finite nonabelian group of order 6 (Exercise
7).
New Groups from Old
The Cartesian product G X Hof sets Gand His defined on page 512 of Appendix B.
Theorem
7.4
on the next page shows that the Cartesian product can be used to pro­
duce new groups from known ones.*
7.1 and assume that you have read Chapter 3, so they are
7.1.A. However, many of the preceding examples are special cases of
these theorems: Example 1 is a special case of Theorem 7.1; Examples 8 and 9 are special cases
ofTheorem 7.2; and Examples 14-16 are special cases ofTheorem 7.3. So you haven't missed
anything crucial for this chapter. You may wish to read Theorems 7.1-7.3 at a later date, after you
have read Chapter 3.
*Theorems
7 .1-7.3
appear in Section
not included in Section
..
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196
Chapter 7
Groups
Theorem 7.4
Let G (with operation*) and H(with operation o) be groups. Define an opera­
tion • on G x H by
(g, h). (g', h') = (g * g', h <> h').
Then G x His a group . If G and Hare abelian, then so is
are finite, then so is G x Hand IG x HI = �llH�
Proof� Exercise 26.
G X
H.
If G
and H
•
EXAMPLE 17
Both Zand "4 are groups under addi tion. In Z X "4 we have (3, 5) • (7, 4)
(3 + 7, 5 +
4) =
(10, 3). The identity is
(0, 0), and the inverse of
=
(7, 4) is ( -7, 2).
EXAMPLE 18
Consider IR* X D4, where IR* is the multiplicative group of nonzero real num­
bers. The table in Example 5 shows that
(2, r1) • (9, v) = (2 9, r1 o v) =
•
The identity element is (1,
r0 ,
)
(18,
d).
and the inverse of (8, r3) is (1/8, r1).
• Exercises
The exercises for this section are the same as those for Section 7.1-see page 180.
Ill
Basic Properties of Groups
Before exploring the deeper concepts of group theory, we must develop some additional
terminology and establish some elementary facts. We begin with a change in notation.
Now that you
are
comfortable with groups, we can switch to the standard multi­
plicative notation. Instead of
a *
b, we shall write ab when discussing abstract groups.
However, particular groups in which the operation is addition (such
as
Z) will still be
written additively.
Although we have spoken of the inverse of an element or the identity element of a
group, the definition of a group says nothing about inverses or identities being unique.
Our first theorem settles the question, however.
Theorem 7.5
Let
G
be a group and let a, b, c E G. Then
{1}
G has a
unique identity element.
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7.2
Basic Properties of Groups
197
(2) Cancelation holds in G:
If ab = ac, then
b=
if ba = ca, then
c;
b = c.
{3) Each element of G has a unique inverse.
Proof• (1) The group G has at least one identity
e and e'
by the definition of a group. If
are each identity elements of G, then
ee' = e
ee' = e'
[Because e' is an identity element.]
[Because e is an identity element.]
Therefore,
e=
'
ee
= e',
so that there is exactly one identity element.
(2) By the definition of a group, the element a has at least one inverse
dsuch that
da =e =ad.
If ab=ac, then d(ab) =d (ac) . By associativity
and the properties of inverses and identities,
(da)b= (da)c
eb = ec
b =c.
The second statement is proved similarly.
(3) Suppose that d and d'
so that
d=
are
d' by (2). Therefore
both inverses of a E G. Then
ad= e= ad',
a has exactly
•
Hereafter the unique inverse of an element
a
one inverse.
in a group will be denoted
a-1• The
uniqueness of a-1 means that
whenever ay
= e =
ya, then y
=
-1
a •
Corollary 7.6
If G is a group and a,
(1)
b E G, then
(abr1 = b-1a-1:
(2) (a-1;-1 =a.
Note the order of the elements in statement
inverse of
ab
as
a-1b-1,
(1). A common
mistake is to write the
which may not be true in nonabelian groups. See Exercise 2
for an example.
Proof of Corollary 7.6 .. (1) we have
(ab)(b-1a-1) = a(bb-1)a-1 = a ea-1 = aa-1 = e
( b-1a-1)(ab)= e. Since the inverse of ab is unique by
b-1a-1 must be this inverse, that is, (ab)-1 =b-1a-1•
and, similarly ,
Theorem 7.5,
(2) By definition, a-1a=e and (a-')(a-1 )-1 =e, so that
a-1a = a-•ca-1}1• Canceling a-1 by Theorem 7 5 shows that
a= (a-1}1. •
.
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.
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...... it.
198
Chapter 7
Groups
Let G be a group and let
integer n,
a E G. We
define
a•=aaa .. ••a
We also define
a2 = aa, a3 = aaa,
and for any positive
(n factors).
JI = e and
(n factors).
These definitions
are
obviously motivated by the usual exponent notation
in
R and
other familiar rings. But be careful in the nonabelian case when, for instance, (ab'j' may
not be equal to d'll'. Some exponent rules, however, do hold in groups:
Theorem 7.7
Let G be a group and let a E G. Then for all m, n in Z,
aman = am+n
Proof.. The proof
case
(amt = amn.
and
consists of a verification of each statement in each possible
(m 2:!:: 0, n 2:!:: O; m 2:!:: 0, n
(Exercise 21).
< O;
etc.) and is left to the reader
•
NOTE ON ADDITIVE NOTATION: To avoid confusion, the operation in cer­
tain groups must be written as addition (for example, the additive group of real
numbers since multiplication there has a completely different meaning). Here is
a dictionary for translating multiplicative statements into additive ones:
Multiplicative
Notation
Operation:
ah
Identity:
e
-1
a
Inverse:
Exponents:
•
Theorem 7.7:
Order of
d'=aa
a (n factors)
a-•=a -1
a-1
d"d' = d"+"
(d"'/' = d""
an
•
•
•
•
•
Additive
Notation
a+b
0
-a
na=a+ a+ +a (n summands)
-a
(-n)a= -a - a (ma) + (na) = (m + n)a
n(ma) = (mn)a
·
·
·
·
·
·
Element
We return now to multiplicative notation for abstract groups. An element a in a group
is said to have finite order if
of the element
a
d<= e for some
positive integer k. * In this case, the order
is the smallest positive integer
"In additive notation, the condition is ka
=
n
such that d' = e. The order of
a is
0.
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� ...... k
7.2
denoted lal. An element
a
B a si c Properties of Groups
is said to have infinite order if
d'
*
e
199
for every positive
integerk.
EXAMPLE 1
In the multiplicative group of nonzero real numbers, 2 has infinite order
because 2k *
1 for all k � 1. In the group L
=
{ ± 1, ± i} under multiplication
i2 = 1, ;3 = - i, and i4 = 1.
of complex numbers, the order of i is 4 because
Similarly,
I-ii=
2
G
1
4. The element
�Y=G
G � �)in
2
�
S3 has order 3 because
2
and
3
The identity element in a group has order
2
1.
EXAMPLE2
In the additive group Zm the element 8 has order 3 because 8 + 8
=
4 and
8 + 8 + 8 = 0.
In the multiplicative group of nonzero real numbers, the element 2 has infinite
-3 ° 5
order and all the powers of 2 (2 , 2 , 2 , etc.) are distinct. On the other hand, in the
multiplicative group L =
{±1, ±i},
the element i has order 4 and its powers are not
distinct; for instanoe,
and
Observe that ;10 =
i2 and 10 =
2 (mod 4). These examples are illustrations of
Theorem 7.8
Let
G
be a group and let a E G.
(1)
If a has infinite order, then the elements a1r, with k EZ, are all distinct.
(2)
If a'=
<I with i * j, then a has finite order.
Proof " Note first that statement (1) is true if and only if statement (2) is true,
because each statement is the contrapositive of the other, as explained
on pages 503-504 of Appendix A. So we need only prove one of them.
We shall prove statement (2):
Suppose that
a'= al, with i > j. Then multiplying both sides by a-J
a1-1 = d' = e. Sinoe i - j > O, this says that a has finite
shows that tt-J =
order.
•
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200 Chapter 7
Groups
Theorem 7.9
Let
G be a group and a E G an element of ft nite order n. Then:
(1)
ti =
e if and only if n
I k;
(2) ti = a1 if and only if I = j {mod n);
(3) If n =
td, with d <i?: 1, then a1 has order d.
Proof"' (1) If n divides k, say k = nt, then ti= d11 = (d')t = i = e. Conversely,
suppose that cl = e. By the Division Algorithm, k = nq + r with
0s
r < n. Consequently,
e = a"= fi'4+• = a1'"a' = (a'')'1a' = e'la' = ea'= a'.
By the definition of order, n is the smallest positive integer with d' = e.
Since r< n, a' = e can oc.cur only when r = 0. Thus, k = nq + 0 and n
dividesk.
(2) First , note that
a1 = a1 if and only if ,;-1 = e. [Proof: if d = al,
then a1-J = e by the proof of Theorem 7 .8(2). Conversely, if a1-1 = e,
then multiplying both sides by al shows that d = al.] But by (1) , with
k = i - j, we have a1-J = e if and only if n I (i - f), that is, if and only
if i !!!:; j (mod n). Therefore, a1 = al if and only if i = j (mod n).
(3) Since la l = n, we have (a'd = a"'= d' = e. We must show that dis the
smallest positive integer with this property. If k is any positive integer s uch
that
(a'"= e, then a11' = e. Therefore, n I tk by part (1), say tk = nr
Hence, k = dr. Since k and dare positive and d I k, we have d s k.
=
(td)r.
•
Corollary 7. 1 O
Let
G be an abelian group in which every element has finite order. If CE G is
G), then the order
an element of largest order in G (that is, I a I s I c I for al I a E
of every element of G divides I c I
•
·
For example, (1,
0) has order 4 in the additive abelian group� X Z2 and every other
element has order 1, 2, or 4 (Exercise 1O(b)). Thus (1, 0) is an element of largest possible
order, and the order of every element of the group divides 4, the order of (1, 0).
ProofofCorollary 7.10 ... Suppose, on the contrary, that a E G and lal does not
divide 14 Then there must be a prime p in the prime factorization of the
lal that appears to a higher power than it does in the prime fac­
14 By prime factorization we can write lal as the product
of a power of p and an integer that is not divisible by p and similarly for
c. Thus there are integers m, n, r, s such that lal = p'm and lcl = p'n, with
(p, m) = 1 = (p, n) and r > s. By part (3) of Theorem 7.9, the element d"
has order p' and cP' has order n. Exercise 33 shows that a"'cP' has order
p'n. Hence, la"'c''I
p'n > p'n
lcl. contradicting the fact that c is an
element of largest order. Therefore, lal divides 14 •
integer
torization of
=
=
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7.2
Basic Properties of Groups
201
• Exercises
NOTE: Unless stated otherwise, G is a
A. 1.
2.
If
e = c in a group, prove that c =
Let a
G � �)
=
and h
""
4. If a, b E Gand ab =
e.
G � �)
3. If a, h, c, d E G, then (abcdf1
5.
group with identity element e.
in S3. Verify that (ab)-1 * a-1&-1.
=?
e, prove that ba = e.
LetfG4 Gbegiven byfi:a)
=
a-1•
Prove that/is a bijection.
2
6. Giveanexample of agroupin which the equationx = ebasmorethan two solutions.
7. Find the order of the given element.
(a) 5 in
(b)
(c)
(d)
U8
(1
)
C -Din
(=i -D
2
2 3 4 5 6 7
3 7 5 I 4 6
GL(2 ,
.
m
S,
ll!)
in GL(_2,
�)
8. Give an example of a group that contains nonidentity elements of finite order
and of infinite order.
9.
10.
(a)
Find the order of the groups U10, U12, and Uu.
(b)
List the order of each element of the group U'11.1.
Find the order of every element in each group:
(a) �
11.
(b) Z4
X
(c)
Z2
S3
(d) D4
Let Gbe an additive group. Write statement
(l}-(3) of
(e) Z
(2) of
Theorem
a, b E Gand n is any integer, show
that (aba-1'f
aH'a-1•
12.
If
13.
If Gis a finite group of order n and a E G, prove thatlal s n.
n+
1
7.8 and statements
Theorem 7.9 in additive notation.
elements e =ti' a,
a2, a3,
..
=
[Hint: Consider the
. , d'. Are they all distinct?] Thus every element
in a finite group has finite order. The converse, however, is false; see Exercise
25
in Section 8.3 for an infinite group in which every element has finite order.
14.
True or false: A group of order n contains
answer.
15. (a)
16.
If
a E Gand a12 =
an
element of order n. Justify your
e, what order can a possibly have?
= e
(b)
If e * b E Gand bP
(a)
If a E Gand lal = 12, find the orders of each of the elements a, a2, a3, •
(b)
for some printe p, what is lbl?
. .
, a11•
Based on the evidence in part (a), make a conjecture about the order of
d'
whenll
a = n.
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202 Chapter 7
17.
Groups
(a)
Let
a,
bE G. Prove that the equations ax = band ya
= beach have a
unique solution in G. [Hint: Two things must be done for each equation:
First find a solution and then show that it is the only solution.]
Show by example that the solution of ax = bmay not be the same as the
(b)
solution of ya
18.
Let G
= b. (Hint: Consider S3.]
{a1, �
, an} be a finite abelian group of order n. Let x = aa
1 i
= e.
=
• • •
· · ·
a,..
Prove that i2
19. If
20.
a, bE G, prove that lbab-11 = lal.
(a)
Show that
a
has order4.
(b)
=
(_� -!)
has order 3in GL(2, R) and b =
(� �)
-
Show that abhas infinite order .
B.21. ProveTheorem7.7.
22. Let G =
{e, a,
b} be a group of order 3. Write out the operation table for G.
[Hint: Exercise28 in Section 7 .l.]
23. Let G be a group with this property: If
a,
b,
cE G and ab
o::o
ca, then b =
c.
Prove that Gis abelian.
24. If (ab)2
=
cl-b2 for all a, b, E G, prove that G is abelian.
25. Prove that Gis abelian if and only if ( ab'r 1
= cc 1 b- 1 for all a, bE G.
26. Prove that every nonabelian group G has order at least 6; hence, every group
of order 2, 3, 4, or 5 is abelian. [Hint: If
elements of the subset H =
cl- it
Hor cl- =
a, bE Gand ab* ba, show that the
{e, a,
b, ab, ba} are all distinct . Show that either
e; in the latter case, verify that aha it H].
27. If every nonidentity element of Ghas order 2, prove that Gis abelian.
[Hint: lal
28. If
=
2if and only if
a
:i:
e
and a
= a-1• Why?]
a E G, prove that lal = la-11.
29. If a, b, c E G, prove that there is a unique element
x E G such that axb =
c.
labl = lbal.
31. (a) If a, hE Gandab = ba, prove that (ab)lallbl = e.
30. If
a,
(b)
bE G, prove that
Show that part (a) may be false if ab*
ba.
32. If IGI is even , prove that Gcontains an element of order 2. [Hint:The identity
element is its own inverse. See the hint for
33. Assume that a, bE Gand ab
ab has order
=
ba. If
Exercise27].
lal and lhl are relatively prime , prove that
lallhl· [Hint: See Exercise 31].
34. Suppose Ghas order 4, but contains no element of order4.
(a)
Prove that no element of Ghas order 3. (Hn
i t: If l&I
of
four distinct elements g, g2, g1
=
e,
= 3, then G consists
d. Now gd must be one of these four
elements. Show that each possibility leads to a contradiction.]
(b)
Explain why every nonidentity element of Ghas order 2.
(c)
Denote the elements of Gby e, a, b,
c and
write out the operation table for G.
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7.3
Subgroups
203
35. If a, b E G, b6 = e, and ab = b4a, prove that b3 = e and ab = ba.
36. Suppose a, b E G with lal = 5, b * e, and aba-1 = fl-. F ind lbl.
37. If (ab)3 = a3b3 and (ab)5 = a5b5 for all a, b E G, prove that G is abelian.
C. 38. If (ab)1 = db1 for three consecutive integers i and all a, b E G, prove that G is
abelian.
39. (a) Let G be a nonempty finite set equipped with an associative oper ation
such that for all a, b, c, d E G:
if ab
"" ac, then b = c and if bd = cd, then b
=
c.
Prove that G is a group.
(b) Show that part (a) may be false if G is infinite.
40. Let G be a nonempty set equipped with an associative operation with these
properties:
(i) There is an element e E Gsuch that ea
(ii)
For each
= a for every a E G.
a E G, there exists d E Gsuch that da = e.
Prove that Gis a group.
41. Let Gbe a nonempty set equipped with an associative operation such that ,
for all
a, b E G, the equations ax = b and ya = b have solutions. Prove that G
is a group.
m
Subgroups
We continue our discussion of the basic properties of groups, with special attention
to subgroups.
Definition
A subset H of a group G is a subgroup of G if H is itself a group under the
operation in G.
{e}, which is
proper subgroups.
Every group Ghas two subgroups: Gitself and the one-element group
called the
trivial subgroup. All other subgroups are
said to be
EXAMPLE 1
The set IR* of nonzero real numbers is a group under multiplication. The group
R** of positive real numbers is a proper subgroup of IR*.
EXAMPLE 2
The set Z of integers is a group under addition and is a subgroup of the additive
group
IQ of
rational numbers.
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204
Chapter 7
Groups
EXAMPLE 3
The subset L
=
{1,-1, i, -i} of the complex numbers is a group under multipli­
cation.* So it is a subgroup of IC*, the multiplicative group of nonzero complex
numbers.
EXAMPLE 4
Recall that the multiplicative group of units in Z8 is U8
=
{1, 3, 5, 7}. The
upper-left quarter of its operation table in Example 14 of Section 7.1or
Section 71.A shows that the subset { 1, 3} is a subgroup of
U8•
EXAMPLE 5
The upper-left quarter of the operation table for D4 in Example 5 of Section 7 .1
or 7.1.A shows that H=
{r°' r1, r2, r3} is a subgroup of D4•
EXAMPLE 6
In the additive group� X Z.,., let H = {(O, 0), (3, 0), (0 , 2), (3, 2)}. Verify that
His a subgroup by writing out its addition table.
When proving that a subset of a group is a subgroup, it is never necessary to check asso­
ciativity. Since the associative law holds for all elements of the group , it automatically holds
when the elements are in some subset H. In fact , you need only verify two group axioms:
Theorem 7.11
A nonempty subset Hof a group G is a subgroup of G provided that
(i) if a, b EH, then ab EH; and
(ii) ifaEH, then a-1 EH.
Proof ...
Properties (i)and (ii)are the closure and inverse axioms for a group.
Associativity holds in H, as noted above.
e EH.
Thus we need only verify that
c EH. By (ii), c-'1 EH,
Since His nonempty , there exists an element
and by (i)cc-1
= e
is in H. Therefore His a group.
•
EXAMPLE 7
Let Hconsist of
1• 1
-
b
·
0
=
all 2 X 2 matrices of the form b
=
(� �)
with b ER. Since
1, His a nonempty subset of the group GL(2 ,
Ill), which was
"See Example 11ofSection7.1 or Section 7.1.A.
..
..
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7.3
defined in Example
15 of
Subgroups
205
Section 7.1 or7.1.A. The product of two matrices in
His in Hbecause
The inverse of.
·
subgroup of
1
0
-h
1
(1 h) ( )
0
1
IS
·
•
· a
wh"ICh IS also m
HIS
· H. There1ore,
"
·
GL(2, R) by Theorem7.11.
When His finite, just one axiom is sufficient to guarantee that His a subgroup.
Theorem 7.12
Let H be a nonempty finite subset of a group
tion in G, then His a subgroup of
Proof "
G.
G. If His closed under the opera­
By Theorem 7.11, we need only verify that the inverse of each element
of His also in H. If
,
aEH then closure implies that akE H for every
positive integer k. Since His finite, these powers cannot all be distinct.
So a has finite order n
by Theorem 7.8 and d' = e. Sincen
-
1
== -1
(modn), we haved'-1 = a-1byTheorem 7.9. Ifn > 1, thenn -
positive and a-1=
that a-1 is in H.
d' -l is in H. If n=
1, then
1 is
a= e and a-1 = e= a, so
•
EXAMPLE 8
Let Hconsist of all permutations in S5 that fix the element 1. In other words,
H= {/E S5If{))= l}. His afinite set since S5 is a finite group. If g, h EH,
then g(l)= 1 and h(l)= 1. Hence, (go h)( 1) = g(h( 1))= g(l)= 1. Thus
g ah EH and His closed. Therefore, His a subgroup of S5 by Theorem7 .12.
The Center of
If
G is a
a
Group
group, then the center of
G is
the subset denoted Z(G) and defined by
Z(G) = {aE GI ag= ga for every gEG}.
In other words,
an element of G is in Z(G) if and only if it commutes with every
G. If G is an abelian group, then Z(G)= G because all elements commute
with each other. When G is nonabelian, however, Z(G) is not all of G
element of
EXAMPLE 9
The center of S3 consists of the identity element alone because this is the only
element that commutes with every element of S3 (Exercise
.......
25).
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206
Chapter 7
Groups
EXAMPLE 10
l or 7 .1.A shows that r1
r1 o r3 =r3 o r 1 ). However,
with every element of D4 because r1 o d <#: d• rp Hence,
The operation table for D4 in Example 5 of Section 7
.
commutes with some elements of D4 (for instance,
it does not commute
r
1
is not in Z(D4) nor is d. Careful examination of the table shows that
Z(D4) =
{ro. r2}
since these are the only elements that commute with every
element of D4• It is easy to verif y that
{r0, r2}
is a subgroup of D4• This is an
example of the following result.
Theorem 7.13
The center Z{G) of a group G is a subgroup of G.
Proof " For every g E G, we have eg =g =ge. Hence, e E Z( G) and Z(G) is non­
empty. If a , b EZ(G), then for any g E G we have ag = ga and bg = gb,
so that
(ab)g = a(bg) = a(gb) = (ag)b =(ga)b =g(ab).
Therefore,
ab EZ(G). Finally,
if
a E Z(G) and g E G, then ag =ga.
a-1
Multiplying both sides of this equation on the left and right by
shows that
1
a-1(ag)a-1 =a- (ga)a-'
ga-• =a-lg
Therefore,
a-1 EZ(G) and Z(G) is a
subgroup by Theorem 7.11.
•
Cyclic Groups
An important type of subgroup can be constructed
a E G, let {a) denote the set of all
powers of
as
follows.
If G is
a group and
a:
{a)= { . . . , a-3, a-2, a�1, d1, a 1, a2, . . } =
.
{a" In EZ}.
Theorem 7.14
If G is a group and
a E G, then {a}= {an In EZ} is a subgroup of G.
Proof " The product of any two elements of {a) is also in {a) because a'al = cf+/.
The inverse of d' is a -k, which is also in {a). By Theorem 7.11, (µ} is a
subgroup of
G.
•
cyclic subgroup generated by a. If the subgroup {a} is the
G is a cyclic group. Note that every cyclic group is abelian
The group (µ}is called the
entire group G, we say that
since dal = a1+J
..
=ala'.
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7.3
Subgroups
207
EXAMPLE 11
The multiplicative group of units in the ringZ15 is U15 = {1, 2, 4, 7, 8, 11, 13, 14}
by Theorem 2.10. In order to determine the cyclic subgroup generated by 7, we
compute
71 = 7
72 = 4
73 = 13
'74 = 1 = 7°.
Therefore, the element 7 has order 4 in lfi.5• We claim that the cyclic subgroup
(?}consists of {7°, 71, 72, 73} = {1, 7, 4, 13}. [Proof' By definition, every ele­
ment of (7}is of the form 71 for some integer i. Since every integer is congruent
modulo 4 to one of 0, 1, 2, 3, the element 71 must be one of 7°, 71, 72 or 73 by
Theorem 7.9(2).] Hence, (7}= {1, 7, 4, 13}. Thus, the cyclic subgroup (7)has
order 4-the order of the element 7 that generates the group.
EXAMPLE 12
Different elements of a group may generate the same cy clic subgroup. For
instance, verify that 13 has order 4 in U 5• Then the same argument used in
1
Example 11 shows that the cyclic subgroup (13}= {136, 131, 132, 133} =
{1, 13, 4, 7} = (7).
The argument used in Examples 11 and 12 \VOrks in general and provides the con­
nection between the two uses of the word "order". It states, in effect, that the order of
an element a is the same as the order of the cyclic subgroup generated by a.
Theorem 7.15
Let G be a group and let a E G.
{1)
If a has infinite order, then (a} is an infinite subgroup consisting of
the distinct elements fl, with
k EZ.
(2) If a has finite order n, then (a} is a subgroup of order n and (a) =
{e
Proof"
=
o 1 2 3
a , a , a , a , . .. 'a"-i}.
(1) This is an immediate consequence of part (1) of Theorem 7.8.
(2) Let a' be any element of (a). Then i is congruent modulo n to one
of O, 1, 2, . . . , n - 1. Consequently, by part (2) of Theorem 7 .9, d must be
equal to one of a0, a1, a?-,
, d'- 1• Furthermore, no two of these powers
• • •
of a are equal since no two of the integers 0, 1, 2, ... , n - 1 are congruent
modulo n. Therefore, (a)= {a", a1, a2,
, a"-1} is a group of order n. •
• • •
NOTE ON ADDITIVE NOTATION: When the group operation is addi­
tion, then, as shown in the dictionary on page 198, we write ka in place
of d'. So the cyclic subgroup(µ)= {na I nEZ}. Theorem 7.15 in additive
notation is shown on the next page.
......
..
......
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it.
208 Chapter 7
Groups
Theorem 7.15
Let
(Additive Version)
G be an additive
group and let
a E G.
(1)
If a has infinite order, then (a) is an infinite subgroup consisting of
the distinct elements ka, with k EZ.
(2)
If a has finite order
(a)=
n,
then
(a} is a subgroup of
{O, 1a, 2a, 3a, 4a, ... , (n
order
n and
- 1 )a}.
EXAMPLE 13
Since Z = {nl In EZ}, we see that the additive group
Z is an infinite cyclic
group with generator 1, that is Z = (1). The set E of even integers is a cyclic
subgroup of the additive group
Z because E = {n2 In E .Z}.
EXAMPLE 14
Each of the additive groups Z,. is a cyclic group of order n generated by 1 because
Z,. consists of the "powers" of 1, namely, 1, 2 = 1+1, 3 = 1 + 1+1, etc. For
instance,Z. =
{1,2,3, O}, that is, {l, 1+l,1+1+1, 1+1+1+1}.
{l, -1, i, -i} of the multiplicative group of nonzero elements of C
(1) because i2 = -1, i3 = -i, and i4 = 1.Similarly, the multipli­
cative group of nonzero elements of Z7 is the cyclic group (3), as you can easily verify.
The subgroup
is the cyclic subgroup
These examples are special cases of the following theorem.
Theorem 7.16
Let F be any one of
0,
R, C, or Zp (with p prime), and let F" be the multiplica­
tive group of nonzero elements of
f.t If G is a
finite subgroup of
F",
then
G is
cyclic.+
Proof �
Let
cE
G be an element of largest order (there must be one since G is
finite), say !cl=
d" = 1 by part
m.
If aE G, then !al divides
(1) of
tion of the equation X"
m
m
by Corollary 7.10, so that
Theorem 7 .9. Thus every element of G is a solu­
-
1 = 0. Since a polynomial equation of degree
has at most m solutions inF(byCorollary4.171), we must have IGJ :Sm.
But {c) is a subgroup of G of order m by Theorem 7.15. Therefore, (c}
must be all of G, that is, G is cyclic. •
tsee Examples 8 and 9 of Section 7.1 or 7.1. A.
*For those who have read Chapters: The theorem and its proof are valid
IJf you haven't read Section 4.4, you'll have to take this on faith for now.
when Fis any field.
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7.3
Subgroups
209
Now that we know what cyclic groups look like, the next step is to examine the
possible subgroups of a cyclic group.
Theorem 7.17
Everysubgroupofa cyclic groupis itself cyclic.
Proof "
Suppose G =
(a) and His a subgroup of
G. If H = (e} , then His the
cyclic subgroup generated bye (all of whose powers
are just e
) . If H #:
{e), then H contains a nonidentity element of G, say a1 with i ¢ 0. Since
His a subgroup, the inverse element a-' is also in H. One of i or -i is
positive, and so Hcontains positive powers of a. Let k be the smallest
positive integer such that ak EH. We claim that His the cyclic subgroup
generated by a". To prove this, we must show that every element of H
is a power of ak. If h EH, then h E G, so that h = d" for some m. By the
Division Algorithm, m
=
kq +
rwith 0 s r < k. Consequently, r = m - kq
and
a' = a'nl-kq = ama-"4 = d"(a''r".
a• E H by closure. Since ak is the
a in Hand since r < k, we must have r = 0.
Therefore, m = kq and h = am = ak'l = (a")'l E(ak). Hence, H = {ak). •
Both d" and ak are in H. Therefore,
smallest positive power of
For additional information on the structure of cyclic groups and their subgroups,
see Exercises 44-46.
Generators of a Group
Suppose G is a group and
a E G. Think of the cyclic subgroup (a} as being constructed
{a} in this way: Form all possible products of a and a-1
from the one-element set S =
in every possible order. Of course, each such product reduces to a single element of
the form d'. We want to generalize this procedure by beginning with a set S that may
contain more than one element.
Theorem 7.18
Let S be a nonempty subset of a group G. Let (S) be the set ofall possible
products, in everyorder, of elements of Sand their inverses.*Then
(1) (S) is a subgroupof Gthat contains set S.
(2) tf His a subgroupof G that contains the set S, then H contains the
entire subgroup (S).
•we allow the possibility of a product with one element so that elements of Swill be in (S',.
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210
Chapter 7
Groups
This theorem shows that (S) is the smallest subgroup of G that contains the set S. In
the special case when S={a}, the group (S) is just the cyclic subgroup (a), which is the
smallest subgroup of G that contains
by
a. The group (S) is called the subgroup generated
S. If (S)is the entire group G, we say that S generates G and refer to the elements of
S as the generators of the group.
Proof of Theorem 7.18 ... (1) (S)is nonempty because the set s is nonempty and
every element of S (considered
(S). If a, b E(S), then
as
a one-element product) is an element of
a is of the form a1a
2
• • •
ak, where k � 1 and each a1
is either an element of S or the inverse of an element of S. Similarly,
b=b1b2
1 and each b1 either an'element of S or the in­
a1� a�1� • • b,
consists of elements of S or inverses of elements of S. Hence, ab E(S),
and(S) is closed. The inverse of the element a=a1a2
ak of (S)is
1
a-1=ak-1
�-la1 - by Corollary 7 .6. Since each a1 is either an element
verse
• • •
b,, with t �
of an element of S. Therefore, the product ab
=
• • •
•
• • •
• •
•
of S or the inverse of an element of S, the same is true of Cit-1• Therefore,
a-1 E (S}. Henoe, (S)is a subgroup of G by Theorem 7.11.
(2) Any subgroup that contains the set S must include the inverse
of every element of S. By closure, this subgroup must also contain all
possible products, in every order, of elements of S and their inverses.
Therefore, every subgroup that contains S must also contain the entire
group(S}.
•
EXAMPLE 15
The group
sinoe
U15= {1, 2, 4, 7, 8, 11, 13, 14} is generated by the set S={7, 11}
71=7
111=11
72=4
73=13
7·11=2
72• 11=14
74=1
73• 11 =8.
Different sets of elements may generate the same group, R>r instance, you can readily
verify that
U15 is also generated by the set {2, 13} (Exercise 9).
EXAMPLE 16
Using the operation table in Example 5 of Section
7 .1 or 7.1.A, we see that in
the group D4,
2
(r1) =r2
r1 o h=t
Therefore, D4 is generated by
(r1)3=r3
(r1)2 o h=v
(r1f=ro
(r1)3 h
o
=
d.
{r1, h}. Note that the representation of group
elements in terms of the generators is not unique; for instance,
(r1)3
o
h=d
and
r1
o
h
o
(ri)2=d.
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7.3
Subgroups
211
• Exercises
A. I. List all the cyclic subgroups of
(a) U1s
2. (a)
(b)
3.
(b) UYl
List all the cyclic subgroups of D4•
List at least one subgroup of D4 that is not cyclic.
List the elements of the subgroup (a}, of 81, where
a=
(
1
2 3 4 56
327651
1\
4/
In ExerciSes 4-8, list (ifpossible) or describe the elements of the given cyclic subgroup.
4. (2) in
the additive group Z12•
5. (2) in the additive group Z.
6. (2) in the multiplicative group of nonzero elements of.Z11•
7. (2} in
the multiplicative group
O* of nonzero rational
numbers.
8. (3} in the multiplicative group of nonzero elements of Z11•
9.
Show that
U15 is generated by the set {2, 13}.
10.
Show that (1, 0) and (0, 2) generate the additive group 7l X
11.
Show that the additive group Z2 X
Z7•
Z3 is cyclic.
12. Show that the additive group Z2 X � is not cyclic but is generated by two elements.
13.
LetHbe a subgroup of a group G. If
e0 is the identity element ofG and en is
en.
the identity element ofH, prove that e0
14.
LetHandKbe subgroups of a group G.
(a)
Show by example thatHU Kneed not be a subgroup ofG.
(b)
Prove thatHU Kis a subgroup of G if and only ifH!;;;Kor
;
K!;;;H.
15. (a)
(b)
16.
:=
LetHandKbe subgroups of a group G. Prove thatH n Kis a subgroup
ofG.
Let {H}
1 be any collection of subgroups ofG. Prove that n H1 is a
subgroup ofG.
Let G 1 be a subgroup of a group G andH1 a subgroup of a group H. Prove
that G1 X H1 is a subgroup ofG X H.
17.
Show that the only generators of the additive cyclic group 7L are 1 and -1.
18.
Show that (3,
19.
Let G be an abelian group and let The the set of elements of G with finite
I), (-2, -1), and (4, 3) generate the additive group Z X Z.
order. Prove that Tis a subgroup ofG; it is called the torsion subgroup. (This
result may not h old if G is nonabelian;
20.
see
Exercise 20 of Section
72
. ).
Let G be an abelian group, k a fixed positive integer, and H =
{aE GI lal divides k}. Prove thatHis a subgroup ofG.
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212
Chapter 7
Groups
21. (a)
If Gis a
group andabEZ(G),is it true that
(b) If Gis a group andabEZ(G), prove that ab
a and
=
bare inZ(G)? [Hint: D4].
ha.
22. If a is the only element of order 2 in a group G, prove that aEZ( G).
23.
Let Gbe a group and leta E G. Prove that (a) = (a-1�
24. Show that{}**,the multiplicative group of positive rational numbers, is not
a cyclic group. [Hint: if 1 *
and r2 ]
r
E{}**,then there must be a rational between r
.
25. Show that the center of S3 is the identity subgroup.
26. (a) LetHand Kbe subgroups of an abelian group Gand letHK=
b EK}. Prove
thatHKis a subgroup of G,
{ab I a EH,
(b) Show that part (a ) may be false if Gis not abelian.
27. Let Hbe a subgroup of a group Gand, for XE G, let�-1Hxdenote the set
{x-1ax I a EH}. Prove that
1
x-Hxis a subgroup of G.
28. Let Gbe an abelian group and n a fixed positive integer.
(a) Prove thatH
=
{aEGI a"= e}is a subgroup of G.
(b) Show by example that part (a) may be false if Gis nonabelian. [Hint: S3.]
29. Prove that a nonempty subsetHof a group G is a subgroup of Gif and only if
whenever a, bEH,then ab- 1EH.
30. LetA(T) be the group of permutations of the set Tand let T1 be a nonempty
subset of T. Prove thatH {f EA(T) lfi.t) I for every t E T1}is a subgroup
=
=
of A(T).
31. Let Tand T1 be as in Exercise 30. Prove that K= {f EA(T) lf(T1) = T1} is a
subgroup of A(T) that contains the subgroupHof Exercise 30. Verify that if
T1 has more than one element, then K * H.
32. LetHbe a subgroup of a group Gand assume that x-1Hx1;;Hfor every XE G
(notation as in Exercise 27). Prove that x-1Hx= Hfor each xE G.
33. Let Gbe a group andaE G. The centralizer of a is the set C(a) =
ga
=
ag}. Prove that
{g E GI
C(a) is a subgroup of G.
34. If Gis a group, prove that Z(G)
Do C(a)(notation as in Exercise 33).
35. Prove that an element a is in the center of a group Gif and only if eta) = G
=
(notation as in Exercise 33).
36. True or false: If every proper subgroup of a group G is cyclic ,then Gis cyclic .
Justify your answer.
37. Suppose thatHis a subgroup of a group Gand that a E Ghas order n. Ifak EH
and(k,n) = 1,prove thataEH.
B. 38.
( a) Let p be prime and let b be a nonzero element of
[Hint: Theorem 7.16.]
z,.
1
Show that bl'- = I.
(b) Prove Fermat's Little Theorem: If pis a prime anda is any integer , then
ff = a (mod p). [Hint: Let b be the congruence class of a in � and use
part(a ).]
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7.3
Subgroups
213
39. If His a subgroup of a group G, then the normalizer of His the set
N(H) =
{x E GI x-1 Hx = H} (notation as in Exercise 27). Prove that N(H) is a
subgroup of G that contains H.
40. Prove that
H
==
{ (� �)I
1 or -1, h E
a =
Z}
is a subgroup of GU.,2, Q).
41. Let Gbe an abelian group and n a fixed positive integer . Prove that H =
{a n I a E G}
is a subgroup of G.
42. Let k be a positive divisor of the positive integer n. Prove that
Hk
=
{a E U,. I a= 1 (mod k)} is a subgroup of Un.
43. List all the subgroups of Z12• Do the same for Z20•
44. Let G =
(a)
(a} be a cyclic group of order n.
Prove that the cyclic subgroup generated by am is the same as the cyclic
subgroup generated by a
d
,
where d =
(m, n). [Hint: It suffices to show that
ad is a power of am and vice versa. (Why?) Note that by Theorem
mu + nv.]
are integers u and v such that d =
(b)
Prove that a"' is a generator of Gif and only if
45. Let G =
(a} be a cyclic group of order
is a divisor of
46. Let G =
(a}
n.
(m, n)
=
1.2, there
1.
If His a subgroup of G, show that IHJ
n. [Hint: Exercise 44 and Theorem 7.17.]
be a cyclic group of order
n. If k is a positive divisor of n, prove
[Hint: Consider the subgroup
that Ghas a unique subgroup of order k.
generated by allfk.]
47. Let G be an abelian group of order mn where (m,
contains an element
a
of order
m
and an element
n) = 1. Assume that G
h of order n. Prove that G is
cyclic with generator ah.
48. Show that the multiplicative group Ill* of nonzero real numbers is not cyclic.
a. Prove that the
49. If G is an infinite additive cyclic group with generator
equation
x + x = a has no solution in G.
50. Show that the additive group Q is not cyclic.
51. Let Gand
[Hint: Exercise 49.]
H be groups. If G X His a cyclic group, prove that Gand Hare
both cyclic. (Exercise 12 shows that the converse is false.)
52. Prove that
{ G �)In Z}
E
is a cyclic subgroup of GL(2,
53. Prove that Zm X .l,. is cyclic if and only if (m,
54. If G "#;
{e} is a group that has no proper
n)
=
R).
1.
subgroups, prove that Gis a cyclic
group of prime order.
55. Is the additive group G =
56. Show that the group
{a+ hv'2 I a, hEZ} cyclic?
U'JJJ of units in Zw is not cyclic.
57. Show ,that the group U18 of units in Z18 is cyclic.
58. If S is a nonempty subset of a group G, show that {S) is the intersection of the
family of all subgroups
H such that S !:,; H.
..
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214
Chapter 7
Ill
Groups
Isomorphisms and Homomorphisms*
H you were unfamiliar with roman numerals and came across a discussion of integer
arithmetic written solely with roman numerals, it might take you some time to realize
that this arithmetic was essentially the same as the familiar arithmetic in Z except for
the labels on the elements. Here is a less obvious example of the same situation.
EXAMPLE 1
Recall the multiplicative subgroup
and the multiplicative group U5
tables are shown below.t
=
L {I, i, -i,-1} of the complex numbers
{l, 2, 3, 4} of units in Z5, whose operation
=
Us
L
2
3
4
2
3
4
2
2
4
1
3
3
3
1
4
2
4
4
3
2
i
-i -1
i
-i -1
i -1
-i
1
i
-i
_,
-1
-1
-1
i
i
1
-i
At first glance, these groups don't seem the same. But we claim that they are
"essentially the same", except for the lablels on the elements. To see this clearly,
relabel the elements of U5 according to this scheme:
Relabel
1
as I;
Relabel 2 as i;
Relabel
3
as -i;
Relabel 4
Now look what happens to the table for U5-it becomes the table for
1
x
,r
i
1.
i
�
1
-i
-i
J
-1
-1
�
I.
-1
�
-i
J
L!
-1
-1
-i
p
1
- 1.
�
i
1.
z
J
J
1
1
J
-i
i
').
as
1
'- 1
�
i
2
�
-i
J
i
1.
1
J.
The rewritten table shows that the operations in U5 and
L
work in exactly the
same way-the only difference is the way the elements are labeled. As far as
"The first few pages of this section explain the concept of isomorphism for groups, which is
essentially the same as the explanation for r ings in Section
3.3.
If you have read that section, feel
free to begin this one at the Definition on page 216.
iTo make the elements of the two groups easily distinguishable, the elements of Lare in
wmlll
.......
.ndkir�l).Bdbmbll....._.
... w......_..:dPLI�...-....
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.........
boldface.
ftonb 11Bam:
m.
7.4
Isomorphisms and Homomorphisms
215
group structure goes Ls is just the group U5 with new labels on the elements . In
more technical terms, U5 and L are said to be
In general,
isomorphic
isomorphic groups are groups that have the same structure, in the sense
that the operation table for one is the operation table of the other with the elements
suitably relabeled. Although this intuitive idea is adequate for small finite groups, we
need to develop a rigorous mathematical definition of isomorphism that agrees with
this intuitive idea
and is readily applicable to large groups as well.
There are two aspects to the intuitive idea that groups G and H are isomorphic:
relabeling the elements of G, and comparing the new operation table with that of H.
Relabeling means that every element of G is paired with a unique element of H (its new
label). In other words, there is a function.f:G-+Hthat assigns to each r E G its new label
f(r) EH. In the preceding example, we used the relabeling function .f: U5...+K given by
/(1)
=
1
/(2)
=
/(3) = -i
i
/(4) = -1.
The function.f:G-+H must have these properties:
(1)
Distinct elements of G get distinct labels in H:
If
r * r' in G, then/(r) * f(r') in H.
(2) Every element of His the label of some element of G:*
For each h E H, there is an
Properties
rE
G such that f(r)
=
h.
(1) and (2)simply say that the function/must be both injective and surjec­
bijection.t
tive, that is,fis a
In order to be an isomorphism, however, the table of G must become the table of H
when/ is applied. If this is the case, then for two elements a and
b of G, the situation
must look like this:
G
H
¥
I
c
a
f(b)
*
f(c)
f(a)
As indicated in the two tables,
a•
Since
a* b
=' c
that/(c) ""/(a)
b
=
c in G
and /( a) * f(b)
in G, we must have/(a
•f(b) in Hwe see that
f(a
*
*
=
f(c)in H
b) = f(c) in H. Combining this with the fact
b)=f(a)* /(b).
This is the condition that/must satisfy in order for/to change the operation tables of
G into those of H. We can now state a formal definition of isomorphism.
*Otherwise we could not get the complete table of H from that of G.
t1njective, surjective, and bijective functions are discussed in Appendix B.
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216
Chapter 7
Definition
Groups
Let G and H be groups with the group operation denoted by �. G is
isomorphic to a group H(in symbols, G = H) if there is a function f:G ...+H
such that
(i) f is injective;
(ii) f is surjective;
(iii) f(a * b)
=
f(a}• f(b) for all a, b E
G.
In this case, the function f is called an isomorphism.
It can be shown that G =Hif and only ifH= G(Exercise 53).
NOTE: In the preceding discussion, we have temporarily reverted to the*
notation for group operations to remind you that in a specific group, the
operation might be addition, multiplication, or something else. In such
cases, condition (iii) of the definition may take a different form; for instance,
Condition (iii)
/(a * b) = /(a) */ (b)
GandHadditive:
fl..a + b ) f(a ) + f(b)
/ (ab) =fl..d)f(b)
=
GandHmultiplicative:
f(a + b) f(a}f(b)
f(a) + f(b)
G additive,Hmultiplicative:
=
Gmultiplicative ,Hadditive:
/(ab)
=
EXAMPLE 2
The multiplicative group U8 = {l, 3, 5, 7} of units in Zg is isomorphic to the
additive group "11.2. X Z2. To prove this, let.f: U8--+ "11.2. X Z2 be defined by
/(1)
=
(0, 0 ) /(3) = (1, 0) /(5)
=
(0, 1) / (7) = (1, 1).
Clearly/is a bijection. Showing thatf(ab) f(a) + f(b) for a, b E U8 is equivalent to showing that the operation table for Z2 X Z2 can be obtained from that
of U8 simply by replacing each a E U8 byf(a) E Z2X Z2 .Use the tables below to
verify that this is indeed the case. Therefore,/ is an isomorphism:
=
Z2 x "11.2.
Ua
0
+
(0, 0)
(1, 0)
(0, 1)
(1 , 1)
(1, 0)
(0, 1)
(1, 1)
(0, 0)
( 1, 1)
(0, 1)
1
3
5
7
5
1
7
7
5
(0, 0)
3
1
3
3
(1, 0)
(0, 0)
(1, 0)
5
5
7
3
(0, 1)
(0, 1)
(1, 1)
(0, 0 )
(1, 0)
7
7
5
1
3
1
( 1, 1 )
(1, 1)
(0, 1)
( 1, 0 )
(0, 0)
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7.4
Isomorphisms and Homomorphisms
217
EXAMPLE3
Let E be the additive group of even integers. We claim that jZ -+ E given by
2a is an isomorphism. Since Z and E are infinite, comparing tab les is
f(a)
not
=
an
option. However, the formal definition of isomorphism will do the job.
We begin by showing that/is injective.* Suppose a,
Then
f(b)
2a
=
f(b)
=
2b
[Definition off]
a=b
[Divide both sides by 2 .]
Hence, f is injective. Now suppose
some integer k. T herefore,f(k)
=
bEZ and f(b)=f(b) in E.
n EE. Since n is an even integer, n= 2k for
""n, and/ is surjective. Finally, for all a,
2k
hEZ,
f(a + b)= 2(a +b)= 2a +
2b= f(a)
+ f(b).
Hence,/is an isomorphism of additive groups.
EXAMPLE 4
The additive group Ill of real numbers is isomorphic to the multiplicative group
R** of positive real numbers. To prove this, let.f:R-+R** be given by f(r)
=
HY.
To show that/is injective, suppose that
/(r)=f(s).
Then
10'
log
=
[Definition off]
10'
10'= log 10'
r=s
[Take logarithms ofboth sides.]
[Basic property oflogarithms]
So/is injective. To prove that/is surjective, let
number, and by the definition of logarithm,
k ER. T hen r
=
log k is a real
f(r)= 10'= lOJogk= k.
Thus,fis also surjective. Finally,
f(r + s)= 10r+1= 10'10'=f(r'Jf(s ).
Therefore,fis an isomorphism and Ill= IR**.
*Injective, surjective, and bijective functions are discussed in Appendix B.
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218
Chapter 7
Groups
EXAMPLES
Two finite groups with different numbers of elements (such as
Zs and Z10) can­
not be isomorphic, because no function from one to the other can be a bijection .
Example 1 presented two groups with the same number of elements that were
isomorphic. However, this is not always the case .
EXAMPLE 6
S3 and the additive group Zo each have order 6, but are not isomorphic. There is
no way to relabel the addition table of Zo to obtain the table of S3 because the
operation in S3 is not commutative, but addition in Zo is . A similar argument in
the general case
(see Exercise
16) shows that for groups G and H,
If G is abelian and His nonabelian, then G and Hare not isomorphic.
EXAMPLE 7
The additive groups � and Z2 X Z2 each have order 4 but are not isomorphic
Z2 has order 2, but Z. has two elements
3). So relabeling the addition table of one cannot
because every nonzero element of Z2 X
of order 4 (namely, 1 and
produce the table of the other. More generally by Exercise 29,
If f is an isomorphism, then a and/(a) have the same order.
If Gis a group, then an isomorphism G-+Gis called an automorphism of the group G.
EXAMPLE 8
If Gis a group, then the identit�· map L(j:G-+ Ggiven by La (r)
morphism of G. It is clear that Lo is bijective, and for any
= r
a, b EG,
is an auto­
i.0(a * b) = a * b = La(a) * L0 (b) .
EXAMPLE 9
Let
c be a fixed element of a group G. Define fG-+ Gby f(g) = c-lgc.
Then
f(b)f(b) = (c-1ac)(r1hc) = c-1a(ce-1)bc
If gEG, then
=
r1abc =/(ab).
cgc- 1 EGand
f(cgir1) = c-1(cgc-1)c = {r1c)g(c-1c) = ege = g.
So /is surjective. To show that/ is injective, supposef(a)
= f(b). Then c-1ac =
c-1bc. Canceling con the right side and c-1 on the left side by Theorem 7.5, we
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...._,..mGJ"�«*d.._ __...,.dlN:t �---.�c.a.�
rir;bl1a-...,,,..��·...,.
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...
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7.4
have
Isomorphisms and Homomorphisms
219
a= b. Hence,/is injective. Therefore,/is an isomorphism, called the inner
automorphism of
Ginduced by c. For more about automorphisms, see Exercises 36,
37, 58, and 59.
The next theorem completely characterizes all cyclic groups.
Theorem 7.19
Let
G
be a cyclic group.
(1) If G is infinite, then
G is isomorphic to the additive group Z.
(2) If G is finite of order n, then G
is isomorphic to the additive group
Z,,.
Proof"" ( 1) Suppose that G={a) is an infinite cyclic group. By Theorem 7.15 G
consists of the elements d' with k E Z, all of which are distinct (meaning
that d=al if and only if i=f). The functionfG--1>.Z defined byf(a�= k
is easily seen to be a bijection (Exercise 17). Since
f(a1a1) =f(a'+ ') = i + j=f(d) + f(a1),
fis an isomorphism. Therefore, G = Z.
(2) Now suppose that G= (b}and b has order n. By Theorem 7.15,
, ll'-1}, and by Corollary 2.5, Z,,= {[O], [1], [2], . . . ,
[n - 1]}. Define g:G--1> .Z,. by g(b� = [i] . Clearly g is a bijection. Finally,
G= {b0, b1, b2,
g(l}ll)
=
•
•
•
g(bi+') = [i +J]
=
[i] +[JI = g(b1) + g(ll).
Hence, g is an isomorphism and G = Z,,.
•
EXAMPLE 10
In multiplicative group O* of nonzero rational numbers, the cyclic subgroup
1
1 1 1
generated by 21s (2) ... , l6' g' 4, '2' l, 2, 4, 8 , 1 6 ,
The
.
{
•
.
group (2) is isomorphic to the additive group Z by Theorem 7.19.
.
}*
.
EXAMPLE 11
The upper left-hand quadrant of the operation table for D4 in Example 5
of Section 7 .1or 7.1.A and Theorem 7.12 show that G= {r0, ri. r2, '3} is
a subgroup of D4• Verify that both G and U5 = {l, 2, 3, 4} are cyclic. By
Theorem 7 .19 each is isomorphic to the additive group�· Hence, they are
isomorphic to each other: G = U5(Exercise 21).
*Exercise 7 of Section 7.3.
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220 Chapter 7
Groups
Homomorphisms
Many functions that are not injective or surjective satisfy condition (iii) of the defini­
tion of isomorphism. Such functions are given a special name and play an important
role in later sections of this chapter.
Definition
Let G and H be groups (with operation*). A function f.·G-+ His said to be
a homomorphism if
f(a *b)
=
f(a) * f(b) for all.,, b E G.
Every isomorphism is a homomorphism, but a homomorphism need not
isomorphism.
be
an
EXAMPLE 12
The functionfR*-+ IR* given byf(x)
tive groups because
f(ab)
=
(ab')'-
=
=
x2 is a homomorphism of multiplica­
a'2b2
=
f(b)f(b).
However,/is not injective because/(1) /( 1) and is not surjective because
f(x) x2 2!:: 0 for all x, so no negative number is an image under/.
=
-
=
EXAMPLE 13
The functionfZ-+ Zs given by f(a)
groups because
f(a
+
b)
=
[a
+
=
b]
[a] is a homomorphism of additive
=
[a]
+
[b] f(b)
=
=
f(b).
The homomorphism/is surjective, but not injective (Why?).
EXAMPLE 14
If Gand Hare groups, the function/GX H-+ Ggiven byf((x, y)) xis a
surjective homomorphism (Exercise 9). If His not the identity group, g is not
injective. For instance, if en i= a EH, then (eq, a) #'- (e°' en) in GX H, but
f((e°' a)) = e0 and/((e0o en)) =err
=
Recall that the image of a function f.G-+ H is a subset of H, namely Im f
{h EHi h =f(a) for some a EG}. The function/ can be considered as a surjective map
from Gto Imf.
=
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7.4
Isomorphisms
and
Homomorphisms
221
Theorem 7.20
Let G and H be groups with identity
f: G-+ His a homomorphism, then
elements eG and
eH,
respectively. If
(1) f(eG) = eH.
(2) f(ll""1) = f(at1 for every a E G.
(3)
Im
(4) If
f is a subgroup
of H.
f is injective, then G fE Im(.
Proof,.. (1) Sinoe/is a homomorphism, e0 is the identity
in
G, and e0is the
identity in H, we have
f(ea)f(ea) =f(eaea)
[/is a homomorphism.]
f(e<i)f(e<i) = f(e0)
[e0 is
f(e<i)f(e0) = enf(e0)
[f(ea) EH and e0 is the identity in H.]
Canoeling/(e0) on the right (by
the identity in G.]
Theorem 7.5) produces/ (e0)
=en.
(2) By (1) we have
f(a-1)/(a) =f(a-1a) =f(e<i) =en =f(a)-1.f(a).
1)
=f(a)-1•
enE Im/by (1), and so Im/is nonempty. Since
f(a)f(b) = f(ah), Imf is closed. The invenie of each/(a)E Im/is also in
Im/because/(a)-1 = f(a-1) by (2). Therefore, Im/is a subgroup of H by
Canoeling/(a) on each end shows that/(a-
(3)
The identity
Theorem 7.11.
(4)
As noted before the theorem,/can be considered
function from
isomorphism.
as
a surjective
G to lmf If /is also an injective homomorphism, then/is an
•
Group theory began with the study of permutations and groups of permutations.
The abstract definition of a group came later and may appear to be far more general
than the concept of a group of permutations. The next theorem shows that this is
not the case, however.
Theorem 7.21
Every group
Proof .,..
G
Cayley's Theorem
is isomorphic to a group of permutations.
Consider the group A(G) of all permutations of the set
A(G) consists of all bijective .functions from
G. Recall that
G to G with composition as
the group operation. These functions need not be homomorphisms.
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222 Chapter 7
Groups
To prove the theorem, we find a subgroup of
A(G) that is isomorphic to
G. *We do this by constructing an injective homomorphism of groups
f:G-+ A(G); then G is isomorphic to the subgroup Im/ of A(G) by
Theorem 7.'lJJ .
If a E G, then we claim that the map <p,,:G-+ G defined by <pJ..x) = ax is
a bijection of sets [that is, an element of A(G)].This follows from the fact
1 = b; hence, 'Pais surjective. If <p6(b)=
that ifb E G, then <p.(a-1h)= a(/l-b)
<pJ..c), then ab= ac.Canceling a byTheorem 7.5, we conclude thatb= c .
Therefore, 'P a i s injective and, hence, a bijection. Thus <p4EA(G).
Now definef:G-+A(G) byf(a) ='Pa· For any a, hEG,f(ab) ='Pab is
the map from Gto Ggiven by <pJ_x) = abx . On the otherhand,f(a) of(b) =
<p,. 0 'Pb is the mapgiven by (q;,, <pn)(x)= <pJ_q;J..x))
q;J..bx)= ab x . Therefore,
f(ab)= f(a) f(b) and/is a homomorphism of groups. Finally, suppose
0
=
0
f(a) =f(c), so that <pj_x)
=
<p,f,x) for all x EG. Then a
= ae =
q;J..e) = <p,f,e) =
ce = c. Hence,/ is injective. Therefore, G == Im/ by Theorem 7.'lJJ.
•
Corollary 7.22
Every finite group G of order n is isomorphic to a subgroup of the symmetric
groups,..
Proof " The group G is isomorphic to a subgroup Hof A(G) by the proof of
n
Theorem 7 .21. Since G is a set of
elements,
A(G) is isomorphic to Sn
by Exercise 38. Consequently, His isomorphic to a subgroup K of Sn by
Exercise22. Finally, byExercise21, G= HandH= Kimplythat G=K. •
Any homomorphism from a group G to a group of permutations is called a
representation of G, and G is said to be represented by a group of permutations. The
homomorphism G-+ A(G) in the proof ofTheorem 7 .21 is called the left regular repre­
sentation of G. By the use of such representations, group theory can be reduced to the
study of permutation groups. This approach is sometimes very advantageous because
permutations are concrete objects that
are
readily visualized. Calculations with per­
mutations are straightfor ward, which is not always the case in some groups. In certain
situations, group representations are a very effective tool .
On the other hand, representation by permutations has some drawbacks. For one
thing, a given group can be represented as a group of permutations in many ways-the
homomorphism G-.+ A(G) of Theorem 7.21 is just one of the possibilities (see Exercises
49, 51, and 54 for others). And many of these representations may be quite inefficient.
According to Corollary 7 .22, for example, every group of order 12 is isomorphic to a
subgroup of S12, but S1
1 has order 12!= 479,001,600. Determining useful information
about a subgroup of order 12 in a group that size is likely to be difficult at best.
Except for some special situations, then, the study of elementary group theory via
the abstract definition (as we have been doing) rather than via concrete permutation
representations is likely to be more effective.The abstract approach has the advantage
of eliminating nonessential features and concentrating on the basic underlying struc­
ture. In the long run, this usually results in simpler proofs and better understanding.
*The group A(G) itself is usually far too large to be isomorphic to G. For instance, if G has order n,
then A(G) has order nl by Exercise ID of Section
7.1.
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7.4
Isomorphisms and Homomorphisms
223
• Exercises
A. 1. (a) Show that the functionfR-+ R given byf(x)
3x is an isomorphism of
=
additive groups.
(b) Let IR** be the multiplicative group of positive real numbers. Show that
fR** -+ R** given byf(x) 3x is not a homomorphism of groups.
=
2. Show that the function g:R**-+ R** given by g(x)
=
v'x is an isomorphism.
3. Show that GL(2, Z2) is isomorphic to S3 by writing out the operation tables
for each group. (Hint: List the elements of GIJ...2, Z2) in this order:
G �). G !)• G �). G �). C �). G �)
G � �). G � D• G � D• G � �).
(1 ) (1 )
and the eleme nts
of S3 in this order:
2 3
2 3
3
1
2
'
1
3 2 .]
4. Prove that the function fill*-+ IR* defined byf(x)
=
x3 is an isomorphism.
5. Prove that the function g:Zg-+ Z9 defined by g(x)
=
2x is an isomorphism.
6. Prove that the function h:Z8 -+ Z8 defined by h (x)
=
2x is a homomorphism
that is neither injective nor surjective.
7. Prove that the functionfill* -+ R**defined by /(x)
=
lxl is a surjective
homomorphism that is not injective.
8. Prove that the function g:R -+ IR*defined by g(x)
=
2x is an injective
homomorphism that is not surjective.
9. If G and Hare groups, prove that the functionfG X H-+ Ggiven byf((a, b))
=
a is a surjective homomorphism.
10. Show that the functionflR-+ Rdefined byf(x)
=
x2 is not a homorphism.
11. Prove that the function g:R* -+ GL(2, IR) defined by g (x)
12. Prove that the function h:R-+ GL(2, Ill) defined by h (x)
injective homomorphism.
C O)
(1 �)
=
injective homomorphism.
=
0
x
x
is an
is an
13. Show that U5 is isomorphic to U10•
14. Prove that the additive group Z6 is isomorphic to the multiplicative group of
nonzero elements in Z7•
15. LetfG-+ Hbe a homomorphism of
each integer n,f(a") f(ar.
groups. Prove that for each
a E Gand
=
16. Iff.G-+ His a surjective homomorphism of groups and Gis abelian, prove
that His abelian .
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224 Chapter 7
Groups
17.
Prove that the function/in the proof of Theorem 7.19(1) is a bijection .
18.
Let G, H, G1o H1 be groups such that G =G1 and H =H1• Prove that
GXH=: G1 XH•1
19.
Prove that a group Gis abelian if and only if the function.f:G-+ Ggiven
by f(x) = x-1 is a homomorphism of groups. In this case, show that f is an
isomorphism.
20.
Let N be a subgroup of a group G and let a E G.
(a) Prove that a-1Na = {a-1na In EN} is a subgroup of G.
(b) Prove that Nis isomorphic to a-1Na. [Hint: Define.f:N-+ a-1Na by
f(n) = a-1na.]
21.
Let G, H, and Kbe groups . If G =Hand H = K , then prove that G = K.
[Hint: If.f:G-+ Hand g:H--+ K are isomorphisms , prove that the composite
function g f:G-+ K is also an isomorphism.]
o
22.
If f:G-+
His an isomorphism of groups and if Tis a subgroup of G, prove
.
that Tis isomorphic to the subgroup/(1) ={j{a) I aE T} of H.
23.
(a)
If G is
f(x)
=
an abelian group , prove that the function.f:G-+ Ggiven by
x2 is a homomorphism.
(b) Prove that part (a) is false for every nonabelian group. [Hint: A counter­
ple is insufficient here (Why?). So try Exercise 24 of Section 7.2.]
exam
B. 24.
Let Gbe a multiplicative group. Let Gop be the set Gequipped with a new
operation * defined by a * b = ba.
(a) Prove that G"Pis a group.
(b) Prove that G = G0P. [Hint: Corollary 76
. may be helpful.]
25.
Assume that a and b are both generators of the cyclic group G , so that G
(a} and G = (b). Prove that the function.f:G-+ G given byf(a1 ) = b1 is an
automorphism of G.
26.
If G =(a} is a cyclic group and .f.G-+ His a surjective homomorphism of
groups, show that/(a) is a generator of H, that is , His the cyclic group (f(a)}.
[Hin t : Exercise 15 .]
27.
Let G be a multiplicative group and c a fixed element of G. Let H be the set G
equipped with a new operation* defn
i ed by a* b acb.
=
=
(a) Prove that His a group.
(b) Prove that the map.f:G-+ Hgiven by f(x) = c-1x is an isomorphism .
28.
LetfG-> Hbe a homomorphism of groups and suppose that aEG has finite
orderk.
(a) Prove that/(a)" = e. [Hint: Exercise 15.]
(b) Prove that l/(a)I divides lal. [Hin t: Theorem 7.9.]
29.
If.f:G-> His an injective homomorphism of groups and aEG, prove that
l/(a)I = lal.
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7.4
Isomorphisms and Homomorphisms
225
30. Letf.G-+H be a homomorphism of groupsand let K be a subgroup of H.
Prove that the set {a E Glf(a) EK} is a subgroup of G.
31. If f.G-'t G is a homomorphism of groups, prove that F =
a subgroup of G.
32. If
A
=
(: �)
{aE G lf(a) =a} is
is a matrix, the number ad - be is denoted det A and called
the determinant of A. Prove that the function.f:GL(2, R)-+ R* given by
f(A) = det A is a surjective homomorphism.
33. Letf.G-+H be a homomorphism of groupsand letKj=
{aEGl/(a) =en},
that is, the set of elements of G that are mapped by f to the identity element
of H. Prove that Kjis a subgroup of G. See Exercises 34 and 35 for examples.
34. The function.f:Z-+Z5 given by
f(x) = [x] is a homomorphism by Example 13.
Find K1(notation as in Exercise 33).
35. The functionf U5-+ Us given by
f(x)
�
Find K1(notation as in Exercise 33).
fl is
a
homomorphism by Exercise 23.
36. Let
G be a group and let Ant G be the set of all automorphisms of G. Prove
that Aut G is a group under the operation of composition of functions.
[Hint: Exercise 21 may help.]
37. Let G be
a group and let Aut G be as in Exercise 36. Let Inn G be the set of
all inner automorphisms of G (that is, isomorphisms of the formf(a) = c-1ac
for some cE G, as in Example 9.). Prove that Inn G is a subgroup of Aut G.
[Note: Two different elements of G may induce the same inner automorphism,
that is, we may have c-1ac = a-1ad for all a E G. Hence, !Inn GI :S IGI.]
38. Let The a set
n elementsand let A(1) be the group of permutations of T.
Prove that A(T) =Sn. [Hint: If the elements of Tin some order are relabeled as
1, 2, . , n, then every permutation of T becomes a permutation of 1, 2, . . . , n.]
..
39. Show that the additive groups Zand 0 are not isomorphic.
In Exercises 40--44, explain why the given groups are not isomorphic. (Exercises 16
and 29 may be helpful.)
40. Z6and S3
41. �XZ2andD4
42. � x Z2 and Z2 x Z2 x Z2
43. U8and U10
44. U10 and U12
45. Is U8 isomorphic to U12? Justify your answer.
46. Prove that the additive group R of all real numbers is not isomorphic to the
-
multiplicative group R* of nonzero real numbers. [Hint: If there were an
isomorphismf:R-+ R*, thenftk) = I for some k; use this fact to arrive at a
contradiction.]
47. Show that D4 is not isomorphic to the quaternion group of Exercise 16 of
Section 7.1.
48. Prove that the additive group
0 is not isomorphic to the multiplicative group
O** of positive rational numbers, even though � and R**are isomorphic.
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226 Chapter 7
49.
Groups
Let G be a group and let A(G) be the group of permutations of the set G.
Define a function g from G to A( G) by assigning to each dE G the inner
automorphism induced by d-1 (as in Example 9 with
c =
d-1). Prove that g is
a homomorphism of groups.
50.
Let Gbe a group andhEA( G). Assume that h o 'Pa
(where 'P· is
as
=
cp. ohfor all a E G
in the proof of Theorem 7.21). Prove that there exists bEG
such thath(x) :o xb-1for all xEG.
51. (a)
Let Gbe a group and
(Jc(x )
(b)
=
c
E G. Prove that the map (J,: G'-+ G given by
xc-1 is an element of A(G).
Prove that h: G4 A(G) given by h(c)
=
9, is an injective homomorphism
of groups. Thus Gis isomorphic to the subgroup Imhof A(G). This is the
right regular representation of G.
52.
Find the left regular representation of each group (that is, express each group
as
53.
a permutation group
H
[Hint:
Letf:G4
as
in the proof of Theorem 7.21):
(b) L.
be an isomorphism of groups. Let
g:Hg
function off as defined in Appendix B. Prove that
groups.
To show that g(ab)
=
'-+
Gbe the inverse
is also an isomorphism of
g(a)g(b), consider the images of
the left­
and right-hand sides underf and use the facts that f is a homomorphism and
Jog is the identity map.]
54. (a)
Show that
D3
[Hint: D3
D3 83.] D3
=
S3•
or 7 .1A
. . Each motion in
function from
(b)
Show that
D4
is described in Example 6 of Section 7.1
permutes the vertices; use this to define a
to
D4,
is isomorphic to a subgroup of
for part (a). This isomorphism represents
subgroup of a permutation group of order 4!
S8,
S4• [Hint:
See the hint
a group of order 8,
=
as
a
24, whereas the left
regular representation of Corollary 7.22 represents Gas a subgroup
of
55. (a)
a group of order 8!
Prove that
H
=
{
�n
(1
multiplication .
(b)
56. (a)
Prove that
H
=
Prove that K = {
multiplication.
(b)
57.
Z.
C
=
40,320.]
1-: n) I n
EZ
}
is a group under matrix
-��n : 2n) I n }
EZ
1
is a group under matrix
Is K isomorphic to Z?
[Hint:
Prove that the additive group Z[x] is isomorphic to the multiplicative group
O**
of positive rationals.
[
Let Po, Pi. p2,
• • •
be the distinct positive
primes in their usual order. Define rp:l' x] '-+ 0** by
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7.5
The Symmetric and Alternating Groups
227
58. Prove that G is an abelian group if and only if Inn G consists of a single
element.
59. (a)
(b)
[Hint:
See Exercise 37.]
Verifythat the group Inn D4 has order 4.
Prove that Inn D4 =
60. Prove that Aut
group
[Hint:
Z2 X Z2•
Z = Z2• [Hint: What are the possible
Z? See Exercises 25 and 26.]
61. Prove that Aut Z..
Section 7.3.]
62. Prove that Aut
=
(Z2
X
See Exercise 37.)
generators of the cyclic
U,.. [Hint: See Exercise 25 above and Exercise 44 of
Z.;J = S3•
APPLICATION: Linear Codes (Section 16.1) maybe covered at this point
if desired.
Ill
The Symmetric and Alternating Groups*
The finite symmetric groups S,. are important because, as we saw in Corollary 7.22,
every finite group is isomorphic to a subgroup of some S,.. In this section, we introduce
a more convenient notation for permutations, and some important subgroups of the
groups S11• We begin with the new notation.
Cons1'der the permutation
.
(
6 .
1 2 3 4
5 5) m s.6• Note that 2 is
. mapped to 4, 4
1 4 3 6 2
is mapped to 6, 6 is mapped to 5, 5 is mapped back to 2, and the other two elements,
1 and 3, are mapped to themselves. All the essential information can be summarized
by this diagram:
It isn't necessary to include the arrows here as long as we keep things in the same order.
A complete description of this permutation is given by the symbol (2465), with the
understanding that
each element is mapped to the element listed immediately to the right;
the last element in the string is mapped to the first;
elements not listed are mapped to themselves.
•Except for a few well-marked examples and exercises, this section is needed only in Sections
9.3-11.5, and 12.3.
8.5,
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228 Chapter 7
Groups
This is an example of cycle notation. Here is a formal definition.
Definition
Leta11 "2, aa
, a1; (with k � 1) be distinct elements of the set {1, 2, 3, . " . , n}.
Then (a.iaA
a.J denotes the permutation in Sn that maps a1 to fi2, 82. to
a3,
, aKo--1 toak• and a1;toa1, and maps every other element of{1, 2, 3, . .. , n}
to itself. (a1aea3
ak) is called a cycle of length k or a k-cycle.
• . • •
•
•
•
• ••
• • •
EXAMPLE 1
(143) is the 3-cycle that maps 1 to 4, 4 to 3, 3 to 1, and 2 to itself; it was
.
1 2 3 4 .
.
written 4
m the old notation. Note that (143) may also be denote d b y
2 1 3
(431) or (314) since each of these indicates the function that maps l t o 4, 4 to 3,
3 to 1, and 2 t o 2.
In S4,
(
)
EXAMPLE 2
According to the definition above, the I-cycle (3) in S,, is the permutation that
maps 3 to 3 and maps every other element of { 1, 2, ... , n} to itself; in other
words,
(3) is the identity permutation. Similarly, for any k in {l, 2,
I-cycle (k) is the identity permutation.
• . .
Strictly speaking, cycle notation is ambiguous since, for example,
note a permutation in S6, in 81, or in any S,, with n �
6.
,
n},
the
(163) might de­
In context, however, this
won't cause any problems because it will always be made clear which group S,, is under
discussion.
Products in cycle notation can be visually calculated just as in the old notation. For
example, we know that
(11
) (
1 2
2 3 4
4
3
4 2 3 ° 2 4 1 3
) = (4
1
)
2 3 4
3 1 2 '
(Remember that the product in S,, is composition of functions, and so the right-hand
permutation is performed first.) In cycle notation, this product* becomes
cf\
3) <{\ 4 3)
'-._.../
=
c1 4 2
3) .
1 is mapped to 2 and 2 is mapped to 4, so
4 is mapped to 3 and 3 is mapped to 2, so
The arrows indicate the process:
that the
product maps
that the
1 to 4.
product maps 4 to 2.
Similarly,
"Hereafter we shal I omitthecomposition symbol •and write the group operation ins. multiplicatively.
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7.5
The Symmetric and Alternating Groups
229
EXAMPLE 3
In the old notation S3 consists of
In the new notation, the elements of S3
(in the same order)
are
(1), (23), (13), (12), (123), and (132).
Two cycles are said to be disjoint if they have no elements i n common. For instance,
(13) and (2546)
both cycles.
are
disjoint cycles in S6, but (13) and (345) are not since 3 appears in
EXAMPLE 4
As shown before
Example 3, (243)(1243)
(1243)(243)
=
=
(1423). Verify that
1
(234 ).
Hence, the cycles (243) and (1234) do not commute with each
other. On the
other
hand, you can easily verify that the disjoint cycles (13) and (2546) do commute:
(13 )(2546)
This is
an
=
(1
3
2
3
4
5
6
5
1
6
4
2
)
=
(2546)(13).
illustration of the following theorem.
Theorem 7.23
If u = (a1a2
• •
·a,) and T = (b1b2
Proof"' Exercise 18.
•
•
•
b,) are disjoint cycles in S111 then UT=
Tu.*
•
It is not true that every permutation is a cycle, but every permutation can be
expressed
as
the product of d isjoint cycles. Consider, for
G � � � � : �)in
example,
the permutation
S1• Find an element that is not mapped to itself, say li and trace
where it is sent by the permutation:
1 is mapped to 5,
2 is mapped
5 is mapped to 4,
to 1
4 is mapped to 2,
and
(the element with which we started).
(a),
(u), and tau (T). For the entire Greek alphabet, see the inside back cover of
*Greek letters are often used to denote permutations. We shall generally use the letters alpha
beta (/J), delta (8), sigma
this book.
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230 Chapter 7
Groups
Thus the
given
(1542) on these four
1, 5, 4, 2 that is not mapped onto itself,
permutation bas the same action as the cycle
elements. Now look at any element other than
say
3. Note that
3 is mapped to 7,
Thus the 2-cycle
7 is mapped to 3.
and
(37) bas the same action on 7 and 3 as the given permutation. The only
6, which is mapped to itsel£ You can now easily verify
element now unaccounted for is
that the original permutation is the product of the two cycles we have found, that �
(1
)
2 3 4 5 6 7
5 1 7 2 4 6 3
=
(l542H37>·
Although some care must be used and the notation is more cumbersome, essentially
the same procedure works in the general case.
Theorem 7.24
Every permutation in Sn is the product of disjoint cycles.*
Proof• Adapt the procedure in the preceding example; see Exercise 44.
•
Theorem 7.25
The order of a permutation T in S,. is the least common multiple of the lengths
of the disjoint cycles whose product is T.t
Proof• Exercise 19.
•
EXAMPLE 5
The permutation T
=
(12)(34)(567) is a product of disjoint c ycles of lengths 2, 2,
2, and 3 is 6. Theorem 7 .25 tells us that
and 3. The least common multiple of 2,
T has order 6. You can verify this directly by computing the powers of T:
T
i'
=
=
(12)(34)(567),
(567),
r
=
Ts
=
(576),
(12)(34)(576),
r3
T
"
=
=
(12)(34),
(l).
•
The Alternating Groups
A 2-cycle is often called a transposition. Transpositions have some interesting properties.
EXAMPLE 6
If
(ab) is a transposition,
verify that (ab)(ab)
=
(1). Hence,
Every transposition is its own inverse.
"As usual, we allow the possibility of a product with just one cycle in it
trhe least common multiple is defined in Exercise 31 of Section
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7.5
The Symmetric and Alternating Groups
231
EXAMPLE 7
(12)(34)( 14)(13) is (13)(14)(34)(12) (the
same transpositions in reverse order). To prove this claim, we use the fact that a
We claim that the inverse of the product
transposition is its own inverse:
(12)(34)(14)(13). (13)(14)(34)(12)
=
(12)(34)(14) . (14)(34)(12)
=
(12)(34) . (34)(12)
=
(12)(12)
=
(1).
A similar argument wurks in the general case and shows that
If u1, u2' u3,
(U1U2U3
• •
•
•
•
and u. are transpositions, then
1
U11-1U,�t = UP11-1 • • • U3U2U1·
, u._i.
'
You can easily verify that
(1)
=
(12)(12),
(123)
=
(12)(23),
(1234)
=
(12)(23)(34).
These are examples of the following theorem.
Theorem 7.26
Every permutation in Sn is a product of (not necessarily disjoint) transpositions.
Proof• Since every permutation is a product of cycles by Theorem 7 .24, we need
only verify that every cycle (a1a2
(a1a,;
•
•
·
a,J
=
•
•
•
a,J is a product of transpositions:
(a1ai)(a2a1)
•
•
•
('2f,
_
1ak) ·
•
This corollary can also be proved directly by induction, without using Theorem 7.24
(Exercise 33).
A permutation in S11 is said to be even if it can be written as the product of an
even number of transpositions, and odd if it can be written as the product of an odd
number of transpositions.
EXAMPLE 8
(132) is even and (1243)(243) is odd because, as you can easily
(132)
=
(12)(13)
and
(1243)(243)
=
verify,
(23)(34)(14).
Since no integer is both even and odd, the even-odd terminology for permutations
suggests that no permutation is both even and odd. This is indeed the
case,
but it
requires proof. The first step is to prove
Lemma 7.27
The identity permutation in Sn is even, but not odd.
Proof• We write the identity permutation as (1). Verify that (12)(12)
=
(1).
Hence, the identity permutation is even. To show that it is not odd, we
use a proof by contradiction. Suppose that
(1)
=
Tk
•
•
•
T2T1 with each T1
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232 Chapter 7
Groups
c be a symbol that appears in at least one
a transposition and k odd. Let
of these transpositions. Let T, be the first transposition (reading from
right to left) in which c appears, say
T,_ 1, •
•
•
T,
=
(cd). Then c does not appear in
T1 and is, therefore, left fixed by these transpositions. If
r =
k,
c is left fixed by all the T's except T,., so that the product-the iden­
tity permutation-maps c to d, a contradiction. Hence, r < k.
then
Now consider the transposition T,+1· It must have one of the follow­
ing forms (where
I.
x, y, c, d denote distinct elements of {1, 2,
(xy)
II.
(xd)
III. (cy)
IV.
· •
n}:
·
(cd).
Consequently, there are four possibilities for the product T,+1T,:
I.
(xy)(cd)
IL
(xd)(cd)
In Case I, verify that (xy)(cd)
III.
(cy)(cd)
IV.
(cd)(cd).
(cd)(xy). Replace (xy)(cd) by (cd)(xy) in
=
the product; this moves the first appearance of cone transposition to the
left. In Case II, verify that (xd)(cd)
= (xc)(xd); if we replace (xd)(cd) by
(xc)(xd), then once again the first appearance of c is one transposition far­
ther left. Show that a similar conclusion holds in Case ill by verifying that
(cy)(cd) (cd)(dy).
=
Each repetition of the procedure in Cases I-III moves the first ap­
pearance of cone transposition farther left. Eventually Case IV must
occur; otherwise, we could keep moving c until it first appears in the last
permutation at the left, Tk, which is impossible, as we saw in the first para­
graph. In Case IY, however, we have T,+1 T, = (cd)(cd) = (1). So we can
delete these two transpositions and write (1) as a product of two fewer
transpositions than before. Obviously, we can carry out the same argu­
ment for any symbol that appears in a transposition in the product. If the
original product contains an odd number of transpositions, eliminating
two at a time eventually reduces it to a single transposition (1)
=
(ab),
which is a contradiction. Therefore, the identity permutation (1) cannot
be written as the product of an odd number of transpositions.
•
Theorem 7.28
No permutation in Sn is both even and odd.
Proof .. Suppose a ES,, can be written as u1u2
•
•
•
u k and as T1T2
•
•
•
T, with
each u,, TJ a transposition, k odd, and r even. Since every transposition is
its own inverse, Corollary 7.6 shows that
(1)
=
aa-1
uk) (T1 • •
=
(u1
=
U1 ''' UkTr -I' ' ' Tt-I
=
U1
Since k is odd and r is even, k +
•
•
•
r
•
•
•
•
T,r1
UkTr • • • 'rt·
is odd, and we have written
(1) as the
product of an odd number of transpositions. This contradicts Lemma 7.27,
and completes the proof of the theorem.
•
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7.5
The Symmetric and Alternating Groups
233
The set of all even permutations in S,, is denoted A11 and is called the alternating
group of degree n; the word "group" is justified by the following theorem.
Theorem 7.29
An is a subgroup of Sn of order nl/2.
Proof .. If a andj:! are in A,., then a= u1u2
• •
·ukandj:! = T1T2
• •
•
T,, with each
u1, T a transposition and k, reven. Thus, aj:I = u1u2 • • • Uif:T1T2 • • • T,.
1
Since k + r is even, aj:I EA,.. So A,, is closed under multiplication. By
1
u2u1• Sincek is even, a- EA,.. Therefore,
A,, is a subgroup by Theorem7.11. Exercise 24 shows that IA,.I = n!/2. •
Example7.•
1
a- = UPk l
-
•
•
•
EXAMPLE 9
The elements of
IA3l =
�
=
S3 are listed in Example 3. Because IS31 = 31, we know that
3. Since (12),
(13), and (23) are obviously
odd, A3 must consist of
(123), ( 132), and (1 ).
• Exercises
A. 1. Write each permutation in cycle notation:
(a)
(c)
(1
)
(b)
(
)
(
)
(d)
(
)
2 3 4 56 7 89
7 2 1 4 56 3 89
1 2 3 4 56 7 8 9
4 8 1 7 5 26 39
1 2 3 4 56 7 8 9
2 4 3 5 76 891
1 2 3 4 56 7 8 9
1 2 5 47693 8
2. Compute each product:
(a) (12)(23)(34)
(b) (246)(147)(135)
(d) (1234)(2345)
(c) (12)(53214)(23)
3. Express as a product of disjoint cycles:
(a)
(c
)
(
)
(b)
(
)
(d)
1 2 3 4 56 7 89
2 1 3 5 4 7 98 6
1 2 3 4 56 7 8 9
3 5 1 2 498 76
(
)
1 2 3 4 56 7 8 9
3 5 1 2 46 8 97
(14)(27)(523)(34)(1472)
(e) (7236)(85)(571)(1537)(486)
4. Write each permutation in Exercise 3 as a product of transpositions.
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....... it.
234
Chapter 7
Groups
5.
6.
7.
Find the order of each permutation .
(12)
(d)
What do you think the order of (123456789) is?
(123)
(c)
(1234)
(c)
(123)(435)
(c)
(12)(123)(1234)
Find the order of each permutation.
(a)
(13)(24)
(b)
(123)(456)
(d)
(1234)(4231)
(e)
(1234)(24)(43215)
Which of these permutations are even:
(a)
8.
(b)
(a)
(b)
(2468)
(246)(134)
List the elements in each group:
(b) A.i
9.
What is the order of each group:
(c) A10
(b) As
I 0. Is the set B,, of odd permutations in S,, a group? Justify your answer.
11.
List the order of each element of A4•
12. Write (12)(34) as the product of two 3-cycles.
a= (123)(234)(567)(78910) has order 10 in S,. (n � 10).
[Hint: Write a as a product of disjoint cycles and use Theorem 7 .25.]
13. Show that
B.
14.
Show that f3
15.
Prove that the cycle (a1�
16.
Show that the inverse of (a1ai
17.
Prove that a k-cycle in the group S,. has order k.
18.
Let a
aT
in
=
{ 1,
=
=
(1236)(5910)(465)(5678) has order 21 in S,,
•
•
•
•
•
10).
ak) is even if and only if k is odd.
·
·
·
akl in
(a1�
ak) and 'T (b1�
[Hint: You must show that
•
(n �
=
•
TU.
•
•
S,. is
(a/Pk- 1
•
•
•
a3�a1).
b,) be disjoint cycles in
S,,. Prove that
aT and Ta agree as functions on each i
2, ... , n}. Consider three cases: i is one of the
a s; i is one of the h's; i is
'
neither.]
19.
Prove Theorem 7.25: The order of a permutation 'Tin S,, is the least common
multiple of the lengths of the disjoint cycles whose product is T.
[Hint: Theorem
20. Let
(a)
(b)
a
7.23 and Exercise 17 may be helpful.]
and f3 be permutations in S,..
Fill the blanks in the table.
a
/3
even
even
even
odd
odd
even
odd
odd
af3a-1
1
a{3a-113even
What conclusions can you draw from the results in part (a).
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7.5
The Symmetric and Alternating Groups
235
.
1 2 3 4 5 6 78
. the permutation
u IS
.
..
.
3789 452 1 6 .
. u as a product of disJOIDt cycles.]
[Hmt: Wnte
·
(
21. Fmd the order of uum,wh
ere
22. Show that
9)
S10 contains elements of orders 10,20,and 30. Does it conta in an
40?
element of order
{(l), (12)(34),(13)(24), (14)(23)} is a
23. Prove that
subgroup of A4•
Bn denote the set of odd permutations in 811• Define a functionfAn-+ Bn
(12)a.
24. Let
by f(a)
(a)
=
Prove that/is injective.
(b) Prove that f is surjective.
[Hint: If f3E Bn, then ( 12)/3E A11.]
So f is bijective. Hence, An and B11 have the same number of elements.
(c) Show that IA11I
= nl/2. [Hint:
both) and IS11I =
See Exercise
=
(24)
Sn is in A11 or Bn (but not
.
39(a) and (b) for a
25. Show that the subgroup G of
T
Every element of
nl ]
generalization of this exercise.
S4 generated by the elements u
26. Prove that the center of
(a)
(1234) and
S11 (n > 2) is the identity subgroup.
27. If U is a k-cycle with k odd, prove that there is a cycle
28. Let
=
has order 8.
T such that r = U.
u be a k-cycle in S11•
Prove that
(b) If k
=
29. Let u and
u2 is a cycle if and only if k is odd.
2t,prove that there are t-cycles T and f3 such that u2
=
Tf3.
T be transpositions in S11 with n 2: 3. Prove that UT is a product of
(not necessarily disjoint) 3-cycles.
30. Prove that every element of An is a product of 3-cycles.
31. Let
u be a product of disjoint cycles of the same length. Prove that u is a
power of a cycle.
32. Prove that the decomposition of a permutation as a product of disjoint cycles
is unique except for the order in which the cycles are listed.
33. Use induction on n to give an alternate proof of Theorem 7.26: Every
element of S,, is a product of transpositions.
for
n =
k
-1
and if
[Hint: If the statement is true
TE Sk, consider the transposition (kr), where r = T(k).
(kr)T fixes k
{1,2,.. .,k-1}.]
Note that
and hence may be considered as a permutation of
34. If n 2: 3; prove that every element of Sn can be written as a product of at most
n
-1
transpositions.
uE Sn. Prove that UTU-1 is a transposition.
1
ak) and if u E Sn, prove that UTu-
35. Let T be a transposition and let
T is the k-cycle (a1a2
(u(a1)u(av
u(tJt)).
36. If
·
•
·
·
•
=
•
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236 Chapter 7
Groups
37. Let H consist of all permutations in Sn that fix 1 andn, that is,
H=: {aESnlaQ):o landa(n) =n}.
Prove that His a subgroup of S11•
38. Show that D4 is isomorphic to the group G in Exercise 25. [Hint: Note that
every element of D 4 produces a permutation of the vertices of the square
(see Example 5 in Section 7.1or7.LA.). If the vertices are numbered 1, 2,
3, 4, then this permutation can be considered as an element of S4• Define a
functionf:D4 � S4 by mapping each element of D4 to its permutation of the
vertices. Verify that/is an injective homomorphism with image G .]
39. Let G be a subgroup of S11 that contains an odd permutation
T.
(a) Prove that the number of even permutations in G is the same as the
number of odd permutations in G.
(b)
Explain why 2 divides
(c)
If K is a subgroup of S11 of odd order, prove that K is actually a subgroup
IGI·
of A�.
C. 40. Prove that every element of An is a product of n-cycles..
41. Prove that the transpositions (12), (13), (14), ... , (ln) generate Sn.
42. Prove that (12) and (123 •
·
·
n) generate S,,.
43. If /is an automorphism of S3, prove that there exists u ES3 such that
f(T)
=
ara-1 for every TE S3.
44. Use the following steps to prove Theorem 7.24: Every permutation Tin Sn is a
product of disjoint cycles.
(a) Let a1 be any element of {l, 2, ... ,n} such thatr(a1) *
a1. Let a2 T(aJ,
T(a3), and so on. Let k be the first index such that T(a.J is
one of ah
, ak-I· Prove that T(ak)
a1• Conclude that T has the same
effect on ah ••• , a,, as the cycle (a1a2 • • • ak)·
a3
=
T(aj, a4
• • •
(b)
=
b1 be any element of { 1, 2, •.. ,n} other than ai, ••. , ak that is not
b2 T(b1), b3 T(b-ZJ, and so on. Show that
T(bJ is never one of a1, ••• , ak. Repeat the argument in part (a ) to find a
b, such that T(b,) b1 and T agrees with the cycle (b1b2
b,) on the h's.
Let
mapped to itself by T. Let
=
=
:o:
(c)
•
•
•
Let
c1 be any element of {l, 2, ... , n} other than the a's or h's above such
T(c1) * c1• Let c2 T(c1), and so on. As above, find c, such that T
agrees with the cycle (c1c2 • • • c,) on the e's.
that
(d)
=
=
=
Continue in this fashion until the only elements unaccounted for are those
that are mapped to themselves by T.Verify that T is the product of the cycles
(a1 • ai, ) (b 1 • • • b,)(c1 •• • c,) • • •
•
•
and that these cycles are disjoint.
45. Prove that Sn is isomorphic to a subgroup of An+l·
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CHAPTER
8
Normal Subgroups and Quotient Groups
Congruence in the integers led to the finite arithmetics Zm which produced
a number of interesting results. Now we shall extend the concept of congru­
ence to groups, producing new groups and a deeper understanding of algebraic
structure.
•
Congruence and Lagrange's Theorem
In this section we present the analogue for groups of the concept of congruence,
which
was
introduced for integers in Chapter 2 and for rings in Chapter 6.* Except
for some notational changes, the first three results of this section are virtually identical
to those proved earlier for integers and rings. The following chart shows this parallel
development.
INTEGERS
RINGS
GROUPS
Theorem2.l
Theorem 6.4
Theorem8.1
Theorem2.3
Theorem 6.6
Theorem8.2
Corollary2.4
Corollary 6. 7
Corollary8.3
We begin by looking at an example of congruence in Z from a somewhat different
viewpoint.
•chapter 6 is not a prerequisite for this section, but it will be mentioned occasionally. Section 2.1 will
be the model for the presentation here.
237
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...
loll...-.:ilt.
238
Chapter
8
Normal Subgroups and Quotient Groups
EXAMPLE 1
In the integers, a = b (mod 4) means that 4 divides a
-
b, that is, that a
-
b is a
multiple of 4. LetK be the set of all multiples of 4, so that
K=
{O,
:!::4, :!::8, ±12,
. • .
}.
Thus,
a=
b (mod4)
means
a
-
bEK.
Note thatK is actually a subgroup of Z (the additive c yclic subgroup generated
by 4). Instead of thinking of congruence modulo the element 4, we can con­
sider this
congruence modulo the subgroupK:
as
a=
b (modK)
means
a
-
bEK.
Now let G be any group andK a subgroup of G. The last line of the preced­
ing example could be used as a definition of congruence modulo K However, we
normally use multiplicative notation for groups. So we must translate the pro­
posed definition and results from Section 2.1 into equivalent statements in multi­
plicative notation.* The following dictionary may be helpful for this translation.
ADDITIVE NOTATION
MULTIPLICATIVE NOTATION
a+b
ab
0
-c
a
-
b =a+ (-b)
Thus, the additive statement
a
-
bEK is equivalent to the multiplicative state­
ment ab-1 EK, and we have the following definition of congruence.
Definition
LetK
bea
subgroup of a group G and teta, bEG. Then a is congruent to b
K}] provided that ab-1 EK.
modulo K [written a= b (mod
EXAMPLE 2
LetK be the subgroup
{r0, r1, r2, r3}
of D4• Then the operation table in Example 5
of Section 7.1 or7.l.Ashows thatd-1
=
dandh od-1 =hod=
r1 EK. Therefore,
h=d(modK).
"There is a possibility of confusion here since integer multiplication is also defined. In carrying
over congruence from integers to groups, we consider on/ythe additive structure of the integers
and ignore integer multiplication because the integers form an additive group,
but not a
multiplicative one.
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8.1
Congruence and Lagrange's Theorem
239
Theorem 8.1
Let K be
Kis
a
subgroup of a group
G.
Then the relation of congruence modulo
{1} reflexive: a= a {mod K) for all a E
G;
{2} symmetric: if a= b {mod K), then b =a (mod K);
(3) transitive: if a = b {mod K) and b = c {mod K), then a = c (mod K).
The idea is to translate the proof of Theorem 2.1 to the present situation by chang­
ing congruence mod n to congruence mod Kand replacing statements such as "xis
divisible by n" or "n Ix" or ••x = nt" with the statement "xE IC'. We must also change
additive notation to multiplicative notation by using the dictionary above. It's straight­
forward for parts (1) and (3), but a bit trickier for part (2), since integer addition is
commutative,but the multiplicative operation inG may not be.
Proof ofTheorem 8.1 ... (1)
(2)
aa-1
= e and eEK. Hence, a= a (modK).
a = b (mod K) means ab-1 = k for some kEK. Therefore, by
Corollary 7.6,
k-' = ( ab -1)-1 = (h-1r1a-1 = ba-1.
SinceKis a group,the inverse of an element of Kis also inK. Reading
the preceding line from right to left,we see that ba-1 = k-1 EK. Hence,
b = a (modK).
(3) If a= b (modK) and b = c (modK), then by the definition of
1
congruence, there are r, sEK such that ab- = r and bc-1 = s. Therefore,
1
(ab- )(bc-1 ) = rs
ac-1=rs
Thus, ac-1EK(because r and s are inK). Hence, a= c (modK).
•
IfKis a subgroup of a group G and if a EG, then the congruence class of a modulo
Kis the set of all elements of G that are congruent to a moduloK, that is,the set
{bEGlb =a(modK)} = {bEGlba-1EK}
= {bEG I ba-1 = k, with kEK}.
1
Right multiplication by a shows that the statement ba-
=
k is equivalent to b = ka .
Therefore, the congruence class of a moduloKis the set
{bEGlb=ka,withkEK} = {kalkEK},
which is denoted & and called a right coset of KinG . In summary:
The congruence cla!i'> of a modulo K is the right coset Ka
=
{ka I with k E K}.
When the operation in the group G is a ddition, then a right coset is denotedK+a *
.
•For those who have read Section
I+ a.
6.1: Cosets of an ideal I in a ring
were denoted a
+ I instead of
+ i = i +a
It didn't make any difference there because addition in a ring is commutative, so a
for every i E /. However, in Section
Ka* aK, where aK
=
{ak
8.2 we shall see that when G is
I with k eK}.
nonabelian, it is possible to have
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240 Chapter 8
Normal Subgroups and Quotient Group s
Theorem 8.2
Let K be a subgroup of a group G and let a, c E G. Then a = c (mod K) if and
only if Ka = Kc.
Proof• With minor notational changes, the proof
is essentially the same as that of
Theorem 2.3. Just replace "mod n" with "mod K" and
use Theorem 8.1 in place of Theorem 2.1.
"[a]" with "Ka" and
•
Corollary 8.3
Let K be a subgroup of a group G. Then two right cosets of
disjoint or identical.
K
are either
Proof• Copy the proof of Corollary 2.4 with the same notational changes as in
the proof of Theorem 8.2.
•
Lagrange's Theorem
At this point
vve
temporarily leave the parallel treatment of congruence in the integers
and groups and use right cosets to develop some facts about finite groups that have no
counterpart in the integers.
Theorem 8.4
Let
K be a subgroup
of a group G. Then
(1) G is the union of the right cosets of
K:
G
= ae
U Ka.
G
bijection f:K � Ka. Consequently, if K is
f1 nite, any two right cosets of K contain the same number of elements.
{2) For each a E G, there is
a
Proof• (1) Since every right coset consists of elements of G, vve have aeO-U Kn !;.;; G.
If b EG, then b = eh EKh !:::
LJ
4E(j
Ka, so that G !:::
(2) Define f:K �Ka by f(x)
surjective. If f (x)
=f(y),
then
LJ Kn Hence, G = LJ Kn
ae.rfQ·E(/-
= xa. Then by the definition of Ka1 fis
xa =ya, so that x =y by Theorem 7 .5.
Therefore,/is injective and, hence, a bijection. Consequently, if K
is finite, every coset Ka has the same number of elements as K,
namely I.Kl.
•
G, then the number of distinct right cosets of
index of Hin G and is denoted (G:H]. If G is a finite group,
then there can be only a finite number of distinct right cosets of H; hence, the
index [G:H] is finite. If G is an infinite group, then the index may be either finite
If His a subgroup of a group
Hin G is
called the
or infinite.
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8.1
Congruence and Lagrange's Theorem
241
EXAMPLE 3
Let H be the cyclic subgroup (3) of the additive group Z. Then H consists of all
multiples of 3, and the cosets of Hare just the congruence classes modulo 3;
for instance,
H+ 2 = {h + 2 I hEH} = {3z + 2 I zE Z} = [2].
Since there are exactly three distinct congruence classes modulo 3 (cosets of H),
we have
[Z:H] =3.
EXAMPLE 4
Under addition the group
Z of integers
is a subgroup of the group
0 of ratio­
nal numbers. By the definition of congruence and Theorem 8.2,
Z+a=Z+c
if and
only if
a -
cEZ.
1, then Z+a and Z+ c are distinct cosets because
- c cannot be in Z. Since there are infinitely
many rationals between 0 and 1, there are an infinite number of distinct cosets
of Zin Q. Hence, [Q:Z] is infinite.
Consequently, if 0 <
0<
a -
c
c
<
a
<
< 1, which means that a
Theorem 8.5
Lagrange's Theorem
If K is a subgroup of a finite group G, then the order of K divides the order of
G. In particular, IGI
= IKI [G:K],
Proof"' It is convenient to adopt the following notation. If A is a finite set, then IAI
B are disjoint
Bl =IAI+ IBI. Now suppose that [G:K] = n and
n distinct cosets of Kin G by Kc1, Kc2,
, Kc11. By
denotes the number of elements in A. Observe that if A and
finite sets, then IA U
denote the
•
•
•
Theorem8.4
G =Kc1
U
Kc2
U ·
·
·
U
Kc,..
Since these cosets are all distinct, they are mutually disjoint by Corollary 8.3.
Consequently,
IG I =IKc1I + IKc2I +
For each
c,,
however, IKc�
··
·
+ IKc11I·
=IKI by Theorem
8.4. Therefore,
IGI =I.Kl+ I .Kl+···+ IKI =IKln =IKl[G:K].
•
nsummands
Lagrange's Theorem shows that there are a limited number of possibilities for the
subgroups of a finite group. For instance, a subgroup of a group of order 12 must
have one of these orders:
l,
2, 3, 4, 6, or 12 (the only divisors of 12). Be careful,
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242
Chapter
8
Normal Subgroups and Quotient Groups
however, for these are only the possible orders of subgroups. Lagrange's
Theorem does
not say that a group G must have a subgroup of order k for every k that divides G
I .I
For instance, the a lternating group A4 has order 12 but has no subgroup of order 6
(Exercise
44).
Lagrange's Theorem a lso puts limitations on the possible orders of
elements in a group:
Corollary 8.6
Let
G be a finite group.
(1)
If a E G, then the order of a divides the order of
(2)
If
I GI
G.
= k, then ak = e for every a E G.
Proof.,. (1) If aE Ghas order n, then the cyclic subgroup (a) of Ghas order n
by Theorem 7.15. Consequently, n divides 161 by Lagrange's Theorem.
(2) If a E G has order n, then nI k by part (1), say k = nt. Therefore,
cl = a"' = (d')' = et = e. •
The Structure of Finite Groups
A major goal of group theory is the classification of all finite groups up to isomor­
phism; that is, we would like to produce a list of groups such that every finite group is
isomorphic to exactly one group on the list. This is a problem of immense difficulty,
but a number of partial results have already been obtained. Theorem 7.19, for exam­
ple, provides a classification of all cyclic groups; it says , in effect, that every nontrivial
finite cyclic group is isomorphic to exactly one group on this list: Zz., Z3, �•
• •
•
•
All
finite abelian groups will be classified in Section 9.2.
We now use Lagrange's Theorem and its corollary to classify all groups of prime
order and all groups of order less than 8. In the proofs below enough of the necessary
calculations are included to show you how the argument goes, but you should take
pencil and paper and supply all the missing computations.
Theorem 8.7
Let p be a positive prime integer. Every group of order p is cyclic and isomor­
phic to�.
Proof .,. If G is a group of
order p and a is any nonidentity element of G, then
the cyclic subgroup
(a) is
a group of order greater than 1. Since the
{a) must divide p and since p is prime, (a) must be a
group of order p. Thus (a) is all of G, and G is a cyclic group of order p.
order of the group
Therefore, G= Zp by Theorem 7.19.
•
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8.1
Congruence and Lagrange's Theorem
243
Theorem 8.8
Every group of order 4 is isomorphic to either z, or Z2 x Z2•
Proof ,.. Let G be a group of order 4. Either G contains an element of order 4 or
it does not. If it does, then the cyclic subgroup generated by this element
has order 4 by Theorem 7.15 and, hence, must be all of
G. Therefore, G
is a cyclic group of order 4, and G = z, by Theorem 7.19.
Now suppose that G does not contain an element of order 4. Let e, a,
b, c be the distinct elements of G, with e the identity element. Since every
element of G must have order dividing 4 by Corollary 8.6 and since e is
the only element of order 1, each of a, b, c must have order 2. Thus the
operation table of G must look like this:
e
a
b
c
b
c
e
e
a
a
a
e
b
b
c
c
e
e
In order to fill in the missing entries, we first consider the product ab. If
ab = e, then ab = aa and, hence, a = b by cancelation. This is a contra­
diction, and so ab ::/- e, If ab = a, then ab = ae and b = e by cancelation,
another contradiction. Similarly, ab = b implies the contradiction a = e.
Therefore, the only possibility is ab = c. Similar arguments show that
there is only one possible operation table for G, namely,
e
a
b
c
e
e
a
b
c
a
a
e
c
b
b
b
c
e
a
c
c
b
a
e
Letf G-.+Z2 X Z2 be given by j(e) = (0, O),j{a) = (1, O),j{b) = (0, 1),
an d f(c) = ( 1, 1). Show that/is an isomorphism by comparing the
operation tables of the two groups.
•
Theorem 8.9
Every group G of order 6 is isomorphic to either "4, or S3•
Proof.. If G contains an element of order 6, then G is a cyclic group of order 6
and, hence, is isomorphic to Z6 by Theorem 7 .19. So suppose G contains
no element of order 6. Then every nonidentity element of G has order
2 or 3 by Corollary 8.6. If every nonidentity element of G has order 2,
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244
Chapter
8
Normal Subgroups and Quotient Groups
c and d are
c, d, cd} is closed under
multiplication (because c2 = e = rP and cd =de). Hence, His a subgroup
then G is an abelian group by Exercise 27 of Section 7.2. If
nonidentity elements of G, then the set H= {e,
of G by Theorem 7.12. This is a contradiction since no group of order 6
can have a subgroup of order 4 by Lagrange's Theorem. Therefore, the
nonidentity elements of G cannot all have order 2, and G must contain
an element a of order
3. Let N be the cyclic subgroup (a} = {e, a, a2} and
b be any element of G that is not in N. The cosets Ne = { e, a, a1} and
Nb {b, ab, a2b} are not identical since hiN=Ne and, hence, must be
disjoint (Corollary 8.3). Therefore, G consists of the six elements e, a, a2,
b, ab, a1b.
let
=
We now show that there is only one possible operation table for G.
What are the possibilities for ll? We claim that b2 cannot be any of
a, a1,
b, ab,
or a2b. For instance, if b2 =a, then b4 =a1. However, b either has
order 2 (in which case a2= b4=b11l= ee=e, a contradiction) or order 3
(in which case a2 =h4=!:lb =eh=b, another contradiction since b Ii!'. N).
Similar arguments show that the only possibility is ll =e.
ba. It is easy to see that ha cannot
b, e, a, or a2 (for instance, ha= a implies b =e). So the only
possibilities are ha =ab or ba =t?b. If ba =ab, then verify that ba has
Next we determine the product
be any of
order 6 by computing its powers. This contradicts our assumption that
G has no element of order 6. Therefore, we must have
ha=a2b. Using
these two facts:
b2 = e
ba = t?b,
and
we can now compute every product in G. For example,
ht?=(ba)a =
(a2b)a = a1(ba)=t?a2b= a4b =ab.
Verify that the operation table for G must look like this:
e
a
t?
b
ab
ab
a2b
e
e
a
a2
a
a
al
e
b
ab
a2b
b
a2
b
ab
ti-
e
a
crb
b
ab
b
a2b
ab
e
ti-
a
ab
b
t?b
a
e
a2
crb
dlb
ab
b
a2
a
e
a1b
By comparing tables, show that G is isomorphic to S3 under the
correspondence
G
e
a
al
b
ab
,J,
,J,
,J,
,J,
,J,
2
2
:)G
2
3
�)G
2
1
�)G
2
1
:)G
2
2
a1b
�)G
.i
2
3
�). .
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� ...... k
8.1
Congruence and Lagrange's Theorem
245
The last three theorems provide a complete classification of all groups of order less
than 8, as summarized in this table:
If G bas order
then
G is isomorphic to
2
Z2
3
Z3
4
"/4or Z2 X Z2
6
7L6 or S3
5
Zs
7
Z1
The classification of groups is discussed further in Chapter 9, particularly in Section 9.5
where the preceding chart is extended to order 15.
• Exercises
A. 1. Let Kbe a subgroup of a group Gand let a EG. Prove that Ka
if a Ek.
=
Kif and only
In Exen:ises 2---6, Gis a group and Kis a subgroup of G. List the distinct right cosets ofKin G.
2. K = {r0,
v}: G
or 7.1.A.]
=
D4 [fhe operation table for D4 is in Example 5 of Section 7.1
3. K = {r0, r., r,,. r3}; G
4_
=
D4•
= {(123) (123)}·
=
.
K
G S
3
123' 1 32
'
5. K = {1, 17}; G = U32.
6. K =
(3); G
=
U32.
In Exercises 7-1 /, Gis a group and His a subgroup of G. Find th£ index [G:H].
7. H = {ro, r2}; G = D4.
8. H
=
9. H
=
(3); G
=
Z12·
=
Z20•
(3); G
10. His the subgroup generated by 12 and 20; G = Z40•
11. His the cyclic subgroup generated by
12.*
(12 23 43 4);
1
=
4
•
G S
(a)
Let K = {(l), (12 )(34), (13)(24), (14)(23)}. Show that Kis a subgroup of
A
,
4 and hence, a subgroup ofS
4
• [Hint: Theorem 7.12.]
(b)
State the number of co sets of Kin A
4
• Don't list them.
(c)
State the number of cosets of Kin S4• Don't list them.
*Skip this exercise if you haven't read Section 7.5.
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246
Chapter 8
Normal Subgroups and Quotient Groups
In Exercises 13-15, K is a subgroup of G Determine whether the given cosets are
disjoint or identical.
13.
G= Z; K=(7)
(a)
14. * G
(a)
15.
(b) K= 4 and K + 137
K + 4 and K + 3
(c) K + (-4) andK + 59
= S4; K is the subgroup of Exercise 12.
(b)K(l234) and K( 1324)
K(l2) andK(34)
G= U32;K=(9)
(b) K9 and K25
G is the cyclic group (a) and lal = 15. If K = (a3), list all the distinct
cosets of Kin G.
(a) Kl7 andKl9
16. Suppose
17. What are the possible orders of the subgroups of
(a)
Z24
(b) S4
G when G is
(c) D4 X Z10
18. Give examples, other than those in the text, of infinite groups G and H such that
(a)
[G:H] is finite
(b) [G:H ] is infinite
19. Let G be a finite group that has elements of every order from 1 through12.
What is the smallest possible value of
IGI?
G has fewer than 100 elements and subgroups of orders I 0 and 25.
What is the order of G?
20. A group
21. Let Hand K, each of prime order p, be subgroups of a group G. If
prove that
Hn K
=
22. If HandKare subgroups of a finite group G, prove that
divisor of
H oft K,
{e}.
IH n Kl is a common
IHI and IKI.
G is a group with more than one element and G has no proper subgroups,
prove that G is isomorphic to Z for some prime p.
B. 23. If
P
G is a group of order 25, prove that either G is cyclic or else every
nonidentity element of G has order 5.
24. If
25. Let a be an element of order 30 in a group G. What is the index of
(a4) in the
group (a}?
26. Prove that a group of order 8 must contain an element of order 2.
27. If
n > 2, prove that n
28. If
n > 2, prove that the order of the group U11 is even.
-
1 is an element of order 2 in
U,..
29. Let Hand K be subgroups of a finite group G such that Ki;;;;
and [H:K] is finite. Prove that [G:K]= [G:H][H:K].
H, [ G:H] is finite,
[Hint: Lagrange.]
30. Let Hand K be subgroups of an infinite group G such that
K k H, [G:H] is
[G:K] is finite and [G:K] = [G:H][H:K].
[Hint: Let Hai. Ha2,
Ha,, be the distinct co sets of Hin G and let Kb1o
K/Ji., . , Kbm be the distinct cosets of Kin H. Show that Kb1a1 (with 1 s i s m
and 1 s j s n) are the distinct cosets of Kin G.]
finite, and [HK]
:
is finite. Prove that
• • • ,
.
.
•Skip this exercise if you haven't read Se<:tion 7.5.
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8.1
Congruence and Lagrange's Theorem
247
31. If Gis a group of even order, prove that G contains an element of order 2.
32. If G is an abelian group of order 2n, with n odd, prove that Gcontains exactly
one element of order 2.
33.
(a) If a and beach have order 3 in a group and a
[Hint: What are a-1 and b-1?]
2
""
b2, prove that
a
b.
=
(b) If G is a finite group, prove that there is an even number of elements of
order 3 in G.
34. Let Gbe an abelian group of odd order. If
a1, al> a3,
• • •
, a,, are the distinct
an = e.
elements of G(one of which is the identity e), prove that a1aza
3
• •
•
35. If p and q are primes, show that every proper subgroup of a group of order pq
is cyclic.
36. LetH and Kbe subgroups of
a
finite group Gsuch that
[G:H] = p and [G:K]
=
q,
with p and q distinct primes. Prove that pq divides [G:H () K].
37. Let Gbe an abelian group of order n and let kbe a positive integer. If (k,n) =
prove that the functionfG� Ggiven by /(a) =
38. If G is a group of order
n
and Ghas '1!'
d' is an isomorphism.
1
- subgro ups, prove that G
=
1,
(e) or
G=Z2.
C. 39. Let Gbe a nonabelian group of order 10.
(a) Prove that Gcontains an element of order 5. [Hint: Exercise 27 of
Section 7.2.]
(b) Prove that Gcontains five elements of order 2. {Hint: Use techniques
similar to those in the proof of Theorem 8.9.]
40. If a prime p divides the order of a finite group G, prove that the number of
elements of order p in Gis a multiple of p
- 1.
41. Prove that a group of order 33 contains an element of order 3.
a and b such that !al= 4, lhl = 2, and
ha = a3b. Show that Gis a group of order 8 and that G is isomorphic to D4•
42. Let Gbe a group generated by elements
43. Let Gbe a group generated by elements
a
and
b such that la l
=
4, b2
=
c?, and
ha= a3b. Show that Gis a group of order 8 and that Gis isomorphic to the
quaternion group of Exercise 16 in Section 7.1.
44.* (a) Show that A4 (which has order 12 by Theorem 7.29) has exactly three
elements of order 2.
(b) Prove that the elements of order 2 and the identity element form a
subgroup.
(c) Prove that A4 has no subgroup of order 6. Hence, the converse of
Lagrange's Theorem is false. [Hint: If N is a subgroup of order 6, use
Theorem 8.9 to determine the structure of N and use part (b ) to reach a
contradiction.]
*Skip this exercise if you haven't read Section 7.5.
...
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248
Chapter
Ill
8
Normal Subgroups and Quotient Groups
Normal Subgroups
Suppose G is a group and K is a subgroup. Our goal in this section and the next is
to create a new group
(if possible),
whose elements are the right cosets of K (that is,
congruence classes mod K)-much as we created Z.11, whose elements are congruence
classes of integers.
Recall that the definition of addition of congruence classes of integers in Chapter 2
depended on part ( 1) of Theorem 2.2, which states
If
a=
b(modn) and
c=
d(modn), then
a+
c =
b + d(modn).*
If K is a subgroup of a multiplicative group G, then the translation of this statement
to congruence mod K is
If a = b(mod K) and
c =
d (mod K), then
ac =
bd (mod K).
Unfortunately, however, statement(*) is false for some subgroups. (see Exercise 2 for
an example). Nevertheless, there is a class of subgroups for which statement(*) is true.
We shall identify these "special" subgroups in this section and define multiplication of
their right co sets in Section 8.3. t
Recall that if K is a subgroup of G, then the right coset Ka is the set Ka
=
{kalkEK}. Similarly, the left coset aK is defined to be the set
aK
{aklkEK}.
=
EXAMPLE 1
Let K be the subgroup
{r0, v} of D4, whose operation table is shown below. The
{r0 d, v o d}
{ d, r3} and the left coset dK is the set
right coset Kd is the set
{d0 ro, d0 v)
=
o
=
{d, ri}. So Kd :# dK.
D4
rn
ro
r1
rz
r3
ro
r1
r2
r3
ro
r1
'•
d
h
r3
d
h
ro
h
'1
r2
'1
r2
'2
r3
r3
r3
ro
d
d
v
h
h
d
v
t
t
h
d
v
v
v
h
d
v
v
v
d
h
v
d
'2
v
d
h
h
ro
r3
'1
ro
r1
r3
'2
r3
r1
r2
ro
r1
rl
'2
r3
ro
*We don't deal with integer multiplication here because the integers form a group under addition, but
not under multiplication. Similarly in Chapter
6, when
developing the basic facts about congruence
and cosets in rings, we dealt only with the additive group of a ring and ignored its multiplication.
6 when we needed to proveTheorem 6.5 (the
2.2 for rings)--the discussion did not apply to every subring, but only to
of which is a special kind of subring.
tEssentially the same thing was done in Chapter
analogue ofTheorem
ideals, each
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8.2
Normal Subgroups
249
EXAMPLE 2
Let Nbe the subgroup
Nv
and the left coset
=
{r0, ri. r2, r3} of D4• Then the right coset Nv is the set
{r0 • v, r1 ov, r2 • v, r3 o v}
=
{v, d, h, t}
vN is the same set:
vN
=
{v or0, v o'" v o rz, v or3}
=
{v, t,
h, d}.
So in this case, Nv
coset of
= vN. * Similar calculations (Exercise 3) show that every right
N is also a left coset, that is,
Nr0
=
roN,
Nr1
=
r1N,
Nr2
=
r�,
Nr3
=
r�
Nd
=
d.V,
Nh
=
hN,
Nt
=
tN,
Nv
=
vN.
Subgroups w ith this property have a special name.
Definition
I
A subgroup N of a group G is said to be normal if Na
=
aNfor every a E G.
EXAMPLE 3
N
=
{r0, r1, r2, r3} is a normal subgroup of D4, but K = {r0, v} is not,
as
shown
in Examples 1 and 2.
EXAMPLE 4
If N is a subgroup of an abelian group G and a E G, then IUl
n
E N, so that the right coset Na is the same
as
= an
for every
the left coset aN. Hence,
Every subgroup of an abelian group is normal.
EXAMPLE 5
Let Mbe the subgroup
{r0, r2} of D4• Then the operation table for D4 in
r2 for every a ED4• So it is
Example 1 shows that r0 •a= a o r0 and r2 o a= a o
certainly true that M
a == aMfor every aE D4• Hence, Mis a normal subgroup
of D4.
In Example 5, the subgroup Mis the center of D4 (see Example 10 of Section 7.3).
So the center of D4 is a n ormal subgroup. The same thing is true in general.
*Remember that the elements of a set may be listed in any order.
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250 Chapter 8
Normal Subgroups and Quotient Groups
EXAMPLE 6
The center Z(G) of a group G is the subgroup
Z(G) =
{cEGI cg= gc for every gEG}
(Theorem 7.13). Since
Z(G)a
=
ca= ac for every cEZ(G) and aEG, we see that
aZ(G) for every aEG. Hence, Z(G) is a normal subgroup of G.
Other examples of normal subgroups appear in Exercises 3--5, 7-9, 14, and 23.
Examples 4-6, though important, are misleading in that the elements of the normal
subgroup N commute with all the other elements of the group in each case. In the gen­
eral case, however, this is not necessarily true. When
The condition Na
=
N is a normal subgroup of G, then,
aN does not imply that '"'
= an
for every 11 E N.
EXAMPLE 7
As we saw in the Example 2,
N= {r0, 1'1r2,1'3} is a normal subgroup of D4• In
Nv = vN. However, v does not commute with all the elements of N.
For instance, r3 o vE Nv and v o r3 E vN, but the operation table for D4 shows that
particular,
r3 o v = t
even though
Thus, if
and
v o r3= d,
so
r3 o v
*
.v o
r3,
Nv= vN.
N is a normal subgroup of G, the elements of N may not commute
G. Nevertheless, you can think of the normal subgroup N
with every element of
as
providing a weak version of commutativity in the following sense.
If
n E N, and a E G, then for some n1, n2 E N,
na =
because na E Na and
an1 and
a11 =
"2"•
Na= aN and similarly, anE aN and aN= Na.
EXAMPLE 8
Once again, consider the normal subgroup
operation table for D4 shows that r3 o
N= {r0, r1, r2, r3) of D4• The
v= t and v o r1= t. Hence,
This is the first part of the preceding boldface statement, with
andn1
n =
r3,
a = v,
= r1•
Our goal at the beginning of this section was to find a class of subgroups for which
statement(•) on page 248 (the group theory analogue of Theorem 2.2) is true. Normal
subgroups are exactly what's needed.
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8.2
Normal Subgroups
251
Theorem 8.10
Let N be a normal subgroup of a group G.
If a
=b
(mod N) and c = d (mod N), then ac
= bd (mod
N).
The proof is essentially a translation into multiplicative notation of the proof
of part (1) of Theorem 2.2, with commutativity of integers replaced by the weak
commutativity in G provided by the normal subgroup N.
Proof ofTheorem 8.10 ... By the definition of congruence, there are elements
m, nEKsuch thatab-1 = mandcat = n. Then
(ac)(bd)-1 = acd-1b-1
[Corollary
7.6]
[Because cd-1 = n.)
= anb-1
Now an E aN and aN = Na by normality, so an = n'lfl for some n2EN. Hence,
(ac)(bd)-1 = anb-1
= n,,ab-1
= nim
[Becau.re ab-t = m EN.]
Therefore, (ac)(bdr' = n2m EN, and ac = bd (mod
N).
•
We close this section with a theorem that provides alternate descriptions of nor­
mality. Verifying condition
(2)
or
that a given subgroup is normal.
(3)
in the theorem is often the easiest way to prove
Theorem 8.11
The following conditions on a subgroup Nof a group G are equivalent:
(1} N is a normal subgroup of G.
(2) 1r1Nai;;;Nfor everyaEG, where a-Wa
=
{a-1nalnEN}.
(3) aNtr1 i;;; Nfor everya EG, where aNa-1
=
{ana-1 In EN}.
(4) a-1Na
= Nfor
every a EG.
(5) aNer1
= Nfor
everyaEG.
Note that in
(4),
a-1Na = N does not mean that a-1na = n for each nEN;
all it means is that a-tna = n for some n EN . Analogous remarks apply to
1
1
(3), and (5).
(2),
Proof ofTheorem 8.11 ... (1) �(2) SupposenENand a-lnaEa-1Na. We must
show that a-lna EN. Note that na is an element of the right coset Na.
Since N is normal by
(1 ), Na = aN. Hence, na = an1 for some n1 EN.
Thus a-1na = a- 1an = en = n1EN . Therefore, a-1Na t;;;N.
1
t
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252
Chapter 8
Normal Subgroups and Quotient Groups
(2) <::>(3) If (2) holds for every element of G, then it holds with a-1 in
place of a, that
But(a- 1) -1
is,
= a,
so that (••) is statement (3): aNa-1 !';;;N. Similarly, i f
( 3 ) holds for every element of G , then it holds with a-1 in place o f a,
which implies statement (2).
(3) =>(4) Since (3) implies (2), we have a-iNa !;;;; N. To prove
N !;;;; a-1Na, suppose nEN. Then n
some 1'2EN. Thus n
Therefore, tr1Na
=
=
= a-1(ana-1}a. By (3) ana-1 = n2 for
a-1n2aEa-1 Na, which proves that N !;;;; a-1Na.
N.
(4) <::>(5) If(4) holds for every element of G, then it holds with a-1 in
place of a, that is,
N
=
(a-1r1 Na-1
=
aNa-1•
Similarly, if (5) holds for every element of G, then it holds with a-1 in
place of a, which implies statement (4).
(5) =>(1) Suppose nENand anE aN. Then ana-1E aNa-1
so that ana-1
=
Nby(5),
n3 for some n3EN. Multiplying this last equation on the
=
right by a shows that an
= lljaENa. Therefore, aN!';;;Na. Conversely, if
na ENa, then a-1naEa-1Na = Nbecause(5) implies (4). Hence, a-1na =
n4 for some n4EN. Multiplying on the left by a shows that na
Thus Na!;;;; aN. Therefore, Na
subgroup of G.
=
=
an4EaN.
aN for every aE G and N is a normal
•
EXAMPLE 9
.
Verify that A
=
1 2 3
1 2 3
1 2
1 2 3
2 3 1
3 1 2
{( )(
)( 3)}.
JS a subgroup of S3• You
could show that A is a normal subgroup by calculating the right and left cosets,
but that is cumbersome and time consuming. It's easier to proceed as follows. If
c ES3, then by Exercise 20 of Section 7.4, c- 1 Ac is a subgroup of order 3. But
A is the only subgroup of order 3 in S3(all the other nonidentity elements of
S3 have order 2, and hence, cannot be in a group of order 3 by Corollary 8.6).
Therefore, we must have c-1Ac
=
A. Thus, A is a normal subgroup by part (5)
of Theorem 8.11.
• Exercises
A. 1. Let K be a subgroup of a group G and let aE G. Prove that aK = Kif and only
if aEK.
2. Let K be the subgroup
{r0, v} of D4• Show that r1 = t(mod K) and r2;;;; h
(mod K), but r o r2 -;/: to h(mod K).
1
3. Prove that N = {r0, r o r2, r3} is a normal s ubgroup of D4 by listing all its right
1
and left cosets.
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8.2
4. IfG is a group, show that
5.
(a) Prove that G
=
253
(e) andGare normal subgroups.
{ (� �) I
0}
{ G ) I Ill}
a, b, dE Ill andbad oft
matrix multiplication and that N
(b) Use Theorem 8.11 to show that
6. Prove that
Normal Subgroups
{ G � �), G � !)}
=
l
is a group under
bE
is a subgroup ofG.
N is normal inG.
is a subgroup of S� but not normal.
7. LetG and Hbe groups. Prove that G*
=
{(a, e) I a EG} is a normal subgroup
ofGXH.
8.
(a) List all the cyclic subgroups of the quaternion group (Exercise
16 of
Section 7.1).
(b) Show that each of the subgroups in part (a) is normal.
9. Let
N be a subgroup of a groupG. Suppose that , for each aEG, there exists
Na bN. Prove that N is a normal subgroup .
b E G such that
=
I 0. IfG is a group, prove that every subgroup of
with Exercise 14.]
Z( G) is normal inG. [Compare
11. A subgroup
N of a groupGis said to be characteristic if f{N) !;;;; N for every
automorphism/ of G. Prove that every characteristic subgroup is normal.
(The converse is false , but this is harder to prove.)
12. Prove that for any groupG, the center
13. Let
Z(G) is a characteristic subgroup.
N be a subgroup of a group G. Prove that N is n ormal if and only if
f(N)
=
N for every inner automorphism/ of G.
14. Show by example that if Mis a normal subgroup of N and if
N is a normal
subgroup of a group G, then M need not be a normal subgroup of G; in
other words , normality isn't transitive. [Hillt: Consider M {v, r0} and
=
N
=
{h, v, r2, r0} in D4].
15.* Prove that A,, is a normal subgroup of S,.. [Hn
i t: If
aE S,, and TEA,,, is
u-1 TU even or odd? See Example 7of Section 75
. ].
B. 16. If Kis a normal subgroup of order 2 in a groupG, prove that K!;;; Z(K).
[Hillt: If K
=
{e, k} and a EG, what are the possibilities for aka-1?]
17. LetfG-+Hbe a homomorphism of groups and let K =
Prove that Kis a normal subgroup of
18. If Kand
{aE Gl/(a)
=
eH}·
G.
N are normal subgroups of a groupG, prove that Kn N is a normal
subgroup ofG.
19. Let N and Kbe subgroups o f a group G. I f N i s normal in G, prove that N n K
is a normal subgroup of K.
20.
(a) Let Nand Kbe subgroups of a group G. If Nis normal inG, prove that N K
{nk InEN, kEK} is a subgroup ofG. [Compare Exe�ise 26(b) of Section 7.3.]
=
(b) If both Nand Kare normal subgroups ofG, prove that NKis normal .
•skip this exercise if you haven't read Section 7.5.
...
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254 Chapter 8
Normal Subgroups and Quotient Groups
21. If K
andN
that nk
are normal subgroups of a group G such that Kn
= kn for
N = (e),
prove
every nEN, kEK.
22. If f:G-+ His a surjective homomorphism of groups and if
N is a normal
subgroup of G, prove that.f(N) is a normal subgroup of H.
23. Let N be a subgroup of a group G of index 2. Prove that
N is a normal
subgroup as follows.
(a) If a f1 N, prove that the coset Na consists of all elements of G that are
not inN.
aE G, prove that a-1Na 1;; N and apply Theorem 8.11. [Hint: If
f1 N and nEN, a-1na is either in Nor in Na by part (a ). Show that the
(b) For each
a
latter possibility leads to a contradiction.]
24. LetN
GL(2,
= {AE GL(2, Ill) ldet AEO}. Prove that N is a normal subgroup of
Ill). [Hint: Exercise 32 o f Section 7.4.]
25. Prove that SL(2, R) is a normal subgroup of GL(2, IR). [Hn
i t: SL(2, IR) is
defined in Exercise 23 of Section 7.1 Use Exercise 17 above and Exercise 32 of
Section 7 .4.]
26. Let Hbe a subgroup of order n in a group G. If His the only subgroup of
order n, prove that His normal . [Hint: Theorem 8 .11 and Exercise 20 in
Section 7.4.]
27. Prove that a subgroup
property:
N of a group G is normal if and only if it has this
ahEN if and only if baEN, for all a, b EG.
28. Prove that the cyclic subgroup
each g E G, ga=
(a) of a group G is normal if and only if for
tlg for some kE Z.
29. LetN be a cyclic normal subgroup of a group G, and Hany subgroup ofN.
Prove that His a normal subgroup of G. [Compare Exercise 14.]
30. Let A and
B be normal subgroups of a group G such that An B = (e) and
AB G (see Exercise 20). Prove that A X B = G. [Hint: Define f:A X B -+ G
by f(a, b) =ah and use Exercise 21.]
=
31. Let Hbe a subgroup of a group G and letN(H) be its normalizer (see
Exercise 39 in Section 7.3 ). Prove that
(a) His a normal subgroup ofN(H).
(b) If His a normal subgroup of a subgroup
K of G, then K !;;;N(H).
;
32. Prove that Inn G is a normal subgroup of Aut G. [See Exercise 37 of Section 7.4.]
33. Let T be a set with three or more elements and let A(1) be the group of all
permutations of T. If
aE T,
let Ha=
{jEA(1) jf(a) =a}. Prove that H0 is a
subgroup of A(T) that is not normal .
34. Let G be a group that contains at least one subgroup of order n. Let N
= nK,
K of order n. Prove that N is
1
1
verify that a- Na = na- Ka,
where the intersection is taken over all subgroups
a normal subgroup of G. [Hint: For each
aE G,
where the intersection is over all subgroups K of order n; use Exercise 20 of
Section 7 .4.]
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8.3
35. Let
Hbe a subgroup of a group Gand let N
normal subgroup of
G.
36. If Mis a characteristic subgroup of N and
G, prove that Mis a normal subgroup of
=
Quotient
II
=
255
na-1Ha. Prove that Nis a
<><=11
N is a normal subgroup of a group
G. [See Exercise 11.]
37. Let G be a group all of whose subgroups are normal. If
there is an integer k such that ab
Groups
btl.
a, b E G, prove that
Quotient Groups
Let N be a normal subgroup of a group
G. Then
GINdenotes the set of all right cosets of Nin G.
Our first goal is to define an operation on right cosets so that
G/Nbecomes a
group.
Since right cosets are congruence classes, our experience with Zand other rings suggests
that it would be reasonable to define such
an
operation as follows: The product of the
Na (the congruence class of a) and the coset Nb (the congruence class of b) is the
coset Nab (the congruence class of ab). In symbols, this definition reads
coset
(Na)(Nb) Nab.
=
As in the past, we must verify that the definition does not depend on the elements
chosen to represent the various cosets, and so we must prove
Theorem 8.12
Let N be a normal subgroup of a group G. If Na =Ne and Nb= Nd in G/N,
then Nab=Ned.
Proof"' Na
=
Ne implies that a (mod N) by Theorem 8.2, similarly, Nb Nd
b d (mod N). Therefore, ah cd (mod N) by Theorem 8.10.
= c
implies that
=
=
Hence. Nab=Ned by Theorem 8.2.
=
•
Theorem 8.13
Let N be a normal subgroup of a group G. Then
(1) G/N is a group under the operation defined by (Na)(Nc}=Nae.
(2) If G is finite, then the order of G/N is IGl/M
(3) If G is an abelian group, then so is G/N.
The group
G/N is called the quotient group or factor group of Gby N.
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.
ICDl
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llllnl•_..,.lillll��:Dgbb�...-.:lit.
256
Chapter 8
Normal Subgroups and Quotient Groups
Proof ofTheorem 8.13 � (1) The operation in G/Nis well defined by Theorem 8.12.
N = Ne is the identity element in G/N since (Na)(Ne) =
Nae =Na and (Ne)(Na) =Nea=Na for every Na in G/N. The inverse
of Na is the coset Na-1 since (Na)(Na-1)=Naa -I =Ne and, similarly,
(Na-1)(Na)=Ne. Associativity in G/Nfollows from that in G:
The co set
[(Na)(Nh)](Nc) = (Nab)(Nc) = N(ab)c = Na{hc) = (Na)(Nhc)
= (Na)[(Nb)(Nc)].
Therefore, G/Nis a group.
(2) The order of G/ N is the number of distinct right cosets of N, that is,
the index [GN]. By Lagrange's Theorem,
(3) Exercise 11.
[G.N] = IGl/INI.
•
EXAMPLE 1
In Example 2 of Section 8.2 we saw that N
group of
= (r0, r1., r2, r3} is a normal sub­
D4• The operation table for D4 in Example 1 of Section 8.2 shows that
Nro= {ro 0 ro, r
r , r2 ° ro, r3 ° ro} = {1"1>1 r., r2, '3}
1° o
Nv= {r0 o v, r1 o v, r2 o v, r3o v} = {v, d, h, t}.
Since every element of D4 is in either Nr0 or Nv and since any two cosets of N are
either disjoint or identical (Corollary 8.3), every coset of N must be equal to Nr0
or Nv. In other words, D4fN= {Nr0,
Nv}. Since r0 o v=v= vo r0 and v o v= r0,
the operation table for the quotient group D4fN is
Nr0
Nv
Nr0
Nr0
Nv
Nv
Nv
Nr0
By Theorem 8.7, D4fN is isomorphic to the a dditive group Z2•
EXAMPLE 2
In Example 5 of Section 8.2 we saw that M
= {r0, r2} is a normal subgroup of D4•
Using the operation table for D4, we find that D4fM consists of these four cosets:
Mh = {h,
v} =Mv
Md= {d, t} =Mt.
D4/M
Mr0, Mri. Mh, and Md. When we compute products in D4/M, we express the
answers in terms of these four cosets. For instance, since do r =v in D4, we have
1
We shall choose one way of representing each coset and list the elements of
as
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8.3
(Md)(Mri) = M(do r1) = Mv; but Mv = Mh, so we write
table below. You should fill in the missing entries:
Mr0
Mr0
Mr0
Mr1
Mr1
Mr1
Mh
Mr1
Mh
Mr0
Md
Md
Md
Mh
Mh
Mh
Md
Mr0
Quotient
Groups
257
(Md)(Mr1) = Mh in the
Md
Md
The completed tabel shows that DJM is an abelian group in which every nonidentity
element has order 2 (Exercise 3). So DJM is not cyclic. Hence, D4/M is isomorphic
to Z2 X Z2 by Theorem 8.8.
Examples 3-7 deal with abelian groups. So every subgroup is normal.
EXAMPLE 3
In the additive group Z1:i. let Nbe the cyclic group (4} = {O, 4, 8}. These four
cosets of N contain every element of Z12:
N+0
=
{O, 4, 8}
=
N
N+1 = {l, 5, 9}
N+ 2 = {2, 6, 10}
N+ 3
=
{3, 7, 11}.
Hence, every coset is one of these four. For instance, 5 is in N+ 1 and 5 is also
in N +5 (Why?). So the two cosets are not disjoint. Hence, N+ I = N+5 by
Corollary 8.3. Similarly,
N+4=N+O
and
N+6=N+ 2 .
Using these facts, w e see that the addition table fo r Z12/N i s
N+O
N+l
N+2
N+3
N+O
N+O
N+1
N+2
N+3
N+l
N+l
N+2
N+3
N+O
N+2
N+2
N+3
N+O
N+I
N+3
N+3
N+O
N+l
N+2
Verify that N+1 has order 4. So Zri/N is a cyclic group of order 4 and hence, is
isomorphic to "14, by Theorem 7.19.
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258
Chapter
8
Normal Subgroups and Quotient Groups
EXAMPLE 4
Let Nbe the cyclic subgroup ((1, 2)) of the additive group G =Li X Z4• Since
(1 , 2) + (1, 2) (0,0), we see that N {(O, 0) ,(1, 2)}. Consequently, G/N con­
=
=
sists of these four cosets
N+ (0,0)
=
N+ (1, 0)
=
{(O, 0),(1, 2)}
=
{(1, 0) , (0, 2)}
N + (1 , 2)
N+ (0, 2)
=
{(O,1 ), (1, 3)}
=
N + (1 , 3)
N+ (1,1) = {(1,1), (0, 3)}
=
N + (0, 3)
N+ (0,1)
=
and has the following addition table:
N+ (0, 0)
N+ (1,0)
N+ (0,1)
N+
N+ (0, 0)
N+ (0, 0)
N+ (1,0)
N+ (0, l)
N + (1, 1)
N+ (1, 0)
N+ (l,O)
N+ (0,0)
N+ (1, l)
N+(0, 1)
N+ (0, 1)
N+ (0, 1)
N+ (1,1)
N+ (1,0)
N+(0, 0)
N+ (1, 1)
N+ (1, 1)
N+ (0,1)
N+ (0,0)
N+(l,O )
Use the table to verify that
Therefore,
G/Nis a cyclic group of
G/N= 14 by Theorem 7.19.
order
(1, 1)
4 generated by N+ (O, 1).
It is not always necessary (or even possible) to write out the operation table for a
quotient group
G/ Nin order to determine its structure, as was done
in Examples 1-4.
EXAMPLE 5
By Theorem
2.10, the group U14
.
Mbe the cyclic subgroup (13)
Theorem
8.13. Therefore,
=
=
{l, 3, 5,
{ 1, 13}.
9, 11,13} and
Then IU14/Ml
=
thus has order 6. Let
IU14I
IMI
6
=2=
U1,JMis isomorphic to Z3 by Theorem
3 by
8.7.
EXAMPLE 6
In the additive group Z, let K be the cyclic subgroup
(4) ={O, :t4, ±8, ±12, . . . }.
As we saw in Example 1 of Section
a=
b (mod
4)
8.1,a""' b (mod 4) means a - b EK. Hence,
if and only if
a=
b (mod K).
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Quotient
8.3
Groups
259
a modulo 4 (the congruence class
[aD is exactly the same as the set of integers that are congruent to a modulo K
(the coset K+a). In other words, [a] = K+a. Arithmetic is the same in either
So the set of integers that are congruent to
notation:
Ka+Kh= K(a+ b)
is the same
as
[a] +[b] = [a+b].
Therefore, Z/ K is the group of congruence classes modulo 4, that is, Z/K= Z4• The
same argument works with any positive integer n in place of 4:
If K is the
cyclic subgroup (n) of Z, then Z/K
=
z•.
EXAMPLE 7
The subgroup Z of integers in the additive group Q of rational numbers is
normal since Q is abelian. Example 4 of Section 8.1 shows that there
are
infi­
nitely many distinct cosets of Z in Q. Consequently, the quotient group Q/Z is
an infinite abelian group. Nevertheless, every element of Q/Z has.finite order
(Exercise 25).
The Structure of Groups
If N is a normal subgroup of a group G, then the structure of each of the groups N,
G, and G/Nis related to the structure of the others. 1f we know enough information
about two of these groups,
as
we
can often determine useful information about the third,
illustrated in the following theorems.
Theorem 8.14
Let N be a normal subgroup of a group
abir'b-1 EN for all a, b E G.
G. Then G/N
is abelian if and only if
Proof• G/N is abelian if and only if
Nab= NaNb = NbNa= Nba
for all
a, b E G.
Nab= Nba if and only if (ab)(ba)-1ENby Theorem8.21; and
(ab)(ba)-1 = aba- 1b - 1 by Corollary 7 .6. Therefore, G/Nis abelian if and
only if aba-1b-1ENfor all a, bEG.
•
But
If
G is
a group, Example 6 of Section 8.2 shows that its center Z(G) is a normal
subgroup of
G.
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...
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260 Chapter 8
Normal Subgroups and Quotient Group s
Theorem 8.15
If G is a group such that the quotient group G/Z(G) Is cyclic, then
G is abelian.
Proof� For notational convenience, denote Z(G) by C. Since G/C is cyclic, it
has a generator Cd, and every coset in G/C is of the form (Cdf = Cdk
for some integer k. Let a and b be any elements of G. Since a = ea is in
the coset Ca and since Ca
Cd1 for some i, we have a c1d1 for some
c1 EC. Similarly, b cz.tfl for some c2 EC and integer}. Now d'df
di+J fil+I dld', and c1 and c2 commute with every element of G by the
=
=
=
=
=
=
definition of the center. Consequently,
Therefore,
G is abelian.
•
• Exercises
1. Let Nbe the subgroup
(4) of Z']J). Find the order of 13 +Nin the group
Z1J>!N.
2. Let G be the subgroup (3) of Z, and let Nbe the subgroup
of 6 + Nin the group G/N.
3. Complete the table in Example
D4fMhas order
A. 4. N =
(15).
. Find the order
2 and verify that every nonidentity element of
2.
{ G � �). G � �). G � �)}
is a normal subgroup of S3 by
Example 9 of Section 8.2. Show that S3/ N = Z2•
5. Show that Z18/M= Z6, where Mis the cyclic subgroup
(6).
6. Show that Z6/N = Z3, where Nis the subgroup {O, 3}.
7. Show that U'lfl/(5) is isomorphic to Z3•
8. Let G
that
=
Z,, X Z4 and let Nbe the cyclic subgroup generated by
G/ N = Z,,.
9. Let G
=
� X Z2and let Nbe the cyclic subgroup ((1 , 1)).
(3, 2). Show
Describe the
quotient group G/N.
10.
(a) Let Mbe the cyclic subgroup ((0, 2)) of the additive group G
and let Nbe the cyclic subgroup
isomorphic N.
=
Z2 X Z,,
((1, 2)), as in Example 4. Verify that Mis
(b) Write out the operation table of G/M, using the four cosets M+ (0, 0),
M+ (1,
0), M + (0, 1), M + (1, 1).
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8.3
(c)
Quotient
Groups
261
Show that G/Mis not isomorphic to G/N(the operation table for G/Nis
in Example 4). Thus for normal subgroups Mand N, the fact that M = N
does not imply that G/Mis isomorphic to G/ N.
11.
If Nis a subgroup of an abelian group G, prove that G/N is abelian.
12.
If N is a normal subgroup of a group G and if x2 ENfor every x EG, prove
that every nonidentity element of the quotient group G/Nhas order 2.
13. (a)
(b)
14. (a)
Give an example of a nonabelian group G such that G/Z(G) is abelian.
Give an example of a group G such that G/Z(G) is not abelian.
Show that V =
{ G � � :). G � ! :) G � � �). (! � � ;) }
.
is a normal subgroup of S4•
(b)
Write out the operation table for the group S4/V.
B. In Exercises 15and16,find an element of infinite orckr and an element offinite
order in the given quotient group. There are many correct answers. Remember that Z
is an additive group.
15.
(Z
x Z) /( (5, 5))
16. (Z x Z)/((6, 9))
17.
Let E be the group of even integers and N the subgroup of all multiples of 8.
(a)
Show that E/Nhas order 4.
(b)
To what well-known group is E/N isomorphic? [Hint: Theorem 8.8.]
18.
Show that U32/N = U11,, where N is the subgroup
19.
An element b of a group is said to be a square if there is an element c in the
group such that b
=
{1, 17}.
c2. Let Nbe a subgroup of an abelian group G. If both
N and G/Nhave the property that every element is a square, prove that every
element of G is a square.
20. If G is a group and [G:G/Z(G)]
21.
22.
=
4, prove that G/Z(G)
Z2 X Z2•
Let G be an abelian group and Tits torsion subgroup (see Exercise 19 of
Section
7 .3). Prove that G/ Thas no nonidentity elements of finite
order.
Let R* be the multiplicative group of nonzero real numbers and let N be the
subgroup
{1, -1 }. Prove that R* /Nis isomorphic
lll** of positive real number s .
23.
=
to the multiplicative group
Describe the quotient group R*/R**, where R * and Ill* * are a s in Exercise 22.
24. If G is a cyclic group, prove
25. (a)
(b)
(c)
that G/Nis cyclic, where Nis any subgroup of G.
Fm
. d the order of 9•
8 5;
14 and
4
.
8 mt
. h e add'1tJve group Q / IL.
71
28
Prove that every element of 0./Z has finite order.
Prove that 0./Z contains elements of every possible finite order.
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262
Chapter 8
Normal Subgroups and Quotient Groups
the set ofelements offinite order in the group � /Z is the subgroup
26. Prove that
Q/Z.
27. Let G and Hbe groups and let G* be the subset ofG X Hconsisting ofall
(a, e)
withaEG.
(a)
Show that G* is isomorphic to G.
(b) Show that G* is a normal subgroup ofG X H.
(c)
Show that (G X H)/G*
=
H.
28. Let Mand Nbe normal subgroups ofa group Gsuch
that Mn N
=
(e}.
Prove that Gis isomorphic to a subgroup ofG/M X G/N.
29. IfN is a normal subgroup ofa group Gand if every element ofN and ofG/N
has finite order, prove that every element ofGhas finite order .
30. IfNis a finite n ormal subgroup ofa group G and if G/Ncontains an element
oforder n, prove that G contains an element of order n.
31. Let Gbe a group of order pq, with p and q (not necessarily distinct) primes .
Prove that thecenterZ(G) is either (e) or G.
32. A group His said to be finitely generated ifthere is a finite subset SofHsuch
that H = ($)( see Theorem 7.18). IfNis a normal subgroup ofa group G
such that the groups N and G/Nare finitely generated, prove that Gis finitely
generated.
33. Let G b e
a group and let S be the set ofall elements ofthe form aha-1b-1 with
a, b E G. The subgroup G' generated by the set S(as in Theorem 7.18) is called
the commutator subgroup ofG. Prove
(a)
any g, a, b E G, show thatg-1(aba-1b-1)g
(g-1ag)(g-1bg)(g-1a-1g)(g-1b-1g) is in S].
G' is normal in G. [Hint: For
=
(b) G /G' is abelian.
34. Let Gbe
(a)
the additive group � X �.
Show that N =
{(x, y) IY
=
-x} is a subgroup ofG.
(b) Describe the quotient group G/N.
35. Let Nbe a n ormal subgroup ofa group Gand let G' be the commutator
subgroup defined in Exercise 33. IfNn G'
(a)
N�Z(G)
=
(e}, prove that
(b)Thecenter ofG/NisZ(G)/N.
36. IfGis a group, prove that G/Z(G) is isomorphic to the group Inn G of all
inner automorphisms ofG(see Exercise 37in Section 74
. ).
C. 37. LetA, B, Nbe normal subgroups ofa group G such that N�A, N� B. If
G = ABand A n B = N, prove that G/N =A/N X B/N. ('The special case
N = (e) is
Exercise 30 in Section 8 2
. .)
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8.4
m
Quotient Groups
and
Homomorphisms
263
Quotient Groups and Homomorphisms
There is a close connection between normal subgroups, quotient groups, and homo­
morphisms.* The following definition is crucial for developing this connection.
Definition
Let f:G ....+ H be a homomorphism of groups. Then the kernel off is the set
{a: E GI f(a) ""' eH}.
Thus, the kernel is the set of elements in G that are mapped onto the identity element
in Hby the homomorphism/
EXAMPLE 1
Let R* be the multiplicative group of nonzero real numbers and R** the
multiplicative group of positive real numbers. The functionfR* ....+ R** given
byf(x)
x2 is a homomorphism because/(ab) "" (ab)2
aW
kernel is the set of real numbers x such that x2 = i1 namely, {l,
=
=
f(a)f(b). Its
-1}.
=
EXAMPLE 2
Verify that the function/:R* X R* -1- R* given by f(a, b) = b is a homomor­
phism of multiplicative groups. Its kernel is the set of all pairs (a, b) such that
b
=
1, that is, {(a, 1) la ER*}.
EXAMPLE 3
In Example 13 of Section 7.4, we saw that the functionf::Z ....+Zs given by
f(a) =[a] is a homomorphism of additive groups. Its kernel is the set
K ={aEZ
But [a]
=
[O] if
and only if
jf(a)
=
[O]}
=
{aE:Z i[ a]
=
[O]}.
a = 0 (mod 5) by Theorem 2.3, and a= 0 (mod 5) if
and only if 5 I a by the definition of congruence. Hence, K is the set of all integer
multiples of 5, that is, the cyclic group (5).
You can easily verify that each of the kernels in Examples 1-3 is actually a (normal)
subgroup. The same thing is true in the general case.
"If you have read Chapter 6, this should not come as a surprise. The first part of this section simply
carries over to groups the facts about ideals, quotient rings, and ring homomorphisms that were
developed at the end of Section
6.2.
(pages
154-158).
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264 Chapter 8
Normal Subgroups and
Quotient Groups
Theorem 8.16
Let f.·G-+ H be a homomorphism of groups with kernel
subgroup of G.
K. Then K is a normal
Proof •·'If c, dEK, then f(c) = eHan.df(ti) = eH by the definition of kernel.
Hence,f(cti) = f(c)f(d) = e�H = eH, so that cdEK. If cEK, then by
Theorem 7.'lfJ/(c-1) = f(cr1 = (ea)-1 = eH. Thus c-1 EK. Therefore, K
is a subgroup of GbyTheorem 7.11. To show that K is normal, we must
verify that for anyaEGand cEK, a -1caEK(Theorem 8.11). However,
Therefore, a-1ca EK and K is normal.
•
EXAMPLE 4*
Definej:S,.-+ Z2 as follows:f(u) = 0 if u is even andf(u) = 1 if u is odd.
Then/is a homomorphism (Exercise 7). Clearly, the kernel of /consists of all
even permutations, that is, the kernel is A,.. By Theorem 8.16, A,. is a normal
subgroup of S,..
The kernel of a homomorphism/measures how far fis from being injective.
Theorem 8.17
Let f:G -+ H be a homomorphism of groups with kernel K. Then
K
= (eG) if and only if f is injective.
Prooft • Suppos e K = (ea>· If /(a) = f(b), then
[/ iS a homomorphism.]
[Part (2) of Theorem 7.20]
/(ab-1) = f(a)f(b-1)
= f(a)f(b)-1
= f(a)f(a)-1 =
eH
[f(a) = f(b) by hypothesis..]
Thus, ab-1 is in the kernel, so that ab-1 = e0 and hen ce, a = b. Therefore,
/is injective.
Conversely, suppose fis injective. If c is any element in the kernel K,
then/(c) = eH. By part (1) of Theorem 1.20,f(ea) = eH. Hence,f(c) =
f(ei]), which implies that c = ea since/ is injective. Therefore, e0 is the
only element of K, so K= (e a>· •
"Skip this example
if you
haven't read Section 7.5.
Theorems 8.17-8.20 are
Theorems 6.11-6.13.
ti"he proofs of
simply translations from rings to groups of
...........
..
proofs of
....
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the
8.4
Theorem
8.16
Quotient Groups
Homomorphisms
and
265
states that every kernel is a normal subgroup. Conversely, every
normal subgroup is a kernel:
Theorem 8.18
If N is a normal subgroup of a group
G, then the map 1T:G � G/N given by
1T(a}= Na is a surjective homomorphism with kernel N.
Proof" The map 1T is surjective because given any cosetNa in G/N, we have
1T(a)=Na.
The definition of the group operation in
G/N shows that 1T is
a homomorphism:
1T(ah)=Nab=NaNb=
The identity element of
1T(a)1T(b).
G/N is Ne. So the kernel of 1T is
{aEGf 1T(a)=Ne}= {aEG I Na=Ne}
[Definition of1T]
= {aEG[a= e (mod .N)}
[11ieorem 8.2]
= {aEGjae-1EN}
[Definition of congruence]
= {aEG[aEN} =N
[ae-1= ae= a.]
In order to prove the First Isomorphism Theorem below,
we
•
need this lemma.
Lemma 8.19
Let f:G
�
H be a group homomorphism with kernel K. Let a, bEG. Then
f(a}
=
f(b) if and only if Ka
=
Kb.
Proof ... If f(a)= f(b), thenf(a)f(b)-i=eH. By Theorem 7.20,
f(ab-1)= f(a)f(b-1)=f(a")f(br' = ell"
Hence,
ab-1EKanda = b(modK). So Ka= Kb by Theorem 8.2.
Conversely, suppose
means that
ab-1 EK.
Ka= Kb. By Theorem 8.2, a= b (mod K), which
eH, andby Theorem 7.20,
Hence,f(ah-1)=
Multiplying both ends on the right by f(b) show s
thatf(a)=f(b).
•
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266
Chapter
8
Normal Subgroups and Quotient Groups
Theorem 8.20
Let f:G-+ H
First Isomorphism Theorem
be a surjective homomorphism of groups with
G/K is isomorphic to H.
kernel K. Then the
quotient group
Proof ... We would like to define cp:G/K-+ Hby <p(Ka.)=f(a). However, a coset
can be labeled by many different elements. We need to know that the
cp depends only on the coset, and not on the particular repre­
Ka.=Kb. Then
f(a)=f(b) by Lemma 8.19, which means that cp(Ka)=cp(Kb). Therefore,
the map cp:G/K-+ Hgiven by <p(Ka)=f(a) is a well-defined function,
value of
sentative element chosen to name it. So suppose that
independent of how cosets
are
written.
To prove that cp is surjective, suppose h EH. Then h
= f(c) for some
cEGbecause/is surjective. Thus, <p(Kc)=f(c)=h, and cp is surjective.
To prove that <pis injective, suppose <p(Ka)= <p(Kh). Then/(a)=f(b),
so that Ka.= Kb by Lemma 8.19. Hence, <pis injective. Finally, cp is a
homomorphism because f is
<p(KaKb) =cp(Kab) =f(ab) =f(a)f(b)
Therefore, <p:G/K-+ His an isomorphism.
=
cp(Ka.) <p(Kh).
•
The First Isomorphism Theorem makes it easier to identify certain quotient groups.
EXAMPLE 5
Let
G and H be
groups and definef:G X H-+
G byf(a, b)=a. Then/is a
surjective homomorphism by Exercise 9 of Section 7.4. The kernel of f is
H={(a, b )
lf(a, b)
=ea
}
=
{(a, b) I a= e0} = {(ea. b) I a EH}.
By the First Isomorphism Theorem, (G X H)/ Hs:
G, and it is easy to show
that His isomorphic to H (Exercise 15).
EXAMPLE 6
The functionf:C* -+ R** given by f(a +
bi)= a2 + b2 is a surjective homo­
R**,
morphism of multiplicative groups (Exercise 16). Since 1 is the identity in
the kernel off is N= {a +
Theorem 8.16 and C*/N=
bi I a2 + b2= 1}. Then N is a normal subgroup by
R** by the First Isomorphism Theorem.
EXAMPLE 7
As we saw in Example 1, the functionf:R*-+ R** given byf(x)=x2 is a
homomorphism with kernel K = {1, -1}. Note that/is surjective because
for any positive real number c ,f(Vc)
(Vc)2 c. By the First Isomorphism
Theorem, Ill*/Ks: R**.
=
=
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8.4
Qu otien t Groups and Homomorphisms
267
Subgroups of Quotient Groups
Let N be a normal subgroup of a group G. We now investigate the subgroups of the
quotient group G / N.
Theorem 8.21
Let N be a normal subgroup of a group Gand let K be any subgroup of Gthat
contains N. Then K/N is a subgroup of G/N.
Proof ...
Nis obviously a subgroup of K. By normality, Na= aN for every aE G.
In particular, Na= aN for every aEK. Hence , Nis a normal subgroup
of Kand K / Nis a group by Theorem 8.13. The elements of K/ Nare the
cosets Na with a EK. Since, every such coset is an element of G/ N, we
conclude that K/ Nis a subgroup of G/ N.
•
When Kis a normal subgroup of G, we get a stronger result .
Theorem 8.22
Third Isomorphism Theorem*
Let Kand N be normal subgroups of a group G with N!;;K!;;G. Then K/Nis a
normal subgroup of G/N, and the quotient group (G/N)/(K/N) is isomorphic
to G/K.
Proof ... The basic idea of
the proof is to define a surjective homomorphism
f rom G/ N to G/ K whose kernel is K/ N. Then the conclusion of the
theorem will follow immediately from the First Isomorphism Theorem .
First note that, if Na= Ne in G / N, then ac-1 E Nby Theorem 8.2
and the definition of congruence modulo N. Since N!;;;; K, this means
thatac-1EK. Consequently, Ka= Kc in G/Kby Theorem 8.2 again .
Therefore, the mapf:G/ N-+ G / Kgiven by f(Na) = Ka is a well-defined
function, that is, independent of the coset representatives in G/N.
Clearly /is surjective since any Ka in G / K is the image of Na in G/ N.
The definition of coset operation shows that
f(NaNb)=/(Nab) = Kab = KaKb =f (Na)f(Nb).
Henoe,fis a homomorphism . Since the identity element of G/Kis Ke,
a coset Na is in the kernel of /if and only if f(Na) = Ke, that is, if and
only if Ka= Ke. However, Ka = Ke if and only if a EKby Theorem 8.2.
Thus the kernel offconsists of all cosets Na with aE K; in other words,
K/N is the kernel off. Therefore, K/ Nis a normal subgroup of G/N
(Theorem 8.16), and by the First Isomorphism Theorem, (G/N)/(K/N)=
(G/N)/kernel/= G / K. •
"Yes, Virginia, there is a Second Isomorphism Theorem; see Exercise 40. For more aboutVirginia, go
to
www.stormfax.com/bios.htm
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268
Chapter
8
Normal Subgroups and Quotient Groups
Corollary 8.23
Let N be a normal subgroup of a group
G and
let K be any subgroup of
contains N. Then Kis normal in G if and only if K/N is normal in
G/N.
G that
Proof• If K is normal in G, then K/ N is normal in G/ N by Theorem 8.22.
Conversely, suppose that K/ N is normal in G/ N. Let a be any element of
G and k any element of K. We first prove that a-1kaEK. Since K/N
is normal,
Na-1ka
= (Na-1)(Nk)(Na) = (Na)-1(Nk)(Na)EK/N.
Hence, Na-1/ca
= Ntfor some
Since Nr;;;K , we have a-1ka
=
tEK,
so
ntEK,
that a- 1 ka
as
=ntfor somenEN.
desired. Since a and k were
1
arbitrary, this proves that a- Kar;;;, K. Therefore, K is normal in Gby
Theorem 8.11.
•
We now have complete information about subgroups of G/N that arise from
subgroups of G that contain N. Are these the only subgroups of G/N? The next
theorem answers this question in the affirmative.
Theorem 8.24
If Tis any subgroup of G/N, then T
contains
N.
Proof•LetH
=
= H/N, where
His a subgroup of
G
that
{aEGINaET}. Exercise 23 shows thatHis a subgroup of G.
If aEN, then ae-1
Na
= ae =a EN, so a= e (mod .N). By Theorem 8.2,
= NeET. Hence, a EH. Therefore, N r;;;,H. Finally, the quotient
group HfN consists of all cosets Na with a EH, that is, all Na ET. Thus,
H/N
=
T.
•
Simple Groups
In Section 8.1 we considered the classification problem for finite groups-the attempt
to produce a list of groups such that every finite group is isomorphic to exactly one
group on the list. We now introduce the groups that apparently are the key to solving
the classification problem. Recall that a group G always has two normal subgroups,
the trivial group
{e)
and G itself (Exercise 4 in Section 8.2). A group G is said to be
simple if its only normal subgroups are
(e) and G.
EXAMPLE 8
If p is prime, then any (normal) subgroupHof the additive group ZP must have
order dividing p by Lagrange's Theorem. So Hmust have order 1 or p, so that
H=
(0) orH
=
Zr Therefore, z, is simple.
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8.4
Q u otie n t
Groups and Homomorphisms
269
Theorem 8.25
G is a simple abelian group if and only if G is isomorphic to the additive group
Zp for some prime p.
Proof• The preoeding example shows that any group isomorphic
to 7L1 is
simple. Conversely, suppose G is simple. Since every subgroup of an
abelian group is normal, G has no subgroups at all, except (e) and G.
So if
a
is any nonidentity element of G, then the cyclic subgroup
(a)
must be G itself. Since every infinite cyclic group is isomorphic to Z by
Theorem 7.19 and Z has many proper subgroups, G
=(a) must be a
cyclic group of finite order n. We claim that n is prime. If n were com­
posite, say n =
of order
d by
td with 1 < d < n, then (a') would be a subgroup of G
(3) of Theorem 7.9, which is impossible since G is
part
simple. Therefore, G is cyclic of prime order and, hence, is isomorphic
to someZ, by Theorem 7.19.
Nonabelian simple groups
are
•
relatively rare. There are only five of order less than
1000 and only 56 of order less than 1,000,000. A large class of nonabelian simple
groups, the alternating groups, is considered in Section 8.5.
We now show why simple groups are the basic building blocks for all groups. If G
is a finite group, then it has only finitely many normal subgroups other than itself (and
there is at least one such subgroup since
(other than
G)
{e) is normal). Let
G1 be a normal subgroup
that has the largest possible order. We claim that G/G1 is simple. If
G/Gi had a proper normal subgroup, then
by Theorem
8.24 and Corollary
8.23
this
subgroup would be of the form M/G1, where Mis a normal subgroup of G such that
G1 � M � G. In this case, M would be a normal subgroup other than G with order
larger than IG1� a contradiction. Hence, G/G1 is simple.
(e), let G2 be a normal subgroup of G1 (other than G1) of largest possible
(G:i is normal in Gi. but need not be normal in G.) The argument in the preced­
If G1 '1:
order.
ing paragraph, with G1 in place of G and G2 in place of Gh shows that G1 / G2 is simple.
Similarly, if G2 *
(e}, there is a normal subgroup G3 of
G2 such that G3 '1: G2 and G2/G3
is simple. This process can be continued until we reach some G,. that is the identity
subgroup (and this must occur since the order of G1 gets smaller at each stage). Then
we have a sequence of groups
such that each G1 is a normal subgroup of its predecessor and each quotient group
GtfG,+1 is simple. The simple groups G0/Gh Gif G1,
•
•
•
,
G,,_tf G,. are called the
composition factors of G.
It can be shown that the composition factors of a finite group G are independent
of the choice of the subgroups G1• In other words, if you made different choices of
the G1, the simple quotient groups you would obtain would be isomorphic to the ones
obtained in the previous paragraph. This means that the composition factors of G are
completely determined by the structure of G and suggests a strategy for solving the
classification problem. If we could first classify all simple groups and then show how
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270
Chapter 8
Normal Subgroups and Quotient Groups
the composition factors of an arbitrary group determine the structure of the group, it
would be possible to classify all groups.
The good news is that the first half of this plan has already succeeded. For more
than four decades, a number of group theorists around the world worked on various
aspects of the problem and eventually obtained a list of simple groups such that every
finite simple group is isomorphic to exactly one group on the list.* The complete proof
of this spectacular result runs some 10,000 pages! For a brief history of the search for
simple groups, see Gallian [23] or Steen [25].
• Exercises
NOTE: The
congruence class of a in Z,. is denoted [a]n whenever necessary to avoid
confusion.
A.
In Exercises 1---9, verify tha t the given function is a homomorphism and find its
kernel.
1.
f:C-+ !fl, where/(a
2.
g:R*-+ Z2, whereg(x)
3.
h: IR*--+ !fl*, where h(x)
4.
/: O*-+ O**, wheref(x) = [ x �
5.
g:O X Z-+Z, wheref((x, y)) = y.
+bi)=
6. h:C...+C, whereh(x)
=
=
""'
b.
0 if x > 0 andg(x) = 1if x<0.
x3.
x4•
7.t f:Sn-+ Z2, wheref(u) = 0 if u is even and/(u)
=
1 if u is odd.
8. j:Z12--+ Zm wheref(x) = 3x.
9. j:Z_. Z2 X Z4, wheref(a)
=([ah, [a]4).
10. <p:S,.--+ Sn+h where for each/ES,., <p( f) ES,.+1 is given by
<p(f)(k)
11.
=
{f(k)
n +I
ifl�k�n
ifk=n+l
Suppose that k, n, and rare positive integers such that k In. Show that the
functionf:Zn-+Z1c given byf([a]n)
[ra]1c is well defined (meaning that if
=
[a],,
=
[b],,, then [ ra]1c
=
*The proof was first announced in
took until
2004 forth is
[rb ] J.
1981,
but a few years later a gap in the proof was discovered. It
gap to be fixed.
tskip this exercise if you haven't read Section
7.5.
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8.4
Quotient Groups
and
Homomorphisms
271
In ExerciSes 12-14, verify that the given function is a surjective homomorphism of
additive groups. Then.find its kernel and identify the cyclic group to which the keme/
is isomorphic. [Exercise 11 may be helpful.]
12. h:Z12-+�.where h([a]12) =[a]6·
13. h:Z16-+ �.where h([a]t6) =[3a]6.
14. h:Z18-+ Z3, where h([x]i8)
=
[2x]3•
15. If Hand Hare the groups in Example 5. Show that H => H.
2
16. Prove that the function/: C*-+ R**given byf(a + bi) =d2 + b is a surjective
homomorphism of groups.
17. (a) Produce a list of groups such that everyhomomorphic image of Z12is
isomorphic to exactlyone group on the list. [Hint: See Exercise 26in
Section 7.4.)
(b) Do the same for Z20•
18. Find all homomorphic images of D.,.
19.
Find all homomorphic images of S3•
20. (a) List all subgroupsof Z1JH,where H = {O, 6}.
(b) List all subgroupsof Zw/K, where K = { 0, 4, 8, 12, 16}.
21. Suppose that Gis a simple group and/:G-+ His a surjective homomorphism
of groups. Prove that either fis an isomorphism or H =(e}.
B. 22.
Let Gbe an abelian group.
(a) Show that K
(b) Show that H
=
=
{a E G l!aJ :s; 2} is a subgroup of G.
{x2 Ix E G} is a subgroup of G.
(c) Prove that G/K= H. [Hint: Define a surjective homomorphism from Gto
H with kernel K.]
23. If Nis a normal subgroup of a group Gand Tis a subgroup of G/N,show
that H = {a E GI N a E T} is a subgroup of G.
24. If k [ n andf Un-+ U1cis given by f([xJn)
and find its kernel.
=
[x]1c, show that/is a homomorphism
25. Prove that (Z X Z)/((1, 1)) = Z. [Hint: Show thatf:Z X Z-+Z,given by
f((a, b)) =a - b,is a surjective homomorphism.]
26. Prove that (Z X Z)/((2, 2)) = Z X Z2• [Hin t : Show that h-JL X Z-+ Z X z,..
givenby h((a, b)) (a - b, [b]:z) isa surjective homomorphism.]
=
27. Let Mbe a normal subgroup of a group Gand let N bea normal subgroup
of a group H. Use the First Isomorphism Theorem to prove that M X Nis a
normal subgroup of G X Handthat (G X H)/(MX N) = G/M X H/N.
28. SIJ...2, R)is a normal subgroup of GL(2, R) by Exercise 25of Section 8.2.
Prove that GL(2, R)/SL(2, R) is isomorphic to the multiplicative group R*of
nonzero real numbers.
29. If k In,prove that Z,,/(k) = Z1c. [Exercise 11maybe helpful.]
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272
Chapter 8
Normal Subgroups and Quotient Groups
30. Iff:G-+> His a homomorphism offinite groups, prove that !Im/I divides JGI
andlff� [Im/was defined just before Theorem 7.20.J
31.
Prove that Z12
=
Z3 X �. [Considerf:Z-+> Z3 X �given by/(a)
=
([ah, [a]4).]
32. Let Mbe a normal subgroup ofa group Gand let Nbe a normal subgroup of
a group H. If
f:G-+> His a homomorphism such that/(.M) !;;; N, prove that the
map g:G/ M-+> H/Ngiven by g(Ma) = Nf(a) is a well-defined homomorphism.
33. LetfG-+> H be a surjective homomorphism ofgroups with kernel K. Prove
that there is a bijection between the set ofall subgroups ofHand the set of
subgroups ofG that contain K.
34. (An exercise for those who k now h ow to multiply 3 X 3 matrices.) Let G be
the set ofall matrices of the form
(1 )
where a, b, c E Cl!.
a
b
0
1
c
0
0
1
(a) Show that G is a group under matrix multiplication.
(b) Find the center C ofG and show that C is isomorphic to the additive
group Cl!.
(c) Show that G/ C is isomorphic to the additive group Q X Cl!.
35. Let G and H be the groups in Exercises 33 and 34 ofSection 7.1. Use the
First Isomorphism Theorem to prove that His normal in G and that G/ His
isomorphic to the multiplicative group Iii* ofnonzero real numbers.
[Hint: Consider the map f: G-+> Iii* given byf(Ta,b) a.]
=
36. Let Nbe a normal subgroup ofa group G and let f:G-+> H be a
homomorphism ofgroups such that the restriction offto Nis an
isomorphism N = H. Prove that G = N X K, where K is the kernel off
[Hint: Exercise 30 in Section 8.2.]
37.
Prove that 0*
=
0** X Z2• [Hint: Exercises 4 and 36.]
38. Let Nbe a normal subgroup ofa group G. Prove that G/Nis simple if and
only ifthere is no normal subgroup K such that N � K � G.
[Hint: Corollary 8.23 and Theorem 8.24.]
39.* The additive group Z[x] contains Z
(the set ofconstant polynomials) as a
normal subgroup. Show that Z[x]/Z is isomorphic to Z[x]. This example
shows that G/ N= G does not necessarily imply that N (e). [Hint: Consider
the map T:Z[x]-+> Z[:X]/Z given by 1(/(x)) Z + xf(x).]
=
=
C. 40. (Second Isomorphism Theorem) Let Kand N be subgroups ofa group G, with
N normal in G. Then NK {nk In EN, k E .K} is a subgroup ofG that contains
both Kand Nby Exercise 20 ofSection 8.2.
=
(a) Prove that Nis a normal subgroup ofNK.
"Skip this exercise if you have not read the first part of Section 4.1.
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-...d.'lm:mJ"��
...
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8.5
The Simplicity of An
(b) Prove that the function.f:K-+ NK/N given by f(k)
homomorphism with kernel Kn N.
(c) Conclude that K/(N n
=
273
Nk is a surjective
K) = NK/N.
41. Cayley's Theorem 7.21 represents a group G as a subgroup of the permutation
group A(G). A more efficient way of representing G as a permutation group
arises from the following generalized Cayley's Theorem. Let K be a subgroup
of G and let The the set of all distinct right cosets of K.
(a) If
a E G, show that the mapf.:T ...+ T given by f.(Kh)
permutation of the set T.
=
Kha is
a
(b) Prove that the function cp:G-+ A(T) given by cp(a)
/.-•, is a
homomorphism of groups whose kernel is contained in K.
=
(c) If K is normal in G, prove that K
=
kernel cp.
(d) Prove Cayley's Theorem by applying parts (b) and (c) with K
=
(e}.
metabelian if it has a subgroup N such that N is
abelian, N is normal in G, and G/N is abelian.
42. A group G is said to be
(a) Show that S3 is metabelian.
(b) Prove that every homomorphic image of a metabelian group is metabelian.
(c) Prove that every subgroup of a metabelian group is metabelian.
APPLICATION: Decoding Techniques (Section 16.2) may be covered at
this point if desired.
II
The Simplicity of An*
As we saw at the end of Section 8.4, simple groups appear to be the key to solving the
classification problem for finite groups. This fact and the following theorem are one
reason that the alternating groups An are im portant.
Theorem 8.26
For each
n * 4,
the alternating group An is a simple group.
The group� is not simple (Exercise 7). Although the entire proof of Theorem 8.26
is rather long, it requires only basic facts about the s ymmetric groups and normal
subgroups. There will be many instances in the proof where we will deal with permuta­
tions such as (abed) or (alb) or (ab)(cd). In all such cases,
distinct letters represent distinct elements
of
{1, 2, ... , n}.
The proof of the theorem requires two lemmas.
•section
7.5 is n prerequisite. This section
is not used in the sequel nnd
mny be
omitted if desired.
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274
Chapter
8
Normal Subgroups and Quotient Groups
Lemma 8.27
Every element of An (with n � 3) is a product of 3-cycles.
Proof" Every element of A" is by definition the product of pairs of transposi­
tions. But every such pair must be of one of these forms: (ab) (cd) or
(ab) (ac) or (ab) (ab). In the first case verify that (ab) (cd)= (adb) (adc),
in the second that (ab) (ac) = (acb), and in the last that (ab) (ab)= (1)=
(abc) (acb). Thus every pair of transpositions is either a 3-cycle or a
product of two 3-cycles. Hence, every product of pairs of transpositions
is a product of
3-cycles.
•
Lemma 8.28
If N is a normal subgroup of An {with n 2! 3} and N contains a 3-cycle, then
N=An.
Proof• For notational convenience, assume that ( 123) EN [the argument when
(rst)EN is the same; just replace 1, 2, 3 by r, s, t, respectively]. Since
(123) EN, we see that (123)(123)= (132) is also in N. Fork;';!: 4, let
x = (12)(3k) and verify that x-1= (3k)(l2). The normality of N implies
that x(l32}x-1 EN by Theorem 8.11. But
x(l32)x-I = (12)(3k}(l32)(3k)(12) = (12k).
Therefore,
( *)
N contains all 3-cycles of the form (12k) with k <!:: 3.
Verify that every other
(la2),
where a,
3 -cycle can be written in one of these forms:
(lab),
(2ab),
(abc)
b, c 2! 3. By (*) and closure in N,
(la2) = (12a)(l2a) EN;
(lab) = (12b)(l2a)(12a) EN;
(2ab) = (12b)(12b)(l2a) EN;
(abc) = (12a)(12a)(l2c)(l2b)(l2b)(l2a) EN.
Thus N contains all 3-cycles, and, hence, N contains all products of
3-cycles by closure. Therefore, N= An by Lemma
8.27.
•
We are now ready to prove Theorem 8.26. The following fact will be used frequently:
(••)
The inverse of the cycle (a1a2"3
For example,
•
•
•
ai) is the cycle
(12345)-1 = (15432) and (678r1
=
(a a.alo-t
1
•
•
;
a3'1z).
(687), as you can easily verify.
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8.5
The Simplicity of An
275
Proof ofTheorem 8.26 ... A1 and A3 are simple abelian groups (Exercise 2). So
assume n �
5. We must prove that An has no proper normal subgroups.
(1). We need only
Let N be any normal subgroup of A,,, with N #
show that N = A,,. When all the nonidentity elements of N are written
as products of disjoint cycles, then there are three possibilities for the
lengths of these cycles:
1. Some cycle has length � 4.
2. Every cycle has length s 3, and some have length 3.
3. Every cycle has length s 2.
We shall show that in each of these cases, N = A,,.
Case 1 N contains an element a that is the product of disjoint cycles, at least
4. For notational convenience we assume that
r)'r, where Tis a product of disjoint cycles, none of which
invol ve the symbols l, 2, 3, 4,
, r.t Leto = (123) EA,,. Since N is a
normal subgroup and a EN, we have a-1(Baa-1) EN by Theorem 8.11.
one of which has length r �
a
=
(1234
·
·
•
• . .
An easy computation shows that
a-1(8a8-1)
[(1234
""
r-1(1234
·
·
r-1(1r
·
432)(123)(1234
=
=
=
Therefore,
r)'T]-1 (123)[(1234
=
·
·
T-1T(lr
·
·
·
·
·
·
·
r)-1(123)[(1234
·
432)(123)(1234
·
·
·
·
r}r](l23)-1
r)r](l23r1 [Corollary 7.6]
·
•
·
·
r)T(l32)
·
·
·
r)(l32)
[Statement(**)]
[111eorem 7.23]
(l)( l3r) = (13r).
(l3r) EN, and hence, N
=
A,, by Lemma
8.28.
Case 2A N contains an element <J that is the product of disjoint cycles, at least
two of which have length 3. For convenience we assume that <J =
( 123)( 456) T, where T is a product of disjoint cycles, none of which in­
volve the symbols 1, 2, ... , 6. Leto
(124) EA,,. Then, as in Case 1,
N contains u-1(8u8-1), and we have a similar calculation:
=
u-1(8u8-1)
=
[Corollary 7.6]
r-1(465)(132)(124)(123)(456)7 (142)
[Statement(**)]
=
T-1T(465)(132)(124)(123)(456)(142)
[111eorem 7.23]
=
(14263).
=
Therefore,
[(123)(456)'rr1(124)(123)(456}r(124r1
r-1(456r1c123)-1(l24)(123)(456)r(124)-1
=
(14263) EN, and N =An by Case 1.
tThe same argument works with an arbitrary r-cycle (abed·· · t) in place of (1234 · · · r); just replace
1 by a, 2 by b, etc. Analogous remarks apply in the other cases, where specific cycles wil I also be
used to make the argument easier to follow.
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276 Chapter 8
Normal Subgroups and Quotient Groups
Case
2B N contains an element a that is the product of one 3-cycle and some
2-cycles. We assume that a = (l23)r, where T is a product of disjoint
transpositions, none of which involve the symbols l, 2, 3. Since a
product of disjoint transpositions is its own inverse(Exercise 5),
Theorem 7.23 shows that
al= (l23)r(l23)T = (l23)(123)TT = ( 123)(123) = (B2).
2
But u EN since aEN. Therefore, (132)EN, and N=
Case 2C Ncontains a
3-cycle. Then N
=
A,. by Lemma 8.28.
An by Lemma 8.28.
N is the product of an even number of disjoint
2-cycles. Then a typical element u of Nhas the form ( 12)(3 4)1, where
T is a product of disjoint transpositions, none of which involve the
symbol s 1, 2, 3 , 4. Let B
(123)EA11• Then, as above, u-1(aua-1)EN.
Case 3 Every element of
=
Using Corollary 7.6, Theorem 7.23, and statement(••), we see that
a-1(8a8-1) = T-1(3 4)(12)( 123)( 12)(34)r(l32) = (B)(24).
{l , 2,
, n} distinct from 1, 2, 3 , 4.
(Bk) EAir Let f3 = ( 13)(24), which was just shown to be in N.
1
Then by the normality of N and closure, {3(a{3(F )EN. But
Since n � 5, there is an element kin
. . .
Let a=
{3(af3a-1)
Therefore,
Theorem
=
(l3)(24)( 13k)(l3)(24 )(lk3) = ( 13k).
(Bk) EN, and N = A,. by Lemma 8.28.
•
8.26 leads to an interesting fact about the normal subgroups of S,.:
Corollary 8.29
If n;;::; 5, then (1), Am and Sn are the only normal subgroups of Sn.
Sketch of Proof
N is a normal subgroup of Sn. Then N n A,. is
19 of Section 8.2). Theorem 8.26
shows that N n A,, must either be A,, or ( 1). If N n An = A,,, then N= A,,
or Sn (Exercise 10) . If N n An= (l), then all the nonidentity elements of
..
Suppose that
a normal subgroup of A11 (Exercise
N are odd. Since the product of two odd permutations is even, that is, an
element of
An,
N n An= ( 1 ), the product of any two elements of N
N = ( 1) (Exercises 8 and 9). •
and
is (1). Therefore,
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8.5
The Simplicity of An
277
• Exercises
A. 1.
(a) List all the 3-cycles in S4•
(b) List all the elements of A4 and express each as a product of 3-cycles.
2.
(a) Verifythat .A2= (1).
(b) Show that A3 is a cyclic group of order 3 and hence simple by Theorem 8.25.
3. Find the center of the group .A4•
4. If
n � 5, what is the center of A11?
B. 5. If a E S,, is a product of disjoint transpositions, prove that a2= (1).
6. Prove that As has no subgroup of order 30.
7. Prove thatN
=
[Hint: Exercise 23 of Section 8.2.]
{(l), (12)(34), (13)(24), (14)(23)} is a normal subgroup of .A4•
Hence, A4 is not simple.
[Hint: Exercise 23 of Section 7.5. For normality, use
Exercise l (a) and straightforward computations.]
8. Prove that no subgroup of order 2 in
Sn (n � 3) is normal. [Hint: Exercises 26
of Section 7.5 and 16 of Section 8.2.]
9. Let Nbe a subgroup of S,, such that <TT= (1) for all nonidentityelements
a, TEN. Prove thatN= ( l ) orN is c yclic of order 2. [Hint: If N:F- (1), let
a be a nonidentity element of N. Show that a has order 2. If Tis any other
nonidentity element of N, show that a= T.]
I 0. IfN is a normal subgroup of S,, and N n A,, = A11, prove that N
=
A,, or S,,.
[Hint: Why is An!;;;N
;
!;;;; S,,? Use Theorem 7 .29 and Lagrange's Theorem.]
11. Prove that .A11 is the only subgroup of index 2 in S,,.
[Hint: Exercise 23 of
Section 8.2 and Corollary 8.29.]
12. If f:S,.-+ S,. is a homomorphism, prove thatf(A,.) �A,..
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P A R T
2
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CHAPTER
9
Topics in Group Theory
This chapter takes a deeper look at various aspects of the classification problem
for finite groups, which was introduced in Section 8.1. After the necessary pre­
liminaries are developed in Section 9.1, all finite abelian groups are classified up
to isomorphism in Section 9.2. The basic tools for analyzing nonabelian groups are
presented in Sections 9.3 and 9.4. Applications of these results and several other
facts about the structure of finite groups are considered in Section 9.5, where
groups of small order are classified.
Sections 9.3 and 9.4 are independent of Sections 9.1 and 9.2 and may be read
first if desired. Sections 9.1-9.4 are prerequisites for Section 9.5.
•
Direct Products
If G and Hare groups, then their Cartesian product G
X
His also a group, with the
operation defined coordinatewise (Theorem 7.4). In this section we extend this notion
to more than two groups. Then we examine the conditions under which a group is
(isomorphic to) a direct product of certain of its subgroups. When these subgroups are
of a particularly simple kind, then the structure of the group can be completely deter­
mined,
as
will be demonstrated in Section 9.2. Throughout the general discussion, all
groups are written multiplicatively, but specific examples of familiar additive groups
are
written additively
If G1, G'b
product G1
. • .
X <h X
•
• •
(ato "2•
It is easy
to
as
usual.
, G11 are groups, we define a coordinatewise operation on the Cartesian
•
X
G,, as follows:
•
,
.
verify that G1
a,J(hu h2,
X G2 X
X
•
• •
•
•
•
,
b,J
=
(a1b" "2h2,
• • •
,
a,P,J.
G,, is a group under this operation: If e1 is the
identity element of G1, then (e1, f?:l, • • • , e,,) is the identity element of G1 X G2 X · • • X G,,
-1
and (a1 -1, a2 , • • • , a,,-1) is the inverse of (a" "2· .. . , a,J. This group i s called the direct
product of Gi, G2, • • • , G,,. *
•when each G; is an additive abelian group, the direct product
direct sum and denoted G1 EB G1 EB
· · ·
EB G..
of G1' ... ,
G. is sometimes called the
281
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282 Chapter 9
Topics in Group Theory
EXAMPLE 1
Recall that U,. is the multiplicative group of units in Zn and that U4
and U6
of the
=
=
{ 1, 3}
{1, 5} (see Theorem 2.10). The direct product U4 X U6 X .l3consists
12 triples
1, 0),
(3, 1, 0),
(I, 1, 2),
(3, 1, 2),
(1, 1, 1),
(3, 1, 1),
( l,
(1, 5, 0),
( 3, 5, 0),
(1, 5, 2),
(3, 5, 2).
(1, 5, 1),
(3, 5, 1),
Note that U4 has order 2, U6 has order 2, Z3 has order 3, and the direct product
U4 x U6 X Z3 has order
2 2 3=
•
12. Similarly, in the general case,
•
if Gj, G1,
•
G1 X G2 X • • •
•
,
•
G. are finite groups, then
G. has order IG11 IG�
x
•
·
· ·
IG.�
In the preceding example it is important to note that the groups U4, U6, and
Z3 are not contained in the direct product U4 X U6 X Z3• For instance, 5 is an
element of u(., but 5 is not in l'4 x U6 x Z3 because the elements of U4 x U6 x Z3
are triples. In general, for 1 s i s
n
G; is not a subgroup of the direct product G1
x
G2 X • • • x
G••*
This situation is not entirely satisfactory, but by changing our viewpoint slightly
we can develop a notion of direct product in which the component groups may
be considered as subgroups.
EXAMPLE 2
It is easy to verify that M= {O,
3}
and N= {O,
2, 4}
are normal subgroups of
�(Do it!). Observe that ever y element of Z6can be written as a sum of an ele­
ment in Mand an element in Nin one and only one way:
O=O+O
1=3+4
2=0+2
3=3+0
4=0+4
5= 3 +2.
Verify that, when the elements of Z6are written as sums in this way, then the
addition table for�looks like this:
o+o
3+4
0+2
3+0
0+4
3+2
o+o
o+o
3+4
0+2
3+0
0+4
3+2
3+4
3+4
0+2
3+0
0+4
3+2
o+o
0+2
0+2
3+0
0+4
3+2
o+o
3+4
3+0
3+0
0+4
3+2
o+o
3+4
0+2
0+4
0+4
3+2
o+o
3+4
0+2
3+0
3+2
3+2
o+o
3+4
0+2
3+0
0+4
•1t is true, however, that an isomorphic copy of G; is a subgroup of G, x
G2 X
• · •
X G. (see Exercise 12).
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9.1
Direct Products
283
Compare the l6 table with the operation table for the direct product M X N:
(0, 0)
(3, 4)
(0,2)
(3, 0)
(0, 4)
(3,2)
(0, 0)
(3, 4)
(0,2)
(3, 0)
(0, 4)
(3, 2)
(0, 0)
(3, 4)
(0,2)
(3, 0)
(0, 4)
(3, 2)
(3, 4)
(0, 2)
(3, 0)
(0, 4)
(3, 2)
(O, 0)
(0,2)
(3, 0)
(0, 4)
(3, 2)
(0, 0)
(3, 4)
(3, 0)
(0,4)
(3, 2)
(0, 0)
(3,4)
(0, 4)
(3,2)
(0, 0)
(3, 4)
(0,2)
(3, 0)
(3, 2)
(0, 0)
(3, 4)
(0, 2)
(3, 0)
(0, 4)
(0, 2)
The only difference in these two ta bles is that elements are written a + bin
the first and (a, b) in the second. Among other things, the tables show that the
direct product M X Nis isomorphic to "4, under the isomorphism that assigns
each pair (a, b) EM X Nto the sum of its coordinates a + b E�.
Consequently,
we
can express Z6 as a direct product in a purely internal fashion ,
without looking at the set M X N, which is external to Z6: Write each element uni quely
as a sum a + b, with
a
EM and b EN. We now develop this same idea in the general
case, with muhiplicative notation in place of addition in "4,.
Theorem 9.1
Let N1, N2 ••• , NA be normal subgroups of a group G such that every element
in G can be written uniquely in the form a1a2 • • • ak, with a1EN1• *Then G is
isomorphic to the direct product N1 x N2 x
·
·
·
x N1<..
The proof depends on this useful fact:
Lemma 9.2
Let Mand N be normal subgroups of a group G sue h that Mn N = (e) . If a EM
and b EN, then ab = ba.
a-1b-1ab. Since Mis normal, b- 1 abEMhy Theorem 8.11.
Closure in M shows that a-1b-1ab = a-1(b- 1ab) EM. Similarly, the
Proof• Consider
normality of Nimplies that a-1b-1 aENand, hence,a-1b-1ab =
(a- 1b-1 a) b EN. Thus a-1b-1 abE Mn N = (e). Multiplying both sides
of a- 1b-1ab = eon the left byhashows that ab =ha. •
Proof ofTheorem 9.1 ·Guided
by the example preceding tlte theorem (but using
multiplicative notation ), we define a map
"Uniqueness means that if a1a2
•
•
•
ak
=
b1bt .. . bk with each 110 b; EN;, then i!;
=
b; for every i.
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284 Chapter 9
Topics in Group Theory
Since every element of G can be written in the form a1az ·
· •
ak (with
OiENJ by hypothesis,/is surj ective. If/(a1o az, . . . , a,J= f(b1 o b,., . . . , bk),
then a10z · • • ak = b1 h2 • • • bk. By the uniqueness hypothesis, a1= b, for
each i(l
s i s k). Therefore,
(ai, a:i,
•
•
.
,
a ,J= (b1, b,.,
.
.
.
, b,J in
N1 x N2 X
•
•
x N"'
•
and/is injective.
In order to prove that/is a homomorphism we must first show that
� = (e) when i =I= j.
the N's are mutually disjoint subgroups, that is, N1 ()
If a EN1 n �. then a can be
written as a product of elements of the N's
in two different ways:
ee
· •
·
eae
t
t
·
·
·
e
· ·
t
.. e= a =
t
ee
·
e
· ·
t
· ·
·
t
eae
· ·
e.
·
t
t
The uniqueness hypothesis implies that the components in N1 must be
equal: a= e. Therefore, N1 n �= (e) for i =I= j. In showing that/is a
homomorphism,
we shall make repeated use of
with Lemma 9 .2 , implies that a,b1=
this fact, which together
b1a, for a1EN1 and b1EN;
, a �,J
/[(ah ... , a,J(bt> ... , bk)] = f(a1b1,
= a1b1 aJJ2 a.jl3
a�k
. • .
=
"'�)'t'
.
.
.
.
•
•
= a1a2 b1a3 b,P3
= a1a2
�t
Continuing in this way we successively move
until we obtain
f[(a1o
•
.
.
,
a,J(hh
.
.
•
, b k)]= (a1az
= f(ai.
·
•
• •
•
.
•
. a,b,
a�k
•
b2b3 ... a�k·
a4, a5,
• • •
ak)(b1 h2
, ak}f(bi.
•
,
•
•
.
ak to the left
•
.
b,J
, hi).
Tb:erefore,/is homomorphism and, hence, an isomorphism.
Whenever G is a group and N1 o
, Nk are subgroups satisfying the hypotheses
is the direct product of Ni,
, N1; and write
XN1r.. Each N, is said to be a direct factor of G. Depending on the con­
.
.
•
of Theorem 9 . 1 we shall say that G
G
=
N1 X ·
·
·
•
.
.
•
text, we can think of Gas the external direct product of the N1 (each element a k-tuple
(a1, ••• , a,J EN1
X
uniquely in the form
· X N,J or as an internal direct product
a1a:i · · · akEa,,EG).
·
·
The next theorem is often easier to
use than Theorem 9.1
(each element written
to prove that a group is
the direct product of certain of its subgroups. The statement of the theorem uses the
following notation. If Mand N are subgroups of a group G, then MN denotes the set
of all products mn, with m EM and n EN.
......
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... ..
......
.......
9.1
Direct Products
285
Theorem 9.3
If Mand N are normal subgroups of a group G such that G=MN and M n N=(e),
then G=MXN.
For the case of more than two subgroups,see Exercise 25.
Proof ofTheorem 9.3 ... By hypothesis every element of G is of the form mn, with
m EM,n EN. Suppose that an element had two such representations,say
nm= m1nto withm, m1 EM andn,n1 EN. Then
mn = m1n1
1
1
m1- mn = ml m1n1
1
m1- mn = n1
1
1
m1- mnn- = n1n-1
1
m1-1m = n n1
[Left multiply both sides by m1-�.]
[Right multiply both sides by 11-1.]
1
1
But m1-1m EM and n1tC EN and Mn N = {e). Thus m1- m
m=m1;
=
e and
similarly,n =n1• Therefore,every element of Gean be written
uniquely in the form mn (m EM,n EN), and,hence,G=M X Nby
Theorem 9.1.
•
EXAMPLE 3
By Theorem 2.10,the multiplicative group of units in Z15 is U15 =
{l, 2,4,8} are
{l}. Every element of N is in MN (for
{1,2, 4, 7, 8, 11, 13,14}. The groups M = {1, 11} and N =
normal subgroups whose internection is
instance, 2= 1·2),and similarly for M. Since 11·2=7,11·8=13,and
11
•
4=14,we see that U15=MN. Therefore, U15=M X NbyTheorem 9.3.
Since N is cyclic of order 2 and M cyclic of order 4 (2 is a generator),we con­
clude that Uis is isomorphic to Z2 X�(see Exercise 10 and Theorem 7.19).
• Exercises
NOTE: Unless stated otherwise, GI>
• . .
,G,. are groups.
A. 1. Find the order of each element in the given group:
(a) Z2XZ.
(b)
Z3 XZ3 XZ2
(c) D4 XZ2
2. What is the order of the group UsX U6X U1X U8?
3. (a) List all subgroups of Z2 XZ2•
(b)
(There are more than two.)
Do the same for Z2XZ2 XZ2•
4. If G and Hare groups, prove that G XH = HX G.
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286 Chapter 9
Topics in Group Theory
5. Give an example to show that the direct product of cyclic groups need not be
cyclic.
6.
(a)
Write Z12 as a direct sum of two of its subgroups.
(b)
Do the same for .l1s·
(c)
Write Z30 in three different ways
as
subgroups.[Hint: Theorem 9.3.]
a direct sum of two or more of its
7. Let Gi, . .. , G11 be groups. Prove that G1 X• • · XGn is abelian if and only if
every G1is abelian.
8. Let i be an integer with 1 s i s
Prove that the function
n.
'1T1:G1XG2X · · • X Gn-+ G1
given by 'TT!.ab
._ a,;, a3, . .., an)
= a; is a surjective
homomorphism of groups.
9. Is Z8 isomorphic to Z., X Z2?
B. 10.
(a)
If fG1-+ H1 and g:G2-+ H2 are isomorphisms of groups, prove that
the map6:G1XG2...+H 1 X H2given by6(a,b) = (f(a),g(b))is an
isomorphism.
(b)
If G1 = H1for i
=
1, 2, ... , n, prove that
G1 X • •
•
.X G,.
=
H1 X ·
•
•
XHn.
11. Let H, K , M, N be groups such that K =MXN. Prove that H XK =
HXMXN.
12. Let i be an integer with 1 s i s n. Let G1 be the subset of G1X· • •XG,.
consisting of those elements whose ith coordinate is any element of G1 and
whose other coordinates are each the identity element, that is,
G1""
{(et> .•. , e1_1,
a,.
e1+1> ••• , en) I
a1EG1}.
Prove that
(a) G1 is a
normal subgroup of G1X • • •XG11•
(b) G, =GI.
(c)
G1 X• • •XGn is the (internal) direct product of its subgroups Gi. ... ,
Gn. [Hint: Show that every
element of G1X• • • XGn can be written
uniquely in the form a1a 2• • •
an,
w ith a1 E G1; apply Theorem 9.1.]
13. Let G be a group and let D :=:: {(a, a, a) I aE G}.
(a)
Prove that Dis a subgroup of G XGXG.
(b)
Prove that Dis normal in GXGXG if and only if G is abelian.
14. If G" ... , Gn are finite groups, prove that the order of (ab a,;, ...
, a,,) in
. . , la..I·
G1X• · • XG11 is the least common multiple of the orders la11, la�, .
15. Let ii. i2'
• • •
, i,, be a permutation of the integers 1, 2,
• . .
, n. Prove that
G11 X G1,. X • · • X Gr.
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9.1
Direct Products
287
is isomorphic to
[Exercise 4 is the case
16.
n
=
2.]
If N, Kare subgroups of a group Gsuch that G =
N X Kand Mis a normal
subgroup of N, prove that Mis a normal subgroup of G. [Compare this with
Exercise 14 in Section 8.2.]
17.
Let Q* be the multiplicative group of nonzero rational numbers,
subgroup of positive rationals, and Hthe subgroup
O*
18.
19.
20.
=
O** XH.
lfi.6 is isomorphic to Z2 X Z-t [Hint:
Prove that
Let Gbe a group andfi:G� G,.,h.:G� G2, •
Theorem
• •
9.3.]
,f,,:G� G, homomorphisms.
For i = 1, 2, ... , n, let 'iT1 be the homomorphism of Exercise 8. Let
f*:G�· G1 X • • • X G,, be the map defined by f*(a) = (Ji(a1),f,_(aj),
(a)
Prove that/* is a homomorphism such that 'iTr of*
(b)
Prove that/* is the unique homomorphism from Gto G1 X
that 'iT1 of*
Let N1
,
• • •
,
whenever
G=
N1
. •
.
,f,,(a,.)).
=ft for each i.
· ·
= f; for every i.
·
X G,. such
Nk be subgroups of an abelian group G. Assume that every
element of Gcan be written in the form a1 •
21.
O** the
{l, -1}. Prove that
a1a2
• • •
X N2 X •
a,, = e, then a1 =
X Nk.
e
•
•
a,, (with a1 E NJ and that
for every i. Prove that
· •
Let Gbe an additive abelian group with subgroups Hand K. Prove that
G= H X Kif and only if there are homomorphisms
H
such that 81(1T1(x)) +
'iTi
82
0
=
tj G� K
�('iT2(x))
82
81
=
Gand 'iTt 0 81 =in, 'iT2 o Bi = 'K•
the identity map on X, and 0 is the map
x for every x E
0, and 'iT2 ° 81 ""0, where
ix is
that sends every element onto the zero (identity) element.
Exercise
22.
24.
IHD = I.
(a)
Show by example that Lemma
(b)
Do the same for Theorem 9.3.
9.2 maybe false if Nis not normal.
N and Kare
N and K?
Let N, Kbe subgroups of a group G, with N normal in G. If
abelian groups and G= NK, is Gthe direct product of
25.
[Hint: Let 'iTt be as in
Let Gand Hbe finite cyclic groups . Prove that GXHis cyclic if and only if
( IGI,
23.
8.]
Let N1,
• • •
,
Nk be normal subgroups of a group G. Let N1N2 •
the set of all elements of the form
G"" N 1N2
•
• •
• • •
Nk and that
N,n
for each
a1a2
(N1
• • .
N1-1N1+1
i(l sis n). Prove that
· •
Nkdenote
ak with a1E Nt Assume that
• .
.
Nk)
=
G= N1 X N2 X •
(e)
· ·
X Nk.
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288 Chapter 9
Topics in Group Theory
26. Let N1,
• • •
, Nk be normal subgroups of a finite group G. If
(notation as in Exercise 25) and IGI
N1 X N2 X
27. Let N,
· · •
=
IN11 IN21
•
· · ·
IN�
x N1r
,
G
=
N1N2 •
prove that G
• •
Nk
=
H be subgroups of a group G. G is called the semidirect product of N
NH, and N () H {e). Show that each of the
and H if N is normal in G, G =
=
following groups is the semidirect product of two of its subgroups:
28. A group G is said to be indecomposable if it is not the direct product of
two of its proper normal subgroups. Prove that each of these groups is
indecomposable:
(c) Z
29. If p is prime and n is a positive integer, prove that Zr is indecomposable.
30. Prove that 0 is an indecomposable group.
31. Show by example that a homomorphic image of
an indecomposable group
need not be indecomposable.
32. Prove that a group G is indecomposable if and only if whenever H and K are
normal subgroups such that G =
HX K,
then
H
=
(e}or K
=
(e}.
33. Let /be the set of positive integers and assume that for each i E/, G1 is a
group.* The infinite direct product of the G1 is denoted
of all sequences (a1o ai, ...) with a1 E G1• Prove that
coordinatewise operation
C. 34. With the notation as in Exercise
IT
1 el
IT
tel
G1and consists
G1 is a group under the
33, let :L G1 denote
consisting of all sequences (ch cl>
.
the subset of IT G1
tel
tel
..) such that there are at most a finite
number of coordinates with e1 '# ef' where e1 is the identity element of GJ'
Prove that
:L G1 is a normal subgroup of IT G1• :L G1 is called the infinite
tel
le/
le/
direct sum of the G1•
35. Let G be a group and assume that for each positive integer i,
N1 is a normal
subgroup of G. If every element of G can be written uniquely in the form
n1I
•
n,_·
•;
n,_,
with i1 < i2 <
< ik andn1 EN0 prove that G = :L N1 (see
'*
1
1
le[
34).t [Hint: Adapt the proof of Theorem 9.1 by definingf(a1, az, •..)
• •
Exercise
· · ·
to be the product of those a1 that are not the identity element.]
36. If (m, n)
=
1, prove that u,,,,,
*Any infinite index set
notation.
I may
um x ult.
be used here, but the restr iction to the positive integers simplifies the
tun iqueness means that if a;, .. · 111, =
and for r =
=
1, 2, ... , k: i, = j,and il;,b1,.
b1;
· •
b;,•
with i1 <
i2 <
.. · < ik and j, < j2 <
· · ·
< j1,
then II =
t
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9.2
37. LetHbea groupandT1:H-+ Gi,T1:H-+
�
•• .
Finite Abelian Groups
.
289
, T,.:H-+ G,.homomorphisms
with this property: Whenever G is a group and g1:G-+ G1, g2:G...+ G2,
• • •
,
g11:G-+ G,. are homomorphisms, then there exists a unique homomorphism
g*:G-+H such that T1 o g* = g1 for every i. Prove thatH = G1 X G2 X
·
• •
X G,,.
[See Exercise 19.]
m
Finite Abelian Groups
All finite abelian groups will now be classified. We shall prove that every finite abe­
lian group G is a direct sum of cyclic subgroups and that the orders of these cyclic
subgroups are uniquely determined by G. The only prerequisites for the proof other
than Section 9.1 are basic number theory (Section 1.2) and elementary group theory
(Chapters 7 and 8, omitting Sections 7.5 and 8.5).
Following the usual custom with abelian groups, all groups are written in additive
notation in this section. The following dictionary may be helpful for translating from
multiplicative to additive notation:
MULTIPLICATIVE NOTATION
ADDITIVE NOTATION
a+b
ab
e
0
d<
ka
ti= e
ka=O
MN= {mnlmeM,neN}
M+N= {m+nlmeM,neN}
direct product M X N
direct sum M 81
direct factor M
N
direct summand M
Here is a restatement in additive notation of several earlier results that will be used
frequently here:
Theorem 7.9
Let
G be an additive group and let a E G.
(1} If a has order n, then ka
= 0 if and only if n I k.
If a has order td, with d >
(3)
0, then ta has order d.
•
Theorem 9.1
If N1,
•
•
,
, Nk are normal subgroups of an additive group G such that every
element of
a1 E N1, then
G can be written uniquely in the form a1 +
G = N1 EfJ N2 Etl
E0 Nk.. •
·
·
a2
+
· ·
·
+ ak with
·
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290 Chapter 9
Topics in Group Theory
Theorem 9.3
If Mand N are normal subgroups of an additive group
and Mn N
= (O}, then G = M(BN.
Finally we note that Exercise
•
G such that G = M +N
11 of Section 9.1 will be used without explicit mention
at several points.
If G is an abelian group and p is a prime, then G(p) denotes the set of elements in
G whose order is some power of p; that is,
G(p) = {aE GI lal = p• for some n oi=: O}.
It is easy to verify that G(p) is closed under addition and that the inverse of any element
in G(p) is also in G(p) (Exercise 1). Therefore, G(p) is a subgroup of G.
EXAMPLE 1
If G = Z m then G(2) is the set of elements having orders 2°,
21, 22, etc. Verify
that G(2) is the subgroup {O, 3, 6, 9}; similarly, G(3) = {O, 4, 8}. If G = Z3 EB Z3,
then G(3) = G since every nonzero element in G has order 3.
The first step in proving that a finite abelian group G is the direct sum of cyclic
subgroups is to show that G is the direct sum of its subgroups G(p), one for each of the
distinct primes dividing the order of G. In order to do this, we need
Lemma 9.4
Let
a
G
be an abelian group and a E G an element of finite order. Then
= a1 +a:, + ·
· ·
+ Bt. with a, E G(p,), where Pt• .. . , Pt are the distinct positive
primes that divide the order of a.
Proof" The proof is by induction on the number of distinct primes that divide the
order of a. If lal is divisible only by the single prime p., then the order of
a is a power of p1 and, hence, aEG(p1). So the lemma is true in this case.
Assume inductively that the lemma is true for all elements whose order
is divisible by at most k
-
1 distinct primes and that lal is divisible by the
distinct primes Pto
•Pk· Then lal = p1 Ti·
· p{" and n = pt'\ so that lal =
m = p{'
• • .
· ·
•
•
Theorem 1.2 there are integers u, v such that
a=
la =
r
Pk ', with each r1 > 0. Let
Then (m, n) = 1 and by
mn .
(mu + nv)a= mua
1 = mu + nv. Consequently,
+
nva.
But mua EG(pi) because a has order trUI, and, hence, p {' (mua) = (nm')u.a =
u(mna) = uO = 0. Similarly, m(nva) = 0 so that by Theorem 7.9 the order of
nva divides m, an integer with only k - 1 distinct prime divisors. Therefore,
by the induction assumption nua= a2 + tlJ + · + elk, with a;-E G(pJ. Let
a1 = mua; then a= mua + nva = a1 + a2 + · + ak, witha1EG(pJ. •
·
·
·
·
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F i n ite Abelian Groups
9.2
291
Theorem 9.5
If
G is a finite abelian group, then
G = G(p1) EB G(p2) EB··· EB G(,p1),
where Pt.
...
, p1 are the distinct positive primes that divide the order of
G.
Proof... If a E G, then its order divides IGI by Corollary 8.6. Hence, a =
a1 + · ·
+ a" with a1 E G(p)
1 by Lemma 9.4 (where a1
·
p1 does not divide
that
a1 + ai + ·
· ·
= 0 if the prime
laD. To prove that this expression is unique, suppose
+ a, = b1 + b2 + ·· · + b" with a1, b1E G(pJ. Since G is
abelian
a1 - b1 = (b,, - ai) + (b3
For each i,
m = p{•
·
- a:J +
·
·
·
+
(b,
- a,).
b1 - a1 E G(pJ and, hence, has order a power of p1, say p/•. If
· p1\ then m(b1 - a1) = 0 for i 2!: 2, so that
•
m(a1 - b1) = m(b,, - a!} + · · + m(b, - a,) = 0 + · · · +
·
0
= 0.
a1 - b1 must divide m by Theorem 7 .9. But
a1 - b1 E G(p1), so its order is a power of p1• The only power of p1 that
divides m = p{'
p,'• is p18 = 1. Therefore, a1 - b1 = 0 and a1
b1•
Similar arguments for i = 2, .
, t show that ai = b1 for every i. Therefore,
every element of G can be written uniquely in the form a1 + · · · + a" with
aiE G(p)1 and, hence, G = G(_p]) EB· ·· EB G(pJ by Theorem 9.1. •
Consequently, the order of
· ·
·
=
. .
If pis a prime, then a group in which every element has order a power of pis called
a p-group. Each of the G(pJ in Theorem 9.5 is a p-group by its very definition. An
element a of a p-group B is called an element of maximal order if lbl s lal for every
b EB. If lal = P' and b EB, then b has order pl with} s n. SinceJI' =plp"-1 we see that
Jf'b = rl(pfb) = 0. Hence,
If
a
is an element of maximal order p• in a p-group B, then Jl'b
=
0 for every heB.
Note that elements of maximal order always exist in a.finite p-group.
The next step in classifying finite abelian groups is to prove that every finite abelian
p-group has a cyclic direct summand, after which
we
will be able to prove that every
finite abelian p-group is a direct sum of cyclic groups.
Lemma 9.6
Let
G be a finite abelian p-group and a an element of maximal order in G. Then
G such that G =(a} Ei3 K.
there is a subgroup K of
The following proof is more intricate than most of the proofs earlier in the book.
Nevertheless, it uses only elementary group theory, so if you read it carefully, you
shouldn't have trouble following the argument.
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292 Chapter 9
Topics in Group Theory
Proof of Lemma 9.6 ... Consider those subgroups Hof G such that (a) n H= (0).
There is at least one (H = {O)), and since G is finite, there must be a largest
subgroup K with this property. Then (a} n K=(0), and by Theorem 9.3
we need only show that G=(a) +K. If this is not the case, then there
is a nonzero
b
such that
� (a) +K. Let
hence , plb =
0 = 0+0E{a) +K for some positive j). Then
b
k be the smallest positive inte­
ger such that JibE(a) +K ( there must be one since G is a p-groupand,
is not in
=/-1b
(1)
c
(a)+K
andpc=P'"b is in (a}+K, say
pc=
(2)
If
ta+k
(tEZ, kEK).
a has order JI', then jl'x=0 for all xE G because a has maximal order.
Consequently, by (2)
1
= -p<-1 k E {a) n K=(O}andp"� ta = 0. Theorem 7.9
shows that?' (the order of a) dividesjl'-11, and it follows thatp It,
say t=pm. Therefore,pc=ta+k=pma+k, and consequently,
k =pc - pma=p(c - ma) . Let
Therefore,p"-1ta
d=
(3)
c-ma.
Then pd=p(c
- ma)= kEK, but d � K(since c - ma= k' EK would
=ma+k'E(a)+K, contradicting (1)). Use Theorem 7.12
to verify that H= {x+zd I xEK, z E Z} is a subgroup of G with
K � H. Since d = 0 + ldEH and d � K, His larger than K. But K is the
largest group such that {a} n K = {O) , so we must have (a) n H #: (0). If w
imply that
c
is a nonzero element of
(a) n H, then
=Sa= k1+rd
(4)
(k1EK; r, SEZ).
w
We claim thatp ./' r; for if
r =
0 #: w = sa = k1 +
r) = 1, and by
u, v with pu+rv= 1. Then
py, then sincepdE K,
ypdE(a} n K, a contradiction. Consequently, (p1
Theorem 1.2 there are integers
= le=(pu
c
+
rv)c=u(,pc) +v(rc)
u(ta+k)+v(r(d+ma)) [by (2) and (3)]
= u(ta+k) +v(rd + rma)
= u(ta+k)+v(sa - k1+rma) [by (4)]
= (ut +vs+rm)a+(uk - vk1)E(a}+K.
=
This contradicts
Theorem 9.3.
..
(1). Therefore,
G
=(a)+K, and, hence, G ={a) EB Kby
•
.......
..
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9.2
Finite Abelian Groups
293
Theorem 9.7 The Fundamental Theorem of Finite
Abelian Groups
Every finite abelian group G is the direct sum of cydic groups, each of prime
power order.
Proof• By Theorem9.5, Gis the direct sum of its subgroups G(p) , onefor each
primep that divides IGI. Each G(p) is ap-group. So to complete the
proof, we need only show that every finite abelianp-group His a di­
rect sum of cyclic groups, each of order a power of p. We prove this by
induction on the order of H.The assertion is true whenHhas order2
by Theorem 8.7.Assume inductively that it is truefor all groups whose
order is less thanIHI and leta be an element of maximalorderJi' inH.
ThenH (a)El3 Kby Lemma9.6. By induction, Kis a direct sum of
cyclicgroups, each with order a power of p. Therefore, the same is true
of H= (a) El3 K. •
=
EXAMPLE 2
The number 36 can be written as a product of prime powers injustfour
ways: 36 = 2 2 3 3 = 2 2 32 = 22 3 3 = 22 32• Consequently, by
Theorem 9.7 every abelian group of order 36 must be isomorphic to one of the
following groups:
•
•
•
•
•
•
·
•
You can easily verify that no two of these groups are isomorphic (the number
of elements of order2or3 is different for eachgroup) . Thus we have a com­
plete classification of all abeliangroups of order 36 up to isomorphism.
You probably noticed that a familiar group of order 36, namely Z.36, doesn't appear
explicitly on the list in the precedingexample. However, it is isomorphic to L. El3 Z9,
as we nowprove.
Lemma 9.8
If (m, k}
=
1, then Z,,, E13 Zk
=
Zm1c-
Proof .. The order of (1, 1) inZm El3 zk isthesmallest positive integer t such that
(0, 0) = 1(1, 1) = (t, t) . Thus t = 0 (modm) and t = 0 (modk), so that
m It andk It. But (m, k) = 1 implies thatmk It by Exercise 17in
Section 1.2.Hence, mks t. Sincemk(l , 1) = (mk, mk) = (0, 0) and
t is the smallest positive integerwith this property, we must have mk =
t 1(1, 1) 1.Therefore, Z,., El3 "4 (a group of ordermk) is the cyclic group
generated by (1, 1) and, hence, is isomorphic tozmk by Theorem 7.19. •
=
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294 Chapter 9
Topics in Group Theory
Theorem 9.9
If n =
p,n•p2n•
•
•
·Pt', with p11
•
•
•
,
p1 distinct primes, then
Proof" The theorem is true for groups of
order
2. Assume
inductively that it
is true for groups of order less than n. Apply Lemma 9.8 with m = p1"•
and k
=
Pi"'
•
•
•
p,'"· Then Z,,
shows that 4 = z ,,.. EB
•
•
·
=
z,,..
Ef) z,,.,,
(f) Z,:--·
and the induction hypothesis
•
Combining Theorems 9.7 and 9.9 yields a second way of expressing a finite abelian
group as a direct sum of cyclic groups.
EXAMPLE 3
Consider the group
Arrange the prime power orders of the cyclic factors by size, with one row for
each prime:
2
2
22
23
3
3
3
5
52
Now rearrange the cyclic factors of G using the columns of this array as a guide
(see Exercise 15 of Section 9.1) and apply Theorem 9.9:
G
=
G=
(Zi) (f) (Z2 EB Z.J'j Ef) (Z4EB Z3 Ef) ZS) EB (Z8 (f) Z3 (f) Z:?SJ
Z2 (f)
�
E9
E9
Zoo
�·
This last decomposition of G as a sum of cyclic groups is sometimes more
convenient than the original prime power decomposition: There are fewer
c yclic facto� and the order of each cyclic factor divides the order of the next
one. Although the notation is a bit more involved, the same process works in
the general case and proves the following Theorem.
Theorem 9.10
Every finite abelian group is the direct sum of cyclic groups of orders
m1, m2,
•
•
•
,
m11 where m1 I m2, m2 I m3, m3 I m4,
•
•
•
, and
m1_1 I m1•
We pause briefly here to present an interesting corollary that will be used in
Chapter 11. A version of it was proved earlier as Theorem 7 . 16 .
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F inite Abelian Groups
9.2
295
Corollary 9.11
If G is a finite subgroup of the multiplicative group of nonzero element s of a
field F, then G is cyclic.*
Proof .. Since G is a finite abelian group, Theorem 9.10 implies that
G = Z,,., ®
® Zm" where each 111t divides m,. Every element b in
·
Zm, ®
·
·
·
·
® Zm, satisfies mp
·
=
0 (Why?). Consequently, every element
g of the multiplicative group G must satisfy g"'•
solution of the equation x"'•
and x!"•
we
-
-
1F = 0 has at most
0). Since
1F =
m,
=
lF (that is, must be a
G has order
m1m2
•
•
•
m,
distinct solutions in F by Corollary 4.17,
must have t = 1 and G = z,,.,
•
If G is a finite abelian group, then the integers
m1,
•
•
•
,
in T heorem 9.10 are
m,
called the invariant factors of G. When G is written as a direct sum of cyclic groups
of prime power orders, as in Theorem 9.7, the prime powers
called the elementary
are
divisors of G. Theorems 9.7 and 9.10 show that the order of G is the product of its
elementary divisors and also the product of its invariant factors.
EXAMPLE 4
All abelian groups of
order 36 can be classified up to isomorphism in terms
of their elementary divisors (as in Example
2) or in terms of
their invariant
factors (using the procedure in Example 3):
GROUP
Z1©Z2@Z3@Z3
Z1©Z2@Z9
ELEMENTARY
DIVISORS
INVARIANT
FACTORS
ISOMORPHIC
GROUP
2,2,3,3
2, 2, 32
6,6
Z6©Z6
2, 18
Z2©Z11
3,
Z3@Z12
'1!, 3, 3
'1!, 32
Z4©Z!©Z3
l4(B.lg
12
36
ZJ6
The Fundamental Theorem 9.7 can be used to obtain a list of all possible abelian
groups of a given order. To complete the classification of such groups, we must show
that no two groups on the list are isomorphic, that is, that the elementary divisors of a
group are uniquely determined.t
Theorem 9.12
Let G and H be finite abelian groups. Then G
G and H have the same elementary divisors.
*If you have
ti'he
is isomorphic to H if and only if
not read Sections 3.1 and 4.4, skip this corollary until you have.
remainder
of
this section is optional. Theorem
Fundamental Theorem
of Finite Abelian
9.12
is often considered to be part
of
the
Groups.
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296 Chapter 9
Topics in Group Theory
It is also true that G = H if and only if G and H have the same invariant factors
(Exercise 24).
Proof ofTheorem 9.12 ... If Gand
Hhave the same elementary divisors, then both G
and Hare isomorphic to the same direct sum of cy clic groups and, hence,
are isom orphic to each other. Co nversely, if fG-+ His an isomorphism ,
then a andf(a) have the same order for each a E G. It follows that for
each prime p,f(G( p) ) = H(p) and, hence, G(p) = H(p) . Theelementary
divisors of G that are powers of the prime p are precisely the elementary
divisors of G(p) , and similarlyfor H. So we need only prove that isomor­
phic p-groups have the same el ementary divisors. In other words, we need
to prove this half of the theorem only when G and Hare p-groups.
Assume G and Hare isomorphic p-groups. We use induction on the
order of G to prove that G and H have the same elementary divisors.
All groups of order 2 obviously have the same elementary divisor, 2, by
Theorem 8.7. So assume that the statement is true for all groups of order
less than IG� Suppose that the elementary divisors of G are
with
P"',Ji", . . . ,p'",p,p, ... ,p
T
n 1 2:!': n2 2:!':
•
•
•
2:!': n , > 1
copies
and that the elementary divisors of Hare
p"'', p"'>,
• • •
'
p"'>,p,p, ... 'p
s copies
Verify thatpG = {px Ix E G} is a subgroup of G (Exercise 2). If Gis the
directsum ofgroups C" verifythatpGis the directsum of the groupspC1
(Exercise 4). If Ciis cyclic with generator aof order ti', then pC1is the cyclic
group generatedby pa. Since pahas order Jl'-1 by part (3)of Theorem 7.9,
pC1is cyclic of order �1• Note that when n = 1 ( that is, when £;is cyclic of
orderp) , then pC1 = (0). Consequently, the elementary divisorsofpG are
.,n,-1 ......-1
..n,� 1 .
p
'" - ' ... ,p
A similar argument shows that the elementary divisors ofpH are
j/"1-1' 1""-1. . . . ' /l""-1.
If f:G-+His an isomotphism, verify that/(pG) =pH so that pG =pH.
Furthermore,pG*G (Exercise9), so that lfJGI< IG� HencepGandpH
havethe same elementary divisors by the induction hypothesis; that is,
t=kand
ff"-1 =JI"'�\
so that n1
-
1 = Int
-
1 for i = 1, 2,
.
..
, t.
Therefore, n, = m1 for each i. So the only possible difference in elemen­
tary divisors of G and His the number of copies of p that appear on
each list . Since !GI is the product of its elementary divisors, and similarly
for IHI, and since G = H, we have
P"rl"
·
· ·
Jl'p' = !GI = IHI = Jl"'JJ"'•
· ·
•
JJ"'"fl.
Since m1 = n, for each i, we must have p' = p' and, hence,
and Hhave the same elementary divisors. •
r
=
s.
Thus G
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9.2
Finite Abelian Groups
297
• Exercises
NOTE: All groups are written additively, andpalways denotes a positive prime, wzless
noted otherwise.
A. I. If Gis an abelian group, prove that G(p) is a subgroup.
2. If G is an abelian group, prove thatpG
=
{px I x E
G} is a subgroup of G.
3. List all abelian groups ( up to isomorphism) of the given order:
(a) 12
(b) 15
(c) 30
(d)
(e) 90
(f) 144
(g) 600
(h) 1160
72
4. If G and G,(1 sis n) are abelian groups such that G
® pGn.
show that pG p G1 ®
=
·
·
=
G1 ®
·
·
·
E8
Gn,
•
5. Find the elementary divisors of the given group:
(b) Z6 ® Z12 ® Zu
(d) Z12 E8 Z30 ® Zuio EEJ .l240
(a) z2SIJ
(c) Z10 ® Zw ® ZJo El Z.40
6. Find the invariant factors of each of the groups in Exercise 5.
B. 7.
Find the elementary divisors and the invariant factors of the given group. Note
that the group operation is multiplication in the first three and addition in the last .
(a) Us
(b) U11
(c)
U1s
(d) M(Z2)
8.
If G is the additive group O/Z, what are the elements of the subgroup G(2)?
Of G(p) for any positive primep?
9.
(a) If Gis a finite abelianp-group, prove thatpG * G.
(b) Show that part ( a) may be false if Gis infinite. [Hint: Consider the group
G(2) in Exercise 8.]
IO. If Gis an abelianp-group and (n,p)
f(a) na is an isomorphism.
=
1
prove that the mapfG-+ Ggiven by
=
11. If Gis a finite abelian p-group such thatpG ·{O), prove that G= Zp EB
for some finite number of copies of Z,=
• • •
Et> zp
12. (Cauchy's Theorem for Abelian Groups) If Gis a finite abelian group and pis
a prime that divides I GI, prove that Gcontains an element of orderp.
[Hint: Use the Fundamental Theorem to show that G has a cyclic subgroup
of order Ji'; use Theorem 7 .9 to find an element of orderp.]
13. Prove that a finite abelian p-group has order a power of p.
14. If Gis an abelian group of order p1m, with (p, m)
orderp.'
=
1, prove that G(p) has
15. If G is a finite abelian group andpis a prime such that p" divides IGI, then
prove that Ghas a subgroup of order p".
16. For which positive integers n is there exactly one abelian group of order n (up
to isomorphism)?
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298 Chapter 9
Topics in Group Theory
17. Let G, H, K be finite abelian groups.
(a) If
G ® G = H ® H, prove that G = H.
(b)If Gff)H= Gff> K, prove thatH= K.
18. If G is an abelian group of order
n
and k In, prove that there exist a group H
of order k and a surjective homomorphism G � H.
19. Let G be an abelian group and Tthe set of elements of finite order in G. Prove
that
(a) Tis
a subgroup of G(called the torsion subgroup).
(b) Every nonzero element of the quotient group G/Thas infinite order.
20. If G is an abelian group, do the elements of infinite order in G (together with
0) form a subgroup? [Hint: Consider ZEE> .Z3.]
C. 2 1 . If G is an abelian group and/: G � .l a surjective homomorphism with kernel
K, prove that G has a subgroup H such that H
=
.l and G
""
K ® H.
22. Let G and H be finite abelian groups with this property: For each positive
integer
m
the number of elements of order m in G is the same as the number
of elements of order m in H. Prove that G = H.
23. Let G be finite abelian group with this property: For each positive integer
m such that m I IGI, there are exactly m elements in G with order dividing m.
Prove that G is cyclic.
24. Let G and H be finite abelian groups. Prove that G = H if and only if G and H
have the same invariant factors.
25. If G is an infinite abelian torsion group (meaning that every element in G has
finite order), prove that G is the infinite direct sum k G(p), where the sum is
taken over all positive primes p. [Hint: See Exercises
and adapt the proof of Theorem
Ill
9.5.]
34 and 35 in Section 9.1
The Sylow Theorems
Nonabelian finite groups are vastly more complicated than finite abelian groups,
which were classified in the last section. The Sylow Theorems are the first basic step
in understanding the structure of nonabelian finite groups. Since the proofs of these
theorems are largely unrelated to the way the theorems are actually used to analyze
groups, the proofs will be postponed to the next section.* In this section we shall try
to give you a sound understanding of the meaning of the Sylow Theorems and some
examples
of their applications.
Throughout the general discussion in this section all groups are written multiplica­
tively and all integers are assumed to be nonnegative.
"Puritans who believe that the work
must
come before the fun should read Section
9.4
before
proceeding further.
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9.3
Once again the major theme
The Sylow Theorems
299
is the close connection between the structure of
a group G and the arithmetical properties of the integer IGI. One of the most im­
portant results of this sort is Lagrange's Theorem , which states that if G has a
subgroup H, then the integer IHI divides IGI. The First Sylow Theorem provides a
p artial converse:
Theorem 9.13
First Sylow Theorem
Let G be a finite group. If
subgroup of order ,I.
p is a prime and rf divides I GI 1 then G has a
•
EXAMPLE 1
The symmetric group S6 has order 6!
=
720
=
24
•
32
•
5. The First Sylow
Theorem (with p = 2) guarantee s that s6 has subgroups of orders 2, 4 , 8, and
16. There may well be more than one subgroup of each of these orders. For
instance, there are at least 60 subgroups of order 4 (Exercise 1). Applying the
theorem withp = 3 shows that S6 has subgroups of orders 3 and 9. Similarly,
S6 has at least one subgroup of order 5.
If p is a prime that divides the order of a group G, then
G contains a subgroup K
of order p by the First Sylow Theorem. Since K is cyclic by Theorem 8. 7 , its generator
is an element of order p in G. This proves
Corollary 9.14
Cauchy's Theorem
If G is a finite group whose order is divisible by a prime p, then G contains an
element of order p.
•
Let Gbe a finite group and p a prime. If JI' is the largest power of p that divides IGI,
then a subgroup of
G of orderp" is called a Sylow p-subgroup. The existence of Sylow
p-subgroups is an immediate consequence of the First Sylow Theorem.
EXAMPLE2
Since S4 has order 4! = 24 = 23 • 3, every subgroup of order 8 is a Sylow
2-subgroup. You can readily verify that
{(l),
(1234), (13)( 24), (1432),
(24), (1 2)(34), (13), (14)(32) }
is a subgroup of order 8 and, hence, a Sylow 2-subgroup. There are two other
Sylow 2-subgroups (Exercise 2 ). Any subgroup of S4 of order 3 is a Sylow
3-subgroup. Two of the four Sylow 3 -subgroups are {(123), (132), (1)} and
{(13 4), (143), (l)}.
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300 Chapter 9
Topics in Group Theory
EXAMPLE 3*
Let
p be a
prime and Ga finite
G(p)
=
ahelian group of
{aE GI lal
=
order p"m,where p
i/ for some
k;;:::
.r m. Then
O}
is a Sylow p-subgroupof Gsince G(p)has order P' by Exercise 14of Section 9.2.
As we shall see, G (p) is the unique Sylow p-subgroup of G. Theorem 9.5 shows
that G is the direct sum of all its Sylow subgroups (one for each of the distinct
primes that divide IGD.
Let Gbe a groupand x E G. Example 9 of Section 7.4 shows that the map f:G-+ G
given by f(a) x-•axis an isomorphism. If Kis a subgroup of G, then the image of K
under/isx-1.Kx x
{ -1kxlkEK}. Hence,x-1.KX isasubgroupof Gthat iS isomorphic
to K. In particular,x-1.Kx has the same order as K. Consequently,
=
=
if Kis a Sylow p-subgroup of G, then so is x-1Kx.
The next theorem shows that every Sylow p-subgroup of Gcan be obtained from Kin
this fashion .
Theorem 9.15
Second Sylow Theorem
If P and Kare Sylow p-subgroups of a group G, then there exists
that P
=
x-'Kx.
XE G
such
•
Theorem 9.15, together with the italicized statement in the preceding paragraph,
shows that
any two Sylow p-subgroups of G are isomorphic.
Corollary 9.16
Let G be a finite group and Ka Sylow p-subgroup for some prime p. Then K is
normal in G if and only if K is the only Sylow p-subgroup in G.
Proof.,.we know thatx-1.KX is a Sylow p-subgroup for everyxEG. If Kis the
only Sylowp-subgroup of G, then we must have x-1KX Kfor every
xE G. Therefore, Kis normal by Theorem 8.11. Conversely,suppose
K is normal and let P be any Sylow p-subgroup. By the Second Sylow
Theorem there exists xE Gsuch that P x-1Kx. Since Kis normal,
P
x-1Kx
K. Therefore, K is the unique Sylow p-subgroup. •
=
=
=
=
"Skip this example if you haven't read Section 9.2.
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9.3
The Sylow Theorems
301
The preceding theorems establish the existence of Sylow p-subgroups and the rela­
tionship between any two such subgroups. The next theorem tells us how many Sylow
p-subgroups a given group may have.
Theorem 9. 17
Third Sylow Theorem
The number of Sylow p-subgroups of a finite group G divides
I GI
and is of the
form 1 + pk for some nonnegative integer k.
Applications of the Sylow Theorems
Simple groups (those with no proper normal subgroups)
are
the basic building blocks
for all groups. So it is useful to be able to tell if there are any simple groups of a partic­
ular order. The Third Sylow Theorem, together with appropriate counting arguments
and Corollary 9.16, can often be used to establish the existence of a proper normal
subgroup of
a
group G, thus showing that G is not simple.
EXAMPLE4
If G is a group of
order 63
32 • 7, then each Sylow 7-subgroup has order 7 and
=
the number of such subgroups is a divisor of 63 of the form 1 + 1k by the Third
Sylow Theorem. The divisors of 63
form 1 + 1k (with k <?:
0) are
are
1, 3, 71 9, 21, 63 and the numbers of the
1, 8, 15, 22, 29, 36, 43, 50, 57, 64, etc. Since 1 is the
only number on both lists, G has exactly one Sylow 7-subgroup. This subgroup is
normal by Corollary 9.16. Consequently, no group of order 63 is simple.
EXAMPLE 5
We shall show that there is no simple group of order 56
=
3
2
•
7. The only
divisors of 56 of the form 1 + 7k are 1 and 8. So G has either one or eight
Sylow 7-subgroups, each of order 7. If there is just one Sylow 7-group, it
has to be normal by Corollary 9.16. So G is not simple in that case. If G has
eight Sylow 7-groups, then each of them has six nonidentity elements, and
each nonidentity element has order 7 by Corollary 8.6. Furthermore, the
intersection of any two of these subgroups is (e) by Exercise 21 of Section 8.1.
Consequently, there are 8
•
6
=
48 elements of order 7 in G. Every Sylow
2-subgroup of G has order 8. Each element of a Sylow 2-subgroup must have
order dividing 8 by Corollary 8.6 and, therefore, cannot be in the set of 48
elements of order 7. Thus there is room in G for only one group of order 8.
In this case, therefore, the single Sylow 2-subgroup of order 8 is normal by
Corollary 9 .16, and G is not simple.
In the preceding examples, the Sylow Theorems were used to reach a negative con­
clusion (the group is not simple). But the same techniques
can
also lead to positive
results. In particular, they allow us to classify certain finite groups.
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302 Chapter 9
Topics in Group Theory
Corollary 9.18
Let G be a group of order pq, where p and q are primes such that p > q. If
q ..r
(p
- 1 ), then G = Zpfl
Proof� By the Third Sylow Theorem, the number of Sylow p-subgroups must divide
IGI
=
pq, and hence, must be one of l,p, q, or pq. However, the number
must also be of the form 1 + pk for some integer k. Since p > q, we cannot
have q = 1 +pk. Furthermore, both p = 1 +pk and pq = 1 +pk imply that
p 11, which is impossible. Therefore, there is exactly one Sylow p-subgroup
H of
orderp, which is normal by Corollary 9.16. A similar argument (using
the fact that q .t (p - 1)) shows that there is a unique Sylow q-subgroup K
of order q, which is also normal. Since H n K is a subgroup of both Hand
K, its order must divide both IHI = p and IKI = q by Lagrange's Theorem.
Hence, H n K = (e). Exercise 15 shows that G = HK. Therefore,
G = H X K by Theorem 9.3. But H= �and K =�by Theorem 8.7.
Consequently, by Lemma 9.8, G H x K = z, x Z" = Zpq.* •
=
EXAMPLE 6
It is now easy to classify all groups of order 15 = 5 3. Apply Corollary 9.18
with p = 5, q = 3 to conclude that every group of order 15 is isomorphic to Z15•
•
Similarly, there is a single group (up to isomorphism ) for each of these orders:
33
=
11·3, 35
=
7
•
5, 65
=
13
•
5, 77 = 11
•
7, and 91
=
13
•
7.
Other applications of the Sylow Theorems are given in Section 9.5.
• Exercises
NOTE: Unless stated otherwise, G is afinite group andp is a positive prime.
A. 1. Show that S6 has at least 60 subgroups of order 4.
[Hint: Consider cyclic
subgroups generated by a 4-cycle (such as ((1234))) or by the product of
a 4-cycle and a disjoint transposition (such as (( 1234)( 5 6) )); also look at
noncyclic subgroups, such as {(l), ( 12), (34), (12 )(34)}.]
2.
(a) List three Sylow 2-subgroups of
84•
(b) List four Sylow 3-subgroups of S4•
3. List the Sylow 2-subgroups and Sylow 3-subgroups of A4•
4. List the Sylow 2-subgroups, Sylow 3-subgroups, and Sylow 5-subgroups of
Z12 X
.l12
X Z10• [Section 9.2 is a prerequisite for this exercise.]
"The proof of Lemma 9.8 is independent of the rest of Section
9.2 and may be read
now if you skipped
that section.
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Q.3
5.
6.
8.
(b)
115
(c)
143
391
Prove that there are no simple groups of the given order:
(a)
B.
(b) p = 5 and JGI = 60
p = 3 and IGI =72
Classify all groups of the given order:
(a)
7.
303
G possibly have when
How many Sylow p-subgroups can
(a)
The Sylow Theorems
(c)
(b) 200
42
(d)
231
255
Use Cauchy's Theorem to prove that a finite p-group has
9. If
order p" for some n � 0.
N is a normal subgroup of a (not necessarily finite) group G and both N
GIN are p-groups, then prove that G is a p-group.
and
10.
If His a normal subgroup of
G and IHI =,I', show that His contained in
G. [You may assume Exercise 24 in Section 9.4.]
every Sylow p-subgroup of
11.
12.
If f is an automorphism of G and K is a Sylow p-subgroup of G, is it true that
f(K) = K?
Let
K be a Sylow p-subgroup of G and Hany subgroup of G. Is Kn Ha
[Hint: Consider S4.]
Sylow p-subgroup of In
13.
If every Sylow subgroup of
G is normal,
prove that
G is the direct product of
its Sylow subgroups (one for each prime that divides IGD. A group with this
property is said to be
14.
nilpotent.
If p is prime, prove that there are no simple groups of order 2p.
15. (a)
If Hand Kare subgroups of G, then
HK denotes the set
{hkEG I hEH,kEK}. If HnK= (e), prove that IHKI =IHI· IKI.
[Hint: If hk = h1kb then h1 -th k1k-1 .]
=
(b)
If
H and Kare
any subgroups of
G,
prove that
HKI =I HI ·I K I.
I
I HnKI
16.
If
17.
If
G is a group of order 60 that has a normal Sylow
G also has a normal Sylow 5-subgroup.
3-subgroup, prove that
G is
a noncyclic group of order 21, how many Sylow 3-subgroups does
G is
a simple group of order 168, how many Sylow 7-subgroups does
Ghave?
18.
If
Ghave?
19.
If p and q are distinct primes, prove that there are no simple groups of order pq.
20.
If
21.
Prove that there are no simple groups of order 30.
22.
If p and q are distinct primes, prove that there is no simple group of order p2q.
23.
(a)
If 161
=
105, prove that
G has
a subgroup of order 35.
(b)
If 161
=
375, prove that
G has
a subgroup of order 15.
G has order /fm with m
< p, prove that
G is not simple.
..
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304 Chapter 9
Topics in Group Theory
24. Let K be a Sylow p-subgroup of Gand Na normal subgroup of G. Prove that
K n Nis a Sylow p-subgroup of N.
C
25.
If p, q, r are primes with p < q <
r,
prove that a group of order pqr has a
normal Sylow r-subgroup and, henoe, is not simple.
Ill
Conjugacy and the Proof of the Sylow Theorems
Appendix D (Equivalence Relations) is a prerequisite for this section. The proofs of the
Sylow Theorems depend heavily on the concept of conjugacy, which we now develop.
Let Gbe a group and a,
such that
b E G. We say that a is conjugate to b if there exists x E G
b = x-1ax. For example, (12) is conjugate to (13) in S3 because
(123)-1(12)(123) = (132)(12)(123) = (13).
The key fact about conjugation is
Theorem 9. 19
Conjugacy is an equivalence relation on
Proof• We write a -b if
G.
a is conjugate to b. Reflexive: a -
Symmetric: If a - b, then
a since
a= eae= e-1ae.
b = x-1ax for some x in G. Multiplying on the
left by x and on the right by x-1 shows that a= xbx-1
(x-1r1bx-1•
,-1 by
Henoe, b - a. TransitiVe: If a - band b - c, then b = x-1ax an.de
for some x, y E G. Henoe, c= y-1cx-1ax)y= (y-1x-1) a(xy )= (xy)-1a(xy).
=
=
Thus
a
- c; therefore, - is an equivalence relation.
•
The equivalence classes in Gunder the relation of conjugacy are called
classes. The discussion of
conjugacy
equivalence relations in Appendix D shows that
The conjugacy class of an element
a
consists of all the elements in Gthat are
conjugate to a.
Two conjugacy classes are either disjoint or identical.
The group Gis the union of its distinct conjugacy classes.
EXAMPLE
1
(12) in S3 consists of all elements x-1(12)x, with x E S3•
A straightforward computation shows that for any x E S3, x-1(12)x is one of
The conjugacy class of
(12), (13), or (23); for instance,
(23)-1(12)(23)= (23)(12)(23) = (13)
(132)-1(12)(132) = (123)(12)(132) = (23).
Thus the conjugacy class of
(12) is {(12), (13), (23)}. Similar computations show
that there are three distinct conjugacy classes in S�:
{(1)}
{(123 ), (132)}
{(12), (13), (23)}.
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Conjugacy and the Proof of the Sylow Theorems
9.4
305
Although these conjugacy classes are of different sizes, note that the number of
elements in any conjugacy class (1, 2, or 3) is a divisor of 6, the order of S3• We
shall see that this phenomenon occurs in the general case as well.
Let
G be a group and aE G. The centralizer of a is denoted C( a) and consists of
G that commute with a, that is,
all
elements in
C(a)
If
=
{gEGlga
=
ag}.
G
S5 and a
(123), for example, you can readily verify that C(a)
{ (1 ), (123), (132)} and that C(a) is a subgroup of S3• If a is a nonzero rational number
in the multiplicative group 0*, every element of Q* commutes with a, so C(a) is the
entire group O*. These examples are illustrations of
=
=
=
Theorem 9.20
If G is a group and aE G, then C(a) is a subgroup of G.
Proof•
Since ea
then
=
ae,
we have e E C(a), so that C(a) is nonempty. If
(gh)a
=
g(ha)
=
g(_ah)
=
(ga)h
=
(ag)h
So ghE C(a), and C(a) is closed. Multiplying ga
and right by g-t shows that ag-1
g-1a. Hence,
=
=
g, h E C(a),
a(gh).
ag on both the left
gEC(a) implies that
=
g�1E C(a). Therefore, C(a) is a subgroup by Theorem 7.11.
•
The centralizer leads to a very useful fact about the size of conjugacy classes:
Theorem 9.21
Let G be a finite group and aE G. The number of elements in the conjugacy
class of a is the index [G:C(a)] and this number divides I GI·
Proof•
For notational convenience, we shall sometimes denote C(a) by C in this
proof. Let Sbe the set of distinct right cosets of Cin G, and let The the
conjugacy class of
a in G (which consists of the distinct conjugates of a).
Define afunctionf:S� Tby the rule:/(Cx}
x-1ax. We shall show
below that/is a well-defined bijection of sets, which means that Sand
=
Thave the same number of elements. The number of elements in Sis
the number of distinct right cosets of C(a), namely (G:C(a)], and the
number of elements in Tis the number of distinct conjugates of
a. This
proves the first part of the theorem. As for the final part, the number
[G:C(a)] divides IGI by Lagrange's Theorem 8.5.
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306 Chapter 9
Topics in Group Theory
Now for the details: Reading each of the following "if and only if"
statements in the direction=> shows that/is well defined (meaning that
Cx =Cy implies f(Cx)=f( <'.))):
Cx =Cy <::>.xy-1EC
[11ieorem 8.2]
<::> (xy-1)a =a(xy-1)
[Definition of C]
<=>a =(xy�1)-ia(xy��
[Left multiply by (xy-1r1 .]
<::>a =yx-1axy-1
<::> y-1ay =x-1ax
[Corollary 7.6]
<=>/(Cy)=f( Cx)
[Definition of/]
[Left multiply by y-1 and
right multiply by y.]
Reading these same statements in the direction *= from bottom to top
shows thatf(Cx)=f( Cy) implies Cx =Cy, so that/is injective.* Finally,
/is surjective because, given any conjugate u-1au of a, it is the image of
the coset
Cu. Therefore, f is bijective and the proof is complete.
•
C1o Cl> ... , C, be the distinct conjugacy classes of G.
Ct. Since distinct conjugacy classes are mutually disjoint,
Let G be a finite group and let
Then G =Ci U
Ci U
(1)
IGI
·
=
U
· ·
IC1
U
C2
U
·
•
•
U
c,i
=
IC1I
IC2I +
+
· · · +
IG,I,
where ICil denotes the number of elements in the class C1• Now choose one element,
say a1, in each class C,. Then C1 consists of all the conjugates of a,. By Theorem 9.21,
ICJ is precisely [G:C(aJ], a divisor of 161· So equation (I) becomes
IGI
(2)
=
(G:C(a1)1
+
(G:C(ai)I + ·
·
· +
(G:C(a,)(.
This equation (in either version (I) or (2)) is called the cla� equation of the group G. It
will be the basic tool for proving the Sylow Theorems. Other applications of the class
equation are discussed in Section 9.5.
EXAMPLE 2
In Example 1 we saw that S3 has three distinct conjugacy classes of sizes 1, 2,
and 3. Since IS31 =6, the class equation of S3 is 6 =1 + 2 + 3.
If
c
and x are elements of a group G, then
ex =xc if and only if x-1cx =c. Thus c is
in the center of G [ex=xc for every x E G] if and only if c has exactly one conjugate, itself
(x-1cx = c for every xEG]. Therefore, the center Z(G) of G is the union of all the one­
element conjugacy classes of G, so that the class equation can be written in a third form:
IGI
(3)
=
IZ(G)I
+
IC1I
+
IC2 I
+ · · · +
IC,(,
, C, are the distinct conjugacy classes of G that contain more than one
element each and each ICil divides IGJ.
where c:;,
. . .
In addition to the class equation, one more result is needed for the proof of the
Sylow Theorems.
*The reasons in the right-hand column above must
be adjusted
when reading from bottom to top
(Exercise 4).
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Conjugacy and the Proof of the Sylow Theorems
9.4
Lemma 9.22
307
Cauchy's Theorem for Abelian Groups
If G is a finite abelian group and pis a prime that divides the order of G, then
G contains an element
of order p.
The lemma is an immediate consequence of the Fundamental Theorem of Abelian
Groups (Exercise 12 in Section 9.2). The following proof, however, depends only on
Chapters 7 and 8.
Proof of Lemma 9.22 ... The proof is by induction on the order of G, using the
Principle of Complete Induction.* To do this, we must first show that
the theorem is true when IGI = 2. In this case, ifp divides I GI, thenp = 2.
The nonidentity element of G must have order 2 by part (1) of
Corollary 8.6, and so the theorem is true.
Now assume that the theorem is true for all abelian groups of order
less than n and suppose IGI = n. Let a be any nonidentity element of
G. Then the order of a is a positive integer and is therefore divisible by
some prime q (Theorem 1.8), say lal = qt. The element b = a1 has order
q by Theorem 7 9 If q = p, the theorem is proved. If q "# p, let N be the
cyclic subgroup (b). N is normal since G is abelian and N has order q by
Theorem 7.15. By Theorem 8.13 the quotient group G/Nhas order
IGl/l.NI = n/q < n. Consequently, by the induction hypothesis, the theorem
is true for G/N. The primep divides IGI, and IGI = I NI IG/NJ = q IG/NJ.
Since q is a prime other thanp,p must divide IG/NI by Theorem 1.5.
Therefore, G/N contains an element of order p, say Ne. Since Ne has
orderp in G/N, we have NcP = (Nef = Ne and, hence, c' EN. Since N
has order q, cpq = ( cP)'I = e by part (2) of Corollary 8.6.
Therefore, c must have order dividingpq by Theorem 7.9. However,
e cannot have order 1 because then Ne would have order 1 instead of p
in G/ N. Nor can e have order q because then (Ne)"= Ne" = Ne in G/N,
so thatp (the order of Ne) would divide q by Theorem 7.9. The only
possibility is that e has orderp or pq; in the latter case, c" has orderp by
Theorem 7 .9. In either case, G contains an element of order p. Therefore,
the theorem is true for abelian groups of order n and, hence, by induc­
tion for all finite abelian groups. •
.
.
Proofs of the Sylow Theorems
We now have all the tools needed to prove the Sylow Theorems.
Proof of the First Sylow Theorem 9.13 ... Tue proof is by induction on the order
of G. If IGI 1, thenp0 is the only prime power that divides IGI, and G
itself is a subgroup of order p0• Suppose IGI > 1 and assume inductively
that the theorem is true for all groups of order less than IGI· Combining
the second and third forms of the class equation of G shows that
=
IGI = IZ(G)I
+
[G:C(aJ]
+
[G:C(ai)]
+
·
·
·
+
[G:C(a,.)],
•see Appendix C.
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308 Chapter 9
Topics in Group Theory
where for each i, [G:C(aJ] > 1. Furthermore, IZ(G)I � 1 (since eEZ(G)),
and IC(aJI < IGI (otherwise, [G:C(aJ] = 1).
Suppose there is an indexjsuch thatp does not divide [G:C(a1)]. Then
by Theorem l.5 pc must divide IC(aj)I because JI divides IGI by hypothesis
and IGI = IC(�)I [ G: C(ap] by Lagrange's Theorem. Since the subgroup
C(ai) has order less than IGI, the induction hypothesis implies that C(a1),
and, hence, G has a subgroup of order/'.
On the other hand, ifp divides [G:C(a.i)] for every i, then since p
divides IGI, p must also divide IGI - [G:C(ai)]
- [G:C(a,.)] =
IZ(G)f. Since Z(G) is abelian, Z(G) contains an element c of order p by
Lemma 9.22. Let N be the cyclic subgroup generated by c . Then Nhas
orderp and is normal in G (Exercise 8). Consequently, the order of the
quotient group G/ N, namely fGf /p, is less than I GJ and divisible by 1-1•
By the induction hypothesis G/Nhas a subgroup Tof order J1-1• There
is a subgroup Hof G such that N k: Hand T =H/NbyTheorem 8.24.
Lagrange's Theorem shows that
•
-
·
·
·
IHI = INI IH/NJ = INJ ITI =pp--1 =JI.
·
•
So G has a subgroup of order/' in this case, too.
•
The basic tools needed to prove the last two Sylow Theorems are very similar to those
usedabove, except that we will now deal with conjugate subgroups rather than conjugate
elements. More precisely, let Hbe a fixed subgroup of a group G and let A and B be any
subgroups of G. We say that A is H-amjugate to B if there exists an xEHsuch that
B
= x-1Ax = {x�1ax I a E A}.
In the special case when His the group G itself, we simply say that A is conjugate to B,
or that Bis a conjugate of A.
Theorem 9.23
Let H be a subgroup of a group
G. Then H-conjugacy is
an equivalence rela­
tion on the set of all subgroups of G.
Proof•Copytheproof ofTheorem 9.19, using subgroups A, B, Cin place of
elements a, b,
c.
•
Let A be a subgroup of a group G. The normalizer of A is the set N(A)
defined by
N (A) = {gEGfg-1Ag =A}.
Theorem 9.24
If A is a subgroup of a group
G, then N(A) is a subgroup of G and A
is a normal
subgroup of N(A).
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9.4
Conjugacy and the Proof of the Sylow Theorems
Proof• Exercise 7 shows that A
309
� N( A)and that g E N(A)if and only if Ag = gA.
Using this fact, the proof of Theorem 9.20 can be readily adapted to prove
that N(A)is a subgroup. The definition of N(A)shows that A is normal
inN(A). •
Theorem 9.25
Let H and A be subgroups of a finite group G. The number of distinct
H-conjugates of A (that is, the number of elements in the equivalence class
of A under H-conjugacy) is [H:H n N(A)) and, therefore, divides IHI.
Proof• The proof
of Theorem 9.21 carries over to the present situation if you
replace GbyH, a by A, and C by H n N(A).
•
Lemma 9.26
Let Q be a Sylow p-subgroup of a finite group G. If x E G has order a power of
p and x-1Qx =Q, then xEQ.
Proof• Since Q is normal in N(Q) by Theorem 9.24, the quotient group N(Q)/Q is
defined. By hypothesis, x E N(Q). Since lxl is some power of p, the coset
Qx in N(Q)/Q also has order a power of p. Now Qx generates a cyclic
subgroup T of N(Q)/Q whose order is a power of p. By Theorem 8.24,
T =H/Q, where His a subgroup of Gthat contains Q. Since the
orders of the groups Q and Tare each powers of p and IHI = IQI ITI
by Lagrange's Theorem, IHI must be a power of p. But Q i;; H, and IQI
is the largest power of p that divides IGI by the definition of a Sylow
•
p-subgroup . Therefore, Q =H, and, hence, T =H/Q is the identity
subgroup. So the generator Qx of Tmust be the identity coset Qe. The
equality Qx =Qe implies that x E Q. •
Proof of the Second Sylow Theorem 9.15 ... Since Kis a
Sylow p-subgroup , Khas
order p", where IGI =/I'm and p .t m. Let K = K1, K2,
, K, be the dis­
tinct conjugates of Kin G. By Theorem 9.25 (withH =Gand K =A),
t =[G:N(K)]. Note that p does not divide t [reason: P'm =IGI =
IN(K)I [G:N(K)] =IN(.K)I t and p" divides IN(.K)I because Kis a subgroup
of N(.K)]. We must prove that the Sylow p-subgroup Pis conjugate to K,
that is, that Pis one of the Ki· To do so we use the relation of P-conjugacy.
• • •
•
·
Since each K1 is a conjugate of K1 and conjugacy is transitive, every
conjugate of K1 in Gis also a conjugate of K•1 In other words, every con­
jugate of K1 is some Kr Consequently, the equivalence class of K1 under
P-conjugacy contains only various Kr So the set S = {K1> K2 ,
; K}
,
of all conjugates of Kis a union of distinct equivalence classes under
• • •
P-conjugacy. The number of subgroups in each of these equivalence
classes is a power of p because by Theorem 9.25 the number of sub­
groups that are P-conjugate to K1 is [P: P n N(K1)] , which is a divisor of
IPI =fl' by Lagrange's Theorem. Therefore, t (the number of subgroups
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310
Chapter Q
Topics in Group Theory
in the set S) is the sum of various powers of p (each being the number of
subgroups in one of the distinct equivalence classes whose union is S).
Since p doesn't divide
t, at least one of these powers of p must be [11= 1.
Thus some K, is in an equivalence class by itself, meaning that
x-1K1x= K1forevery xEP. Lemma 9.26 (with Q= K,) implies that
x EK1 for every such x, so that P !:;;; K1• Since both P and K1 are Sylow
p-subgroups, they have the same order. Hence, P= K;. •
Proof of the Third Sylow Theorem 9.17 � Let s= {Kl>
.
.
•
'
Kr} be the set of all
are all the
Sylow p-subgroups of G. By the Second Sylow Theorem, they
distinct conjugates of K1• The proof of the Second Sylow Theorem shows
that
t= [G: N (K1)], which divides the order of G by Lagrange's Theorem.
Let P be one of the K1 and consider the relation of P-oonjugacy. The
only P-conjugate of Pis P itself by closure. The proof of the Second Sylow
Theorem shows that the only equivalence class consisting of a single sub­
group is the class consisting of P itself. The proof also shows that S is the
union of distinct equivalence classes and that the number of subgroups in
each class is a power of p. Just one of these classes contains P, so the num­
ber of subgroups in each of the others is a positiVe power of p. Hence, the
number
t of Sylow p-subgroups is the sum of 1 and various positive powers
of p and, therefore, can be written in the form 1 + kpfor some integer k.
•
• Exercises
NOTE:
Unless stated otherwise, G is afinite group and p is a positive prime.
A. 1. List the
distinct conjugacy classes of the given group.
a E G, then show by example that C(a) may not be abelian. [Hint: If
a= (12) in S5, then (34) and (345) are in C(a).]
2. If
3. If His
of a in
subgroup of G and a EH, show by example that the conjugacy class
H may not be the same as the conjugacy class of a i n G.
a
4. Write out the part of the proof of Theorem
9.21
showing that/is injective,
including the reasons for each step. Your answer should begin like this:
-1
/(Cy)= f(C x)�y ay= .x-1ax
�a= yx-1axy:-1•
5. List
all con jugates of the
Sylow
[Definition offl
[L<dl' multiply by y and right multiply by y-1.]
3-subgroup ((123)) in S4•
6. If Hand Kare subgroups of Gand His normal in
K, prove that K is a
subgroup of Gin which
subgroup of N(ll). In other words, N(Jl) is the largest
His a normal subgroup.
7. If A is a subgroup of G, prove that
(a) A!;;N(A);
(b)
g E N(A) if and only if
Ag= gA.
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Conjugacy and the Proof of the Sylow Theorems
9.4
8.
B.
311
If Nis a subgroup of Z(G),prove that Nis a normal subgroup of G.
9. If Cis a conjugacyclassin Gand/is an automorphism of G, prove that/( Cjis
also a conjugacyclass of G.
10.
Let Gbe an infinitegroup and Hthe subset of allelements of Gthat have only
a finite number of distinct conjugates in G. Prove that His a subgroup of G.
11.
If Gis a nilpotent group (see Exercise 13of Section 9.3),prove that Ghas
this property: If mdivides IGI, then Ghas a subgroup of order m. [You may
assume Exercise 22.]
12.
Let Kbe a Sylow p-subgroup of Gand Na normal subgroup of G. If Kis a
normal subgroup of N,prove that Kis normal in G.
13. Prove Theorem 9.23.
14.
Let Nbe a normal subgroup of G, a EG,and Cthe conjugacy class of
(a) Prove that
a
EN ifand only
if
in G.
a
Cs;; N.
(b) If C1is any conjugacy class in G, provethat C1\;; Nor C1 n N
==
0.
(c) Use the class equation to show that INI= ICil +···+IC�. where C1o ... ,
Ckare all the conjugacy classes of Gthat are contained in N.
15. If N :¢: {e)is a normal subgroup of Gand IGI
(Hint: Exercise 14(c) maybe helpful.]
16.
=
p", prove that N n Z(G) :¢: (e).
Completethe proof of Theorem 9 .24.
17. Prove Theorem 9.25.
18.
If Kis a Sylow p-subgroup of Gand Hisa subgroup that contains N(K),
prove that[G:H]
1 (mod p).
==
19. If Kis a Sylow p-subgroup of G,prove that N(N(K)) = N(K).
20.
If His a proper subgroup of G, prove that Gis notthe union of all the
conjugates of H. (Hint: Remember that His a normal subgroup of N(ll);
Theorem 9.25 maybe helpful.]
21.
If His a normal subgroup of Gand His asubgroup of Gwith WI
/'·,
prove that His containedin every Sylow p-subgroup of G.[You may assume
Exercise 24.]
C. 22.
""
If IGI = p",prove that Ghasa normal subgroup of order Jl'-1•[Hint: You may
assume Theorem 9.27 below. Use induction on n. Let N = (a) , where a E Z( G)
has order p (Whyis there such an a?); then G/N hasa subgroup of order p"-2;
use Theorem 8.24.]
23.
If !GI= p",prove that everysubgroup of Gof order Jl'-1is normal.
24.
If His a s ubgroup of Gand Hhas order some power of p, prove that
His contained in a Sylow p-subgroup of G. [Hint: Proceed as in the
proofs of the Second and Third Sylow Theorems but use the relation of
H-conjugacy instead of P-conjugacyon the set {Ki. .. . , K,} of all Sylow
p-subgroups.]
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312
Chapter 9
II
Topics in Group Theory
The Structure of Finite Groups
The tools developed in Sections 9. l-9.4 are applied here to various aspects of the
classifci ation problem. In particular, all groups of orders s 1 5 are classified. We begin
with some useful facts about p-groups.
Theorem 9.27
If G is a group of order p", with p prime and n
�
contains more than one element. In particular, I Z( G )I
Proof� By Lagrange's Theorem, ]Z(G)I
1, that is, that IZ(G�
shows that
k 2!::
IZ(G)I
=
2!:: p.
1, then the center Z( G)
=
p" with 1 s ks n.
p1• with 0 s k s n. We now show that
Form (3) of the class equation (page 306)
=
I GI - le.I - IC2I - ... - IC.I
where each ICd is a number larger than 1 that divides IG� Since IGI p",
the divisors of IGI larger than 1 are positive powers of p. Therefore, each
ICil is divisible by p. Since IGI is also divisible by p, it follows that p divides
I Z(�and, hence, !Z(G)l 2!:: p. •
=
Corollary 9.28
If p is a prime and n > 1, then there is no simple group of order p".
Proof� If G is a group of order p", then Z(G) is a normal subgroup. If Z(G) :;:
G, then G is not simple.
Theorem 8.25. •
If Z(G)
=
G, then G is abelian and not simple by
Corollary 9.29
If G is a group of order p2, with p prime, then G is abelian. Hence, G is
isomorphic to
Zr or Z,,
x
ZP'
EXAMPLE 1
By Corollary 9.29, every group of order 9 is isomorphic to Z9 or Z3 x Z3•
Similarly, the only groups of order 169 132 (up to isomorphism) are Z169 and
Z13 x Z13•
=
......
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9.5
The Structure of Finite Groups
313
Proof of Corollary 9.29 ... Z(G) has order p or p2 by Lagrange'sTheorem and
Theorem 9.27. If Z(G) has order p2, then G = Z(G), which means that
G is abelian. If Z( G) has order p, then the quotient group
orderlGl/IZ(G )I
=
p2/p = pbyTheorem 8.13. Hence,
G/Z(G) has
G/Z(G) is cyclic by
Theorem 8. 7. Therefore, Gis abelian by Theorem 8.15.The last state­
ment of the theorem now follows immediately from the Fundamental
Theorem of Finite Abelian Groups.
•
In Corollary 9.18 certain groups of orderpq (withp, q prime ) were characterized.
We can now extend that argument to some groups of order p2q.
Theorem 9.30
Let p and q be distinct primes such that q =I= 1 {mod p) and p2 =/= 1 (mod q). If G
is a group of order p2q, then G is isomorphic to "Zpiq or Zp X Zp x
Zq.
EXAMPLE 2
Theorem 9.30 allows us to classify all groups of order 45. Note that 45
=
32
•
S,
and that 5 ;(!; 1 (mod 3) and 32 ;;e 1 (mod 5). So if Gis a group of order 45,
then by Theorem 9.30 (withp
Z3 X
3 and q
5), Gis isomorphic to Z;.5 or to
Z3 X Z5• Similar arguments may be used to classify groups of many differ­
=
=
ent orders, including
99
=
9. 11,
153
=
9. 17,
325
=
25. 13,
175
=
539
25
=
•
7,
245
=
49. 5,
49. 11.
Proof of Theorem 9.30 ... By theThird SylowTheorem, the number of Sylow
p-subgroups of G is congruent to 1 modulopand divides IG� Since the
divisors of !Gl are l,p,p2, q,pq, andp1q , the only possibilities are 1 and
q. There cannot be q of them because q ';/= 1 (modp). Hence, there is a
unique Sylowp-subgroup H, which is normal by Corollary 9.16. Similarly,
Ghas 1,p, or# Sylow q-subgroups, and neither p nor pi is possible since
pi ;!jE 1 (mod q). Hence, there is a unique normal Sylow q subgroup K.
The order of the subgroup H n K must divide both IHI p1 and IA'.!
q by
Lagrange'sTheorem. Hence, H n K = (e). Furthermore, HK
G
-
=
=
=
by Exercise 15 in Section 9.3.Therefore, G
=
HX K by Theorem 9.3.
Now His isomorphic to Zp' or ZP X Z, by Corollary 9.29 and K = Z9
by Theorem 8.7. Consequently, by Lemma 9.8, G
Zr X Zq = Zp'qor G
=
HX K=Z, X z X Zq.
,
=
HX K=
•
Corollary 9.31
If p and q are distinct primes, then there is no simple group of order p2q.
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314
Chapter 9
Topics in Group Theory
Proof... Suppose G is a group of order p2q. If either p2 !ji!!
1(mod
q) or q
$ 1
(mod p ), then the proof of Theorem 9 .30 shows that G has a normal
Sylow subgroup and, hence, is not simple. If both p2
= 1(mod q) and
- 1)and p I (q - 1), which implies that p s
q - 1or, equivalently , q � p + 1. Since p1 - 1 = (p - l)(p + 1), we
know that q I (p - 1) or q I (p + 1) by Theorem 1.5. The former is impos­
sible because q O?: p + 1, and the latter implies that q s p + 1, so that
q = p + 1. Since p and q are primes, the only possibility is p = 2 and
q =3. Exercise 2 shows that no group of order 22 3 =12 is simple. •
q
=
1(mod p), then q I (p2
•
Dihedral Groups
We now introduce a family of groups that play a crucial role in the classification of
groups of order 2p. Recall that the group D4 consists of various rotations and reflections
of the square (see Section 7.1or7.1.A). This idea can be generafu.ed
be a regular polygon of n sides
as
follows. Let P
(n O?: 3).* For convenient reference, assume that P has its
center at the origin and a vertex on the negative x-axis , with the other vertices numbered
counterclockwise from this one, as illustrated here in the cases n
y
6
5
y
=5
and
n = 6.
5
2
Think of the plane as a thin sheet of hard plastic. Cut out P, pick it up, and replace it, not
necessarily in the same position, but so that it fits ex:actly in the cut-out space. Such a motion
is called a symmetry of P.t By considering a symmetry as a function fium P to itself and
using composition of functions as the operation (gfmeans motion/ followed by motion g),
the set D,. of all symmetries of P forms a group, called the dihedral group of degree n.
Theorem 9.32
The dihedral group Dn is a group of order
such that
lrl =n,
ldl=2,
2n generated
by elements r and
d
and
Proof"' The proof that Dft is a group is left to the reader. Let r be the counter­
clockwise rotation of 360/n degrees about the center of P; r sends
vertex 1to vertex 2, vertex 2 to vertex 3, and so on. Note that
r
has
•"Regular" meansthat all sides of Phavethe same length and all its vertex angles (each formed by
two a djacent sides) a r ethe same size. It can
be shownthatthe perpendicular bisecto rs ofthen sides
all in tersect at a single point, which is called the center of P.
tAll motions that result in the same final position for Pare considered to be the same.
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9.5
The Structure of Finite Groups
315
order n because r" is a 360° rotation that returns P to its initial position
d be the reflection in the x-axis. As shown in
d "reverses the orientation" of P: vertices that were
(the identity symmetry). Let
the following figure,
formerly numbered counterclockwise from vertex l are now numbered
clockwise:
The element
d has order 2 because reflecting twice in the x-axis also
returns P to its initial position.
Since adjacent vertices of P remain adjacent under any symmetry,
the final position of P is completely determined by two factors: the
new orientation of P (whether the vertices
are
numbered clockwise
or counterclockwise from vertex 1) and the new location of vertex 1.
Consequently, every symmetry is the same as either
(0 s
[Cmmterdockwise rotation of i(360/n)
degrees that preserves orientation and moves
vertex 1 to th£ position originally occupied by
vertex i + J]
i < n)
or
(0 s
i < n)
[Reflection in the x-axis that reverses
orientation followed by a counterclockwise
rotation that moves vertex 1 to th£ position
originally occupied by vertex i + J]
Therefore
D,.
= {e = r0, r, ,:i,
.
•
.
, yr-•; d = ,Pd, rd, ,:id, ... , yr-14.
Furthermore, the 211 elements listed here are all distinct (r' and 1' move
vertex
1
to different positions and 1' =
rid is impossible since t preserves
the vertex orientation, but rid reverses it). Hence, D,. is a group of order 211.
Finally, verify that drd moves vertex
1
to the position originally
n and leaves the vertices in counterclockwise order.
In other words, drd is the rotation that moves vertex 1 to vertex n, that
is, drd = 1'-1• Since r has order n, ,-1 = t"-1 and, hence, drd = ,-1•
Multiplying on the right by d shows that dr = ,-1d. •
occupied by vertex
We can now classify another family of groups.
Theorem 9.33
If G is a group of order 2p, where p is an odd prime, then
the cyclic group Z'l/J or the dihedral group Dp.
G
is isomorphic to
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316
Chapter 9
Topics in Group Theory
EXAMPLE 3
Theorem 9.33 can be used to classify all groups of orders 6, 10, 14, 22, 26, 34,
D11,
D19• Theorem 9.33
etc. For instance, every group of order 22 is isomorphic either to Z'Zl. or
and every group of order 38 is isomorphic either to Zs8 or
also provides a second proof that there are exactly two nonisomorphic groups
of order 6. (See Theorem 8.9 for the first proof.)
Proof of Theorem 9.33 � G contains an element a of order p and an element b of
order 2 by Cauchy's Theorem (Corollary 9.14). Note that b1 = e implies
b-1 =
b. Let Hbe the cyclic group (a). Since IGI = 2p, the subgroup
Hhas index 2 and is., therefore, normal by Exercise 23 of Section 8.2.
Consequently,
bah= bah-1 EH. Since His cyclic, bah=d for some t.
Using this and the fact that b2= e, we see that
ar= (at) t= (bab)t= (bab)(bab)(bah) ···(bah)= bath= b(bah)b
Hence,
t2 = 1 (modp) by part (2) of Theorem 7 .9.
=a
Consequently,
p divides t1-1=(t- l)(t + 1), which implies thatpi(tby Theorem 1.5. Thus t = 1 (modp) or t = -1 (modp).
1) orpl(t +
1)
If t = 1 (modp), thenbab =at=a by Theorem 7.9. Multiplying
both sides by
b shows that ha= ab. It follows that ab has order 2p=I GI
(Exercise 33 of Section 7 .2). Therefore, G is cyclic and isomorphic to Z2p
by Theorem 7.19.
bah=a-1• Exercise 9 shows that the map
fDp-+ G given by f(ldl) =dbl is a homomorphism. Let K be the
subgroup (b). Since IHI = p (withp odd) and !Kl= 2, H n K = (e) by
If t = -1 (modp), then
Lagrange's Theorem and G = HK by Exercise 15 in Section 9.3. Thus
every element of G can be written in the form dbl, which implies thatf
is surjective. Since D 1 and G have the same order.fmust be injective and,
hence, an isomorphism.
•
Groups of Small Order
We are now in a position to complete the classification of groups of small order that
was begun in Section 8.1, where groups of orders s 7 were classified. We already
know three abelian groups of order 8 (Z2 X Z2 X Z2, � X Z2, and Z8) and one nona­
belian one
(D4). Another nonabelian group
of order 8, the quaternion group Q, was
introduced in Exercise 16 of Section 7.1. It is not isomorphic to
D4 by Exercise 47
of
Section 7.4. These five groups are the only ones:
Theorem 9.34
If G is a group of order 8, then G is isomorphic to one of the following groups:
�. � x Z2, L; x Z2 x Z2, the dihedral group D4, or the quaternion group
..
.......
..
Q.
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:dgbt.,___,_�OOllll!m·a;J'timlo1f..._...._:ligl:UllWtrimml-....k
9.5
The Structure of Finite Groups
317
Proof" If G is abelian, then G is isomorphic to Z8, � x Zz, or Z1 X Z x Z2 by the
2
Fundam ental Theorem of Finite Abelian Groups. So suppose G is a nona­
belian group of order 8. The nonidentity elements of G must have order 2 ,
4, or 8 b y Lagrange's Theorem. However, G cannot contain a n element of
order 8 (because then G would be cyclic and abelian), nor can all the non­
identity elements of Ghave order 2
(see Exercise 27 of Section 7.2). Hence,
Gcontains an element a of order 4. Let b be any element of Gsuch that
b it (a)= {e, a, a2, a3}. Then the eight elements e, a, a2, a3, b, ab, a2b, a3b
all distinct because lal = 4 and d = alb implies b = d-1 E (a), contrary
to the choice of b. Thus G = {e, a, a2, a3, b, ab, a2b, a3b}.
The subgroup (a) has order 4 and index2 in G. Hence, (a) is normal by
Exercise 23 of Section 8.2. Now the element bab-1 has order 4 by Exercise 19
of Section 7.2 and bab-1 E (a) by normality. Therefore, bab-1 is either a or a3
(becausee has order 1 anda2has order2). If bab-1 =a, however, then
ha= ab, which implies that Gis abelian. Therefore, bab-1 = J a-1 so that
ha= a-1b. This fact can be used to construct most of the multiplication table
of G. For instance, (ab)a2 = a(ba)a = a(a-1b)a = ba = a-1b = a3b. You can
are
=
use similar arguments to verify that the table must look like this:
e
a
a2
a3
b
ab
a2b
a3b
e
e
a
al
a3
b
ab
a2b
Jb
a
a
a2
a3
e
ab
a2b
a3b
b
a3b
b
ab
b
ab
a2b
a2
a2
a3
e
a
a2b
d3
b
a3
e
a
al
a3b
b
a3b
ab
ab
ab
b
alb
a1b
a2b
a2b
ab
b
a3b
d3b
Jb
a2b
ab
b
a2b
b2•
Since b
1
a1b implies b
a' E (a}, which is a contradiction, b2 must be one of e, a, a2, or a3• If b2
1
a, however, then ab = b2b = bb
ba, which implies that G is abelian.
Similarly, fl-= a3 implies that G is abelian (Exercise 15). Therefore, b2 =
e or b2 = a2. Each of these possibilities leads to a different table for G.
Completing the table when fl-= e and comparing it to the table for D4 in
In order to complete the table, we must find
=
=
=
=
Example 1 of Section 8.2 shows that G = D4 under the correspondence
a1----+ r,,
b----+d,
ab ----+ h,
a2b----+ t,
a1b ----+ v
(Exercise 4). Similarly, completing the table when b2 = a2 and comparing it
to the table for the quaternion group
Q shows that G = Q (Exercise 5).
•
According to the Fundamental Theorem of Finite Abelian Groups there are two
abelian groups of order 12:
� X Z1 = Z 11 and Z 1 X Zl X Z1. We have
also seen two
nonabelian groups of order 12: the alternating group A4 and the dihedral group D<,. It
can be shown that there is a third nonabelian group T of order 12, which is generated
by elements
a and b such that lal = 6, fl-= a3, and ba a-1b and that no two of
16).
=
these
three nonabelian groups are isomorphic (Exercise
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...
....... it.
318
Chapter 9
Topics in Group Theory
Theorem 9.35
If G is a group of order 12, then G is isomorphic to one of the following groups:
Z12, Z2 x Z2 X Lg, the alternati ng group Ai, the dihedral group 06, or the
group T described in the preceding paragraph.
Proof• An argument similar to the proofofTheorem
93
. 4 can be used to prove
the theorem. See Theorem 11.6 . 4 in Hungerford [5].
•
The preceding results provide a complete classification ofall groups oforders :S 15,
that is, a list ofgroups such that every group oforder :S 15 is isomorphic to exactly one
group on the list.
ORDER
2
3
4
5
6
7
8
9
10
11
12
13
14
15
GROUPS
Z2
Z3
Z.,Z2
Zs
REFERENCE
Theorem8.7
Theorem8.7
x Z2
�,S3
Z1
Z8,� X Z2,Z2 X Z2 X Z2,D 4,
Z9,Z3 x Z3
Z10.Ds
Zn
Zu.Z2 X Z2 X Z3, A", D6, T
Z13
Z14,D1
Z1s
Theorem88
.
Theorem8.7
Theorem8. 9
Theorem8.7
Q
Theorem 9.34
Corollary 9. 29
Theorem 93
. 3
Theorem8.7
Theorem 9.35
Theorem8.7
Theorem 9.33
Corollary 91
. 8
This list could be continued to order 100 and beyond. For more than half of the
orders between 2 and 100, the techniques presented above provide a complete clas­
sification ofgroups of that order (Exercise 6). For other orders, however, a great deal
of additional work would be necessary. For instance, there
are14
different groups of
order16 and 267 oforder 64. There is no known formula giving the number of distinct
groups oforder n.
• Exercises
A. 1. If p and q are primes with p < q and q siE 1 (mod p) and G is a group oforder
2
p q, prove that G is abelian.
2. Prove that there is no simple group of order12. [Hint: Show that one of the
Sylow subgroups must be normal.]
3. Prove that D3 is isomorphic to S6•
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...
� ...... k
9.5
The Structure of Finite Groups
319
4. (a) In the proof of Theorem 9.34, complete the operation table for the group
Gin the case when b2 = e.
(b) Show that G= D4 under the correspondence
a1--+ r1,
b--+ d, ab--+ h, a2b --+ t, a3b--+ v
by comparing the table in part (a) with the table for
D4 in Example 1 of
Section 8.2.
5.
(a) In the proof of Theorem 9.34, complete the operation table for the group
Gin the case when b2 = a2•
(b) Show that G= Q under the correspondence
db'--+ i'J'
(0 s rs 3, 0 s ss 1)
by comparing the table in part (a) with the table for
Q (see Exercise
16 in
Section 7.1).
6. Theorems 8.7, 9.7, 9.30, and 9.33, and Corollaries 9.18 and 9.29 are sufficient
to classify groups of many orders. List all such orders from 16 to 100.
B. 7. If Gis a group such that every one of its Sylow subgroups (for every prime p) is
cyclic and normal, prove that Gis a cyclic group.
8. Let
n � 3 be a positive integer and let Gbe the set of all matrices of the forms
or
withaEZ,,.
(a) Prove that Gis a group of order 2n under matrix multiplication.
(b) Prove that Gis isomorphic to D,,.
bah= a-1, the
--+
Ggiven by j(rdf) "" a'bl is a homomorphism. [Hint: bah = a-1 is
p
equivalent to ba = a-1b. Use this fact and Theorem 9.32 to compute products
9. Complete the proof of Theorem 9.33 by showing that when
mapfD
in GandD,.]
10. Prove that the dihedral group D6 is isomorphic to S3 X Z •
2
11. (a) If n = 2k, show that ,J< is in the center of D,,.
(b) If n is even, show that Z(D,,) = {e, ,I<}.
(c) If n is odd, show that Z(D,,) = {e}.
12. In Theorem
9.32, r is used to denote a rotation. To avoid confusion here, r will
r will denote the 120° rotation in D3• The
D6 can be written in the
form rd!, and the elements of D3 in the form r1dl.
denote the 60° rotation inD6 and
proof of Theorem 9.32 shows that the elements of
(a) Show that the function rp:D6--+ �given by rp(rdf) = r'dl is a surjective
homomorphism, with kernel {r°, r3}.
(b) Prove that D6/Z(D6) is isomorphic to D3• [Hint: Exercise 11.]
13. What is the center of the quaternion group Q?
14. Show that every subgroup of the quaternion group Q is normal.
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320 Chapter 9
Topics In Group Theory
15. If G is a group of order 8 generated by elements
b � (a), and b2 = a3, then
a
and
b such that Jal==
4,
G is abelian. [This fact is used in the proof of
Theorem 9.34, so don't use Theorem 9.34 to prove it.]
16. Let G be the group
(a) Show that rl
(b)
=
S
3
6,
X
Z4 and let a=
((123), 2) and
b=
((12), 1).
b2 = a3, and ha= a-1b.
Verify that the set T =
{e = tf, a1, a2, a3, a4, a5, b, ab, a2h, a3b, a4b, a5b}
consists of 12 distinct elements.
(c)
Show that Tis a nonabelian subgroup of G.
[Hint:
Use part (a) and
Theorem 7.12.)
(d) Show that Tis not isomorphic to D6 or to A4•
17. Let n be a composite positive integer and p a prime that divides n. Assume
that 1 is the only divisor of
n
that is congruent to 1 modulo p. If G is a group
of order n, prove that G is not simple.
18. If G is a simple group that has a subgroup Kof index n, prove that IGI
divides n!.
[Hint: Let Tbe the set of distinct right co sets of Kand consider
the homomorphism cp:G-+ A(1) of Exercise 41 in Section 8.4. Show that cp is
injective and note that A.(1)
=-
S" (Why?).]
C. 19. Classify all groups of order 21 up to isomorphism.
20. Classify all groups of order 66 up to isomorphism.
21. Prove that there is no simple nonabelian group of order less than 60.
[Hint:
Exercise 18 may be helpful.]
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C H A P T E R
10
Arithmetic in Integral Domains
In Chapters 1and4 we saw that the ring Z of integers and the ring F[x]of polynomi­
als over a field F have very similar structures: both have division algorithms, great­
est common divisors, and unique factorization into primes (irreducibles). In this
chapter we find conditions under which these properties carry over to arbitrary
integral domains, with particular emphasis on unique factorization.
Unique factorization turns out to be closely related to the ideals of a domain. On
the one hand, unique factorization is not possible unless the principal ideals of the
domain satisfy certain conditions (Section 10.2). On the other hand, ideals can be
used to restore a kind of unique factorization to some domains that lack it. Indeed,
ideals were originally invented just for this purpose, as we shall see in Section 10.3.
Section 10.4 (The Field of Quotients of an Integral Domain) is independent of
the rest of the chapter and may be read at any point after Chapter 3. Sections 10.2
and 10.3 depend on Chapter 6, but the rest of the chapter may be read after
Chapter4.
The interdependence of the sections of this chapter is shown below. The
dashed arrows indicate that Sections 10.2, 10.3, and 10.5 depend only on the first
part of Section 10.1 (pages 322--324) and that Section 10.5 uses only three results
in Section 10.2, all of which can be read independently of the rest of that section.
�
,,...10.2_ -
--
10.1 <=--
10.4
-- ,,... 1
--
-....
0 .S
/'
A shortened version of Sections 10.1 and 10.2 that contains all the basic informa­
tion may be obtained by omitting the last parts of each of these sections (see the
notes on pages 325 and 337).
321
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322 Chapter 10
Im
Arithmetic in Integral Domains
Euclidean Domains
In early chapters we analyzed the structure of Zand the polynomial ring F[x] by using
divisibility, units, associates, and primes (irreducibles). We begin by defining these con­
cepts in the more general setting of an integral domain.*
Throughout this chapter, R is an integral domain.
Let a, b ER, with a nonzero. We say that a divides b (or a is a factor of b) and write
a I b if b ""' ac for some c ER. Recall that an element u in R is a unit provided that
uv IR for some vER. Thus the units in R are precisely the divisors of IR.
=
EXAMPLE 1
The only units in Z are 1 and -1. If Fis a field, then the units in the polyno­
mial ring
F[x] are the
nonzero constant polynomials (Corollary 4.5).
EXAMPLE 2
The setZ[v'2J
The element
{r+ Mjr,sEZ} is a subring of
1 + v'2 is a unit in Z[v'2] because
=
(1
+
v'2)(-1
+
the real numbers(Exercise 1).
v'2) =
1.
The ring in the preceding example is one of many similar rings that will frequently be
as examples later. If dis a fixed integer, then it is easy to verify that the set Z[W]
{r + sv'd!r,sEZ} is an integral domain that is contained in the complex numbers. If
d 2!: 0, then Z[ Yd] is a subring of the real numbers(Exercise 1) . When d
-1, then the
ring Z[v=TJ is usually denoted Z[1] and is called the ring of Gall'>..">ian integers.
used
=
=
Let
Remark
have u(vb)
=
(uv)b
uER be a unit with inverse v,
lRb
b. Therefore,
=
so that
uv
=
lR.
For any b ER we
=
a unit divides every element of R
ER is an associate of bER provided a bu for some unit u. Now, u
uv = IR, and vis also a unit. Multiplying both sides of a = Im by v
= lmv
blR b. Use these facts to verify that
An element
a
=
has an inverse, say
shows that
av
=
a
=
is an associate of b if and only if b is an associate of
a
and
a nonzero element of R is dMsible by each of its associates.
"The basic definitions apply in any commutative ring with identity.
We
restrict our attention to
integral domains because most of the theorems fail in nondomains.
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10.1
Euclidean D o m a i n s
323
EXAMPLE 3
Every nonzero integer n has exactly two associates in .£;
n
and -n. If Fis a
field, the associates of f(x) E F[ x] are the nonzero constant multiples of f(x).
In the ring
v'2 = (2
Z[v'2], the elements v'2 and 2 v'2 are associates because
v'2)0 + v'2) and 1 + v'2 is a unit by Example 2.
-
-
A nonzero element p ER is said to be irreducible provided thatp is not a unit and
the only divisors of p are its associates and the units of R.
EXAMPLE 4
The irreducible elements in Z are just the prime integers because the only divi­
sors of a prime p are ±p (its associates) and ±1 (the units in Z). The definition
of irreducible given above is identical to the definition of an irreducible polyno­
mial in the integral domain F[x], when Fis a field (see Section 4.3). In Section 10.3
we shall see that1 + i is irreducible in the ring Z[i].
The next theorem is usually the easiest way to prove that an element is irreducible
and is sometimes used as a definition. Theorem 4.12 is the special case when R = F[x].
Theorem 10.1
Letp bea nonzero, nonunitelement inan integral domainR.Then p is irreducible
if and only if
whenever p =rs, then r ors is a unit.
Proof .. Ifp is irreducible and p = rs, then r is a divisor of p. So r must be either
a unit or an associate of p. If
r
is a unit, there is nothing to prove. If r is
an associate of p, say r = pv, thenp =rs=pvs. Cancelingp on the two
ends {Theorem 3. 7) shows that lR = vs. Therefore,
s
is a unit.
To prove the converse, suppose p has the stated property. Let c be any
divisor of p, sayp
=
ed. Then by hypothesis either cordis a unit. If d
is a unit, then so is a1• Multiplying both sides of p
that
c =
=
cdby a1 shows
a1p. Thus in every case c is either a unit or an associate ofp.
Therefore,p is irreducible.
•
Euclidean Domains
The Division Algorithm was a key tool in analyzing the arithmetic of both
F[x].
Z
and
So we now look at domains that have some kind of analogue of the Division
Algorithm. To see how to describe such an analogue, note that the degree of a poly­
nomial in F[x] can be thought of as defining a function from the nonzero polynomials
in
F[x]
to the nonnegative integers. By identifying the key properties of this function
we obtain this
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.
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324
Chapter
Definition
10
Arithmetic in Integral Domains
An
integral
domain R is a Euclidean domain if there is a function 8
from
the nonzero elements of Rto the nonnegative integers with these properties:
(i) If a and b are nonzero elements of R, then B(a) s 6{ab).
(ii}
ff a, b ER and b *OR, then there exist q, r ER such that a= bq + r
<8(b).
and either r =OR or 8(r)
EXAMPLE 5
F[x] is a Euclidean domain with
B(f(x)) = degree of f(x). Property (i) foll ows from
If Fis a field, then the polynomial domain
the function
8 given
by
Theorem 4. 2 because
B(f(x)g(x)) = degf(x)g(x) = de gf(x) + deg g(x)
:2: deg/(x)
B(f(x)),
'=
and property (ii) is just the Division Algorithm (Theorem 4.6).
EXAMPLE 6
7L is a Euclidean domain with the function 8 given by B(a) = lal. Property (i)
holds because labl = lallbl � lal for all nonzero a and b. If a, b ElL, with b > 0,
then by the Division Algorithm (Theorem 1.1) there are integers q and r such that
a = bq + r and 0 s r < b. Either r = 0, or rand b are both positive, in which
case, B(r)
lrl = r < b lhl = B(b). Therefore, property (ii) holds when b > 0.
For the case when b < 0, see Exercise 9.
=
=
EXAMPLE 7
lL[i] = {$+ti I s, tElL} is a
8 (s + ti) s2+ t2• Since s + ti 0 if
and only if both s and tare 0, we see that B(s + ti) ;'.;?: 1 whens + ti # 0. Verify
that for any a= s + ti and b = u +vi in lL[i], 8(ab) = 8(a) B(b) (Exercise 17).
Then when b * 0 we have
We shall prove that the ring of Gaussian integers
Euclidean domain with the function 8 given by
==
=
B(a) = B(a) 1 s 8(a)8(b) = B(ab),
•
so that property (i) holds. If
b #= 0, verify that a/bis a complex number that can
+di, where c, dE Q (Exercise 11) . Since c E Q, it lies
between two consecutive integers; and similarly for d. Hence, there are integers
m and n such that Im - cl s 1/2 and In - dis 1/2. Since a/b = c +di,
be written in the form c
a = b[c +di]= b[(c - m + m) + (d - n + n)i]
b[(m + ni) + ((c - m) + (d - n)1)]
= b [m + ni] + b[(c - m) + (d - n)i]
:=
= bq + r,
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.......
....
10.1
Euclidean
Domains
325
where q = m + niEZ[i] and r = b[(c - m) + (d- n)i]. Since r =a - bq and a,
b , qEZ[i], we see that rEZ[i]. Property (ii) holds because
.S(r) = .S(b).S[(c - m) + (d - n)i] = <l(b)[(c - m)2
s <> (b)[(l/2)2 + (1/2)2] = (1/2). <>(b) < 8( b).
+
(d - n)�
NOTE: The remainder of this section is optional. The development here is
elementary and assumes only the basic facts about rings in Section 3.1. A
more sophisticated approach is presented in Section 10.2, where ideals are
used to develop the key facts about a wider class of domains that includes
Euclidean domains as a special case. Thus this section develops some re­
markably strong results with a minimum of mathematical tools, whereas
Section 10.2 obtains the same results more efficiently in a wider setting.
I� is possible that a given integral domain may be made into a Euclidean domain
in more than one way by defining the function 8 differently (see Exercises 12 and 13) .
Whenever the Euclidean domains in the preceding examples are mentioned, however,
you may assume that the function S is the one defined above.
In Ftx], the units are the polynomials of degree 0 (Corollary 4.5), that is, the poly­
nomials that have the same degree as the identity polynomial lp Furthermore, if k is
a constant (unit in F[x]), then/(x) and kf(x) have the same degree. Analogous facts
hold in any Euclidean domain.
Theorem 10.2
Let R be a Euclidean domain and
conditions are .equivalent:
(1)
u
a nonzero element of R. Then the following
is a unit.
u
(2) 8(u) = S(1R}·
(3) S(c) = 8(uc) for some nonzero cER.
Proof,. (1) � (2) Exercise 15.
(2) � (3) Statement (3) holds with c = lR because S(l� = 8(u ) = S(u 1�.
(3) � (1) According to (ii) in the definition of a Euclidean domain (with c
•
and uc in place of
c
= (uc)q +
r
a
and b), there exist q, r ER such that
and either
r =OR
or
8 (r) < 8(uc).
8(uc), then by part (i) of the definition (with c and lR - uq in
place of a and b) and statement (3),
If S(c ) s
S(c)
s
8(c(1R - uq)) = 8(c - ucq) = 8(r)
<
8(uc) = 8(c),
so that S(c) < 8(c), a contradiction. Hence, we must have r =OR. Thus
c = (uc)q, which implies that l R = uq. Therefore, u is a unit. •
....
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326 Chapter 10
Arithmetic in Integral Domains
In the remainder of this section we shall develop the basic facts about greatest com­
mon divisors, irreducibles, and unique factorization in Euclidean domains. The devel­
opment here parallels the ones given in Chapter 1 for Z and in Chapter 4 for F[x] and
most of the arguments are the same ones used there, with appropriate modifications.
Alternatively, the major results in Sections 1.2-1.3 and 4.2-4.3 may be considered as
special cases of the theorems proved here.
Greatest Common Divisors
The integers are ordered bys and polynomials in F[x] are partially ordered by their
degrees. This made it natural to define greatest common divisors in these domains in
terms of size or degree. The same idea carries over to Euclidean domains, where "size"
is measured by the function
Definition
B.
Let R be a Euclidean domain and a, b ER (not both zero).
common divisor of a and bis an element d such that
{I) d I
a and
{ii) if c I
a
A
greatest
d I b;
and c I b, then a(c) s B{d).
Any two elements of a Euclidean domain R have at least one common divisor,
namely lR. If
divisor
c
c I a, say a = ct, then 8(c) :S B(ct) = B(a). Consequently, every common
b satisfies B(c) :S max {B(a), 8(b)}, which implies that there is a
of a and
common divisor of largest possible B value. In other words, greatest common divisors
always exist.
When gcd's were defined in 71.. and F[x], an extra condition was included in each
case: The gcd of two integers is the positive common divisor of largest absolute value
and the god of two polynomials is the monic common divisor of highest degree. These
extra conditions guarantee that greatest common divisors in Z and F[x] are unique.
In arbitrary Euclidean domains there are no such extra conditions and greatest com­
mon divisors are not unique. Thus the preceding definition is consistent with, but not
identical to, what was done in Z and F[x].
EXAMPLE 8
71.. is a Euclidean domain with B(a) "" lal. Under the preoeding definition, 2 is
the god of 10 and 18 just as before. However, -2 also satisfies this definition
because
-2 divides both 10 and 18 and any common divisor of 10 and 18 has
absolute values l-21. Note that the greatest common divisors 2 and -2 are
associates in Z.
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10.1
Euclidean Domains
327
Theorem 10.3
Let R be a Euclidean domain and a, bER (not both zero) .
{1)
If dis a greatest common divisor of a and b, then every associate of
dis also a greatest common divisor of a and JJ,
(2)
Any two greatest common divisors of a and bare associates.
(3) If dis a greatest common divisor of a and b, then there exit u,
such that d
VER
=au+ bv.
Proof... (1) Exercise 16.
We now find a particular greatest common divisor of a and b that will
then be used to prove statements
S
(2) and (3). Let
= {8(w) I OR :F wER and w=as+ bt for some s, tER}.
Since at least one of a=a1R + bOR and b=aOR+
blR is nonzero by
hypothesis, Sis a nonempty set of nonnegative integers. By the Well­
Ordering Axiom, S contains a smallest element , that is, there are
elements
(A)
d*, u*, v* of R such that d* =au* + bv* and
for every nonzero
w
of the form as + bt (withs, t ER),
8(d*) s 8(w).
We claim that d* is a greatest common divisor of a and b. To prove
this we first show that d*
there
are
elements q,
r
I a. By the definition of Euclidean domain,
= d*q + r and either r =OR or
such that a
8(r) < 8(d*). Note that
r
=a - d*q =a - (au*+ bv*)q
=a - aqu* - bv*q =a(lR - qu*) + b(-v*q).
Thus
r is a linear combination of a and b, and, hence, we cannot have
8(r) < 8(d*) by (A). Therefore, r=OR, so that a= d*q and d* I a. A similar
argument shows that d* I b and, hence, d* is a common divisor of a and b.
Let c be any other common divisor of a and b. Then a = cs and b = ct
for some s, tER and hence
(B)
d* = au* + ht? = (cs)u* + (ct)v* = c(ru* + tv*).
Thus by part (i) of the definition of Euclidean domain 8(c) s
8(c(su*+ tv*)) =8(d*). Therefore, d* is a greatest common divisor of
a
and b. Note that (B) also shows that
(C)
every common divisor
c
of
a
and b divides
d*.
This completes the preliminaries. We now prove the rest of the theorem.
(2) Let d be any greatest common divisor
of a and b. Since d divides
both a and band d* is a greatest common divisor, we must have 8(d) s
8(d*)
by part (ii) of the definition. The same definition with the roles of d and
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328 Chapter 10
Arithmetic in I n tegra l D om ai n s
d* reversed shows that IJ(d*) s B(d). Henre,
know that d I d*, say d*
=dk. Therefore,
IJ(d) =B(d*). By ( C) we
B(d*) =B(dk). Henre, k
B(d)
=
is a unit by Theorem 10.2 and dis an associate of d*. Since every gcd is
an associate of d*, any two of them must be associates of each other by
Exercise 6.
(3)If dis a greatest common divisor of a and b, then as we saw in the
previous paragraph d*
d =d*k-1
=dk, with k a unit. Sinre d* =au* + bv*, we have
=(au * + bv*)k-1 =a(u*k-1) + b(v*k-1).
Hence, d =au+
bv,
with
u =u*k-1and v =t.i*!c.-1•
•
Corollary 10.4
Let R be a Euclidean domain and a, bER {not both zero). Then dis a greatest
common divisor of a and b if and only if d satisfies these conditions:
I a and d I b;
if c I a and c I b, th en c I
(i) d
(ii)
d.
Proof'" If dis a greatest common divisor of a and b, then dsatisfies (i) by defini­
b. Let d * be as in ( *** )in
d*, say d* = ct. Furthermore, d* is
an associate of d by Theorem 10.3 so that d* =dk, with k a unit. Hence,
d =d*k-1 =(ct)k-1 = c(tk-1), so that c I d. Therefore, condition (ii)holds.
tion. Suppose c is a common divisor of a and
the proof of Theorem 10. 3. Then c I
The proof of the converse is E xercise 18.
The Euclidean Algorithm
(Exercise
•
15 of Section 1.2) provides the most efficient
way of calculating the greatest common divisor of two integers. With minor modifica­
tion its proof carries over to Euclidean domains and provides a constructive method
of finding both greatest common divisors and the coefficients needed to write the gcd
of
a
and bas a linear combination of a and b. See Exercise 31.
Unique Factorization
Elements
a and b of a Euclidean domain are said
to
be relatively prime if one of their
greatest common divisors is lR. In any domain the units are the associates of lR. Thus
by Theorem 10.3,
a
and
b
are relatively prime if and only if one of their greatest
common divisors is a unit.
Theorem 10.5
Let R be a Euclidean domain and a, b, c ER.
prime , then
a
I
If a [ be and a and bare relatively
c.
Proof... Copy the proof of Theorem 1.4, using Theorem 10.3 in place of
Theorem 1.2.
•
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10.1
Euclidean Domains
329
Corollary 10.6
Let p be an irreducible element in a Euclidean domain R.
(1) If p
(2) If p
I be, then p I b or p I c.
I a1a2 • Bni then p divides at least one of the a1•
•
•
Proof• (1) Let dbe a greatest common divisor of p and b. Since ddivides p, we
know that dis either an associate of p or a unit. If dis an associate of
p, then p is also a greatest common divisor of p and b by Theorem 10.3;
in particular, p
hence, p I
c
I b. If
dis a unit, then p and b are relatively prime and,
by Theorem 10.5.
(2) Copy the proof of Corollary 1.6, using (1) in place of
Theorem 15
. .
•
Theorem 10.7
Let R be a Euclidean domain. Every nonzero, nonunit element of R is the prod­
uct of irreducible elements,* and this factorization is unique up to associates;
that is, if
P1P2 · · ·Pr
=
Q1Q2 • • • Qs
with each p1 and q1 irreducible, then r = sand, after reordering and relabel­
ing if necessary,
p1
is an associate of q1 for i = 1, 2, . . . , r.
Proof• Let S be the set of
all nonzero nonunit elements of R that are not the
product of irreducibles. We shall show that S is empty, which proves that
every nonzero nonunit element has at least one factorization as a prod­
uct of irreducibles. Suppose, on the contrary, that S is nonempty. Then
the set
{B(s) Is ES} is a nonempty set of nonnegative integers, which
contains a smallest element by the Well-Ordering Axiom. That is, there
exists
a ES such
(•)
that
B(a) s B(s)
for every
SES.
Since a ES, a is not itself irreducible. By the definition of irreducibility,
5(b) s B(bc) by the definition of
B(bc), then b would be a unit by Theorem 10.2,
which is a contradiction. Hence, B(b) < B(bc) = B(a), so thatbftSby(•).A
similar argument shows that c ft S. By the definition of S, both b and c are
a =be with bothband
c nonunits. Now
Euclidean domain. If B(b)
=
the product of irreducibles and, hence, so is a
•we
allow the possibility
of
=
be. This contradicts the fact
a product with just one factor in case the original element is itself
irreducible.
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330 Chapter 10
Arithmetic in Integral Domains
that a ES. Therefore, S is empty, and every nonzero nonunit element of R
is the product of irreducibles. To show that this factorization is unique up
to asrociates, copy the proof of Theorem 4.14, replacing constant by unit
and Corollary4.13 by Corollary 10.6.
•
• Exercises
NOTE:
Unless stated otherwise, R is an integral domain.
A. 1. Show that Z[Vd] is a subring of C. If d � 0, show that Z[Vd] is a subring of R.
2. Let d * ±1 be a square-free integer (that is, dhas no integer divisors of the
form c2 exoept ( ±1)2). Prove that in Z[Vd], r + sv'd = r1 + .r1 v'd if and only
if r =
r1 and s = s1• Give an example to show that this result may be false if d
is not square-free.
3. If the statement is true, prove it; if it is false, give a counterexample:
(a) If a I band
c
Id in R, then ac I bd.
(b) If a I band c Id in R, then (a + c) I(b + d).
4. Prove that
5. If
a
=
c and dare associates in R if and only if c Id and d I c.
be with
a
* 0 and band
6. Denote the statement
"a
c nonunits, show
that a is not an associate of b.
is an associate of b" by a - b. Prove that- is an
equivalence relation; that is, for all
(iii) If r-s ands- t, then r- t.
r,
s,
t ER: (i) r- r. (ii) If r- s, then .r- r.
7. Prove that every associate of an irreducible element is irreducible.
8. If
u
and v are units, prove that u and v are associates.
9. Show that the function 8 in Example 6 has property (ii) in the definition
[Hint: Apply the Division
a as dividend and lbl as divisor. Then modify the result.]
of a Euclidean domain in the case when b < 0.
Algorithm with
10. Is 2x + 2 irreducible in Z[x] ? Why not?
11. If
c
a = .r + ti and b = u. + vi are in Z[i] and b :F 0, show that a/b = c +di, where
=;:�andd= �: �-
12. (a) Show that
(b) Is
Z is a Euclidean domain with the function 8 given by 8(n)
Q a Euclidean domain when 8 is defined by fJ(r)
=
=
n2•
11
13. Let R be a Euclidean domain with function 8 and let k be a positive integer.
(a) Show that R is also a Euclidean domain under the function 8 given by
8(r)
=
fJ(r) + k.
(b) Show that R is also a Euclidean domain under the function f3 given by
{3(r) k8(r).
=
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10.1
Euclidean Domains
14. Let Fbe a field. Prove that Fis a Euclidean domain with the function
by ll(a)
==
0 for each nonzero a E F.
15. LetRbe a Euclidean domain and
331
8 given
uER. Prove that u is a unit if and only if
8(u) = 8(1.R).
16. If dis the greatest common divisor of
a and bin a Euclidean domain, prove
a and b.
that every associate of dis also a greatest common divisor of
17.
(a) If a= s + ti and b= u +viare nonzero elements of Z[i], show that
8(ab) = 8(a)8(b), where 8(r +s{) = r + ;,
(b) If Ris a Euclidean domain, is it true that
a,
8(ab) = 8(a)8(b) for all nonzero
bER?
18. Complete the proof of Corollary 10.4 by showing that an element dsatisfying
conditions (i) and
(ii) is a greatest common divisor of a and b.
r in the definition of a Euclidean domain are
[Hint: In Z[i], let a = -4 + i and b 5 + 3i; consider
-1 + i.J
19. Show that the elements q and
not necessarily unique..
q == -1 and q
=
=
B. 20. If any two nonzero elements of Rare associates, prove thatRis a field.
21. If every nonzero element of Ris either irreducible or a unit, prove thatRis a
field.
22.
(a) Show that
l
+ iis not a unit in Z[i]. [Hint: What is the inverse of 1 + iin C?]
(b) Show that 2 is not irreducible in
Z[i].
23. Let p be a nonzero, nonunit element of Rsuch that whenever p I cd, then p I
c
or p Id. Prove that p is irreducible.
24. If fR�Sis a surjective homomorphism of integral domains, p is irreducible
inR, andf(p) * 05, isf(p) irreducible in Sl
25. LetRbe a Euclidean domain. Prove that
(a) ll(l.R)
(b) If
a
::s;
8(a) for all nonzero a ER.
and bare associates, then
(c) If a I band 8(a)
=
8(a)= 8(b).
8(b), then a and bare associates.
26. Show thatZ[\/=2]is a Euclidean domain with
8(r + sV=l)= r2 + 2s2.
w= (-1 + \1-3)/2 and Z[w]= {r + sw Ir, sEZ}. Prove that Z[w] is
8(r + sw)= (r + sw)(r +sol) = r'- - rs+ s2•
[Hint: Note that w3 1 and al + w + 1 = 0 (Why?).]
27. Let
a Euclidean domain with
=
28. Prove or disprove: LetRbe a Euclidean domain; then
I= {aER I8(a)> 8(1.R)} is an ideal inR.
29. LetR be a Euclidean domain. If the function 8 is a constant function, prove
thatRis a field.
30.
(a) Prove that l - i is irreducible in Z[i]. [Hint: If a I(1 - i ) ,
see Exercises 17( a ) and 25.]
(b) Write
then 1 - i =ab;
2 as a product of irreducibles in Z[i]. [Hint: Try 1 - i as a factor.]
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332 Chapter 10
Arithmetic in Integral Domains
C. 31. State and prove the Euclidean Algorithm for finding the gcd of two elements
of a Euclidean domain.
32. Let R be a Euclidean domain such that
nonzero
8(a + b) s max{!J(a), !J(b)} for all
a, bER. Prove that q and r in the definition of Euclidean domain are
unique.
•
Principal Ideal Domains and Unique Factorization
Domains
A Euclidean domain is, in effect, a domain that has an analogue of the Division
Algorithm. Consequently, all the proofs used for the integers and polynomial rings,
most of which ultimately depended on the Division Algorithm, can be readily carried
over to Euclidean domains. We now consider domains that may not have an analogue
of the Division Algorithm but do have the other important arithmetic properties of
such
Definition
as
Z,
unique factorization and greatest common divisors.
A principal
ideal domain (PID)
is an integral domain in which every ideal
is principal.
The next theorem shows, for example, that
Z, O[x], and Z[i]
are
all principal ideal
domains because all of them are Euclidean domains (see Examples 5-7 of Section 10.1).
Example 8 of Section 6.1 shows that the polynomial ring
Z[x] is not a PID.
Theorem 10.8
Every Euclidean domain Is a principal ideal domain.
Proof• Suppose I is a nonzero ideal in a Euclidean domain R. Then the set
{8 (i) I i El} is a nonempty set of nonnegative integers, which contains a
bEI
smallest element by the Well-Ordering Axiom. That is, there exists
such that
!J(b) :S !J(r)
for every
iEl.
We claim that/is the principal ideal (b) =
{rb I rER}. Since b Eland/
rbE/for every rER; hence, (b) �I. Conversely, suppose cEl.
Then there exist q, rER such that
is an ideal,
c=bq+r
and
or
!J(r) < 8(b).
Since
r = c - bq and both c and b are in 1, we must have rEI. Hence, it is
!J(b) by(•). Consequently, r =OR and c = bq +
r = bqE(h). Thus/�(b) and, hence, J= (b). Therefore, Ris aPID. •
impossible to have !J(r) <
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10.2
Principal Ideal Domains and Unique Factorization Domains
The converse of Theorem
333
10.8 is false: There are principal ideal domains that are
[21]). Thus the class of Euclidean
not Euclidean domains (see Wilson and W illiams
domains is strictly contained in the class of principal ideal domains.
In our development of the integers, polynomial rings, and Euclidean domains we
first considered greatest common divisors and used them to prove unique factoriza­
tion. Although this approach could also be used with principal ideal domains, it is
just as easy to proceed directly to unique factorization.* We begin by developing the
connection between divisibility and principal ideals in any integral domain.
Lemma 10.9
Let a and b
be
elements of an integral domain R. Then
{1) (a) i= (b) if and only if b I a.
(2) (a) = (b) if and only if b I a and a I b.
(3) (a) � (b) if and only if b I a and b is not an associate of a.
Proof• (1) Note first that the principal ideal (b) consists of all multiples of b,
that is, all elements divisible by
aE(b)
b. Hence,
bla.
if and only if
(a)�(b), then a is in the ideal (b), so that b I a. Conversely, if
b J a, then a E (b), which implies that every multiple of a is also in the
ideal (b). Hence, (a)�(b).
Now i f
(2) (a)= (b) if and only if (a)�(b) and (b)�(a). By (1), (a)�(b) and
(b)�(a) if and only if b I a and a I b.
use (1), (2),and Exercise 4 in Section 10.1, which
a I b and b I a if and only if b is an associate of a. •
(3) To prove this,
shows that
To understand the origin of the next definition, it may help to recall the typical
a1 as a product of primes. Find a prime divisor p1 of
a1 and factor: "1 = p1az.. Next find a prime divisor p2 of a2 and factor: a2 = p1fl3, so
that a1 = P1P2a3. Now find a prime divisor P3 of a3 and factor again: a3 = P3"4 and
a1 = p1Pip3a4• Continue in this manner. Since a1 has only a finite number of prime
divisors, we must eventually have some ak prime so that ak = Pk
1 and a1 =
PlPlPk
·Pk l, The only way to continue factoring (with positive factors and with­
out changing the p's) is to use the fact that 1 = 1 1 repeatedly to write a1 as
process for factoring an integer
•
•
•
•
•
a, = PiPlPJ · · · Pk 1 1 1 · · · 1.
•
·
•
Now look at the same procedure from the point of view of ideals. We have az. I a0 a3 I az.,
a4 I a3,
•
•
•
, 11ak,111,
111, and so on. Consequently, by Lemma 10.9 this factorization
process leads to a chain of ideals
(a1) i= (aj i= ("1) i=
•
•
•
i=
(ak) i= (1) !:;: (1) i= (1) i=
•
•
•
"Greatest common divisors are discussed at the end of this section; also see Exercises 20-22.
......
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it.
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334 Chapter 10
Arithmetic in Integral Domains
in which
all the ideals are equal
after some point. This suggests that factorization as
a product of irreducibles is somehow related to chains of principal ideals in which all
the ideals
Definition
are
equal after some point and motivates the following definition.
An integral domain R satisfies the ascending
chain condition (ACC) on
principal ideals provided that whenever {a1) � (a2) � (a3) � ·,then there
•
exists a positive integer n such that (a1)
=
•
(an) for all IO?: n.
Note that in this definition the identical ideals beginning with (a,J may not be the
ideal (1iJ. Nevertheless, the preceding discussion suggests the possibility that Z has the
ACC on principal ideals. This is indeed the case as we now prove.
Lemma 10.10
Every principal ideal domain R satisfies the ascending chain condition on
principal ideals.
Proof"' Jf (a1 ) � (aj !;;;;
•
theoretic union
then
•
•
is an ascending chain of ideals in R, let A be the set­
LJ (a.). We claim that A is
1;,,1
an
ideal. Suppose
a,
bEA;
a E ("J) and bE (ak) for somej, k O?: 1. Eitherj s k or k s j, say j s k.
Then (a)!;;;; (akl, so that a, bE (ak)· Since (aJ is an ideal,
we
know that
a - bE (tlJ,) �A and ra E(ak) !;;;; A for any rER. Therefore, A is an ideal by
Theorem
6.1. Since R is a PIO, A= (c) for some cER. Since A= '"'
LJ' (a,),
we know that
cE (a,J for some n.
Consequently,
(c) !;;; (a,,) and for each
i ';;!:. n
(a,,)!;;;; (aJ !;;;; LJ (a,)
f::!:l
Therefore,
As
we
=
(a,) = (a,,) for each i ';;!:. n.
shall see, Lemma
10.10 is
A=
(c) !;;;; (a,.).
•
the key to showing that every nonzero nonunit
element in a PIO can be factored as a product of irreducibles. The fact that this fac­
torization is essentially unique is a consequence of the next lemma .
Lemma 10.11
Let R be a principal ideal domain.
p
If p is irreducible in Rand p I be, then p I b or
1c.
Proof*"' If p I be, then be is in the ideal (p). If (p) were known to be a prime
ideal, we could conclude that bE(p) or cE (p), that is, thatp I b or p I c.
Since every maximal ideal is prime by Corollary 6.16, we need only show
"For an alternate proof using greatest common divisors in place of Corollary 6.16, see Exercise 23.
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10.2
Principal Ideal Domains and Unique Factorization Domains
that
336
(p) is a maximal ideal. SupposeI is any ideal with (p) !;;;I!;;;
;
; R. Since
R is a PID, I= (d) for some dr=R. Then (p)!::(d) =I implies that
dip.
d
Since p is irreducible, d must be either a unit or an associate of p. If
is a unit, then I=
(d) = R by Exercise
9 of Section 6.1. If dis an
associate of p, say d = pu, thenp I dand, hence, (d) !:: (p). In this case,
(p) !;;;; (d) !:: (p), so that (p) = (d) = I. T herefore, (p) is maximal, and
the proof is complete.
•
Theorem 10.12
Let R be a principal ideal domain. Every nonzero, nonunit element of R is
the product of irreducible elements,* and this factorization is unique up to
associates; that is, if
P1P2 · ··Pr = q,q.,. · · Qs
with each p1 and q1 irreducible, then r =s and, after reordering and relabeling
if necessary,
p, is an associate of q1 for i =
1, 2,
. .
. , r.
Proof• Let a be a nonzero, nonunit element in R. We must show that a has at
least one factorization. Suppose, on the contrary , that a is not a product
of irreducibles. Then a is not itself irreducible. So a =a1b1 for some
nonunits a1 and b1 (otherwise every factorization of a would include a
unit and a would be irreducible by Theorem 10.1). If both a1 and b1 are
products of irreducibles, then so is a. Thus at least one of them, say a1o is
not a product of irreducibles. Since b1 is not a unit, a1 is not an associate
of a (Exercise 5 in Section 10.1). Consequently, (a)� (a1) by part (3) of
Lemma 10.9.
Now repeat the preceding argument with a1 in place of
to a nonzero nonunit
a2
such that (a1)�
a. This leads
(a,;) and a2 is not a product of
irreducibles. Continuing this process indefinitely would lead to a strictly
ascending chain of principal ideals (a1)�
(ai) � (a:J �
•
•
-,
contradict­
ing Lemma 10.10. Therefore, a must have at least one factorization as a
product of irreducibles.
Now we must show that this factorization is unique up to associates.
To do this, adapt the proof of Theorem 4.14 (the case when R = F[xD
to the general situation by replacing the word constant by unit and using
Lemma 10.11 and Exercise 2 in place of Corollary 4.13.
•
To appreciate the importance of Theorem 10.12, it may be beneficial to examine a
domain in which unique factorization fails.
•we allow the possibility of a product with just one factor In case the original element is itself
irreducible.
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336 Chapter 10
Arithmetic in Integral Domains
EXAMPLE 1
Let
CMx]
denote the set of polynomials with rational coefficients and integer
constant terms. For instance, x,
not. Verify that
is irreducible in
ix
=2
x, and 2 are in
Oz[x], but x'- +
k �
and
are
O.z[x] is an integral domain and that the constant polynomial 2
Oz[x] (Exercise 16). The irreducible element 2 is a factor of
xEOz[x] because x
because
i
•
=
2
G}
x
•
(� }
x
Similarly, 2 is an irreducible factor of
Hence, x = 2 2
•
•
(± }
x
k
x
In fact, the process of
factoring out irreducible 2's never ends because
(*)
x = 2·
G)
x
= 2·2·
=
2·2
·
(� )
x
·
·
2
·
= 2·2·2·
(�x)
=
•
G)
x
·
= ··
·
·.
In view of this, it should not be surprising that x cannot be factored as a prod­
uct of irreducibles of
Cl!z[x] (Exercise
17).
C.Ompare this situation with the prime factorization of
a1
in Z
as
described on
page 333. In Z the factorization becomes trivial after a finite number of steps (the
only remaining factors are 1 's), and all the ideals in the corresponding chain are equal
after that point. In the factorization
(*) in Oz[x],
h owever, thingi;
are
different. The
remaining factors each time a 2 is factored from x are the elements
No two of these elements
are
associates (Exercise 3) and each element is 2 times
the following one, that is, each element is divisible by the following one. Therefore,
by part (3) of Lemma 10.9
Hence, the ACC for p rincipal ideals does not hold in
Oz[x].
Unique Factorization Domains
In our study of Euclidean domains and principal ideal d omains, the main result was
that unique factorization held. Now we reverse the process and consider domains in
which unique factorization always holds to see what other p roperties from ordinary
arithmetic they may have.
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10.2
Definition
Principal Ideal Domains and Unique Factorization Domains
An integral domain R is a
337
unique factorization domain (UFO) provided
that every nonzero, nonunit etement of R is the product of irreducible
elements,* and this factorization is unique up to associates; that is, if
P1P2
'
'
·
Pr
=
Q1qz
'
1
,;.qs
with each p1 and q1 irreducible, then r = sand, after reordering and re-label­
ing if necessary,
p, is an assoctate of q1 for i
=
1, 2,
•
.
•
, r.
EXAMPLE 2
Theorem 10.12 shows that every PID is a unique factorization domain. In
particular, the ring .Z[i] of Gaussian integers is a UFD.
EXAMPLE 3
As noted in Example 1, Oz[x] is not a unique factorization domain because the
element x has no factorization as a product of a finite number of irreducibles. In
Section 10.3 we shall see that .Z [ v=-5] fails to be a UFD for a different reason:
Every element is a product of irreducibles, but this factorization is not unique.
EXAMPLE 4
A proof
that the polynomial ring .Z[x] is a UFD is given in Section 10.5. Since
.Z[x] is not a principal ideal domain (see Example 8 of Section
6.1),
we see that
the class of all unique factorization domains is strictly larger than the class of
all principal ideal domains.
NOTE: The remainderof this section is optional and is not needed for the sequel.
When working with two integers, you can always arrange things so that the same
primes appear in the factorizations of both elements. For instance, consider the prime
factorizations -18 = 2 · 3• ( -3) and 40 = 2 · (-2) · ( -2) · 5. The list of all primes that
appear in both factorizations is 2,
3, -3,
2,
-
2,
-
2, 5, but several of these primes are
associates of each other. By eliminating any prime on the list that is an associate of an
earlier number on the list we obtain the list 2, 3, 5 in which no two numbers are associ­
ates. We can write both 18 and 40
-18 = 2.
40
=
2.
3.
(
-3)
=
(
(- 2) • (-2) .
as
products of these three primes and the units ± 1:
-1). 2 . 3. 3
5
=
=
( -1) .
2°. 32•
(-1)(-1). 2 . 2. 2 . 5
=
s<>
(1). 23
•
30 . 51
Essentially the same procedure works in any UFD.
*We allow the possibility of a product with just one factor in case the original element is itself irreducible.
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338 Chapter 10
Arithmetic in Integral Domains
Theorem 10.13
If c and dare nonzero elements in a unique factorization domain R, then
there exist units
u
and
v
and irreducibles p1, p2,
•
•
•
, Pk• no two of which are
associates, such that
where each m1 and n1 is a nonnegative integer. Fur thermore,
cld
if and only if
for each
i= 1, 2,"
•
I
k,
In the example preceding the theorem, with c= -18 and d = 40, we had u= -1, v= 1,
Pi=
2,p2 =
3, andp3= S.
Proof of Theorem 10.13. Since R is a UFD, both c and d can be factored, say
c = q1q2 •
q1 and d= r1r2 • r1 with each q1 and 'J irreducible. In the list
qt> q2, • • • , q4, r., r:z, . . . , r, delete any element that has an associate appear­
•
•
•
•
ing earlier on the list and denote the remaining elements by Pi. p2,
• • • ,
Pk· Then each p, is irreducible, no two of them are associates of each other,
and each one of the q's and r's is an associate of some p1• Consequently, in
the factorization c
=
q1q2
•
• •
q,each q1is of the form wp1 with w a unit.
By rearranging terms, c can be written (product of units) (product of p's).
The product of these units is itself a unit, call it u.
By rearranging thep's
in this product and inserting otherp's with zero exponents if necessary,
we can write c
=
up1m1p2m•
•
•
•
pk''\ with each m1 � 0. A similar procedure
works for d and proves the first part of the theorem.
To prove the first half of the last statement of the theorem, suppose
c I d. Then d=
cb for some
b ER. Since the irreducible p1 appears exactly
n, times in the factorization of d, it must also appear exactly n1 times in the
factorization of cb. But p1 already appears m, times in the factorization of
and may possibly appear in the factorization of
Conversely, suppose that m1 s
a =
Therefore,
c I d.
n1
c
b, so we must have Int s n1•
for every i. Verify that d
(u-1v) (pt''1-""P2"•-nr,.
·
·
= ca,
where
Pkn,.-mo).
•
Corollary 10.14
Every unique factorization domain satisfies the ascending chain condition on
principal ideals.
Proof•First, suppose
(c) and (d) are principal ideals in a UFD R such that
(d) � (c). Then c Id and c is not an associate of dby Lemma 10.9. If c and
d are written in the form given by Theorem 10.13, then each m, s n1• If
1
m1 = n, for every i, then c = uv - d, which means that c is an associate of
d, a contradiction. Hence, there must be some index} for which� < n1
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10.2
Principal Ideal Domains and Unique Factorization Domains
339
Suppose (a1) <;;;:(Oz) <;;;: (a3) <;;;: • • is a chain of principal ideals in R.
Lemma 10.9 shows that each at divides a1• By Theorem 10.13 we
may assume that a1 = vp{"1p2.., ·Pk,,. and that each at is of the form
p/'"', where thep1 are nonassociate irreducibles. If
ai = upr'P1m, •
there are just a finite number of strict inclusions (�) in the chain of
ideals, then there are only equalities after a certain point and the ACC
holds. There cannot be an infinite number of strict inclusions because
the first paragraph shows that each time a strict inclusion occurs, one
of the exponents on one of thep's must decrease. Consequently, after
a finite number of strict inclusions, there would be an a,. of the form
°
o
a,, = up1 · · · = Pk = u. Thus an is a unit, which implies that (a,J = R by
Exercise 9 of Section 6.L For each i <2: n we have (a,,) c;; (aJ c;,; R (a,J, so
that (a,J (ai.). Therefore, R satisfies the ACC on principal ideals. •
•
•
•
•
•
=
=
Irreducibles in a unique factorization domain have a property that we have
used frequently in the special cases of Euclidean domains and principal ideal
domains.
Theorem 10.15
Let p be an Irreducible element in a unique factorization domain R. If Pl be,
then pI b or p I c.
Proof" If bor c is OR, then there is nothing to prove becausep I OR. If e is a unit
and p I be, then pt = be for some t ER and ptc�1 = b. Hence, p I b; simi­
larly, if bis a unit, thenp I c. If both band e are nonzero nonunits, then
b q1 • • • qk and c = qk+I • • • q1 with the q1 (not necessarily distinct)
irreducibles. Sincep I be, we havepr = be q1 • • • q8 for some r ER. The
irreduciblep must be an associate of some q1 by unique factorization .
Therefore,p divides q1 and, hence, divides b or c. •
=
=
We are now in a position to characterize unique factorization domains.
Theorem 10.16
An integral domain Ris a unique factorization domain if and only if
(1) Rhas the ascending chain c ondition on principal ideals; and
(2) whenever pis irreducible in Rand pied, then
pie or pld.
As the proof of the theorem shows, condition (1) corresponds to the existence of
an irreducible factorization for each nonzero nonunit element and condition (2), to
the uniqueness of this factorization. The two conditions are independent: (1) fails and
(2) holds in Oz[x] (see Example 1 and Exercise 33), whereas (1) holds and (2) fails in
Z[ ·\l=S] (as we shall see in Example 4 and Exercise 21 of Section 10.3).
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340 Chapter 10
Arithmetic in Integral Domains
Proof ofTheorem 10.16 �If Ris a UFD, then Rsatisfies (1) and (2) by Corollary
10.14
and Theorem 10.15. Conversely, assume R satisfies (1) and (2) and let
a
be a nonzero nonunit element of R. T he argument used in the proof of
Theorem 10.12, which depends only on the ACC, is valid here and shows
that a can be factored as a product of irreducibles. To show that this
factorization is unique, adapt the proof of Theorem 4.14 (the case when
R = F[xD to the general situation by replacing the word constant by unit
and using (2) and Exercise 2 in place of Corollary 4.13.
•
Greatest Common Divisors
Greatest common divisors were a useful tool in our study of Z,F[ x], and other Euclidean
domains. In each case the gcd of two elements was defined to be a common divisor of
"largest size," where size
was
measured by absolute value in Z, by poly nomial degree
in F[x], and by the function 8 in an arbitrary Euclidean domain. Unfortunately, there
may be no similar way to measure "size" in an arbitrary integral domain, so greatest
common divisors must be defined in terms of divisibility properties alone:
Definition
Let a1, a2,
•
•
•
, an be elements (not all zero) of an integral domain R. A
greatest common divisor ofa1, a2
•
•
,
•
Bn Is an element d ofR such that
(i) d divides each ofthe a1;
{ii} if c ER and cdivides each ofthe a1, then
cJd.
Corollaries 1.3, 49
. , and 10.4 show that this definition is equivalent to the definitions
used previously in Z, Ftx], and other Euclidean domains. The only difference is that great­
est common divisors in Z andF[x], are no longer unique (see the discussion on page 326).
Theorem 10.17
Let d be a greatest common divisor ofa1, a2,
•
•
•
, an in an integral domain R.
Then
{1} Every associate of d is also a gcd ofa1,
•
(2) Any two greatest common divisors ofa1,
•
, an.
•
•
•
•
,
an
are associates.
Proof �c1) Exercise 7.
(2) Suppose both d and t are gcd's of
a1,
ai.
. • .
, a,.. Then I divides each
and, therefore, t I d by (ii) in the defini tion of the greatest common
divisor d. But d also divides each
ar,
and, hence, d I t by
(ii) in the defini­
tion of the gcd t. Since t I d and d It, we know that d and t are associates
by Exercise 4 of Section 10.1.
•
WARNING: In some integral domains a finite set of elements may not
have a greatest common divisor (see Exercise
13 in Section 10.3).
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10.2
Principal Ideal Domains and Unique Factorization Domains
341
Theorem 10.18
Leta1, a2,
Then a1,
•
•
, an (not all zero) be elements in a unique factorization domain R.
,
•
•
•
, an have a greatest common divisor in
Proof• The gcd of any set of elements is the g c d of
set,
so
R.
the nonzero members of the
we may assume that each t1i is nonzero. By Theorem 10.13 there are
irreducibles P1>
• •
•
, p1 (no two of which
an d nonnegative integers
are
associates), units
u1,
• • •
,
u,,,
mg such that
Let k1 be the smallest exponent that appears on p1; that is,
minimum of
m11,
mm "'31>
• • •
,
k1 is the
m,,1• Similarly, let k1 be the smallest
exponent that appears on p2, and so on. Use Theorem 10.13 to verify
that d =
pltp1"2
• • •
p, k, i s a gcd of
ah
. • •
, a,,.
•
In an arbitrary unique factorization domain, it may not be possible to write the
gcd of elements a and bas a linear combination of
Section 10.5, for example,
a
and bas it was in Zand F[x]. In
x and2 in
1 is not a linear combination of x and2 in Z[x] (Exercise 6). In a
principal ideal domain, however, the gcd of a and h can always be written as a linear
combination of a and h (Exercise 20).
we
shall see that 1 is a gcd of the polynomials
the UFD Z[x], but
• Exercises
A. 1. If a, b
are
that (ab)
2.
nonzero elements of
� (h).
an integral
Suppose p is an irreducible element in an
p
I be, thenp I b or p I c. If p I a1�
3. (a) Prove that the only units
(b)
• •
domain and a is a nonunit, prove
integral domain R such that whenever
• a,,, prove that p divides at least one flt·
in Clz[x] are 1 and
-1.
[Hint: Theorem 4.2.]
If f(x) EClz[x], show that its only associates are/(x) and
-f(x).
4. Is a field a UFD?
5.
Give an example to show
need not be a UFD.
6. Prove that
1
that a
subdomain of a unique
factorization domain
is not a linear combination of the polynomials2 and x in Z[x], that
is, prove it is impossible to find/(x), g(x) EZ[x] such
that2/(x) + xg(x)
=
1.
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342 Chapter 10
Arithmetic in Integral Domains
7. Let d be a gcd of
a1o
of dis also a gcd of
8.
• . . ,
ah
ak
• • •
in an integral domain. Prove that every associate
, ak.
Letp be an irreducible element in an integral domain. Prove that lRis a gcd of
p and a i f and only ifp ,r a.
B. 9.
IO.
Let Rbe a PID. If (c)is a nonzero ideal in R,then show that there areonly
fn
i itelymany ideals in Rthat contain (c). [Hint: Consider the divisors of c.]
Prove that an ideal (p)in a PID is maximal
if and
only
ifpis irreducible.
11. Prove that every ideal in a principalideal domain R (exoept Ritself ) is
contained in a maximal ideal. [Hint: Exercise 10.]
12. Prove that an ideal in a PID is prime if and only ifit is maximal.
[Hint: Exercise 10.]
13. Let f.R � S be a surjective homomorphism of ringswith identity.
(a)
If Ris a
PID, prove that every ideal in Sis principal.
(b) Show by example that Sneed not be an integral domain.
14. Letp be a fixed prime integer and let R bethe set of all rational numbers that
can bewritten in the form a/b with
bnot divisible
(a) Ris an integral domain containing Z. [Note n
(b)
If a/b ERandp ..r
(c)
If /is
a,
by p. Prove that
=
n/1].
then a/bis a unit in R.
a nonzero ideal in Rand I:# R., then Icontainspt for some t > 0.
(d) Ris a PID. (If Iis an ideal, show that I= (If), where ifis the smallest
power of p in /.)
15. Let /be a nonzero ideal in Z[i]. Show that the quotient ring Z[i]/Iis finite.
16. (a) Ifpis prime in Z,prove that the constant polynomialpis irreducible in
CMx]. [Hint: Theorem 4.2 and Exercise 3.]
(b)
Ifp and q are positive primes in Z with p
associates in Oz[x].
if:. q, prove
thatp and
q are
not
17. (a) Show thatthe only divisors of x in Oz[x]are the integers (constant poly­
nomials) and first-degree poly nomials of the form .!_ x with 0
n
if:.
n
EZ.
(b) For each nonzero n EZ, show that the polynomial lx isnot irreducible
n
in Oz[x]. [Hint: Theorem 10.1.]
(c) Show that x cannot be written as a finite product of irreducible
in Oz[x].
elements
18. A ring
R is said to satisfy the ascending chain condition (ACC) on idealsif
whenever 11 !;;;; 12 � 13 �
is a chain of ideals in R (not necessarily principal
ideals), then there is an integer nsuch that 4 =I,, for allj � n. Prove that if
every ideal in a commutative ring Ris finitely generated,then R satisfei s the
ACC. [Hint: See Theorem 6.3 and adapt the proof of L emma 10.10.]
• • •
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10.2
Principal Ideal Domains and Unique Factorization Domains
343
19. A ring R is said to satisfy the descending chain condition (DCC) on ideals if
whenever 11 212 213 2
is a chain of ideals in R, then there is an integer n
such that � = In for allj 2!:: n.
·
·
·
(a) Show that Z does not satisfy the IX:C.
(b) Show that an integral domain R is a field if and only if R satisfies the
DCC. [Hint: If 0 '#a ER is not a unit, what can be said about the chain
of ideals (a) 2 (a") 2 (a3) 2
?]
·
20.
21.
·
·
Let R be a PID and a, b ER, not both zero. Prove that a, b have a greatest
common divisor that can be written as a linear combination of a and b.
[Hint: Let /be the ideal generated by a and b (see Theorem 6.3); then I= (d)
for some d ER. Show that dis a gcd of a and b.]
Let R be a PID and S an integral domain that contains R. Let a, b, d ER.
a gcd of a and b in R, prove that dis a gcd of a and b in S.
[Hint: See Exercise 20.]
If dis
22.
Extend Exercise 20 to any finite number of elements.
23.
Give an alternative proof of Lemma 10.1 l as follows. If p I b, there is nothing to
prove. Ifp k b, then lR is a gcd of p and b by Exercise 8. Now show that p I c by
copying the proof of Theorem 1.4 with pin place of a and Exercise 20 in place
of Theorem 1.2.
24.
Let R be an integral domain. Prove that R is a PID if and only if (i) every
ideal of R is finitely generated (Theorem 6.3) and (ii) whenever a, b ER, the
sum ideal (a)+ (b) is principal. [Sum is defined in Exercise 20 of Section 6.l.]
25.
Let R be an integral domain in which any two elements (not both OR) have
a gcd. Let (r,s) denote any gcd of r ands. Use - to denote associates as in
Exercise 6 of Section 10.1. Prove that for all r, s, t ER:
(a) Ifs - t, then rs - rt.
(b) Ifs- t, then (r, s)-(r, t).
(c) r(s, t)- (rs, rt).
(d) (r, (s, t)) -((r, s), t). [Hint: Show that both are gcd's of r, s, t.]
26.
Let R be an integral domain in which any two elements (not both 00 have a
gcd. With the notation of Exercise 25, prove that if (b, c) - � and (b, d)- la.
then (b, cd)-1R. [Hin t: By Exercise 25(a) and (c), d- (bd, cd) , so that
lR -(b, d)-(b, (bd, cd)). Apply parts (d), (c), and (a) of Exercise 25 to show
that (b, (bd, cd)) -(b, cd).
27.
Let R be an integral domain in which any two elements (not both zero) have a
gcd. Let pbe an irreducible element of R. Prove that whenever pI cd, then pI c
or p I d. [Hint: Exercises 8 and 26.]
28.
If R is a UFD, if a, b, and c are elements such that a I c and b I c, and if IR is a
gcd of a and b, prove that ab I c.
29.
Let R be a UFD. If a I be and if lR is a gcd of a and b, prove that a I c.
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344 Chapter 10
Arithmetic in Integral Domains
30. A
least common multiple (lcm) of the nonzero elements
element
then b I
ah
, ak is an
each a1 divides band(ii) if each a, divides an element c,
Prove that any finite set of nonzero elements in a UFD has a least
.
•
.
bsuch that(i)
c.
common multiple.
a and bi n R have a least common multiple if and
only if the intersection of the principal ideals (a)and (b)is also a principal ideal.
31. Prove that nonzero elements
rWJ is finitely generated(Theorem 6.3)
C, 32. Prove that every ideal IinZ
follows. Let/0 =/()Zand let 11 =
as
{bEZ I a+ bVdElforsome aEZ}.
(a) Prove that /0 and /1 are ideals inZ. Therefore,/0 =(ro) and/1 =(r1 ) for
some r1EZ.
(b) Prove that /0 r;;;. /1•
( c) By the definition of 11 there exists a1 EZsuch that a1 + r1
that !is the ideal generated byr0 and
a1 + r ['Vd.
thens El1 so thats= r1s1• Show that(r +
this to write r +
Ydis in I. Prove
+s W El,
[Hint: If r
sv'd) -s1(
a1
sVdas a linear combination of r0 and
v'd) E/0; use
+ r1
a
1
+ r1Yd.]
33. Prove that p(x) is irreducible in Clz[x] if and only if p(x) is either a prime
integer or an irreducible polynomial in Q[x] with constant term ± 1.
Conclude that every irreducible p(x) in Ozx
[ ] has the property that
whenever p(x) I c ( x)d(x), thenp(x) I c(x) or p(x) Id (x).
34. Show that every nonzero/(x) in Ozx
[ ] can be written in the form
cX'p1(x)
pk(x), with c E O!, n 2!:: 0, and eachp1(x) nonconstant irreducible
in Oz[x] and that this factorization is unique in the following sense: If f(x) =
d:X"q1(x)
q.(x) with dE Q, m Oi?: 0, and eachq1(x) nonconstant irreducible
in Oz[x], then c = ±d, m = n, k = t, and, after relabeling if necessary, each
p1(x) = ±q,(x).
·
·
•
•
·
·
35. Prove that any two nonzero polynomials in Clz[x] have a god.
36.
(a) Prove that/(x) is irreducible inZx
[ ] if and only if f(x) is either a prime
integer or an irreducible polynomial in Qx
[ ] such that the gcd inZof the
coefficients of f(x) i s 1.
(b) Prove thatZ[x] is a UFD. [Hint: See Theorems 4.14 and 4.23.)
11111
Factorization of Quadratic Integers*
In this section we take a closer look at the domainsZ[W]. Because unique factoriza­
tion frequently fails in these domains, they provide a simplified model of the kinds of
difficulties that played a crucial role in the historical origin of the concept of an ideal.
These domains also illustrate how ideals can be used to "restore" unique factorization
in some domains that lack it. We begin with a brief sketch of the relevant history.
*The prerequisites for this section are pages 322-324 of Section
10.1
and the definition
of
unique
factorization domain (page 337).
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10.3
Factorization of Quadratic Integers
345
Early in the last century, Gauss proved the "Law of Biquadratic Reciprocity,"
which provides a fast way of determining whether or not a congruence of the form
x4 = c (mod n) has a
solution. Although the statement of this theorem involves only
integers, Gauss's proof
was
set in the larger domain Z[i]. He proved and used the fact
that Z[i] is a unique factorization domain.
Since Gauss's proof involved Z[i) and i is a complex fourth root of 1, the German
mathematician E. Kummer thought that analogous theorems for congruences of
degree p might involve unique factorization in the domain.
Z[w]
where
=
{Do+ a1w + a,.w2 +
cos(27r/p) + i
w =
sin
(27r/p)
·
·
·
+ tlp-iw'-11 a1EZ},
is a complex pth root of 1. He
was
develop higher-order reciprocity theorems because he discovered that
unable to
Z[w]
may not
be a UFD.*
Later in the century questions about unique factorization arose in connection
with the following problem. It is easy to find many nonzero integer solutions of the
equation x2 +
y2
=
$1, such as 3, 4, 5, or 5, 12 , 13. But no one has ever
y3 z3 or x4 + y4 z'<, which suggests that
integer solutions for x'3 +
x11 + y11
=
=
z" has no
found nonzero
=
nonzero integer solutions when n > 2.
This statement is known as Fermat's Last Theorem because in the late 1630s Fermat
wrote it in the margin of his copy of Diophantus'
Arithmetica
and added "I have
discovered a truly remarkable proof, but the margin is too small to contain it." Fermat's
"proof" has never been found. Most mathematicians today doubt that he actually had
a valid one.
In 1847 the French mathematician G. Lame thought he had found a proof of
Fermat's Last Theorem in the case when n is prime.t His proof used the fact that for
any odd positive prime p,
x' + yP
x' + yP can be factored in the domain Z[w] described above:
=
(x + y)(x
+
wy)(x + w2y)
· ·
·
(x + wP-1y).
Lame's purported proof depended on the assumption that
Z[w] is a unique factoriza­
tion domain. When he became aware of Kummer's work, he realized that his proof
could not be carried through.
Kummer had already found a way to avoid the difficulty. He invented what he
called "ideal numbers" and proved that unique factorization
does hold for these ideal
numbers. This work eventually led to a proof that Fermat's Theorem is true for a large
class of primes, including almost all the primes less than 100. This was a remark­
able breakthrough and deeply influenced later work on the problem.• But it had even
greater significance in the development of modern algebra. For Kummer's "ideal num­
bers" were what we now call ideals.
We shall return to ideals at the end of the section. Now we consider factorization
in the domains
•The domain
Z[W]. These domains are similar
Z[w] is a
t1t the theorem
to the ones that Kummer used and
UFO for every primep less than 23and fails to be a UFO for every larger prime.
is true for prime exponents, then it is true for all exponents; see Exercise
•Fermat's Last Theorem was finally proved in
1994
1.
by Andrew Wiles. His proof uses results and
techniques not available until relatively recently.
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346 Chapter 10
Arithmetic in Integral Domains
illustrate in simplified form the problems he faced and his method of solution. We
shall assume that the integer dis square-free, meaning that d :/: 1 and d has no integer
factors of the form c2 except (±1)2• The following function is the key to factorization
inZv'l
[
.lj
Definition
The function
N: Z[ W]-+Zgiven by
N(s + tVd} = (s + tV<i)(s-iVd) = s2-df­
is called the norm.
For example,
inZ[v'3J,
N(5 + 2v'3) = 52 - 3 22 = 13
·
and
N(2 - 4v'3) = l2 - 3(-4)2 =
-
44 .
Note that
when d < 0, the norm of every element is nonnegathe.
For instance, inZv'=S]
[
,
N(s + tv'=S) = s2 - (-5)f = s2 + sf � o.
In Example 7 of Section 10.1, we saw that the norm makes Z[i]
Euclidean domain. This is not true in general , but we do have
=
Z\/=I
[
] into a
Theorem 10.19
If dis a square-free integer, then for all
(1) N(a) =
O if and only if a
a, b
E
Z[W]
= 0.
{2} N(ab) = N(a)N(b).
Proof.. (1) If a = s + tv'd, then N(a) = s2 dt2 so that N(a) = o if and only if
-
s2 = df. If d = -1, then ; = -r can occur in Zif and only ifs = 0 = t,
that is, if and only if a = 0. So supposed -1. Every prime in the
factorization of rand t2 must occur an even number of times. But the
prime factors of ddo not repeat becausedis square-free. So if p is a prime
factor of d, it must occur an odd number of times in the factorization
of df By unique factorization inZ, the equation s2 = df is impossible
unless s = 0 = t, that is, unless a = 0.
.
(2) Let a = , + sva and b = m + nva. The proof is a straightfor­
ward computation (Exercise 3). •
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10.3
Factorization of Quadratic Integers
347
Theorem 10.20
Letd be a square-free integer. Then u EZ[Vd] is a unit if and only if N(u) = ±1.
Proof,.. If u is a unit, then uv = 1 for some vEZ[Vd]. By Theorem 10.19,
N(u}N(v) = N(uv) = N(l) = 12 - d 02 = 1. Since N(u) and N(v)
are integers, the only possibilities are N(u) = ± 1 and N(v) = ± 1.
Conversely, if u = s + t'\/d and N(u) = ±1, let ii= s- t'\/d eZ['\/d].
Then by the definition of the norm, uu N ( u)
± 1. Hence,
u(±:.U) 1 and u is a unit. •
•
=
=
=
EXAMPLE 1
In Z[v'2] the element 3 + 2v'2 is a unit because N (3 + 2Vl)
2 22
1. Verify that the inverse of 3 + 2Vi is 3 - 2v'2. Every
power of a unit is also a unit, so Z[v'2] has infinitely many units, including
(3 + 2v'2), (3 + 2V'2)2, (3 + 2v'2)3,
=
.32· -
•
=
•
•
•
According to Theorem 10.20 we can determine every units+ tv'd in Z[v'd] by
finding all the integer solutions (for sand t) of the equations SJ. - dt2 = ±1. When
d > 1, these equations have infinitely many solutions (see the preceding example and
Burton [12D. When d = -1, the equations reduce to ;. + t2 = 1.* The only integer
solutions ares= ±1, t = 0, ands= 0, t = ±L So the only units in Z[i] = Z[v'=I] are
±land ±i. If d< - 1, say d = -kwithk> 1, then the equations reduce to?+ kr = l*
.
Since k > 1, the only integer solutions ares= ±1, t = 0. Thus we have
Corollary 10.21
Let d be a square-free integer. If d > 1, then Z[W] has infinitely many units.
The units in Z[v=:f] are ±1 and ±i. Ifd < -1, then the units in Z[W] are ±1.
Corollary 10.22
Let d be a square-free integer. If pEZ[Vd] and M,p) is a prime integer In Z,
then p is irreducible in Z[ Vd].
Proof.. Since N(p) is prime, N(p)
* ±1, sop is not a unit in Z['\/d] by
Theorem 10.20. Ifp =ah in Z['\/d], then by Theorem 10.19, N(p) =
N(a)N(b) in Z. Since N(a), N(b), N(p) are integers and N(p) is prime,
we must have N(a) = ±1 or N(b) = ±1. So a or b is a unit by Theorem
10.20. Therefore,p is irreducible by Theorem 10.1. •
"Since the left side
of the equation is always
.....
nonnegative,
�1 cannot be on
the right side.
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it.
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348 Chapter 10
Arithmetic in Integral Domains
EXAMPLE 2
The element 1
- i is irreducible in Z[i] because N(l - v=l) = 2. Similarly, 1 + i
is also irreducible. Therefore, a factorization of 2
as
Z[i] is given by 2 = (1 + i)(l - f).
10.22
The converse of Corollary
1 + v=5
a product of irreducibles in
is false. For instance, in
Z[v=5]
the norm of
is 6 , which is not prime in Z. But the next example shows that
1
irreducible in Z[v=5}.
+
v=5 is
EXAMPLE 3
1 + '\'-5 is irreducible in Z[v'=S], suppose 1 + '\'-5 = ah. By
10.1 we need only show that a orb is a unit. By Theorem 10.19,
To show that
Theorem
N(a)N(b) = N(ah) = N(1
+
Y-5)
=
6. Since N(a) and N(b)
integers, the only possibilities are N(a) =
1, 2, 3, or
6. If
are
nonnegative
a= s + tv=5
and
N(a) = 2, then s2 + sf= 2. It is easy to see that this equation has no integer
solutions for s and t; so N(a) = 2 is impossible. A similar argument shows
that N(a) = 3 is impossible. If N(a) = 1, then a is a unit byTheorem 10.20. If
N(a) = 6, then N(b) = 1 and bis a unit.Therefore, 1 + '\'-5 is irreducible.
We have seen an example of an integral domain in which a nonzero, nonunit element
could not be factored as a product of irreducibles (Exercise 17 in Section
10.2). We shall
now see that Z[W] may fail to be a UFO for a different reason: Although factorization
as a product of irreducibles is always possible in Z[Vd], i t may not be unique.
Theorem 10.23
Let d be a square-free integer. Then every nonzero, nonunit element in Z[Vd]
is a product of irreducible elements.*
Proof" Let S be the set of all nonzero, nonunits in Z[Vd] that are not the product
of irreducibles. We must show that Sis empty. So suppose, on the con­
trary, that Sis nonempty. Then the set W = {IN(t) 11 t ES} is a nonempty
set of positive integers. By the Well-Ordering Axiom, W contains a small­
est integer. Thus there is an element aESsuch thatjN(a)
Is I N(t) jfor
every tE S. Since aESwe know that a is not itself irreducible. So there
exist nonunits b, cE Z[Vd] such that a
= be. At least one of b, c must
be in S(otherwise a would be a product of irreducibles and, hence, not
b ES. Since band care nonunits, JN(b) I> 1 and IN(c) I> 1 by
ButlN(a) I= IN(h) llN(c)lbyTheorem 10.19, so we must
have 1 < IN(b) I <I N(a) � But bES, so I N(a) Is IN(b) I by the choice of a.
in S), say
Theorem 10.20.
This is a contraction.Therefore, Sis empty, and the theorem is proved.
•
"As usual, we al low a "product" with just one factor.
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k
Factorization of Quadratic Integers
10.3
349
EXAMPLE 4
The domain Z[v'=S] is not a unique factorization domain. The element 6 in
Z[v=-5] has two factorizations:
6
2 3
and
6
(1 + VCS)(l - v'=S).
=
·
=
The proof that 1 +
v=5 is irreducible was given in Example 3. The proofs that 2, 3,
and 1 - v=5 are irreducible are similar. R>r instance, if 2
ab, then N(a)N(b)
N(ab) M,2) 4 so that N(a) 1, 2, or 4. But N(a) 2 is impossible because the
equation s2 + 5t2
2 has no integer solutions. So either N{a) 1 and a is a unit ,
=
=
=
=
=
or
N(a)
=
Theorem
=
=
=
4. In the latter case N{b)
=
1 and bis a unit Therefore, 2.is irreducible by
10.1. Since the only units inZ[v'=S] are ±1, it is clear that neither 2 nor 3
v=5 or1 - v=-5. Thus the factorization of 6 as a product of
is an associate of I +
irreducibles is not unique up to associates and .l(v'=S] is not a UFO.
(1
The preceding example demonstrates that the ir reducible
2
+ v=-5)(1
+ v'=5 or 1 - Y-5.
-
V=s) in Z[v'=S] but does not divide either 1
divides the product
So when unique factorization fails,
an
that when p I cd, then p
Another consequence of the failure of unique fac­
I c or p I d. *
irreducible element p may not have the property
torization is the possible absence of greatest common divisors (Exercise 13).
Unique Factorization of Ideals
We are now in the position that Kummer was in a century and a half ago and the
question is: How can some kind of unique factorization be
restored
in domains such
as Z[v'=S]? Kummer's answer was to change the focus from elements to ideals.t The
product JJ of ideals I and J is defined to be the set of all sums of elements of the form
ab, with a El and bEJ; that is,
IJ
=
{a1bi + azh2 + • · · + a,,b,. In 2: 1,
akEI, bkEJ}.
Exercise 36 in Section 6.1 shows that JJ is an ideal. Instead of factoring an element a
as a product of irreducibles, Kummer factored the principal ideal
(a) as a product of
prime ideals.
EXAMPLE 5
We shall express the principal ideal (6) inZ[v'=5] as a product of prime ideals.
The irreducible factorization of elements 6
=
2
·
3 seems a natural place to start,
(2)(3) (Exercise 16).
(2) is not a prime ideal (for instance, the product (1 + V=5) (1 - v'=-5) 6
is in (2) but neither of the factors is in (2)). So we must look elsewhere. Let P be
the ideal in Z[v'=5] generated by 2 and 1 + v'=S, that is,
and it is easy to prove that the ideal (6) is the product ideal
But
=
P
=
{2a
+
(1
+
V-5)b I a, bEZ[\f-5]}.
*This is not particularly surprising in view ofTheorem
10.16.
tKummer used different terminology, but the ideas here are essentially his. We use the modern
terminology
of ideals that was introduced
by
R.
Dedekind, who generalized Kummer's theory.
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350 Chapter 10
Arithmetic in Integral Domains
Then Pis an ideal by Theorem
only if
r
6.3. Exercise 17
shows that r +
sVCSE
Pif and
and s are both even or both odd. This implies that the only distinct
cosets in Z[\/=5]/ Pare 0
+ Pand 1 + P, as we now see: If m+ n'\f-5
and n even, then (m + nv'=S) - 1 =(m - 1) + n'\f-5 E P because
m - 1 and n are even. Hence, (m+ nv'=S} + P = 1 + P. Similarly, if mis
even andn is odd, then (m - 1) + n "\/=SE Pbecause m - 1 and n are odd. It
follows that the quotient ring Z['\1'=5]/Pis isomorphic to Z 2 • Therefore, Pis
a prime ideal in Z[\/=5] by Theorem 6.14. A similar argument (Exercise 19)
shows that Q1 and Q2 are prime ideals, where
has m odd
Q1 =
Q2 =
v=s)h I a, hEZ['\f-5)},
{3a + (1 - v'=S)h I a, hEZ["\/=S]}.
{3a + (1
+
2
18 and 19 show that the product ideal P =PPis precisely
ideal (2) and that Q1Q2 =(3). Therefore, the ideal (6) is a product of
2
prime ideals: (6) =(2)(3) = PQQ
1 2•
Exercises
the
four
Kummer went on to show that in the domains he was considering, the factorization
of an ideal as a product of prime ideals is unique except for the order of the factors.
This result was later generalized by R. Dedekind. In order to state this generalization
precisely,
we need to fill in some background.
An algebraic number is a complex number that is the root of some monic polyno­
mial with rational coefficients. If t is an algebraic number and
t is the root of
a poly­
nomial degree n in O[x], then
O(t)
=
{"o + a1t + a1/2 +
is a subfield of C and every element in
·
O(t)
·
·
+
a,.._1f'-1 I a,EO}
is an algebraic number.* An algebraic
integer is a complex number that is the root of some monic polynomial with
integer
coefficients. It can be shown that the set of all algebraic integers in O(t) is an integral
domain. If
w is
a complex root of
xP -
1, then the domain
is in fact the domain of all algebraic integers in
Q(w)
(see
Z[w] that Kummer used
Ireland and Rosen [13;
page l99D. So Kummer's results are a special case of
Theorem 10.24
Lett be an algebraic number and R the domain of all algebraic integers in
Q(t). Then every ideal in R (except O and R) is the product of prime ideals
and this factorization is unique up to the order of the factors.
For a proof
see Ireland and Rosen [13; page 174].
Most of the rings Z[ W] are also special cases of Theorem 10.24. For if dis a square­
t = Yd is an algebraic number (because it is a root of x1 - d) and
+ a1W I a1E0}. The algebraic integers in the field Q(W) are called
free integer, then
O(W)
={ao
"For a proof see Theorems 11.7 and 11.9.
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10.3
quadratic integers. Every element
r
+
Factorization of Quadratic Integers
sv'd of Z[Vd] is
a quadratic integer in
351
Q(Vd)
because it is a root of this monic polynomial in Z[x]:
i1-
-
2rx + (r2
-
d?) = (x
-
(r
+
sVti))(x - (r
-
sVd)).
When d = 2 or 3 (mod 4), then Z[Vd] is the domain R of all quadratic integers
Q(Vd), but when d = 1 (mod 4), there are quadratic integers in R that are not
Z[Yd] (see Exercise 22). *
in
in
Theorem 10.24 has proved very useful in algebraic number theory. But it does not
answer many questions about unique factorization of elements, such as: If R is the
domain of all quadratic integers in C(W), for what values of dis Ra UFD? When
d < 0, Ris a UFD if and only if d =
-1, -2, -3, -7, -11, -19, -43, -67, or -163
[19]). Whend> 0, Ris known to be a UFD ford== 2, 3, 5, 6, 7, 11, 13, 17,
19, 21, 22, 23, 29, and many other values. But there is no complete list as there is when
(see Stark
dis negative. It is conjectured that Ris a UFO for infinitely many values ofd.
• Exercises
A. 1. If i' + I = :J< has no nonzero integer solutions and k In, then show that
r' + y"' = :z!' has no nonzero integer solutions.
2. Let
w
be a complex number such that oi'
Z[w] = {Do+ a1w + apJ2 +
is an integral domain.
3. If
a
=
r
+ s'\/d and b
[Hint: ui' =
= m
·
·
=
·
1. Show that
+ a1_1�1 Ia,EZ}
1 implies J+1
+ n'\/d in Z[v'd],
= w,
J+2
= <1i, etc.]
show that N(ab)
=
N(a)N(b).
4. Explain why Z[v'=SJ is not a Euclidean domain for any function a.
5. If a E 0 is an algebraic integer,
[Hint: Theorem 4.21.]
as defined on page 350, show that a E Z.
B. 6. In which of these domains is 5 an irreducible element?
(a)
Z
(c) Z[v=-2]
(b) Z[i]
7. In Z[Y-7), factor 8 as a product of two irreducible elements and as a product
of three irreducible elements.
[Hint: Consider ( 1 + Y-7)(1
-
"V'=?).]
8. Factor each of the elements below as a product of irreducibles in
Any factor of
(a) 3
9.
(a)
(b) 7
(c)
4
+ 3i
Verify that each of 5 +'\/2, 2
in
Z[i], [Hint:
a must have norm dividing N(a).]
(d) 11 + 7i
-
Vi, 1 1 - 7'\/2, and 2 + \/2 is irreducible
Z['\/2].
*Since dis square-free, d""' 0 (mod 4).
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352 Chapter 10
Arithmetic in Integral Domains
(b) Explain why the fact that
(5
+
\/2)(2 - Vi)
=
(11 - 7\/2)(2 + \/2)
does not contradict unique factorization in.Z[V2).
10. Find two different factorizations of 9 as a product of irreducibles in.Z['\/=5].
11. Show that .Z[V-6] is not a UFD. [Hint: Factor
10
in two ways.]
12. Show that.Z[VIO] is not a UFD. [Hint: Factor 6 in two ways.)
13. Show that 6 and 2 + 2'\/=5 have no greatest common divisor inZ[v'=S].
[Hint: A common divisor a of 6 and 2 + 2 '\/=5 must have norm dividing
both N(6) 36 and N(2 + 2v'=5) 24; hence, a = r +�with r2 +
5s2 N(a) 1, 2, 3, 4, 6, or 12. Use this to find the common divisors. Verify
that none of them is divisible by all the others, as required of a gcd. Also see
Example4.]
=
=
=
""
14. Show that 1 is a gcd of 2 and 1 + Y-5 in zcA. but 1 cannot
the form 2a + (1 + Y-5)b with a, b EZ[V-5].
be
written in
15. Prove that every principal ideal in a UFD is a product of prime ideals
uniquely except for the order of the factors.
16. Show that (6) = (2)(3) inZ[v=5]. (The product of ideals is defined on page 349.)
17. LetP be the ideal {2a + (1 + �5)bla, bEZ[v=5]} inZ[Y-5]. Prove that
r + s'\/=5 EP if and only if r "" s (mod 2) (that is, rands are both even or
both odd).
18. LetP b e as in Exercise
17. Prove that p2 is
the principal ideal (2).
19. Let Q1 be the ideal {3a + (1 + Y-5 )b I a, b EZ[Y-5]} and Q2 the ideal
{3a + (1 - v'=S)bla, hEZ[v'=5]} in.Z[v'=5].
(a) Prove that r + s� E Q1 if and only if r = s (mod 3).
(b) Show that .Z['\/=5]/Q1 has exactly three distinct cosets.
(c) Prove thatZ['\/=5]/Q1 is isomorphic to.Z3; conclude that Q1 is a prime ideal.
(d) Prove that Q2 is a prime ideal. [Hint: Adapt (a)-(c).]
(e) Prove that Q1Q2
(3).
20. If r + s� E .Z[�] withs ¥: 0, then prove that 2 is not in the principal
ideal (r + J�).
=
21. If dis a square-free integer, prove thatZ[v'd] satisfies the ascending chain
condition on principal ideals.
C. 22. Let dbe a square-free integer and let O(Vd) be as defined on page 350. We
know thatZ[Vd] <;;; O:(v'd) and every element of Z[Vd] is a quadratic integer.
Determine all the quadratic integers in O(v'J) as follows.
(a) Show that every element of O(W) is of the form (r + s Vd)/t, where
r, s, t EZ and the gcd (r, s, t) of r, s, tis I. Hereafter, let a = (r + sv'd)/t
denote such an arbitrary element of C(Vd).
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10.4
The Field of Quotients of an Integral Domain
353
(b) Show that a is a root of
p(x) = x2-
(�) (r2 � )
x+
dr2
EQ[x].
[Hint: Show that p(x) = (x - a)(x - a), where a= (r - sv'd)/t.]
(c) Ifs* 0, show that p(x) is irreducible in O[x].
(d) Prove that a is a quadratic integer if and only if p(x) has integer
coefficients. [Hint: Ifs to 0, use Exercise 5; ifs * 0 and a is a root of a
monic polynomialf(x)E Z[x], use Theorem 4.23 to show that a is a root
of some monic g(x) EZ[x], with g(x) irreducible in O[x]. Apply (c) and
Theorem 4.14 to show g(x) = p(x).]
(e) If a is a quadratic integer, show that ti2r and t2i4ds2. Use this fact to prove
that t must be 1 or 2. [Hint; dis square-free, (r, s. t) 1; use (b) and (d).]
=
(f) If d= 2 or 3 (mod4), show that a is a quadratic integer if and only if
t = 1. [Hint: If t = 2, then ,:i = ds2 (mod 4) by (b) and (d). Ifs is even,
reach a contradiction to the fact that (r, s, t) = 1; ifs is odd, use Exercise 7
of Section 2.1 to get a contradiction.]
(g) If d = 1 (mod4) and aE Q(Vd), show that a is a quadratic integer if and
only if t = 1, or t = 2 and both randsare odd. [Hint: U se (d).]
(h) Use (f) and (g) to show that theset of all quadratic integers in O('\/d) isZ[Vd]
if d= 2 or 3 (mod4) and
{m
+
2nVii Im, n,EZ andm
=
n(mod 2)
}
if d= l (mod4).
Ill
The Field of Quotients of an Integral Domain*
For any integral domain R we shall construct a field F that contains R and consists of
"quotients" of elements of R. When the domain R is Z, then F will be the field 0 of
rational numbers. So you may view these proceedings either as a rigorous formaliza­
tion of the construction of Q from Z or as a generalization of this con struction to
arbitrary integral domains. The field Fwill be the essential tool for studying factoriza­
tion in R[x] in Section 10.5.
Our past experience with rational numbers will serve as a guide for the formal
development. But all the proofs will be independent of any prior knowledge of the
rationals.
A rational number ajb is determined by the pair of integers a, b (with b to 0). But
3
4
.
.
.
21 = 6 = g' and
diflierent paJIS may detenmne the same rational numbe r; ...
ior mstance,
.
.
m general
a
c
b
d
if and only if
ad= be.
•This section is independent of the rest of Chapter 10. Its prerequisites are Chapter3 and Appendix D.
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354 Chapter 10
Arithmetic in Integral Domains
This suggests that the rationals come from some kind of equivalence relation on pairs
of integers (equivalent pairs determine the same rational number). We now formalize
this idea.
Let R be an integral domain and let S be this set of pairs:
S ={(a,
b) la, bER and b *OR}·
Define a relation - on the set S by
(a, b)- (c, d)
ad= bcinR.
means
Theorem 10.25
The relation
- Is an equivalence relation on S.
Proof• Reflexive: Since r is commutative ab = ba, so that (a, b)-(a., b) for every
(a, b) in S. Symmetric: If (a, b)-(c, d), then ad= be. By commutativ­
d) -(a, b). Transitive: Suppose that (a, b)-(c, d)
and (c, d) -(r, s). Then ad= be and cs= dr. Multiplying ad= be bys and
using cs = dr we have ads = (be)s = b(cs) = bdr. Since d OR by the defini­
tion of Sand R is an integral domain we can cancel dfrom ads
bdr and
conclude that as = br. Therefore, (a, b) - (r, s). •
pair
ity cb = da, so that (c,
=
The equivalence relation -partitions S into disjoint equivalence classes by C.orollary D.2
in Appendix D. For convenience we shall denote the equivalence cl� of (a., b) by [a, b] rather
than the more cumbersome [(a.,
b)]. Let F denote the set of
all equivalence cl�s under-.
Note that by Theorem D.l,
[a, b] = [c, d]
in F
(a, b) -(c, d)
if and only if
in S.
Therefore, by the definition of -,
[a, b] = [c, d] in F
if and only if
ad=
be in R.
We want to make the set Finto a field. Addition and multiplication of equivalence
classes are defined by
[a, b] + [c, d] = [ad+ be, bd]
[a, b][c, d] = [ac, bd].*
In order for this definition to make sense, we must first show that the quantities on
the right side of the equal sign are actually elements of the set F. Now
[a, b] is
the
*These definitions are motivated by the arithmetical rules for rational numbers (just replace the
fraction 1/s by the equivalence class [1, s]):
a
c
ad+bc
a
c
ac
b+d=bd IJ"Ci=bd
......
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-...ed.
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.....
...........
..
......
The Field of Quotients of an Integral Domain
10.4
365
pair (a, b) in S. By the definition of S we have b 'I: OR; simi­
d *OR. Since R is an integral domain, bd t= OR. Thus (ad+ be, bd) and (ac , bd)
are in the set S, so that the equivalence classes [ad+ be, bd] and [ac, bd] are elements
equivalence class of the
larly,
of F. But more is required in order to guarantee that addition and multiplication in
Fare well defined.
. 1
In ordinary an"thmetic,
2
4
because 8
•
3
5
12
=
40
5
3
=
4
. 1by - pro duces the same answer
and replacmg
10
2
8
3 =3
-
•
.
10. The answer doesn't depend on how the fractions are repre-
sented. Similarly, in F we must show that arithmetic does not depend on the way the
equivalence classes are written:
Lemma 10.26
Addition and mu ltiplication in Fare independent of the choice of equivalence
class representatives. In other words, if [a,
b] =[a', b'] and [c, d] =[c', d'],
then
[ad + be, bd] =[a'd' + b'c', b'd']
and
[ac, bd] =[a'c', b'd'].
Proof"' As noted above [ad+ be, bd] =[a'd' + b'e', b'd'] in F if and only if
(ad + bc)b'd' =bd(a'd' + b'e') in R. So we shall prove this last state­
[a, b] =[a', b'] and [c, d] = [e', d'] we know that
ment. Since
and
ah' =ba'
(*)
cd' =de'.
Multiplying the first equation by dd' and the second by
the results show that
ab'dd'
cd'bb'
ab'dd' + cd'bb'
(ad+ bc)b'd'
ba 'dd'
=
dc'bb'
=
=
ba'dd' + dc'bb'
bd(a'd' + b'c').
[ad+ be, bd] [a'd' + b'c', b' d'].
For the second part of the proof multiply the first equation in(*) by
Therefore,
ed'
=
bb' and adding
=
and the second by ba' so that
ab'cd' =ba'cd'
and
cd' ba' =de'ha'.
By commutativity the right side of the first equation is the same as the
left side of the second equation so that the other sides of the two equa­
tions are equal:
ah'cd' =de'ba'. Consequently,
(ac)(b'd') =ab'ed' =dc'ba' = (bd)(a'c').
The two ends of this equation show that
[ac, bdj =[a'c', b'd'].
•
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356 Chapter 10
Arithmetic in Integral Domains
Lemma 10.27
If R is an in tegral domain and F is as above, then for all nonzero a,
b, c, d, k ER:
(1) [OR, b] = [OR, d];
(2) [a, b]= [ak, bk];
(3} [a, a]= [c, cJ.
Proof... Exercise 1.
•
Lemma 10.28
With the addition and multiplication defined above, Fis a field.
Proof ... Closure of addition and multiplication follows from Lemma
10.26 and
the remarks preceding it. Addition is commutative in Fbecause addition
and multiplication in R are commutative:
[a, b] +[c, d] =[ad+be, bd] =[cb +da, db] = [c, dJ +[a, b].
Let 0p be the equivalence class [ OR• b] for any nonzero b ER
Lemma 10.27 all pairs of the form
equivalence class). If
(by ( 1) in
OR are in the same
Lemma 10.27 (with k =b):
(OR, b) with b
[a, b] E F, then by (2) in
:F
[a, b] +Op=[a, b] +[OR, b] =[ab+bOR, bb] =[ab, bb] =[a, b].
Therefore, Op is the zero element of F. The negative of [a, bJ in Fis [-a, b]
because
[a, b] +[-a, b] = (ab - ba, b2] =[OR, b� = Op.
T he proofs that addition is associative and that multiplication is associa­
tive and commutative are left to the reader (Exercise 2), as is the verifica­
tion that [lR, la] is the multiplicative identity element in F. If [a, b] is a
nonzero element of F, then a * OR. Hence, [b, a] is a well-defined element
of F and by (3) in Lemma 10.27
[a, b][b, a] =[ab, ba] =[lp, lRab] =[IR, la].
T herefore,
[b, a] is the multiplicative inverse of [a, b]. To see that the dis­
F, note that
tributive law holds in
[a, b]([c, d] + [r, sD =[a, b][cs+dr, drJ
=[a(c3+ dr), b(ds)]
=[acs +adr, bdr].
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10.4
The Field of Quotients of an Integral Domain
On the other hand, by (2) in Lemma
10.27
[a, b][c, d] + [a, b][r, s] = [ac, bd]
357
(with k = b)
+
[ar, bs]
= [(ac)(bs) +(bd)(ar), (bd)(bs)]
= [(acs + adr) b, (bdr) b]
= [acs+adr, b�].
Therefore,
[a, b]([c, d] + [r, sD = [a, b][c, d] +[a, b][r, s].
•
We usually identify the integers with rational numbers of the form
a/l. The same
idea works in the general case:
Lemma 10.29
Let R be an integral domain and F the field of Lemma 10.28. Then the subset
f?'< ={[a,
1R] I a ER} of Fis an Integral domain that is isomorphic to R.
Proof" Verify that R* is a subring of
F (Exercise 3). Clearly [lR, lRI, the identity
element of F, is in R*, so R* is an integral domain. Define a map
f:R--+ R* byf(a) = [a, 1R). Then/is a homomorphism:
f(a) +f(c) = [a, la]+ [c, lRJ = [alR+ l Rc,
lRl RI
= [a+ c, l RJ =f(a +c)
f(a)f(c) = [a, IRJ[c, lRJ = [ac, lRJ = f(ac).
= f(c), then [a, lRJ = [c, lRJ, which implies that a l R = lRc by the
boldface statement following Theorem 10.25. Thus a = c and/is injec­
Iff(a)
tive. Since/is obviously surjective,/is an isomorphism.
•
The equivalence class notation for elements of F is awkward and doesn't convey the
promised idea of "quotients". This is easily remedied by a change of notation, Instead
of denoting the equivalence class of
(a, b) by [a, b] ,
denote the equivalence class of
(a, h) by a/h.
If we translate various statements above from the brackets notation to the new quotient
notation, things begin to look quite familiar:
Theorem 10.30
Let R be an integral domain. Then there exists a field Fwhose elements are of
the form afb with a, b ER and b ¢ OR, subject to the equality condition
�
b
=
£
d
inf
ad= be in R.
if and only if
Addition and multiplication in Fare given by
a c ad+bc a c ac
b (j = bd'
b + (j = bd
The set of elements in F of the form af1R(a ER) is an integral domain isomor­
phic to R.
'
.
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...
....... it.
358 Chapter 10
Arithmetic in Integral Domains
Proof... Lemmas 10.28 and
10.29 and the notation change preceding the
•
theorem.*
It is now clear that if R
Z, then the field Fis precisely O. So Theorem 10.30 may
be taken as a formal construction of Q from Z. In the general case, we shall follow the
same custom we use with 0: The ring R will be identifi£d with its isomorphic copy in
F. Then we can say that Ris the subset of F consisting of elements of the form a/lR.
The field Fis called the field of quotients of R.
=
EXAMPLE 1
Let F be a field. The field of quotients of the polynomial domain F[x] is
denoted by F(x) and consists of allf(x)/g(x), where f(x), g(x) EF[x] and g(x) :¢:
Ox. The field F(x) is called the field of rational functions over F.
The field of quotients of an integral domain R is the smallest field that contains R
in the following sense. t
Theorem 10.31
Let R be an integral domain and Fits field of quotients. If Kis a field containing
R, then Kcontains a subfield E such that R !::: E !::: Kand Eis isomorphic to F.
Proof.,. If a/hEF, then a, b ER and bis nonzero. Since R!::K, b-1 exists. Define a
mapf:F-+Kby f(a/b) ab-1• Exercise 9 shows that/is well defined, that
is, a/b
c/d in Fimplies.f(a/b) f(c/d) in K. Exercise 10 shows that/is
=
=
=
an injective homomorphism. If Eis the image of Funder f, then F=
For each a ER,
a
=
ala -I
=
f(afla)EE,
so Rr;; Er;; K.
E.
•
• Exercises
NOTE:
Unless noted otherwise, R is an integral domain and Fitsfield of quotients.
A. 1. Prove Lemma 10.27.
2. Complete the proof of Lemma 10.28 by showing that
(a) Addition of equivalence classes is associative.
(b) Multiplication of equivalence classes is associative.
(c) Multiplication of equivalence classes is commutative.
3. Show that R*
=
{[a, lRI
I
a
ER} is a subring of F.
*At this point you may well ask, "Why didn't we adopt the quotient notation sooner?" The reason is
psychological rather than mathematical. The quotient notation makes things look so much I ike the
familiar rationals that there is a tendency to assume everything works like it always did, instead of
actually carrying out the formal (and tiresome) details of the rigorous development.
"'Theorem
10.31
is not used in the sequel.
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10.5
Unique Factorization in Polynomial Domains
359
B. 4. If R is itself a field, show that R = F.
5. If R
=
Z[i], then show that F = {r + si I r, s E Q}.
6. If R = Z[W], then show that F =
{r + sva I r, s E 0).
7. Show that there are infinitely many integral domains R such that Z � R !;;;;; 0,
each of which has Q
as
its field of quotients. [Hint: Exercise 28 in
Section 3.1.]
8.
LetfR--+ R1 be an isomorphism of integral domains. Let F be the field of
quotients of R and F1 the field of quotients of R1• Prove that the map
f*:F--+ F1 given by f*(a/b) = f(a)/f(b) is an isomorphism.
9. If R is contained in a field K and
[Hint: a/b = c/d implies ad
10.
=
a/b = c/d in F, show that ab-1
=
cd-1 in K.
be in K.]
(a) Prove that the map/in the proof of Theorem 10.31 is injective.
[Hint: f(a/b) = f(c/d) implies ah-1 = cd-1; show that ad= be.]
(b) Use a straightforward calculation to show that/is a homomorphism.
11. Let a,
b ER. Assume there are positive integers m, n such that am = If", d' =
ll', and (m, n) = 1. Prove that a = b. [Remember that negative powers of a and
b are not necessarily defined in R, but they do make sense in the field F; for
instance, a-2
=
1R/a2.]
12. Let R be an integral domain of characteristic 0 (see Exercises 41-43 in
Section 3.2).
(a) Prove that R has a subring isomorphic to Z [Hint: Consider {nlR In EZ}.]
(b) Prove that a field of characteristic 0 contains a subfield isomorphic to Q.
[Hint: Theorem 10.31.]
13. Prove that Theorem 10.30 is valid when R is a commutative ring with no
zero divisors (not necessarily an integral domain). [Hint: Show that for any
nonzero a ER, the class [a, a] acts as a multiplicative identity for F and the set
{[ra, a] I r ER} is a subring of F that is isomorphic to R. The even integers are
a good model of this situation.]
Im
Unique Factorization in Polynomial Domains*
Throughout this section R is
a
unique factorization domain. We shall prove that the
polynomial ring R[x] is also a UFD. The basic idea of the proof is quite simple: Given
a polynomial /(x), factor it repeatedly
until/(x) is written
as
as
a product of polynomials of lower degree
a product of irreducibles. To prove uniqueness, consider/(x) as
*The prerequisites for this section are pages
322-324 of Section 10.1, the definition of unique
factorization domain (together with Theorems 10.13, 10.15, and 10.18), and Section 10.4. Theorems 10.13,
10.15, and 10.18 depend only on the definition of
UFO and may be
read independently of the rest of
Section 10.2.
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360 Chapter 10
Arithmetic in Integral Domains
a polynomial in F[x], where F is the field of quotients of R. Use the fact that F[x] is
a UFD (Theorem 4.14) to show that factorization in R[x] is unique. There are some
difficulties, however, in carrying out this program.
EXAMPLE 1
The polynomial 3x2 + 6 cannot be factored as a product of two polynomials of
lower degree in Z[x] and is irreducible in O[xJ. But 3x2 + 6 is reducible inZ[xJ
because 3x2 + 6
3(x2 + 2) and neither 3 nor x2 + 2 is a unit inZ[x].
=
So the first step is to examine the role of constant polynomials in R[x). By
Corollary 4.5 and Exercise I
the units in Rix) are the units in R
and
the irreducible constant polynomials in Rlxl are
the irreducible elements of R.
For example, the units of Z[x] are± l. The constant polynomial 3 is irreducible inZ[xJ
even though it is a unit in Q[x].
The constant irreducible factors of a polynomial in R[xJ may be found by factoring
out any constants and expressingthem as products of irreducible elements in R.
EXAMPLE 2
InZ[x],
6x2 + 1 8x + 12
=
6(x2 + 3x + 2)
=
2 3(x2
·
+ 3x +
2).
Note that x2 + 3x + 2 is a polynomial whose only constant divisors in Z[x] are
the units :t 1. This example suggests a strategy for the general case.
Let R be a unique factorization domain. A nonzero polynomial in R[x] is said to be
primitive if the only constants that divide it are the units in R. For instance, x2 + 3x +
2 and 3x4 - 5x� + 2x are primitive inZ[x]. Primitive polynomials of degree 0 are units.
Every primitive polynomial of degree I must be irreducible by Theorem 10.l (because
every factorization includes a constant (Theorem 4.2) and every such constant must be
a unit). However, primitive polynomials of higher degree need not be irreducible (such
as x1 + 3x + 2
(x + l )(x + 2) inZ[xD· On the other hand, an irreducible polynomial
of positive degree has no constant divisors except units by Theorems 4.2 and 10.1. So
=
an irreducible polynomial of positive degree is primitive.
F urthermore, as the example illustrates,
every nonzero polynomial /(x) E Rix)
factors as/(x) cg(x) with g{x) primitive.
=
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U n i q u e Factorization In Poly n o m ia l Domains
10.5
361
To prove this claim, let c be a greatest common divisor of the coefficients of f(x).* Then
f(x)
=
cg(x) for some g(x). Now
then g(x)
=
dh{x) so that.f(x)
=
we
show thatg(x) is primitive. If dER divides g(x),
cdh(x). Since cd is a constant divisor of f(x), it must
divide the coefficients of f(x) and, hence, must divide the gcd c. Thus cdu
u
= c
for some
E R. Since c #: OR we see that du = 1R and dis a unit. Therefore, g(x) is primitive.
Using these facts about primitive polynomials, we can now modify the argument
given at the beginning of the section and prove the first of the t wo conditions neces­
sary for R[x] to be a UFD.
Theorem 10.32
Let R be a unique factorization domain. Then every nonzero, nonunit
R[x] is a product of irreducible polynomials.t
f(x) in
Proof" Letf(x) = cg(x) with g(x) primitive. Since R is a UFD c is either a unit
or a product of irreducible elements in R (and, hence, in R[x]). So we
need to prove only thatg(x) is either a unit or a product of irreducibles
in R[x]. If g(x) is a unit or is itself irreducible, there is nothing to prove.
If not, then by Theorem 10.1 g(x)
= h(x)k(x)
with neither h(x) or k(x)
a unit. Sinceg(x) is primitive, its only divisors of degree 0 are units, so
we must have 0 < degh(x) < degg(x) and 0 < deg k(x) < degg(x).
Furthermore, h(x) and k(x) are pr imitive (any constant that divides one
of them must divide g(x) and hence be a unit). If they are irreducible,
we're done. If not, we can repeat the preceding argument and factor
them as products of primitive polynomials of lower degree, and so on.
This process must stop after a finite number of steps because the degrees
of the factors get smaller at each stage and every primitive polynomial
of degree 1 is irreducible. Sog(x) is a product of irreducibles in R[x] .
•
The proof that factorization in R[x] is unique depends on several technical facts
that will be developed next. But to get an idea of how all the pieces fit together, you
may want to read the proof of Theorem 10.38 now, referring to the intermediate re­
sults
as
needed and accepting them without proof. Then you can return to this point
and read the proofs, knowing where the argument is headed.
Lemma 10.33
Let R be a unique factorization domain and g(x), h(x) ER[x]. If p is an irreduc­
ible element of R that divides g(x)h(x), then p divides g(x) or p divides h(x).
Proof• Copy the proof
of Lemma 4.22, which is the special case R
=
Z. Just
replace Z by R and prime by irreducible and use Theorem 10.15 in place
of Theorem 1.5.
*The gcd c exists byTheorem
•
10.18.
tAs usual we allow a "product" with just one factor.
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362 Chapter 10
Arithmetic in I n te g ral Doma i ns
Corollary 10.34
Let
Gauss's Lemma
R be a unique factorization domain. Then the product of primitive
R[x] is primitive.
polynomials in
Proof.. If g(x) and h(x) are primitive and g(x)h(x) is not, then g(x)h(x) is
divisible by some nonunit c ER. Consequently, each irreducible factorp
of c divides g(x)h(x). By Lemma 10.33, p divides g(x) or h(x), contradict­
ing the fact that they are primitive. Therefore, g(x)h(x) is primitive. •
Theorem 10.35
Let R be a unique factorization domain and r, s nonzero elements of R. Let f(x)
and g(x) be primitive polynomials in R[x] .such that rf{x) = sg(x). Then rand s
are associates in Rand f(x) and
g(x) are associates Jn R[x].
Proof.. If ris a unit, then/(x) = r-1sg(x). Since r-1sdivides the primitive
polynomial/(x), it must be a unit, say (r-1s)u = lR· Hence,Jlx) and g(x)
are associates in R[x]. Furthermore, u is a unit in R and su = r so that r
ands are associates in R.
If ris a nonunit, then r = piJJ2
Pk with eachp1 irreducible. Then
PiP2
pkf(x) = sg(x), so Pt dividessg(x). By Lemma 10.33 Pt divides
s or g(x). Sincep1 is a nonunit and g(x) is primitive, Pi must divides, say
s =p1t. ThenptP2
pk,f(x) =sg(x) =p1tg(x). Cancelingpt shows
that P2
pJ(x) = tg(x) . Repeating the argument withp2 shows that
p3
p,J(x) = zg(x), where P2Z = t and, hence,piJJ2z= p1t = s. After
k such steps we have/(x) = wg(x) ands= PIP2
Pkw for some w E R.
Since w divides the primitive polynomial/(x), w is a unit. Therefore,
f(x) and g(x) are associates in R[x]. Sinces =Pt · PkW = l'W, rands
are associates in R. •
•
•
•
•
•
•
•
•
•
•
.
•
•
.
•
•
•
•
•
·
Corollary 10.36
Let R be a unique factorization domain and Fits field of quotients. Let f(x),
g(x) be primitive polynomials in R[x]. If f{x) and g(x) are associates in F[x],
then they are associates in R[x].
Proof.. If f(x) and g(x) are associates in F[x], then g(x) = !_f(x) for some
s
r
nonzero -eFby Corollary 4.5. Consequently, sg(x) = rf(x) in R[x].
s
Therefore,/(x) and g(x) are associates in R[x] by Theorem 10.35. •
Corollary 10.37
Let R be a unique factorization domain and Fits field of quotients. If f(x) E R[x]
has positive degree and is irreducible in
R[x], then f{x) is irreducible in F[x].
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10.5
Unique Factorization In Polynomial Domains
363
Proof" Iff(x) is not irreducible in F[x], then/(x) = g(x)h(x) for some g(x), h(x)
E F[x] with positive degree. Let b be a least common denominator of the
coefficients of g(x). Then bg(x) has coefficients in R . So bg(x) = ag1(x) with
a ER
and g1(x) primitive of positive degree in R[x]. Hence, g(x) = �g1(x).
Similarly h(x) =
� h1(x) with
�
d E R andh1(x) primitive of positive degree
�
in R[x]. Therefore,/(x) = g(x')h(x) = ig1(x) h1(x) = :g1(x)h1(x),
so that bdf(x) = acg1(x)h1(x) in R[x]. Nowf(x) is primitive because it is
irreducible and g1(x)h1(x) is primitive by Corollary 10.34. So bd is an as­
sociate of ac by Theorem 10.35, say bdu = ac for some unituER.
Therefore,/(x) = :g1(x)h1(x) = ug1(x)h1(x). Sinoe ug1(x) and h1(x) are
polynomials of positive degree in R[x], this contradicts the irreducibility
of f(x). Therefore,f(x) must be irreducible in F[x]. •
Theorem 10.38
If Risa unique factorization domain, then so is R[x].
Proof" Every nonzero nonunitf(x) in R[x] is a product of irreducibles by
Theorem 10.32. Any such factorization consists of irreducible constants
(that is, irreducibles in R) and irreducible polynomials of positive degree.
Suppose
c1
• • •
C,,.P1(x)
·
·
·
p,.(x) = di •
• •
dnqi(x) •
• •
qi(x )
with each c,, '4 irreducible in R and each pi.x), qj._x) irreducible of posi­
tive degree in R[x] (and, hence, primitive).* Then p1 (x)
p,/._x) and
q 1(x) · • • qi_x) are primitive by Corollaryl0.34. So Theorem 10.35 shows
that c1 • • • cm is an associate of d1 • • • d,. in R and p1(x) • • • p,/._x) is an
associate of q1(x)
qi(x) in R[x). Hence, c1
c,,, = ud1d2 •
d,, for
some unit uE R. Associates of irreducibles are irreducible (Exercise 7 of
Section 10. l), so ud1 is irreducible. Since R is a UFD, we must have m = n
and (after relabeling if necessary) c1 is an associate of ud1 (and hence of
d1), and c1 is an associate of d1 for i 2!:: 2. Let F be the field of quotients
of R. Each of the p/x), qfx) is irreducible in F[x] by Corollary 10.37.
Unique factorization in F[x] (Theorem 4.14) and an argument similar to the one just given for R show that k = t and (after relabeling if
necessary) each pf..x ) is an associate of qf..x) in F[x]. Consequently, pf..x)
and q1(x) are associates in R[x] by Corollary 10.36. Therefore, R[x] is a
UFD. •
·
•
•
·
•
•
·
·
•
•
•
"It may bethat neither factorization contains constants, but this doesn't affect the argument It is not
possible to have irreducible constants in one factorization but not
..........
in the other (Exercise5).
..
..
.......
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364 Chapter 10
Arithmetic in Integral Domains
An immediate consequence of Theorems 1.8 and 10.38 and Example 8 of
Section
6.1 is
Corollary 10.39
Z [x ] is
a unique factorization domain that is not a principal ideal domain.
As illustrated in the preceding discussion, theorems about Z [x] and O![x] are quite
likely to carry over to an arbitrary UFD and its field of quotients. Among such results
are
the Rational Root Test and Eisenstein's Criterion (Exercises 9-11).
• Exercises
NOTE:
A.
Unless stated otherwiseR is a UFD and Fitsfield of quotients.
1. LetR be any integral domain and pER. Prove that p is irreducible inRif and
only if the constant polynomial pis irreducible inR[x].
[Hint:
Corollary
may be helpful.]
4.5
2. Give an example of polynomialsf(x), g(x)ER[x] such that/(x) and g(x)
associates in
3. If
c1
•
•
•
F [x] but not inR[x].
,,J(x)
c
=
g(x) with
are
Does this contradict Corollaryl0.36?
c1 ERand
g(x) primitive in R[x], prove that each
c1is a unit.
4. If g(x) is primitive inR[x], prove that every nonconstant polynomial inR[x]
that divides g(x) is also primitive.
B. 5. Prove that a polynomial is primitive if and only if lR is a greatest common
divisor of its coefficients. This property is often taken as the definition of
primitive.
6. If f(x) is primitive in R[x] and irreducible in F[x], prove that/(x) is irreducible
inR[x].
7. If Ris a ring such thatR[x] is a UFD, prove thatRis a UFD.
8. If Ris a ring such thatR[x] is a principal ideal domain, prove thatRis a field.
9. Verify that the Rational Root Test (Theorem
4.21) is valid with Z and Q
replaced byR and F.
I 0. Verify that Theorem 4.23 is valid with Z and
0 replaced byRand F.
11. Verify that Eisenstein's Criterion (Theorem 4.24) is valid with Z and
replaced byR and Fand prime replaced by
12. Show that
[Hint:
-...ed.
x3 - 6x2 + 4ix + 1
11.]
+
Q
i"educible.
3i is irreducible in (Z[iD[x].
Exercise
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11
C H A P T E R
Field Extensions
High-school algebra deals primarily with the three fields
0, R, and C and plane
geometry, with the set R x R. Calculus is concerned with functions from R to Ill.
Indeed, most classical mathematics is set in the field C and its subfields. Other
fields play an equally important role in more recent mathematics. They are used in
analysis, algebraic geometry, and parts of number theory, for example, and have
numerous applications, including coding theory and algebraic cryptography.
In this chapter we develop the basic facts about fields that are needed to prove
some famous results in the theory of equations {Chapter 12) and to study some of
the topics listed above. The principal theme is the relationship of a field with its
various subfields.
Ill
Vector Spaces
An essential tool for the study of fields is the concept of a vector space, which is
introduced in this section. Vector spaces are treated in detail in books and courses
on linear algebra. Here we present only those topics that are needed for our study of
fields. If you have had a course in linear algebra, you can probably skip most of this
section. Nevertheless, it would be a good idea to review the main
results, particularly
Theorems 11.4 and 11.5.
Consider the additive abelian group* M(lll) of all 2 X 2 matrices over the field R
of real numbers. If r is a real number and A
=
(: !)
is an element of M(R), then the
*Except for the last two results in the chapter, group theory is not a prerequisite for this chapter. In
this section you need only know that an additive abelian group is a set with an addition operation
that satisfies Axioms 1-5 in the definition of a ring (page 44).
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,,,__
....
.......
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_
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366 C h a pter 11
Field Extensions
product of the number rand the matrix A is defined to be the matrix rA
This operation, which is called
scalar multiplication,
=
e: ;!).
takes a real number (field ele­
ment) and a matrix (group element) and produces another matrix (group element).
This is an example of a more general concept. Let F be a field and Gan additive abe­
lian group.* Then a scalar multiplication is an operation such that for each
each
Definition
v
a
E F and
E Gthere is a unique element av E G.
Let F be a field. A vector space over F is an additive abelian group* V
equipped with a scalar multiplication such that for all a, a,, a2 E F and v,
V11 V2E V:
(i) a(v1 + v2) = av1 + av2;
(ii) (a1 + a2)v a1v + a2v:
=
(iii) a1(a2v) = (a1a2)v:
(iV) 1FV = V.
EXAMPLE 1
Scalar multiplication in M(lll), as defined above, makes M(R) into a vector
space over R (Exercise 1).
EXAMPLE 2
Consider the set 02 = 0 X 0, where 0 is the field of rational numbers. Then
02 is a group under addition (Theorem 3.1 or 7 .4); its zero element is (0, 0) and
the negative of (s, t) is (-s, -t). For a EO and (s, t) E02, scalar multiplication
is defined by a(s, t)
(ar, at). Uoder these operations Cf is a vector space over
Q (Exercise 2).
=
EXAMPLE 3
The preceding example can be generalized as follows. If Fis any field and n :2: 1
an integer, let F"
:
FX FX
·
•
·
X F (n
summands). Then F" is a vector space
over F, with addition defined coordinatewise:
(si. s2,
•
•
•
, s,,)
+
(ti. t,_,
•
.
•
,
t,,) = (s1
+
ti. s2
+
12,
•
•
•
,
s,,
+
t,,)
and scalar multiplication defined by:
(see Exercise 5).
"See the preceding footnote.
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11.1
Vector Spaces
367
EXAMPLE 4
The complex numbers C form a vector space over the real numbers R, with
addition of complex numbers (vectors) defined
as
usual and with scalar mul­
tiplication being ordinary multiplication (the product of a real number and a
complex number is a complex number).
Special terminology is used in situations like the preceding example. If F and Kare
F � K,
fields with
we say that K is an extension field of
F. For
instance, the complex
numbers C are an extension field of the field � of real numbers. As the preceding
example shows, the extension field I[; can be considered as a vector space over�- The
same thing is true in the general case.
If Kis an extension field of F, then Kis a vector space over F, with
addition of vectors being ordinary addition in Kand scalar
multiplication being ordinary multiplication in K
(the product of an element the subfield
F and an
element of K is an element of K).
For the purposes of this chapter, extension fields are the most important examples of
vector spaces.
If Vis a vector space over a field F, then the following properties hold for any v E V
and a E F (Exercise 21 ):
Opv
=
Ov; aOy
=
Oy,
-(av)
=
(-a)v
=
a(-v).
Spanning Sets
F and that w and v., Vi• • . . , vn are elements
is a linear combination of vi> v�, • . . , v11 if w can be written in
Suppose Vis a vector space over a field
of V. We say that
w
the form
for some
Definition
ai E F.
If every element of a vector space Vover afield Fis a linear combination of
v11
v2,
•
•
•
,
v11, we say that the set {v1, v2,
•
•
•
,
v0) spans V over F.
EXAMPLE 5
03 over 0 because
b, c) of 03 is a linear combination of these three vectors:
The set {(1, 0, 0), (0, 1, 0), (0, 0, I)} spans the vector space
every element (a,
(a, b, c) =a (1, 0, 0) + b (0,
1, 0) +
c
(0, 0, 1).
EXAMPLE6
Every element of C (considered
as
a vector space over�) is a linear combina­
tion of 1 and i because every element can be written in the form al
+bi, with
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368
Chapter 11
Field Extensions
a, b E It Thus the set { 1, i} spans C over Ill. The set { 1 + i, 5i, 2 + 3i} also
spans C because any a+ bi EC is a linear combination of these three elements
with coefficients in R:
b
a+ bi= 3a(l + i) + -(51) + (-a)(2 + 31).
5
Linear Independence and Bases
i} not only spans the extension field C of R but it also has this property: If
bi 0, then a 0 and b 0. In other words, when a linear combination of 1 and
i is 0, then all the coefficients are 0. On the other hand, the set { 1 + i, 5i, 2 + 3i} does
not have this property because some linear combinations of these elements are 0 even
The set { l,
al +
=
=
=
though the coefficients are not; for instance,
1
2(1+i)+5(5i)- 1(2 + 3i)
=
0.
The distinction between these two situations will be crucial in our study of field
extensions.
Definition
A subset {v1, v2, • • • , Vn} of a vector space V over a field F is said to be
linearly independent over F provided that whenever
with each c1EF, then �1 =OF for every i. A set that is not linearly indepen­
dent is said to be linearly dependent
{:ui, � . . , u,,.} is linearly dependent over F if there exist elements
, bm of F, at least one of which is nonzero, such that b1u1+�1"1. +
+ b,,.'Um Oy.
Thus, a set
bl>�
•• • •
•
· · ·
=
EXAMPLE 7
The remarks preceding the definition show that the subset {I, i} of C is linearly
independent over IR and that the set
{1 + i, 5i, 2 + 3i} is linearly dependent.
Note, however, that both of these sets span C.
EXAMPLE 8
Consider the subset
pose
Ct.
{(3, 0, 0), (0, 0, 4)} of the vector space 03 over Cl! and sup­
c1(3, 0, 0) + c2(0, 0, 4) (0, 0, 0). Then
c2EQ are such that
(0, 0, 0)
c1(3, 0, 0) + c2(0, 0, 4)
=
(3ch 0, 4ci),
c1 0 c2• Hence, {(3, 0, 0), (0, 0, 4)} is linearly indepen­
Q. However, the set {(3, 0, 0), (0, 0, 4)} does not span 03 because
which implies that
dent over
=
=
=
=
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11.1
Vector Spaces
369
there is no way to write the vector (0, 5, 0), for example, in the form a1(3, 0, 0)
+ a2(0, 0, 4) (3a" 0, 4a2J with CJiE Q.
=
Let V be a vector space over a field F. The preceding examples s how that linear
independence and spanning do not imply each other; a subset of V may have one,
both, or neither of these properties. A subset that has both properties is given a special
name.
Definition
A subset {v1, v2,
, v11} of a vector space V over a field F ls said to be a
basis of V If it spans Vand is linearly independent over F.
•
•
•
EXAMPLE9
Example 5 shows that the subset {(1, 0, 0), (0, I, 0), (0, 0, l )} spans the vector
space 03 over Q. This set is also linearly independent over Q (Exercise 8) and,
hence, is a basis.
EXAMPLE 10
Examples 6 and 7 show that the set { 1, i} is a basis of C over�. We claim that
the set { 1 + i, 2i} is also a basis of Cover �- If c1Q + 1) + c2(21) "" 0, with c1,
c2 E Iii, then c1 l + (c1 + 2cz)i 0. This can happen only if c1 "" 0 and c1 + 2ez. 0.
But this implies that 2c2 0 and, hence, c2
0. Therefore, {l + i, 2i} is linearly
independent. In order to see that {l + i, 2i} spans C, note that the element
=
=
=
=
a + bi ECcan be written as a(l + i) +
(b-a)
2-
2i.
One situation always leads to linear dependence. Let Vbe a vector space over a field
F and Sa subset of V. Suppose that v, ui. u2,
, ut are some of the elements of Sand
that v is a linear combination of uh ui, . . . , ur, say v = a1u1 +
+ a,u., with each
Oi E F. If w1,
, w, are the rest of the elements of S, then
• • •
•· ·
•
•
·
•
v
=
a1u1 +
·
·
·
+ a,u1 + O.rw1 +
·
·
•
+
OJIW,
and, hence,
-lpV + a1u1 +
·
·
·
+
a, u1 + O;w1 +
·
•
•
+
Opo,
=
Ov.
Since at least one of these coefficients is nonzero (namely -1p), Sis linearly dependent.
We have proved this useful fact:
If
v
E Vis a linear combination of u1, u2,
containing
v
•
•
, , u1
E V, then any set
and all the U; is linearly dependent.
In fact, somewhat more is true.
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3 70 Chapter 11
Field Extensions
Lemma 11.1
Let Vbe a vector space over a field F. The subset{u1, u2,
• • •
, Un} of Vis linearly
dependent over F if and only if some uk is a linear combination of the preced­
ing ones, u1, U2, • • • , uk-1·
Proof• If some 'Uk is a linear combination of the preceding ones, then the set
is linearly dependent by the remarks preceding the lemma. Conversely,
suppose {uh . .. , u,.} is linearly dependent. Then there must exist e le ment s
Ci.
'C11 EF, not all zero,, such that C1U1 + C2U2 + . . . + C11U,, = 017. Let k
be the largest index such that ck is nonzero. Then c1 Op for i > k and
• • •.
=
C1u1 + C2t12 + · · · + CkUk
=
Or
ckuk = -c1u1 - c2u.i -
•
•
•
- ck-luk-1·
Since Fi s
a field and ck* 0, ck-I exists; multiplying the prece ding equa­
tion by ck-t shows that uk is a linear combination of the preceding u's:
The next lemma gives an upper limit on the size of a linearly independent set. It
says, in effect, that if Vcan b e spanned by n elements over F, then every linearly inde­
pendent subset of V contains at most n elements.
Lemma 11.2
Let V be a vector space over the field
{v1, v2,
•
•
•
,
Vn}·
If {u1, u2,
•
•
•
,
F
that is spanned by the set
Um} is any linearly independent subset of V, then
ms n.
Proof• By the definition of spanning, every element of
V (in particular u.1) is a
combination of V[,
, v,,. So the set {u11 v1, 'V1;
, 1111} is linearly
dependent. Therefore, one of its elements is a linear comb inatio n of the
preceding ones by Lemma 11.1, say v1 = a1u1 + b1v1 + · · · + b1-1v;...1·
If v1 is deleted, then the remaining set
linear
•
•
•
•
•
• •
still spans Vsince every element of Vis a linear combination of the v 's
and any appearance of v1 can be replaced by a1u1 + b1v1 + · · · +
b1.,.1 v,_ 1. In particular, u2 is a linear combination of the elements of the
set (*) . Consequently, the set
is linearly dependent. By Lemma 11.1 one of its elements is a linear
combination of the preceding ones. This element can't be one of the u's
because this would imply that the u's were linearly dependent. So some
CapJriliM 20t2C..-.i.
.
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11.1
v1 is a linear combination of
u1, u2,
Vector Spaces
371
and the v's that precede it . Deleting v1
produoes the set
This set still spans Vsince every element of Vis a linear combination of
the v's and v1, v1 can be replaced by linear combinations of u1o u,;, and the
other v's. In particular, 'UJ is a linear combination of the elements in this
new set. We can continue this process, at each stage adding a u, deleting
a v, and producing a set that spans V. If m >n, we will run out of v's be­
fore all the u's are inserted, resulting in a set of the form {u1o u2, • • ,
•
u,,}
that spans V. But this would mean that u,,. would be a linear combination
of u0
• • •
,
{ui;
u,,, contradicting the linear independence of
Therefore, ms n.
.
.
. , um}
•
•
Theorem 11.3
Let V be a vector space over a field F. Then any two finite bases of V over F
have the same number of elements.
Proof• Suppose {u., ... , u,,.} and {v., . .. , v,,) are bases of
v's span Vand the u's
are
Vover F. Then the
linearly independent, so ms
Now reverse the roles: The u's span Vand the
n
by Lemma 11.2.
1/s are linearly indepen­
dent, son s m by Lemma 11.2 again. Therefore, m
= n.
•
According to Theorem 11.3, the number of elements in a basis of V over Fdoes not
depend on which basis is chosen. So this number is a property of V.
Definition
lfa vector space Vover a field f has a finite basis, then Vis said to be finite
dimensional over F. The dimension of V over Fis the number of elements
in any basis of V and is denoted [V:f]. If V does not have a finite basis, then
Vis said to be infinite dimensional over F.
EXAMPLE 11
The dimension of
03 over 0 is 3 because {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis.
F is an n-dimensional vector space over F
More generally, if Fis a field, then
(Exercise 27).
EXAMPLE 12
[C:lll]
=
2 since { 1, i} is a basis of IC over Ill. On the other hand, the extension
field R of Q is an infinite-dimensional vector space over
0. The proof
of this fact
is omitted here because it requires some nontrivial facts about the cardinality of
infinite sets.
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3 72 Chapter 11
Field Extensions
Applications to Extension Fields
In the remainder of this section, K is an extension field of a field F. We say that K is
a finite-dimensional extension of F if K, considered as a vector space over F, is finite
dimensional over F.
Remark If [K:F] = 1 and { u} is a basis, then every element of K is of the form
for some c E F. In particular, lF =cu, and, hence, u =c-1 is in F. Thus, K=F. On
the other hand, if K= F, it is easy to see that {lF} is a basis and, hence, [K:FJ =1.
Therefore,
cu
(K:Fl =I
if and only if
K=F.
If F, K, and L are fields with F !;::; K !;::; L, then both K and L can be considered as
vector spaces over F, and L can be considered as a vector space over K. It is reason­
able to ask how the dimensions [K:F], [L:K], and [L:F] are related. Here is the answer.
Theorem 11.4
Let f, K, and L be fields with F !;::; K � L. If [K:f] and [L:K] are finite, then L is a
finite-dimensional extension off and [L:F] = [L:K][K:f].
Proof• Suppose [K:F] = m and [L:K] = n. Then there is a basis
{u11.,
u,,.} of
Kover Fand a basis {v1,
, v,.} of L over K. Each u1 and v1 is nonzero
by Exercise 19; hence, all the products Uf/J; are nonzero. The set of all
products {u1v111 s i s m, 1 s j s n} has exactly mn elements (no two
of them can be equal because u(IJ1 = ukvt implies that u(IJ1 - UkVt = Ox
with u1, uk EK, contradicting the linear independence of the v's over K).
We need to show only that this set of mn elements is a basis of L over F
because in that case [L:K][K:FJ=nm=[L:F].
If w is any element of L, then w is a linear combination of the basis
elements v., ... , v,,, say
.
•
• •
.
with each
Each b1 EK is a linear combination of the basis elements uh
there are agEF such that
b1 = a11U1 + a21U2 +
b2 =a12u1 + On� +
·
·
'
·
•
·
b1 E K.
• • •
, u,.. so
+ a...1um
+ 11,,aU,..
Substituting the right side of each of these expressions in(*) shows that
is a sum of terms of the form agu1v1 with aq EF. Therefore, the set of
all products u1v1 spans L over F.
w
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11.1
Vector Spaces
373
To show linear independence, suppose cv E F and
�cgn,v1
(**)
=
l,J
1 1
+
CuU V
+
C12U1V2
·
·
+
·
c,,.,.umvn =OF.
By collecting all the terms involving Vto then all those involving
so
on,
(c11U1
v2'
and
we can rewrite (**) as
+
C21U2
+
. .
.
+
c,,,,.u,,,)v,
(c1:zU1 + c22� +
+
+ · · ·+
. . .
(ci..u1 +
+
c..au,,,)vz
Czxtt:z +
·
·
·
+
c,..,,u,,J v,, =OF.
The coefficients of the v's are elements of K, so the linear independence
of the v's implies that for each j = 1, 2, . . . , n
'Vu1
+
c1J� +
·
·
·
CmJUm = OF.
+
Since each c11 EF and the u's are linearly independent over F, we must
have elf=OF for all i,j. This completes the proof of linear independence,
and the theorem is proved. •
The following result will be needed for the proof of Theorem 11.15 in Section 11.4.
Theorem 11.5
Let Kand L be finite dimensional extension fields of F and let f:K-1> L be an
isomorphism such that f{c) = c for every c EF. Then [KF
: ] = [L:F].
Proof• Suppose [K:F] = n and {u1 ,
, u,,} is a basis of K over F. In order to
prove that [L:F] = n also, we need only show that {f(u1),
,/(u,,)} is
a basis of Lover F. Let v EL; then since/is an isomorphism, v = f(u)
for some u EK. By the definition of basis, u = c1u1 +
+ c,,u,, with
each c1EF. Hence, v = f(u) = /(c1u1 +
+ cnu,,) =f(c1)f(u1) + · · · +
f(cJf(u,,). Butf(cJ = c1 for every i, so that v = cJ(u1) +
+ c,f(u,J.
Therefore, {f(u1), . . . ,/(u,,)} spans L. To show linear independence,
suppose that
• • •
• • •
·
·
·
·
·
·
·
dtf(ut)
with each
+
·
·
·
+
·
·
dnf(u,,) =Op
d1 E F. Then sincef(dJ = d, we have
f(diu1
+ • · · +
d,,u,,) =f(d1lf(u1) +
= dtf(Ui)
+
·
· · · + f(dnlf(u,,)
·
·
+
d,/(u,,) = O_p.
Since the isomorphism/is injective, d1u1 +
+ d,.u,, = Op by Theorem 6.11.
But the u's are linearly independent in K, and, hence, e\'efY d, = Op Thus
{/(ui), ,/(u,,)} is li nearly independent and, therefore, a basis. •
·
·
·
. • .
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....... it.
374 Chapter 11
Field Extensions
• Exercises
NOTE: V denotes a
vector space over afield F. and K denotes an extension.field of F.
A. I. Show that M(R) is a vector space over R.
2. Show that 02 is a vector space over
Q.
3. Show that the polynomial ring R[x} (with the usual addition of polynomials
and product of a constant and a polynomial) is a vector space over R.
4. If n � 1 is an integer, let
R,.[x] denote the set consisting of the constant
IR[x} of degree$ n. Show that Rn[x]
polynomial 0and all polynomials in
(with the usual addition of polynomials and product of a constant and a
polynomial) is a vector space over Ill.
5. If n 2!:: 1 is an integer , show that Fn is a vector space over F.
6. If {v1, v2,
,
• • •
{w, vi, v:i,
• • •
7. Show that
,
v11} spans Kover Fand w is any element of K, show
vn} also spans K.
that
{i, I + 2i, I + 3i} spans Cover R.
8. Show that the subset {(1,
overO.
9. Show that { v'2, v'2 +
0, 0), (0, 1, 0), (0, 0,
i,v'3
-
I)} of 03 is linearly independent
i} is linearly dependent over IR.
IO. If vis a nonzero element of V, prove that {v} is linearly independent over F.
any subset of
11. Prove that
Vthat contains 0vis linearly dependent over F.
{u, v, w} of Vis linearly independent over F, prove that
{u, u + v, u + v + w} is linearly independent.
12. If the subset
13. If S =
{vi,
. . .
, Vk}
is a linearly dependent subset of V, then prove that any
subset of Vthat contains S is also linearly dependent over F.
14. If the subset T = { ut. .
.
.
, U:t}
of Vis linearly independent over F , then prove
that any nonempty subset of Tis also linearly independent.
15. Let b and
d be distinct nonzero real numbers and c any real number.
{b, c + di} is a basis of C over !ft
Prove that
16. If K is an n-dimensional extension field of "11..1" what is the maximum possible
number of elements in K?
17. Let {vi,
. . •
,
vn} be a basis of
{c1v1o c2v:z,
of F . Prove that
18. Show that
19. If {vi, v2 •
.
Vover Fand let cl>,
• •
{1, [x]} is a basis of "ll..2[x}/(x2 +
• •
, v,,}
•
.
, en be nonzero elements
, c,.v,, } is also a basis of Vover F.
x
+ 1) over Z2•
is a basis of v, prove that v1 of; Ov for every i.
20. Let F, K, and L be fields such that Fr;;_ K <;;;; L. If S
o:
{vb v:i,
over F, explain why S also spans L over K.
B. 21. For any vector
(a)
v
=
=
, v,,}
spans L
E V and any element a E F , prove that
Ov. [Hint: Adapt the proof of Theorem
(b) 40v= Ov.
(c) -(av) (-a) v a(-v).
0.i;V
. • •
3.5.)
=
eap,ngm.20:12�1..umiq.A:l.lliala 11--4.....,-aatn. t:IDJllilrd,. llC...t,, ar�io.wmlliarls,_,. 0.1"�dpll.-mkd.�lrlDlllllm�M ....... Jion1M•Bam:.ndkir�•).Bdbarbll._....._
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Vector Spaces
11.1
v'2} of IR is linearly independent over Q.
22. (a) Prove that the subset {l,
v'3 is not a linear combination of
(b) Prove that
O. Conclude that {l, v'2}
does not span
1 and V2 with coefficients in
R over Q.
23. (a) Show that
{l,
v'i, W} is linearly independent over Cl!.
(b) Show that
{1,
v'i, v'3, W}
is linearly independent over
24. Let v be a nonzero real number. Prove that
0 if and only if v is irrational.
25. (a) Let k 2:!:
375
Q.
{1, ff} is linearly independent over
1 be an integer. Show that the subset {1, x, J?, x:1,
is linearly independent over
. • •
, �} of IR[x]
R (see Exercise 3).
(b) Show that lli[x] is infinite dimensional over IR.
26. Show that the vector space ll,. [x] of Exercise 4 has dimension
n + 1 over
IR.
27. If Fis a field, show that the vector space Fn has dimension n over F.
28. Prove that
K has exactly one basis over F if and
only
if K
=
F;:;; Z2•
1F + l F :/: OF. If {u, v, w} is a basis of Vover F, prove that the set
{u + v, v + w, u + w} is also a basis.
29. Assume
30. Prove that
{v1o ...
, vn} is a basis of
Vover Fif and only
if every
means that if
w"" c1v1 +
·
·
·
·
+
+
c11vn and w
=
d1v1 +
· ·
·
element of V
v1,
, v,, ("unique"
+ d,,v,,, then c1 d,
can be written in a unique way as a linear combination of
• • •
=
for every 1).
31. Letp(x)
=
ao +
1x +
a
·
·
a,,x' be irreducible in F[x] and let L be the
n over F.
extension field F[x)/(p(x)) of F. Prove that L has dimension
[Hint: Corollary 5.5, Theorems 5.8 and 5.10, and Exercise 30 may
32. If S =
over F.
be helpful.]
{vi. . .. , v,} spans V over F, prove that some subset of Sis a basis of K
[Hint: Use Lemma 11.1 repeatedly to eliminate v's until you reduce to a
set that still spans Vand is linearly independent.)
33. If the subset {u1,
• . •
, u,} of Vis linearly independent over Fand wE Vis not a
linear combination of the u's, prove that
{uh . . . , u,, w} is linearly independent.
34. If Vis infinite-dimensional over F, then prove that for any positive integer k,
[Hint: Use
1, and Exercise 33 can be used to prove
Vcontains a set of k vectors that is linearly independent over F.
induction; Exercise 10 is the case k
=
the inductive step.]
{vi> . .. , vn} of Vis linearly independent over F and that
+ c,.v,,, with c1 EE Prove that the set {w -11, w - Vi•
, w - v11}
is linearly independent over Fif and only if c1 +
+ Cn * 1F·
35. Assume that the subset
w ""= c1v1 +
· ·
·
• . •
·
·
·
36. Assume that Vis finite-dimensional over F and S is a linearly independent
subset of V. Prove that Sis contained in a basis of V.
and S =
{u1,
• • •
[Hint: Let
[ V:F]
=
n
, um}; then m s n by Lemma 11.2. If S does not span V,
u's. Apply
33 to obtain a larger independent set; if it doesn't span , repeat the
then there must be some w that is not a linear combination of the
Exercise
arg ument. Use Lemma
11.2 to show that the process must end with a basis that
contains S.]
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3 76 Chapter 11
Field Extensions
37. Assume that
(i) {v1>
•
(ii) {tJi,
•
(iii) {v"
.
•
.
•
.
•
[V:F] = n and prove that the following conditions are equivalent:
,
v.,} spans Vover F.
,
v.,} is linearly independent over
•
F.
, v.,} is a basis of VoverF.
38. LetF, K , and
L be fields such thatF!:;:K� L. If [L:F] is finite, then prove that
[L:K] and [K:F] are also finite and both are :5 [L:F]. [Hint: Use Exercises 20
and 32 to show that [ L:K] is finite . To show that [K:F] is finite, suppose
[L:F] = n. The set {Ix} is linearly independent by Exercise IO; if it doesn't
span K, proceed as in the hint to Exercise 36 to build larger and larger linearly
independent subsets of K. Use Lemma
11.2 and the fact that [L:F] = n
to show that the process must end with a basis of Kcontaining at most
n
elements.]
39. If [K:F]
= p, with p prime, prove that there
[Hint: Exercise 38 and Theorem 11.4.]
1111
is no field E such thatF�
E � K.
Simple Extensions
Field extensions can be considered from two points of view. You can look upward from
a field to its extensions or downward to its subfields. Chapter
of the upward point of vie w . We took a field F and
an
5
provided an example
irreducible polynomial Jl(.x) in
F[x] and formed the field of congruence classes (that is, the quotient field)F[x]/(p(x)).
Theorem 5.11 shows that F[x]/(p(x)) is an extension field of F that contains a root
of Jl(.x).
In this section we take the downward view, starting with a field Kand a subfieldF.
If u EK, what can be said about the subfields of Kthat contain both u and F? Is there
a smallest such subfield? If u is the root of some irreducible
Jl(.x)
in
F[x],
how is this
smallest subfield related to the extension field F[x]/(p(x)), which also contains a root
of p(x)?
The theoretical answer to the first two questions is quite easy. Let Kbe an extension
EK. Let F(u)denote the intersection of all subfields of K that contain
field of F and
u
bothF and
(this family of subfields is nonempty since K at least is in it). Since the
u
intersection of any family of subfields of Kis itself a field (Exercise
By its definition,
F(u) is
1), F(u)is a field.
contained in every subfield of Kthat contains Fand u, and,
hence, F(u)is the smallest subfield of K containingFand u . F(u)is said to be a simple
extension of F.
As a practical matter, this answer is not entirely satisfactory. A more explicit
description of the simple extension field F(u) is needed. It turns out that the structure
of
F(u) depends on whether or not u
is the root of some polynomial in F[x]. So we
pause to introduce some terminology.
Definition
An element u of an extension field Kot Fis said to be algebraic over F if u is
polynomial in F[x]. An element of K that is not the
polynomial In F[x] is said to be transcendental over F.
the root of some nonzero
root of any nonzero
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11.2
Simple Extensions
377
EXAMPLE 1
In the extension fe
i ld C of Ill, i is algebraic over IR because i is the root of :x?- + 1 E
n[x]. You can easily verify that element 2 + i of C is a root of x1 -
:x? - 7x + 15 E
O[x]. Thus 2 + i is algebraic over Q. Similarly, � is algebraic over Q since it
is a root of x5 - 3.
EXAMPLE 2
Every element c in a field F is algebraic over Fbecause c is the root of x -
c EF[x].
EXAMPLE 3
The real numbers
1T
and e are transcendental over
0 (proof omitted). Hereafter
we shall concentrate on algebraic elements. For more information on transcen­
dental elements, see Exercises 10 and 24-26.
If u is an algebraic element of an extension fe
i ld K of F, then there may be many
polynomials in F[x] that have u as a root. The next theorem shows that all of them
are multiples of a single polynomial; this polynomial will enable us to give a precise
description of the simple extension field Ji(u).
Theorem 11.6
Let K be an extension field of F and u EK an algebraic element over F. Then
there exists a unique manic irreducible polynomial p(x} in F[x] that has u as a
root. Furthermore, if u is a root of g(x) ef{x], then p(x) divides g(x).
Proof "" Let S be the set of all nonzero polynomials in F[x] that have u as a root.
Then Sis nonempty because u is algebraic over F. The degrees of poly­
nomials in S form a nonempty set of nonnegative integers, which must
contain a smallest element by the Well-Ordering Axiom. Letp(x) be a
polynomial of smallest degree in S. Every nonzero constant multiple
of p(x) is a polynomial of the same degree with u as a root. So we can
choose p(x) to be monic (if it isn't, multiply by the inverse of its lea ding
coefficient).
Ifp(x) were not irreducible in F[x], there would be polynomials k(x)
and t(x) such that p(x) = k(x)t(x), with deg k(x) < degp(x) and deg t(x) <
degp(x). Consequently, k(u)t(u) = p(u) =OF inK. SinceKis afield either
k(u) =OFor t(u) =Op, that is, either k(x) or t(x) is in S. This is impossible
since p(x) is a polynomial of smallest degree in S. Hence,p(x) is irreducible.
Next we show thatp(x) divides every g(x) in S. By the Division
Algorithm, g(x) =p(x)q(x) + r (x), where r(x) =OF or deg r(x) < degp(x).
Since u is a root of both g(x) and p(x),
r(u) = g(u) - p(u)q(_u) = OF+ Opj(u) = OF-
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3 78 Chapter 11
Field Extensions
Sou is a root of r(x). If r(x) were nonzero, then r(x) would be in S,
con tradicting the fact thatp(x) is a polynomial of smallest degree in S.
Therefore, r(x) O� so thatg(x) p(x)q(x). Henoe,p(x) divides every
polynomial in S.
To show thatp(x) is unique, suppose t(x) is a monic irreducible
polynomial in S. Thenp(x) I t(x). Sincep(x) is irreducible (and, hence,
nonconstant) and t(x) is irreducible, we must have t(x) cp(x) for some
c E F. Butp (x) is monic, so c is the leading coefficient of cp(x) and ,
hence, of t(x). Since t(x) is monic, we must have c = lp Therefore,p(x)
t(x) and p(x) is unique. •
=
=
=
=
If K is an extension field of F and u EK is algebraic over F, then the monic, irre­
ducible polynomialp(x) in Theorem 11.6 is called the minimal polynomial of u over F.
The uniqueness statement in Theorem 11.6 means that once we have found any monic,
irreducible polynomial in F[x] that has u as a root, it must be the minimal polynomial
of u over F.
EXAMPLE 4
i2 - 3 is a monic, irreducible polynomial in Q[x] that has V3 ER as a root.
Therefore, x'-·- 3 is the minimal polynomial of V3 overQ. Note thatx'- - 3 is
reducible over� since it factors as (x - V3)(x + V3) in ll[x]. So the minimal
polynomial of v'3 over R is x - v'3, which is monic and irreducible in R[x].
EXAMPLE 5
Let u '\13 + V5 ER. Then u2 3 + 2v'3V5 + 5 8 + 2v'I5. Hence,
u2 - 8 2v'I5 so that (u2 - 8)2
60, or, equivalently, (u2 - 8)2 - 60 0.
Therefore, u v'3 + V5 is a root of (x'-- 8)2 - 60 x' - 16x'- + 4 E O[x].
Verif y that this polynomial is irreducible in Q[x] (Exercise 14). Hence, it must
be the minimal polynomial of v'3 + \/5 over Cl!.
=
=
=
=
=
=
=
=
The minimal polynomial of u provides the connection between the upward and
downward views of simple field extensions and allows us to give a useful description
of F(u).
Theorem 11.7
Let K be an extension field of F and u EK an algebraic element over F with
minimal polynomial p(x) of degree n. Then
(1) F(u) = F[x]/(p(x)).
(2) {1f, u, u2,
, un-1} is a basis of the vector space F(u) over F.
(3) [F(u) : F] = n.
•
•
•
......
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:dJbb�
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.....
...........
.......
......
11.2
Simple Extensions
379
Theorem 11.7 shows that when u is algebraic over F, then F(u) does not depend on K
but is completely determined by F[x] and the minimal polynomialp(x). Consequently,
we sometimes say that F (u) is the field obtained by adjoining u to F.
Proof ofTheorem 11.7 ... (1) Since F(u) is a field containing u, it must contain
every positive power of u. Since F(u) also contains F, F(u) must
contain every element of the form h0 + h1u + h111?· +
+ h,u'
with b1 EF, that is, F( u) contains the element/(u) for everyf(x) EF[x].
Verify that the map <p:F [x]-+ F(u) given by <p(f(x)) =f(u) is a
homomorphism of rings. A polynomial in F[x] is in the kernel of <p
precisely when it has u as a root. By Theorem 11.6 the kernel of <p
is the principal ideal (p(x)). The First Isomorphism Theorem 6.13
shows that F[x]/(p(x)) is isomorphic to Im <p under the map that sends
congruence class (coset) [f(x)] tof(u). Furthermore, sincep(x) is
irreducible, the quotient ring F[x]j(p(x)), and, hence, Im cp, are fields
by Theorem 5.10. Every constant poly nomial is mapped to itself by <p
and <p(x) = u. So Im <pis a subfield of F( u) that contains both F and
u. Since .F(u) is the smallest subfield of K containing F and u, we must
have F(u) =Im cp;;;;; F[x]/(p(x)).
·
·
·
(2) and (3) Since F(u) =Im cp, every nonzero element of F(u) is
of the formf(u) for somef(x)EF[x]. If degp(x) = n, then by the
Division Algorithm/(x) p(x)q(x) + r(x), where r(x) = h0 + h1x +
1
+ b3_1X-- EF [x]. Consequently,f(u) =p(u)q(u) + r(u) =OFq(u) +
r(u) = r(u) = holF + b1u +
+ b,._1u"-1• Therefore, the set
{lp, u, u1, ... , u"-1} spans F(u). To show that this set is linearly
independent, suppose c0 + c1u +
+ Cw-1u•-1 = OF with each
c1EF. Then u is a root of c0 + c1x +
+ c11_1x"�1, so this poly­
nomial (which has degree s n - 1 ) must be divisible by p(x) (which
has degree n). This can happen only when co + c1x +
+ c11_1,xa-1
is the zero polynomial; that is, each c1 =Op Thus {lp, u, u.2,
, U--1}
is linearly independent over F and, therefore, a basis of F(u).
Hence, [F(u) : F ] = n. •
=
·
·
·
·
·
·
·
·
·
·
·
·
·
·
·
• . •
EXAMPLE 6
The minimal polynomial of v'3 over 0 is x2 - 3. Applying Theorem 11.7 with n =2
we see that {1, v'3} is a basis of o( VJ) over 0, whence [o( VJ) : Q] =2. Similarly,
Example 5 shows that v'3 + VS has minimal polynomial x4 - 16x1 + 4 over Q so
that [O(v'3 + VS): O] =4 and {1, v'3 + VS, ( v'3 + '\/5)2, ( v'3 + '\/5)3} is a
basis.
An immediate
consequence of Theorem 11.7 is that
if u and
v have the same minimal polynomial p(x)
in F(xJ, then F(u) is isomorphic to F(v).
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380 Chapter 11
Field Extensions
The reason is that both F(u) and F(v) are isomorphic to F[x]/(p(x)) and, hence, to
each other. Note that this result holds even when u and v are not in the same extension
field of F. The remainder of this section, which is not needed until Section 11.4, deals
w ith generalizations of this idea. We shall consider not only simple extensions of the
same field, but also simple extensions of two different, but isomorphic, fields.
Suppose F and E are fields and that rr:F-+ E is an isomorphism. Verify that the
map from F[x] to E[x] that maps/(x) = ao + a1x + "2X2 +
+ a,rx!' to the polyno­
mial uf(x) = u(ao) + u(a1)x + u(a2)x2 +
+ u(a,.)X' is an isomorphism of rings
(Exercise 21 in Section 4.1). Note that if /(x) = c is a constant polynomial in F[x]
(that is, an element of F), then this isomorphism maps it onto u(c) EE . Consequently,
we say that the isomorphism F[x]-+E[x] extends the isomorphism u:F-+ E, and we
denote the extended isomorphism by u as well.
·
·
·
·
·
·
Corollary 11.8
Let
u:F -+ E be an isomorphism of fields. Let u be an algebraic element in
some extension field of F with minimal polynomial p(x) EF[x]. Let v be an
algebraic element in some extension field off, with minimal polynomial
up(x) Ef[x]. Then u extends to an Isomorphism of fields fi:F(u )-+ E(v) such
that u(u) = v and u(c) = u{c) for every c E F.
The special case when u is the identity map F-+ F states whenever u and v have
the same minimal polynomial, then F{u) = F{v) under a function that maps u to v and
every element of F to itself.
Proof of Corollary 11.8 .. The isomorphism u extends to an isomorphism (also
denoted u) F[x]-+ E[x] by the remarks preceding the corollary. The proof
of Theorem 11.7 shows that there is an isomorphismT":E[x]/(up(x))-+E(v)
given byT([g(x)D g(v). Let 1T be the surjective homomorphism
=
E[x]-+ E[x]/(up(x))
that maps g(x) to fg(x)] and consider the composition
F[x] � E[x] � E[x]/(up(x)) �E(v)
f(x) -----+ uf(x)---+[uf(x)] -----+ uf(v).
Since all three maps are surjective, so is the composite function. The
kernel of the composite function consists of all h(x) EF[x] such that
uh(_v) OE> Since 'f is an isomorphism, uh(v) = OE if and only if [uh(x)]
is the zero class in E[x]/(up(x)), that is, if and only if uh(x) is a mul­
tiple of up(x). But if uh(x) = k(x) up(x), then applying the inverse of
the isomorphism u shows that h(x) u-1 (k(x))p(x). Thus the kernel of
the composite function is the principal ideal (p(x)) in F[x]. Therefore,
F[x]/(p(x)) = E(v) by the First Isomorphism Theorem 6.13; the proof
=
•
=
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11.2
Simple Extensions
of that theorem shows that this isomorphism (call it
O([f(x)D = uf( v) . Note that 9([x]) =
381
9) is given by
c E F, O([c]) =
v and that for each
u(c). So we have the following situation, where "qi is the isomorphism of
Theorem 11.7:
F[u] �
---..!..+ E(v)
�tij(v)
F[x]/(p(x))
f[u] +--- [J(x)]
c +---- [c]
--+
---
u(c)
cE F.
The composite function (J • -rp-1 : Ji(u)
tends u and mapsu to v. •
-+ E(v) is an isomorphism that ex­
EXAMPLE 7
The polynomial
x!
root in R, namely
w
=
l
� '\/3;
2 is irreducible in
-
O[x] by Eisenstein's Criterion. It has a
V'i. Verify that '\Ylw is also a root of X1 - 2 in C, where
is a complex cube root of
1. Applying Corollary 11.8 to the
0-+ 0 we see that the real subfield Q(V'2) is isomorphic to
identity map
the complex subfield
0(-V-2m) under a map that sends V'2 to -V-2m and each
0 to itself.
element of
• Exercises
NOTE:
Unless stated otherwise, K is an extensionfield of thefield F.
A. 1. Let {E1liE 1} be a family of subfields of K. Prove that nE'1 is a subfield of K.
lel
2. If u E K, prove that Ji(u2) r;;;. Ji(u).
3. If u EK and cEF, prove that F(u +
4. Prove that
0(3 + 1) = 0( 1 - i).
c) = F(u) = Ji(cu).
5. Prove that the given element is algebraic over Q;
(a)
3 + Si
(b)
Vi - V2
6. If u EK and u1 is algebraic over
7. If Lis
a
(c) 1 +
V'2
F, prove that u is algebraic over F
field such that Fr;;. Kr;;;. LanduELis algebraic over F, show that u is
algebraic over K.
8. If u,
v
E K and u +vis algebraic over F, prove thatu is algebraic over Ji(v).
9. Prove that
10. If
u
and
Vii is algebraic over 0('11').
E K is transcendental over F and 0F * cE F, prove that each of
u2
11. Find
is transcendental over F.
u
+ 1F>
cu,
[0({/2): Q).
.......
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...
382 Chapter 11
Field Extensions
12.
If a+ biE C and b :¢: 0, prove that C
13.
If [K:F ] is prime and u EK is algebraic over F, show that either F(u)
F(u) F.
=
ll(a + h1).
=
Kor
=
14.
B. 15.
Prove that x" - 16x2 + 4 is irreducible in C[x].
Show that every element of C is algebraic over IR [Hint: See Lemma 4.29.]
16.
If u EK is algebraic over F and c EF, prove that u + 1 F and cu are algebraic
over F.
17.
Find the minimal polynomial of the given element over Q:
ca> v1 +vs
(b) v'3i
+
v'2
18.
Find the minimal polynomial of Vi + i over Q and over R.
19.
Let u be an algebraic element of Kwhose minimal polynomial in F[x] bas prime
degree. If Eis a field such that F o:;;; E !;;;; Ji(u), show that E For E Ji(u).
=
20.
=
Let u be an algebraic element of Kwhose minimal polynomial in F[x] has odd
degree. Prove that F(u) F(u2).
=
1111
21.
Let F 0(1T4) and K
of Kover F.
22.
If rands are nonzero, prove that O(Vr)
some tEO.
23.
If K is an extension field of C such that [KQ
: ] 2, prove that K Q (\/'J)for
some square-free integer d. [Square-free means dis not divisible by y for any
primep.]
24.
If u EK is transcendental over F, prove that F(u) = F(x), where F(x) is the
field of quotients of F[x] , as in Example 1 of Section 10.4. [Hint: Consider the
map from F(x) to F(u) that sendsf(x)/g(x) to f(u)g(ur1.]
25.
If u EK is transcendental over F, prove that all elements of Ftu), except those
in F, are transcendental over F.
26.
Let F(x)be as in Exercise 24. Show that
x+
over F.
=
=
Q(1T). Show that 1T is algebraic over Fand find a basis
=
Q(Vs)if and only if r
=
�
t2s for
=
E
F(x) is transcendental
Algebraic Extensions
The emphasis in the last section was on a single algebraic element. Now
extensions that consist entirely of algebraic elements .
Definition
=
we
consider
An extension field Kofa field f is said to be an algebraic extension of F if
every element of Kis algebraic over f.
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11.3
Algebraic Extensions
383
EXAMPLE 1
If
a + bi EC, then a + bi is a root of
(x - (a + b{))(x - (a - bi))= x1 - 2ax + (0: + b2) ER[x].
a + bi is algebraic over Ill, and, hence, C is an algebraic extension
0 since
there are real numbers (such as 'ii and e) that are not algebraic over Q.
Therefore,
of R. On the other hand, neither C nor R is an algebraic extension of
Every algebraic element
of F, namely
F(u),
u
over F lies in some finite-dimensional extension field
by Theorem
11.7.
On the other hand, if we begin with a finite­
dimensional extension of F we have
Theorem 11.9
If K is a finite-dimensional extension field of F, then
sion off.
K
is an algebraic exten­
Proof.,. By hypothesis, K has a finite basis over F, say {vi. v2,
, v,.}. Since
thesen elements span K, Lemma 11.2 implies that every linearly inde­
•
•
•
pendent set in K must haven or fewer elements.
u EK, there are two possibilities: (1) 1.l= 11/ with 0 � i < j; and
u are distinct. In Case (1), u is a root of
the polynomial JI - x1 EF[x] and hence, is algebraic over F. In Case (2),
{lp, u, u2,
, u"} is a set of n + 1 elements in K and must, therefore, be
If
(2) all nonnegative powers of
•
•
•
linearly dependent over F. Consequently, there are elements c1 in F, not
all zero, such that cJp +
c1u +
2
c2u
the root of the nonzero polynomial
and, hence, algebraic over F.
+
c0
·
·
+
·
+ c,.u"= Op. Therefore, u is
+ c2x1 +
+ c,.x' in F[x]
c1x
·
·
·
•
If an extension field K of F contains a transcendental element
be infinite dimensional over F (otherwise
u
u,
then K must
would be algebraic by Theorem
11.9 is
dimensional algebraic extensions (Exercise 16).
Nevertheless, the converse of Theorem
11.9).
false since there do exist infinite­
Simple extensions have a nice property. You need only verify that the single ele­
ment
u
is algebraic over F to conclude that the entire field
.F(u)
is an algebraic
extension (because F(u) is finite dimensional by Theorem 11. 7 and, hence, algebraic
11.9).
by Theorem
This suggests that generalizing the notion of simple extension
might lead to fields whose algebraicity could be determined by checking just a finite
number of elements.
If
ul>
•
•
•
,
u,. are elements of
an extension field K of F, let
.F(uh u.z,
.
•
•
'u,,)
denote the intersection of all the subfields of K that contain F and every 'Ut· As in the case
of simple extensions, F(u1,
'Ut· F(uh
•
•
•
•
•
•
, u,,) is the smallest subfield of K that contains F and all the
, u,J is said to be a finitely generated extemion of F, generated by u1o
•
•
•
,
u,..
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_
384
Chapter 11
Field Extensions
EXAMPLE 2
The field 0('\/3, i) is the smallest subfield of C that contains both the field 0
and the elements V3 and i.
EXAMPLE 3
A finitely generated extension may actually be a simple extension. For instance,
the field Q(i) contains both i and - i, so Q(i, - 1) = Q(1).
EXAMPLE 4
Every finite-dimensional extension is also finitely generated. If {uh ... , u,.] is
a basis of Kover F, then all linear combinations of the u1 (coefficients in F) are
in F{u1,
, uJ. Therefore, K .ftu1,
, u,.).
• • •
=
• • •
The key to dealing with finitely generated extensions is to note that they can be
obtained by taking successive simple extensions. For instance, if K is an extension
field of F and u, v EK, then F(u, v) is a subfield of K that contains both F and u
and, hence, must contain F(u). Since v is in F(u, v), this latter field must contain
F(u)(v), the smallest subfield containing both F(u) and v. But F(u)(v) is a field
containing F, u, and v and, hence, must contain F(u, v). Therefore, F(u, v) F(u)(v).
Thus the finitely generated extension F(u, v) can be obtained from a chain of simple
extensions:
=
F<;;. F(u) !;:;:; F{:u)(v)
=
F(u, v).
EXAMPLE 5
The extension field o(v'3, i) can be obtained by this sequence of simple
extensions:
As we saw in Example 4 of
Section 11.2, X1 - 3 is the minimal polynomial
of v'3 over Q, so that [ 0( v'3): Q] "" 2 by Theorem 11. 7. Similarly, x2 + 1
[whose coefficients are in o( V3) ] is the minimal polynomial of j over a( \/3)
because its roots ± i are not in o{ v'3), so x2 + 1 is irreducible over o(v'3) by
Corollary 4.19. By Theorem 1 1. 7 again, (0( v'3)(1):0( \/3)] 2. Consequently,
by Theorem 11.4,
=
(O(v'J, i):O]
=
[O{V3)(i):O (V3)][0{V3):0]
=
2
•
2
=
4.
Thus, the finitely generated extension o( v'3, i) is finite dimensional and, hence,
algebraic over Q by Theorem 11.9.
....
... ...
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�a--.�c.a.�
riBfit1D....,,,..��·...,.
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11.3
x ens i ons
Algebraic E t
386
Essentially the same argument works in the general case and provides a useful
an extension is algebraic:
to determine that
way
Theorem 11.10
If K
F(u1,
, Un) is a finitely generated extension field of F and each u1 is
=
•
•
•
algebraic over F, then K i s a finite-dimensional algebraic extension of F.
Proof... The field K can be obtained from this chain of extensions:
F!;;F(ut) !;;.F(u1o � !;;;; F(ul> �. UJ) !,; • • •
!,; .F(u1o . .. , Un-t) !,; F(ui, .
Furthermore,
. . , u,,)
=
K.
.F(u1)(�), F(uto u:z, UJ) = .F('U(, u2)(u3), and in
F(uh
, u1_1)(uJ. Each Ut
is algebraic over Fand, hence, algebraic over F(ui. ... , u,_1) by Exercise
general
F(u" .
.F(u1o ui)
.
.
=
, u,) is the simple extension
7 of Section 11.2. But every simple extension by
. . •
an algebraic element is
finite dimensional by Theorem 11.7. Therefore,
[.F(u i. ... , u,):.F(u1o
••.
, 'tl.1-1)]
is finite for each i = 2 , ... , n. Consequently, by repeated application of
T heorem 11.4, we see that
[K:F(ub
Thus
. • .
, U,,....1)]
•
·
[K:F] is the
product
[Ji(ui, ui, UJ):F(uh �)][.F(uh ul):.F(ui)][.F(u1):F).
·
[K:F) is finite, and, hence, K is algebraic over Fby Theorem 11.9.
•
EXAMPLE 6
'\13 and VS are algebraic over Q, so 0('\13, v'5) is a finite-dimensional
Q by Theorem 11.10. We can calculate the dimen­
sion of 0( '\13, v'5) over Q by considering this chain of simple extensions:
Both
algebraic extension field of
o !,; o(V3) !:;; o(v'J)( v'5) o( \13, v's).
=
We know that
(0( v'3):0]
=
2. To determine
find the minimal polynomial of
r - 5 ; it i s irreducible i n Q[x],
(10(\/3)(\1'5):0( v'3)] we shall
v'5 over 0('\13) .The obvious
candidate i s
but w e must show that i t i s irreducible over
0(\13), in order to conclude that it is the minimal polynomial. If VS or -VS
is in o( v'3), then ±v'5
+ b'\13, with b Q, Squaring both sides shows
= a
that 5
=
d- + 2ab'\/3 + 3fil, whence v3
a,
=
E
5-d--3fil
i contradicting
2ab
V3 is irrational; a similar contradiction results if a = 0 orb = 0.
± v'5 are not in o( \13), and, hence, x2 - 5 is irreducible over o( \13)
by Corollary 4.19. Sor - 5 is the minimal polynomial of VS over 0( v'3), and
the fact that
Therefore,
[O(V3)(v'5) : 0(\/3)] 2 by Theorem 11.7.Consequently, by Theorem 11.4
[O(v'3, VS):O] [O(v3)(VS):Cl!( v3)][Cl!( v'3):0] = 2 2 4.
=
=
...
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·
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=
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386 Chapter 11
Field Extensions
The remainder of this section is not used in the sequel. Theorem 11 .4 tells us that
the top field in a chain of finite-dimensional extensions is finite dimensional over the
ground field. Here is an analogous result for algebraic extensions that may not be finite
dimensional.
Corollary 11.11
If L is an algebraic extension field of Kand Kis an algebraic extension field of
F, then Lis an algebraic extension of F.
Proof"'" Let u EL. Since u is algebraic over K, there exist a1 EK such that
Clo + a1u + atu2 +
+ amtl"
Ox. Since each of the a, is in the field
f\a1o
, a,,.), u is actually algebraic over F{a1o
, a,,J. Consequently,
·
·
·
=
• • •
•
.
•
in the extension chain
Fr;;, Ji(ah
.
.
•
,
a,,J r;;;;, Ji(a"
.
•
•
, a,,,)(u)
=
F{a1o ,
•
•
,
a,,,, u)
, a,,.)(u) is finite dimensional over F{a"
, a,,.) by Theorem 11.7.
[F(ah
, a,,,):F] is finite by Theorem 11.10 since each a, is
algebraic over F. Therefore, Ji(a11
, a,,,, u) is finite dimensional over F
F{a,
. • •
• . •
Furthermore,
• • •
• • •
by Theorem 11.4 and, hence, is algebraic over Fby Theorem 11.9. Thus
u is algebraic over F. Since u was an arbitrary
braic extension of F.
element of L, L is an alge­
•
Corollary 11.12
Let K be an extension field of F and let E be the set of all elements of K that
are algebraic over F. Then Eis a subfield of Kand an atgebraic extension field
of F.
Proof... Every element of Fis algebraic over F, so F<;;,.E. If u, v EE, then u and v
v) is an algebraic ex­
F(u, v) c.=.. E. Since F(u, v) is
are algebraic over Fby definition. The subfield F(u,
tension of Fby Theorem 11.10, and, hence,
a field, u + v, uv, -u, -v E F(u, v) r;;;;,E. Similarly, if u is nonzero, then
u-1 EF(u, v) r;;;E.
;,
Therefore, Eis closed under addition and multiplica­
tion; negatives and inverses of elements of E are also in E. Hence, Eis a
field.
•
EXAMPLE 7
If K
=
C and F =
Q in Corollary
11.12, then the field Eis called the field of
algebraic numbers. The field Eis an infinite-dimensional algebraic extension
of
0 (Exercise 16). Algebraic numbers were discussed in a somewhat different
context on page 350 .
..
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11.3
Algebraic Extensions
387
• Exercises
NOTE : Unless stated otherwise, K is an extension field of thefield F.
A. I. If u, vEK, verifythat F(u)(v)
=
F('ll)(u).
2. If Kis a finite field , show that Kis an algebraic extension of F.
3. Find a basis of the given extension field of Q.
o(vs, v'7) (c) o(Vi, v'3, V5)
Find a basis of 0( Vi, + v'3) over 0( v'3).
Show that [O(v'3, i):O]
4.
Verify that [0( Vi, VS, VIO}:O)
4.
(a)
4.
5.
6.
o(vs, ; )
(h)
(d)
o(\o/2, v'3)
=
=
7. If [K:F] is finite and u isalgebraic over K, prove that [K(u):K] $ [F(u):F].
8.
If [K:F ] is finite and u is algebraic over K, prove that [K(u):F(u)]
[Hint: Showthat any basis of Kover Fspans K(u)over F(u).J
$ [KF
: ].
9. If [K:F ] is finiteand u is algebraic over K, prove that [F(u):F ]divides [K(u):FJ.
B. 10. Prove that [K:F]is finite if and only if K =
F(u1,
• • •
, 'I.lit), with each
u1
algebraic over F. [This is a stronger version of Theorem 11.10.)
11. Assume that
u, v EKare algebraic over F, with minimal
polynomials p(x)and
q(x), respectively.
(a) Ifdegp(x)
=
m and deg
q(x)
=
n and (m, n)
=
1, prove that [F(u, 'll):F]
(b) Show by example that the conclusion of part (a)may
mn.
=
be false if m and
n
are not relatively prime .
(c) What is
[Q( v'2, �:O!]?
12. Let D be a ring such that Fr;;;,. Dr;;;: K. If Kis algebraic over F, prove that Dis a
fe
i ld. [Hint: To find the inverse of a nonzero
that F(u) !;;D.J
u
ED, use Theorem 11.7 to show
13. Letp(x)and q(x)be irreducible in F[x] and assume that degp(x) is relatively
prime to deg q(x). Let u be a root ofp(x)and
v a root
extension field of F. Prove that q(x)is irreducible over
14.
(a) Let F1 r;;;,_F2 r;;;,_ F3 !;:;; •
of q(x)in some
F(u).
• • be a chain of fe
i lds. Prove that the union of all the F1
is also a field .
(b) If each F, is algebraic over F., show that the union of the F, is an algebraic
extension ofF1•
15. Let Ebe the fe
i ld of all elements of Kthat are algebraic over F, as in Corol­
lary 11.12. Prove that everyelement of the set K - Eis transcendental over E.
16. Let Ebe the field ofalgebraic numbers
(see Example 7). Prove that Eis an
infinite dimensional algebraic extension ofQ. [Hint: It suffices to show that
[ E: Q] � n for every positive integer n. Consider roots of the polynomial
x" - 2 and Eisenstein's Criterion.]
..
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..
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388 Chapter 11
17.
Field Extensions
Assume that IF+ IF:¢: Op If u E F, let Vu denote a root of x?-'- u
in K.Prove that F( Vu + v'V) F( Vu,\IV). [Hint: 1, (v'U + v'v),
(Vu + \/V)1, (Vu + Vv )3, etc., must span F( Vu + Vv) by Theorem 11.7.
Use this to show that Vu and Vv are in F(Vu + Vv).]
=
18.
If n1,
•
•
•
, 11, are distinct positive integers, show that
[O{Vni. .. . , Vii;): OJ� 21•
C.19.
Ill
If each n,is prime in Exercise 18, show that� may be replaced by=.
Splitting Fields
Let F be a field and /(x) a polynomial in F[x]. Previously we considered extension
fields of F that contained a root of f(x). Now we investigate extension fields that
contain all the roots off (x).
The word "all" in this context needs some clarification. Supposef(x) has degree n.
Then by Corollary 4.17,f(x) has at most n roots in any field. So if an extension field
K of F contains n distinct roots off(x), one can reasonably say that K contains "all"
the roots off(x), even though there may be another extension of F that also contains
n roots off(x). On the other hand, suppose that K contains fewer than n roots of f(x).
It might be possible to find an extension field of K that contains addi tional roots of
f(x). But if no such extension of K exists, it is reasonable to say that K contains "all"
the roots. We can express this condition in a usable form as follows.
Let K be an extension field of F andf(x) a nonconstant polynomial of degree n in
F[x]. Iff(x) factors in K[x] as
f(x) = c(x - u1)(x - ui)
•
·
•
(x - u,J
then we say that/(x) splits over the field K. In this case, the (not necessarily distinct)
elements u" . .., un are the only roots off(x) in K or in any extension field of K. For
if vis in some extension of K and/(v) =OF> then c(v - u1)(v - ui)
(v - u,J =Op
Now cis nonzero sinoe/(x) is nonconstant. Hence one of thev - u1must be zero, that
is, v u1• So if f (x) splits over K, we can reasonably say that K contains all the roots
of f(x). The next step is to consider the smallest extension field that contains all the
roots off(x).
•
•
·
=
Definition
If Fis a flefd and f(x) EF[x], then an extension field K of Fis said to be a
splitting field (or root field) of f(x) over F provided that
(i) f(x} splits over K, say f(x)
{ii)
K
=
f(u,, u21
•
• •
=
c(x - u1)(x - u2)
•
• •
(x - un);
, UiJ).
EXAMPLE 1
If x2 + 1 is considered as a polynomial in lll[x], then C is a splitting field since
X- + 1 = (x + iXx - 1) in Qx] and C = R(i) = R(i, -1). Similarly, o{ v'2) is a splitting
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rir;bl1a-...,,,..��·...,.
...
........
11.4
field of the polynomial x? - 2 in O[x] since r - 2
o('\12)
=
°' ('\12,
- '\12).
=
Splitting
Fields
389
(x + '\12)(x -'\12) and
EXAMPLE2
The polynomial.f(x) x4 - x? - 2 inQ[x ] factors as (x1 - 2)(r + 1 ), so its
roots in Care±\12 and± i. Therefore, O(v'2, i) is a splitting field of f(x)
overQ.
=
EXAMPLE 3
Every first-degree polynomial ex + din F[x] splits over F since ex + d =
c(x - (-c-1d)) with -e-1dEF. Obviously, Fis the smallest field containing both
F and c-1d, that is, F = l{e-1d). So Fitself is the splitting field of ex + dover F.
EXAMPLE 4
The concept of splitting field depends on the polynomial andthe base field. For
instance, C is a splitting field of x2 + 1 over R but not over Q because C is not
the extension O(i, -i) O(i). See Exercise 1 for a proof.
=
At this point we need to answer two major questions about splitting fields: Does
every polynomial inF[x] have a splitting field over Fl If it has more than one splitting
field over F, how are they related?
The informal answer to the first question is easy. Given/(x) EF[x], we can find an
extension F(u) that contains a root u of f(x) by Corollary 5.12. By the Factor Theorem
in Ji(u)[x], we know that/(x) = (x - u)g(x). By Corollary 5.12 again there is an exten­
sion F(u)(v) of F(u) that contains a root v of g(x). Continuing this, we eventually get a
splitting field of .f(x). We can formalize this argument via induction and prove slightly
more:
Theorem 11.13
Let F be a field and f(x) a nonconstant polynomial of degree n in f{x). Then
there exists a splitting field K of f(x) over F such that [K:f] :5 nl.
Proof.. The proof is by induction on the degree of f(x). Iff(x) has degree 1,
then Fitself is a splitting field of f(x) and [F:F] = 1 :5 11. Suppose
the theorem is true for all polynomials of degree n - 1 and thatf(x)
has degree n. By Theorem 4.14f(x) has an irreducible factor in F[x]
Multiplying this polynomial by the inverse of its leading coefficient
produces a monic irreducible factor p(x) of f(x). By Theorem 5.11
there is an extension field that contains a root u of p(x) (and, hence,
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390 Chapter 11
Field Extensions
of /(x)). Furthermore, p(x) is necessarily the minimal polynomial of u.
11.7 [F(u): F] = degp(x) :!;; degf(x) =n.
- u)g(x) for some g(x)
E F(u)[x] . Since g(x) has degree n - 1, the induction hypothesis guar­
antees the existence of a splitting field K of g(x) over F(u) such that
[K:F(u)] :!;; (n - l)!. In K[x],
Consequently, by Theorem
The Factor Theorem 4.16 shows thatf(x) = (x
g(x) =c(x - ui)(x - ul)
and, henoe,f(x) =c(x
- u)(x - u1)
K = F(u)(u1,
•
•
•
•
·
• •
·
·
(x - u,,_1)
(x - u,._i). Since
, u,._i) =F(u, ui, ... , Ua-V
we see that K is a splitting field of f(x) over F such that
[F(u):F] :!;; ((n -
[K:F] =[K:F(u)]
l)!)n =n!. This completes the inductive step and the
proof of the theorem.
•
The relationship between two splitting fields of the same polynomial is quite easy
to state:
Any two splitting fields of a polynomial in Ffxf are isomorphic.
Surprisingly,
the easiest way to prove this fact is to prove a stronger result of which
this is a special case.
Theorem 11.14
Let u:F � E be an isomorphism of fields, f(x) a nonconstant polynomial in
f[x], and uf(x) the corresponding polynomial in f[x]. If K is a splitting field of
f(x) over F and L is a splitting field of uf(x) over £, then u extends to an
isomorphism K
If F
=
=
L.
E and u is the identity
map F � F, then the theorem states that any two
splitting fields of j(_x) are isomorphic.
Proof of Theorem 11.14 ... The proof is by induction on the degree of f(x ) . If
degf(x)= 1, then by the definition of splitting fieldf(x) =c(x - u) in
K [x] andK= F(u). Butf(x)=ex - roisinF[x], so we must have c
and cu in F. Hence, ''U = c-1cu is also in F. Therefore, K=F(u) =F. On
page 380 we saw that a extends to an isomorphism F[x] = E[x]; hence,
uf(x) also has degree 1, and a similar argument shows that E=L. In
this case, u itself is an isomorphism with the required properties.
Suppose the theorem is true for polynomials of degree n - 1 and that
f(x) has de gree n. As in the proof of Theorem 11.13,/(x) has a monic
irreducible factor p(x) in Ji{ x] by Theorem 4.14. Since u extends to an
isomorphism F[x] = E[x], (page 380), up(x) is a monic irreducible factor
of uf(x) in E[x]. Every root of p(x) is also a root of f(x), so K contains
all the roots of p(x), and similarly L contains all the roots of up(x). Let
u be a root of p(x) in K and v a root of up(x) in L. Then u extends to an
·
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...
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11.4
Splitting
Fields
391
isomorphism F(u) -+ E( v) that maps u to v by Corollary 11.8, and the
situation looks like this:
L
K
UI
UI
J<tu)
UI
F
=,
E{v)
UI
...J4
E.
The Factor Theorem 4.16 shows that/(x)
hence, in E{v)[x]
uf(x)
=
u(x - u)u g(x)
=
=
(x
(x - u u)ug(x)
Now f(x) splits over K, say f(x)
- u)g(x) in J<tu)[x] and ,
=
(x
- v)ug(x).
c(x - u)(x - ul)
(x - u,J.
(x - u,.). The
- u)g(x), we have g(x) c(x - �
smallest s ubfield containing all the roots of g(x) and the field F(u) is
F(u, u:z, .. . , u,,) K, so K is a splitting field of g(x) over F(u). Similarly,
Lis a splitting field of ug(x) over E(v). Since g(x) has degree n - 1, the
induction hypothesis implies that the isomorphism F(u) = E(v) can be
Since/(x)
=
=
(x
•
=
·
• ·
·
·
=
extended to an isomorphism K = L. T his completes the inductive step
and the proof of the theorem.
•
A splitting field of some polynomial over F contains all the roots of that poly­
nomial by definition. Surprisingly, however, splitting fields have a much stronger
property, which we now define.
Definition
An algebraic extension field Kof Fis normal provided that whenever an
irreducible polynomial in F[x] has one root in K, then it splits over K (that
is, has all its roots in K}.
Theorem 11.15
The field K is a splitting field over the field Fof some polynomial in F[x] if and
only if K is a finite-dimensional, normal extension of F.
Proof"' If K is a splitting field of f(x) EF[x], then K
F(u1o ... , u,,), where the
u1 are all the roots off(x). Consequently, [K:F] is finite by Theorem 11.10.
Letp(x) be an irreducible polynomial in F[x] that has a root v in K.
Consider p(x) as a polynomial in K{x] and let L be a splitting field of
p(x) over K, so that Fr,;;,. Kr,;;,. L. To prove thatp(x) splits over K, we n eed
=
only show that every root of p(x) in L is actually in K.
E
Let w ELbe any root of p(x) other than v. By Corollary 11.8 (with
F and u the identity map), there is an isomorphism F(v) = F(w) that
=
.......
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392 Chapter 11
Field Extensions
maps
v to w and maps every element of Fto itself. Consider the subfield
K(w) of L; the situation looks like this:
K
UI
F{v)
K(w)
UI
=
F{ w)
UI
F.
UI
F
=
Since
K(w)
we
=
F{u., ... , u,.)(w)
Ftu1o
=
.
.
,
•
Un,
w)
=
.F{w)(u1o ..., u,.)
K(w) is a splitting field of f(x) over Ftw). Furthermore, since
K is a splitting field of f(x) over F, K is also a splitting field
of f(x) over the subfield F(v). Consequently, by Theorem 11.14 the iso­
morphism F{v) = F(w) extends to an isomorphism K � K(w) that maps
v tow and every element of Fto itself. Therefore, [K:F]
[K(w):F] by
Theorem 11.5. In the extension chain Fl: Kl:K(w), [K(w):K] is finite by
Theorem 11.7 and [K:F] is finite by the remarks in the first paragraph of
the proof. So Theorem 11.4 implies that
see
that
v E Kand
=
[K:F]
[K(w):F]
=
=
[K(w):K][K:F].
[K:F] on each end shows that [K(w):K] 1, and, therefore,
K(w) K. But this means that w is in K. Thus every root ofp(x) in Lis
in K, andp(x) splits over K. Therefore, Kis normal over F.
Conversely, assume K is a finite-dimensional, normal extension of F
with basis { uh ..., u11} Then K
Ji(11.J,
, u,,). Each u.1 is algebraic
over F by Theorem 11.9 with minimal polynomial p,(x).Since eachpi(x)
splits over Kby normality,.f(x) p1(x)
p,,(x) also splits over K.
Therefore, K is the splitting field of f(x). •
Canceling
=
=
•
•
=
=
·
·
• •
·
EXAMPLE 5
The field
o("'2) contains the real root � of the irreducible polynomial
x3 - 2 E Cl![x] but does not contain the complex root "3'lw (as described in
Example 7 of Section 11.2). Therefore, 0( 4¢'2) is not a normal extension of Cl!
and, hence, cannot be the splitting field of any polynomial in Cl![x].
At this point it is natural to ask if a field
every polynomial in
Flx]
F has
an extension field over which
splits. In other words, is there an extension field that
contains all the roots of all the polynomials in
F[x]?
The answer is "yes," but the
proof is be yond the scope of this book. A field over which every nonconstant
polynomial splits is said to be algebraically closed. For example, the Fundamental
Theorem of Algebra and Corollary
4.28
show that the field C of complex numbers
is algebraically closed.
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11.4
Splitting Fields
393
If K is an algebraic extension of Fand K is algebraically closed, then K is called the
algebraic closure of F. The word "the" is justified by a theorem analogous to Theorem
11.14 that says any two algebraic closures of Fare isomorphic . For example, C is the
algebraic closure of� since C
R( i) is an algebraic extension of R that is algebraically
closed. The field C is not the algebraic closure of Q, however, since C is not alge­
braic over Q. The subfield E of algebraic numbers (see Example 7 of Section 11.3)
is the algebraic closure of Q (Exercise 20).
=
• Exercises
NOTE: Fis a field
v'2 is not in Q(i) and, henoe, C * 0(1). [Hint: Show that Vi
b E 0, leads to a contradiction.]
A. 1. Show that
with
a,
2. Show that
x2 - 3 and x2 -
splitting field, namely
=
a
+ bi,
- 2 are irreducible inO[x] and have the same
2x
o(\13).
x4 - 4x2 -
3. Find a splitting field of
4 overO.
5 over
0 and show that it
has dimension
4. If f(x) E R[x], prove that �or C is a splitting field of/(x) over R.
5. Let
K be a
show that
6. Let
f(x) over
splitting field of
K be a splitting field
of f(x) over F. If [K:F] is prime, u EK is a root of
f(x), and u ft F, show that K
7. If
u is
F. If Eis a field such that Fr;; Er;; K,
field off(x) over E.
K is a splitting
algebraic over Fand
=
K
=
F(u).
F(u)is a normal extension of F, prove that
K
is a splitting fe
i ld over F of the minimal polynomial of u.
8. Which of the following are normal extensions of Q?
(a)
0(\13)
o(�
(h)
(c)
o(vs, ;)
9. Prove that no finite field is algebraically closed.
a1
[Hint:
If the elements of the
... , a"' with a1 nonzero, consider
+ (x - a1)(x - ai)
(x - a,.) E F[x] .]
field Fare
ai.
·
·
·
B. 10. By finding quadratic factors, show that
x4
+ 2x3
- sx2 -
6x -
1 over
0.
0(V2, V3) is a splitting field of
11. Find and describe a splitting field of x4 + l overO .
12. Find a splitting field o f x4
(a)
over
0.
(b)
-2
over R.
13. Find a splitting field of x6 +
14. Show that
x3 + 1 over
0.
0(v'2, ;) is a splitting fei ld of x2 - 2V2x + 3 over 0(Vi).
15. Find a splitting field of x2
+ 1 over
16. Find a splitting field of x3
+
x
Z3•
+ 1 over
Z2•
...
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394 Chapter 11
Field Extensions
17. If Kis an extension field of Fsuch that [K:F]
=
2, prove that Kis normal.
18. Let F, E, Kbe fields such that F� E !;;;; Kand E =
Ji(u1,
• . •
,
)
u, ,
where the
u1
are some of the roots of f(x) E F[x]. Prove that Kis a splitting field of f(x)
over F if and only if Kis a splitting field of f(x) over E.
19. Prove that the following conditions on a field Kare equivalent:
(i) Every nonconstant polynomial in K[x] has a root in K.
(ii)
Every nonconstant polynomial in K[x] splits over K (that is, Kis
algebraically closed).
(iii) Every irreducible polynomial in K[x]has degree I.
(iv) There is no algebraic extension field of Kexcept Kitself.
20. Let K be an extension field of Fand Ethe subfield of all elements of Kthat
are algebraic over F, as in Corollary 11.12. If Kis algebraically closed, prove
that Eis an algebraic closure of F. [The special case when F = Q and K ::: C
shows that the field E of algebraic numbers is an algebraic closure of Q.]
21. Let Kbe an algebraic extension field of F such that every polynomial in Ji(x)
splits over K. Prove that Kis an algebraic closure of F.
C. 22. If Kis a finite-dimensional extension field of Fand cr:F-+ Kis a homomorphism
of fields, prove that there exists an extension field L of Kand a homomorphism
r:K...+ L such that r(a)
=
er(a) for every aEF.
23. Prove that a finite-dimensional extension field Kof Fis normal if and only if
it has this property: Whenever L is an extension field of Kand u:K-+ L an
injective homomorphism such that u(c)
Ill
= c
for every cEF, then u(K) s;;; K.
Separability
Every polynomial has a splitting field that contains all its roots . These roots may all be
distinct, or there may be repeated roots.* In this section we consider the case when the
roots are distinct and use the information obtained to prove a very useful fact about
finite-dimensional extensions .
Let F be a field . A polynomial/(x)EF[x] of d egree n is said to be separable if it
t
n distinct roots in some splitting field. Equivalently,/(x) is separable if it has no
has
repeated roots in any splitting field. If K is an extension field of F, then
uEKis
said to be separable over Fif
u
an element
is algebraic over Fand its minimal polynomial
J(x) EF[x]is separable . The extension field Kis said to be a separable extension (or to
be separable m•er
F) if every element of Kis separable over F. Thus a separable exten­
sion is necessarily algebraic.
= (x - u1) • • • (x - u.) in the splitting field and some u; = u1
with i4'j.
tsince any two splitting fields are isomorphic, this means that f(x) has n distinct roots in every
splitting field.
•A repeated root occurs when f(x)
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11.5
Separabl lity
395
EXAMPLE 1
The polynomial x2 + 1 E Q[x] is separable since it has distinct roots i and -i
in C. Butf(x) x"- x3 x + 1 is not separable because it factors as
(x - lf(x2 + x + 1). Hence,f(x) has one repeated root and a total of three
distinct roots in C.
=
-
There are several tests for separability that make use of the following concept. The
of
derivative
f(x) =Co+ CtX + c,.x1 +
.
.
+
.
c,,x"EF[x]
is defined to be the polynomial
f'(x)
=
c1
+
2c x + 3c3x2 +
2
·
·
·
+
nci.X"-1 EF[x]*.
You should use Exercises 4 and 5 to verify that derivatives defined in this algebraic
fashion have these familiar properties.
(f + g)'(x)
(fg)'(x)
=
=
f' (x) + g'(x)
f(x)g' (x) + f'(x)g(x).
Lemma 11.16
Let F be a field and f(x} EF[x]. If f(x) and f'(x) are relatively prime in F[x], then
f(x} is separable.
Note that the lemma operates entirely in F[x] and does not require any knowl­
edge of the splitting field to determine separability. For other separability criteria , see
Exercises 8-10.
Proof of Lemma 11.16 ... We shall prove the contrapositive: If f(x) is not separable,
thenf(x) and/'(x) are not relatively prime (which is logically equivalent
to the statement of the theorem).f Let Kbe a splitting field of f(x) and
suppose thatf(x) is not separable. Thenf(x) must have a repeated root u
in K. Hence,f(x) (x - u)2g(x) for some g(x) E K[x] and
=
f'(x)
=
(x
-
u)2g'(x) + 2(x - u}g(x).
Tberefore,f'(u) = OJ'K'(u) + On:(u) =Op and u is also a root of f '(x). If
AX) EF[x] is the minimal polynomial of u, then Ax) is nonconstant and
divides both/(x) andf'(x). Tberefore,f(x) andf'(x) are not relatively
prime. •
•when F =ii!, this is the usual derivative of elementary calculus.
But our definition is purely algebraic
be defined in
and applies to polynomials over any field, whereas the limits used in calculus may not
some fields.
tsee Appendix A (pages 503, 504 and 506) for the definition and use of the contrapositive in proofs.
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396 Chapter 11
Field Extensions
Recall that for a positive integer
nc is the
element
n and c E F,
c
+
c+
·
·
A field Fis said to have characteristic 0 if n1F
(},
+ c (n summands).
:#OF
for every positive
R, and Call have characteristic 0, but Z3 does not (since 3
field of characteristic 0 is infinite (Exercise
positive
•
n. For example,
1 = 0 in Z3J. Every
3). If Fhas characteristic 0,
then for every
n and c E F,
nc = c +
So
·
nc = OF if and
·
·
·
+ c = (lF
only if
+
·
·
·
+
IF)c = (nlF)c
c = OF- This fact is the key to separability in fields
of char­
acteristic 0:
Theorem 11.17
Let F be a field of characteristic 0. Then every irreducible polynomial in F[x] is
separable, and every algebraic extension field K of F is a separable extension.
The theorem may be false if F does not have characteristic 0 (Exercise 15).
Proof of Theorem 11.17 ... An irreducible p(x) EF[x] is nonconstant and, hence,
p(x) =ex" + (lower-degree terms),
with
c :# OF
and
n � 1.
Then
p'(x) = (nc)x"-1 +
(lower-degree terms),
w ith
nc :#Op.
Therefore, p'(x) is a nonzero polynomial of lower degree than the
irreducible p(x). So p(x) and p'(x) must be r elatively prime. Hence, p(x)
is separable by Lemma 11.16. In particular, the minimal polynomial of
each
u EK is separable. So K is a separable extension.
•
Separable extensions are particularly nice because every finitely generated (in
particular, every finite-dimensional) separable extension is actually simple:
Theorem 11. 18*
If K is a finitely generated separable extension field off, then K
= F(u) for
someuEK.
Proof ... By hypothesis K = F{u1,
, u,,). The proof is by induction on n. There
n = 1 and K = F{u1). In the next paragraph we
shall show that the theorem is true for n = 2. Assume inductively that it
is true for n = k - 1 and suppose n = k. By induction and the case n = 2,
there exist t, u EK such that
• • •
is nothing to prove when
K = F(uu
•
•
.
,
uk) = F(u1,
•
•
•
, uk_1)(u,J = F(t)(u.J = F(t, ukl
=
F(u).
*This theorem wil I be used only in Section 12.2.
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11.5
To complete the proof, we assume K
Separability
397
= F(v, w) and show that K is
a simple extension of F. Assume first that Fis infinite (which is always
the case in characteristic 0 by Exercise 3). Let p(x) E F[x] be the minimal
polynomial of v and q(x) E F[x] the minimal polynomial of
splitting field of p(x)q(x) over F. Let w
w. Let L be a
, w,, be the roots of
the w1 are distinct. Let
= wi. w2,
• • •
q(x) in L. By the definition of separability, all
v = v11 v2,
, vm be the roots of p(x) in L. Since Fis infinite, there exists
• • •
c E Fsuch that
Vt - V
c# --­
w-w1
for all
1 s i s m, 1
< j s n.
Let u = v + cw. We claim that K = F(u). To show that wE F(u), let
h(x) = p(u - ex) EF(u)[x] and note that w is a root of h(x):
h(w) = p(u -
cw) = p(v)
=
Op
:f. 1) is also a root of h(x). Then p(u - cwj) =
- cw1 is one of the roots of p(x), say u - cw1 = vr Since
Suppose some w1 (with/
0"' so that
u
u = v + cw,
v
+ cw
we would have
- cw1 = v1
or, equivalently,
c
v,- v
w-w,
= ---.
This contradicts(•). Therefore, w is the only common root of q(x) and h(x).
Let
r(x) be the minimal polynomial of w over F(u). Then r(x)
q(x), so that ever y root of r(x) is a root of q(x). But r(x) also
divides h(x), so all its roots are roots of h(x). By the preceding para­
graph, r(x) has a single root w in L. Therefore, r(x) EF(u)[x] must have
degree 1, and, hence, its root w is in F{u). Since v = u - cw, with u,
wEF{u), we see that vEF(u) and, hence, K = F{v, w) !;F(u). But
u = v +cw E K, so F{u) !; K, whence K = F{u). This completes the
proof when Fis infn
i ite. For the case of finite F, see Theorem 11.28 in
divides
the next section.
•
EXAMPLE2
( v'3, v'5), we have v = v'3, Vz = -v'3,
- VS, so we can choose c = 1. Then u = v'3 + v'5 and
o(v'3, VS) is the simple extensiono( v'3 + v'S).
Applying the proof of the theorem too
w = VS, ·Wi =
• Exercises
NOTE: K is an
extensionfield of the field F.
A. 1. If K is separable over Fand Eis a field with F!;; E!; K, show that K is
separable over E.
2. If F has characteristic 0, show that K has characteristic 0.
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398 Chapter 11
Field Extensions
3. Prove that every field of characteristic 0 is infinite. [Hint: Consider the elements
nlp with nEZ, n > O.]
B.
4. If f(x), g(x)EF[x], prove
(a) (f + g)'(x) ""f'(x) + g'(x).
(b)
IfcEF, then (cf)'(x)
=
cf'(x).
5. (a) If f(x) =ex!'EF[x] and g(x) b0 + b1x + · · + bkx"EF[x], prove that
(fg)'(x) f(x)g'(x) + f'(x)g(x).
=
·
=
(b)
f(x)g'(x) +
If f(x) , g(x) are any polynomials in F[x], prove that (fg)'(x)
f'(x)g(x). [Hint: If/(x) = ao + a1x + · · · + a,.X', then (fg)(x)
aog(x) +
a1 xg(x) + · · · + a,.X'g(x); use part (a) and Exercise 4.]
=
=
6.
If f(x) EF[x] and n is a positive integer, prove that the derivative of f(x'f is
nf(x)H-1f'(x). [Hint: Use induction on n and Exercise 5.]
7.
(a)
If Fhas characteristic O,f(x) EF[x], andf'(x)
somecEF.
=
Op, prove that/(x)
=
c for
(b) Give an example in Z2[x] to show that part (a) may be false if Fdoes not
have characteristic 0.
8.
Prove that u EK is a repeated root of f(x)EF[x] if and only if u is a root of
both/(x) and/ '(x) . [Hint:f(x)
(x - uf'g(x) with m � 1, g(x) E K[x], and
g(u) =F Op, u is a repeated root of f(x) if and only if m > 1. Use Exercises 5 and
6 to computef'(x).]
=
9.
IO.
Prove thatf(x)EF[x] is separable if and only if/(x) and/'(x) are relatively
prime. [Hint: See Lemma 11.16 and Exercise 8.]
Let p(x) be irreducible in F[x]. Prove that p(x) is separable if and only if
p'(x) ¢Op.
11. Assume Fhas characteristic 0 and K is a splitting field of f(x) EF(x]. If d(x)
is the greatest common divisor of f(x) andf'(x) and h(x) =f(x)/d(,x) EF[x] ,
prove
(a) f(x) and h(x) have the same roots in K.
(b) h(x) is separable.
12. Use the proof of Theorem 11 .18 to express each of these as simple extensions
ofQ:
(a)
o(v'2. \/3)
Ch>
o('\/3, ;)
(c)
o(v'2, '\/3, v5)
13. If p and q are distinct primes, prove that o('\/i, "\/q)
=
a( Vi + v'q).
14. Assume that Fis infinite, that v, w EK are algebraic over F, and that w is the
root of a separable polynomial in F[x]. Prove that F{v, w) is a simple extension
of F. [Hint: Adapt the proof of Theorem 11.18.]
15. Here is an example of an irreducible polynomial that is not separable. Let
F Z2(t) be the quotient field of .li[t] (the ring of polynomials in
=
........ftom.1M•Bam:.ndkir�.Bdbmbll_...._
..--il......._..:dPLI�........
eap,ngm.20:12�1...umiq.A:l.lliala a--a.....,-aa1n. t:IDJllilrd,. IC....t,, ar�-.-... arm,_,. 0..1"�dpll.-mllnl.�1r1C11Hm.�M
........ q-��
fld.�d'5ct-�...-.......,..c...,.�._.... rigbl1D...,,,..��-
...
..
11.6
the indeterminate
Section
(a)
t
Finite Fields
with coefficients in Z.J), as in Example
1
399
of
10.4.
Prove that
Xl -
tis an irreducible polynomial in F[x].
[Hint: If Xl - t
has a root in F, then there are polynomials g(t), h(t) in Z2[t] such that
fg(t)/h(t)]2 = t; this leads to a contradiction; apply Corollary 4.19.]
(b)
Prove that
zero
Ill
x2 - t EF[x] is not separable. [Hint:
Show that its derivative is
and use Exercise 10.]
Finite Fields
F inite fields have applications in many areas, including projective geometry, combina­
tories, experimental design, and cryptography. In this section, finite fields are charac­
terized in terms of field extensions and splitting fields, and their structure is completely
determined up to isomorphism.
We begin with some definitions and results that apply to rings that need
not
be
fields or even finite. But our primary interest will be in their implications for finite
fields.
R be a ring with identity. Recall that for a positive integer m and c ER, me is
c +
+ c (m summands). The ring R is said to have characteristic 0 if
mlR #:OR for every positive m. On the other hand, if mlR =OR for some positive m,
then there is a smallest such m by the Well-Ordering Axiom. Then R is said to have
characteristic n if n is the smallest positive integer such that nlR = OR.* For example, Q
has characteristic 0 and Z3 has characteristic 3.
Let
the element c +
·
·
·
Lemma 11.19
If R is an integral domain, then the characteristic of R is either O or a positive
prime.
Proof'" If R has characteristic 0, there is nothing to prove. So assume R has
0. If n were not prime, then there would exist positive
k, t such that n = kt, with k < n and t < n. The distributive laws
characteristic n >
integers
show that
k summands
= lRlR
+
•
•
•
+
tsummands
lRlR = lR
+
•
•
'
+
lR
[kt summands]
= (kt)lR = nl R = OR-
"If
you have read Chapter
7,
you will recognize that when the characteristic
simply the order of the element 1R in the additive group
of R
is positive, i t is
of R.
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...
400 Chapter 11
Field Extensions
klR =ORor tlR =OR,contradict­
nis the smallest positiveinteger such that nlR =OR.
Since Ris anintegral domain either
ing the fact that
Therefore,
nis
prime.
•
Lemma 11.20
Let R be a ring with identity of characteristic n
if n lk.*
> 0. Then
k1R =OR if and only
Proof•Ifn lk,say k =nd, thenklR = ndlR = (nlR)(dl.R) =OR ( dl.R) =OR.
klR = OR. By the Division
n. Now nlR =OR,so that
Conversely,suppose
with 0 :$ r <
Since r < nand
nis the smallest positive
k = nq + r
nlR = ORby
= 0. Therefore, k = nq
integersuch that
the definition of characteristic,we must have
andn lk.
Algorithm,
r
•
Theorem 11.21
Let R be a ring with identity. Then
(1) The set P = {k1RlkEZ} is a subring of R.
(2) If R has characteristic 0, then P = Z.
{3} If R has characteristic
n >
0, then P = Zn .
Proof ... De:fine f :Z __., R by f(k) = klR. Then
f(k + I) = (k + t)lR = klR + tlR =f(k)
+ f(t).
The distributive laws (as in the proof of Lemma 11.19)show that
f(kt) =(kt) lR = (kl.R)(tl .R) = f(k){(t).
Therefore,/is a homomorphism. The image of
f ispreciselythe set P,
,fcan be con­
from Zonto P. Then P = Z/�f)
and,therefore, Pis a ring by Co rollary 3.11. C
sidered as a surjectivehomomorphism
onsequen
tly
by the First Isomorphism Theorem 6.13. If Rhas characteristic 0, then
ksuch that klR =ORis k =0. So the kernel of fis the
(0)in Z,and P = Z/(O) = Z. If Rhas characteristic n > 0, then
Lemma 11.20 shows that the kernel of fis the principal ideal (n)consist­
in
g ofall multiples of n. Hence, P = Z/(n ) = Z,.. •
the onlyintege r
ideal
•This lemma is just a special case (in additive notation) of part (1) of Theorem 7.9, with a = 1R and
e=OR.
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....
...
......
...
11.6
According to Theorem
11.21
Finite Fields
401
a field of characteristic 0 contains a copy of Zand,
hence, must be infinite. Therefore, by Lemma 11.19 we have
Corollary 11.22
Every finite field has characteristic p for some prime p.
The converse of Corollary
characteristic p (Exercise
8).
11.22 is false,
however, since there are infinite fields of
K is a field of prime characteristic p (in particulai; if K is finite), then Theorem 11.21
Zr This field P is called the prime
subfield of K and is contained in every subfield of K (because every subfield contains
1 x and, hence, contains tlx for every integer t). * See Exercise 4 for another description
of P. We shall identify the prime subfield P with its isomorphic copy Z1; then
If
shows that K contains a subfield P isomorphic to
every field of characteristic p contains z,.
The number of elements in a finite field K is called the
order of a finite field
prime subfield
K of
characteristic p, we consider
order of K. To determine the
K as an extension field of its
Z1:
Theorem 11.23
A finite field Khas order pn, where p is the characteristic of Kand n
Proof" There is certainly a finite set of
elements that spans K over
= [K: Zp].
z, (the
set K
itself, for example). Consequently, by Exercise 32 of Section 11.1, K has
a fi nite basis {ut,
u2,
•
•
•
,
u,.}over z,. Every element of
K can be written
uniquely in the form
with each c1E Z1 by Exercise 30 of Section 11.1. Since there are exactly p
possibilities for each
of the form(•). So
basis =
Theorem
[K:Z1].
11.23
c1,
there
are
precisely p" distinct linear co mbinations
K has order p",
with n = number of elements in the
•
limits the possible size of a finite field. For instance, there can­
not be a field of order 6 since 6 is not a power of any prime. It also suggests several
questions: Is there a field of order JI' for every prime p and every positive integer n?
"If K
has characteristic
0,
then
K
P of Z. Since K contains the
K contains a copy of the field
prime subfield) is contained in every
contains an isomorphic copy
multiplicative inverse of every nonzero element of P, it follows that
Cll.
p, this field (called the
10.31 (with R = P "'Zand F"' Q) for a more precise statement and proof.
As in the case of characteristic
subfield of K. See Theorem
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402 Chapter 11
Field Extensions
How are two fields of order p" related? The answers to these questions are given in
Theorem 11.25 and its corollaries. In order to prove that theorem, we need
a
techni­
cal lemma.
Lemma 11.24
The Freshman's Dream*
Let p be a prime and Ra commutative ring with identity of characteristic p.
Then for every a, b ER and every positive integer n,
Proof" The proof is by induction on n. If n =
1, then the Binomial Theorem in
Appendix E shows that
(a+ b)P = d' +
+
...
�Y + (�y__,11
( )a/:l"""t
'h
+
Each of the middle coefficients
.. .
+
p
+ Y.
p-1
e)
=·rl (pp
� r)! is an integer by
Exercise 6 in Appendix E. Sinoe every term in the denominator is strictly
less than the prime p, the factor of p in the numerator does not cancel, and,
therefore,
e)
is divisible by p, say
(�)
=
tp. Sinoe R has characteristic p,
Thus all the middle terms are zero and (a + hY
d' + lf'. So the theo­
rem is true when n = 1. Assume the theorem is true when n = k. Using
=
this assumption and the case when n = 1 shows that
(a+ h)r'
=((a+ b),-.)P
= (al + bty' = (a")P + (al')P = al"' +bl'".
Therefore, the theorem is true when n = k + 1 and, hence, for all
induction.
•
n
by
*Terminology due to Vincent 0. Mc Brien.
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11.6
Finite Fields
403
Theorem 11.25
Let K be an extension field of Zp and n a positive integer. Then K has order
11
p if and only if K is a splitting field of x'1' - x over Zp.
Proof" Assume Kis a splitting field of f(x) = f - x EZ,(x). Since
J' (x) = p"x'"-l - 1 = o:r-1 - 1 = -1,.f(x) is separable by
Lemma 11.16. Let Ebe the subset of Kconsisting of the p" distinct roots
of
;/' -
x.
Note that
c
EE if
(a
Therefore,
a
+
b EE,
a
=
ti'
+
II'
= a
+
b.
and Eis closed under addition. The set Eis closed
under multiplication since
in E. If
b)1"
+
cp' = c. We shall show that the
b EE, then by Lemma 11.24.
and only if
set Eis actually a subfield of K. If a,
(ab)P" = r/ll" =ab. Obviously, Ox and lxare
is a nonzero element of E, then
-a
and a-1 are in Ebecause,
for example,
The argument for
-a
is similar (Exercise 7). Therefore, Eis a subfield of
K. Since the splitting field Kis the smallest subfield containing the set E
of roots, we must have K = E. Therefore, Khas order p".
Conversely, suppose Khas order p". We need only shmv that every ele­
ment of Kis a root off
-
x,
for in that case, the p" distinct elements of
*
ff - x. Clearly Ox
is a root, so let c be any nonzero element of K. Let c., c2,
, ck be all the
nonzero elements of K(where k p" - 1 and c is one of the cJ and let u be
the product u = c1c2c3
ck. The k elements cc1o cc2,
, eek are all dis­
tinct (since cc1 = cc1 implies c1 = c1) , so they are just the nonzero elements
Kare all the possible roots and Kis a splitting field of
•
• •
=
•
•
•
• • •
of Kin some other order, and their product is the element
u
= (cc1)(cc0
·
•
•
( ccx) = d'(c1c2c3
•
•
•
u.
Therefore,
CAJ = d'u.
t!' = 1 x and, hence, d'+1 = c, or equivalent
- c =Ox. Since k + 1 p", c is a root of xt" - x. •
Canceling u shows that
ck+1
Theorem
11.25
=
has several important consequences; together with the theorem
they provide a complete characterization of all finite fields.
Corollary 11.26
For each positive prime p and positive integer n, there exists a field of order pl1.
Proof" A splitting field of x" - x over z, exists by Theorem 11.13; it has order
p" by Theorem 11.25 •
"A short proof, using group theory, is given in Exercise 22.
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404 Chapter 11
Field Extensions
Corollary 11.27
Two finite fields of the same order are isomorphic.
Proof• If Kand L are fields of
order JI', then both
over z, by Theorem 11.25 and, hence,
(with a the identity map on
Z,).
are
are
splitting fields of
xP" -
x
isomorphic by Theorem 11.14
•
According to Corollary 11.27, there is (up to isomorphism) a unique field of order
JI'. This field is called the Galois field of order JI'. We complete our study of finite fields
with two results whose proofs depend
on
group theory.
Theorem 11.28
Let K be a finite field and Fa subfield. Then K is a simple extension of F.
Proof • By Theorem 7 .16 the multiplicative group of
nonzero elements of
K is cyclic. If u is a generator of this group, then the subfield Ji(u)
contains Op and all powers of
Therefore,K=
Ji(u).
u
and, hence, contains every element of K.
•
Corollary 11.29
Let p be a positive prime. For each positive integer n, there exists an
irreducible polynomial of degree n in
Zp[x].
Proof• There is an extension field K of Zp of order JI' by Corollary 11.26. By
Theorem 11.28, K = Zp(u) for some u EK. The minimal polynomial of u
in Z,[x] is irreducible of degree [K:Z,] by Theorem 11.7. Theorem 11.23
shows that [K:Zp] =
n.
•
• Exercises
A. 1. If Ris a ring with identity and
m, n EZ, prove that (ml.R)(nl.R) = (mn)lR.
ffhe case of positive m, n was done in the proof of Lemma 11.19.]
2. What is the characteristic of
(b) Z2 X Z6
(a) Q
(d) M(R)
3. Let R be a
every
a
(e) M(Z3)
ring with identity of
ER.
characteristic
n ;:?:
0. Prove that na = OR for
4. If K is a field of prime characteristic p, prove that its prime subfield is the
intersection of all the subfields of K.
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11.6
Finite Fields
405
5. Let Fbe a subfield of a finite field K. If Fhas order q, show that Khas order
tj', where n = [K:F].
6. Show that a field Kof order JI' contains all kth roots of
7. Let Ebe the set of roots of xP"
prove that -a EE.
B.
-
1 K• where k
=
JI'
-
1.
x EZ,[x] in some splitting field. If aEE,
8. Letp be prime and let Z,(x) be the field of quotients of the polynomial ring
Z,[x](as in Example 1 of Section 10.4). Show that Zp(x) is an infinite field of
characteristic p.
9. Let R be a commutative ring with identity of prime characteristicp . Ifa,
b ER and n::?: 1, prove that
(a -
b )p"
=
al"'
-
bf1".
I 0. Let Kbe a finite field of characteristic p. Prove that the map /:K--+> K given by
f(a)
d' is an isomorphism. Conclude that every element of Khas apth root
=
in K.
11. Show that the Freshman's Dream (Lemma 11.24) may be false if the
characteristicp is not prime or if R is noncommutative. [Hint: Consider Z,,
andM(Z:z).]
12. If c is a root off(x) E Z,,[x], prove that d' is also a root.
13. Prove Fermat's Little Theorem: If pis a prime and aEZ, then cf =a (mod p) . If
a is relatively prime top, then d'-1 = 1 (modp).[Hint: Translate congruence
statements in Z into equality statements in z, and use Theorem 11.25.]
14. Let Fbe a field and/(x) a monic polynomial in F[x] , whose roots are all
distinct in any splitting field K. Let Ebe the set of roots of f(x) in K. If the set
E is actually a subfield of K, prove that Fhas characteristicp for some prime p
and that f ( x )
x1' - x for some n � 1.
=
15. (a) Show that x3 + x + 1 is irreducible in Z2[x]and construct a field of
order 8.
(b)
Show that X3
-
x + 1 is irreducible in Z3[x]and construct a field of order 27.
(c) Show that x4 + x + 1 is irreducible in Z2[x] and construct a field of
order 16.
16. Let Kbe a finite field of characteristic p, Fa subfield of K, and m a positive
integer. If L
{a E K I aP� E.F}, prove that
=
(a)
(b)
Lis a subfield of Kthat contains F.
L
=
g(a )
F.
[Hint: Use Exercise 10 to show that the mapg:K--+> K given by
=
d""i s an isomorphism such thatg( F)
=
F. What isg-1(F )?]
17. If E and Fare subfields of a finite field Kand E is isomorphic to F, prove that
E=F.
18. Let Kbe a field and k, n positive integers.
(a)
Prove that JI<
lx divides x" - lx in K[x]if and only if k In in Z.
r by the Division Algorithm; show that x" - lx
+ x"-fk.]
(.xk - lx)h(x) + (x' - lx), where h(x) x"-k + x"-'Jk +
[Hint: n
=
-
kq +
=
=
·
· ·
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406 Chapter 11
Field Extensions
(b)
Ifp � 2 is an integer , prove that (I' - 1) I (p" - 1) if and only if k I n.
[Hint: Copy the proof of part (a) withp in place of x.]
19. Let Kbe a finite field of order p".
(a)
20.
'
If Fis a subfield of K, prove that Fhas orderp for some d such that d In.
[Hint: Exercise 18 may be helpful.]
(b} If d In, prove that Khas a unique subfield of order p". [Hint: See Exercise 17
and Corollary 11.27 for the uniqueness part.]
Letp be prime and/(x) an irreducible polynomial of degree 2 in Z,,[x]. If Kis
3
an extension field of z, of orderp , prove that/(x) is irreducible in K[x].
21. Prove that ever y element in a finite field can be written as the sum of two
squares.
22. Use part
(2) of Corollary 8.6 to prove that every nonzero element c of a finite
field Kof order p" satisfies ci'- l
=
lx. Conclude that
and use this fact to prove Theorem 11.25.
Application
[
BCH
c
is a root of :xi'"
-
x
codes (Section 16.3) may be covered at this point if desired.
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C H A P T E R
12
Galois Theory
A major question in classical algebra was whether or not there were formulas for
the solution of higher-degree polynomial equations (analogous to the quadratic
formula for second-degree equations). Although formulas for third- and fourth­
degree equations were found in the sixteenth century, no further progress was
made for almost 300 years. Then Ruffini and Abel provided the surprising answer:
There is no formula for the solution of all polynomial equations of degree n when
n <?: 5. This result did not rule out the possibility that the solutions of special types
of equations might be obtainable from a formula. Nor did it give any clue as to
which equations might be solvable by formula.
It was the amazingly original work of Galois that provided the full explanation,
including a criterion tor determining which polynomial equations can be solved
by a formula Galois' ideas had a profound influence on the development of later
mathematics, far beyond the scope of the original solvability problem.
The solutions of the equation f(x)
O lie in some extension of the coefficient
field of f(x). Galois' remarkable discovery was the close connection between such
=
field extensions and groups (Section 12.1). A detailed description of the connec­
tion is given by the Fundamental Theorem of Galois Theory in Section 12.2. This
theorem is the principal tool for proving Galois' Criterion for the solvability of
equations by formula (Section 12.3).
Im
The Galois Group
The key to studying field extensions is to associate with each extension a certain group,
called its Galois group. The properties of the Galois group and theorems of group
theory can then be used to establish important facts about the field extension. In this
section we define the Galois group and develop its basic properties. Throughout this
section Fis afield
407
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408 Chapter 12
Definition
Galois Theory
Let Kbean extension field of F. An f-automorphism of K is an isomor­
phism u:K ...+KthatfixesF elementwise(that is,
u(c) = cforeverycEF).
The set of all F-automorphisms of K is denoted Gal.J( and is called the
Galois group of Kover f.
The use of
the word "group" in the definition is justified by:
Theorem 12.1
If K is an extension field of
F, then GalFK is a group under the operation of
composition of functions.
Proof� GalpKis nonempty since the identity map i:K...+Kis
phism.* If
u, 'TE GalpKthen u
0
an automor­
'T is an isomorphism from Kto K
by Exercise 27 of Section 3.3. For each c E F,
u(c) =
c.
Hence ,
(u o 'T)(c) = u(T(c)) =
u o 'TE Ga!FK, and GalpKis closed. Composition of
functions is associative, and the identity map i is the identity element of
GalFK. Every bijective function has an inverse function by Theorem B. l
in Appendix B. If
by Exercise 29 of
(Exercise
u E GalFK, then u-1 is an isomorphism from Kto K
1
Section 3.3. Verify that u- (c) = c for every c EF
1). Therefore, u-1 EGalFK, and GalpKis a group.
•
EXAMPLE 1.At
The complex conjugation map u:C-+ C given by u(a + bl) = a - bi is an auto­
morphism of C, as shown in Example 3 of Section 3.3. For every real number a,
u(a)
So
u
= u(a + Ot) = a - Oi = a.
-i are the roots of i1' + 1 ER and that u maps
u(1) = i and u(-() = i. This is an example of the
is in Ga!RC. Note that i and
these roots onto each other:
-
next Theorem.
Theorem 12.2
f(x) Ef{x].
of f(x).
Let K be an extension field of F and
u EGalFK, then u(u) is also a root
If u EK is a root of
f(x)
and
*Throughout this chapter,• denotes the identity map on the field under discussion.
ti'hroughoutthis section and the next, three basic examples appear repeatedly. The first appearance
of Example
1
is labeled
labeled 2.A, and so on.
1.A,
its second appearance
1.B,
etc.; the first appearance
of
Example
2
is
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-..d.1111my��"'*-001.-.d.n,'dl9cl.b�.--.....---.��---ftgbl:ID-�ICllllMll:- .. tim9��:Dgb&l� ...... it.
12.1
Proof .. If f(x) = c0+ c1x + c2i'-+
·
·
·
u is
409
+ c,.r', then
c0 + c 1u + c2u2 +
Since
The Galois Group
·
·
·
a homomorphism and u( cJ
"
c,,u
+
= OF'
= ci for each c1EF,
OF= u(Op) = u(c0 + c1u + c1u2 +
+ c,.u")
= u(co) + u(c1)u(u) + u(c2Ju(u)2 +
+ u(c,.)u(u'f
2
+ c11u(u'f = f(u( u)).
co+ c1u(u) + c2u(u) +
•
·
·
·
=
Therefore,
Let
·
u(u) is a root off(x).
·
·
·
·
•
u EK be algebraic over F with minimal
polynomial p(x)E.F[x]. Theorem 12.2
states that every image of u under an automorphism of the Galois group must also be
a root of p(x). Conversely, is every root of p(x) in
morphism of GalpK? Here is
one
K the image of u under
some auto­
case where the answer is yes.
Theorem 12.3
Let
K be
the splitting field of some polynomial over F and let u,
there exists
uEGalFK such
vEK. Then
that u(u) = v if and only if u and v have the same
minimal polynomial in F[x].
Proof• If u and v have the same minimal polynomial, then by Corollary 11. 8
u:F(u) = F(v) such that u(u) = v, and u fixes
K is a splitting field of some polynomial over F,
it is a splitting field of the same polynomial over both F (u) and F(v).
Therefore, u extends to an F-automorphism of K (also denoted u) by
Theorem 11.14. In other words, uEGalpK and u(u) = v. The converse is
there is an isomorphism
F elementwise. Since
an immediate consequence of Theorem 12.2.
•
EXAMPLE 1.B
Example l .A shows that GalnC has at least two elements, the identity map i and
the complex conjugation map u. We now prove that these are the only elements
T be any automorphism in GalnC. Since i is a root of x2 + 1,
T(i) = ±i by Theorem 12.2. If T(t) = i, then since T fixes every element of Ill,
in Gal0C. Let
T(a + bi) = T(a) + T(b)T(1) = a + bi,
and, hence, T = i. Similarly, if T(t ) =
- i,
then
T(a + bi) = T(a) + T(b)T(i) = a + b(-1) = a - bi,
and, therefore,
T = u. Thus GalRC =
{i,
u}
is a group of order 2 and, hence,
isomorphic to Z2 by Theorem 8. 7.
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410
Chapter 12
Galois Theory
The preceding example shows that an Bl-automorphism of C=R(i) is completely
determined by its action on i. The same thing is true in the general case:
Theorem 12.4
Let K=f(u1, ... , Un) be an algebraic extension field of F. tf u, TE GalfK and
u(u1)
'T(u1} for each i= 1, 2, . . . , n, then u =T. In other words, an auto­
morphism in GalFK is completely determined by its action on u11
, Un.
=
•
•
•
T-1 o uEGalp[(. We shall show that f3 is the identity map i.
Since u( u;) = T(u;) fo r every i,
Proof.. Let f3
=
1
1
{3(u1) =('T-t 0 u) (ut) =T- (u(ui)) =T- (T(uJ) = (t-1 0 T)(u i)
=
Let vEF(u1). By Theorem 11.7 there exist
c1EF such that v = c0 +
1
· · · + cm_1ut"'- , where mis the degree of the minimal polynomial of
homomorphism that fixes u1 and every element of F,
i(u;}=u1•
+ Cffe-12 +
Since f3 is a
c1u1
u1•
{3(v) ={3(c0 + c1u1 + c2u12 + · · · + c,._1ut'�1)
{3(c0) + {3(c1){3(u 1) + {3(c-2){3(u i2) + • · + f3(c,._1)f3(u 1m-I)
+ c,._1u1 •-1
c0 + c1 u1 + c7:ul· +
v
=
•
=
·
·
·
=
Therefore, {3(v)=vfor every vE F (u1). Repeating this argument with F(u1) in place of
vfor everyvEF(u1)("7) F ( ui. ui). Another
repetition, with F(uh t,1.'2) in place of F and u3 in place of uh shows that {3(v)
v for
every v EF(tti 1 u,, U]). After a finite number of repetitions we have {3(v)
v for every
1
v EF(u1, u,, . . . , u,J
K, that is, i=f3=T- o u. Therefore,
Fand uiin place of u1 shows that{3(v)
=
=
=
=
=
'T='Toi='To ('T-l
o a) =(To 'T-1) o U =i o U=u.
•
EXAMPLE 2.A
0(v'3,V'S) over Q
v'3 to v'3 or -\13, the roots of x2 - 3. Similarly, it must take VS to
±VS, the roots of x2 - 5. Since an automorphism is completely determined by
its action on \13 and '\15 by Theorem 12.4, there are at most four automorphisms
in GaloO(\13,v'S), corresponding to the four possible actions on v'3 and VS:
By Theorem 12.2 any automorphism in the Galois group of
takes
We now show that G�O
(v'J,v'5) is a group of order 4 by constructing non­
identity automorphisms T, a,[3 with these actions. To construct T, note that x2 - 3
v'3 and -'\/3 over Q. By Corollary 11.S,
= -'\13, and u
Example 6 of Section 11.3 shows that :x?- - 5 is the mini­
is the minimal polynomial of both
there is art isomorphism u:C(
fixes Q elementwise.
mal polynomial of
...
v'3) Q(-v'3) such that a( v?)
=
v'5 over 0(V3). By Corollary 11.8 again, u extends to
......
....
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-----..,.��dou.ad........UU,-.dlM:l.
�
�c.-g..p�----rlgbtlD....,,,.�o:ldlllllt.it���...-. ..
12.1
Q(
V's).·
u(
TheGaloisGroup
411
v'5)
a 0-automorphism T of
'\13)(\/S) 0( \/'3,
such that T(
= v's.
Therefore, TE GalQO(v'3, v'5) and T(v'3)=
v'3) = -v'3 and T(v'5) = vs.
A similar two-step argument produces automorphisms and f3 with the actions
listed above. Furthermore, each of T, f3 has order 2 in GalQO( v'3, VS); for
instance,
=
a
a,
v'J=
(PT)('\13)= T(T('\13))= T(-'\13) = -T(\1'3) = -(-'\13) =
and (7' o
T)(v'S) =
VS
= i (v'S).
Therefore,
To T = i
Use Theorem 8.8 to conclude that Galq0(\/'3,\/5)
=
i('\13)
by Theorem 12.4.
Z1 X Z or compute
2
the operation table directly (Exercise 4). For instance, you can readily verify
that (T <>
= f:J(v'3) and (To
= f:J(v'S) and, hence, To = f3 by
Theorem 12.4.
a)(V3)
In
the
a
a)(v'5)
preceding example, 0(\1'3,v's)
is the
splitting field
of f(x)
=
(x2 - 3)(x 2 - 5), and every automorphism in the Galois group permutes the four roots
\/'3, -'\13, VS, -v's of/(x). This is an illustration of
Corollary 12.5
If K i s the splitting field of a separable polynomial f(x) of degree n in F[x], then
GalFK is isomorphic to a subgroup of Sn.
Proof.. By separabilityf(x)
has n distinct roots in K, say
Sn to be the group of permutations of the set
u1o
•
.
R = {u1o
.
,
un. Consider
, un}· If a E
• . •
Galp/(, then u(u1), a(u,_),
, u(u,,) are roots of f(x) by Theorem 12.2.
Furthermore, since a is injective, they are all distinct and, hence, must be
ui. u2, , un in some order. In other words, the restriction of a to the
set R (denoted a IR) is a permutation of R. Define a map 8:Ga1pK-+- Sn
by
= IR. Since the operation in both groups is composition of
functions, it is easy to verify that 8 is a homomorphism of groups.
K = F (uh ... , u,,) by the definition of splitting field. If aIR = TI R, then
a(uJ = T(u� for every i, and, hence, a= T by Theorem 12.4. Therefore,
8 is an injective homomorphism, and thus GalJc is isomorphic to Im 8, a
subgroup of Sn, by Theorem 7.20. •
.
•
.
• • •
fJ(u) u
If
K is the splitting field of f(x), we shall usually
identify GalpK with its isomorphic subgroup in s.
by identifying each a utomorphism with the permutation it induces on the roots of
f(x) .
EXAMPLE 3.A
Let K be the splitting field of x3 - 2 over O. Verify that the roots of x3 - 2 are
'\o/2, �2w, �2w2, where w = (-1 + v'3i)/2 is a complex cube root of 1. Then
�is a subgroup of S3• By Theorem 12.3, there is at least one automorphism
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.
ICOl
dllklDlil.
lll!ml•_..,.limlo��:Dgbb�...-.:lit.
412
Chapter 12
Galois Theory
u that maps the first root
°\Y2ru2to itself
tation (12)
or
or to
"312to the second "3'2ru; it must take the thirdroot
thefirst root ef2by Theorem 12.2. So u is either the permu­
(123) in S3•
WhenKis thespil ttingfield of apolynomialflx)EF[x,] then by
Corollary 125 every element of GalFK produces a permuta­
CAUTION:
tion of the roots off(x), but not vice versa: A permutation
of the roots need not come from an F-automorphism of K.
For example, 0( v'J, v'5) is a splitting fe
i ld of flx)
(x1 - 3Xx1 - S), but by Example 2A there isno 0-automorphism
=
of
o(v'3,VS) that gives this permutation of the roots
v'3
J.
-v'3
J.
\/5
-\/5
Let Kbe an extension field of F. A field Esuch that Fr;:;Er;:; Kis called an interme­
as an extension of E. The
diate field of the extension . In this case, we can consider K
Galois group GalEK consists of all automorphisms of K that fixEelementwise. Every
such automorphism automatically fixes each element of F since Fr;:; E. Hence, every
automorphism in GalEK is in GalFK, that is,
if E is an intermediate field, GalEK is a subgroup of GalpK.
EXAMPLE 2.B
0(v'3) is an intermediate field of the extension 0( V3,
v'5)
shows that GaluO( v'3,
=
v'5) of
0. Example 2.A
{ i, T, a, f:I}. The automorphisms that
element of 0(v'3) are exactly the ones that map
Therefore,
v'3 to itself by
fix every
Theorem 12.4.
GalQ(�Q(v'J,\/5)
is the subgroup {i, a} of {i, T, a, f:I}.
We now have a natural way of associating a subgroup of the Galois group with
each intermediate field of the extension. Conversely, if His a subgroup of the Galois
group, we can associate an intermediate field with Hby using
Theorem 12.6
Let K be an extension field of F. If His a subgroup of GaliJ{ , let
EH
=
{k EK I u(k)
=
k forevery u EH}.
Then EH is an intermediate field of the extension.
The field EH is called the fixed field of the subgroup H.
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dlremad.'lmm,-��._Gd.-.m.lly.n.ctbl.---....-.��l..Amiiog...- .. :dgbtm-__,_�roollm·a;J'tlmlo1f�:ligl:U�:NlpiNit.
12.1
The Galois Group
413
Proof ofTheorem 12.6 ... If c, dEEHand uEH, then
u(c + d) = u(c)
+
u(d) =
c
+
d
and
u(cd)
=
u(c)u(d) = ed.
Therefore, EHis closed under addition and multiplication . Since
u(Op) =Op and
u(lp)
= lpfor every automorphism, Oyand
Theorem 3.10 shows that for any nonzero
u(-c) = -u(c)
Therefore,
=
-c E EH and
u(c-1)
in EHand any
=
lyare in EH.
u in H,
u(cr1 = c-1•
c-1 EEn. Hence, EH is a subfield of K. Since
His a subgroup of GalpK,
Therefore, F� EH.
and
-c
c
•
u(c) = c for every c E F and every u EH.
EXAMPLE 2.C
Consider the subgroupH=
over
{i, a} of
the Galois group
{i, T, a,p} of o{v'3, VS)
O. Since ( \13) = \13, the subfield 0(\13) is contained in the fixed field
a
EHof H. To prove that EH= 0( V3), you must show that the elements of 0( v'J)
are the
only ones that are fixed by i and a; see Exercise 14.
EXAMPLE 1.C
As we saw in Example LB, GalnC = {t, u}, where
u is the complex conjuga­
tion map. Obviously, the fixed field of the identity subgroup is the entire field
C. Since
a
fixes every real number and moves every nonreal one, the fixed field
of GalnC is the field R.
Unlike the situation in the preceding example, the ground field F need not always
be the fixed field of the group Galp[(.
EXAMPLE 3.8
Vl to
-\Y2 is the only
Every automorphism in the Galois group of 0( V'2) over 0 must map
a root of xl
- 2 by Theorem
12.2. Example 3.A shows that
real root of this polynomial. Since
O(V'i) consists entirely of real numbers
G�C( V2) must map '\Yi to itself.
by Theorem 11. 7, every automorphism in
Therefore, Gal00(V'2) consists of the identity automorphism alone by
Theorem 12.4. So the fixed field of Galq0(-\Y2) is the entire field 0( �·
• Exercises
NOTE:
Unless stated otherwise,
K is an
extension field of thefield F.
A. 1. If u is an F-automorphism of K, show that u-1
is also an F-automorphism of
K.
2. Assume [K:F] is finite. Is it true that every F�automorphism of K is completely
determined by its action on a basis of K over Fl
CopJftglli.20t2�l...umlill.g.Al.1li9iiba_...a.Uqootbeo::iped.ICUfllld.nr�iawfdil«blJll"l.0.10� .......... tinl.p:dJCCIGl mAJM._....fmn. flBcd:udhr�1).Bdlaftlll........
....... my�mmal._oot..mu;,-dkl.baftml.lmmliag-.m---�l...Amiof;--•rilht1u_,,,.��-..,.1imllljf....:Dgbl.!lllWtrktkJas
...
....... it.
..
..
414
Chapter 12
Galois Theory
3. If [K:F] is finite , uE GalpK, and u EKis such that u(u) = u, show that
UEGalF(u�·
4. Write out the operation table for the group
[See Example 2.A.]
5. Let/(x)EF[x] be separable of degree n and Ka splitting field off(x). Show
that the order of GalpKdivides n!.
6. If Kis an extension field of Q and u is an automorphism of K, prove that u is
a 0-automorphism. [Hint: u(I) = l implies that u(n) = n for all n EZ.]
B. 7. (a) Show that Gal00( Vi) has order 2 and, hence, is isomorphic to Z2.
[Hint: The minimal polynomial is x'2 - 2; see Theorem 1 1.7 .]
fl. 0, show that GalQQ( \I'd) is isomorphic to Z2•
Show that Gal0Q I ( �) * (i ).
(a) Let w = (-1 + '\/3i)/2 be a complex cube root of 1. Find the minimal
(b) If dE Q and Vd
8.
9.
polynomial p(x) of w over Q and show that ai is also a root of p(x).
[Hint: w is a root of x1 - l.]
(b) What is GalQO(w)?
I 0.
(a)
Find GaloO( V'l, VJ). [Hint: See Example 2.A.]
(b) If p, q are distinct positive primes, find GalQO( Vp, yq).
11. Find GalQQ( V'l, i). [Hint: Consider 0 !;;; Q ( '\1'2) !;;;; 0 ( '\1'2,
in Example 2.A.]
12. Show that Gal00( V'l, VJ,
i) and proceed as
'\/5) = Z2 x Z2 x Z2•
13. If Fhas characteristic 0 and Kis the splitting field off(x) EF[x], prove that
the order of GalpKis [K:F] . [Hint: K F(u) by Theorems 11.17 and 11.18.]
=
H be the subgroup {'• a} of GalQO( VJ, '\/5) {'• T, a, f3}. Show that
the fixed field of His 0( v'3). [Hint: Verify that 0( v'3) r;;. EHr;;. 0( \/3, v's);
what is [Q( VJ, '\/5) :0 ( VJ)J?J
14. Let
=
15. (a) Show that every automorphism of R maps positive elements to positive
elements. [Hint: Every positive element of ll is a square.]
(b) If a, b ER, a< b, and u E Galolll, prove that u(a) < u(b).
[Hint: a < b if and only if b - a > O.]
(c) Prove that GalQR
(i). [Hint: If c < r < d, with c, dEO, then c < u(r) <ti,
show that this implies u(r) = r.]
=
C. 16. Suppose(, {
2
, ••• , ("
1 are n distinct roots of x!'
of Q. Prove that Gal0Q(C) is abelian.
=
-
1 in some extension field
17. Let Ebe an intermediate field that is normal over Fand uEGalpK. Prove that
u(E) E.
=
llC...t,,
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12.2
Ill
The Fundamental Theorem of Galois Theory
415
The Fundamental Theorem of GaloisTheory
The essential idea of Galois theory is to relate properties of an extension field with
properties of its Galois group. The key to doing this is the Fundamental Theorem of
Galois Theory, which will be proved in this section.
Throughout this section, K is afinite-dimensional extension field ofF. Let S be the
set of all intermediate fields and T the set of all subgroups of the Galois group GalFK.
Define a function fP:S--+ Tby this rule:
For each intermediate field E,
The function rp is called the Galois correspondence. Note that K (considered as a
subfield of itself) corresponds to the identity subgroup of GalFK, and the subfield F
corresponds to the entire group GalFK (considered as a subgroup of itself).
EXAMPLE 2.D*
( VS) of 0 and the
0(v'3). By the preceding remarks and Example2.B on
Consider the Galois correspondence for the extension O v'3,
intermediate field
page 412,
we
have
o(v'3:,VS)
O( v'3)
GaIQ(V3,VS}O(v'3,V5) {i}.
Ga�MJO(v'3, VS)
{ }
Q
GalQQ( VJ, VS)
{i, T, /:!}.
Example2.C shows that E
0(v'3) is the fixed field of the subgroup H (i, }
GalQCv'JJO( \/3, VS). Furthermore, K 0(\/3, VS) 0( v'3)( VS) is a normal,
separable extension of the fixed field E 0(v'3) because it's the splitting field of
----i'
=
---+'
=
----i-
=
,,a .
a,
=
=
=
a
=
=
=
x2-
5 (Theorem11 .15) and has characteristic 0(Theorem11.17.)
We now construct the tools necessary to show that, under appropriate assump­
tions, the Galois correspondence is a bijective map from the set of intermediate fields
to the set of subgroups of GalFK.
Lemma 12.7
Let K be a finite-dimensional extension field of F. ff His a subgroup of the
Galois group Gal
and E is the fixed field of H, then K is a simple, normal,
,K
separable extension off.
Example 2.D above (with K
=
0(\/3,'\1'5), E
=
O(v'3), and H
=
{i, a}) is an
illustration of Lemma12.7.
*The numbering scheme for examples in Sections 12.1 and 12.2 is explained on page 408.
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416
Chapter 12
Galois Theory
Proof of Lemma 12.7 ... Each uEK is algebraic over F by Theorem 11.9 and, hence,
algebraic over E by Exercise 7 in Section 11.2. Every automorphism in
H must map u to some root of its minimal polynomialp(x) EE[x] by
Theorem 12.2. Therefore, u has a.finite number o f distinct images under
u1o tt:z,
, u1EK.
(withrEH), thena(aj = a(T(u)).SinceaoTEH,
we see that a(aj is also an image of u and, hence, must be in the set
{ui. u2'
, u1}. Since a is injective, the elements a(u1),
, u(u,) are t
distinct images of u and, hence, must be the elements u1, u:z, ... , u1 in some
order. In other words, every automorphiSm in H permutes u1, u:z, .. . , ur Let
automorphisms in H, say u =
• . •
If(]' EHandu1 = T(u)
• • •
• • •
- u2)
f(x) = (x - uJ(x
·
·
(x
•
- u,).
Since the u1 are distinct,/(x) is separable. We claim that/(x) is actually
in E[x]. To prove this, let aEH and recall that (]'induces an isomor­
phism K[x] = K[x] (also denoted a), as described on page 380. Then
af (x) = (x - u(u1))(x
- a(ui))
·
·
•
(x
- a(uJ).
Since (]' permutes the Uj, it simply rearranges the factors of f(x), and, hence,
uf(x) = f(x). Therefore, every automorphism of H maps the coefficients
of the separable polynomial f(x) to themselves, and, hence, these coeffi­
cients are in E, the fixed field of H. Since u = u1 is a root of f(x) E E [x],
u
is separable over E.Hence, Kis a separable extension of E.
The field Kis finitely generated over F (since [K:F] is finite; see
Example 4 in Section 11.3). Consequently, Kis finitely generated over E,
and, hence, K= E(u) for someuEKbyTheorem 11.18.1.et/(x) be as in
the preceding paragraph. Then/(x) splits in K[x], and, hence, K = E(u)
is the splitting field of f(x) over E.Therefore, Kis normal over E by
Theorem 11.15.
•
Theorem 12.8
Let
K be
a finite-dimensional extension field off, If His a subgroup of the
Galois group Ga/FK and Eis the fixed field of H, then H=
[K:f] . Therefore, the Galois correspondence is surjective.
Ga/EK and I HI =
Proof... Lemma 12.7 shows that K = E(u) for some uEK. Ifp(x), the minimal
polynomial of u over E, has degree n, then [KE
: ] = n by Theorem 11.7.
Distinct automorphisms of GalEKmap u onto distinct roots of p(x) by
Theorems 12.2 and 12.4. So the number of distinct automorphisms in
GalEKis at mostn, the number of roots of p(x). Now H1;;GalEKby the
definition of the fixed field E. Consequently,
IHI
s IGalEKI s n = [K:E ].
1.etf(x) be as in the proof of Lemma 12.7. Then H contains at least t
u under H). Since
automorphisms (the number of distinct images of
u = u1 is a root of f(x),p(x) divides/(x). Hence,
IHI
<:!:: t = deg/(x) <:!:: degp(x) = n = [KE
: ].
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..
..
12.2
The Fundamental Theorem of Galois Theory
417
Combining these inequalities, we have
IHI s IGalEKI s [K:E] :S IHI.
Therefore, IHI
=
IGalEKI
=
[K:E], and, hence, H
=
GalEK.
•
EXAMPLE 3.C
G�C(V'2) {i) by Example 3.B, so both of the intermedi­
o(-¢'2) and c are associated with (i} under the Galois correspondence.
Note that C( \Y2) is not a normal extension of Q [it doesn't contain the com­
plex roots of x3
2, so this poly nomial has a root but doesn't split in o( Vi)].
The Ga lois group
=
ate fields
-
Galois Extensions
Although the Galois corresp ondence is surjective by Theorem 12.8, the preceding
example shows that it may not be injective. In order to guarantee injectivity, additional
hypotheses on the extension are necessary. The preceding proofs and example suggest
that normality and separability are likely candidates.
Definition
If K is a finite-dimensional, normal, separable extension field of the field F,
we say that K is a Galois extension of For that K is Galois over F.
A Galois extension of characteriStic 0 iS simply a splitting field by Theorems 11.15
and 11.17.
Theorem 12. 9
Let K be a Galois extension of Fand E an intermediate field. Then E is the fixed
field of the subgroup GalEK.
If
E and L are intermediate fields with Gal�
=
Galt,K", then Theorem 12.9 shows
that both E and L are the fixed field of the same group, and, hence, E
=
L. Therefore,
the GaloiS co"espondence iS injectlVefor GaloiS extensions.
Proof ofTtworem 12.9 .. The fixed field l1i of GalEK contains E by definition. To show
that .&, !::.: E, we prove the contrapositive: If u $. E, then u is moved by some
automorphism in Gal�, and, hence, u $. E:,. Since K is a Galois extension
of the intermediate field E (normal by Theorem 11.15 and Exercise 5 of
Section 11.4; separable by Exercise 1 of Section 11.5), it is an algebraic
extension of E. Consequently, u is algebraic over E with minimal polyno­
mial p(x) E E[x] of degree :l'!: 2 (if degp(x)
=
1, then u would be in E). The
roots of p(x) are distinct by separability, and all of them are in K by normal­
ity. Let v be a root of p(x) other than u. Then there exists u E Ga l_,,K such
that u(u)
= v
by Theorem 12.3. Therefore, u $. &, and, hence, J1i
=
E.
•
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...
...... it.
418
Chapter 12
Galois Theory
Corollary 12.1 O
Let K be a finite-dimensional extension field of F. Then K is Galois over F if and
only if Fis the fixed field of the Galois group GalfK.
Proof" If K is Galois over F, then Theorem 12.9 (with E
=
F) shows that Fis
the fixed field of GalFK. Conversely , if Fis the fixed field of GalpK, then
Lemma 12. 7 (with E
=
F) shows that K is Galois over F.
•
In view of Corollary 12.10, a Galois extension is often defined to be a finite­
dimensional one in which Fis the fixed field of GalFK. When reading other books on
Galois theory, it's a good idea to check which definition is being used so that you don't
make unwarranted assumptions.
EXAMPLE 2.E
The field
0(V3,Y5) is a Galois extension of Q because it is the splitting
(x2 - 3)(x2 - 5). So the Galois correspondence is bijective by
field of f(x)
=
Theorem 12.8 and the remarks after Theorem 12.9. The Galois group
( v'3, v'5)
Gal0Q
{i, T, a, (3}
=
by Example 2.A. Verify the accuracy of the
chart below, in which subfields and subgroups in the
same
relative position cor­
respond to each other under the Galois correspondence. For instance,
corresponds to
{i, a}
Cl!( '\/3)
by Example 2.B.
Subgroups
Intermediate Fields
Note that all the intermediate fields are themselves Galois extensions of Q
(for instance,
o( v'S) is
the splitting field of
x?
-
5). Furthermore, the corre­
sponding subgroups of the Galois group are normal. A similar situation holds
in the general case, as
Theorem 12.11
we now see
.
The Fundamental Theorem of Galois Theory
If K is a Galois extension field of F, then
(1) There is bijection between the set S of all intermediate fields of the
ex tension and the set T of all subgroups of the Galois group GalFK,
given by assigning each intermediate field E to the subgroup
GalEK. Furthermore,
[K:E]
=
......
!Galt:KI
and
[E:F]
=
[GalFK:GalEK].
......
......
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ftnnb118om:.ndlat�1).BdlmiM...,,.._.._
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k
12.2
The Fundamental Theorem of Galois Theory
419
(2) An intermediate field E is a normal extension of F if and only if the
corresponding group GalEK is a normal subgroup of Gal,K, and in
this case Gal,£= Gal,K/GalEK.
Proof" Theorem 12.8 and the remarks after Theorem 12.9 prove the first state­
ment in part (1) . Each intermediate :field Eis the fixed field of
by Theorem
12.9. Consequently, [K:E]
particular, if F =
Theorem
E,
then [K:F] =
=
GalEK
IGalEKI by Theorem 12.8. In
IGalFKI. Therefore, by Lagrange's
8.5 and Theorem 11.4,
[K:E][E:F]
=
[K:F]
=
IGalFKI
=
IGal�I [GalFK:Gal�.
Dividing the first and last terms of this equation by
[K:E]
=
IGal�I
shows that
[E:.F]
=
[GalFK:GalEK].
To prove part (2), assume first that GalEK is a normal subgroup of
Gal pK. Ifp(x) is an irreducible polynomial in F[x] with a root u in E, we
must show that Ji..x) splits in E[x] . Since K is normal over F, we know that
Ji..x) splits in K[x]. So we need to show only that each root vof p(x) in K is
actually in E. There is an automorphism u in GalFK such that u(u) = v by
Theorem 12.3. If T is any element of GalEK, then normality implies
T 0 O' = u T1 for some Tt E GalEK. Since u EE, we have T(v)
T(u(u)) =
u(T1(u)) = u(u) = v. Hence, vis fixed by every element Tin GalEK and,
therefore, must be in the fixed field of GalEK, namely E (see Theorem 12.9).
Conversely, assume that Eis a normal extension of F. Then Eis finite
0
=
dimensional over Fby part (1). By Lemma 12.12, which is proved below,
there is a surjective homomorphism of groups 8:GalpK-+ GalpE whose ker­
nel is GalEK. Then GalEK is a normal subgroup of
GalFKby Theorem 8.16,
and GalpKfGalEK = GalpE by the First Isomorphism Theorem 8.20.
•
EXAMPLE 3.D
The splitting :field
K of X3 - 2 is
a Galois extension of Q whose Galois group is
a subgroup of S3 by Example 3.A. * Note that Q i:: OV'i) i:: K. Since
is the minimal polynomial of
V'2, [0("312):0]
=
X3 - 2
3 by Theorem 11.7. Neither
("312w and V'iw2) is a real number, and, hence, neither is in
O(V'i). So [K:Q] > 3. Since [K:Q] s 6 (Theorems 11.13, 11.14) and [K:O] is
of the other roots
divisible by 3 (Theorem
6 by Theorem
11.4), we must have [K:Q]
12.11 and is S3•
=
6. Thus Gale{( has order
((123)} of order 3
2: (( 12) ), (( 13)}. {(23)}. Verify that the Galois
The only proper subgroups of S3 are the cyclic group
and three cyclic groups of order
correspondence is as follows, where subgroups and subfields in the same rela­
tive position correspond to each other. The integer by the line connecting two
•we consider 58 as the group of permutations of the roots
(12) interchanges
V'2, *"'• �2 in this order. For instance,
..;'2 and �and fixes \Ww1.
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.
ICDl
dllklDlii.
llllnl•_..,.lillll��:Dgbb�...-.:lit.
420 Chapter 12
Galois Theory
subfields is the dimension of the larger over the smaller. The integer by the line
connecting two subgroups is the index of the smaller in the larger.
Intermediate Fields
Subgroups
3
�;�
<(23)>
<{13)>
The field Q(w) is an intermediate field because w
<(12)>
=
G)(oV2)2(V'2.tt.1)
E K.
Cl!(w) is the splitting field of r + x + 1 (Exercise 3) and, hence, Galois over Cl!.
The corresponding subgroup is the normal subgroup ( ( 123)). On the other
hand, Example
3.C shows that 0(� is not Galois over 0; the corresponding
subgroup ( (23)) is not normal in S3•
The preceding example illustrates an important fact:
The Galois correspondence Is Inclusion-reversing.
For instance, Q s;;; Q(w), but the corresponding subgroups satisfy the reverse inclusion:
S1 ;:1 (( 123 )}.
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12.2
The Fundamental Theorem of Galois Theory
421
Finally, we complete the proof of the Fundamental Theorem by proving
Lemma 12.12
Let K be a finite-dimensional normal extension field of F and E an intermedi­
ate field, which is normal over F, Then there is a surjective homomorphism
of groups O:GalFK-+ Galff whose kernel is GalEK.
Proof.,.
Let u EGalp&'." and u E E. Then
u
is algebraic over F with minimal
polynomial p(x). Since Eis a normal extension of F, p(x) splits in .ETx],
that is, all the roots of p(x) are in E. Since a(u) must be some root of
p(x) by Theorem 12.2, we see that u(u) EE. Therefore, a(E) s;;Efor
every u EGalp&'.". Thus the restriction of a to E(denoted u IE) is an
F-isomorphism E= a(E) . Hence, [E:Fl = [a(E):.F] by Theorem 11.5.
11.4,
Since F<;,. a (E) i;; E, we have [E:.F] = [E:a(E)] [a(E):.F] by Theorem
which forces [E:a(E)] = 1. Therefore, E= a(E), and a IEis actually an
automorphism in Galp£.
Define a function
6 :GalpK-+ GalpE by 6(a) = a IE. It is easy to
verify that 0 is a homomorphism of groups. Its kernel consists of the au­
tomorphisms of K whose restriction to Eis the identity map, that is,the
subgroup GalEK .
To show that 0 is surjective, note that K is a splitting field over F
by Theorem
and, hence,K is a splitting field of the same poly­
11.15,
nomial over E. Consequently, every TE Gal FE can be extended to an
11. 14.
F-automorphism a in GalFK by Theorem
This means that
a IE= T, that is, 6(a) = T. Therefore, 6 is surjective. •
In the preceding proof, the normality of K was not used until the last paragraph.
So the first paragraph proves this useful fact:
Corollary 12.13
Let K be an extension field of F and E an intermediate field that is normal over
F. If aEGalFK, then a IEEGal�.
• Exercises
NOTE: K is an
extension.field of the field F.
A. 1. If K is Galois over F, show that there are only finitely many intermediate fields .
2. If K is a normal extension of
0 and [K :O] = p, with p prime, show that
GaloK=Z,.
3.
(-1 + Wi)/2 is a root of x3 - 1.
Show that wand w2
roots of x2 + x + 1. Hence, Q(w) is the splitting
field of x2 + x + 1.
(a) Show that w =
(b)
are
.......
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:ligllll�...-.:lit.
...
.......
422 Chapter 12
Galois Theory
4. Exhibit the Galois correspondence of intermediate fields and subgroups for
the given extension of Q:
(a) Q(va) , wheredEQ, but Wfl. Q.
(b) Q(w), where w is as in Exercise 3.
5. If Kis Galois over F and GalpKis an abelian group of order
10, how many
intermediate fields does the extension have and what are their dimensions
over F?
6. Give an example of extension fields Kand L of F such that both Kand L are
Galois over F, K :F L, and Galp\'."= GalpL.
B. 7. Exhibit the Galois correspondence for the given extension of Q:
(a) Q( v'2, V3)
(b)
O(i,'\/2)
8. If Kis Galois over F, GalpKis abelian, and E is an intermediate field that is
normal over F, prove that Gal,&K and GalpE are abelian.
9. Let Kbe Galois over F and assume GalpK = Z11•
(a)
If Eis an intermediate field that is normal over
F, prove that GalEK and
GalpE are cyclic.
(b) Show that there is exactly one intermediate field for each positive divisor
of n and that these are the only intermediate fields.
IO. Two intermediate fields E and L are said to be
conjugate if there exists
u E GalpK such that u(E)
L. Prove that E and L are conjugate if and
only if GalEKand Gair$ are conjugate subgroups of GalpK(as defined on
page 308).
=
11.
(a) Show that K
=
0(�, i) is a splitting field of x4
-
2 over Q.
(b) Prove that [KO
: ]
8 and conclude from Theorem 12.11 that Ga� has
o rder 8. [ Hint: Q !;;; 0{'¢2) s;; 0(-¢'2, i).J
=
(c) Prove that there exists <TE GalctK such that u(�)= (�)i and u(1)
and that <T has order 4.
=
i
(d) By Corollary 12.13 restriction of the complex conjugation map to Kis an
element T of GalciK. Show that
[Hint: Use Theorem 12.4 to show these elements are distinct.]
(e) Prove that GalciK= D4• [Hint: Map <T to r1 to T to v.]
12. Let Kbe as in Exercise
11. Prove that G�i/{ = �·
C. 13. Let Kbe as in Exercise 11. Exhibit the Galois correspondence for this extension.
i) '¢'2) and Q( ( 1 i) '¢2).J
Exhibit the Galois correspondence for the extension Q( v'2, v'3, v'S) of O.
[Among the intermediate fields are o( ( 1 +
14.
-
[The Galois group has seven subgroups of order 2 and seven of order 4.]
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12.3
1111
Solvability by Radicals
423
Solvability by Radicals
The solutions of the quadratic equation ax1
+ bx + c = 0 are given by the well-known
formula
x=
This fact
was
-b ±
y,; - 4ac.
2a
known in ancient times. In the sixteenth century, formulas for the solu­
tion of cubic and quartic equations were discovered. For instance, the solutions of
X' + bx +
c
=
0 are given by
x
=
-\f.1(-c/2) + Vil+-\fl(-c/2) - Vil
td(-\f/( -c/2) + W) + ltl2(-\f/( -c/2) - v'd)
x
=
td2(-\f/(-c/2) +Vil)+ w(-\f.1(-c/2) - Vil),
x=
where d
=
(IJl /27)
+ (c2/4), w
(-1 + v'3i)/2 is a complex cube root of 1, and the
=
other cube roots are chosen so that
N1(-c/2)
+
Vd)(-v'(-c/2) - v'd)
=
-b/3.*
In the early 1800s Ruffini and Abel independently proved that, for n 2: 5, there
is no formula for solving all equations of degree n. But the complete analysis of the
problem is due to Galois, who provided a criterion for determining which polynomial
equations
are
solvable by formula . This criterion, which is presented here, will enable
us to exhibit a fifth-degree polynomial equation that cannot be solved by a formula. To
simplify the discussion, we shall assume that all.fields have
characteristic 0.
As illustrated above, a "formula" is a specific procedure that starts with the coefficients
of the polynomialf(x)EF[x] and arrives at the solutions of the equation/(x)
=
Op by
using only the field operations (addition, subtraction, multiplication, division) and the
extraction of roots (square roots, cube roots, fourth roots,
root of an element
etc.).
In this context, an nth
c in Fis any root of the polynomial X' - c in some extension field of F.
If f(x) E F[x], then performing field operations does not get you out of the coef­
ficient field F (closure!). But taking an nth root may land you in an extension field.
Taking an mth root after that may move you up to still another extension field. Thus
the existence of a formula for the solutions of f(x) = Op implies that these solutions lie
in a special kind of extension field of F.
EXAMPLE 1
Applying the cubic formula above to the polynomial
the solutions of
x3 +
x3 +
3x + 2 shows that
3x + 2 = 0 are
Y/-1
wY/-1
+
+
V2 +-\f/-1 - V2,
v'2 + (w2)"¢'-1
( w2) Y/-1 +
V2 +
-
V2,
w"1-1 - "2.
•The formulas for the general cubic and the quartic are similar but more complicated.
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424 Chapter 12
Galois Theory
All these solutions lie in the extension chain:
0 t;;;CJ!(w)
t;;; O(w. Vi)!;; O(w, Vi, \Y-1 +v'2) i;;;;Q(w,Vi, -v'-1 + '\/2, -v'-1- v'2)
II
II
Fo � F1
s;
II
F2
II
F3
�
II
F4.
!;;;
Each field in this chain is a simple extension of the preceding one and is of the form FJ..u),
where ti' EJij for some n (that is, u is an nth root of some element of F'):
F1 = Fo (w),
Jii = F1 ( \12),
where
w3
==:
I
EFo-
( V2")2 = 2 EF0 t;;; F1.
where ('\o/-1 + \1'2)3 = -1 + Vl E F2•
F3 = Jii.(-v'-1 + '\/2),
F4=1'J(-v'-l - '\/21
where ('o/°-I - '\/2)3= -l-\12EF21:;l'J.
Since
where
F4 contains all the solutions of x3
3x + 2
+ 3x
+2=
0, it also contains a splitting
field of x3 +
The preceding example is an illustration of the next definition.
Definition
A field
K is said to be a radical extension of a field F if there is a chain of
fields
F
=
F0i;;;F
; 1t;;;f2t;;;·
•
·t;;;ft
=
K
such that for each i = 1, 2, .. , t,
,
F1 = Fi-1(u/) and some power of u, is in h--t·
Letf(x) E F[x]. The equation/(x) = OF is said to be solvable by radicals if there is a
radical extension of F that contains a splitting field of/(x). The example above shows
that x3 + 3x
+2
""'
0 is solvable by radicals.
The preceding discussion shows that if there is a for mula for its solutions, then the
equation/(x)
=
Op is solvable by radicals. Contrapositively, if f(x)
:=:
OF is not solvable
by radical, then there cannot be a formula (in the sense discussed above) fo r finding its
solutions.
Solvable Groups
Before stating Galois' Criterion for an equation to be solvable by radicals, we need to intro­
duce a new class of groups. A group G is said to be solYable if it has a chain of subgroups
G = G0 2 G1 2 G2 2
·
·
·
2 Gn-t ;;i Gn =
(e}
such that each G1 is a normal subgroup of the preceding group G1_1 and the quotient
group G,_if G1 is abelian.
EXAMPLE 2
Every abel ian group G is solvable because every quotient group of G is abelian,
so the sequence G2
(e) fulfills the conditions in the definition.
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Solvability by Radicals
12.3
425
EXAMPLE 3
Let {(123)} be the cyclic subgroup of order 3 in S3• The chain S3;;i {(123)} ;;i ((1))
shows that S3 is solvable. But for other symmetric groups we have
Theorem 12.14
For n
� 5 the group Sn is not solvable.
Proof• Suppose, on the contrary,
that s,. is solvable and that
(rst) be any
{ 1, 2, .. . , n} other than
r, s, t (u and v exist because n 2! 5). Since S,,/G1 is abelian, Theorem 8.14
(with a =(tus), b =(s r v)) shows that G must contain
1
is the chain of subgroups required by the definition. Let
3-cycle in S,. and let u, v be any elements of
(tus)(srv)(tus)-1(srvr1 =(rus)(srv)(tsu)(svr) =(rst).
Therefore, G1 contains all the 3-cycles. Since Gif G2 is abelian, we can
repeat the argument with G in place of S,, and G2 in place of G1 and
conclude that
1
Gi contains all the 3-cycles. The fact that each G1_1/ G1 is
abelian and continued repetition lead to the conclusion that the iden­
tity subgroup G, contains all the 3-cycles, which is a contradiction.
Therefore, S,. is not solvable. •
Theorem 12.15
Every homomorphic image of a solvable group
Proof•
G
is solvable.
Suppose thatfG-l> His a surjective homomorphism and that G =
G0 ;;i G1 ;;i Gi ;;i
;;i Gi ={ea) is the chain of subgroups in the defini­
tion of solvability. For each i, let H1 =f (GJ and consider this chain of
• •
•
subgroups:
H
=H0 ;;i H1 ;;i H2 ;;i
•
•
•
;;i H,
=f((ea)) ={eH}·
Exercise 22 of Section 8,2 shows that H1 is a normal subgroup of H1_1
=I, 2, ... , t. Let a, b EH1_1• Then there exist c, dE G1_1 such
for each i
thatf(c) =a and/(d) b. Since G1_tf G1 is abelian by solvability,
cdc-1d-1 EG, by Theorem 8.14. Consequently,
1
aba-1b- =f(c)f(d)f(c-1)/(d-1) =f(cdc-1d-1) Ef(G1) =H1•
=
Therefore, H1--1/H1 is abelian
.......
by Theorem 8.14, and His solvable.
..
•
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426 Chapter 12
Galois Theory
Galois' Criterion
If f(x) E F[x], then the Galois group of the polynomial/(x) is GalFK, whereK is a splitting
field off(x) over F.* Galois' Criterion states that
f(x)
= OF iS
solvable by radicals if and only if the Galois
group off(x) is a solvable group.
In order to prove Galois' solvability criterion, we need more information about
radical extensions and nth roots. If Fis a field and ( is a root of x!' - IF in some
extension field of F(so that {!' = 1F), then (is called an nth root of unity. The deriva­
tive nX-1 of X' - 1 Fis nonzero (since Fhas characteristic 0) and relatively prime to
- lF. Therefore, x!' - lpis separable by Lemma Il.16. So there are exactly n distinct
x!'
nth roots of unity in any splitting fieldK of x!' -
lp If (and Tare nth roots of unity
inK, then
((Tt =
("7"
= IF lF = lFl
so that ( T is also an nth root of unity. Since the set of nth roots of unity is closed under
multiplication, it is a subgroup of order n of the multiplicative group of the field K
(Theorem 7.12) and is, therefore, cyclic by Theorem 7.I6 or Corollary 9.11. A genera­
tor of this cy clic group of nth roots of unity inK is called a primitive nth root of unity.
Thus (is a primitive nth root of unity if and only if (, (2, (3,
, (" = lp are the n
•
•
•
distinct nth roots of unity.
EXAMPLE 4
The fourth roots of unity in Care 1 , -1,i, -i. Since i2 = -1, ;3 = -i,andi4 = 1,
i is a primitive fourth root of unity. Similarly , -i is also a primitive fourth root of
unity . DeMoivre's Theorem shows that for any positive n,
cos(21T/n) +
When n
=
i sin(21T /n) is a primitive nth root of unity in C.
3, this states that
w = cos(21T/3) + i sin(2'lT/3) = (-1/2) + ('\/3/2)i
is a primitive cube root of unity.
Lemma 12.16
Let F be a field and ( a primitive nth root of unity in F. Then F contains a
primitive
dth
root of unity for every positive divisor d of n.
Proof.,. By hypothesis (has order n in the multiplicative group of F. If n
= dt,
d by Theorem 7 9 So r! generates a subgroup of order
d, each of whose elements must have order dividing d by Corollary 8.6.
In other words, ((!!ft= lpfor every k. Thus the d distinct powers (1,
then (' has order
"Since any two splitting fields
.
.
of f(x) are isomorphic by Theorem 11.14, it follows that the corre­
of f(x) is independent of the choice of K.
sponding Galois groups are isomorphic. So the Galois group
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12.3
Solvability by Radicals
427
, (l�a1, (£'f = 1F are roots of x' - lF. Since x' - 1F has at most
d roots and every dth root of unity is a root of x' - IF> C is a primitive dth
(l�2,
• • •
root of unity.
•
We can now tie together the preceding themes and prove two theorems that are
special cases of Galois' Criterion as well
as
essential tools for proving the general case.
Theorem 12.17
Let F be a field of characteristic O and la primitive nth root of unity in some
F. Then K = F( l) is a normal extension of F, and GalFK is
extension field of
abelian.
Proof• The field K = F(() contains all the powers of C and is, therefore, a split­
ting field of
X' - lF. *
Hence, K is normal over Fby Theorem 11.15.
Every automorphism in the Galois group must map lonto a root of
x!' - lF by Theorem 12.2. So if u, TE GalFK, then u(C) = lk and
T(()
= C for some positive integers k, t. Consequently,
(u T)(C) = u(r(C)) = u(C') = u(l)' = (£1')' = l'" ·
(T 0 u)(() = T(u(C)) = T(l� = T(t/' = (l')k l"'·
0
=
Therefore,
u o T = r o u by Theorem 12.4, and Galp{ is abelian.
•
Theorem 12.18
Let F be a field of characteristic O that contains a primitive nth root of unity.
If u is a root of xn - cEf{x] in some extension field of F, then K
F(u) is a
=
normal extension of F, and GalFK is abelian.
Proof t •By hypothesis, tl' = c. If l is a primitive nth root of unity in F, then for
anyk,
f, .. . , (," = 1Fare distinct elements of F, the ele­
(u, [2u, (3u, . .. , l"u = u are then distinct roots of X' - c. Hence,
K = Ji\'.u) is a splitting field of x!' - cover F and is, therefore, normal
over FbyTheorem 11.15.! If u, 'T, EGalpK, then u(u) = f'u and T(u) =
Cu for some k, tby Theorem 12.2. Consequently, since r.1' and (,' are in F,
Consequently, since(,
ments
*The field K = F(() is a radical extension of F since Cn = 1,. Thus Jt' - 11 =OF is solvable by radicals.
So the theorem, which says that Gal,K (the Galois group of Jt' -1F), is abelian (and hence, solvable),
is a special case of Galois' Criterion.
tfor an alternate proof showing that Gal,K is actually cyclic, see Exercise 22.
trhe field K = F(u) is also a radical extension of F since un = CEF, so Jt'
radicals. Hence, the theorem is another special case of Galois' Criterion.
..........
..
- c = 0,, is solvable by
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428 Chapter 12
Galois Theory
(u 0 T)(u) = U(T(u}) = u((u) = u((�u(u) = C'((/'u) = (1+ku.
(T 0 u)(u} = T(u(u)) = T(tu) = T(C1)T(u) = ("(('u) = r+ku.
Therefore,
u o
T =Touby Theorem 12.4, and GalpK is abelian.
Theorem 12.19
•
Galois' Criterion
Let F be a field of characteristic O and f(x) Ef{x]. Then f(x) =OF is solvable by
radicals if and only if the Galois group off(x) is solvable.
We shall prove only the half of the theorem that is needed below; see Section V.9 of
Hungerford [5] for the other half.
Proof of Theorem 12.19 .. Assume thatf(x) = OF is solvable by radicals. The proof,
whose details are on pages 429-431, is in three steps:
1. Theorem 12.21: There is a normal radical extension K of F that con­
tains a splitting field E off(x).*
2. The field Eis normal over Fby Theorem 11.15.
3. Theorem 12.22: Any intermediate field of K that is normal over Fhas
a solvable Galois group; in particular, GalpE (the Galois group of
f(x)) is solvable.
•
Before completing the proof of Theorem 12.19, we use it to demonstrate the insol­
vability of the quintic.
EXAMPLES
We claim that the Galois group of the polynomialf(x)
=
2x5 - lOx + 5 E Q[x]
is S5, which is not solvable by Theorem 12.14. Consequently, the equation
2x5 - 10x + 5 = 0 is not solvable by radicals by Theorem 12.19. So, as
explained on page 424,
there is no formula (involving only field operations and
extraction of roots) for the solution of all fifth-degree
polynomial equatiom.
To prove our claim, note that the derivative off(x) is 10x4 - 10, whose only
real roots
are
±1 (the others being ±1). ThenI'(x)
=
40x3, and the second­
derivative test of elementary calculus shows thatf(x) has exactly one relative
maximum at x = -1, one relative minimum at x = 1, and one point of inflec­
tion at x = 0. So its graph must have the general shape shown on the next page.
In particular,f(x) has exactly three real roots.
•This is a crucial technical detail. The definition of solvability by radicals guarantees only a radical
extension
of F containing E. But a radical extension need
be used.
not be normal over F (Exercise 19), and if
it is not, the FundamentalTheorem 12.11 can't
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12.3
Solvability by Radicals
429
It'\.
I \
I \
'
2
-I
1
l
\
\
2
\ J
'"
I
Note thatf(x)is irreducible in O[x] by Eisenstein's Criterion (with p = 5).If
K
[K:O] by the Fundamental
[K:O(r)] [O(r):O] by Theorem 11.4
is a splitting field of f(x) in C, then Ga.lctK has order
Theorem. If r is any root off(x), then
and
[Q(r):O]
=
[K:O]
=
5by Theorem 11. 7. So the order of GalaK is divisible by 5. It
follows that GaloK contains an element of order 5.*
The group GaloK, considered
as
f(x), is a subgroup of S5 (Corollary
Ss
are the
a group of permutations of the roots of
12.5).But the only elements of order 5 in
5-cycles (see Exercise 19 in Section 7.5). So GaloK contains a 5-cycle .
Complex conjugation induces an automorphism on
K (Corollary
12.13). This
automorphism interchanges the two nonreal roots of f(x) and fixes the three
real ones. Thus GaloK contains a transposition. Exercise 8 shows that the only
subgroup of S5 that contains both a 5-cycle and a transposition is S5 itself.
Therefore, Galof( = S5 as claimed.
We now complete the proof of Galois' Criterion, beginning with a technical lemma
wh ose import will become clear in the next theorem.
Lemma 12.20
Let F, E,
L be fields of characteristic O with
f<;;E<;; L =E(v)
and
If L is finite dimensional over F and E is normal over F, then there exists
an extension field M of L,
which is a radical extension of E and a normal
extension of F.
Proof• By Theorem
11.15, Eis the splitting field over Fof some g(x)EF[x].
Let p(x) E F[x] be the minimal polynomial of
v over F and let M be a
splitting field of g(x) p(x) over F.Then Mis normal over Fby Theorem 11.15.
Furthermore , F<;; E<;; L <;; M (since L
E(v) and Eis generated over
, v, be all the roots of
i there exists u1E GalFM such that u1 (v) = v1 by
Fby the roots of g(x)). Let
p(x) in
M. For each
v = vh f.1:2,
=
• • •
•1f you have read Chapter 9 use Corollary 9.14; otherwise, use Exercise 9 in this section.
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430 Chapter 12
Galois Theory
Theorem 12.3. Corollary 12.13 shows that u1 (E) !;;;; E. By hypothesis,
vk=
b E E; so for each i,
(v.if= u,(vf= ui(v�= ul.._b) EE!;; E(vi. .•• , '111-1 )·
Consequently,
E !;;;; L = E(vJ 1;;;E(v1> t12)1;;;E (vb
is a radical extension of E. •
v:i. v:J !;;;;
• •
•
!;;;; E (v11 v2,
• •
.
,
v,) = M
Theorem 12.21
Let F be a field of characteristic O and
f(x) Ef[x]. If f(x) = OF is solvable by
radicals, then there is a normal radical extension field of F that contains
a splitting field of
f(x).
Proof• By definition some splitting field K off(x) is contained in a radical
extension
F= Fo1;;Fi !;;;; F1 !;;;; F31;;
•
•
• 1;; Fts
F1= Fj_1 (aj and ( Ui)"' is in Fj_1 for each i= 1, 2,
, t. Applying
E = F, L = F., and v u1 produoes a normal radical
extension field M1 of Fthat contains Fj. By hypothesis ( u1)'°'EF1 !;;M1•
Applying Lemma 12.20 with E= M1o v = �. and L = M1(u:J produces
where
. •
Lemma 12.20 with
.
=
a normal extension field M2 of Fthat is a radical extension of M1 and,
hence,
a
radical extension of F. Furthermore, M2 contains
F2 = Fi(-u:J.
Continued repetition of this argument leads to a normal radical exten­
sion field
M, of F that contains F, and, hence, contains K.
•
Theorem 12.22
Let K be a normal radical extension field of F and E an intermediate field, all
of characteristic 0. If Eis normal over F, then Gal� is a solvable group.
Proof• By hypothesis there is a chain of subfields
F= F01;;;F11;;F21;;;F31;;
where F1=
F1_1(uJ and
•
•
•
1;; Ft=
K,
(u1)"' is in F1_1 for each i = 1, 2,
the least common multiple of nh n2,
• • •
root of unity. For each i � 0, let E1=
, nt
.
. •
, t. Let n be
and let l be a primitive nth
F1((). Then for each i '2:
1
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12.3
Solvability by Radicals
431
is a radical extension of Fthat contains K(and, hence, E).* The normal
extension K F1 is the splitting field of some polynomial p(x) E F[x]
by Theorem 11.15, and, hence, L E1
Fi((J is the splitting field of
p(x)(X'
1p) over F. Therefore, Lis Galois over F by Theorems 11.15
and 11.17.
Consider the following chain of subgroups of GalpL:
=
=
=
-
GalpL;;;i GalE,L;;;i GalE,L;;;i GalE,L;;;i
•
•
•
;;;i Gale,_,L;;;i GalLL =
(i}.
We shall show that each subgroup is normal in the preceding one and
that each quotient is abelian.Since each n1 divides n, E0 contains a primi­
tive n1th root of unity by Lemma 12.16.Consequently, by Theorem 12.18
each E1 (with i '2: 1) is a normal extension of Ei-1, and the Galois group
GalE,_,Ei is abelian. Since Lis Galois over F, it is Galois over every�·
Applying the Fundamental Theorem 12.11to the extension L of .Ei-1> we
see that Gali;;Lis a normal subgroup of Gali;;_,L and that the quotient
group GalE_, ,L/GalE,Lis isomorphic to the abelian group GalA\_,E,.
Similarly by Theorems 12.11and12.17, E0 is normal over F, GalEoLis
normal in GalpL, and GalpL/Gal4L is isomorphic to the abelian group
GalFEo· Therefore, GalpLis a solvable group.
SinceEis normal over F, the Fundamental Theorem shows that
Galnf, is normal in GalpL and GalpL/GalELis isomorphic to GalpE.
So GalpE is the homomorphic image of tlte sol vable group GalpL
(see Theorem 8.18 ) and is, therefore, solvable by Theorem 12.15.
•
• Exercises
NOTE: Fdenotes a field, and allfields have characteriStic 0.
A. 1. Find a radical extension of 0 containing the given number:
(a) �1 + Vt - ..Y2
+
VS
(b)
2.
(�) /(-.VS)
(c) ('¢'3 - \12)/(4 + \12)
Show that x2 - 3 and
- 2x -
x2
2 E C[x] have the same Galois group.
[Hint: What is the splitting field of each?]
3. If K is a radical extension of F, prove that [K:F] is finite.
[Hint: Theorems 11.7and 11.4.]
*The construction
of L does
not use the hypothesis thatK is normal over F, and, as we shal I see
below, every field in the chain is a normal extension of the immediately preceding one.
But this is not
enough to guarantee that Lis normal (hence Galois) over F(Exercise 19). We need the hypothesis
thatK is normal over Fto guarantee this, so that we can use the FundamentalTheorem on L.
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432 Chapter 12
Galois Theory
4. Prove that for n
Theorem
5.
<?::
5, A,. is not solvable.
[Hint: Consider the subgroup H =
{(12)(34), (13)(24), (14)( 23), (l)} of A4.]
(a)
Show that S4 is a solvable group.
(b)
Show that D4 is a solvable group.
6. If
[Hint: Adapt the proof of
12.14.)
G is a simple nonabelian group, prove that G is not solvable. [This fact and
8.26 provide another proof that A,. is not solvable for n <?:: 5.)
Theorem
7. List all the nth roots of unity in C when n =
(a) 2
B. 8. Let
(e) 6
( d) 5
G be a subgroup of S5 that contains a transposition fJ = (rs) and a 5-cycle a.
Prove that G
(a)
(c) 4
(b) 3
=
Ss as follows.
Show that for some k,
ak
is of the form (rsxyz). Let T =a* E G; by
relabeling we may assume that fJ
(b)
Show that (12),
(c)
Show that (13), (14), (15) E G.
=
(12) and T = (12345).
(23), (34), (45) E G. [Hint: Consider T*<TT-k fork<?:: 1).
[Hint: (12)(23)(12) :: ?]
(d) Show that every transposition is in G. T herefore, G
=
S5 by Theorem 7 .26.
G be a group of order n. If Sin, prove that G contains an element of order
5 as follows. Let Sbe the set of all ordered 5-tuples (r, s, t, u, 11) with r, s, t, u,
9. Let
11E G and rstuv
(a)
(b)
=
e.
Show that S contains exactly n4 5 -tuples. [Hn
i t: If r, s,
(rstu)-1,
then
(r, s, t, u, 11) ES.)
Two 5-tuples in Sare said to be
equivalent if one
t, u, E G and v
=:
is a cyclic permutation of
the other.* Prove that this relation is an equivalence relation on S.
(c)
Prove that an equivalence class in S either has exactly five 5-tuples in it or
consists of a single 5-tuple of the form (r, r,
r, r, r).
(d) Prove that there are at least two equivalence classes in S that contain
a single 5-tuple. (Hint: One is {(e, e, e, e, e)}. If this is the only one,
4
show that n -""' 1 (mod 5). But 5 In, so n4
0 (mod 5), which is a
==
contradiction.]
(e)
If
{(c, c, c, c, c)}, with c
c has order 5.
"¢
e, is a single-element equivalence class, prove
that
10. If N is a normal subgroup of
G, N is solvable, and G/N is solvable, prove that
G is solvable.
11. Prove that a subgroup Hof a solvable group
G1 ;;;i
•
•
•
2 G,.
=:
G is solvable. (Hint: If G = G0 ;;;2
(e} is the solvable series for G, consider the groups H1 = H n G1•
To show that Hi-if H1 is abelian, verify that the map
by
H1_ifH1->; G1_1/G1 given
H1x->; G1x is a well-defined injective homomorphism.]
*For instance, (r, s, t, u, r) is equivalent to each of (s, t, u,
(r, s, t, u, v) and to no other 5-tuples in S.
v,
r), (t, u, v, r, s), (u,
v, r,
s, t), (v, r, s, t, u),
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12.3
Solvability by Radicals
433
12. Prove that the Galois group of an irreducible quadratic polynomial is
isomorphic to Z2•
13. Prove that the Galois group of an irreducible cubic polynomial is isomorphic
to 71_3 or 81•
14. Prove that the Galois group of an irreducible quartic polynomial is solvable.
[Hint: Corollary 12.5 and Exercises 5 and 11.]
15. Letp(x), q(x) be irreducible quadratics. Prove that the Galois group of f(x)
p(x)q(x) is isomorphic to Z2 X Z2 or Z2• [Hint: If u is a root of p(x) and v a
root of q(x), then there are two cases: vrFF(u) and 11EF(u).]
=
16. Use Galois' Criterion to prove that every polynomial of degrees 4 is solvable
by radicals. [Hint: Exercises 12-15.]
17.
Find the Galois group G of the given polynomial in Q[x]:
(a) x6 - 4x3 + 4 [Hint: Factor.]
(b)
x
4
(c) ;;f
- 5x2+ 6
+
6x3 + 9x
(d) x4 + 3 x3 - 2x - 6
(e) xi - lOx
- 5 [Hint: See Example 5.]
18. Determine whether the given equation over Q is solvable by radicals:
(a) x6 + 2x3 + l
=
(c) 2x5 - 5x4 + 5
(b) 3xi
- 15x + 5 0
5
(d) x - x4 - 16x + 16
0
=
0
=
=
0
19. (a) Prove that O(v'2i) is normal over Q by showing it is the splitting field of
x2+ 2.
(b) Prove that 0(-t'l( 1 - i)) is normal over O(v'2i) by showing that it is the
splitting field of x2 + 2v'2i.
(c) Show that Q !;;; O(v'2i) s;; 0(-¢'2(1 - i)) is a radical extension of Q with
[0(-¢2( 1 - i) ):0] 4 and note that Q contains all second roots of unity
(namely ± 1).
=
(d) LetL 0(-¢'2(1 - i)). Show thatv "¢2(1 + i) isnotinL.
[Hint: If vELand u V2( l - i) EL, show thatv/u iand(v - u)/2i
-¢'2EL, which implies that [L:Q ] � 0(-¢'2, i):O,] contradicting (c) and
Exercise 12(b) in Section 12.2.]
=
=
=
=
=
(e) Prove thatL 0( "¢2(1 - i) ) is not normal over Q [Hint: u and v (as in
(d)) are roots of the irreducible polynomial x4 + 8.]
=
20. Let (be a primitive fifth root of unity. Assume Exercise 21 in Section 4.5 and
prove that GalQQ((), the Galois group of xs - l, is cyclic of order 4.
21. What is the Galois group of xs + 32 over Q? [Hint: Show that 0(() is a
splitting field, where ( is a primitive fifth root of unity; see Exercise 20.]
22. Prove that the group GalpK in Theorem 12.18 is cyclic. [Hint: Define a map
/from GaIFK to the additive group Zn by f(u) k, where u(u) ('u. Show
that/is a well-defined injective homomorphism and use Theorem 7.17.]
=
.. ....
=
..
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..
434 Chapter 12
Galois Theory
C. 23. If p is prime and G is a subgroup of SP that contains a transposition and a
p-cycle, prove that G =Sr [Exercise 8 is the case p = 5.]
24. Iff(x) E O[x] is irreducible of prime degree p andf(x) has exactly two
nonreal roots, prove that the Galois group off(x) is Sr [Example 5 is
essentially the case p
5.]
==
25. Construct a polynomial in O[x] of degree 7 whose Galois group is S7•
eap,ngm.20:12�1..umiq.A:l.lliala 11--4.....,-aatn. � m:...t,, arda(lticlbld.io.wmlliarls,_,. 0.1"�dpll.-mkd.�lrlDlllllm�M ....... Jion1M•Bam:.ndkir�.Bdbmbll_...._
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..
P A R T
3
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C H A P T E R
13
Public-Key Cryptography
Prerequisites: Section 2.3
Codes have been used for centuries by merchants, spies, armies, and diplomats to trans­
mit secret messages. In recent times, the large volume of sensitive material in government
and corporate computeriz.ed data banks (much of which is transmitted by satellite or
over telephone lines) has increased the need for efficient, high-security codes.
It is easy to construct unbreakable codes for one-time use. Consider this "code pad":
Actual Won:/:
Code Word·
morning
evening
Monday
Tuesday
attack
bat
glxt
king
button
figle
If I send you the message FIGLE BUTTON BAT, there is no way an enemy can know
for certain that it means "attack on Tuesday morning" unless he or she has a copy of
the pad. Of course, if the same code is used again, the enemy might well be able to
break it by analyzing the events that occur after each message.
Although one-time code pads are unbreakable, they are cumbersome and inef­
ficient when many long messages must be routinely sent. Even if the encoding and
decoding
are
done by a computer, it is still necessary to design and supply a new pad
(at least as long as the message) to each participant for every message and to make all
copies of these pads secure from unauthorized persons. This is expensive and imprac­
tical when hundreds of thousands of words must be encoded and decoded every day.
For frequent computer-based communication among several parties, the ideal code
system would be one in which
1. Each person has efficient, reusable, computer algorithms for encoding and
decoding messages.
2. Each person's decoding algorithm is not obtainable from his or her encoding
algorithm in any reasonable amount of time.
437
°'l'Jrilll:!O
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_,.....,_.... ,,__ ... _.., _ ... _......,...,-c.g,..1.o1m1o&--1Mriglltto___ .. ..,_ll..-.-�-· ...... ll.
438 Chapter 13
Public-Key Cryptography
A code system with these proper ties is called a public-key system. Although it may not
be clear how condition 2 could be satisfied, it is easy to see the advantages of a public­
key system.
The
encoding algorithm of each participant could be publicly announced-perhaps
published in a book (like a telephone directory)-thus eliminating the need for couriers
and the security problems associated with the distribution of code pads. This would not
compromise secrecy because of condition 2: Knowing a person's
encoding algorithm
would not enable you to determine his or her decoding algorithm. So you would have no
way of decoding messages sent to another person in his or her code, even though you
could send coded messages to that person.
Since the encoding algorithms for a public-key system are available to e\lecyOne, forgery
appears to be a poSSibility. Suppose, for
example,
that a bank receives a coded message
claiming to be from Anne and requesting the bank to transfer money from Anne's account
into Tom's account. How can the bank be sure the message was actually sent by Anne?
The answer is as simple as it is foolproof. Coding and decoding algorithms are in­
verses of each other: Applying one after the other (in either order) produces the word
you started with. So Anne first uses her secret
decoding algorithm to write her name;
say it becomes Gybx. She then applies the bank's public encoding algorithm to Gybx
and sends the result (her "signature") along with her message. The bank uses its secret
decoding algorithm on this "signature" and obtains Gybx. It then applies Anne's pub­
lic
encoding algorithm to Gybx, which turns it into Anne. The bank can then be sure
decoding algorithm to
the message is from Anne, because no one else could use her
produce the word Gybx that is encoded as Anne.
One public-k ey system was developed by R. Rivest, A. Shamir, and L. Adleman
in 1977. Their system, now called the RSA system , is based on elementary number
theory. Its security depends on the difficulty of factoring large integers. Here
are
the
mathematical preliminaries needed to understand the RSA system.
Lemma 13.1
Let p, r, s, c EZ with p prime.
If p ./' c
and re= sc (mod p), then r = s (mod p).
Proof• Since re= sc (modp),p divides re - sc (r - s)c. By Theorem 1.5
Pl(r- s) orplc. Sincep ./' c1 we havep l(r - s), and, hence, r= s(modp).
=
Lemma 13.2
If pis
Fermat's Little Theorem
prime, a EZ, and p
Proof*• None of
•
,r a, then
the numbers
a,
aP-1 = 1 {mod p).
2a, 3a,
.
• .
,
(p - l)a is congruent to 0 modulo
p by Exercise 1. Consequently, each of them must be congruent to one
of 1, 2 , 3, ... , p - 1 by Corollary 2.5 and Theorem 2.3. If two of them
were
congruent to the same one, say ra =
1 Si, r, S Sp
i = sa (mod p) with
- 1,
*A proof based on group theoryis outlined in Exercise 38 of Section 7.3, and one based on field theory
is in Exercise 13 of Section 11.6.
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13
Public-Key Cryptography
haver = s (mod p) by Lemma 13.l
1, 2, 3,
then we would
439
c = a). This is
, p - 1 are con­
(with
impossible because no two of the numbers
• . .
gruent modulo p (the difference of any two is less than p and, hence,
not divisible by p). Therefore, in some order a, 2a, 3a,
congruent to
a•
2a
•
, (p - l)a are
.
1, 2, 3, . . , p - 1. By repeated use of Theorem 2.2,
. .
.
3a ... (p - l)a = 1 2 3 . .
•
.
·
(p -
1)
(mod p).
Rearranging the left side shows that
l 2 3 . . . (p - 1) = 1·2 3
a· a • a . . . a·
•
•
•
r1(1 2. 3 . . . (p - 1)) = 1(1 2
•
Now p ,t
•
•
. . .
3
(p - 1) (mod p)
(p - 1)) (mod p).
. . •
(p - 1)) (if it did,p would divide one of the fac­
1.6. Therefore, d'-1 = 1 (mod p) by Lemma 13.1 (with
. (p - 1)).
•
(1·2 3
•
. . .
tors by Corollary
c =
1 • 2. 3 .
.
Throughout the rest of this discussion p and q are distinct positive primes. Let
n = pq and k = (p - l)(q - 1). Choose d such that (d, k)
1. Then the equation
dx
1 has a solution in Zk by Theorem 2.9 (with n k). Therefore, the congruence
dx = 1 (mod k) has a solution in Z; call it e.
=
=
=
Theorem 13.3
Let p, q, n, k, e, d be as in the preceding paragraph. Then
every
bed= b (mod
n) for
bEZ.
Proof > Since e is a solution of dx = 1 (mod k), de - 1 = kt for some t. Hence,
ed = kt + 1, so that
b"" = b1t+I
If
p
,t
=
lftbl
=
!f.P-IXq-l)tb
=
(lJP-l'jq-1')tb,
b, then by L emma 13.2,
Ir= c1r1)C,-1>b
=
(l)c,-1>1 b
=
b (mod p).
p I b, then b and every one of its powers are congruent to 0 modulo p.
b (mod p). A similar argument shows that
Ir= b (mod q). By the definition of congruence,
If
Therefore, in every case, Ir=
q I (b"' - b).
and
Therefore, pq I
=
divides
(fr - b) by Exercise 2. Since pq
•
W" - b), and, hence, b = b (mod n).
•
The least residue modulo
n
of an integer
by n. By the Division Algorithm,
(mod
n,
n).
c
= nq + r,
is the remainder
so that
c
-
r
=
r
when
c
is divided
nq, and, hence, c
= r
Since two numbers strictly between 0 and n cannot be congruent modulo
the least residue of
modulo
c
n, this means that n
c
is the only integer between 0 and n that is congruent to
c
n.
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440 Chapter 13
Public-Key Cryptography
We can now describe the mechanics of the RSA system, after which we shall show
how it satisfies the conditions for a public-key system .The message to be sent is first
converted to numerical form by replacing each letter or space by a two-digit number:*
space= 00, A= 01, B= 02, ..., Y= 25, Z= 26.
For instance,, the word GO is written as the number 0715 and WEST is written
23051920, so that the message "GO WEST" becomes the number 07150023051920,
which we shall denote by B.
Let p, q, n, k, d,
e,
be as in Theorem 13.3, with p and q chosen so that B < pq= n.
To encode message B, compute the least residue of B' modulo n; denote it by C. Then
C is the coded form of B. Send C in any convenient way.
The person who receives C decodes it by computing the least residue of Cd modulo
n.
This produces the original message for the following reasons. Since B', is congruent
modulo n to its least residue C, Theorem 13.3 shows that
c' = (Ir)d = B" = B (mod n).
The least residue of
Cd is
modulo n and 0 < B <
n.
the only number between 0 and
n
that is congruent to C'1
So the original message B is the least residue of Cd.
Before presenting a numerical example, we show that the RSA system satisfies the
conditions for a public-key system:
1. When the RSA system is used in practice,p and q are large primes (several hun­
dred digits each).Such primes can be quickly identified by a computer. Even
though B, e, C, d are large numbers, there are fast algorithms for finding the
d
least residues of B' and C modulo n. They are based on binary representation
of the exponent and do not require direct computation of B' or Cd (which would
be gigantic numbers). See Knuth [31] for details. So the encoding and decoding
algorithms of the RSA system are computationally efficient.
2. To use the RSA system, each person in the network uses a computer to choose
appropriate p, q, d and then determines n, k,
e.
The numbers
e
and
n
for the
encoding algorithm are publicly announced, but the prime factors p, q of n and
the numbers d and k
sages by using
e
are
kept secret. Anyone with a computer can encode mes­
and n. But there is no practical way for outsiders to determine
d (and, hence, the decoding algorithm) without first findingp and q by factoring
n.t W ith present technology this would take thousands of years! So the RSA
system appears secure, as long as new and very fast methods of factoring are
not developed.
Even when
n
is chosen as above, there may be some messages that in numerical
form are larger than n. In such cases the original message is broken into several blocks,
each of which is less than n. Here is an example, due to Rivest-Shamir-Adleman.
*More numbers could be used for punctuation marks, numerals, special symbols, etc. But this will be
sufficient for illustrating the basic concepts.
t Alternatively, one might try to find k and then solve the congruence ex =
1
(mod k) to get d. But this
can be shown to be computationally equivalent to factoring n, so no time is saved.
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13
Public-Key Cryptography
441
EXAMPLE 1
Letp = 47 and q = 59. Thenn = pq = 47 59 = 2773 and k = (p - l )(q- 1) =
58 = 2668. * Letd= 157. A graphing calculator or computer quickly veri­
fies that (157, 2668) = l and that the solution of 157x = 1(mod2668) is e = 17.t
We shall encode the message "IT'S ALL GREEK TO ME." We can encode only
numbers less than n ""2773. So we write the message in two-letter blocks (and
denote spaces by #):
·
46
·
IT
0920
S#
1900
AL
0112
L#
1200
GR
0718
EE
0505
K#
1100
TO
2015
#M
0013
E#
0500.
Then each block is a number less than 2773. The first block, 0920, is encoded by
using e = 17 anda computer to calculate the least residue of 92011 modulo 2773:
92017
=
948 (mod 2773).
The other blocks are encoded similarly, so the coded form of the message is
0948
2342
1084
1444
2663
2390
0778
0774
0219
1655.
A person receiving this message would used= 157 to decode each block. For
instance, to decode 0948, the computer calculates
948 151 = 920 (mod 2773).
This is the original first block 0920
""
IT.
For more information on cryptography and the RSA system, see Hoffstein, Pipher,
and Silveman [33], Rivest-Shamir-Adleman [34], Simmons [35], and Trappe and
Washington [36].
• Exercises
A. I.
Let p be a prime and k, a E Z such that p ./' a and 0 < k < p. Prove that ka ¥= 0
(mod p). [Hint: Theorem 1.5.]
2. If p and q are distinct primes such that p I c and q I c, prove that pq I c. [Hint:
If c =pk, then q lpk; use Theorem 1.5.]
*T hese numbers will illustrate the concepts.
But they
are too small to provide a secure code since
2773 can be factored by hand.
tTo solve the congruence on a calculator, use the Technology Tip on page 12 to find u and v such that
157u + 2668v= 1.Then 157u -1=2668v, which means that 157u
=
1(mod2668).
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442 Chapter 13
Public-Key Cryptography
3. Use a calculator and the RSA encoding algorithm withe=
3, n = 2773 to
encode these messages:
(a) GO HOME
(b) COME BACK
(c) DROP DEAD
[Hint: Use 2-letter blocks and don't omit spaces.]
4. Prove this version of Fer mat's Little Theorem: Ifpis a prime and
d'
==a
(modp).
a
EZ, then
[Hint: Consider two cases,p I a andp ,r a; use Lemma 13.2 in
the second case.]
B. 5. Find the decoding algorithm for the code in Exercise 3.
6. Let C be the coded form of a message that was encoded by using the RSA
algorithm. Suppose that you discover that C and the encoding modulus n
are not relatively prime. Explain how you could factor n and thus find the
decoding algorithm. [The probability of such a C occurring is less than
10-99
when the prime factors p, q, of n have more than 100 digits.]
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14
C H A P T E R
The Chinese Remainder Theorem
Prerequisites: Section 21
. and Appendix C for Section14.1; Section 3.1
for Section14 .2; Section 6.2 for Section14.3.
The Chinese RemainderTheorem (Section 14.1) is a famous result in number theory
that was known to Chinese mathematicians in the fl rst century. It also has practical
applications in computer arithmetic (Section 14.2). An extension of the theorem
to rings other than Z has interesting consequences in ring theory (Section 14.3).
Although obviously motivated by Section 14.1, Section 14.3 is independent of the
rest of the chapter and may be read atany time after you have read Section 6. 2.
Ill
Proof of the Chinese Remainder Theorem
A congruence is
"
=
an
equation with integer coefficients in which"=" is replaced by
(mod n)." The same equation can lead to different congruences, such as
6x + 5 = 7 (mod 3)
or
6x +5 = 7 (mod5).
Only integers make sense as solutions of congruences, so the techniques of solving
equations are not always applicable to congruences. For instance, the equation 6.x +5 = 7
has x = 1/3 as a solution, but the congruence 6x + 5
=
7 (mod
3)
has no solutions
(Exercise )
3 , and 6x + 5 = 7 (mod5) has infinitely many solutions (Exercise4).
A number of theoretical problems and practical applications require the solving of
a system of linear congruences, such as
x= 2 (mod4)
x=5 (mod 7)
x= O(mod11)
x= 8(mod15)
443
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444 Chapter 14
The Chinese Remainder Theorem
A solution of the system is an integer that is a solution of
every congruence in the
sys­
tem. We shall examine some cases in which a system of linear congruences must have
a solution.
Lemma 14.1
Ifm and
n
are relatively prime positive integers and a, b EZ, then the system
x = a(modm)
x = b {mod n)
has a solution.
Proof" Since (m, n) = 1, there exist integers u and v such that mu + nv = 1 by
Theorem 1.2. This equation and the definition of congruence lead to
four conclusions:
(ii) nv
(i) mu s 0 (mod m)
(iii)
nv =
O (mod
n)
(iv)
s
mu
Let t = bmu + anv Then by (i),
.
t = bmu + anv = b
•
1
(mod m)
[Becawe 1 - nv =mu.]
n)
[Because 1 - mu = nv.]
=1 (mod
(ii),
and Theorem 2.2,
0+a·1 =a
(mod m),
so that t =a (mod m). Similarly, by (iii), (iv), and Theorem 2.2,
t= bmu + anv = b
· 1 +a · 0= b
(mod n),
so that t = b (mod 11). Therefore, tis a solution of the system.
•
The proof of Lemma 14.1, provides the
Solution Algorithm for the System in Lemma 14.1
1. Find u and v such that mu
+ nv = 1.*
2. Then t = bmu + anv is a solution of the system
EXAMPLE 1
To solve the system
x = 2 (mod4)
x
apply the algorithm with m = 4, n
= 5 (mod 7),
=
7, a= 2, b = 5:
1. It is easy to see that u = 2, v = -1 satisfy 4u + 1v = 1.
2. Therefore, a solution of the system is
t= bmu +
anv = 5
•
4
•
2+ 2
•
7
•
(-1) = 26.
"This can be done by hand by using the Euclidean Algorithm; see Exercise
15 in
Section
also be done on a computer or graphing calculator; see the Technology Tip on page
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1.2.
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14.1
Theorem 14.2
Let m1, m2,
that (m1, m1)
•
•
=
•
Proof of the Chinese Remainder Theorem
445
The Chinese RemainderTheorem*
be pairwise relatively prime positive integers (meaning
whenever i :f:. j). Let a1, a2,
, a, be any integers.
, m,
1
•
•
•
(1) The system
x=
a1 (mod m1)
x=
a2 (mod m2)
x=
a3 (mod m3)
x =a, (mod
m,)
has a solution.
(2) If t is one solution of the system, then an integer z is also a solution
if and only if z = t (mod m1 m2 m3
m,).
•
•
•
For reasons that will become apparent below, we shall
use
induction to prove the
first part of the theorem. For a proof that does not use induction, see Exercise 21.
Proof ofTheorem 14.2 ... (1) Tue proof is by induction on the number, of congru­
ences in the system. If
(with
m
=
r =
mh n = lnz., a
is a solution when
r =
=
2, then there is a solution by Lemma 14.1
ah b
=
al). So suppose inductively that there
k and consider the system
x = a1(mod m1)
x = a2(mod Tnz.)
x = a3(mod lnJ)
x = a1c(mod m,J
x=
ak+I(mod ffllc+1)
By the induction hypothesis, the system consisting of the first k congru­
ences in(•) has a solutions. F urthermore,
m11nz.ffl3
• • •
relatively prime (Exercise 5). Consequently, by Lemma
x=s
(mod
m1m2!'13
•
•
•
mk and mk+l are
14.1, the system
m,J
*So named because it was known to Chinese mathematicians in the first century.
........
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mm-.111&om:udlar�a).B:blrilf_...._
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446 Chapter 14
The Chinese Remainder Theorem
has a solution
t. The number tnecessarily
t = s(mod m1m2m3
Consequently, for each i = 1, 2, 3,
...
satisfies
m,J.
•• •
, k,
t = s(modmJ.
t - sis divisible by m1m,.m3
mk, then it is divisible by each
mJ. Nows is a solution of the first k congruences in(•*), so for each is k
(Reason: If
•
t = s(mod mJ
•
•
s = a1 (mod m;).
and
By transitivity(Theorem 2.1),
t = a,(mod m1)
for i = 1, 2, ... , k.
Since tis a solution of(**), it must also satisfy t = ak+t (mod
m1<+J·
H ence, tis a solution of the system( * ) , so that there is a solution
when
k + 1. Therefore, by induction, every such system has a
r =
solution.
(2) If z is any other solution of the system, then for each i
z =ai(modmJ
=
1, 2, ... , r,
t = a1 (mod mJ.
and
By transitivity(Theorem 2.1), z = t(mod
mJ. Thus
m1 I(z - t), m,. I (z - t), "'3 I(z - t), ... , m, I(z - t).
Therefore,
m1m.p13
• • •
m, I(z
- t) by Exercise 7. Hence,
z = t(mod
m1m,.m3
Conversely, if z = t(mod m1m.p13
for each i
=
1, 2,
. .. , r.
• •
•
m,).
m,.), then, as above, z = t(mod mJ
Since t = a,(modmJ, transitivity shows that z =a,
• •
•
(mod m;) for each i. Therefore, z is a solution of the system.
•
The proof of Theorem 14.2 actually provides an effective computational algorithm
for solving large systems: Solve the first two by Lemma 14.1, then repeat the inductive
step
as
often as needed to determine a solution of the entire system.
EXAMPLE 2
We shall solve the system
x
= 2(mod4)
x
= 5 (mod 7)
x= 0(mod 11)
x = 8(mod 15).
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-....ed.--
.. ��1*-Ml........,dllcl....�......��Lamaloa ........riBbtla-....,,..�IDllllll.-..,....jf......._.:ligbb�........
14.1
Proof of the Chinese Remainder Theorem
447
Example l shows that x = 26 is a solution of the system consisting of the first two
congruences:
x = 2(mod4)
x = 5 (mod 7).
Next we solve the system
x = 26(mod 4
7)
•
x= (
0 mod11).
First, note that u = 2 and
v
= -5 satisfy 28u + 11v =1. * Then the Solution
Algorithm preceding Example 1 (with a
=
26, m = 4 · 7 = 28, b = 0 , n = 11) shows
that a solution is
bmu + anv = 0
•
28
·
2 + 26 • 11
·
(-5) = -1430.
You can readily verify that x=-1430 is also a solution of the system consisting of the
first three congruences:
x = 2(mod4)
x = 5 (mod 7)
x = (mod
0
11).
Finally, we solve this system:
x = -1430(mod 4
·
7
·
11)
(mod 15).
x=8
Note that u=2 and v = -41 satisfy 308u + 15v = 1.* So by the Solution Algorithm
(with a=-1430,m=4
bmu +
·
7
•
anv = 8
You can verify that x
=
11=308, b =8,n=15), a solution is
·
308
•
2 +(-143)
0
•
15
•
(-41) = 884,378.
884,378 is a solution of the entire system
x = 2(mod4)
x = 5 (mod 7)
x = O(mod11)
x = 8(mod 15).
Since 4 7
·
·
11 • 15 = 4620 and 884,378 = 1958 (mod 4620), as you can easily
verify, x = 1958 is also a solution of the system by Theorem1 4.2 . When work­
ing by hand, the smaller solution is easier to use. So we say that the solutions
of the system are all numbers that are congruent to 1958 modulo 4620.
*The values for u and v we re found with a graphing calculator program; see the Technology Tip on
page
12.
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..
448 Chapter 14
The Chinese Remainder Theorem
Technology Tip: Systems such as the one in Example 2 can be solved by the
Chinese Remainder Theorem program for TI graphing calculators that can be
downloaded from our website (ADDRESS TBA). In Example 2, when asked,
you enter the list of constants {2, 5,
{4,7,
0, 8}
and the corresponding list of moduli
11, 15}. The program then produces the solution, as shown in Figure 1.
SOLUTIOH
19:58
t100ULO
462.0
Done
FIGURE1
To solve the same system with Maple, use the comm.and
chrem ([2, 5, 0,
8], [4, 7,
11, 15]);
•
• Exercises
A. 1. If u = v(mod n) and u is a solution of 6x + 5 = 7 (mod n), then show that vis
also a solution. [Hint: Theorem22
. .]
2. If 6x + 5 = 7 (mod n) has a solution, show that one of the numbers 1, 2, 3,
n-
1 is also a solution.
[Hint: Exercise
... ,
1 and Corollary2.5.]
3. Show that6x + 5 = 7 (mod 3) has no solutions.
[Hint: Exercise2.]
4. Show that6x + 5 = 7 (mod 5) has infinitely many solutions.
[Hint:
Exercises 1 and2.]
m., fn:2,
, m"' mk+t are pairwise relatively prime positive integers (that is,
m1) 1 when i #: j), prove that m1fn:2
mk and mk+I are relatively prime.
[Hint: If they aren't, then some prime p divides both of them (Why?). Use
5. If
(m,,
• • •
=
•
•
•
Corollary 16
. to reach a contradiction.]
(m, n) 1 and m Id andn I d, prove that mn Id. [Hint: If d
n I mk; use Theorem 1.4.]
6. If
=
=
mk, then
mi. m2,
, m, be pairwise relatively prime positive integers (that is,
(m,mj) 1 when i #: j). Assume that m1 Id for each i. Prove that
m1mtn3
m, Id. [Hint: Use Exercises 5 and 6 repeatedly.]
7. Let
• • •
=
•
•
•
In Exercises 8-13, solve the system of congruences.
8. x= 5 (mod 6)
x= 7(mod 11)
9. x= 3 (mod 11)
x= 4(mod 17)
IO. x = 1(mod2)
11. x= 2(mod 5)
x= 2(mod3)
x= 0 (mod6)
x= 3 (mod 5)
x= 3 (mod7)
......
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to:.J.._t1&dl::udkx'�.BimorW......-._
-..d.1lllmy��
ao1.--.n,dltcl.n.�.--.....---.��---ftgbt1D__,,,.�mallllll:-..,.m..��:Dgb&l�
k.
...
......
14.1
Proof of the Chinese Remainder Theorem
13. x = 1(mod7)
12. x = 1(mod5)
x= 6 (mod11)
x = 0(mod12)
x = 3 (mod 6)
x
5(mod11)
=
x = 10
449
x = 9(mod13)
(mod13)
x=O(mod17)
B.14. (Ancient Chinese Problem) A gang of17 bandits stole a chest of gold coins.
When they tried to divide the coins equally among themselves, there were
three left over. This caused a fight in which one bandit was killed. When the
remaining bandits tried to divide the coins again, there were ten left over.
Another fight started, and five of the bandits were killed. When the survivors
divided the coins, there were four left over. Another fight ensued in which
four bandits were killed. The survivors then divided the coins equally among
themselves, with none left over. What is the smallest possible number of coins
in the chest?
15. If(a,
n) = d and d I b, show that ax= b(mod n) has a solution. [Hint: b = de
e, and au + nv = d for some u, v (Why?). Multiply the last equation
by c; what is aue congruent to modulo n?]
for some
16. If (a,
n) = d and d .t b, show that ax= b(mod n) has no solutions.
17. If(a,
n) = 1 ands, tare solutions of ax=b(mod 11), prove thats=t(mod n).
[Hint: Show that n I (as - at) and use Theorem 1.4.)
18. If (a, n)
= dands, t are solutions of ax= b(modn), prove thats= t (modn/d).
19. If (m, n)
=
d, prove that the system
x =a(modm)
x=b(modn)
has a solution if and only if a= b(mod d).
20. Ifs,
tare solutions of the system in Exercise 19, prove thats=t(mod r),
r is the least common multiple of m and n.
where
21. (Alternate Proof of part ( 1) of the Chinese Remainder Theorem) For each
i = 1;2 , .. .
, r,
let N1 be the product of all the m1 except m1, that is,
N1
=
m1m2
•
•
•
m1_1m1+1
• •
•
m,.
(a) For each i, show that(N1, m1) = 1, and that there are integers u1 and
that N(IJ.1 + m (V1 =
v1
such
1.
(b) For each i andjsuch that i ;f:.j, show that N1u1=0(mod m1).
(c)
For each i, show that N,.u; = 1 (modm1).
(d) Show that t = a!f1u1 + a2N2ui + a1N1u3 +
the system.
· ·
·
+ a,N,u, is a solution of
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it.
....
......
450 Chapter 14
1111
The Chinese Remainder Theorem
Applications of the Chinese Remainder Theorem
Every computer has a limit on the size of integers that can be used in machine arith­
metic, called the word size. In a large computer this might be235• Computer arithmetic
with integers larger than the word size requires time-consuming multiprecision tech­
niques. In such cases an alternate method of addition and multiplication , based on the
Chinese Remainder Theorem, is often faster.
For any numbers
r, s,
t, n less than the word size, a large computer can quickly
calculate
r
+ sand r
·
s(even
when the answer is larger than the word size);
the least residue of t modulo n*(including the case when t exceeds the word size­
see
Exercise2);
sums and products in
Z,,.
Finally, a computer can use a slight variation of the Chinese Remainder Theorem
solution algorithm (Theorem 14.2) to solve systems of congruences. But this may
involve numbers larger than the word size and, hence, require slower multiprecision
techniques.
To get an idea of how the alternate method works, imagine that the word size of
our computer is 100 , so that multiprecision techniques must be used for larger num­
bers. The following example shows how to multiply two four-digit numbers on such a
computer, with minimal use of multiprecision techniques.
EXAMPLE 1
We shall multiply 3456 by7982 by considering various systems of congruences
and using the Chinese Remainder Theorem. We begin by choosing several
numbers as moduli and finding the least residues of 3456 and7982 for each
modulus:t
3456
= 74(mod89)
7982
3456
=
36(mod95)
7982
=2
3456
= 61(mod97)
7982
= 28(mod97)
3456
= 26(mod98)
7982
=
3456
= 90 (mod99)
7982
= 62(mod 9 9).
Then by Theorem 2 2.
least residue of 74
·
we
= 61(mod89)
know that 3456
(mod95)
44(mod 98)
•
7982
=
74
·
61 (mod 89). Taking the
61 modulo 89 and proceeding in similar fashion for the other
congruences, we have
*The least-residue modulo n of a number tis the remainder rwhen tis divided by n.
Algorithm, t
=
nq + r so that t - r
trhereason why89, 95, 97,
98,
and
=
nq and
By the Division
t == r (mod n).
99 were chosen
as moduli will be explained below.
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14.2
Applications of the Chinese Remainder Theorem
3456
•
7982
=
74
·
61
=
64 (mod 89)
3456
·
7982
=
36
·
2
=
72 (mod 95)
3456
·
7982
=
61
·
28
=
59 (mod 97)
3456
•
7982
=
26
•
44
=
66 (mod 98)
3456
·
7982
=
90
•
62
=
36 (mod 99).
451
Therefore, 3456 · 7982 is a solution of this system:
( ***)
x =
64 (mod 89)
x =
72 (mod 95)
x =
59 (mod 97)
x =
66 (mod 98)
x =
36 (mod 99).
The Chinese Remainder Theorem* shows that one solution of(•••)is 27,585,792
and that every solution (including 3456
89 • 95
•
97
•
98
·
99
=
7982)is congruent to this one modulo
•
7,956,949,770(which we denote hereafter by M). Since no two
numbers between 0 and
tion between 0 and
3456
·
M can be congruent modulo M, 27,585,792 is the only solu­
M. We know that 0 < 3456 7982 < 10" 10" 108 < M. Since
·
7982 is a solution, we must have 3456
Now look at this example
residue
of
a
number modulo
·
7982
•
=
=
27,585,792 .
from a different perspective. If you think of the least
as an element of Z"' then the congruences in (•)say
n
that the integer 3456 may be represented by the element (74, 36, 61, 26, 90)in the ring
Za9 x °LJs x Z97 x Z98 XZ99•
that 74
·
61
=
Similarly, 7982 is represented by (61, 2, 28, 44, 62).Saying
64 (mod 89)in(**) is the same
as
saying 74
·
61
=
64 in
Z89•
So the
congruences in (••)are equivalent to multiplication in Za9 X°LJs XZ91XZ98X1'.99:
(74, 36, 61, 26, 90). (61, 2, 28, 44, 62)
=
=
(74. 61, 36 . 2, 61 . 28, 26. 44, 90 . 62)
(64, 72, 59,
66, 36).
The solution of (•**)shows that the element (64, 72, 59, 66, 36) of the ring
Z89 X � s X Z97 X �8 XZ99 represents the integer 27 ,585, 792.
The procedure in the case of a realistic word size is now clear. Let m1, •••
pairwise relatively prime positive integers:
1 . Represent each integer t as an element of Zm,
ence class of t modulo each m1•
2. Do the arithmetic in Z,,,, X ·
·
·
X
•
• •
, m, be
Xz,,., by taking the congru­
Xz,,.,.
3. Use the Chinese Remainder Theorem to convert the answer into integer form.
The m1 must be chosen so that their
product
M is larger than any number that will
result from the computations. Otherwise, the conversion process in Step 3 may fail
(Exercises
3-5). This is sometimes
done, as in the example, by taking the
m1
to be as
"Up to this point, all computations have been quickly pertormed by our imaginary computer. This is
the first place where slower muHiprecision calculations may
be needed because of numbers that
exceed the word size.
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.
CD111111:•_..,...._��:ligl!U�....-.it.
452 Chapter 14
The Chinese Remainder Theorem
large as possible without exceeding the word size of the computer. If smaller moduli
are chosen, more of them may be necessary to ensure that Mis large enough.
The conversion process from integer to modular representation and back (Steps 1
and 3) requires time that is not needed in conventional integer multiplication (espe­
cially Step 3, which may involve multiprecision techniques). But this need be done only
once for each number, at input and output. The modular representation may be used
for all intermediate calculations. It is much faster than direct computation with large
integers, especially in a computer with parallel processing capability, which can work
simultaneously in each Zm,- Under appropriate conditions the speed advantage in Step 2
outweighs the disadvantage of the extra time required for Steps 1 and 3. For more
details,
see
Knuth [31].
It is sometimes necessary to find an exact solution (not a decimal approximation)
of a system of linear equations. When there are hundreds of equations or unknowns
in the system and the coefficients are large integers, the usual computer methods w ill
produce only approximate solutions because they round off very large numbers dur­
ing the intermediate calculations. The Chinese Remainder Theorem is the basis of a
method of finding exact solutions of such systems.
Very roughly, the idea is this. Let
pairwise relatively prime).* For each
mi,
m1,
...
, m,
be distinct primes (and, hence,
translate the given system of equations into
a system over ll.m, by replacing the integer coefficients by their congruence classes
modulo
Then solve each of these new systems by the usual methods (Gauss­
m1•
Jordan elimination works equally well over the field z..., as over Ill, and round-off is
not a problem with the smaller numbers in Z,,.). F inally, use the Chinese Remainder
Theorem and matrix algebra to convert these solutions modulo
m1
into a solution of
the original system.t
• Exercises
A. 1. Assume that your computer has word size 100. Use the method outlined in
the text to find the sum 123,684 + 4 13,456, using m1
m4
=
=
95,
m2
=
97,
m3
=
98,
99.
2. (a) F ind the least residue of 64,397 modulo 12, using only arithmetic in Z12•
[Hint: Use Theorems 2.2 and 2.3 and the fact that 64,397 =
(((6 . 10 + 4)10 + 3)10 + 9)10 + 7.]
(b) Let n be a positive integer less than the word size of your computer and
t any integer (possibly larger than the word size). Explain how you might
find the least residue of I modulo n, using only arithmetic in Zn (and thus
avoiding the need for multiprecision methods).
•considerations of size similar to those discussed above play a role in the selection of the m1•
'This
conversion is a bit trickier than may first appear. For instance, the system
Bx+ Sy = 12
�+� = W
You can verify that
becomes
x + Sy = 5
�+� = 3
x = 4 , y = 3 i s a solution o f the Z 7 system. I t is
1j2, y
from this to the solution of the original system, which is x
=
over Z1.
not immediately clear how t o get
=
8/5.
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14.3
The Chinese Remainder Theorem for Rings
453
3. Use the method outlined in the text to represent 7 and 8 as elements of Z3 X Z5•
Show that the product of these representatives in Z3 X Z5 is (2, 1).If you use the
Chinese Remainder Theorem as in the text to convert (2, 1) to integer form, do
you get 56? Why not? This example shows why the method won't work when the
product of the 111ti s less than the answer to the arithmetic problem in question.
Also
see
Exercise 5.
"14 X Zs be given by fi..t)
B. 4. Let/Z
: -+Z3 X
=
({t]3, [t1, [t ]5), where [ t],, is the
congruence class oft in Z,.. The function/may be thought of as representing
as an element of Z3 X "14 X Z5 by taking its least residues .
(a)
(b)
If 0 s r, s < 60, prove thatf(r)
[Hint: Theorem 14.2.]
=
Give an example to show that if
f(s) if and only if r
=
t
s.
r ors is greater than 60, then part (a) may
be false.
5. Let
m1o m2,
• • •
,
m, be pairwise relatively prime positive integers and
/Z
: -+Zm, x z,,,. x
·
· ·
x Zm" the function given by
f(t)
([t],,,,, [t],,,,, ..., [t],,,),
=
where [ t],,,, is the congruence class of tin z,,., Let M
0 s r,
s
case.]
< M, prove thatf(r) = f(s) if and only if r
m 1m2
m,. If
s. [Exercise 4 is a special
=
=
• • •
35
6. Assume Exercise 7(c). If your computer has word size 2 , what
m1 might you
choose in order to do arithmetic with integers as large as 2184 (approximately
5
2.45 X.10 5)?
C. 7.
(a)
If
a
and bare positive integers, prove that the least residue of 2" - 1
modulo
(b)
If
a
'l!'
-
1 is 2'
1, where r is the least residue of
-
and b are positive integers, prove that the greatest common divisor of
2a - 1 and 2b - 1 is 2'
-
1, where tis the gcd of
Euclidean Algorithm and part (a).]
(c)
a
and b. [Hint: Use the
Let a and bbe positive integers. Prove that 2" - l and 2b - 1 are relatively
prime if and only if
Ill
a modulo b.
a
and b are relatively prime.
The Chinese Remainder Theorem for Rings
The Chinese Remainder Theorem for two congruences can be extended from Z to
other rings by expressing it in terms of ideals.The key to doing this is the definition of
congruence modulo an ideal (Section 6.1) and the following fact: When A and
ideals in a ring R, the set of sums
B are
{a + bI a EA, b E B} is denoted A + B and is itself
an ideal (Exercise 20 of Section 6.1).
Let m and n be integers.Let Ibe the ideal of all multiples of
minZ and J the ideal
n. Then congruence modulo m is the same as congruence modulo the
i<kal I. If (m, n) l , then mu + nv 1 for some u, v EZ.Multiplying this equation by
any integer r shows that m(ur) + n(vr)
r. Thus every integer is the sum of a multiple
of m and a multiple of n, that is, the sum of an element of the ideal I and an element
of the ideal J.Therefore, I+ J is the entire ringZ. So the condition (m,n)
l amounts
to saying I+ J Z.
of all multiples of
=
=
=
=
=
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454 Chapter 14
The Chinese Remainder Theorem
When
(m, n) = 1, the intersection of
the ideals I and J is the ideal consisting of all
14.1). So two integers are congruent modulo mn
precisely when they are congruent modulo the ideal In J.
The italicized statements in the preceding paragraphs tell us how to translate the
Chinese Remainder Theorem for two congruences into the language of ideals. By
replacing the ideals in that discussion by ideals in any ring R, we obtain
multiples of mn (Exercise 6 of Section
Theorem 14.3
Let
Chinese RemainderTheorem for Rings
I and J be ideals in a ring R such that I+ J = R. Then for any a, b ER , the
system
x= a (mod/)
x =
b (mod J)
has a solution. Any two solutions of the system are congruent modulo
I n J.
an identity, the theorem can be extended to the case of r ideals I" /i, ... ,
� = R whenever i * j
(see Exercise 6 and Hungerford [5; p. 131D.
When R has
I, and congruences x = '4 (mod IJ, under the hypotheses that I, +
Proof of Theorem 14.3 ... Since I+ J = R and b - a ER, there exist i E l,j EJ
such that i + j
= b - a. Hence, a + i = b - j. Let t = a + i; then
t - a = (a + 1) - a = i El,
so that t = a (mod J). Similarly, since a + i
=
b-j
b =(a+ 1) - b = (b - j)-
b
= -jEJ.
t-
Hence, t= b (mod J), and t is a solution of the sy stem. Ifz is also
a
solution, then
z =a (mod/)
by Theorem
and
t = a (modi)
imply that
z = t(mod/)
6.4. Similarly,z = t(mod J). This means thatz - t EI and
= t ( mo d /n J).
•
z - tEJ. Therefore,z - tEI n Jandz
One consequence of the Chinese Remainder Theorem is
a
usef ul isomorphism of
rings.
Theorem 14.4
If I and J are ideals in a ring
of rings
R and I + J = R, then there is an isomorphism
R/(I n J)
=
R/I X R/J.
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14.3
The Chinese Remainder Theorem for Rings
455
Proof • Define a map f:R-+ R/ Ix R/J by f(r)= (r + I, r +J). Then/is a
homomorphism because
f(r) +f(s) = (r +I, r +J) +(s + I, s +J)
= ((r + s) +I, (r +s) + J)= f(r +s)
and
f(r)f(s) = (r
I, r +J)(s + I, s +J)
= (rs + I, rs +J)= f(rs).
+
To show that/is surjective, let (a +I, b +J) ER/ IX R/J. We must find
an element of R whose image under f is (a + I, b +J). By Theorem 14.3
there is a solution
t ER for this system:
x s
a(modl)
x =
b (modJ).
Butt = a (mod I) implies that t +I= a+ !by Theorem 6.6. Similarly,
t s b (mod J) implies t + J= b +J, so that
f(t) = (t + I, t +J) = (a + I, b + J).
Therefore, f is surjective.
Let Kbe the kernel off. By the First Isomorphism Theorem 6.13,
R/K
R/ IX R/J. Now K consists of all elements r ER such
thatf(r) is the zero elementin R/ IX RfJ, that is, all r such that
is isomorphic to
(r +I, r + J) = (OR+ J,
OR
+J),
or equivalently,
r +I= OR+ I
and
r+ J= OR +J.
But
r +I= OR + I means that r =OR (mod I), and, he