Galois Theory
for Beginners
A Historical
Perspective
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Volume
MATHEMATICAL
LIBRARY
35
Galois Theory
for Beginners
A Historical
Perspective
Bewersdorff
Jérg
Translated
by David
Kramer
Q
e‘
"04
5
2
in
u
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%"A'nEn
AMERICAN
MATHEMATICAL
Providence, Rhoda
Island
SOCIETY
Editorial
as
Gerald
B.
Robin
Forman
Board
Folland
Brad
Osgood
Michael
(Chair)
Starbird
Originally published in the German
language by
Friedr.
Vieweg 85 Sohn Verlag, 65189 Wiesbaden, Germany,
Algebra fiir Einsteiger. 2. Auflage(2ndedition)”.
“Jorg Bewersdorff:
@ Friedr. Vieweg & Sohn Verlag|GWV Fachverlage GmbH,
Wiesbaden, 2004
Translated
by David
with
Kramer
by the
information
additional
For
corrections
author.
Subject Classification.
Primary
Secondary 12F10.
Mathematics
2000
and
additions
and
updates
on
this
12-01;
book,
visit
/ Jorg
Bewersdorff
www.ams.org/bookpages/stml-35
of
Library
Cataloging-in—Publication Data
Congress
Bewersdorff,
Jorg.
[Algebra fiir Einsteiger.
English]
Galois
:
a
theory for beginners
translated
by David Kramer.
p.
cm.
Includes
ISBN-13:
perspective
library,
ISSN
1520-9121
', V.
;
35)
index.
978—0—8218—3817-4
(acid-free paper)
2. Polynomials.
I. Title.
theory.
1. Galois
QA214
mathematical
(Student
-—
historical
.B49
2006
512’.32—dc22
2006048423
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111009080706
the
guidelines
Contents
Preface
to
the
English Edition
vii
Prefaces
to
the
German
ix
Chapter
1.
Cubic
Equations
Chapter
2.
Casus
Irreducibilis:
Editions
The
Birth
of the
Complex
Numbers
Chapter
3.
Biquadratic
Chapter
4.
Equations
The
Fundamental
Chapter
5.
and
Proof
The
Search
Equations
of
Degree
Theorem
23
n
and
of
Their
Properties
Algebra: Plausibility
32
for Additional
Solution
Formulas
Permutations
The
37
42
Fundamental
Theorem
on
Symmetric
Polynomials
Ruffini and
27
the
General
47
Equation
of Fifth
Degree
50
i
Contents
Vi
6.
Chapter
The
Can
That
Equations
of
Decomposition
7.
The
Construction
Constructions
Classical
The
8.
Chapter
with
The
The
Construction
57
63
Polygons
and
69
Compass
Problems
of Equations
74
of the
of Tschirnhaus
Fifth
and
Degree
9.
The
Computing
A
89
Galois
the
Group
Galois
Quick Course
10.
Chapter
The
in
and
of
an
93
Equation
114
Group
Calculating
Algebraic Structures
Groups
with
Polynomials
119
Galois
Theory
125
and
Fields
Fundamental
130
Theorem
of Galois
Theory:
An
Example
Artin’s Version
Galois
The
81
of Bring and
J errard
Chapter
55
60
of Regular
Transformations
Degree
Criterion
Straightedge
Solution
in
Polynomials
Integer
Eisenstein’s Irreducibility
Chapter
Be Reduced
144
of the
Theorem
of the
of
149
Theory
Unsolvability
Problems
Fundamental
Classical
Construction
161
Epilogue
165
Index
177
Preface
to
the
English
Edition
This
book
is
translation
a
of the
second
edition
of my
German
book
Von dew" Gleichungsaufliisung
zur
Galoz's- TheoAlgebra fiir Einsteiger.‘
edition
has been expanded
rie, Vieweg, 2004. The original German
of exercises.
The goal of the book
is described
in
by the addition
the original preface.
In a few words
it can
be sketched
as
follows:
Galois theory is presented in the most
elementary way, following the
historical
evolution.
The main focus is always the classical
application
to algebraic equations
and their
solutions
I am
grateful
by radicals.
to David
than
translate
the present
Kramer, who did more
book,
having also offered several suggestions for improvements.
My thanks
are
also directed
to Ulrike
of Vieweg, who
Schmickler—Hirzebruch,
first proposed
a translation
to the American
Mathematical
Society,
and to Edward
Dunne, of the AMS, for managing the translation.
Jorg Bewersdorff
Translator’s
I wish
to
for his helpful
Jorg Bewersdorff
on
the translation
and to the following individuals
at
Mathematical
Dunne
for entrusting
Society: Edward
express
collaboration
the
Note
American
my
appreciation
to
vii
7
Preface
viii
me
with
this
project,
’IEXnicalsupport,
the
and
Barbara
Arlene
Beeton
for her
O’Sean for
the
to
English
friendly
careful
her
Edition
and
intelligent
copyediting of
translation.
Kramer
‘David
E>
the
to
Pregfaces
German
Editions
Math
is like
love; a simple idea, but
it
get complicated.
can
—
Preface
to
the
First
German
subject
of this
book
is the
history
bra.
We will recount
the
search
for formulas
of fundamental
importance
Let
brieflyat
best
us
look
mathematicians
reader,
recall
you
of
of
a
problem in alge—
describing the solutions
classical
a
unknown
one
finallyto knowledge
led
ures
in
equations
Drabek
Edition
The
of polynomial
R.
and
how
a
quite unexpected
succession
of fail—
sort, and
indeed,
in mathematics.
object that enticed many of the world’s
over
a period
of three
centuries.
Perhaps, dear
from your
school
days quadratic equations of the
the
form
zn2—6:1c+1=O
as
well
as
the
formula”
“quadratic
2
a;1,2=—§i
3:for the
solution
of the
“general”
quadratic
2:2 + pm + q
=
equation
0.
IX
Prefaces
X
If
formula
apply this
we
to
example,
our
Editions
German
the
to
obtain
we
the
two
solu-
1
tions
$1
If you
interested
and
in
51:2
:3~—2\/§.
numerical
solution, you can pull out
to comhandy pocket calculator
(or perhaps you knowlhow
roots
square
by hand) and obtain the decimal
representations
your
pute
are
5.828427.
=
:01
=3+2\/i
culator
.
and
.
332 =
a
0.171572.
.
.
.
You
could
also
use
cal-
your
equation.
to the original
verify that these Values are in fact solutions
A skeptic who wished
to verify that
the solutions
derived
from
formula
the
to
the
are
the
solutions
exact
would
have
to
substitute
roots
into the equation
and
containing the square
demonstrate
that the quadratic polynomial 11:2 6:1: + 1
0 actually
vanishes—that is, assumes
the value zero—at the values
:1:
$1 and
expressions
2
—
=
33
=
(I32.
The
Solution
known
how
of
of Higher
Equations
solve cubic equations such
to
Degree.
It has
long been
as
:1:3—3x2—3:1:—1=0
by means
of
have
use
formula
similar
the
quadratic formula.
Indeed,
such formulas
were
first published in 1545 by Cardano
(1501-1676)
in his book
Ars Magna.
However, they are quite complicated, and
little
a
for
numerical
to
calculation.
In
of
practically
explicit formulas
an
age
unlimited
such
computing power, we can do without
in practical
applications, since it sufiices completely to determine
solutions
of numeric
by means
algorithms.
Indeed, for every
methods
equation in a single variable there exist approximation
iteratively, that is, step by step, compute the desired solution
and more
precisely. Such a procedure is run until the solution
reached
an
suitable
for the given application.
accuracy
not
However
such
only the
numerical
3.847322.
.
.
in the
iterative
numeric
Value
previous
of
procedures
a
solution
is
but
the
example,
are
unsuitable
sought,
such
the
such
that
more
has
when
as
931
=
“exact” value
$131+?/§+\3/Z.
It is not
aesthetic
only that
quality,
such
but
an
in
a certain
algebraic representation
possesses
solution
is insufficient
if
addition, a numeric
Preface
the
to
First
German
Edition
xi
0
one
hopes
solution
numeric
derive
to
mathematical
knowledge and principles from
hypothesize, for example, based
of the
equation. Let us
calculation, the following identities:
the
on
6/w_1=§(enva+
V13),
163
e"
=
262537412640768744,
and
27r
1
2 cos
=
17
8
1
:
1
1
8\/7
:
2 \/71
34
8
+%\/17+3\/fi—\/34—
Without
if indeed
identities,
A direct
check
merely the
But
into
going
to
result
back
to
it
detail,
they
determine
lie
correct,
are
whether
of chance
numeric
Cardano.
In addition
that
plausible
seems
behind
such
mathematical
some
laws.
or
are
they are in fact correct
approximation would be difficult.1
the
to
for cubic
solution
equa-
for
tions, Cardano
published in his Ars Magna a general formula
quartic equations, that is, equations of the fourth degree, also known
as
biquadratic equations. Using such formulas, the equation
a:4—8m+6=0
11 will
reveal
that
are
identities
correct.
The
first
was
only the first and third
discovered
mathematician
be easily
by the Indian
Ramanujan
(1887-1920) and can
checked.
The
in Chapter
it a proof that
within
third, which will be discussed
7, contains
the regular
with
heptadecagon
straight(seventeen—sided
polygon) can be constructed
edge and compass.
The
second
is not
The
exact.
actual
value
of the right—hand
side is
equation
262537412640768743.9999999999992501
this
However,
approximate
identity
deep number—theoretic relationships.
coincidences
320.
in
mathematics?
is
more
For
American
more
.
than
mere
on
this,
Mathematical
.
.
chance.
see
Philip
Monthly
.
It is based
J. Davies,
on
some
Are
88
pp.
there
311~
(1981),
‘
Prefaces
xii
be shown
can
to
the
have
Editions
German
the
to
solution
_€(
+\/~€/4+2\/if/4-x/§+
f/4+2«/5+
\3/4~2\/§
:
$1
With
third-
the
almost
simultaneous
fourth—degree
equations
and
of formulas
discovery
the
came
for
solving
problem of
inevitable
similar
formulas
for equations of higher degree. To accomplish
finding
this, the techniques that were used for the cubic and quartic equations
were
systematized, already in Cardano’s time, so that they could be
to
applied
of
failure,
no
such
formulas
This
question
fifth degree.
of the
equations
mathematicians
began
after
to
But
after
three
suspect
that
perhaps
hundred
years
there
were
all.
was
resolved
in
intermediate
Values
that
1826
by Niels Henrik Abel (18021829), who showed that there cannot exist general solution formulas
for equations of the fifth and higher degree that involve
only the usual
arithmetic
of roots.
One says that
such
operations and extraction
be solved in radicals.
The heart of Abel’s proof is
equations cannot
that
for the
existing formula,
the
various
could
prove
of the
equation
one
solutions
would
appear
corresponding
that
would
in
a
hypothetically
symmetries
lead
to
a
among
contradic-
tion.
A generalization
of Abel’s approach,
Theory.
plicable to all polynomial equations, was found a few
Galois
the
Evariste Galois
twenty~year—old
results
of his
researches
before
he
killed
one
can
in
of the
duel.
was
ap-
years
later
by
down
the
He wrote
(1811—1832).
previous
few
months
In these
on
the
evening
writings are criteria that allow
to investigate any particular
whether
it
equation and determine
be solved in radicals.
For example, the solutions
to the equation
was
a
:35
cannot
which
be
so
expressed,
while
—
zv
the
—
1
=
0
equation
a35+15:n—44=0
Preface
has
the
to
the
First
German
Edition
xiii
solution
ml:5/—1+fl+{/3+2fl+{/3-2\/§+€
.
Of much
is the method
significance than such solutions
that
Galois discovered, which was
unorthodox, indeed revolutionary,
at the time, but today is quite usual
in mathematics.
What
Galois
did was
to establish
a relationship
between
two completely
different
In this way he
types of mathematical
objects and their properties.
was
able to read off the properties
of one
of these
objects, namely
the solvability of a given equation and the steps in its solution, from
those
of the corresponding
object.
greater
only the principle of this approach that benefited
future
mathematics.
In addition, the class of mathematical
objects
that Galois created
for the indirect
investigation of polynomial equa—
tions became
an
important mathematical
object in its own right, one
with many
important applications. This class, together with similar
of modern
objects, today forms the foundation
algebra, and other
have also progressed along analogous
subdisciplines of mathematics
paths.
But
it
not
was
The
object created
tion, called today the
Galois
relations
solutions
tities
between
such
the
cc?
as
by Galois
=
:32 +
2.
that
be
can
group,
to
corresponds
of the
the
given equa-
defined on
the
in the
form
equation
Concretely,
a
Galois
group
basis
of
of idenof
consists
j‘J1P;\;\\4
9f the solutions.,,_Such,a
ren11mhering_lgeloncs-,§o
Galois
group
precisely if every relationship is transformed
by this
renumbering into an already existing relationship. Thus for the case
of the relation
3;? 52:2 + 2 in our example, the renumbering corre~
sponding to exchanging the two Solutions
121 and
3:2 belongs to the
Galois
group
only if the identity 23% $1 + 2 is satisfied. Finally,
every
renumbering belonging to the Galois group corresponds to a
the solutions
of the equation.
symmetry among
Moreover, the Galois
can
be determined
without
group
knowledge of the solutions.
renumberings
=
=
The
Galois
group
can
be
mentary
but
not
particularly
table, and
it
can
be looked
described
elegant.
upon
as
a
by a finite table
Such
sort
a
of
table
that
is called
multiplication
is elea
group
table,
in
Prefaces
xiv
is the
result
in succession.
An
each
which
entry
to
of operating
on
the
German
Editions
two
elements
of the
is shown
example
Ga-
in
Figure 0.1. What is
the Galois group,
about
and its corresponding group
significant
table,
it always contains
the information
is that
about
whether, and if so,
how, the underlying equation can be solved in radicals. To be sure,
the proof of this in a concrete
application can be quite involved; nevof steps
ertheless, it can always be accomplished in a finite number
according to a fixed algorithm.
lois
group
“~«’
k.~E
'1jk'~
Q’1
mC
'~+
U.
QU
eu
D
::
0.1.
Figure
represented
icals
The
a
will
equation
tions
of the
tables
of size
of the
group
:35
equation
—
5:1: + 12 is
table
of which
the solvability
by means
determined
means.
by purely combinatorial
as
be
can
Galois
be
considered
in
detail
fifth
degree that
are
not
60
X
60
or
120
X
in
Section
solvable
in rad-
This
9.17.
Equa-
in radicals
have
120.
in textbooks
Today, Ga1ois’s ideas are described
stract
setting. Using the class of algebraic objects that
mentioned, it became possible at the beginning of the
tury
to
reformulate
indeed
in such
of such
objects.
a
way
More
what
has
that
the
to
come
be called
problem itself
precisely, the properties
solution
can
be characterized
in terms
whose
common
characteristic
is that
basic
arithmetic
operations.
These
sets
are
a
we
previously
very
twentieth
Galois
ab-
cen-
theory,
and
be
posed in terms
of equations
and their
can
of associated
they
in
sets
closed
of numbers
of numbers
under
are
called
the
four
fields.
Preface
the
to
Thus
First
with
starting
forms
such
as
the
Edition
a,,_1ac"'1 +
smallest
(12
be obtained
cessive
basic
of numbers
from
—I—
a0
+ alzr
-
that
0,
=
all
contains
coeflicients of the
the
that
using suc~
an
enlarged set
given equation
operations. Then one obtains
of particular
use
in studying the
is
all numbers
that
can
quantities,
equation
arithmetic
by allowing in one’s calculations,
the equation, the solutions
m1,:L'2,
example
-
ai + ao,
—
<10
can
-
of numbers
set
~—
that
xv
given equation
a
23" +
one
German
in
.
.
be obtained
.
coefficients of
addition
to
This
is therefore
.
from
set
the
formed
of the
expressions
of
form, for
,
(lo
2
—a:1
—
0.21132 + a1.
(12
If it
of the given equation by
possible to represent the solutions
nested
expressions involving radicals, then one can obtain additional
fields of numbers
to the coefficients
some
of
by allowing in addition
these nested
radicals.
Thus every solution
of an equation corresponds
to a series of nested
fields of numbers, and these can be found, accordof Galois
ing to the main theorem
theory, by analysis of the Galois
Thus by an analysis of the Galois group
group.
alone, one can answer
the question Whether
the solutions
of an equation
can
be expressed
now
in radicals.
,.
p
,
L
6
’£""z"/7’
em
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yr
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z«.._.4..:.... Jr4Jr/c‘/aL-43v‘/
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J‘/_f-/'.LQ’as%u://~
A.‘/'/fir-se»
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ta
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‘inf:
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.
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i‘“*‘5."’//7"77/‘/7’/V’f+
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'7‘
55"‘;
'/ii,-z~r.‘7-:..~..>,--L»:-v/.
{‘3««'-AD
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936.
52;./' I
-»~
1/
`\xF76
E?‘.::t7.u)4/L;-la.‘
-«.‘»;...22.
K
Figure
0.2.
Evariste
In
this
passage
ter.
composed
with
the
Galois
and
a
from
fragment
he
describes
how
help
of the
subgroup
a
his
last
let-
can
be
de-
group
G
H.
Section
See
10.4.
Prefaces
Xvi
This
achieved
abstraction
the
at
to
Editions
German
the
of the
beginning
twentieth
cen-
today basically unchanged marks both the end of a historical
in the problem that
we
have described
process
during which interest
has shifted
in focus:
For Cardano
and his contemporaries
the main
solutions
to explicit problems using proproblem was to find concrete
cedures
of general applicability. But soon
the point of View shifted
and
the focus was
on
the important
properties of the equations. Beginning with Galois, but in full force only after the turn of the twentieth
classes
of objects
century, the focus shifted
drastically. Now abstract
tury
and
such
as
groups
host
of
problems,
and
fields became
the
basis
those
that
inspired the
including
objects
in the
first place.2
About
This
Book.
In order
to reach
as
wide
for
the
an
formulation
a
of these
creation
audience
of
as
possible
from college courses
in
(assumed is only general knowledge obtained
mathematics), no attempt has been made to achieve the level of gen-
erality, precision,
matical
textbooks.
and
completeness
The
focus
will
that
be
are
the
rather
on
hallmarks
of mathe-
ideas, concepts, and
techniques, which will be presented only insofar as they are applicable
to some
concrete
application and make further
reading in the exten—
sive literature
possible. In such a presentation,
complicated proofs
have no place. However, proofs are
without
doubt
the backbone
of
with mathematics.
In the spirit of compro—
any serious
engagement
mise, difficult proofs, except those in the last chapter, are set off from
the main
text
so
that
in the logic can
be avoided
the
without
gaps
flow of the narrative
being interrupted.
Considerable
emphasis is placed on the historical
development
the subject, especially since the development of modern
mathematics
in
recent
ral
it can
be
sciences, and also because
give a time-lapse view of false starts
to
centuries
is much
2In particular,
many
mation
theory, in particular
less
well
known
very
and
than
that
interesting
important
of the
to
of
natu-
be
able
discoveries.
have
been
found
in modern
infor~
important
applications
in cryptography,
as
in, for example, the public key codes
realized
in 1978.
In these
the key for encoding
is
asymmetric
encryption
procedures,
made
the
risk
of unauthorized
The
mathematical
public without
creating
decoding.
basis
for such
as
RSA and E1Gama1
is computations
public key encryption
algorithms
carried
out
in special
of elalgebraic
objects with a very large-—butfinite—number
ements
class
defined over
(precisely, the objects are residue
rings and elliptic curves
finite fields). An introduction
to this
can
be obtained
from
Johannes
Buchsubject
Introduction
to Cryptography,
mann,
Springer, 2004.
Preface
And
to
the
First
German
Edition
xvii
furthermore, a presentation that follows the historical
develophas the advantage of making many
mathematical
abstractions
the natural
of individual
consequence
investigations, so that one
ment
seem
gets the
never
somehow
At the
that
descended
would
from
time,
same
of
impression
we
heaven
able
are
with
starting
be necessary
to
to
make
this
The
topics
of the
individual
The
reader
who
in
to
include
definition
unmotivated
an
completely arbitrary
a
leave
out
in
work
a
great
deal
manner.
of material
seeking great generality.
a significant
drawback
to our
However, we must mention
approach:
will be necessary,
even
if they are of
Many complicated calculations
an
elementary nature, whose results would be more
simply derived
from a qualitative
point of View on the basis of general principles.
In
order
book
a
distinct
possible from mathematical
as
in
textbooks, I have chosen the same
style of presentation
my book
Luck, Logic, and White Lies. Every chapter begins with a
or
less rhetorical, question that
simple, usually more
gives the reader
an
idea of the nature
and level of difficulty
of the chapter ahead, even
if the chapter usually goes far beyond simply answering the ques—
tion posed. This structure
should
also offer the more
mathematically
sophisticated reader, for whom the overview offered here will often
be too superficial
and incomplete,
a quick way
of determining
which
parts of the book are of particular interest, after which the references
to the literature
will indicate
a path
of additional
reading.
as
as
tochapters are too closely woven
gether to make it possible to read the chapters independently of one
another.
in only a partic—
Nevertheless, the reader who is interested
ular aspect of the subject is encouraged to plunge directly into the
relevant
a reference
to another
chapter. Even if one then encounters
carried
out there
will
chapter, at least the details of the calculations
be unnecessary
for an understanding
of the following chapters.
Of
the beginning of every
course,
chapter offers the opportunity to start
if the details
over
of the previous chapter became
too difficult.
greater
0
In
be
distance
Chapters
skipped.
wishes
might
adhere
1 through
to
keep the very abstract
to the following plan:
6 the
proofs
in the
set-off
passages
at
sections
may
a
Xvjfi
o
Prefaces
For
understanding
ter
7 that
the
is necessary
is
Chapter
o
In
be
0
8
Chapter
skipped.
Chapter
be omitted
can
which
part,
deals
with
the
the chapter
may
entirely.
set—offsections
9 the
10 and
first
the
only part of Chap-
(17-gon).
regular heptadecagon
0
the
following chapters,
Editions
German
the
to
the
at
the
end
also
epilogue may
of
be omitted.
I” course
should
“Algebra
place Chapters 9 and 10, which deal with Galois theory, as well as
the epilogue, at the center
of their reading. For a deep understanding
of the subject the following are
of particular
the main
importance:
theorem
on
of
symmetric polynomials (Chapter 5), the factorization
polynomials (Chapter 6), and the ideas around cyclotomy (the diviReaders
sion
who
of the
wish
follow
to
given to the remaining
and prior knowledge.
Following
on
typical
circle) (Chapter 7). How
be
tation
a
the
Classical
the
chapters
historical
of
solvability
methods
much
relative
depends
development
on
of the
is divided
equations
of solution, based
should
attention
reader’s interests
the
subject,
the
into
parts:
three
presen—
less
0
complicated
of equations, were
used historically for
equivalent reformulations
for quadratic,
deriving the general formulas
cubic, and quartic
equations (Chapters 1 through 3).
0
Systematic investigation
possible
comes
the
individual
when
one
calculational
of the
on
more
or
discovered
solution
the
intermediate
terms
of the
expresses
steps in
solutions
formulas
be-
results
totality
of
of the
being sought (Chapters 4 and 5). This leads to the
solution
of equations
in special forms, namely, those
that
are
less complex than those in the general form in that they exhibit
that can
particular
be formu—
relationships among the solutions
lated as polynomial identities.
In addition
to equations
that can
be broken
down into equations of lower degree (Chapter
6), the
so—calledcyclotomic equations 3:”
1
0 are
examples of such
less—complex
equations (Chapter 7). Finally, in this part should
be included
the attempt,
described
in Chapter 8, at findinga
~
=
Acknowledgments
solution
general
which
0
xix
is
a
fifth—degree
equations,
works
only in special cases.
for
formula
formula
that
the
result
of
findingsolution
formulas, we
finallyarrive at the limits of solvability of equations in radicals.
These
limits, as recognized and investigated by Abel and Galois, are dealt with, aside from a brief preview in Chapter 5, in
Chapters 9 and 10. The focus here is on Galois groups.
Based
systematic
on
With
attempts
at
of Galois
we
reach
a
level
investigation
groups
of difiicultywell beyond that
of the first chapters.
Therefore,
two different
are
presentations
given. In Chapter 9 a relatively
is given, supplemented
examelementary overview
by numerous
is reduced
as
ples, in which the scope of the concepts introduced
much
as
possible. The resulting holes are filled in Chapter 10,
which
leads
to the
the
mathematical
are
closed
cussion
lois
the
under
basic
arithmetic
deepen
The
reader
who
what
modern
four
wishes
theory.
textbook
fields
will
Galois
theory beyond
called
objects
to
of Galois
objects
the
of these
theorem
main
to
be limited
is contained
his
or
in this
which
theory,
referred
involves
earlier, which
The disoperations.
to
to
those
her
understanding
book
can
relevant
aspects
move
on
of Gato any
One
as
might mention
representatives
of these
books
the two classics
van
der
Algebra, by Bartel Leendert
Waerden
(1903-1996), and Galois theory, by Emil Artin (1898~1962),
whose first editions
appeared in 1930 and 1948. But conversely, the
as
an
extension
of the usual
present book can be seen
algebra textbooks
in the direction
of providing examples and historical
1notiva—
on
algebra.
tion.
Acknowledgments
I would
like
book:
I received
J iirgen
to
thank
all
considerable
those
who
advice
shared
about
in
errors
the
creation
and
infelicities
of this
from
to
Behrndt, Rudolf Ketter, and Franz Lemmermeyer. Thanks
their
the number
of errors
help I was able to reduce
considerably,
the errors
that remain
are
though of course
my fault
entirely. I thank
and its program
director
Ulrike
Sch1nickler—Hirzebruch
Vieweg—Verlag
for having accepted this book
for publication.
Finally, I thank my
Prefaces
XX
been
have
Preface
The
whose
without
wife, Claudia,
often
the
to
tried
patience
German
Editions
this
could
book
not
written.
the
to
circumstance
pleasant
only two
years
Preface
to
Second
German
that
Edition
book’s first
this
edition
sold
in
out
the
gives me
opportunity to expand the bibliography
and to correct
some
errors
spotted by several alert readers, partic—
ularly Daniel Adler, Ulrich Brosa, Kurt Ewald, Volker Kern, Ralf
Krawczyk, and Heinz Liineburg.
the
Third
German
Edition
I
to thank
for the discovery of errors:
Eragain have alert readers
win Hartmann,
Alfred
Moser, and David Kramer, who is also the
translator
of the English~language
edition.
Finally, I have fulfilled
mentioned
desire
to provide the book
with
a
set of
my frequently
exercises
Coordinates.
will
rections
also
be answered
will be
published
The
on1ine.de.
URL
can
Readers
encouraged
are
to
via e—mailto mai1@bewersdorff-online
infelicities
or
tions
The
reader.
Author’s
The
rors
for the
AMS
be found
the
to
extent
on
my
website:
will
also
maintain
on
the
back
.
possible.
http:
a
report
Additions
//www
web
page
.
de.
and
er-
Quescor-
bewersdorff-
for
this
book.
cover.
J org Bewersdorff
Chapter
1
Cubic
Equations
Find
a
number
that
when
added
to its
cube
root
yields
6.
given above have “entertained” generations
of schoolchildren.
Such problems are
at least
several
hundred
old. They appear
as
the first thirty problems that were
years
posed to
Niccolo Fontana
better
known
as
(1499 or 1500—1557),
Tartaglia (the
His challenger was
Antocompetition.
stutterer), in a mathematical
nio Fior (1506—?),
to whom
Tartaglia also posed thirty problems}
1.1
Problems
like
the
one
usual, the path to a solution
begins with findingan equation
represents the problem. In our example, with a: representing the
root
in question, we obtain
the equation
As
that
cube
a:3+a2—6=0.
But
how
to solve
it?
Quadratic equations can always be solved
the square.”Then
one
root
by “completing
simply takes the square
and out pops the solution.
That
is, in the general case of a quadratic
are
we
equation
m2+px+q=0,
1A complete
listing
of
the
set
can
problems
by Fior
di matematica
disfida.” Der Strait
thirty
be
found
ReNicole
in
“Die Cartelli
zwischen
Acampora,
und
Ludovico
Institut
fiir die Geschichte
der Naturwissenschaften
Tartaglia
Ferrari,
Die
(Reihe Algorismus,
35), Munich, 2000, pp. 41—44. See also Friedrich
Katscher,
Icubischen
bei Nicola
die relevanten
Temtstellen
aus
seinen
Gleichungen
Tartaglia:
diverse” auf deutsch
flbersetzt und lcammentiert,
“Quesitiet inventioni
Vienna, 2001.
nato
"-1
1
2
the
quantity
other
side
(3)2
of the
is added
both
to
sides,
and
Cubic
.
the
Equations
q is moved
the
to
yielding
equation,
2
2
m2+m+<§>
42>
P!\xFD
Now
the
left—handside
of the
equation
be
can
represented
as
a
square:
2~q.
2=
And
by taking
roots,
square
obtains
one
Note
the
formula
Z'—q.
:—”2‘
Z:
following important
solution
general
H32
19
3312
the
of
equations: if
forms the negative sum
of the two solutions
and their
product,
obtains
the coeflicients of the original equation, namely,
one
one
$1
Such
+ :02
and
—p
2
of the
completions
property
2
mm
in the
square,
quadratic
q.
form
of
geometric
manip-
known
ulations, were
1700 BCE.
already to the Babylonians around
treated
Quadratic equations were
systematically in the works of the
which were
later transBaghdad scholar al—Khwarizmi (ca. 780—850),
lated
into Latin, inspiring mathematical
in Europe for cenprogress
turies.
His name
is the origin of the word
algorithm. Moreover, the
word
algebra is derived from the title of one of his works, al—Jab7".
of hanpoint of view, al—Khwarizmi’s method
All statements
and
dling quadratic equations is quite cumbersome.
proofs are expressed in Words, Without
algebraic symbols, which had
From
the
modern
yet been
not
invented.
is of
Furthermore, all the argumentation
a
And finally,since negative
geometric nature.
numbers
had not
been
discovered——andno wonder, given the geometric context——alKhwarizmi
had to distinguish various
types of equations, which today
would
we
$2
2
are
notate
pm +q,
less
than
as
$2
2
and
since
we
or
equal
zero,
pm,
902
have
no
we
2
q,
.102+
q
2
pm,
9:2 +1223
2
q, and
difiiculty
accepting coefficients that
can
easily reduce all these to a single
type.
Figure
used
side
1.1
gives
an
impression
of the
method
by al—Khwarizmi. From the figureone can see
can
be calculated
length as of the inner square
of
argumentation
that
from
the
area
desired
q
2
l
l
l
1
1.
Cubic
Equations
9:2 -1- pm of the
the
3
hatched
region using
calculation
a
to
corresponding
formula
2
T
.
(73)
q + 4
=
—
4
3.
2
B.
4
x
B
4
:z:2+pa:
tion
=
term
1.2
Al-Khvvarizmi’s treatment
1.1.
Figure
The
q.
hatched
of the
corresponds
area
quadratic
to
the
equa-
constant
q.
this
our
attention
digression on quadratic equations, we shall turn
again to the question posed by Fior. When Fior posed
his
problems,
mathematics
dred
years
Following
since
had
been
mathematical
the
introduced
tions, negative
cal notation
the
out
was
little
progress
in the
hun-
seven
numbers
slow
notation
was
marked
European mathematics
knowledge gained in earlier periods of
the
Arabs
and
R3
ancient
Greeks.
Arabic
Europe, which served the need
Such calculations
required in trade.
was
used
remained
to
the
into
by professional
mathematics
Although
century
to
calculations
carried
generally
made
al—Khwarizmi.
by its static adherence
activity, notably from
merals
had
masters
in
unknown.
many
nu-
of the
were
of calculation.
commercial
applica~
Moreover, mathematidevelop. Thus, for example, in the fifteenth
used for the expression
V31 m R16 was
{/31 x/1'6.
~
4
Cubic
1.
first great
Equations
was
beyond Arabic’ mathematics
taken
by Scipione del Ferro (1465?~1526), who taught at the UniverHe was
sity of Bologna at the beginning of the sixteenth
century.
the first,as reported
some
decades
later by Cardano
in his book
Ars
Magna,2 to solve general cubic equations, namely those of the type
233+ pacz q. Without
making his method
public, del Ferro revealed
it to his student
Antonio
Fior.
At this time, Niccolo
Fontana, alias
Tartaglia, was also working on solving cubic equations.
Tartaglia,
who was
calculator
in Venice, was
one
of the
working as a master
best
mathematicians
in Italy?’And in fact, he figured
out
how to
solve cubic equations
of the type $3 + 19202 q. However, his method
seems
to have been
less a general solution
algorithm than a way of
could be easily found.4
setting up special equations whose solutions
of the
One
advances
=
=
At the
mentioned
competition
posed thirty problems of the
at
the
beginning
of the
chapter,
Fior
type $3 + pa:
q for Tartaglia
to solve, while
conversely, Tartaglia posed thirty somewhat
atypical
At
problems, including cubic equations of the type :r3 + pm
q.
contestant
cold solve any of the problems that
he had
first,neither
been
posed. But shortly before the end of the competition,
on
13
February 1535, Tartaglia figuredout how to solve equations of the
form :33 + pm
and Fior, he kept his method
of
q. Like del Ferro
=
=
2:
solution
And
now
sociated
today
secret.
a
today
there
enters
upon
with
the
solution
the
stage the
formula.
whose
man
Girolamo
name
Cardano,
is
as-
known
for his
wave
and
discovery of what is called the Cardano
Cardano
suspension and by profession actually a physician, managed to convince
Tartaglia to reveal to him his formula, under the
to Tartaglia
afterward—that he would keep it
guarantee—according
more
2Girolamo
The
Great
with
additions
Art or the Rules
of Algebra, the English transla~
tion
of the 1545
edition
from
the editions
of 1570 and 1663
(Cambridge,
Massachusetts,
1968); see the beginning of Chapter 1 and Chapter 11.
3An idea of the accomplishments
of a master
and
in particular
of the
calculator,
of Tartaglia,
can
be found
in the historical
person
novel
Der
Rechenmeister,
by Dieter
A substantial
Jorgensen,
Berlin, 1999.
part of the novel deals with the discovery of the
solution
formula
for cubic
and
the resulting
conflict.
equations
4See the work cited by Renato
32-34.
On the other
Acampora,
pp.
hand, based
on
the
fact
that
is known
to have
studied
the work
of Archimedes,
Tartaglia
Phillip
Schultz
and
cubic
speculates
Australian
Mathemat(Tartaglia, Archimedes
equations,
ical Society
Gazette
11 (1984), pp.
could
have
used
a geometric
81-84) that Tartaglia
method
in which
he determined
the intersection
:32 and the
point of the
hyperbola
y
=
Cardano,
—q/(:5+ P).
parabola
y
=
1
.
Cubic
5
Equations
secret.
Nonetheless,
Ars Magna, a book
Cardano
the
published
the
describing
solution
current
state
procedure in his
of algebraic knowl-
edge.5
The
of cubic
solution
is based
equations
binomial
cubic
the
on
formula
(u -1- v)3
Cardano
which
able
was
(u3 123),
3uv(u + U) +
=
to
derive
in
to
analogous manner
for the quadratic
an
the
used by al-Khwarizmi
equation,
geometric method
of course
used three—dimensional
the argument
though in this case
figuresand volumes
(see Figure 1.2). However, this identity can also
u + '1) yields
a
as
a
cubic
be interpreted
equation, where the sum
:13 of the
cubic equation
solution
.
if the
:33 +p:1c + q
conditions
32m
U3 + v3
satisfied. The
are
if
0
=
simple task. Since
v3 are known, one
—q,
=
equation
quantities
$3 +p:1: + q
both
the
and
can
solve
cubic
find suitable
can
one
—p,
=
sum
the
1;,
and
’U.
=
But
can
then
that
is
0
product of the
quadratic equation
be solved
relatively
quantities 11,3and
a
0 \x96
to
u3 and
obtain
113 as the
two
solutions
<§>%<§>i
so
that
and
u
1) can
3
works
to
the
3
two
equations
0X\x9F
oh <2)?
led to a great
of his Word
Ca.rda.no’s alleged
falling out
breaking
around
to the
this
and
It is thanks
Cardano.
dispute
publications
referred
to in an
earlier
footnote) that we have knowledge of the history
5The result
between
Tartaglia
up
from
<;>2+
<2)?
`t\xAC
(see the
leading
be determined
the
of
Ars
Magna.
Cubic
1.
5
desired
the
Finally,
is obtained
solution
of the
at
cubic
Equations
$3 + pcc +
equation
'
q
0
=
Ca7"dano’sformula,
from
w=Hm/<§>2+<§
in‘
.!
§§...
ifiim
!.
\
Figure
1.2.
be
decomposed
all
parallelepipeds,
can
apply this
is
the
in his
Ars
two
subcubes
with
side
lengths
to
our
1.3
decimal
In
his
Value
Ars
and
u,
terms.6
6/lrs
Chapter
We have
XXIII.
three
U, and
problem
approximately
Magna Cardano
ing quadratic
Magna,
is
The
Magna.
u
903+ m
m_33+233 61:33
whose
of the
bi-
=
into
result
basis
geometric
11)?’ 3uv(u + 1)) + (n3+ U3), similar
Cardano’s presentation
to
we
here
Depicted
equation
(11,+
nomial
If
LI
2
+
—
large cube
rectangular
11.
=
0, we
obtain
involv-
61
33’
1.634365.
also
solved
cubic
equations
already
seen,
in the
introduction,
an
1.
Cubic
Equations
of such
example
7
with
equation
an
term:
quadratic
a
m3—3a:2~3a:—1-—-0.
To solve
such
transformed
them
equations, Cardano
using a generally
0.
applicable procedure into equations of the form y3 + py + q
Starting with a cubic equation in the general form
=
:I:3+a,m2+b:c+c=0,
the
transformation
solution
consists
cc, which
allows
in
the
—g
adding the summand
quadratic and cubic
the
desired
to
terms
to
be
com-
i
bined:
:c3+aa:2
G3
@293
(a:+(1)3
3
3
27
—
obtain
To
equation,
the
(anQ)?’
E
=
3
of the
of
occurrence
.7:
2
G)
3
transformation
complete
replaces every
one
G2
3
+
a3.
27
coefiicients of this
in the
equation
via
the
substitution
CI:
13:?/mg»
after
obtaining,
collecting
in like powers
terms
of y, the
identity
:v3+aa;2+ba:+c=y3—|—py+q,
with
1
2
p:_§a
2
+b>
1
3
q—2—7a
—~§ab+c.
Once
with
has
solved
the
reduced
Cardano’s formula,
the
solution
obtained
one
with
51:3 3:32
—
—
33:
the
transformation
1
—
2
0,
0
equation 3/3+ py + q
of the original equation can
be
In
the
concrete
y
example
\xC0
cubic
:13 =
—
transformation
the
=
;v
=
y +
1 leads
to
the
,xmwx
equation
3/3 61/
—~
whose
the
6
=
from
=
Cardano’s formula
\3/i + 3/1
leads
original equation:
cc=1+\3/5+3/Z.
»
0,
solution
31
obtained
~
to
the
following
solution
of
Cubic
1.
In
addition
Ars
Magna,
would
aid
the
to
two
fundamental
the
future
in
of the
and
complex numbers.
then
numbers
various
in the
Ars
of numbers
include
to
Cardano
Magna, which
types of cubic
evidenced
in Cardano’s
developments are to be found that
development of mathematics, namely the
extension
set
in calculation
progress
Equations
equations
did
first negative
not
would
have
as
$3 + pm
such
in
quantities
fact use negative
allowed
=
q and
him
$3
to
=
solve
pm -1- q
single type. But he did show a greater openness
to negative numbers by listing in addition
to the “true” solutions
to an equation
the
negative solutions, which he called “false” solutions.
For Cardano, a
as
a
“false” solution
to a “true” solution
of another
corresponded
equa—
tion, namely one with as replaced by —m. For example, for Cardano
-4 is a false solution
to the equation $3 -1- 16
12:12,while 4 is a true
solution
of the equation 1:3
12$ + 16.7
=
::
Exercises
(1) Find
solution
a
to
the
cubic
equation
:v3+6m2+9a:—2=O.
(2) The
cubic
equation
:33 + 61:
has
2
as
a
solution.
formula?
7A1's Magna,
Chapter
I.
How
~
20
is this
=
0
solution
given by Cardano’s
2
Chapter
Casus
Irreducibilis:
The
Birth
of the
Complex
Numbers
x3
solve
the
cubic
Cardano’s formula,
the
formula
to
the
no
If you
does
to
try
indicate
not
clearly
m
=
3 is
that
a
equation
appears
equation has
:
solution.
8:1: + 3
using
fail. But this
solution, for
‘
Like
book
the problem that introduced
Chapter 1, the above equation
also be considered
since
it comes
from Cardano’s
“classical,”
Ars Magnal However, Cardano, who simply gives 3 as a solu-
tion
and
2.1
may
then
calculates
two
deeply into the difficulties
hardly have been unknown
q
=
but
Let
us
-3
of the
look
at
the
additional
presented
to him.”
details.
solutions, does not go more
by his formula, though they can
From
equation, one obtains
the complicated expression
not
the
the
m_33_|_19/5+33
6
3
2
2
coefiicients p
=
and
-8
expected solution
:3
=
19/5
6
3’
1Chapter XIII.
2Cardano mentions
the problem
in a 1539
letter
to Tartaglia,
thus
six years
the
of A’/‘s Magna.
See the
work
cited
publication
by Acampora
earlier, pp.
numbers
were
not
Moreover, because
negative
frequently
used, such situations
arise
not
in equations
of the type 933 + 133:
q.
before
62-63.
could
=
-9
3
10
2.
Cardano’s time
bers,
even
some
tentative
have
been
calculations
when
sum
is 10 and
find the
he seeks
whose
quadratic
place in the Ars M agna
Cardano
the
such
is 40.
question
of negative
num-
roots
square
of
roots
square
makes
negative
num—
of finding
two numbers
problem
product
if in fact
solutions,
Numbers
Complex
of the
with
to solve
the
practically
of the
another
at
of
out
account
on
though
bers
the
Birth
would
simplification
Whose
in
The
Of course,
Cardano
they existed, namely,
knew
the
as
whose
how
to
solutions
of
equation
:t2—10ac—|-40:0,
with
the
result
that
obtains
one
the
5 + \/-15
both
Although
Cardano’s time
dared
to
of the
would
the
out
carry
and
numbers
have
following two
5
numbers:
x/— 5.
—
found
hardly represent
thought of as numbers,
been
what
in
Cardano
calculation
(5+\fl—T5)(5—\/‘—T5)=25+15=4o
thereby performing
call
the
first
known
calculation
is such
calculations, using
meaningful results, that doubtless
the
ing Cardano
of
termediate
in their
was
to
allow
values
in
own
roots
square
a
calculation,
right, for which
today
we
important role in this
played by the so-called
mathematicians
negative numbers,
and
later
as
mathematical
objects
set
of allowable
numbers
irreducibilis, which
casus
roots
of
in which
numbers.
negative
of reduced
case
within
cubic
root
square
is negative:
W (2)3
(2
+
<
3
Chapter
XXXVII,
Rule
II.
o
.
denoted
the
case
Ca.rdano’s formula
In
equations
:r3+p:I:+q=0
3Ars Magna,
in-
of the
square
of the
first as
extension
appeared
radicand
at
follow-
interest.
there
the
achieve
independent
equations
in the
motivated
to
an
of cubic
occurs
of arithmetic
developed
solution
situation
rules
there
in the
when
what
complex numbers.3
It
An
with
particular,
of the
this
form
The
2.
Is
and
Birth
it
end
calculate
with
the
correct
by Rafael
in the
year
such
roots
results?
The
first steps
Bombelli
of
negative
numbers
in this
direction
There
913
15m +4,
with
the
he solved
radical
L’/llgebm,
book
(1526-1572) in his
of his death.
calculating
11
with
=
by courageously
Numbers
Complex
to
up
published
the
possible
taken
were
of
the
equation
expression
m={/2+\/1+?/2—\/E—{/2:11\/-1:
§/211\/‘T
produced by Cardano’s formula.
Finally, with the known solution
4 staring him in the face, he obtained
:3
values
for the two cubic
roots, namely, by calculating
=
\/—l)3
8
12¢ 1
(2 \/—1)3
(2
+
thereby obtaining
It
thus
6
—
seemed
the
desired
that
:3
=
\/-1
——
\/
6+
~—
—
p\x92
12‘/—
8 +
=
1
=
2 +
llx/-1,
=
2
11¢ 1,
2 + \/—1 + 2
—
x/—
=
4.
the
complicated expression involving the
roots
of negative numbers
was
commented
square
equal to 4. Bombelli
thus:
“An extravagant
thought, according to many. I myself was for
a long time
of the same
seemed
to rest
more
on
opinion. The matter
until I found a proof.”These
sophistry than on truth, but I searched
had provided an explanation for an already known
daring calculations
it is certain
that
similar
result, just as in the history of mathematics
when
were
first viewed
as
developments occurred
negative numbers
a useful
addition
to the set of permissible
numbers.
In comparison
to the negative numbers, the square
roots
of negative numbers
introduced
a greater
level of abstraction, since there is no obvious
analogy
in our
to the negative balance
in a
everyday experience, in contrast
bank account,
which
can
be represented
Thus
by a negative number.
it took almost
another
two hundred
years before the objects hesitantly
introduced
were
by Bombelli
accepted into general mathematical
use,
under
the name
What
was
needed
was
a descripcomplex numbers.
>b»vm
‘9:x~\‘r
§=‘
\x83\xF1
,
.
of their
tion
properties
how
as
to
or
less
Berlin,
fundamental
they could be used.
philosophical question,
4Quoted by
Moritz
1900-1908,
Band
Cantor,
2,
p.
625.
This
so
that
could
there
occur
“what actually
Vorlesungen
fiber
are
would
be
no
doubt
only when the more
complex numbers?”
Geschichte
der
Mathematik,
12
was
aside
in favor
of
The
first
decisive
steps
Caspar Wessel
of
definingthem
put
ties.
Birth
The
2.
(1745-1818)
in
this
the
Complex
the
basis
on
direction
Numbers
of their
taken
were
proper-
in
1797
by
.
However, Wessel’s formal definition by no means
eliminated
all
doubt, not least because his writings were not widely circulated.
Thus
it
for almost
was
sible
numbers
another
eked
half
out
existence
an
those
infinitesimals,
current
century
that
these
similar
to
infinitelysmall
imaginary
that
of the
then
could
use
analysis (calculus): Mathematicians
them
and elegantly to obtain
“correct” results
quickly, results
method
without
might well be obtained
use
by another
intermediate
Thus
steps.
(1777-1855), on
the
occasion
mental
of
algebra,
theorem
even
the
of his
1796
which
I will
out
carry
of the
sus-
Gauss
formulation
of the
funda-
decisively,
proof without
my
the
of
use
imaginary
myself that
Perhaps
Gauss
Leibniz
had spoken in 1702 of a “wonder of analysis,
(1646—1716)
monstrosity
he had
of
then
Friedrich
quantities, although I could well have allowed
luxury employed by all modern
analysts.5
a
that
thus:
wrote
even
effectively
Carl
great
complex numbers
uses
still
of mathemat-
quantities
ical
pect
impos-
or
hit
of the
meanwhile
the
upon
of the
essence
reservations
of many—
imagination”—thirty—six
years
later, after
often compelled to mention
the “metaphysics”
human
been
complex numbers:
The
difficulties that
believes
one
to
surround
the
of
theory
imaginary
have their
basis
in large measure
quantities
in
the less than
If one
had
optimal nomenclature.
called
positive numbers
and
“direct,”
negative numbers
“inverse,”
.
there
imaginary numbers
“lateral,”
of confusion, clarity
plicity instead
This
remark
of Gauss
should
mathematical
from
later
interpretation
pretations:
A mathematical
and
have
instead
of darkness.
that
in isolation
sim-
to
indicate
and
freed
reflects such
inter-
object “lives” only because
from
of this
been
generally
be viewed
nomenclature
5This and following quotations
are
taken
1999.
The
last chapter
Zahlen, Frankfurt/M.,
tation
of the
history of complex numbers.
.
would
be understood
definitions should
that
.
of its freedom
Herbert
Pieper, Die komplezen
book
offers
an
extensive
presen-
The
2.
from
Birth
the
The
2.2
whose
of complex
coordinates
line
numbers
can
be
viewed
representing
the
set
the
defines
a
serve
a
of real
a.nd
conceived,
purpose.
numbers.
(a, b) is to interpreted
With
the
idea
that
as
bx/-1,
+
mathematical
for it is
use
by definition all pairs (a, 17)
numbers.
Geometrically, the set
as
a plane,
in analogy with the
real
a
one
13
includes
b are
pair of numbers
a
to
and
a
numbers
complex
number
it continues
long as
set
Numbers
Complex
It is “created” when
contradiction.
“tended” as
of
of
operations
complex numbers
on
fol-
as
lows:
(a,b) +
(c,d)
(a+c,b+d),
:
(cL,b) (c, d)
X
(ac~ bd,bc+ ad).
=
to so—calledinoperations are explained in reference
verse
elements.
That
is defined as the addition
of a
is, subtraction
as
negated value, and division
multiplication
by the multiplicative
The
inverse
The
inverse.
inverse
elements
_(a'>b)
_1_
(afib)
Of course,
in the
failing
the
familiar
laws
of
laws, which
the
reader
will
0
have
All the
—(a2+b27a2+b2
~b
a
definitions accomplish
for
we
("av -17)
:
definition it is assumed
last
Indeed, these
to
defined thus:
are
have
order
relation,
operation
of the
an
some
associative,
notation
common
among
and
Furthermore,
real
zero
goal, since
numbers.
and
(0,0) and
except
numbers
have
We shall
list
complex
all
these
so
that
definitions,
the
nomenclature:
numbers, such as
law properties,
distributive
the
real
desired
easily verify from
can
relations
the
the
(a, b) gé(0,0).
that
the
one
the
commutative,
continue
to
(1,0) have
the
hold.
famil—
iar
in addition
and
properties of serving as identity elements
and division
are
in fact the
multiplication.
Finally, subtraction
inverse
and multiplication.5
operations of addition
a
6Taken all together,
field. We shall return
a
to
set
with
this
concept
two
operations
in
Chapters
these
satisfying
9 and
10.
conditions
is called
2.
14
The
o
of
subset
these
Birth
like the
the
of
numbers
complex
operations
of the
set
of real
the
set
Numbers
Complex
form
(a, 0) behaves
under
and
it
with
denominator
complex
numbers
can
real numbers.
For
of the
simplicity, we will often write a complex number
(a, 0) simply as a, and for a complex number (a, b) we call
form
tified with
them, just
as
be identified with
can
therefore
real
be looked
the
on
as
numbers,
of fractions
The
integers.
of the
extension
an
be iden-
can
1
the
a
part.
We have
0
The
(0,1) ><(0,1)
corresponds
the
to
(0,-1)><(0,-1)
=
real
number
-1.
:
a result
(—1,0),
Therefore,
interpreted
numbers
the
two
that
complex
as
roots
(0,1) and (O,-1) can be
square
of -1.
The number
(0,1) is given the special notation
2', called
the imaginary unit.
In a complex number
(a, b), we call b the
imaginary
part.
We have
0
called
the
equation
(a, b) X (a, -b)
a2 + b2, where
=
(a, -b) is
the
conjugate of the complex number
(a, b). It is denoted
by (a, b). We call \/a2 + b2 the absolute value or modulus
of the
number
(a, b). Within
resentation
of the
complex number
the
complex plane, as the geometric
is called, the modulus
complex numbers
the
represents
distance
of the
number
rep-
of
from
a
the
origin.7 An example is displayed in Figure 2.1. Finally, complex
conjugation
(taking the conjugate of a complex number) possesses
the following property:
((a, b) X (c, d)) (a, b) X (c, cl).
=
All these
properties
ofthe
form
in fact
defined the
(a, b)
make
together
(a, 0)+(b, 0) X (O,1)
=
set
certain
us
that
with
a+bz', where 1'2
objects of the form
:
of mathematical
=
pairs (a, b)
1, We have
a
b\/-_1
+
This
of the real numbers
was
achieved
resulting extension
without
using the previously undefined expression \/-—1,
whose use, moreover,
is not
always completely unproblematic, since it can easily lead to
calculations
erroneous
Another,
very
will
be
to
ric
representation
of
use
7The definition
their
on
and
difference
the
complex
are
integral
makes
such
important,
us
in what
of the
of the
as
x/-1\/-1
follows
of
such
numbers, whereby notions
defined with
similar
properties
two
a
function
to
to
the
We first note
complex
as
\/1
=
1.
=
complex numbers
closely related
numbers.
between
creation
is
\/(-1)(-1)
of the
example
complex
distance
the
possible
=
numbers
theory,
as
geomet-
that
the
that
every
modulus
of
complex analysis,
convergence,
continuity,
derivative,
those
of classical
analysis.
or
.
2.
The
Birth
of the
2.1.
Figure
its conjugate
complex
number
numbers
have
Complex
The
complex
1
2i.
—~
The
located
Numbers
plane
modulus
the
number
of both
numbers
1 + 21' a.nd
is
@\x80\x89
circle, that is, the circle of radius 1 with center
at the origin, can
be represented
in terms
of the
functions
sine and cosine.
To be precise, such complex
trigonometric
a
the
with
15
on
unit
representation
coscb + isin (1),
'
‘
"
from the positive horgbis the angle that runs counterclockwise
izontal
axis (i.e., the positive real axis) to the line from the origin to
the complex number
in question
(see Figure 2.2). If We now multiply together two such numbers
lying on the unit circle, we can do so
merely by adding their angles together. A proof of this follows easily
where
from
the
addition
laws
for
and
sine
cosine:
,
,
(cos gb-1- isin q5)(cosip+ isinip)
+ 2'(cos¢sinz/1
+ sinqficos
(cos ¢cos1[2 si11¢>sin1/1)
1p)
=
—
+ i sin(gz5
+ P\xCF\
cos(¢+ @F_
=
If
We
now
introduce
the
is 1 for
the
complex numbers
eralize
this
result
if
a
complex
to
number
all
has
modulus
nonzero
nonzero
of
a
complex
number
lying on the unit circle), we
Note
complex numbers.
(hence positive) modulus
(which
can
gen-
first
that
m,
we
16
2.
Figure 2.2. Representation
cos
+ isin :15located
on
the
¢>
can
set
a
complex
z
3
lnm
2
(the natural
number
with
2
of
e5 (cos ¢+z'sin
The
under
ham
number
of the
form
circle.
logarithm) and
angle g1)and
write
modulus
m
es
65.
=
be
can
Thus
written
see8 that
=
of
case
raising
de M oiwe
name
de Moivre
’sformula,
(1667-1754) never
(es(cos ¢ + i sin q5))
2.3
Before
the
knowledge we
to the
return
have
number
complex
a
n
we
Numbers
Complex
qb)>< et(cos¢+z'sin 1,0) es” ( cos(¢+1/2)
+1’sin(¢+i/2
special
the
complex
a
unit
eS(cos q5+ 2' sin (b). We then
=
the
of
Birth
The
formulated
it
a
power
goes
Abra—
its namesake
though
even
to
explicitly:
+ 2’sin(n¢)).
(eS)n( cos(n¢>)
=
irreducibilis,
casus
just gained
the
to
would
we
like to
apply
equation
m3—1=0.
field of real
In the
If we
to the
move
suggests
that
of which
lie
the
positive
8The
series
for
numbers
reason
the
real
for
y).
331 =
1 is the
numbers,
then
equation
have
additional
unit
must
circle
two
of
the
and
validity
so
of
this
by Leonhard
:1: + iy,
one
three
solutions, both
form angles with
solutions
form
will
equation
functions
the
only solution.
de Moivre’s formula
(see Figure 2.3) and
and %’5,that
277'
axis
sine, cosine,
exponential
which
was
first accomplished
in 1748
see
that
for arbitrary
complex numbers
i sin
that
field of complex
the
on
it is clear
are
become
clear
when
the
power
extended
to the
complex
numbers,
Euler
(1707-1783). One can then
has
the
identity
em+iy
=
em (cos y +
2.
The
Birth
of the
an
equilateral
triangle.
solutions
Complex
It
Numbers
is thus
17
that
apparent
the
two
additional
are
and
:1c2=—l+—\/gt’
:v3=—1—£i.
2
2
2
These
three
solutions
2
called
are
third
roots
of unity. Equations of the
form 23” 1
O, which form the topic of Chapter 7, will be called
of their geometric significance.
cyclotomic equationsg because
=
—
Figure
2.3.
31:3
equation
2.4
The
cube
three
—
roots
are
1
three
=
solutions
with
them
extend
can
always obtainable.
recognizing that
the two
—«
'"
We
2
ter
solutions,
1, the solutions
9From
the
Greek
in
addition
Cardano’s formula
circle,
so
that
begin by defining
underlying Cardano’s formula,
to
=
-12,
=
—~q,
the
pair (u, 11)introduced
(Cu,(21))and ((211,
(2)),
kuklos,
general cubic
2
equations
ug + v3
as
the
_$ + \x80\xB8\xE2
31m
have
(2 of the cyclotomic
for
significance
of
are
one
C
and
1, Q, and
0.
of unity
roots
in that
equation
The
and
-tomia,
cutting.
so
that
in
Chap-
altogether,
one
18
The
2.
obtains
the
three
following
Birth
of
solutions
to
the
the
Numbers
Complex
reduced
cubic
equation
W me\/*+<§>i
-l-
M“:
»Q
<>2+<>?
was
to
<>2+<>?
<;>2+<
co’‘3
marl3+ <;>2+<’;>%
Q
NJ
This
tions
formula
holds
assumed
in
general,
in its
derivation.
since
restrictions
no
on
solu—
the
However, in the case of casus
irreducibilis, one must note that for the two complex numbers
u3 and
113,which are conjugate to each other, pairs of cube roots u and v are
selected
that
are
also conjugate to each other; only in this
way can
the two equations
that determine
u a.nd 12 be satisfied. Not
only have
we
shown that the calculations
carried
out by Bombelli
are
justified,
were
but
furthermore,
$1,
$2,
and
3:3
we
all
are
that
see
in the
general
the
case,
three
solutions
real, since
+ €——(j—1)U
€'j—l,u/
+ €'.7"1u
<"‘(.7“1)fl+Cj‘1“
C“"(j‘1),U
wj,
3?],
=
:
for
j
to
calculate
=
solutions
For
the
:
with
is, in employing Cardano’s formula, it is necessary
even
in the case
that
all three
complex numbers
real
and
1, 2, 3. That
are
=
problem
Cardano’s Ars
distinct.
:03
=
Magma, one
8:0 +
3, which
obtains
the
the
at
appears
beginning
solution
$_33+Z,19\/E+33
16
.19\F
1
643
2
1
=3.
,
5
2
1
3
,
5
of
2.
The
Birth
of
the
The
other
two
solutions,
Numbers
Complex
which
19
Cardano
knew,
are
p\xA1
a32=%(——1+’i\/3)<
::<+«a>
and
:1:3=%l(—1—z'\/g)
+}l(—1+z\/.3)
<3+z'\/E) <34E\xCC
=%(—3+fi).
of the
So, was all this effort worth it? In any case, the extension
to the complex numbers
has converted
the
underlying field of numbers
solution
to
algorithm into a unified process.
Moreover, the extension
the complex numbers
has removed
the uncertainty
that we might obtain incorrect
results
in calculating with nonreal
intermediate
results.
there
is still one
However, in actual calculations
problem: Our procedure
does not provide an effective
way of simplifying expressions of
m».-«
eaesfi
the
type
33
Zm
_
2
_
6
3
numerically. At least the latter option
is relatively easy to accomplish if one
begins with a number in polar
coordinates.”In the case
of casus
irreducibilis, that is, in the case
or
at least
5
.19
to
approximate
\¥
them
fir (Br
(2
,
+
where
coefiicient p must
the
complex conjugates
are
of the
be
<
3
negative,
0’
the
two
numbers
1/,3 and 123
form
c\x97—<;>2—<§>‘”<
10
On
that
of simplifying
cube
of comroots
hand, the former
problem,
the
real
and
with
plex numbers,
parts
is, to express
imaginary
separately
expressions
be completely
resolved:
If a cubic
with
rational
involving roots, cannot
equation
none
of which
is rational,
such
coefficients has three
real
for example,
solutions,
as,
:33
6:1: + 2
there
is no
for the
roots
nested
radicals
expression
0, then
involving
whose
are
all real.
See
B. L. van
der Waerden,
intermediate
values
Algebra, vol. I,
Section
2003.
64, Springer,
—
the
other
that
=
20
The
2.
The
of each
modulus
of these
of the
Birth
numbers
Numbers
Complex
is
3
2_
2
9
\/(2) (2)1(1)3)31/(_23)3_
’
_:3
”"
3
The
can
its
made
angle
be obtained
from
modulus.
cf)
that
reduced
be
To
=
so
cubic
the
two
numbers
quotient
of the
with
real
the
of the
part
real
axis
number
and
positive
“upper”
angle is give by
the
precise,
3g
\xA0(\xC4
(V?)
<21“/:3P)
arccos
obtain
we
by the
3
=
the
arccos
following formula
$3 + pa: + q
equation
,
for the
three
of the
solutions
O:
=
§cos<3¢+j
gr),
1
a:j+1—2,/
Such
solution
.2
.
g~0,1,2.
based
functions
trigonometric
actually has
nothing to do with algebra. However, the question of the solutions
of
a cubic
equation is certainly an algebraic problem. Furthermore, the
solution
method
of casus
irreducibilis
is excellently suited
for learning
how to work with complex numbers.
a
on
were
first discovered
in 1591
just described
by Francois Viete (1540-1603). They were published posthumously in
1615. However, Viete did not use complex numbers
in his derivation,
the triple-angle formula
for the cosine:
using instead
The
formulas
we
have
31¢ 4cos3 zl)
cos
Then, using
an
=
of the
equation
y3
a
solution
equation
can
be found
of the
transformation
=
sy,
—«
3cos¢.
form
33/icosiizb
—
form
where
9:3 + pm +
the
O,
=
using the relation
reduced
:2:
—
cubic
0, one
first makes
the
is chosen
such
the
=
cos
q
=
parameter
3
resulting equation
33
3g
1
=
8P
-p/ 3
1/). To solve
a
y
0.
that
21
Exercises
One
then
with
by starting
solution
of the
angle 1/)such
that
obtains
a
an
—E
31/)
the
making
the
0
1
substitution
works
procedure
satisfied. The
are
=
21’ ~19/3
3:
Where
333+ pa: + q
cubic
W
cos
and
reduced
=
2,/—§cos1/2,
only if the
p<0
and
second
inequality
conditions
g
1
21>
-29/3
is
31
to
equivalent
<;>2+<§>2o»
Literature
Paul
J
.
the
on
An
Nahin,
of
History
Tale:
Imaginary
Complex
The
Numbers
Princeton,
Story of \/-1,
1998.
Lutz
der
Fiihrer, Kubische
komplemen Zahlen,
Gleichungen und die widerwillige Entdeckung
43 (2001), pp. 57—67.
Praxis
der Mathematik,
Exercises
(1) Derive
the
formulas
square
root
of
analogous
root
(2) Show
of
a
that
formulas
complex
the
real
complex number
for the
number.
and
a
imaginary
+ bi.
Also
real
of the
three
of the
to
find
imaginary parts of the cube
Explain the problem that arises.
real
and
coefficients is also
a
of
«E
a
polynomial
solution.
following complex numbers
'_:\/2
131\/7l2':\/5,
parts
attempt
complex conjugate of a solution
with
equation
(3) Which
a
the
for
are
roots
of unity?
i:\/2+\/3,gx/6+¢§\/is
3
Chapter
Biquadratic
We seek
Equations
of the equation
solution
a.
(I74+ 6:02 -1- 36
=
60$.
classical, coming as well
from Cardano’s Ars Magma (Chapter XXXIX, problem V). However,
difficulties for Cardano, because
such problems caused
they offered
in the forward
Thus
he mentions
to
no
geometric interpretation.
with a line, quadmtum with a
his book, “While positio is associated
surface, and cubum with a solid, it would be foolish to attempt to
does not permit it.”
extrapolate. Nature
The
3.1
problem
However,
Cardano
was
able
thanks
able
this
to
to
his
chapter
is also
Ludovico
student
to
transform
equations
Ferrari
ATS Magna, the
in the
describe
equations.
(quartic, fourth—degree)
quadratic
was
for
of the
In
(1522-1569),
solution
particular,
to
bi-
Ferrari
form
a;4+pa;2+q:v—|—r=0
of two terms
by the addition
is obtained
a perfect
square
only slightly from the method
adds
to
2,2332+ 22 to both
be chosen
later,
and
sides
in powers
both
of m and
sides
51:2in such
a
way
that
of the
equation. Deviating
described
by Cardano, one most simply
of the equation, where
the value for z is
on
thereby
obtains
1:4+2zw2+z2=(2z~p)a:2—q:13+(z2—r).
Although the left—handside of the equation is already in the form
is not necessarily the case
for
the
of a perfect square,
such
+
(:32 z)2,
23
3.
24
side.
However,
1-ight—hand
as
z
can
be
now
Biquadratic
Equations
suitably chosen, namely,
so
satisfy the condition
to
—p\/z2
2x/22
Squaring both
sides
of this
condition
—q.
=
—
leads
to
2
Q
(2z——p)
(z2—-7“)
Z
=
and
thus
the
to
cubic
equation
2
z3~—§z2—rz+I*)2t:—
Once
has
one
solvent, the
determined
solutions
to the
solution
a
this
to
2
original biquadratic
so-called
cubic
results
equation
re-
from
a:2+z=i(«/2z~—p:n+\/z
where
each
of the
the
quadratic
four
solutions:
two
possible signs yields
formula.
Altogether,
one
1
a:3,4—2
to
sometimes
chapter,
pi
that
note
Ferrari’s procedure
are
with
obtains
the
obtains
the
in
his
cubic
resolvent
problem
23 ——3z2
—36z—342
which
be
can
reduced
transformed
via
the
-7“,
22
1".
the
=
illustrates
numbers
posed
the
at
calculated
start
of this
0,
substitution
2
=
y +
1 into
the
cubic
g3 -393/-380
Using the
resolvent
z
one
V22
Magma, Cardano
in which
the
following
1
Ars
For
of
1
—2—z
4p
examples
incorrect.
one
therefore
1
22
remains
by virtue
1
1
It
solutions
E‘/2z—pd:
——2—z—
110+
=
221,2
two
obtains
biquadratic
=
1 +
towers
equation.
:0.
solution
{‘/
190 +
of
3\/3767
expressions
+
3x/3767,
\3/190
in radicals
—
that
solve
the
original
3.
Biquadratic
Equations
25
in the introduction,
By now it should be clear that as indicated
these algebraic formulas
are
completely useless if the goal is simply
to obtain
numerical
values, for these can be more
quickly and easily
of iteration
computed by means
procedures. Thus for the equation at
the beginning of this chapter one obtains
the solutions
and
3.09987
0.64440.
as
well as the pair of complex conjugates —1.87214. :|:
.
.
.
.
.
2‘ 3.81014.
.
.
-
.
.
.
from
Nevertheless,
mathematical
point of view, the algebraic
is impressive.
result
Who would have guessed a priori that cube roots
would
in the solution
of a fourth—degree
appear
equation? However, if
viewed
at
correctly, this result is not as surprising as it might appear
first glance. In fact, we encountered
something comparable in the case
of cubic equations:
Just
as
Cardano’s formula
contains
roots
square
in addition
to cube roots, a general formula
for biquadratic equations
must
be similarly constituted.
Otherwise, the general formula would
not
be applicable
to a special equation
like 234
2:3
O, with the
a
=
—
solution
3.2
\xC0\x9C
(131 =
Ferrari’s formula
Since
biquadratic equations
wwmmwmxmave
third
we
power,
biquadratic
in which
the
describe
a
must
equation
presented
as
of the
here
can
as
does
Variable
of
method
be used
not
only for
to
appear
converting
the
the
general
be
done
form
w4+aa:3+b:t2+cm+d=0
into
an
equation
in the
reduced
form
y"+py2+qy+r=0In
of the cubic
analogy to the case
equation,
replacing the variable av via the substitution
this
can
by
a
‘T:ywZ>
with
the
result
that
the
two
terms
in
that
ye‘
arise
cancel
each
other:
a:4+cLa:3+bac2+c$+d=y4+py2-l-qy+7*.
Of course,
equation
as
can
in the
case
be calculated
polynomial expressions.
of the
cubic,
the
from
those
of the
coefficients
of the
original equation
reduced
using
Literature
Ludwig
der
Biquadratic
3.
26
Equations
Matthiessen,
littemlen
Heinrich
Biquadratic
on
Gmmdziige dew antiken
Gleichungen, Leipzig, 1896.
Dérrie,Kubische
und
und
biquadmtdsche
modemen
Exercises
all four
solutions
3:4
presented
(2) Determine
at
the
—
beginning
all four
of the
8m + 6
equation
0
=
of the
chapter.
of the
equation
solutions
31:4+ 8953+ 242:2
~
112$
+ 52
Algebra
Gleichungen, Munich,
1948.
(1) Determine
Equations
=
0.
Chapter 4
of
Equations
Degree
and
TheirProperties
We seek
equation
an
2, 3, 4, and
The
4.1
in
success
quadratic
equations
equations
of
a
better
ter
stated
was
work
of
In artem
In addition
extensively the
ble without
case
equation,
:33
—
can
a
creating
of two
search
included
a
wish
bi-
for
obtain
to
be found
Francois Viete’s 1591
in
useful
symbolic notation, Viete discussed
of transformations
of equations that are permissi-
possess
solutions.
He also
given numbers
given solutions
found
$1,332,
:31, 2:2,
a
.
.
way
of construct-
,:1cn
as
.
needs
one
solutions.
only a quadratic
namely
—
case
It
to
that
1,
equations through systematic
the problem posed at the start
of this chapisagoge.
sorts
numbers
polynomial
analyticem
302
In the
This
solved.
changing the
ing equations
In the
connection,
and
the
are
5.
higher degree.
In this
solutions
findingsolution
procedures for cubic and
led inevitably to the desire
to do the same
understanding
study.
whose
77,
of three
(:01+ $2 +
(321+ :c2)a:+ 231232
prescribed
solutions
2
0.
the cubic
.’1?1,.’L'2,£lI3,
+ (:r1a:2+ 531533 +
$3)1E2
m2:v3)m arirciwa
—
equation
=
0
27
of
Equations
4.
28
meets
the
required
conditions.
:33, :34
are
solutions
of the
9:4
~
Degree
and
n
Their
Analogously, the
biquadratic
four
Properties
numbers
91:1, 032,
equation
(:61+ 902 + 933 + m4)a:3
2
+
(E11173 + (E21133 + (131934 + £I32£C4
-i-((L’1.'132
+ CL‘3£l34).’I3
—
+ £L'1.’132ZB4
-I- 231233334 + m2:c3a:4)a3
(IL‘1$‘2$'3
+ £L‘1.’L‘2$3.T40.
=
Finally, Viete produced an equation
given numbers
:31, 1:2, $3, 5104,935:
$5
—
+
whose
+ 161132565 + 1131333135
w3964035)1L‘2
+ :I71$2€E3€B5
+ .’E1$2iE4$5+ 0311133154935 +
(CU1372153334
:I32933$4-T5)33
:v1:r2a:3:z74m5
Viete’s last
of the
0.
=
example solves
for which
chapter,
$5
~
one
the
problem
obtains
the
159:4 + 85903
the
=
0.
theorem
substituting
(root is a frequent
one
of the
values
beginning
equationl
225$? + 274:3
~
at
posed
120
~
Only the obvious symmetry prevents Viete’s formula
one
completely confusing. Of course,
may easily check
root
five
£U4375)~”U3
+ 632333135 + 331134505 + 032934505 +
——
the
($1152+ 1131933 + 132553 + 131334 + 032934 + 963134 + 932135 + $3$5
+ 931102934 + 3511331134 + $2$3iF4
($1302-‘E3
+
are
(901+ :22 + 2:3 + 0:4 + a:5)w4
+
—
solutions
synonym
for
in for the
variable.
from
that
being
Vz'éte’s
solution) is correct
by
However, of greater
interest
is the question
of how one
obtains
such results, including the
for more
than
analogous ones
five solutions.
This, too, is not difli—
for the first time in 1637 by Rene Descartes
cult, and was described
in his work
(1596—1650)
solutions
are
the
given numbers
1In Viete’s notation,
1Qo
A facsimile
together
Reich, Der doppelte
Erhard
Scholz
(ed.),
La Géometrie: If an
~
with
the
German
zur
—
.
.
225Q + 274N,
translation
can
friihneuzeitlichen
der
.
whose
,a:,,,
then
one
simply
may
is
equation
15QQ + 850
Auftakt
Geschichte
331, $2,
is sought
equation
Algebra,
be
120.
equatur
found
in
Henk
Viéte
Algebra:
Mannheim, 1990,
pp.
J.
und
M.
Bos, Karin
in
Descartes,
183~234.
4.
of
Equations
the
take
Degree
and
n
Their
equation
(0: m1)(a: mg)
—
In this
no
others.
an
equation
in the
are
—
-
that
it is obvious
form,
there
29
Properties
One
is
5131,I132,
—
.
.
.
,
that
and
solutions
are
:37,
simply to multiply
required
familiar
more
(:1: \xA0
«
-
obtain
to
out
form.
made
explains an observation
particular, Viete’s root theorem
in his Ars Magna (Chapter I, equation 2:3 +72
already by Cardano
In
:
11:32).He had
:33 + ba:
found
three
M32 + c and
=
solutions
for
some
equations
that
the
sum
of the
observed
coeflicient a of the
of the
form
solutions
agrees
An
explanation of this
since it
to have discovered,
fact would have been diflicult for Cardano
to bring the equation
of negative numbers
presupposed the existence
into a form in which
the right~hand side is equal to zero.
with
the
4.2
Descartes
circumstances
quadratic
also
discusses
the
left—handside
term.
problem of Whether and under
of an equation of the form
the
what
a:”+a,,-1:n"_1+---+a1a:+a0=0
can
be
into
decomposed
If there
solutions
of the
product
is such a decomposition
into
linear
are
But
known.
conversely, says
provides
one
a
form
(r
3:1)
-
—
-
-
\xE0\x8C&
clearly the
then
factors,
(tr
—
Descartes, each solution
facof the equation into linear
step in the factorization
:0 on
the left
example, if :31 is a solution, then the variable
then
be replaced by 931 + (:1: 221). If one
side of the equation can
of $1 and (cc :31),
into powers
develops the powers
($1 + (:3 :z:1))’°
tors.
For
—
—
——
then
one
:12" +
finds that
the
+
an_1rc"‘1
-
-
-
+
(:6 $1)”+ bn_1(w a:1)"‘1
=
out:
+ an
+ aim
-
be factored
(93 ml) can
term
—
»
—
-
-
+
b1(cc $1) + bo,
—
with
:0.
bo =:z:”+a,,n1:I:"_1+-«-+a1:r+a0
One
thus
has
the
9:”+
a,,_1:c"_1 +
result:
desired
-
-
-
+ a1:I: + (10
=
bn—1($-T1)"_2
1L’1)"_1
(€13501)\xD0m1
=
(9: 331)(a:"‘1
—
—
+
*“
+
+
—
+
+
c,,_g:v”“2
-
~
-
co)
.
-
-
-
+
51)
4.
30
Finally,
the
have
we
factor
succeeded
(:1: :21), where
If
and
the
in
determined
solution
ml
via
can
be
continued.
there
are
at
most
n
linear
Therefore,
most
n
4.3
If the
n
tions?
the
number
Here
In
of
then
is the
n,
at”
what
the
mean
Thus
be broken
we
off
number
those
of the
original
and
addition.
.
.
of
breaking off a
of the nth degree,
process
broken
off.
have
can
a
polynomial
at
of
degree
possible
solu-
dijferent solutions,
only the single solution
since
least
number
number
of
of
rather
mean
equation
the
number
an
is.
If
linear
a
solution
factor
is called
with
appears
a
than
more
multiple solution,
zero.
of linear
of linear
asking for the number
factors
nth—degree
equation, and what the smallest
counting solutions
are
of the
.
are
corresponding
we
,c,,_2
equation
0, for example, has
=
By “number of solutions” we
such
the
an
of solutions
do not
we
equation
tors.
multiplication
by
solutions.
be at most
can
Properties
polynomials that can be thus
asserted, an nth—degree
equation
Descartes
as
from
solutions,
factor
Their
coeflicientsC0,C1,
be
linear
and
n
dividing the original polynomial
can
finds additional
one
Degree
all the
—
resulting polynomial
equation
of
Equations
once,
and
we
fac-
that
can
possible
then
the
say
that
multz'plz'citg/.2
The
possibility of breaking off factors corresponding to solutions
allows us to say something about the minimum number of solutions:
If there is no nth—degree
equation without
any solutions, then a linear
factor
be
can
broken
resulting equation,
have a solution, the
off of any polynomial.
Furthermore, since the
provided that its degree is at least 1, must again
process
be
can
continued, and indeed, it can be
continued
until the polynomial has been
entirely decomposed into linear
factors.
That is, if it can be proved that
every nth—degree
equation
has at least one
solution, then there will always be n solutions, where
solutions
are
counted
Already before
jectured
in
1629
that
with
multiplicity.
Descartes,
an
Albert
equation
Girard
with
had con(1590—1632)
complex
coefficients
always
the number
possesses
of solutions
equal to its degree. Despite the
efforts of many
mathematicians, a complete proof of this conjecture,
now
called the fundamental
theorem
of algebra, was achieved only in
2Cardano
in
Chapter
dealt
with
multiple
I of the Ars
Magna.
solutions,
for
example
the
equation
1:3 + 16
=
2:1:
4.
of
Equations
Degree
71,
and
Their
31
Properties
Gauss.
This proof justified—at
least from an
1799, by Carl Friedrich
algebraic point of view—the use of complex numbers, since a further
numbers
is not necessary.
enlargement of the set of available
of algebra”
is to be un—
designation “fundamental theorem
derstood
mishistorically. From today’spoint of View it is somewhat
is fundamentally not algebraic in
leading, since in fact, this theorem
nature.
That
is, it is based only tangentially on the properties of the
related
to the four basic
arithmetic
Of
complex numbers
operations.
much
are
the properties
of the complex numbers
greater importance
that
are
related
to distance,
that
to converis, properties related
gence,
continuity, and so on. A comparison with a similar result in
the realm
of the real numbers
The graph of
might make this clearer:
the polynomial
(133 2 viewed as a function
of a real variable
runs,
The
—
in the
standard
“upand
coordinate
system,
from
“down and
to
the
left” to
the :1: axis at least
right.”Therefore, it “must” cross
once.
That
is, the polynomial under investigation, and indeed any
polynomial of odd degree, has at least one zero (that is, a value of as
that
makes
the polynomial equal to zero). What
to be simappears
ple and obvious is in fact due to fundamental
properties of the real
intermediate
value
numbers, which find expression in the so-called
theorem.
Two properties
are
crucial:
to
the
o
A function
0
The
defined by a
polynomial is continuous, that is, there
are
no
holes or jumps in its graph; rather, its value
changes
at every
point by less than any prescribed bound, provided the
small.
change in :1: is sufficiently
such as exist, for exam“holes,”
numbers.
ple, in the rational
Indeed, for each point on the number line one
can
find infinitely
rational
numbers
within
many
distance
of the given point.
any given small
Nonetheless, the
of approximating
the point by rational
numbers
takes
process
one
outside
the set of rational
numbers.
For example, if we aproot
of 2 by more
and more
terms
of its
proximate the square
decimal
expansion,
set
of real
1,
numbers
1.4,
1.41,
has
no
1.414,
1.4142,
\x90\x9B\x81
4,
32
this
sequence
not
rational,
of rational
of the
complex
ness) and
Their
in the
example
the
existence
of real
set
contains
an
sketch
a
of the
of
for the
argument
complex
key properties
satisfies the
fundamental
equation
theorem
of
alge-
theorem,
as
well
of the
Theorem
Algebra:
proof.
a
Fundamental
of
and
Proof
text, it suflices to prove the following theorem:
complex coeflicients whose degree is at least one has
A polynoleast
at
one
zero.
We
begin
with
of the
properties
a
plausibility argument,
absolute
value
function
which
makes
crucial
[Z122] @x\xA3This
has
f(z)
=
a
=
an—1z"”1
+
ann
1,
are
approximated
complex coefficients
taken
on
by the function
-
.
.
value) z. Concretely, for
Izl 2 R ==
the
that
consequence
2" +
with
a
-
:10, the
,a1,
.
of the
complex numbers, [a + bi]
complex numbers
and 22, the
zl
b2, namely, that for two arbitrary
trmngle inequality holds, that is, |z1+z2| 3
|z1]+]z2|, as
=
use
for
V a2 +
has
given based
in the
with
absolute
be
plausibility
Plausibility
As stated
no
numbers.
7},which
the
on
is
of “holes” (called complete-
lack
number
following section
The
one
Properties
numbers
numbers, namely, the
——1.The
=
bra
mial
and
77,
A proof of the fundamental
theorem
of algebra can
the continuity of polynomial functions
and the two
on
as
Degree
approaches a number
that
The point here is that
namely P\xF9\xB5
there
is
comparable
2'2
of
Equations
for
-
well
the
identity
a given
polynomial
as
+ (1127 + ao,
absolute
values
of the
values
by |z”[for sufliciently
large (in
complex number
z
with
+2(Ian—:I+---+Ia1I+laoI),
1
inequality
]an—1z"_1+ aiz
+
5
-
-
-
+
aol
Ian—1Ilz"”1I+---+Ia1IIzl+Iaol
1
S -lzl
S(lan~1l+--‘+|<11l+laol)lzl"
$—&
1
,,
-
We
would
now
like
to
consider
how
the
motion
of
number
complex
z about
a circle
of radius
R with
center
the origin is affected
by the polynomial
For the term
z" it is clear, since by de Moivre’s
f
one
formula,
circuit
around
the circle of radius
R is mapped
to n circuits
around
the
circle of radius
R”. This circle is depicted in the middle
part of Figure
a
4.
Equations
4.1.
The
of
Degree
remaining terms
for large R, as we have
of the
annulus
within
and
outer
radius
§R". For
bounding
circles
for
a
dashed
in both
line
Figure
about
4.1.
the
that
an
circuit
Their
33
Properties
have
polynomial
so
n-fold
as
and
seen,
an
the
71,
little
effect
function
we
obtain
for
with
0
center,
as
a
inner
on
this
result
value
f
radius
%R",
example, the middle of Figure 4.1 shows two
is shown
remaining terms; the bounding annulus
the center
and right parts of the figure.
A revolution
of
a
circle
origin (left) is mapped
of sufficiently
large radius
by an nth-degree
polynomial
to an n-fold
revolution
within
an
annulus
centered
at the origin
of the
(right). The middle
image shows the curve
highest
at two
power,
2", and the maximal
“perturbation”
sample
of lower
points due to the terms
degree.
What
happens if we vary the radius of the initial circle? Ignoring the
is continuous.” Since
details, what is important for us is that “everything
we
are
only arguing for plausibility, we will now offer a verbal description
of the consequences:
Independent of the radius of the initial circle that the
without
point z traverses, the image points f
always form a closed curve
any discontinuities
any changes experienced
by the
(holes). Furthermore,
curve
due to a change in the radius
can
be kept arbitrarily
small by keeping
the change in radius
small.
sufficiently
There
is one
more
obvious
fact:
For zero
radius
there is but a single
lies within
the inner
image point, namely (10, which by our construction
circle
of radius
Now
%R".
in
analogy to the case of a real polynomial of odd degree,
the decisive
If one
the circle on
continuity argument:
gradually contracts
which z is moving toward
the origin beginning at radius
R, then the image
to the point
curve,
beginning with its n-fold circuit of the origin, contracts
Thus
be crossed,
at some
and therefore
the
ao.
point the origin must
must
contain
within
the circle of radius
R at least one—
polynomial f
and indeed, for R sufficiently
large, n—compleXzero.
comes,
34
4.
Since
Equations
of
Degree
all
to
make
is not
it
watertight,
proof, one
at
will
we
simple
take
and
71.
the
Their
Properties
heuristic
previous
argument
difierent tack
a
altogether in presenting a formal
taken
first in 1815 by Jean
Robert
Argand (1768~1822) and
simplifieda few years later by Augustin—Louis
Cauchy (1789—l857).
We have already seen
that the function
values
of the polynomial
f
exceed
in absolute
value [f
jag] outside a sufficiently
large circle. The
=
minimum
the
real—valuedfunction
of the
circle, where
it
is
|f
is therefore
be
to
found
inside
taken,
about
the extreme
according to a theorem
values
of continuous
real~valued
functions, at some point, which we call zo.
Developing the polynomial as a function
of z
to
zo leads
——
f(z)
where
be +
=
the
index
bm(z
zo)'" +bm+1(z
—
2 1 is such
m
be 75 0, since
otherwise,
we
We
determine
a
for
now
have
-
-
bn(Z
+
-
bm 75 0. Furthermore,
found
already
complex
the property
with
example,
that
zo)'"+1
+
—
number
a
we
zo)",
—
assume
may
zero.
Moivre’s formula,
de
11), using
be
m’—-.__._
w
—
bm,
and
then
0 <
e
now
obtain
form
the
argument
be chosen
later.
1, to
<
f(z1)
be
:
——
=
zl
For
(1
This
bmem%—
|f(Z1)l S (1
-
-
“
B
cients
bo, bm,
to
us
estimate
the
value
-
f (21), we
-
-
-
[f(z1)]:
Gm)lbol+ em“ (lbm+1wm+1l
+
+ lbnwnll
‘
—
'
'
—
+
.
.
.
,
»
-
-
bn and the choice
of the
number
w.
One
sufficiently small
that
1 —eB is positive.
Then
with
such
found
value
than
minimum:
If
smaller
contradiction
Finally,
value
+ |b,,w"|)/Ibo] depends only
([b,,,,+1w"‘+1l
where
The
number,
Ibo[(1 em(1 eB)),
=
a
small
a
Em)bo + bm+1€m+1’U)m+1
+
+ b,,e"w".
allows
equation
=
-
the
+ ew, where
e is
associated
function
+
bm+1e"”+1wm+1
+ b,,e"w"
+
Tn
1‘
zo
the
assumed
is eliminated
observe
we
fundamental
theorem
of functions
of
a
that
of
if we
the
algebra
complex variable.
give
shortest
are
based
may
a
on
coeffi-
the
choose
now
selection, we
<
Ibo]
=
have
If (zo)].
up
the
and
most
beautiful
proofs of the
on
basic
theorems
of the
assumption
6
b0 75 0.
theory
Exercises
35
Exercises
(1) Show
that
in
come
the
=
of a polynomial
1,
the
at
.
.
.
,n
distinct
polynomials
that
points
331,
of the
form
(13
9J(93)
H
“Az=1....,n
_
real coefficients
with
numbers.
nth—degree
polynomial
an
y1,...,yn
for j
zeros
pairs of complex conjugate
(2) Construct
ues
nonreal
—
takes
.
.
.
,a:n.
on
the
given val-
Hint:
Consider
(Ez-
mj _mi
#2’
at
the
problem
points
:3
is called
=
The
polynomial that solves
the Lagrange interpolation formula.
1121,. ..,m,,.
this
5
Chapter
The
Search
Additional
Formulas
for
Solution
Is
there
to
equations
“blueprint”
for the solution
the fourth degree?
common
a
up
to
formulas
The
5.1
procedures that Cardano
published for solving cubic and
the beginning
of a historical
biquadratic
equations marked
period
in which
a Variety of attempts
were
made
to find a general formula
for solving equations
of the fifth degree. In pursuit
of this goal, it
seemed
a good idea
to search
for similarities
in the solution
procedures
In the case
of equations of fourth
already discovered.
degree, various
alternatives
to Ferrari’s solution
method
were
considered, which with
other
and other
intermediate
led
results
equivalence transformations
to
the
resu1ts.1
same
To be
existence
of
offers
clue
no
the
sure,
n
fundamental
complex
as
to
how
roots
those
theorem
for
of
the
algebra guarantees
nth-degree equation. However, it
an
solutions
can
be calculated.
Neverthe-
theorem
of algebra, we can reformulate
less, based on the fundamental
the problem of findingthe solutions
of an nth—degree
equation: Since
1The most
Matthiessen,
gen,
Leipzig,
complete
description
der antiken
Grundziige
1896.
of
such
und
methods
modernen
is
to
Algebra
be
found
der
in
litteralen
Ludwig
Gleichi/.n—
38
in
The
5.
of the
equation
every
Additional
for
Search
Solution
form
a:"+an_193"“1
+---+a1:n+ao
the
left side
ac" +
be
can
decomposed
a,,_1:1:"_1 +
-
-
-
Formulas
+a1a:
into
+a0
linear
:0
factors
(3: m1)(:c 11:2) (J2 a:,,),
=
—
—
~
-
-
—
given nth—degree
equation be transformed
into an equivalent
system of equations
corresponding to Viete’s root theorem?
That
is, for given complex coefficients
we
seek complex
a,,_1,
,a1,a0,
the
can
.
numbers
:31, 1:2,
.
.
,:1:,, that
.
.
.
satisfy the system
531 +032+'-'+-Tn
+ 031133 +
$1$2
-
-
-
of equations
—Gn—1,
=
+ 33n—~1a7n a'n—2:
=
£lI1£E2"'{13n_1£I3n
(—1)"a0.
=
The
symmetric
in :z:1,cz:2,...,ccn
expressions
appearing on the left
sides of the equations
are
called
elementary symmetric polynomials.
But it is not only the solution
of given equations with
explicitly known
coefiicientsthat
We
orem.
ables, so
the
problem
mentary
called
reinterpreted on the basis of Viete’s root
certainly view the solutions
:c1,m2,
,a:,, as
may
that
.
the
of
the
formula
is
the
.
.
.
variables
,a1,
Clo.
This
+ £I72)2
(931
4931$2
§(331+$2):l:—2~
noting that
in terms
5.2
Similar
and
biquadratic
computations
cubic
from
,m,,
interpretation
the
is
to
ele-
usually
1
=
1
£121--$2
§(fU1+zl72)=l
§(
in
is
surely
a simple
—
equations
equation,
.
place of the square
root, which
value in solving the equation, we have
of the solutions, namely
($1 $2).
expressions
are
.
corresponds
quadratic
—
expression
.
vari-
.
equation, the well—knownquadratic
following interpretation:
a
given the
key intermediate
formula
91:1, $2,
1
It is Worth
the
a,,_1,
general solution
a
.
the-
equation.
of
case
1
the
for
determining
general
For
931,2
search
polynomials
the
=
be
can
in terms
can
solutions
also be found.
correspondingly
for whose
of the
more
reduced
of the
To be sure,
complicated.
form
general cubic
the requisite
We
:33 + pm + q
=
begin with
0 the
three
5.
The
Search
solutions
the
is clear
(22+
Formulas
39
help of Cardano’s formula
the
with
computed
these
from
z
using
=u+11,
$2
:
$3
=
€u+€2v7
C2u+(v.
equations, using the identity (2 +(—|—1O, which
1
third—degree
cyclotomic equation Written 23
1), one obtains expressions for u and 1) in terms of
the
1) (z
+
:01
three
=
=
~
—
solutions:
14
oo|v—*c.o
+ C21?+ C$3)»
(501
=
(£81
‘U:
Again there
in
Solution
values
From
the
be
can
Additional
for
is
concise
a
Cardano’s formula
"1-(21133)
-1- {I132
expression
in terms
of the
.
for the
three
square
solutions
root
appearing
(:31, 932, $3:
(3)2 (2)3 5(U3~'~3>
+
=
1
=
—
1
+
+ 4x2 + c2m3)3
(371"l‘<2372
(371
54
54
—
—
1
C)
’1‘g(C2
“
—
x
0մ
1
=
As
the
one
can
see
—
931503(03303 932133$1.123 08\xDB
+
-
+
—
.
pI\xD9
:c2)(m2
ac3)(:c1
$3).
~~i§
—
from
the
presence
~
of i in the
—
last
expression,
for
particular case of three real roots the expression under the radical
is the observation
that the
sign is always negative. More significant
expression is equal to zero
precisely when there is a multiple root.
One can
also write
the analogous product
for the general equation
of every
other
of pairs of roots
degree in which all the differences
of differences, Wl10se square
Such a product
is called
the
appear.
5.
discriminant,
irrespective
to
when
precisely
zero
If
we
given
are
Solution
Formulas
of the
degree of the equation,
is a multiple root.
there
cubic
a
Additional
for
Search
The
40
equation
with
quadratic
is then
equal
term
:c3+a:v2+bx+c=0,
then
the
solution
begins, as presented
process
in
Chapter1, with
the
substitution
at
3”‘ Z
so
that
it,
1), and
be
determined
reduced
a
3/
from
is to
the
solutions
replace
each
just derived
ga
=
:10,
~
formulas
three
values
u,
In Ferrari’s procedure
tion
$4 + p222+ qzv + r
cubic
All that
original equation.
1,2,3, in the three
933-, j
=
§(9c1
+ C132+933),
unchanged.
formulas
for the
hold
un-
equation.
for
solving the
O, the crucial
=
The
thus
x/(q/2)? + (p/3)?’
12, and
cubic
general
5.3
of the
ca.n
1
three
2
values
Ca.rdano’sformula
in
solution
thereby leaving the
for the
intermediate
with
:3, +
changed
The
appear
of the
1
intermediate
3)
is obtained.
equation
x/(q/2)? + (p/3)3 that
is necessary
formulas
cubic
_
reduced
biquadratic
step is determining
a
equa—
solution
resolvent
P
612
pr
Z3—§Z2—TZ+—2‘—+§=
on
the
from
basis
two
of which
quadratic
the
Using Viéte’s root
two
these
two
solutions
theorem
the
equations
for quadratic
equations,
following Values
for the
pairs of solutions:
=z+
x3a34=z~—
this
one
determined
pairwise
2z~p:rI}:\/z2—7"+z=:O.
$1932
From
be
can
equations:
$2?
from
four
immediately
x/22-7‘,
\/z2—
obtains
1
z
.
=
ac3m4).
§(x1:r2
+
can
derive
products
of the
we
5.
The
Search
for
For
the
sake
of
,2, the
selection
Additional
Solution
Formulas
41
completeness we should note that the solution
z
cubic resolvent
21 of the
corresponds to a possible, but by no
means
Since
prescribed, numbering of the solutions
:31, mg, 233, 934.
Ferrari’s procedure
is a consequence
of equivalence
transformations,
with respect
resting on the condition
prescribed by the cubic resolvent
—
the
value
to the
correct
to
the
%
and
This
as
solutions.
of the
5
solutions
resolvent
has
of a different
resolvent
therefore
affect
only the numbering of
that
the
the
solutions
a
consequence
be determined
can
can
from
solution
also
other
two
leads
solutions
3:1, (I32, 933, 3:4
as
follows:
1
E
z2
E
932934)»
§(:U1rv3
=
--
‘i
1
Z3
With
this,
solutions
tor,
--
can
express
the
in the
solution
of the
we
dano’s formula
11321133).
§(l31£L‘4
:
931, {B2, $3, :04 of the
the
square
single factor
cubic
of this
Up to
—
—
1
(21
with
32)
the
($1
2(:c1:v2
result
that
z2)(z2
*'
a
Carof the
fac—
constant
23), Where
~—
a
1
4 923924
=
in terms
equal
(251 z2)(z2 z3)(z1
looks like
product of differences
to
in
appears
resolvent
original equation.
is
root
that
root
square
for the
024)(:v2
ms),
§(m1
$2934)
ccirva
—
—
complete product
—
of differences
obtain
We
Z3)(Z1 Z3)
—
—
1
a:4)(m2
a:3)(a:1
a:2)(:r3
m4)(m1
:v3)(m2
:34).
\xAF\xFF
=
—
Therefore,
up
equation
coincides
For
proceed
—
to
in the
—
»~
constant
factor,
the
discriminant
with
that
of the
cubic
resolvent.
equations
that
a
biquadratic
as
—
case
of
a
cubic
not
are
equation:
of the
in reduced
First,
a
—
original
form, one
biquadratic
may
equa-
tion
a:4+am3+bm2+ca:+d=0
with
a
cubic
term
is transformed
Via the
substitution
(1
into
a
for the
reduced
intermediate
biquadratic
values
equation.
that
arise
In
order
in the
to
process
obtain
in terms
formulas
of the
=
for
of the
original equation,
just
1, 2, 3, 4, in the formulas
solutions
j
Search
The
5.
42
1
3:, +
we
solution
replace each
should
derived
Formulas
Solution
Additional
:3],
by
1
Z(:c1+a:2—l—:1:3+m4
—
Zazrj
take
of the general biquadratic
care
polynomials thus obtained
of the cubic
In particular,
we
obtain
the
“first” solution
equation.
The
resolvent
as
Z1
+ 332 + $3 + 334)2£B3$4)
§(C'31172 EECE1
+
=
5.4
What, then,
solving quadratic,
*
the
are
similarities
the
among
for
methods
three
In all three
biquadratic
equations?
intermediate
the crucial
values, that is, the expressions for
cases,
as
the roots
formulas, are representable
appearing in the solution
that
£131,532,
is, polynomial, expressions in the solutions
“simple,”
form of such expressions depends on how the
Of course,
the actual
and
cubic,
.
solutions
the
to
ability
solutions
glance?
.
.
numbered.
are
Is the
.
Since
931, (I12,
.
really
..
solution
a
surprising at
always starts
as
method
polynomials in
at first
it might seem
values
intermediate
express
as
with
the
coefficients
general equation with
in the solutions, it is actually
the
of
be expressed in terms
that
all the intermediate
values can
obvious
the solutions
operations and nested roots.
using the usual arithmetic
However, What is not a priori obvious is that polynomials alone sufiice,
That
for degrees two, three, and four.
the case
which
was
is, in the
of the intermediate
values, no expression of the form,
representation
equation, that is, in reference
elementary symmetric polynomials
of the
to
the
Say,
V‘/$1+ \xF0\x88c
may
appear.
Permutations
Changing
the
order
Since
it doesn’t matter
with
the
natural
of
numbers
a
number
finite
what
objects
objects are called,
the
1, 2,
of
.
.
.
,n.
is called
one
a
usually
permutation.
names
them
5.
Search
The
for
Additional
Solution
It is easy to determine
the number
is nl, read
“n factorial,”
defined by n!
so,
observe
which
the
that
leaves
number
the
72
number
1 positions
—
1 can
be
for the
number
placed
elements,
Symbolically,
that
is, the locations
one
has
1
3
1)
—
0
.
.
-
-
see
is
this
2
n
1.
-
list
the
images of
the exchange.
after
>
simply,
more
(a(1) 0(2)
In
special
will
use
such
the
numbers
it
cases
be useful
can
for the
notation
a
from
1 to
n
to
.
use
so-called
moved
are
cr(n))
0(3)
one
suggestive
cyclic permutations,
notation.
the
order, writing
a
to
more
other
in
in which
some
1—>3—>4—>2—>1insteadof(3
2).
An
important
after
another
all
of
is that
one
can
execute
them
permutations
another
As with other mappings
permutation.
is called
composition and is frequently denoted
property
obtain
to
We
4
1
one
that
’
U01)
\xE4\x8C
-
is to
n
It
positions,
Altogether, then,
3
-
objects.
TL
of the
,a(n)
.
p^N
<0(1) 0(2) 0(3)
To
in any one
2, and so on.
0(1), 0(2),
2
of
«
-
-
the
43
of permutations
:=
1 2 3---n.
of arrangements
is equal to n
(n
simple way of notating a permutation
One
or,
Formulas
and
functions, such a process
by the symbol 0. For example,
(1
2
3
the
order,
3
2
3
4)o<1 4)_(1
4)
2314-3412’
1342
where
2
is usual
for functions
and
from
is read
right
to left.2 What
we
1 is sent to position 2, as
have, then, is that the number
seen
in the second
from the left, and then is sent to position
permutation
in the permutation
on
the far left.
3, as seen
The
collection
of all n! permutations
together with the operation of
composition is called the symmetric group and is denoted
by Sn. Its idenis the identity permutation,
which
leaves every
element
in its
tity element
original place.
The
lution
as
recognition
formulas
for
that
all
intermediate
mappings,
values
of the
known
so-
the
general polynomial equations up to fourth
is due to
degree are given as polynomials in the solutions
331,112,
Joseph Louis Lagrange (1736-1813). Lagrange, who was active in
Berlin
for the twenty years
beginning in 1766 thanks to Friedrich
11,
theorems
for
published in 1771 an investigation into general solution
.
2This perhaps
function
appears
unusual
on
the
order
right,
and
can
be
explained
so
one
writes
by the
(a 0 1')(j)
fact
=
that
o'('r(j)).
the
.
.
argument
of
a
5.
The
Search
of the
nth
degree.
44
equations
for
Formulas
Solution
Additional
starting point was the sysLagrange’s
for equations
formulas
general solution
investigation of the
in the soluSince
the
intermediate
values
fourth
the
to
degree.
up
tion of equations up to degree four are
expressible as polynomials in
tematic
the
solutions
.
arbitrary
express
asks
cisely,
one
in the
solutions
.
.
331, [B2,
,a:,,) is
h(a:1, 0:2,
.
in terms
of the
Lagrange
Via
a
.
.
.
.
that
is, from
how
can
and
recognized
that
of the
form
coefficients
can
be
expressed
polynomials.
equation
an
can
always be found
(z h(ma(1)ama(2)a >ma(n)))
(3 h("I7lam2> `\x95&
-
,a:,,)
coefiicients of
0)
‘H:
_
_
.
elementary symmetric polyfind a simple equation for which
whose
such
the
.
the
symmetric
elementary
construction
one
solution
a
from
be determined
can
.
pre-
.
.
general equation,
nomials.
Concretely,
.
that
for methods
search
to
sense
More
1:1, (122,
polynomials in the solutions
the question how a given polynomial h(:z:1,$2,
the
.
it makes
£131,332, .,
-
‘
-
-
the
from among
product is formed from a suitable selection
that
n! permuations,
1,2,.
is, exchanges of the Variables’ indices
chosen
such that
0
are
every
possible
Concretely, the permutations
the
from
can
arise
that
permuting
through
,:rn)
h(:E1,
polynomial
in the product.
variables
Then, as
exactly once
,a;,,
appears
931,
achieves
that the coeflicients of the equation arising
shall see, one
We
the
where
.
.
.
for
.
.
.
unknown
the
.
..
z
be
can
calculated
in terms
of the
coefficients
of
general equations, that is, the elementary symmetric polynomiTherefore, for the polynomial
operations.
als, using basic arithmetic
the
desired
obtained
will
have
we
equation.
,a:n)
h(a:1,
the
.
.
That
seen
.
this
complicated than
example. The polynomial
all sounds
by means
of
an
more
1
h(ar1,962, $3, 9:4)
appeared
earlier
=
when
Ferrari’s procedure
133$4)
§(m19:2
+
—
were
reality
1
,—6(:v1
investigating
for solving a biquadratic
we
it is in
+ 9:2 + 263 +
the
cubic
equation.
can
be
zv4)2
resolvent
in
Lagrange’s
5.
The
Search
universal
for
Additional
construction
leads
Solution
for this
example
1
z
:c3m4)+
s
1
X
where
we
have
z
used
z
the
equation
+ s
932.104)
§(:1c1a33
+
—
§(:c1r4
+
—
the
to
45
1
§(:v1a:2
+
—
Formulas
+ 3
.’L‘2.’L'3)
=
0,
abbreviation
1
+ £32+ $3 + $4)
8(931,€L‘2,933,904)
1—6(fl71
=
If we
now
but
now
Chapter
multiply the three linear factors together, we
in a generally
applicable way—the cubic
2
.
obtain
again—
resolvent
from
3.
only in this special case, but also in general, one finds with
construction
an
z in which
Lagrange’s
the
equation for the unknown
coeflicients are
Since any
polynomials in the Variables
$1,552,
of the variables
permutation
the linear
131,232,
simply rearranges
factors, the polynomials that form the coefficients of the equation
constructed
for the unknown
2 remain
unchanged. Thus all coefficients
are
And
symmetric
polynomials in the variables
a:1,x2,
such symmetric
polynomials, that is, the coeflicients of the general
of
equation of nth degree, are always able to be expressed in terms
the elementary symmetric
polynomials using addition, subtraction,
and multiplication.
This can
be summarized
in the fundamental
theorem
of symmetric polynomials.
Not
.
.
.
.
..
.
.
Theorem
nomial
This
5.1.
in the
Every symmetric
polynomial in 2:1, .732,
elementary symmetric polynomials.
theorem
.
..
.
is
..
a
poly-
first formulated
by Lagrange. However, the
theorem
was
apparently known a century earlier by the physicist
and inventor of the calculus
Isaac
Newton
(1643-1727), along with
a procedure
for representing
symmetric polynomials, for example, by
determining concretely
was
IE?-1-£L‘§+m§+'
(371+£B2+€U3+--)2—2(£L‘1$2+$1$3+$21E3+
0'|
"
=
-
46
Search
The
5.
for
Additional
Formulas
Solution
and
a:§:v2
=
—
this
How
mfaca :z:§:v1
:c§a;3:v§:n2
+
123961
+
+
+
+
+
-
--
(:v1+m2+:vs+---)(:v1:c2+:v1:va+w2:v3+---)
3(:v1:v2a:s+ a:1:c2a:4 + 5131133534 + :z:2:z:3:n4 +
-
theorem
is
proved
using
algorithm is ex—
of symmetric
theorem
of Lagrange’s
theorem
=Fa)2($2
H(50z'93392 (931 $2)2(-‘E1
=
—
).
-
constructive
a
on
the fundamental
plained in the section
is a specific
There
polynomials.
application
on
symmetric polynomials to the discriminant
'—
-
—
'
—
"
2
i
for which
there
due
symmetry
its
to
be
must
for
polynomial
each degree n
a
expression in the coefficients
of the underlying equation.
Another
$1,552,
polynomial in the variables
importance for Lagrange irrespective of the degree
what is today called the Lagrange res0l1)ent3
.
h’(:E1:"1;27
3:571)
'
Where
C is an
nth
'
:
'
root
+€x2 +
:81
of unity.
h(:z:1,m2,...,:vn)
'
'
had
of the
equation
is
€n—133n>
+
'
central
that
.
Since
C h(ac2,:r3,...,a31)
=
-
C”_1h(:rn,m1,
=
and
+
£2533
.
-
.
.
.
=
\xE0w\x81
,:nn_1)
thus
h((131,£l32,...,1L‘n)n
h(ac2,:v3, .,:c1)"
=
.
.
=
\xC0X\x82
h(:cn,ac1,
=
.
.
.
,mn_1)",
universal
Lagrange’s
procedure for h(:c1,2:2,
run)”
a resolvent
can
be exequation of degree (n
1)! whose coefficients
of those
of the original equation.
If this equation
pressed in terms
were
solvable
using a general formula, the original equation could be
solved, since we have
one
obtains
with
.
.
.
,
—
1
ml
:R(371+“'+517n+hI(371>w2>533>
+'
"‘l‘h4(371a-7771:5132--~a5vn—1))
3However, even
1783) and
degree
Euler
equation.
before
(1707~1783)
Lagrange,
in
their
such
work
expressions
on
solution
were
formulas
used
by
for
the
Bézout
general
(1730nth-
5.
The
and
no
Search
for
Additional
analogous equations
general solution
is successful
in
for the
other
Formulas
solutions.
resolvent
Lagrange’s
for
special
some
Solution
The
cases.
Vandermonde
Alexandre-Théophile
independently of Lagrange, investigated
more
to say about
this in Chapter 7.
The
Fundamental
for
Although
apparent,
his
method
to
achieve
such
success
(1735-1796), who
“his” resolvent.
Theorem
2 5
n
is
first
was
47
in
1770,
We will have
Symmetric
on
Polynomials
We recall
Theorem
5.1:
Every symmetric
polynomial in the variables
$1,332,
expressed as a polynomial in the elementary symmetric
.
.
be
can
.
polyno-
mials.
A proof of this
The
order
in which
is most
the
easily accomplished
induction
of the
ordering
words
theorem
polynomials,
a
language.
in
define
one
will
be taken
related
to
will
the
induction.
by complete
be based
on
lexicographic
a
special
of
ordering
than
the monomial
“bigger”
if in listing the exponents
in order, the first exponent
mil” 931°,"
jl, jg,
jg that differs from the corresponding exponent
ks is larger than
ks. For
the monomial
example, according to this definition,
\xD0x is bigger than
the monomial
based on the lexicographic
:c§:I:§a:§.r4,
order of the two strings
We
-
-
a
:12?@\xA4 to be
monomial
-
.
“251” and
The
induction
step
begins with
now
'
'
'
:
biggest monomial
at?”---:u;"". We
has already been
orem
f(m1,
.
.
where
.
0‘
efficient
could
one
-
-
for
the
proved
for
all
polynomials
nonzero
smaller
than
,mn) is a symmetric
polynomial,
is any
permutation
of the
am1,,,m,,.
It
use
a
$17” cc?" whose
-
J71.
assume
all
follows
suitable
coefficient
that
polynomial
aj1~~‘jn${1H'a737.n
coeflicient amlmmn
induction
hypothesis
with
coefiicients are
nonzero
arbitrary symmetric
an
Z
23371)
.71
mial
.
“2421.”
f($1a
whose
.
the
ml
2
that
mono-
the
monomials
thewith
Since
-:1:,'[‘".
1:27;”
monomial
-
-
monomial
a:ZL"(")
:c;n°(1)
1,2,.
has
every
numbers
whose
is the
2
mg
find
permutation
to
in the
polynomial
\xB0!
a
.
.
-
-
,
,n,
2 mn,
monomial
f(:1:1,
-
.
.
.
the
for
same
co—
otherwise,
bigger
,m,,) is nonzero.
than
48
Now
a
Search
The
5.
for
the
in
special polynomial
very
Formulas
Solution
polyno-
symmetric
elementary
is formed:
mials
g(a;)
=
37j)m1—m2
(ZQ7j-'17k;)m2—Tn3
\xA0
-
am1...m,,
the
with
term
-
($1322
»
-
:cn)'"".
-
-
j
.7
The
Additional
largest
coeflicient is
nonzero
777-2 “"13
.7”?
---$3",
$2
mi“ m’(«’F19«‘2)"'(931932"'iL‘n)m"=001
#
induction
ml
for the
is valid
g.
polynomial f
for the induction
The procedure
step can be put to practical purpose
if for a given symmetric
polynomial, a polynomial in the elementary sym—
of
metric
explicitly. After a finite number
polynomials is to be calculated
polynomial, which is valid both for
steps the procedure ends with the zero
induction
the formal
proof and as the starting point for the induction.
described
shows that for symmetric poly—
the procedure
Furthermore,
one
can
with integer coefficients
nomials
always find integer polynomials in
the elementary
symmetric polynomials.
Finally, we note that with symmetric polynomials the representation
by elementary symmetric polynomials is unique. This is due to the followof
to the case
one
can
restrict
ing uniqueness theorem, in whose formulation
of two equal polynomials,
situation
the zero
polynomial (for the “general”
so
the
that
looks
one
hypothesis
their
at
~
difference).
A polynomial
Theorem
5.2.
symmetric
polynomials,
f(y1,
.
.
.
at the
vanishes
,y,,) that
elementary
is, for which
that
f(Z.'.C_7’,ZCl7jil7]¢,,...,£l31
:0,
j
is
equal to
itself identically
The
f(y1,
.
.
.
yé0.
,yn)
select
We
coefficient
nonzero
zero.
by contradiction.
is
proof
j
a
the
from
among
monomial
that
-1- mg
+
in the
+ TI1n,’ITL2+
lexicographic sort order.
since
for
two
(ml
in
question
-
-
-
-
-
-
+ mm
Such a
equally large n-tuples,
the
are
same.
that
Suppose
It follows
the
have
polynomial
with
3/{"1 y,’{‘"
monomials
the
a
-
-
-
n—tuples
n—tuple
,m,,) has the “largest”
monomial
is uniquely determined,
the
arrangement
exponents
of the
in
.
we
.
of
.
two
monomials
that
f(Z:.'I3j,2CUjI13};,...,(l71(l72
j
j
:
+ g(aJ,1’ VT”),
am;ni+m2+~~+mn$;n2+---+mn,_,m7v:1n
.
wherein
the
polynomial
smaller
than
the
g only monomials
first monomial.
appear
Altogether,
that
then,
as
.
‘
lexicographically
a sum,
expressed as
are
5.
a
The
Search
in the
polynomial
from
the
5.5
We
for
zero
are
vestigations,
variables
Solution
:01,
.
.
obtain
contradicts
the
But
not
going
to
since
aside
from
his theorem
be
needed
in the
will
is worth
noting,
not
go
the
though,
mathematicians
this
such
Formulas
we
,:c,,,
.
polynomial.
his results
later
Additional
on
polynomial
assumption.
a
different
inLangrange’s
symmetric polynomials,
deeply
more
any
49
into
later
great
chapters of this book.
influence of Lagrange’s
work
Abel
and
as
Galois.
It
was
It
on
Lagrange’s
of permuting
significance
the solutions
of an equation.
Lagrange was also certainly the first to
have recognized the main
difficulties
in solving the general equation
of fifth degree. Lagrange was
able to simplify the universal
method
for “his” resolvents
in the case
of the general fifth—degree
equation,
of the sixth degree to be solved, for which
though it led to a resolvent
no
The first attempt
simplification
to establish
presented itself.
the
impossibility of a solution of the general fifth-degree
equation in terms
great
accomplishment
of nested
radicalsficalled
the
Italian
Paolo
and
medicine
at
a
proof
are
tablishing that
fifth degree can
diate
Ruflini
at
have
to
a
noticed
solution
the
in md2'cals~was
(17654822), who
held
undertaken
chairs
by
mathematics
in
the
University of Modena.
Although his attempts
esincomplete, his arguments
go a long way toward
unlike
the general fourth-degree equation, that
of the
have
no
solution
in radicals
for which
the
inter1ne—
polynomials in the variables
$1, $2,
Lagrange’s
methods
of findinga general solution
to the equation
of fifth degree
could not, then, lead to success.
(See the section on Ruflini and the
Values
are
general equation
Ruflini’s work
.
.
..
of fifth degree.)
on
the
impossibility
of
solving
the
general
fifth-
1799 and 1813. A com—
degree equation in radicals
appeared between
plete proof of this impossibility—originally
completely independently
of Ruffini’s Work—Was given in 1826 by the
Dantwenty—four-year-old
ish mathematician
Niels Henrik
Abel. Abel’s proof contains
in particular a proof of the assumption
made by Ruffini Without
proof, namely,
that if a solution
in radicals
of the general fifth—degree
equation were
could
possible, the steps in the solution
always be arranged in such
a
all intermediate
values
are
way that
polynomials in the variables
50
5.
5121,1132,
.
.
.
.
Thus
Search
The
Additional
for
intermediate
values
Formulas
Solution
of the
form,
say,
ts/1+a33+3/:r1+m3a:4
in
formula
general solution
a
Abel’s impossibility
fifth or
with
radicals
always be avoided.
can
proof applies only to
generalequation
the
of
interpreted as
formula” in
in the sense
of a “general
are
to be determined
variables
of special
the solutions
elementary symmetric polynomials. Whether
higher degree,
fifth—degree
equations
in which
the
solutions
such
for
example,
as,
3:1, 3:2,
.
.
.
m5—a:~1:0,
f+%m~nm=a
be
represented
Abel’s proof.
Thus,
can
above
one
in
terms
for
equations cannot
For instance,
can.
$1
From
of nested
radicals
answered
of the
first of the
example, the solutions
while
be so represented,
6/571+ 6/i§+ x5/64
=
is not
Abel’s posthumous
papers,
—-
fact
in
in
second
the
{/144.
that
it is known
in
1828, after
trips to Berlin (1825) and
Norway from research
Paris
(1826), he was working on questions of the solvability in radicals
of special equations of the nth degree. Alas, at the time, Abel was
and with—
circumstances
In modest
seriously ill with tuberculosis.
commensurate
with his mathematical
a position
out having achieved
Abel died in 1829 at the age of twenty-six.4 The
accomplishments,
a particular
circumstances
and under
what
equaproblem of whether
for a solution
had to wait several
in radicals
tion is solvable
by
years
he had
returned
to
Galois.
Ruffini
and
the
for
Ruffini’s argument
arithmetic
computations
and
higher
was
General
Equation
of
Fifth
Degree
consisting of only
of degree five
of roots
of equations
and extraction
of
details.
Moreover, Ruffini’s method
incomplete in some
why there
is
no
general
4For a biography of Abel, see Arild Stubhaug,
Springer, 2000 (Norwegian original, 1996).
Niels
solution
Henrik
Abel
and
His
Times,
The
5.
Search
for
quite unusual
argument
was
heavily on
concrete
from
his
Additional
In
argument.
at
attempt
follows
in
proof
1813
Ruffini’s research
51
relied very
time, in which mathematics
and recognition
Receiving little attention
he attempted
to improve and
simplify his
peers,
what
Formulas
for his
calculations.
mathematical
Solution
we
in
will
look
considers
the
central
in which
ways
idea
of his
last
of
given
form.5
revised
slightly
a
the
at
the
variables
a
polynomial can be permuted without
For exchanging the polynomial.
3z2 remains
ample, the polynomial asy
:1:
unchanged when the variables
and y are
interchanged, while that is not the case for the remaining four
aside from the identity. The following result
permutations
is the basis of
Ruffini’s attempted
proof:
—
Theorem
let
5.3.
f(:1:1,
.
number.
.
,
For
a
polynomial g(a:1,
,a:5) denote
the mth
,:m;) in the variables
where m
g(a31,
5125)“,
power
.
.
.
.
.
identities
f(fl?1,fl72.1v3.m4:035)
f(932»$3,
f(931»£I32,a74,€B5,903)
=
the
for
corresponding permutations
hold for the polynomial g.
identities
A proof
the
begins with
the
=
f(932»$3,
of the variables,
remark
that
.
,
the
if the polynomial f satisfies
Then
:1:1,..
,:c5,
is a natural
.
the
then
the
is
assumption
analogous
equivalent
to
identity
9(=v1.
=
9(=v1.
=
Thus
there
exist
must
two
mth
roots
9(~”02.903,931,f04,f05)m
g(:z:1,cc2,:1:4,:r5,m3)m.
of
unity (1 and
(2 such
that
9($1,-’B2,933;1I«’4,$5)
C19(€B2,973,961,=B4.-'35),
g(.’L‘1,(II2,.'I}3,£‘C4,
(E5) C2g(.’I31,.’I22,£D4,(I35,£II3).
=
=
By repeatedly
applying
the first equation—~in
both
and the remaining two are
the
underlying
equations
three
permutation
variables
are
obtains
unchanged—one
the
of the
variables
permuted
in
cyclically
following result:
9(001,$2,$3.934.w5) C19(fl72.w3,w1,934,fB5)
Cl9(f'3a,f01,$2,=v4»005)
=
=
=Cf9(w1,$2:$3a$4,035)This
tation
together
implies that
with
the
analogous
computation
for the
second
permu-
5See Raymond
G. Ayoub,
Paolo
Ruffini’s contributions
to the
Archive
quintic,
for History Exact
Sciences, 23 (1980), pp. 253~277; Raymond
G. Ayoub, On the nonAmerican
solvability of the general polynomial,
Mathematical
Monthly, 89 (1982), pp.
307-401; Christian
Skau, Gjensen med Abels og Ruffinis bevis for unmuligheten
av
5.
n’ tegradsligningen
lgzise den generelle
nér n 35, Nordislc
algebraisk
Matematisk
Tidskrift (Normat), 38 (1990), pp. 53-84, 192; Ivo Radloff, Abels Unméiglichkeitsbeweis
im Spiegel der modernen
Mathematische
Galoistheorie,
45 (1998),
Semesterbertchte,
pp.
127~139.
Search
The
5.
52
mutation
In
combined.
are
permutations
is followed
by the second:
the
Now
SolutionFormulas
Additional
for
two
first per-
the
particular,
C1g(a22,:v3,:v1,w4,rc5) €1C2
g(032,033,fI34,9J5,IB1)
g(m1,:n2,a:3,:134,:I35)
=
=
first permutation
If the
is executed
then
twice,
obtains
one
C19(~T2,w3,$1,$4,935) CfC29(~’03,931,034,a75
9(»”v1,$2,-T3,764,0»’5)
=
=
the
For
the
two
five variables
the
permutations that underlie
.are
permuted cyclically
last
derived
two
(that is, the
two
equations,
Cycles 331 —>
With
:r2~+:c3-—>cc4—-+m5—>a;1and:I:1—>m3——>:r4—»a
argumentation
analogous
for
that
to
the
obtains
one
three—cycles,
two
1.
<<1<2>“’
(cf<2)5
=
=
the
using
immediately that C?
equations, (1
(C32((3-1
it follows
two
identities
previously
obtained
these
From
=
=
=
1 and
1.
then,
Building
from which
1,
(2
this,
((3)2(C§’)_1
for the polynomial g follow.
identities
finallythe asserted
once
it
is
at
in
five
of
variables,
With this proven
polynomials
property
fifth degree canthe
for
formula
a solution
that
equationof
general
plausible
of lower
for equations
of the formulas
not in the manner
not exist, at least
obtained
are
formulas
those
by
as
described
That
Lagrange,
by
is,
degree.
them
and
through
with
the
polynomials
symmetric
elementary
beginning
in
the
Variables
3:1, 3:2,
g2,
g1,
step
polynomials
step
by
determining
we
on
also
C5‘
obtain
=
1 and
=
=
.
so
each
of them
a
polynomials
obtained
in the
that
the
for
operations.
The
power
jth step therefore
.
that
is found
previous
steps
the
form
has
gj(:1:1,m2,...)mj
=
.
.
can
be
using the
determined
four
.
.
,
from
elementary
‘
..),
fj(CC1,.’1)2,
.
only of the elementary symexpressed in terms
in the
determined
the
and
metric
g1,
g2,
gj_1
polynomials
polynomials
previous steps. If the given general equation is of degree five (or greater),
that
to assert
be applied inductively
can
every
Rufiini’s argument
then
polynomial gj must satisfy the condition
where
the
function
fj
is
.
9a'(iB2,-"33,3?1,
9a‘(9«"1,902,~’vs,=B4,005)
=
.
.
,
9j(fB1,fl32,fl74,i1?5,€
=
9a'(iB2,-"33,3?1,
'
Exercises
53
None
of the
step,
such
steps
for
as,
therefore
can
lead
the
to
example, g(:c1,a;2,
.
.
"k
2
polynomial
of the
last
solution
931.6
Exercises
(1) For
given cubic
a
equation
a:3+aa:2+b:c+c=O,
determine
the
the
cubic
whose
equation
solutions
the
are
of
squares
given equation.
(2) Show
that
the
solution
of the
general biquadratic
equation
:1:4+a:1:3+b:n2+ca3+d=0
be obtained
can
directly, that
biquadratic equation
reduced
structing
a
cubic
in
order
from
the
(3) Carry
the
to
calculate
resolvent
the
out
it into
a
power) by con-
resolvent
9311172 -I- (123124
=
the
solutions
of the
biquadratic
in the
previous
equation
z.
calculations
implied
exercise
for
resolvent
z
6The
incomplete
transforming
(Without a third
for the
equation
Z
is, Without
question
=(a=1 + m2)(ws+ 0:4)arises
natually
deficits
solution
as
the
to
points
at
which
R.uffini’s argument
have
already mentioned
is
fixed.
We
that
Abel
offered
a
a
in radicals
of the general
proof that
if one
equation,
exists,
can
in such
a
that
each
always be obtained
intermediate
to a
way
step corresponds
in the
solutions.
A reproduction
of Abel’s proof with
polynomial
can
commentary
he found
in Peter
An
on
the
sources
Pesic, Abel’s Proof.
and
essay
meaning
of
mathematical
As an
unsolvability,
alternative
Cambridge,
MA, 2003, pp. 155-174.
to Abel’s argument,
it is also
to formal
possible to extend
permutations
and
how
these
can
be
expressions
containing
nested
radicals,
such
as
+ 3/$1 +:z;g:1:4.
15/1+:I:3
One can
find
for beginners,
go
into
this
complete
proof
based
this
in John
approach
Stillwell, Galois theory
American
Mathematical
We will not
Monthly, 101 (1994), pp. 22-27.
to highlight
Galois’s more
further, preferring
general
approach.
a
on
5.
54
(4) For
Search
The
polynomials with
two
900
defines the
--(X
resultant
=
is
resultant
have
21;)fl fi(wz
—
root
j=1
identically
for the
and
that
the
resultant
coeflicients of the
formula
is called
the
the
Give
A permutation
when
Show
from
explicit
precisely
common.
derived
a
zero
in
polynomials
be formally
can
an
$71):
-
by
i=1
Clearly,
factors
linear
into
decompositions
=
R(f,9)
the
Formulas
Solution
Additional
(X
f (X)
one
for
case
n
=
m
cyclic if it permutes
leaves
the
remaining
2:
two
polynomial.
2.
k of the
71
72
numbers
k numbers
in
,n cyclically
1,2,
as
is known
numbers
two
that
A
exactly
exchanges
cycle
place.
the following:
Prove
a transposition.
is the product of cycles.
permutation
Every
(a)
of transpositions.
the
is
product
Every
cycle
(b)
of
is
the
transpositions.
product
Every
permutation
(c)
that
of
is
transpositions
product
permutation
the
(d) Every
other number.
1 with some
exchange the number
that
of
is
the
transpositions
product
(e) Every permutation
j and j + 1.
exchange two adjacent numbers
.
‘
.
.
—
6
Chapter
That
Equations
Reduced
in
In contrast
Can
Be
Degree
to the
equation
1:5—2:v4—~4a:3+2:1:2+11cn+4
=
can
be determined
0, whose solutions
from a quadratic
equation and a cubic equation with integer coeflicients,
nothing is comparable for the equation 21:5 +6ar:2 + 3
0.
=
What
is
the
basis
of this
difference,
and
how
can
it
be
recognized?
6.1
The
techniques for findinggeneral solution
formulas
for equations
of particular
degrees. Now, with
Abel’s proof of the impossibility of finding
a formula
for the solution
in radicals
of the general equation of fifth or higher
degree, we shall
restrict
our
attention
to special equations
of fifth and higher degree.
The
previous
factors,
than
dealt
with
first equation
example
an
chapters
can
of
be
presented at the beginning of this chapter is
equation that though not decomposable into linear
an
decomposed
into
factors
whose
degrees
are
greater
1. Since
a:5—2:r4—4:r3+2:c2+11a3+4=
(m3—31:—4)
(a:2—2:z:—1),
three
of the
five solutions
can
be determined
from
the
cubic
=0,
(:z:3—3m—4)
and
the
remaining
two
solutions
form
332-251;-1:0.
the
quadratic
equation
equation
6.
Equations
That
methods
described
in
56
the
Using
Be
Can
the
previous
Reduced
in
Degree
chapters,
one
obtains
solutions
the
\xB0
\3/2—\/g
\xDE\xE0
(L‘1,2,3=€‘3
C321,
‘
.’1?4)5=1::\/§.
of the
equation, there is no corresponding decomposition
2185+ 63:2 -5- 3 into two polynomials with rational
second
For the
polynomial
such
How
coefficients.
a
assertion
negative
be
can
justified
and
how
of this chapis
the
exist
topic
they
decompositions
in which concrete
to the previous chapters,
computa—
In contrast
ter.
concerned
more
stood in the foreground, here we will be rather
tions
of
to
of
View
polynomials
properties
with a qualitative
relating
point
too long and
are
not
The
coefficients.
or
with rational
proofs
integer
based
are
has
to
what
in
but
they
before,
gone
comparison
difficult,
on
a different
type of argumentation.
be found
can
of
basis
The
when
the
is
applications
following
due
proof,
to
Carl
Gauss:
Friedrich
polynomials whose leading
1
is
the
the
of
(such
power
IL‘)
highest
of
coeflicient
coeflicient
(that is,
are
whose
all
called
rational,
are
coefiicients
of
monic),
polynomials
and whose product g(m)h(;z:)is a polynomial with integer coefiicients.
be
h
must
and
the
Then all the coefificients
g
polynomials
original
of
6.1.
Theorem
h(a:) be two
g(:L')and
Let
integers.
theorem
This
that
be
can
the
generalization of the wellof 2 is irrational.
Namely, the poly-
as
seen
a
root
drastic
known
fact
nomial
{E2
rational
integers,
coefficients,
is obviously impossible.
Moreover, the argumentation in the
of polynomials with integer coefficients
the decomposition
on
in the classical
with that
certain
proof that the
relationship
which
section
has
square
a
not
admit
since
a
such
relations,
contradiction.
where
the
into
decomposition
basis
of both
assumption
is
of such
factors
linear
have
would
coefficients
The
of 2 is irrational.
root
divisibility
a
2 does
—
square
a
a
to
with
be
detailed
check
of
relation
leads
to
6.
That
Equations
The
Recall
Can
Be
Reduced
Decomposition
Theorem
of
in
Degree
Integer
57
Polynomials
6.1:
If \xB0\x8F\x94
and Min) are
coefl°1~
polynomials with rational
cients
whose
product g(:c)h(m) has all integer coefficients,
then
the two polynomials \x80\xEE
and h(:r) have integer coefficients.
two
monic
The
proof of this theorem, which goes back to Gauss, begins with multiplyin g(:r) and \xE0\xB1\xEA
In particular,
ing through by denominators
we
determine
two integers a and b of minimal
size for which
the two polynomials
a
g(m)
and b g(:c) have all their
coefficients
integers. We denote these coefficients
and do, d1,
We now
by co, c1,
investigate the product ab 4\xE5
We will now
show by obtaining a contradiction
that there
is no prime
number
divides
all the coefficients of the product ab
13 that
\xB0c\xE3This
is how we shall prove
the assertion:
since the product \xC0\xE3 has integer
of the asserted
coefficients, the nonexistence
prime p implies at once
that
ab
a
b
1, and therefore
1, so that given how a and b were
chosen,
it follows
that
the given polynomials
have had any
`ѧ and Mac) cannot
denominators
to get rid of, and so their
coefficients
must
have been integers
to begin with.
-
-
.
.
.
.
.
.
«
.
-
=
=
Let
us
therefore
of the
prime number
p
Case
We
a
-
g(:n)
find the
p.
The
can
be
b
nor
that
assume
all coefficients
1.
:
we
smallest
polynomial
consider
two
first
consider
indices
coeflicient of the
expressed
as
the
the
the
while
to
be divisible
choice
of
j
all the
Case
a
g(:v)
-
other
In
2.
or
b
-
j and
I9 such
term
9cj+k of the
by p,
to this
which
neither
that
neither
the
by p.
ab
polynomial
polynomial
Thus
we
dk is divisible
nor
Cj
case,
divisible
-
+
+
Cj+1dk—1
'
to
first
summand
cannot
be divisible
by 13.
must
we
-
in contradiction
the
summands
h(m) are
\xF0k\xDAIn reference
only coeflicients divisible
is, since
this
divides
-
may
by
g(:r)h(ac), which
sum
not
and
-
in
case
‘
seen
ab
prime number
p that
a
cases:
+
Cjdk + Cj—1d/c+1
is
exists
product
h(:13)contains
-
there
assume
that
by p.
Without
all
our
--
,
assumption,
given
be
by p,
divisible
the
coefficients
loss
of
of either
generality we may
assume
that this holds for a- `\xB1\xCB
Since in the polynomial g(:1:)the leading
coeflicient is 1, it must
be the case
that
a is divisible
by p. In particular,
a
> 1. But
that
delivers
our
desired
since
the polynomial
contradiction,
integer coefficients, contradicting the minimal
choice
% g(m) must possess
-
of
a.
now
apply
factor,
we
If we
a
linear
can
such
roots
must
easily
check
only integer
the
divide
of the
roots
to
that
once
Be
the
special
roots
of the
term
of
case
its rational
among
of integers
Since
roots.
polynomial,
one
can
to determine
potential roots
as
Degree
searching for
polynomials with integer
monic
constant
finite number
a
all rational
theorem
at
see
have
coefficients
the
in
Reduced
Can
That
Equations
6.
58
polynomials.
the demonstration
special cases
of the impossibility of a product decomposition into polynomials with
be simplified,
shall look at some
can
we
coefficients
rational
examples.
if
can
be found
We would like to know how a product decomposition
the polynomial 3:5 2x4
As an example, let us consider
one
exists.
If
4a:3 + 251:2+ 1120 + 4 presented at the beginning of the chapter.
be of first or
must
of the factors
then one
is a decomposition,
there
search
tells us that we may limit our
second
degree, and the theorem
see
coefficients.
We
with
to monic
immediately
integer
polynomials
are
that
there is no linear factor, since there
only six factors of 4 to
of the
out to be a root
of which turns
check, namely :|:1, :l:2, :|:4, none
have to be of the
would
any decomposition
polynomial. Therefore
Before
6.2
we
describe
how
in many
—
-
form
:v5—2:v4—4:1:3+2:1c2+11a:+4
=(a:2+a:1:+b)
where
and
a
c
must
be
integers and (2 must
be
b
of the
one
six factors
of
coefficients
the
by evaluating
For example, at
the fifth—degree
polynomial at integer arguments.
:12
2, the polynomial has the value 2, so that, for instance, the
2. Therefore,
hypothetical quadratic factor must divide 2 when w
of 2, namely
of the four factors
be one
the expression 4 + 2a + b must
4. One
can
further
obtain
restrictions
on
=
=
one
that
Already these two
in all, “only”6 4 possibilities
of i1,:|:2.
-
allow
restrictions
for
b need
and
a
us
to
to
be
conclude
tried.1
if one
the five complex
is possible
determines
different
1A completely
approach
one
need
Then
of the polynomial
to be factored
roots
approximation.
using numerical
of linear
factors
produces a polynomial with integer
only check which
possible selection
of rounding
error
the possibility
be done
without
by multiplying
Coefficients, which can
Those
in a hurry
can
obtained.
thus
out
algebra
try a computer
any
polynomials
command
Thus
the
Mathematica
Factor[:z:5 2:134 47:3 + 2:z:2 -1- 11:1; + 4]
system.
—
immediately
yields
the
result
(-1
—
2:1: +
.722)
(-4
—
3:1: +
—
:33).
6.
Equations
That
With
restrictions
on
conclusions,
that
reach
such
negative
eflicients cannot
Can
Be
Reduced
the
be
expressed
lower degree. Such polynomials
in
Degree
coefficients
one
59
of
can
also
course
given polynomial with integer cothe product
of two polynomials of
a
as
called
are
irreducible
over
the
rational
numbers.
As
6.3
easier
promised,
for
that
make
ways
proof of irreducibility
a
decisive
On the
divisibility relations.
at the beginning of the
chapter
irreducibility criterion, named for
so—calledEisenstein
the
we
can
the
mathematician
who
proved this theorem
in
earlier
Schonemann.
years
Ferdinand
by Theodor
Theorem
6.2.
Suppose
Gotthold
0
an_1,
0
a0
Then
.
.
-
is not
the
The
-
divisible
It
a
proof given four
a
=
number
p:
by p.
the
is irreducible
Eisenstein
be found
can
of
by p2.
polynomial f
proof of
difficult.
divisible
are
,a1,ao
.
(1823—1852),
Eisenstein
polynomial f(£L') m” +
that satisfies
the
integer coefiicients
given
are
+
+ alzv + a0 with
an-1a2”‘1
following conditions
for some
prime
-
Max
1850, independently
we
frequently exist
of
use
polynomial 2505+ 6:32 + 3 introduced
use
there
the rational
over
criterion
irreducibility
appropriately
in the
named
numbers.
is not
section
very
of this
chapter.
There
is
trick
no
to
showing
that
the
polynomial
2x5 + 6:122+ 3
is irreducible
means
two
tain
criterion.
using the Eisenstein
(Recall that irreducible
that
the polynomial
cannot
be decomposed nontrivially
into
polynomials of lower degree with rational
To obcoefiicients.)
a monic
polynomial, we begin with the polynomial multiplied by
16, which
can
Eisenstein
criterion
is irreducible
be written
over
(2ac)5+ 24(2:r)2+ 48.
for the
the
prime
rational
carry
over
6.4
An important
is to
the
to
one
for
3, the polynomial
numbers.
2:135+ 6:132+ 3 is also irreducible,
diately
=
p
since
With
respect
to
the
3/5+ 24y2+48
Therefore, the polynomial
a decomposition
would
imme—
y5 + 243/2+ 48.
application of the Eisenstein
0.
cyclotomic equation $7‘ 1
—
=
irreducibility criterion
Since
the
linear
factor
in
Reduced
Be
Can
That
Equations
6.
60
Degree
irreducible
is never
cyclotomic
equation
(ac 1)
factor
for n > 1. However, for a prime exponent
(:1:
n, the linear
rational
with
into
the
is
decomposition
possible
polynomials
only
1)
In other
coefficients.
words, if you factor out the linear factor, what
That
is left is irreducible.
is, it can be shown that the, polynomial
can
—
be
the
removed,
—
:I:"—1
$—1
if
is irreducible
1:
is
n
the
this, make
theorem,
To prove
prime.
using the binomial
1, obtaining,
y +
=
__::L,n—1+$n——2+___+m2+$+1
substitution
y"”+'-~+(Z>y2+(Z>y+
‘“—%’:~1=v"”+
coefficients
are
integers. Furknown, all the binomial
given shows that all the binomial
thermore, the last representation
are
divisible
coefficients that
appear
by n, since the prime factor n
As is well
appearing
the
in
We
nominator.
to
the
to
be irreducible
then
can
~
rational
the
over
Eisenstein’s
Theorem
Recall
by anything
Eisenstein’s criterion
apply
(ac" 1) /(:1:
polynomial
canceled
is not
numerator
—
this
1), whence
de-
the
in
for the
prime
polynomial
is
n
seen
numbers.
Irreducibility
Criterion
6.2:
at" + an~1:c”_1
+
given a polynomial f(a:)
Suppose
that satisfies the following
«+a1:c+ao with integer coefficients
for some
conditions
prime number
p:
o
divisible
an_1,
,a1, a0 are
by p.
a
divisible
a0 is not
by 112.
we
«
=
are
-
.
Then
the
.
.
polynomial
f
is irreducible
rational
the
over
num-
bers.
proof can be carried
opposite of what we are
The
thus
assume
that
we
out
indirectly
trying
have
a
to
prove
once
again;
that
and
arrive
at
decomposition
assume
the
contradiction.
We
is, we
a
g(a:)h(cc), where
\xF0}\x90
=
g(a:)
Exercises
and
61
h(m) are
monic
g(m)
h(:c)
with
least
at
ds
=
cr
equal
From
1. It is of
=
.
.
=
crmr + cr_1mT_1+
=
dsms + ds_1a3s"1+
assumed
course
previous
theorem,
be integers.
do must
d5,ds~1,
ble by the prime number
.
coefiicients:
rational
-
-
-
that
+ co,
-
-
.
the
do,
+
degrees
7"
and
are
s
each
1.
to
the
with
polynomials
,
p,
but
coefiicients cr,c,_.1,
,co and
Since the product
(10
codo is divisinot
p2, exactly one of the coefficients co
all
the
.
.
.
=
and
do must be divisible
that it is co. Thus the coeffi~
by p. Let us assume
cient do is not divisible
index
by p. Since cr
1, we can find the smallest
3' for which Cj is not divisible
by 13. For the corresponding coefficients aj of
2
the
polynomial
f
we
tlj
where
divisible
is
a
first summand
the
by p.
Thus
have
=
the
formula
+
Cycle-1- Cj—1d1
is not
(lj is not
divisible
divisible
-
-
-
Codj,
+
while
by p,
by p, which
on
all of the
account
others
of j S r <
contradiction.
Exercises
(1) Find
factorization
a
of the
$6 + 91? + 19:34
over
(2) Show
the
rational
numbers
the
polynomial
that
polynomial
4a:3 + 5:22
—
into
irreducible
—
13:3
—
3
factors.
m6+41:5—2:n4+:r3—39:2+5a3+1
is irreducible
over
the
rational
numbers.
are
n
7
Chapter
The
Construction
Regular
of
Polygons
With
the
“With concentrated
words,
thought
in
\x90\xE7
the
Carl Friedrich
of bed),”
Gauss
describes
the circumstances
surrounding his dis~
in the year
1796' that the regular seventeen-sided
couery
polygon can be constructed
using straightedge and comHow could
Gauss
have managed to consider
pass.
the
as
an
possibility of a geometric construction
eazercise
of
pure
imagination?
morning
7.1
The
March
scope
(before I got
\x90\xC2\xE5
discovery described
in
heptadecagon
Gauss on
by the eighteen—year—old
whose
beginning of a life in mathematics
above
29, 1796, marks the
and significance
have
described
seldom
literary journal
a
out
his
been
equaled.1 Gauss himself
discovery regarding the regular
thus:
(seventeen—gon)
It is known
to
beginner in geometry that various regular polygons, namely the
triangle, pentaand those
obtainable
gon,
fifteen—gon,
by doubling
the number
of sides, have been known
to be constructible
since the time of Euclid, and it would
ap—
that since that time, mathematicians
pear
have convinced
etry
themselves
was
unable
every
that
to
the
field of elementary
yield further
results;
at
geom-
least
I
1The chronology
mathematical
circle
is based
Mathematisches
of Gauss’s discoveries
is extraordinarily
well
documented
in his
diaries.
The
first entry reads, “fundamentals on which
the division
of the
and
indeed
its divisibility
into
seventeen
etc.” See C. F. Gauss,
parts,
Tagebuch,
Klassiker
Nr.
1796'~1814, Ostwalds
256, Leipzig, 1976.
63
The
7.
64
of
know
successful
no
try’sreach
direction.
more,
it
of the
taken
that
discovery
that
note
in addition
to
to
seems
Regular
extending
at
attempt
in this
All the
of
Construction
me
Polygons
geome-
should
be
those
reg-
to geopolygons, a host of others are amenable
metric
construction, for example, the heptadecagon
ular
.
.
.
.
generstraightedge and compass,
ally of triangles from three given data, are a residue of classical matheThe significance
still a part of the standard
school curriculum.
matics
is less their
of such exercises
practical application than, aside from
to aid the
that
stretches
back to antiquity,
being part of a tradition
with
of thought.
Construction
in developing
student
logical habits
to prescribed
is limited
straightedge (unmarked ruler) and compass
of certain
points,
elementary operations that allow the construction
of unit length. Thus
starting with two points separated by a distance
the following
given a set of points that have been thus constructed,
constructed:
can
be additionally
Geometric
0
constructions
circle
whose
midpoint
structed
and
whose
radius
is the
structed
points.
between
two
Draw
a
0
Draw
a
0
Every
intersection
line
straight
of circles
steps is considered
At
first glance,
have
that
of
circle.
can
in
seen
the
a
is,
regular
there
is
and
and
Chapter 2, the
con-
distance
between
two
con-
constructed
points.
be
equations
roots
connection
no
in
equation
2:"
n—gon, and
indeed
with
unit
that
the
next
previous
two
between
such
the
variable.
However, as
of unity in the complex plane,
one
of the
Figure
in the
drawn
point.
to
nth
been
lines
constructed
seems
has
that
point
a
solutions
n
Consider
show
a
constructions
geometric
we
with
—
1
2
circle
0, are
as
the
vertices
circumscribing
If starting
at the point 1
(1,0)
point of the n—gon in the counterclockwise
7.1.
=
we
be constructed
with
+ isin IM can
direction, namely C cos \xC0*Y
then we will have succeeded
in proving the
straightedge and compass,
regular n—gon to be constructible.
=
Gauss, who was
of complex numbers
well
as
acquainted with the geometric interpretation
one
in his honor
points in the plane~—indeed,
7.
The
Construction
of
Regular
sometimes
speaks of the Gaussian
equations
in radicals.
he first ordered
the
In
nth
find suitable
to
of
unity
in
The
solutions
regular
as
expressed
octagon.
0 form
be
(2
It
7.1.
cos
a
cyclotomic
intermediate
values,
of the
integers.
of the
cyclotomic equation 2:8-1
All eight eighth roots
of unity can
1,(,(2,
powers
.
.
motivated
way,
.,C7 of the primitive
=
root
+ isin
\xC0\xAE\xE0
sensible
seems
to solve
particular
a
by his knowledge of the divisibility properties
Figure
65
able
plane——was
order
roots
Polygons
at this
point
to order
the
roots
according to their
position on the circle, that is, as seen
from Figure 7.1, in the order
where
cos
1,C,(_,"2,...,("”1,
C
@ +z'sin P]\xF0Gauss, however,
realized
that it makes
sense
to list the roots
in quite a different
order,
at least in the case
that n is a prime number.
Considering that Q" 1,
the value
of (j depends only on the remainder
when j is divided
by
=
=
n.
Therefore, we
can
division
by n.
In
possible,
in the
case
remainders
by repeated
2A proof
choose
addition
of
order
any
to
the
of the
obvious
prime number
a
possible remainders
order
n,
to
1,2,.
obtain
..,n
addition
of this
book.
.
fact
can
be
found
in
the
epilogue
to
this
g.2
1, it
—~
all the
1 not
1,2,.
,n
only by repeated
number
multiplication
by a suitable
—
.
This
on
is
nonzero
of
1, but
leads
to
The
7.
66
Construction
of
Polygons
Regular
obtained
g”‘2. The remainder
ordering g0, g1, g2,
divided
by n is called a primitive root modulo n.3
an
starting
case
=
=
=
=
=
=
=
=
the
obtains
order
1.
3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4, 12, 2, 6,
Since
g is
when
,
=
10
E
of n
.
3. Indeed,
17, for example, one can choose g
33
with g0
31
1, after g1
3, g2
9, one has g3
mod 17. Then comes
34 E 3- 10
30 E 13. Altogether, one
In the
27
.
.
the
list
gm E 9, and
In the
of roots
so
of
of unity
could
with
continue
g17 E 3,
ad infinitum.
on
case
gm E 1, we
with
ends
the resulting
(17—gon),
regular heptadecagon
a
takes
the
list
form
(-1 C3 C9 C10 413 Q-54-15(:11 C16 C14 (8 4-7 C4 (12 C2 {-6
)
1
5
7
>
3
)
3
)
)
3
)
1
‘
9
of the roots
of unity,
partial sums
called
calculation
of the roots
periods, which allow for a step—by—step
of unity. One begins with the two periods containing the roots
of unity
that stand
in odd, respectively even,
positions. These are called the
ez'ght~member periods:
The
of this
7
purpose
is to
form
no:€1+4-9+€13+€15+¢-l6__C8+<-4+4-2’
771:§-3+€1O+C5+4-11+;-l4__€7+€l2+€-6.
Next
considers
one
the
four
positions differ by 4 in the list.
called four~member periods:
the
periods containing
These
of four
sums
roots
whose
roots
of unity
are
uo=<1+¢13+<;1“+<4,
u1=c3+<5+c14+c”,
u2=<9+<15+€8+<"’,
us=<1°+<“+<7+<“.
Finally,
of roots
3The
of
we
unity
expression
holds
only
question
17, or in mathematical
mod
both
17, since
when
Equivalently,
consider
a
the
distance
two—memberperiods, which
of eight apart
n” is generally
division
by n. For
“modulo
up
to
a
language,
numbers
the
have
difference
is
12
the
46
congruent
same
—
12
is
in the
used
to
example,
to
46
original
indicate
12
that
is
modulo
are
the
sums
list.
For
the
identity
our
in
“equal”to 46 modulo
12 E 46
17, written
12, on division
by 17.
remainder,
namely
divided
by 17, the remainder
is
zero.
7.
The
Construction
the
purposes
of
Regular
fie
<1+ <16,
fl4
<13+ <4.
real
and
=
All of these
is what
periods
Gauss
are
recognized
67
suffice:
following two periods
=
this
Polygons
have
the
further property—and
by “concentrated
thought”—obtained
by this special construction
that every period can
be obtained
from
the next-longer period
by a quadratic equation. For this the periods
are
paired off, so that each sum and each product of the
pair can be
represented
this
as
a
of periods
of double
begins with
the two
sum
the
length.
Let
us
see
how
works.
The
calculation
771. Their
is not
sum
difficult
too
to
eight—member
periods
770 and
calculate:
no+?71=€1+C2+-~+C16=(1+C1+C2+
where
we
which
follows
to
the
the
that
note
at
of
mass
equation, while
of the
sixty—four
products in
effort, one obtains that
periods
can
of all nth
sum
771171
be calculated
as
-4.
y2 +y
the
origin
of the
4
:
applied
is
clearly
contrast,
determining the
great but elementary
Therefore, the
—
always zero,
theorem
After
solutions
is
Viete’s root
In
770771 is tedious.
=
unity
geometrically,
vertices.
n
of
roots
algebraically from
once
cyclotomic
center
the
two
quadratic
eight—member
equation
0)
yielding
no ’ 1
Now from the
member
periods
the
details:
gory
two
=
-E :l: \xC0*
2
2
eight—member
periods
,ug, ,u1, ,U,2, #1; can
be
770 and
calculated, though we
/1/0+M2:=7lOa
/1o#2=C1+C2+'''+C16=*1,
#1
+#3=7)1a
/-L1/13
=
-1-
171, the four
shall
fouromit
tions,
the
to
Polygons
Regular
two
quadratic
of the
four-member
following
calculation
the
possible
make
which
lead
identities
four
These
of
Construction
The
7.
68
equape~
riods:
yz
-
noy
z?‘
=
The
those
while
second
of the
Finally,
Again, the key is the
can
we
—
mz
1
=
1
=
0,
0.
first equation
of the
solutions
two
—
equation
the
compute
are
,u0 and
=
y1
p1 and
=
zl
=
Z2
y2
of their
periods fil and 34.
and product:
sum
=
this
we
=
=
=
From
the
obtain
solutions
whose
are
If one
wishes,
C from the quadratic
one
=0,
two—memberperiods y1
two
may
[30 and
=
seventeenth
the
calculate
now
m.
equation
quadratic
1/2=/I01/+li1
the
M2,
,u3.
(C1 C16) (C13+C4) M0;
+ C4) (14 +65 + C”+ C3
,80fi4 (<1+ (16)(C13
30 +54
+
4'
=
=
two-member
two
calculation
are
y2
=
64.
of unity
root
equation
C.’/2:/803/+1:0>
whose
two
solutions
are
=
y1
and
‘C1
y2
(16. However,
=
quadratic equation does
construction,
can
into play, since the regular heptadecagon
segment of length ,B0 2 cos \xD0\xD4
this
ric
in
geometneed to be brought
constructed
using a
not
be
a
=
If
the
one
other
the
quadratic
chooses
the
solves
and
approximations,
in the
then
one
equations
in
solutions
obtains
as
result
end
based
order
an
obtained
have
we
the
on
after
one
numerical
identity suggested
introduction:
1
2
1/
1
34
2x/1-7
8+8\/1_7+8
,80=2cos1:=
34+2x/Ti
34—2«/17-2
17+3x/fi—
W‘
W
v
+1
This
expression
regular heptadecagon
in square
is
roots
constructible,
not
only shows
but
also
at
indicates
that
once
how
the
such
a
7.
The
Construction
construction
can
of
be carried
Regular
Polygons
out.4.
reason
The
69
is that
the
constructibil-
ity of a point with straightedge and compass
is equivalent to the point
being expressible by rational
numbers, the four basic arithmetic
oproots.
erations, and the taking of square
For more
on
this, see the
section
on
constructions
with straightedge and compass.
Constructions
Using the system
point
be
can
into
with
of Cartesian
constructed
Theorem
7.1.
to
the point
and
compass
numbers
problem.
Given
only if its
nested
square
subtraction,
We
with
is
begin
indeed
can
said
to
the
be
constructible.)
points.
The
and
three
with
the
left-hand
constructed
and
compass
of what
be translated
can
the
from
measure
point
(0,0)
be constructed
with
be
expressed
in
arithmetic
operations
can
using the four basic
and
that
question
following theorem:
coordinates
observation
Compass
geometric
the
can
multiplication,
constructed
operations
have
plane
two
roots
be
basic
with
in the
if and
of addition,
We
and
the
unit
“primitive”
the
(1,0), a point
and
coordinates,
straightedge
with
purely algebraic
a
Straightedge
straightedge
rational
division.
a
straightedge
In particular,
point
with
and
compass.
we
shall
such
coordinates
(Such a point
show
that
the
four
extraction
of square
roots
lead to constructible
drawings in Figure 7.2 show how- beginning
and b and the unit length 1, one
can
construct
lengths a
the lengths a + b, a
and subtraction
b, ab, and %. Addition
are
easily
carried
out
by transferring one segment to the other using the compass.
and division
are
realized
Multiplication
the parallel
lines
by constructing
indicated
in gray.
The laws of proportions
in similar
triangles guarantee
—~
the
results
shown.
root
is accomplished
Taking a square
in
using the laws of proportion
similar
right triangles. In the picture on the right in Figure 7.2, all three
right triangles (the two smaller triangles form a larger triangle inscribed
in
the
semicircle
relation
The
analyze
1
of diameter
x/5
_
—
a) are
similar.
Observe
that
\/E satisfies the
T.
W
converse
the
1 +
statement
operations
is also
of construction
not
difficult
with
to prove.
To do so, we must
and compass
in
straightedge
terms
of their
effect on the coordinates
of newly constructed
points.
René Descartes
was
the first to make
significantuse of the idea
formulating geometric problems algebraically. Thus on the first pages
of
of
4An explicit description of such a construction
can
be found
in Ian Stewart,
Gauss,
Scientific American,
as
well
as
237, no.
in Heinrich
7, pp. 122-131
Famous
Tietze,
Problems
Solved
and
of Mathematics:
Unsolved
Mathematical
Problems, from Antiquity to Modern
Times, New York, Graylock Press, 1965.
70
7.2.
Figure
of square
tion
7.
The
How
the
basic
roots
of
Construction
and extracoperations
and
with
straightedge
arithmetic
accomplished
are
Polygons
Regular
compass.
Géometrie of 1637
La
work
his
Figure 7.2.5 Descartes
in
powers
a
a
other
way
also
than
surface
in the
is commemorated
name
and
term
first to
those
to
of
areas
interpret geometric products and
and Volumes, which made possible
higher
powers.
the
was
use
of fourth
wide—ranging
finds figurescorresponding
one
Cartesian
Descartes’s contribution
coordinates
(from the
Latinized
Cartesius).
explicit computations to
the regular
for constructing
himself that there is a method
convince
heptadecagon. It sufiiced to realize that using the periods it is possible
of unity using successive
a
root
seventeenth
to calculate
quadratic
equations.
Finally, what is crucial is that for each period, another
length can be found, so that the sum and product
period of the same
the length. We are
in terms
of periods of double
can
be computed
to
see
in
order
detail
this
in
somewhat
to
consider
now
greater
going
with straightedge and
what other regular polygons can be constructed
be
will
calculations
some
rather
complicated
Unfortunately,
compass.
of
for an understanding
which, however, are not necessary
necessary,
be skipped.
the later chapters and may therefore
7.2
by
such
did
not
As
have
already described, the stepwise solution
1
0, where
cyclotomic equation can
we
of the
Gauss
5See
Scholz
have
Gauss
Henk
(ed.),
J.
to
carry
out
—
M.
Geschichte
Bos,
der
=
discovered
n
und
Viete
Descartes,
Reich, Algebra:
1990, pp. 183-234.
Algebra, Mannheim,
Karin
is
a
in:
prime
Erhard
7.
The
Construction
number,
employs
for
which
the
complete
set
the
6
form
unity C
f
g1,g2,
of the
numbers
n
=
1, one
——
Regular
primitive
a
list
.
.
Polygons
modulo
root
divided
can
from
to
the
special
Pf(1)
f, on
=
most
e
k
case
a
of
of
The
Such
member
periods.
.
-
,
(Ckse)
Pf
for which
Pf
are
(<'kg(.f“1)e),
be distinct:
can
(€92), Pr (Gym).
proved relates to the product of two f—member
product can always be expressed as the sum
of fTo see this, observe
that
a
f—1
f—1 f—1
CJ'9z>e+kgqa.
$
2
\xB0\xB5\xFC
Cjsbe)
(icky?!-3)
Pf \xB0\xFD
Pf \x80\xFB
:
:
p=O
summation
obtain,
all periods
--->
f—1
we
4'“-""""“.
+
-
to be
property
periods.
.
¢
f—member
periods
of the
.
-
of
:
Pf CC)» Pf (C9):
then
in
.
O, :i:n,
=
account
Pf (Ck) Pf
If the
results
by n,
g
TL
:
at
integer
1. For every factorization
1, 2,
,n
for each power
of the root
define,
CI“
.
+ (W + CW‘
+
Pf (ch) <’°
equal
is, an
—
.
=
Aside
that
n,
,g”‘1,when
.
71
+ isin @\xE8\xE1
the f—member
\xC0y
periods
cos
=
of
index
as
q=0
of the
inner
p=0
11:0
is transformed
sum
by q
=
p +7‘,
desired,
f—1 f—1
f—1 f—1
Z €(j+kg>~e)gpe
Z Z <'(j‘l‘kg’I‘e)gPe
Pf \xB0\xE5
\xB0}\xF9 Z
:
::
r=O
p=O
'/‘=0
p=O
case
of
\xF0R\xF8
'I‘=O
<
is even,
then,
Pf
:..
If the
member
number
periods
e
can
2 f—member
periods.
>
\xC0\xF2\xE5
be calculated
We
k g'I‘B
I
as
using quadratic
clearly have
Pf (Ck)+ Pf
in the
the
(CW/2)
=
equations
identity
=
77,
P2f (Ck)
-
17, the ffrom
the
the
associated
2 f—member
periods,
it suffices
To
that
see
such
possesses
a
of
Construction
The
7.
72
also
product
to
corresponds
the
that
to show
Polygons
Regular
sum
of the
two
of
sum
a
squares
representation:
+pf
E1
<€Icge/2
(C1c(1+g““))
(Clef/2)2<13),
Pf (Ck)2Pf
+
Z
q=0
f—1
:;
Z132],(Ck(1+g‘”)>
_
q=O
clear
immediately
general
of the
terms
out the sixty—four
who has painfully worked
to anyone
product 770711 in our earlier investigation of the regular heptadecagon.
the result
one
obtains
to the explicit calculation,
In comparison
using
the formula
quickly:
just derived much more
is useful
formula
this
That
be
would
7
%+w?=RKV+RM@f=§:Hs@”“fi=
q:0
Here
it is
only
ferent
from
~1
obtains,
for
the
index
summation
namely,
appears,
q
P16(1)
4 that
=
=
16.
a
dif-
summand
Consequently,
one
desired,
as
1
1
mm=§Qm+mV—@$H@)=§G
shows
period squares
1
0
for solving the cyclotomic equation 1:"
that
Gauss’s method
a
if
n
is
of
a
leads
to
prime
equations
quadratic
sequence
always
are
known
23 + 1. As of today, there
of the form n
number
only
Fermat
five such primes, called
namely 3, 5, 17, 257, and
primesf‘
a regular
to show that
65537. Finally, it is not difficult
n—gon is always
when n has only Fermat
constructible
primes to the first power as its
In
formula
the
general,
for
the
of the
sum
2
—
=
6
Since
-25
(1
the
can
New
+---::2(*“1>3')
29‘+
+2“ ~23”'
2-” is always
=
1
for an odd
composite
prime only if the exponent
number
a
power
Further
prime, since it has 641 as a factor.
in Paulo
be found, for example,
Ribenboim,
York, 1988, 2. VI.
details
on
number
25 + 1
form
not
+ 1
(—1)k+12J"°
a
can
be
s
is
The
k.
Book
’
of the
a number
Therefore,
232 + 1 is
of 2. However,
numbers
of the
of Prime
Number
form
25 + 1
Records,
7.
The
odd
Construction
of
divisors.7
And
prime
Regular
the
Polygons
holds
converse
73
well.
Therefore, a
if and only if the prime
regular n—gon is constructible
decomposition
of n consists, aside from a possible
of 2, of only Fermat
power
primes
to the first power.
Thus n-gons
can
be constructed
for the following
values of n: 2, 3,4, 5, 6, 8, 10, 12, 15, 16,
17, 20, 24, 30,32, 34, 40,
We note finallythat an explicit derivation
of the quadratic
equations in the constructions
of regular 257- and 65537—gons
is not that
difiicult using a computer.8 In both cases,
3 can
again be used as a
as
.
primitive
construction
is
Whether
root.
method
it then
the
on
something greatly
to
basis
is worthwhile
to
of the
quadratic
be doubted.
obtained
derive
However, such
.
..
explicit
an
equations
was
actually
done
acknowledgment of the construction
lar pentagon,
known
since
antiquity, we shall derive
tion
algebraically here.
Starting with the fifth root
of the
regu-
in the
7.3
nineteenth
As
7If 7n
the Euclidean
Since
century.9
of
sort
a
and
relatively
prime, then
algorithm—two integers a and
n
b that
211‘
27r
27r
m
n
nm’
a-—+b~-—=-the
division
of the
circle
into
parts
mn
construc-
of
unity C
=
exist—as
can
be calculated
using
an
satisfy the equation
+ bm = 1.
there
are
the
can
be
the
problem
done
given
the
divisions
into
and
m
71
parts.
'3Today it is hardly imaginable
for
to
the
write
ALGOL
down
on
cyclotomic
his
60.
first
equation
computer
of
that
degree
program
257 was
the
and
to learn
Since
there
was
no
direct
access
to
and
paper
for input.
given to someone
of the periods
that
in the sum
were
appear
a
of
motivation
computer
the
computer,
a
the period
for the author
in
language,
computing
program
was
products
in
this
1975
case
written
at the
first run
the
Indeed,
desired
indices
This
was
computed
much
correctly.
more
intersting than the usual oral exam
for a high-school
diploma.
9F. J. Richelot, De resolutione
$257
algebraica
aequationis
1, sive de divisione
circuli
bisectionam
per
in partes
anguli septies
257
repetitam
inter
se
comaequales
mentatio
Journal
coronata,
and
filr die Reine
Angewandte
IX (1832),
Mathematik,
pp.
12-26, 146-161,
337-356.
Christian
209-230,
Gottlieb, The simple and straightforward
construction
of the
The
regular
Mathematical
257-gon,
Intelligencer,
21/1
31-37.
Johann
Gustav
Hermes
(1999), pp.
(1846-1912), of Lingen, is said, as reported
by Felix Klein
(Vartrblgc iiber aasgewdhlte
Fragen der Elementargeometrie,
Leipzig,
over
1895, p. 13), to have derived
the course
of ten
a
construction
years
method
for
the regular
of this work
65537—gon.An overview
of over
two
hundred
pages,
completed
in 1889
and
at the
deposited
is given by J.
University of Gottingen,
Ueber
Hermes,
die Teilung
des Kreises
in 65537
gleiche Teile, Nachvvlchten
'uon
der
der
Gesellschaft
zu,
Wissenschaften
Géittingen,
Math.-Phys.
Klasse, 3 (1894), pp. 170-186.
Three
phoof
the
work
can
be found
tographs
in Hans—Wolfgang
Henn, Elementare
Geometric
und
Algebra, Wiesbaden,
2003, pp. 33-34.
=
7.
74
cos
On
+ isin
@\xD0\
The
construct
\xE0\x9A`
the
we
of 770711
-1
from the quadratic
account
periods
Construction
=
770
:
711
=
and
710 + 711
1
—
=
1
003%
§Rz denotes
construction
unity other
-1,
one
1
1
obtains
these
two
0.
to
=39C=
the
=
equation
2
Where
periods
two
C2+ C35
312
leads
Polygons
Regular
(1 ‘l’ (4:
+ y
This
of
than
part of the complex number
the
real
can
immediately
1 are
5770 -1-1'?/5:
=
be
made.
The
four
z, from
which
fifth roots
of
follows:
as
1
1
.1
-Z+Z\/5+2?/10+2\/5,
1
—Z
1
—
1
Ex/'5+zZ\/10~2x/5,
1
1
.1
1
1
,1
10—2\/5,
‘“Z—Z\/g—7.Z
—Z+Zx/5—zZ
The
three
The
Classical
famous
problems
Construction
of classical
10+2\/5.
Problems
unsolved
antiquity that remained
into modern
are
times
the squaring of the circle, the doubling of the cube,
and the trisection
of an angle using only straightedge and compass.
The problem of squaring the circle asks for the construction
of a square
whose
area
is equal to that
of a given circle.
On the assumption
that the
circle has area
to constructing
a segment
of length
1, the problem amounts
roots
and squares
of lengths, the
square
\/7_r. Since one can construct
of a segment
of length 7r. Thus
problem is equivalent to the construction
the algebraic equivalent of the problem is to represent
of nested
71' in terms
numbers, and the four basic arithmetic operations. It
square
roots, rational
Lindemann
was
shown
in 1882 that
the number
by Ferdinand
(1852—1939)
71' is
that
transcendent,
is, that it satisfies no polynomial equation with
7.
The
Construction
of
Regular
Polygons
coeflicients. Therefore, 7r cannot
and hence is not constructiblem
rational
roots
The
volume
of the
doubling
is twice
that
be represented
in terms
that
cube,
of the
75
is, the construction
to the
cube, amounts
unit
of
of square
cube
a
With
the methods
length \xE0B<
of Galois
theory one
relatively easily that xi/§cannot
be expressed in terms
of rational
and
nested
The
square
roots.
We
of the
problem
resolution
shall
return
to
this
of
construction
of
segment
whose
can
topic in Chapter
a
prove
numbers
10.
of
is of a similar
angle trisection
nature.
In dealing with the casus
irreducibilis
in Chapter 2 we saw
the close relationship between angle trisection
and cubic
In general, one
equations.
has
the
identity
3
cos3§—Zcos§—Zcos1/2:0.
with
The
relationship
when one
brings
the
the
problem
of doubling the
sine
function
into
the
cube
is seen
more
exist,
it suffices to
clearly
with
picture,
3
7/)
(cos? 0\x94+isin
To
show
show
that
that
a
general
a
number
The
gebraic
algorithm
measure
cannot
full
circle
does
not
be constructed.
Of course,
one
of 360
and
angles.
otherwise, the regular
be
of other
trisected, since
straightedge and
7.4
the
cosx/2 +z'sin1/2.
=
trisection
single angle
easily trisect
can
a
1
2/2
special
compass.
cyclotomic
properties
even
We
equation
shall
degrees, as well as a right angle
However, the angle of 120° cannot
see
:3“
for those
would
nonagon
—
more
1
Values
=
this
on
0 has
of
be constructible
topic in Chapter
10.
interesting
al-
many
for which
n
with
the
regular
n—gon is not constructible.
Gauss himself
recognized, as presented,
along with many other results, in his 1801 work Disquistiones arithmeticae, that all cyclotomic equations are solvable
in radicals.
By
this
simply a “solution”of the form :0
{/T, since such
a symbol
allows a number
of algebraic interpretations.
That
is, the
symbol offers interpretations
that differ greatly as regards the four basic arithmetic
For example, the expression
operations.
{‘/T
comprises
the four complex numbers
of which only 2' and -2’ a.re in1, -1, 2',——z',
distinguishable on the basis of their algebraic properties alone. Thus
1 is uniquely defined as the
multiplicative identity of the complex
is not
meant
10A relatively
well
Paris,
illustrated
book
1997, Chapter
=
discussion
can
be found
elementary
by Jean—Paul Delahaye, Le fascinant
9.
in
the
nombre
very
informative
P72, Editions
and
Belin,
are
——z'
1 and
characterized
One
could
\xB0:8
~—%
has
$2 + 1
Q
0.
=
3 also
+
=
—1 is the
while
numbers,
-2’,this multiplicity
Construction
of
Regular
unique additive
only as the two
inverse
of 1.
The
7.
76
also
than
more
relates
that
argue
one
solutions
Polygons
In
contrast,
of the
equation
form
expression of the
As with
interpretation.
an
12and
that
possess
the identical
symbol
{/5 can
be
only to numbers
algebraic properties.
Thus
one
therefore
as
a
form
That
solvable
(1
~
equation
of an
equation
the
solution
(1,
—
un-
0 is irreducible
=
algebraic properties.
in radicals
can
be interpreted
of irreducible
equations of the
0.
$7
After
1
:39
0 and
formula
solution
the
detail, with the substitutions
respectively 124,we have
for
a
into
In
=
(:1: 1), one
—
substitution
4) via the
(to 3 or
—
factor
linear
the
removing
exponent
use
used
identical
all possess
to
93"
cyclotomic polynomials
can
be shown relatively easily by a method
by radicals
remaining
can
=
the
the
de Moivre.
one
solution
stepwise reduction
as"
the
solutions
its
the
Therefore,
that
when
only
problematically
and
conclude
may
cubic
the
or
—
can
1
0
=
due
halve
are
to
the
a:+a:‘1.Then
biquadratic equation.
y
equations
=
by $3,
divided
m3+a32+:c+1+:r"1+a:_2+a:"3=0
and
:0.
+:v'2+a:"3+:z:“4
:1:4+m3+:n2+m+1+cc”1
One
makes
the
substitutions
:132+:1:_2=y2 -2,
:L'3+:1:“3=y3—3y,
:r4+x‘4=y4—4(y2—2)—6=y4—4g/2+2.
After
the
unknown
y is determined
degree equation, the
quadratic equation
desired
from
unknown
the
resulting
:12 can
932—ym+1=0.
be
third— or
obtained
fourth-
from
the
7.
The
7.5
Construction
In
1:11
the
1
case
0, the
degree equation
—~
=
of
Regular
Polygons
of the
cyclotomic
equation
process
we
have
77
of
degree 11, namely
just described
leads
the
to
fifth-
1/5+y4~4y3~3y2+3y+1=0,
whose
five solutions
That
this
equation,
be solved
11, can
are
given by yj
and
therefore
in radicals
was
=
the
\xF0śj
1,2,3,4,5.
cyclotomic equation of degree
2cos
discovered
for
before
sur
ThéophileVandermonde
(Mémoire
Like Lagrange, Vandermonde
équations).
attempted
Alexandre
lution
to
method
for the
up to the
Gauss, in 1771, by
la résolution des
to
study the sodegree in order
resolvent, now
fourth
degree five. To this end, he used the
Lagrange (see Chapter 5). Although Vandermonde
it to
generalize
named
general equation
=
for
find a general solution
formula—as of course
nized
that
in “special
cases
in which
there are
to
failed
to—he recog—
he had
equations among the
to solve the given equations
roots,”his “method could serve
with—
out having to use
the general solution
formulas.” In the equations
mentioned
that exist among
by Vandermonde
the solutions, there are
identities
for the periods such as
y%=2/2+2,
y§=y5+2,
y%=y4+2,
ZJ§=y1+2,11192:?/1+3/3,
which
with
entire
Vandermonde,
structure
Z/13/3=2/2+2/4,
without
by which
the
yla
3/3:
are
sorted
3/4»
3/5)
exactly corresponding to the general method
thirty years later using 2 as a primitive root
order, for which we shall use the notation
nk
and
y2y3=2/1+3/5,
explicit reference, established
solutions
1/29
yi=y3+2,
into
the
order
discovered
modulo
the
by Gauss
11.
In
this
C-2k
+C—2k,
(€216)
B\x92
:
:
so
77o=y1a
Vandermonde
resolvent,
and
was
indeed
771:-3J2»
able
712:2/4,
773:1/3,
to
determine
the
in the
form
sum
of a
fifth power
774:1/5,
of the
Lagrange
of integer multiples of fifth
In
of unity.
roots
particular,
z(6)
defined for
a
obtains
one
yl
and,
of unity
the
hand,
one
6
as
Lagrange
Polygons
+ 64774
+ 63773
62772
cos
:
sin
+1’
2-733
2-751
for k:
in
discussed
already
1, 2, 3, 4,
Chapter 5,
%<—1+
=
last
these
=
two
(65
11
+
to
note
that
.
finallyobtains,
one
root
with
two—memberperiod y1
identity given
z(e)5 can
for
the
help
fifth roots
for the
representation
for the
the
calculation,
complicated
a
4152 + 1663 + 2664)
equations
already found square
unity, a root representation
remains
\5/z(e4)5
hand, though only after
other
of the
It
resolvent
no + 6771 +
=
z(e)5
From
for the
Regular
{‘/,@+€/z(e2)5+
Q/z(e3)5+
=
the
on
of
fifth root
on
=no
Construction
The
7.
78
\xB0
2 cos
=
be
of
derived
3125 summands
by
completely elementary way from the 55
sorting, grouping, and simplifying on the basis of the period identity
discovered
+ 774
by Vandermonde
-1).
(together with 770 + 771 +
Independently of the actual values of the result, one can see relatively
easily that such a result can be found in the form of a sum of rational
Why
multiples of fifth roots of unity. And this could be the reason
in
=
a
-
First
riod
the
sorted
Vandermonde
of
identities
all, the
solutions
sorting
discovered
has
in the
the
effect
by Vandermonde
way
=
-
described.
that
valid
remains
of the
pe-
when
each
one
every
period 77;, is replaced by nk+1 (taking into account
numbering of the periods is defined so that they
that
is, 715 770, 775 771, etc.). Thus the identity
=
-
the
progress
fact
the
that
cyclically,
=
4
\xF0.
4
4
‘
'
j=o
which
riod
obviously
identities
period
nk is
can
with
for
be derived
any
integers
am
of the
z(e)5 via simplification
and
bj, is Valid
replaced by nk+1:
4
4
‘
j=0
3:0
lc=0
j=0
also
when
pe—
every
7.
The
Construction
If the
then
indices
altogether
of
of the
Regular
periods
Polygons
shifted
are
79
well
as
by 2, 3, and
4,
obtains
one
4
4
5
.
(26%)
5
.
+
4
.
+( \xA0\x810
-
3‘:
i=0
4
4
4
'l‘77k+4)
Z::CLj,k€j(77k
+5Zbj€j
I
=
+
-
-
-
j=0k=O
j=0
4
4
Z X\xFA
ZCLJVV)
=
E]
—
9:0
Here
to
equal
k=0
fifth summand
every
the
on
left
side
of the
last
is
equation
z(e)5, since, for example,
4
4
‘-1
€j’7J’+1
Z
512(6)Z €j+177j+1
:
=
J'=0
The
of
roots
left side
of the
previous
Vandermonde’s result
that
so
i=0
is obvious
unity
in
is therefore
equation
of
a
of rational
sum
hindsight:
4
Z(€)5Z
.
(bj E ZCLj,k)EJ.
—
j=O
k=0
Without
going into
details,
concrete
calculation
of
evaluation
4
1
=
the
that
note
we
z(e)5 can
Although what
there
are
in which
ways
greatly simplifiedover
be
summands.“
of 3125
equal to 5z(e)5,
multiples of fifth
the
to the cycloapplies specifically
1
equaton $11
0, it is not implausible that these consideraapply to every cyclotomic equation of prime degree. The reasons
tomic
have presented
we
=
~
tions
“See Paul Bachmann,
Die
Lehre
van
der Kreistheilung
zur
Zahlentheorie,
Leipzig, 1872 (reprint 1988), pp. 75-98.
For
this
e1eventh—degree
cyclotomic
which
works
equation,
method,
product
und
ihre
Beziehungen
the
special
in general,
case
of the
leads
to
a
representation
2(6)
where
each
of
the last
unity;
similarly
the
fact
roots
of
'
—
W
‘
W
W
the
four
factors
represents
is in fact
an
Here
integer.
calculated
that
each
of the
unity
be
can
in
the
general
four
shown
'
factors
in
the
case,
a
the
as
four
way
integer
as
multiples
in
products
done
was
corresponds
same
of
sum
(Z(€)z(€4)),
to
was
for
a
sum
done
the
period
of
fifth roots
numerators
can
of
be
products.
Moreover,
of integer
of fifth
multiples
for
z(e)5.
that
the
such is indeed
case
be found
can
Which
give general relationships
these
identities
valid
remain
of
Construction
The
7.
80
the
root
Polygons
in Vandermonde’sequations,
periods, and the fact that
the periods are
permuted among
sin
cos
+7}
of unity C
\xD0y\xEB`\x9B\xEA
the
between
when
Regular
replacing
This situation
n.
modulo
root
may
a primitive
denotes
where
g
by C9
of the genIn comparison to the solutions
follows:
as
be characterized
of periods
eral equation of a given degree, the algebraic calculation
since they satisfy additional
equations. This allows, in
is “simpler,”
sim~
of
construction
the
to
general procedure,
Lagrange’s
comparison
under
unaltered
remain
do
not
they
pler resolvents, which, although
of the periods, do so under
permutation arising
every
a permutation
here to
suffices
invariance
»—>
limited
This
from the substitution
(9k.
Q
of
a representation
the
for
for
possibility
example
z(e)5,
demonstrate,
of unity.
of fifth roots
the the sum
as
themselves,
in
as
2
,
from
and
equations
For the
cyclotomic
we
what
construction
a
procedure,
that
general
eventually deriving
suffice.
far should
so
have learned
However, only a general, up to now
be
would
in
completely satisfying
principle
outline,
recognizable only
theGalois
what
is
That
of
view.
from a mathematical
precisely
point
additional
some
have
concepts,
we
after
offer
will
developed
us,
ory
homore
much
be
will
whose
an
argumentation
explanation
namely
of
the
case
many
in
the
equations,
cyclotomic
Moreover,
mogeneous.
of the complicated summation
expressions for periods will become
task
specific
of solving
unnecessary.
Exercises
seventeenth
all the
(1) Express
roots
of unity
in
of square
terms
roots.
(2)
Generalize
the
calculation
for the
case
n
=
17 of the
product
'P(n~1)/2 (C9)
P(n—1)/2(C)
to
the
necessary
of
case
case
a
general prime
distinction
be most
2 3. How
simply characterized?
number
n
can
the
Chapter 8
The
Solution
of the
Equations
Degree
We seek
8.1
of
the
solution
Fifth
of the equation
935
2625:v+61500.
=
This
chapter closely follows a talk given in 1977 by the author
at the Philips Contest
for Young Scientists
and Inventors,
“Special
equations of the fifth degree that are solvable in radicals.” The equa—
tion presented
above is again a classical
example. Already in 1762,
Leonhard
Euler recognized from his studies
of solvability of equations
that this equation belongs to a class of fifth—degree
equations that can
be solved
in radicals.
Like other
mathematicians
of his time, Euler
had attempted
to extend
the methods
for equations
of degree less
than five to those of fifth degree. Even the mountain
of formulas
that
resulted
could not dampen Euler’s optimism, for he wrote,
One
correct
conjecture
approach to
finally
arrive
sult
may
were
an
viously
used
tions] would
would
seem
with
apparent
this
elimination
certainty
that
procedure,
at an
equation of fourth degree.
equation of higher degree, then
.
intermediate
value
itself
roots
to
contain
be unreasonable.
with
would
one
If the
..
for representing
of this degree,
the
re-
[the prethe
solu-
and
that
82
Solution
The
8.
actual
his
in
However,
of
of the
Equations
calculations,
Euler
Fifth
had
Degree
trim
his
makes
this
to
sails
somewhat:
task
so
success,
the
since
However,
of expressions
large number
difiicult
that
it
appropriate to develop some
to such complex formulas.1
seems
one
that
do not
lead
Euler
refers
to
shorten
the
calculations.” In
equations
that
can
the
achieve
cannot
intermediate
results
he
of
measure
any
special
used
cases
“such values
as
Euler
has
avoided
Since
this
class
merely
calculational
difficulties, but the basic impossibility of a general solu—
tion.
Nonetheless, in this way he arrives at a large class of fifth-degree
as
be solved
reality,
in radicals.
not
does
con—
not
fifth—degree
equations, we will look here at the work
time
as
In 1771, thus at almost
the same
mathematician.
of another
the Italian
mathematician
the work of Lagrange and Vandermonde,
Francesco
Malfatti
Giovanni
(1731-1807) was searching for a general
formula
for equations of the fifth degree. Malfatti, who later, in 1804,
at an
commented
unsolvability
critically on Ruifini’s first attempts
to refine
Ruffini
proof based on his own work and thereby motivated
in carrying out extremely complicated
calculahis work, succeeded
tions
of a resolvent
of the sixth
degree. This did not lead to the
that
noticed
However, Malfatti
original goal of a general solution.
a rain the special case
in which
the sixth—degree
resolvent
possesses
tional
solution, the given fifth~degree
equation can be solved. Later,
had characterusing Galois theory, it could be shown that Malfatti
in radicals
ized all equations of the fifth degree that are solvable
(in
to all irreducible
relation
fifth-degree
polynomials over the rational
tain
all solvable
numbers).
Malfatti’s computations
worth
noting that
Euler
had
method
of
not
he continued
been
attack,
are
able
we
will
to
complicated,
successfully from
very
progress.2 To get some
consider
his
calculation,
and
it is very
much
the
point
at
which
idea
of Malfatti’s
beginning
with
the
aller
in:
Leonhard
1Von der Auflifisungder Gleichungen
Euler,
Grade, reprinted
Nr. 226,
Drei
fiber die Aufliisung dar
Ostwalds
Klassiker
Abhandlungen
Gleichungen,
This
and the one
on
Leipzig, 1928.
quotation
following appear
page
45; the equation
in the epigraph
on
50.
page
appears
der Gleichung
2See J. Pierpont,
Zur Geschichte
V. Grades
(bis 1858), Monatshefte
Ma1fatti’s
und
at a solution
fiir Mathematik
Physik, 6 (1895), pp. 15-68.
attempts
36.
are
described
on
33 through
pages
8.
The
Solution
of
Equations
of the
Fifth
Degree
83
equation
$5 + 5a:1:3 + 5bm2 + 5cm + d
only for
the
case
a
b
=
0, that
2
0,
=
is, for equations
of the
type
:1:5+5ca:+d=0.
Furthermore,
does
not
restrict
fact, every
of this
will
we
the
equation
section
the
on
that
eliminates
transformations
further
note
generality as much as it seems
of degree five can be transformed
type using a substitution
See the
that
Ma1fatti’s calculations
that
generality,
the
53,41
for j
begin with
solutions
into
2:
In
equation
an
term.
degree—four
the
of Tschirnhaus
and
of
Bring
the
without
loss
assumption,
in the
represented
are
this
first glance.
at
Jerrard.3
and
of
yé0. We should
cd
assume
form
(ejm+ e2jp+ e3jq+ 64in),
—
O,1,2,3,4 and with 5
+ isin \x80\x94?
This correcos
sponds precisely to the method
employed already by Bézout,
Euler,
Lagrange,4 and Vandermonde.
If one
multiplies the five associated
linear factors together, then one
obtains, along with Euler, the equa~
2
=
tion
:35
—
5(mn + pq)a:3+ 5 (m2p+ n2q + mp2 + nq2)$2
—
n3q + mq3 + 71193 mzng + mnpq
(771319
5
+
—
—
p2q2)
as
+m5+n5+p5+q5+(mn—pq)
(mp2+nq2——m2q—n2
Finally,
the
determine
the
the
unknowns
original equation.
following shorthand:
3/
=
P9
11
+ 7539»
W317
=
4Since
are
771
resolvents
not
=
for
nzp
=
—
(mp2+nq2),
mq3 + np3.
=
transformed
—
however,
into
it
is unfortunate
of the same
equations
+ 64.732
+ e3a:3 + 52934+
(£121
51:5)/5,
the
m,n,p,q
by
We will employ
—m77»»
m2q +
3For specific applications,
coeflicients
-‘=
7“ =
w
grange
to
try
coefficientswith
the
comparing
must
one
values
m5,p5, (15,17,5.
that
equations
type.
etc., at issue
with
here
rational
are
Lan-
84
The
y and
quantities
for the
7'
:32. For the
:33 and
powers
ficients gives the
of
pair
other
Degree
coefiicients
of comparing
result
the
Fifth
definition of the
the
with
together
contain
already
of the
Equations
mentioned
identities
two
of
Solution
The
8.
powers,
comparing
last-introduced
identity
two
coef-
equations
c=—v—w+3y2,
d=m5+n5+p5+q5+20ry.
To be able
in terms
pletely
as
of 7', 1), w, y,
we
=
M2
=
=
=
rw
=
=
thereby obtaining
—
—mn
—
-
=
=
:
the
pair of equations
for the
of the
possible only if two
relations
—
dy
A calculation
the
use
com-
(m2q+ n2p) (mgp-1- n3q)
pq (m5—|—
715)+ (mn)2 (mp2+ nq2)
(m5+ n5)y we
(mp2+ nq2)(mq3+ np3)
+ €15)
(pq)2(mgq+ nzp)
(195
74
W2,
(:05+ <15)
c
mu
well the
formulate
to
form
new
—(12
+112)+ 33/2,
=
=
r(11+ w) + 227"y2.
four
additional
unknown
quantities
identities
are
r,v,w,y
taken
into
will
account:
(m3p+ n3q)(mq3+ 71193)
+ 714192)
+ mn
pq (m4q2
(m2p4+ n2q4)
pg
+
(mgq+ n2p)2
W2 +
=
(~y)(-N2
~
(mp2+ nq2)2 4m2n2p2q2
mn
~
41/‘ -41/4
=
and
-73
=
=
T4
('u
—
=
mn
these
Putting
=
(mp2+ nqz)
(mgq+ 7221))
pq (m3p+ naq) +
(mq3—|—
np3) (21 w)y.
identities
two
w)2y2
=
(1)+
w)2
together,
2
—
obtain
we
4vwy2
—
:=
163/6.
(2)+ w)2y2—|—
be
8.
The
Solution
of
Now, using
Equations
this
and
equation
obtained
from
comparing
7", 11, w, y.
First,
we
that
the
eliminate
’U +
T4
To eliminate
first of these
into
the
the
two
result
least
c
0.
=
we
an
From
y
y
7é0, we
z
shall
253/6 6cy4+ c2y2.
=
—
7“ as
and
well, we take
then
the fourth
substitute
the
second
equation
(z3
in radicals.
m, n, p, q would
would
can
now
also
be equal to zero, resulting in
253/2 c 7E0, since otherwise
have
——
multiply
so
that
253,12,
sixth
—
a
the
last
bicubic
equation
resolvent
by 253/"2. We
results, that is,
degree:
c)4 (:52 (302 + 2502) d4z.
—
it is sometimes
=
useful
to
use
the
bicubic
resolvent
form
equivalent
course,
=
we
=
see,
—
solved
of the
remain:
(253/2 c)4(252/4 603/2+ c2)yz.
(2
Of
power
—
0.
of the
equation
in the
values
——
values
=
substitute
we
the
—
Furthermore,
have
As
determine
3312 a,
=
of the
of the
must
then
previously
0 helps us in what follows
special case cd
complications: First, we have 3; 7E0, since otherwise, at
some
three
equations
obtain
to
exclusion
avoid
of
pair
—
equations
=
to
85
(253/2 C)7°,
=
variable
d4y4
Our
Degree
via
w
w
following equations
dy
the
Fifth
we
coefficients,
may
12 +
so
of the
5cz2 + 15c2z +
in its
general
If it were,
5c3)2 (d4+ 25605)
=
z.
form, the bicubic resolvent
cannot
be
then beginning with the variable
2, the
values
Solution
of
n, p, q could
then
The
8.
86
y, 7*, 12, w, m,
of the
Equations
Fifth
Degree
in turn:
be calculated
1
y:g\/2»
dy
7‘:
253,/2-0’
_3y3—cy—r2
_
,
2y
_3y3——cy+7“2
_
,
2y
s'u+y2
mm:
Paq
v+y2
< 7")
rd:
=
+y,
2y
2y
5
2
w—1—y2
7‘ i
w+y2
(
2y
5
2
5.
C’!
—
7"
2y
idenderived
the
from
previously
directly
equation
tities, in the case of the last two equations with the help of Viete’s
arbe chosen
the sign of the unknown
Note that
theorem.
root
y can
and
the
the
since
pairs
q)
exchanges
(p,
merely
sign
changing
bitrarily,
n
of the variables
that
the
note
p,
q,
m,
ordering
Furthermore,
(m,
+ n3q is satisfied.
such that the equation 1)
is always taken
77131)
Each
almost
comes
=
radicals.
resolvent
bicubic
equation
d
bicubic
the
in the
can
be found.
this
to
epigraph
Here
we
chapter
shall
with
take
the
as
he
that
resolvent
recognized
can
be used to solve special equations of the
In particular, this is possible when a rational
obtained
and
that
himself
Malfatti
8.2
fifth degree in
to
the
example
the
solution
an
coefficients
c
=
—525
—61500.
=
Since
the
bicubic
resolvent
is
a
monic
polynomial
with
integer coChapter 6, must
in
all rational
solutions, as demonstrated
efiicients,
infor~
additional
2506. One obtains
be integers dividing the number
Since
resolvent:
of the bicubic
from the second representation
mation
must
solution
rational
37809000002 is a square,
d4 + 256c5
every
And finally,division
of an integer.
be the square
by 56 shows that
=
is
also
§~
that
z
a
solution
is divisible
of
by 5.
that
is,
equation with integer coefficients,
of possible integer
Having limited the number
an
8.
The
Solution
solutions
to
that
y
out
=
of
of the
Fifth
the
solution
2
1)
—150,w
Equations
obtains
112, one
15, 7"
-150,
=
=
Degree
87
5625.
=
It then
turns
1350, and finally,for
=
j:0a172>3>41
cc,-+1
5
ej
=
(5 4x/E)
75
+
Malfatti’s attempt
8.3
classical
tutions
tradition,
and
a
namely,
hindsight,
extent
that
the
fied if one
the
relevant
.
.
,:v5.
.
1
using
we
from
2
:c2+e
(a31+e
192-3
4
the
substisuccess
of
possible, is clarivalues
as
polynomials
success
the
in the finest
suitable
that
see
was
intermediate
Thus
—
methodology
a
equations
In
to
:31,
shows
solve
to
transformations.
solutions
~
541’
,5/75
+
solution
Malfatti’s approach,
in the
(35 11x/16)
225
+
at
expresses
623'
(35 11x/1’0)(5 4m).
639‘
H225
+
+
5
identities
two
:v3+ea:4+e
5
:35)
and
1
3
4
:1c2+e:r3+e
(:v1+e
q=———5
one
m4+e
2
$5)
obtains
5
253/
=
25pq
=
p:\xFF (62+ 53)(mlacg+ 582.133 + :1:3:c4 + $4335 + 935231)
+
j=1
+
In the
special
(6+ 64)($1933+ 132934 + $3935 + 5841131 + 115932)considered
case
here, a
=
b
5
Z-or
y'=1
=
Z
1SJ'
W
=0
=
0, since
we
have
Solution
The
8.
88
of
—\/5,we obtain
particularly simple representation5
and
—e
the
+ e2 + 63
Z’
54
—~
1
2
~
this
with
Furthermore,
of Vandermonde,
resolvent
5A
I $2033
of
of
bicubic
the
—
relation
resolvent
based
the
between
in
the
of the
solution
rational
a
that
is clear
it
a
as
z
2
representation
interpreted
derivation
solution
resolvent
1 1133534 I £349351 005561)
,
existence
the
be
can
Degree
5($1flU2
_
253/
—
for the
=
Fifth
of the
Equations
sense
bicubic
solutions.
approach
Lagrange’suniversal
der
die
auflosbaren
Gleichungen
on
Uber
Runge,
also
Heinrich
see
7 (1885), pp.
173-186;
Form
1:5 + um + 11
Mathematica,
0, Acta
One
first
670—676:
Volume
der
1898,
pp.
Braunschweig,
I,
Algebra,
Weber, Lehxrbuch
altered
of the slightly
behavior
the
representation
polynomial
investigates
found
be
5) can
(see Chapter
C.
in
=
«/5
:
: 23531
$4025
5(w1av2+w2ws+ws:c4
y=
the
under
120
permutations
that
is, they
square
root
possible
permutations
the
polynomial
leave
of
unchanged.
of sixty
the collection
discriminant:
to
belong
of the
five
the
\/5:
H($i
$4921
903025
202024
$1903
solutions
:01,
are
All of these
that
permutations
.
.
Ten
.,ar5.
$5922)
of
these
permutations;
the
unchanged
even
leave
ma‘)-
"
i<.7
effect
whose
are
ten
odd permutations
there
Furthermore,
transform
the sixty even
permutations
change its sign. Thus
and
the
different
sixty odd
ys,
3;, yz,
y1
polynomials
six polynomials,
an
additional
into
—y1,.
namely y7
solutions
of the sixth—degree
are
thus
equation
polynomials
:
.
.
.
,
=
..
polynomial y is to
six
y into
polynomial
the
on
the
permutations
—y5.
=
,y12
transform
The
first
six
in
the
y
Z/6+>\5y5+'~+A1y+)\o=0»
Whose
coefiicients
A0,
ya.
yl,
polynomials
:35 + 5ca; + d
equation
.
.
.
.
,
A5 arise
To
obtain
.
.
,
from
these
the
coeflicients
polynomials
symmetric
elementary
in terms
of
and
c
d of the
original
in terms
of the
expressed
are
solutions
only “almost” symmetric;
2:1,
y1,.
,y5) are symmetric,
namely, the
and
altered
of odd
those
while
by a sign change for odd permutations
degree are
on
theorem
symmetric
are
Using the fundamental
permutations.
unchanged
by even
in the
of c,d, \/5,/\o,
the
and
functions
.,/\5 as polynomials
degrees
considering
rational
must
exist
variables
23' for Aj), there
1:5
:21,
(namely 4, 5, 10, and 12
numbers
#4 satisfying
[.t(),[,l.1,]J.2,
y1,.
,y5
0, the polynomials
the resulting
polynomials
LE5. However,
of even
degree (in the variables
polynomials
.
.
.
=
.
are
.
,
.
,
.
.
—
.
.
.
,
+ #063
:16+ #463/4+ #2-22212
=
I41\/BM
obthe
equation
finally obtains, after squaring
in the
main
route
derived
bicubic
resolvent
text;
by a different
tained, the form
be representable
D must
that
the discriminant
one
determines
here
\/D by observing
oz and
ozcs +[3d4 with two constants
20 of the form
of degree
a symmetric
as
polynomial
One
be found
can
constants
finally obtains
the
equations.
using particular
,6, where
D
55 (256125
+ or‘).
After
=
determining
the
of the
constants,
one
The
8.
Solution
The
of
of
Equations
Transformations
of
of
first systematic
attempt
five
undertaken
degree
was
Fifth
Degree
Tschirnhaus
and
The
the
and
89
of
Bring
Jerrard
at
in
solution
general
a
method
for
equations
1683
by Ehrenfried Walther, Count of
Tschirnhaus
(165l~1708). Tschirnhaus’s idea is based on the hope that one
could generalize the well-known
substitutions
that cause
the second—highest
coefficient
to disappear
so
that
additional
coefficients would
disappear as
well.
Instead
of
transforming
—1
a:"+an_1:I:"
given equation
a
-2
+an_2w"
+-~+a1m+ao=0
using the substitution
into
an
of reduced
equation
form
y" -I-bn—2y"‘2
+"'+b1?J
Tschirnhaus
began
his
investigations
2;
with
the
yj
are
1) and
parameters
original
=
both
equation
=
0,
substitution
a
of the
form
9:2 + pa: + q
=
be determined.
q to
are
with
+ b0
transformed
The
into
the
77, solutions
n
+ pasj + q, where
the coefficients of the
zero
precisely when the two conditions
:31,
solutions
y1,.
.
.
.
of
,a:,,
with
,y,,
y"_1 and y"”2
of
powers
.
.
Z?Jj=Zy§=0
are
satisfied. If one
starts
with
reduced
a
in which
equation
the
second-highest power is already 0, then
19 and q the following conditions
that
must
one
obtains
the
coefficient
for the
of
parameters
be satisfied:
0=Zyj =2:(cc,2~+pm,~+q)
+nq
=Zx§+PZmj
=Za:§+nq,
°=Zy§=Z(w?+zwj+q)2
Z33?-F2p::ac?+(p2+2q)Zw§+nq2.
=
The
first of the
mination
of the
for
the
q into
two
parameter
second
conditions
q.
condition,
If
immediately
one
then
then
one
permits
substitutes
obtains
quadratic equation (except in the special case
the third~highest
is already zero). Thus
power
the
for
the
a
unique
obtained
deter—
value
parameter
p‘ a
in which
the coefficient of
the so-called
Tschirnhaus
transfomnation
of
that
the
ized
such
Solution
The
8.
90
of
of
Equations
given nth-degree
resulting equation
the
Fifth
Degree
be
parameter-
to
zero
always
has coeflicients equal
equation
a
can
for
the
3/""2.
y"_1 and
powers
Tschirnhaus
believed
of higher degree,
using transformations
contain
more
allow further
to be chosen, would
which of course
parameters
of the equations,
so
that every
simplification
equation could be solvable in
his idea with
did not succeed
in supporting
radicals.
Although Tschirnhaus
of
concrete
calculations, it is nevertheless
possible to use a transformation
the
now
that
form
y=:n4+pm3+qm2+r:v+s
for his
special
of
case
a
fifth-degree
equation
91:5+ a4m4+ a3:r3 + a2a:2+ ala:
resulting
in
equation
an
of the
+ ao
0,
=
form
y5+b1y+bo=0The
parameters
equation. This
be
can
fact
was
determined
by solving
first discovered
quardratic
a
mathemati-
Swedish
by the
in 1786
and
cubic
a
the mathematical
Bring (1736—1798),
though without
note
of his achievement.
world taking proper
Only much later, in 1864, afthe transformation,
ter George Birch
J errard
(1804-1863) had rediscovered
recalled.
The transformation
is today generwere
Bring’sinvestigations
are
so
However, its details
ally called the B7"ing—Jerra7'd
transformation.
out.6
are
difficult to carry
complicated that the actual calculations
Erland
Samuel
Literature
on
R. Bruce
King, Behind
Samson
Breuer,7 Uber’die
cian
of the
Equations
the
Fifth
Degree
1996.
Quartic Equation, Boston,
trinomischen
auflésbaren
irredukttblen
1918.
Gleichungen fiinften Grades, Borna—Leipzig,
Sigeru Kabayashi, Hiroshi Nakagawa,
Japonica, 5 (1992), pp. 882-886.
6A description
Zur
Geschichte
der
of the
Bring~Jerrard
Gleichung V. Grades
Physik, 6 (1895), pp. 18-19.
7The sad fate of the victims
recall
here
the
of
1933 expulsion
Schultze, Mathematiker
auf der
transformation
Flucht
equation,
can
be
found
(bis 1858), Monatshefte
of racial
Samson
of
Resolution
and
Breuer
var
fiir
J. Pierpont,
Mathematik:
und
in
demands
See Reinhard
(1891—1978).
political
Hitler,
persecution
Braunschweig,
Math.
1998,
pp.
that
we
Siegmund
109, 292.
Exercises
91
Daniel
Lazard, Solving quintics in radicals, in: Olav Arnfinn Laudal,
Ragni Piene, The Legacy of Niels Henrik Abel, Berlin,
2004, pp. 207~
225.
Blair
K.
Spearman, Kenneth
quintics 3:5 +aa:+b,
Amemcan
S. Williams,
Characterization
Mathematical
Monthly, 101 (1994), pp.
986-992.
Blair
K.
Spearman, Kenneth
X 5 + aX2
aX
+ b and
26
(1996), pp.
+
S. Williams, On solvable
b, Rocky Mountain
Journal
753-772.
Exercises
(1) Solve
the
equation
:I:5+15m+12=0.
(2) Solve
the
equation
$5 + 330:7;
—
4170
=
of solvable
0.
quintics
X
5
+
of Mathematics,
Chapter 9
The
Galois
of
Group
an
Equation
How
can
tell whether
one
degree is solvable
9.1
The
previous
types
question
thus
if there
of special
equations
is
of the fifthor higher
a
natural
continuation
to
the
in radicals?
formulated
results:
equation
an
is
solution
no
solvable
are
in
general
radicals?
Galois
in
Galois, who
grew
up
in
the
our
equation, what
This question
French
by the twenty—year—old
mathematician
1832, shortly before his death in a duel.1
answered
was
of
lilvariste
post-Napoleon
restoration, appears
to have
studied
the solution
of equations
in radicals
entirely as a
self—taught
mathematician, although he did obtain a good education
for the time, first at the College
in Paris, and then
Louis-le—Grand,
at
twice
and
Ecole Préparatoire,
later
the
failed
he
was
the
entrance
examination
initially refused
early 1831 on account
the Republican Guard
the
entrance
of his
republican
later
resulted
Ecole Normale.
However, he
for the Ecole Polytechnique,
to the Ecole Préparatoire
in
agitation.
in several
His
membership in
months’ imprisonment.
1The dramatic
circumstances
of Galois’s discovery and the
duel
have
mysterious
frequently led to a romanticization
of his life. An example
of this
is the novel
Tom
by
The
French
Pertsinis,
1997.
Those
more
Mathematician,
interested
in the
cold facts
should
look at the article
The
by A. Rothmanx
short
life of Evariste Galois,
Scientific
American,
April 1982, pp. 112-120.
See also
T. Rothman:
Genius
and
biographers:
the fictionalization of Evariste
Math.
Galois, Amer.
Monthly 89 (1982), 84-106, or
the
Toti
biography by Laura
Rigatelli, Eh1am'ste Galois
1996
1811—1832,
Blrkhauser,
(Italian original, 1993).
93
9.
94
Galois’s attempts
to
Galois’s
begins
The
Galois
Group
of
an
Equation
publish his ideas failed due to the lack of
understanding
by the reviewers, resulting in part from the terse preThe first significant
sentation,
only much later recognized as correct.
after Galois’s death, on the
only fourteen
years
publication occurred
of Joseph Liouville
recommendation
(1809~1882).
thinking
with
the
then
current
state
of knowl-
or
less with the content
of the previous
edge, which corresponds more
to
have been awakened
chapter. From that point, his interest
may
ask to what
extent
relations
based
on
polynomial identities
among
the solutions
arise that reduce
the complexity of the equation in comcase.
Thus
in Chapter
we
have seen
7 how
parison to the normal
Gauss and Vandermonde
solved cyclotomic equations using such re—
in
lationships. And Lagrange, too, whose work has been described
this book relatively briefly,
investigated polynomial expressions in the
solutions
not
only for the case of the general equation, but also for
special equations.
Galois’s central
idea, for which there was no precedent at the
time, but which later proved to be extremely fruitful in application to
other problems in mathematics,
consists
in going beyond the current
of investigation
state
by studying a characteristic
object of greater
for each equation, Galois
associated
a mathsimplicity? Specifically,
ematical
object called a group in general, and for an equation the
Galois
group,
in honor
of Galois.
The
Galois
group
consists
of a sub-
of the
of the solutions
permutations
together with the operation of
as
described
in Chapter 5. The usefulcomposition of permutations,
ness
of this association
is that
it is possible to classify Galois
groups
in such a way that offers a classification of equations with respect to
their
solvability. In particular:
set
0
solvproperties of a given equation—irreducibility,
ability in radicals, and in the case of solvability the degree of the
be determined
reference
without
required root operations~can
to the equation
from properties
of the Galois
group.
All important
2A telling example from
a
different
area
of mathematics
is that
of knots.
For
an
see
Alexei
a
Mathematics
with
elementary
introduction,
Sossinsky, Knots:
Twist,
Harvard
The
Cambridge,
University
Press, 2004; Lee Neuwirth:
theory of knots, Sci240
can
be
entific Ameiican
(June, 1979), pp. 110-124.
Less—spectacular
examples
found
in
almost
every
mathematical
area.
9.
The
Galois
Group
Moreover, the
0
Equation
an
number
of different
95
Galois
groups
equations.
Thus
complete understanding of all the Galois
equations of low degree.
groups
than
the
number
As stated
9.2
of
ourselves
of
possible
in the
preface to this book, we
“modern point of view,”that
in the
is n1uch
smaller
one
gain
can
associated
are
define the
that
Galois
in
group
is, in Galois
theory,
century.3 However, we first shall
“elementary”
way using the terminology
an
have
we
with
going to interest
developed in the early twentieth
as
developed thus far. We shall follow Galois’s path in
broad outline, though without
going into detail.‘‘Furthermore, we are
not going to offer
complete proofs of anything, which would make little
sense
given the development thus far, which has been
primarily of a
motivational
nature.
Instead, we shall offer some concrete
examples.
In the next
chapter we will return
to fill in some
of the gaps in this
chapter’s
presentation.
In their
tions
on
for
general
a
Abel, and before
dealing apparently
dure
a
solution
formula.
Galois
designed for the general equation
special
To
conditions
of a particular
degree,
the root
operations,
places within
of
of the
analysis
this
equations
end, he
terms
of
the
collection
and
called
to
a
the
Ruflini,had focused
with
the most
striking
be
how this
applied
representation
known
quantity
of equa-
him
recognized
could
desired
solution
if it
proce—
to the
solution
of their
roots.
be represented
can
in
already known quantities using the four arithmetic
operations.
One begins by
considering the coefficients of the equation as
known, in analogy to the general equation, whose coeflicients correspond to the elementary symmetric
One can
polynomials.
enlarge
ticular,
but
Galois
called
ties
B.
of known
not
such
exclusively, the
adjoined quantities.
Melvin
quantities
numbers
by adding
of
roots
added
The
a
process
to
the
itself
certain
values,
already
known
collection
of known
he called
in par-
quantities.
quanti-
adjunction.
We
Kiernan, The development
of Galois
theory from Lagrange to Artin, Archive
for History of Exact
Sciences, 8 (1971/1972), pp. 40-154, as well as in an annotated
revision
of B. L. van
der
Waerden, Die Galois—Theorievon Heinrich
Weber
bis Emil
Archive
Artin,
for History of Exact
Sciences, 9 (1972), pp. 240~248.
4An extensively
commented
translation
of Galois’s original
can
be found
paper
in Harold
M. Edwards,
Galois
Theory, New York, 1984.
An overview
of the
history
can
be found
in Erhard
Scholz, Die Entstehung der Galois-Theorie, in: Erhard
Scholz
Geschichte
der Algebra,
(ed.),
Mannheim, 1990, pp. 365~398.
The
9.
96
note
that
bers
that
more
to
this
use
Galois’s concept
of known
of
Group
leads
quantities
to
A subset
9.1.
is closed
under
the
of the
four
numbers
complex
we
num~
have
shall
would
is called
like to
a
field
is, if the
sum,
difference, product, and quotient (aside from division
by zero)
of any two elements
(not necessarily distinct) of the field is again in
the field.5
The
smallest
arithmetic
of
sets
called
are
Equation
an
fields,a concept about which
say in the following chapter.
However, since we
notion, we give the following definition.
today
Definition
if it
Galois
of known
collection
operations,
that
that
results
quantities
from
an
coefficients
is the field of rational
numbers, deequation with rational
noted
of an equation,
by Q. Then building on this field,the solutions
such as, for example,
:c3—3av—4=O,
represent
Thus
stepwise
a
for the
enlargement
field of known
of the
quantities.
solution
ac1=\3/2+\/5+?/2—x/§
first adjoin
we
of known
collection
with
the
\/.3 to the
rational
quantities
\/§ the
of
adjunction
numbers.
We
resulting
from
thereby
the
obtain
rational
as
our
numbers
set
e(¢§)={a+w§|a,be@},
which
it is easily shown
that
is in fact
Q \xD0Mi
fieldof the
rational
to be closed
under
the
four
basic
operations,
so
field. One
a
numbers
speaks of this field as an ezvtension
obtained
by adjoining to Q the number
x/3
Now
as
to obtain
second
a
step,
the
solution
since
then
301, one
3/2_\/g:
5The
are
given
not
notion
subsets
here will
of
a.
of the
suflice.
field
needs
to
adjoin only \3/2 + x/3
1
32+«/3‘
to include
usually defined in greater
generality
numbers.
for our
the
complex
However,
purposes,
is
that
definition
sets
9.
The
is also
a
Galois
Group
of
known
quantity.
The
result
97
is the
field denoted
extension
by
0c\xB6\/5)-6
6/2 +
Q
9.3
Building on
ering an
complex
lutions.
The
CLn_1,
coefficients
is, we
.
.
.
,
.
that
,
and
cL1,a0
by consid-
group
all the
+ as
solutions
n
the
0
=
without
multiple
that
so-
distinct.7
are
extent
it is possi—
arbitrary intermediate
an
contains
beginning of the chapter
simply than
the
among
to
coefiicients
the
in the
solutions
general
231,
.
.
.
that
an
case
if there
a
$3
corresponds
the
to
equation
the
can
poly-
are
Every such
mn.
,
polynomial whose value is zero for
Thus, for example, the polynomial relation
,:rn.
.
.
+ aim
-
“known quantities”that
more
relations
corresponds
.
.
of the
the
at
be represented
.
an_1,
-
ap—
may
CL1, (L0.
We noted
nomial
of
we
nth—degree
equation
the
assume
quantities,
of Galois
equation, to
analyzed by associating with
field K
a
notion
-
process
ble, is now
step
central
an_1m”"1
+ an_29c”“2
+
That
solution
of known
terminology
equation, namely,
as”+
with
the
definition of the
the
preach
ml,
Equation
an
relation
arguments
3:2 + 2
=
polynomial
h(X1,.,.,X,,)=X12—X2——2,
6However,this
extension
field does
To get all three
solutions
into
start
with
the field
+
the case
considered
here of
equation.
not
contain
an
extension
the
other
two
solutions
field with
the second
of the
cubic
adjunction,
3
which
contains
Q(§1—
the cube
\xD0u\xA0
roots
of unity.
7For
complex coefiicients,
this condition
is not
a
really
since
restriction,
multiple linear factors
can
be eliminated
them
by
out
factoring
using
the
four basic
only
This
operations.
is done
with
the help of the Euclidean
algorithm,
described
later, by findingthe greatest
common
divisor
of the
given polynomial
and
its derivative.
Since
could
one
,
x
(<2—w1>J' m2)"W)’ (w ~a:1>J'“1~a:1>J'“1
z2>’°‘1
~
—w1>J'
s
\xC0'\x97
(j(.'1:—mg)(m—:n3)---+k(ar:—a21
><
the
greatest
divisor
common
of the
(ac
Through
obtains
no
dividing the
a
polynomial
multiple
mations
of
roots.
The
polynomials
original
that
—
Jan
polynomial
m1>J"1
polynomial
has the
first person
was
given
same
to
Hudde
—
m1>J"1
and
its
derivative
is
equal
to
m2>‘°*1
-
--.
by this
greatest
common
set of roots
as
the original
employ such considerations
later
the
(1628—1704),
divisor, one
polynomial,
mayor
in
thereby
but
with
transfor-
explicit
of Amsterdam.
here
lutions
and
letters
uppercase
field K,
underlying
each
follows
what
in
and
where
coeflicients in K that
Galois
The
9.
98
for
by BK the
denote
we
the
have
of
variables
the
Value
set
the
0 at
Equation
an
for the
letters
lowercase
use
we
of
Group
so—
polynomial. For
of polynomials with
a
arguments
ml,
.
.
.
,a3,,.
totality of all polynomials in BK is much too great
for a detailed
listing. Even a complete description is not a trivial task.
took a route
Galois himself
by which he used only a single polynomial
this polynomial
He constructed
for his purposes.
created
specifically
using what is now called the Galois resolvent, which is a special quanbe
can
the
solutions
all
in
terms
of
which
expressed
,a:,,
:31,
tity
We will go into this very
explicit
using the four basic operations.
the
on
in
the
section
in
not
computing
detail,
great
though
approach,
the
Of course,
.
Galois
group
resolvent
Galois
the
that
the
defined set
the
symmetric
solutions.
are
:33
the
three
the
to
4
—
0,
=
equation
BQ. However,
polynomials,
in the
reduction
X3 +X§ +X§ -6
X1X-2X3 -4,
+X3,
set
belong
nonsymmetric
a
3a:
»—
polynomials
X1 +X2
for
without
always polynomials that obviously belong to
BK. Such examples can most easily be found among
polynomials. For the earlier example
there
Of course,
sufiice to note
it will
characterized
sufliciently
be
can
the
explicitly calculating
.
chapter). Here
of this
end
(at the
.
since
complexity
is
what
really
only they reflect
of the equation.
of interest
that
relations
For
are
the
allow
Vandermonde’s
(see Chapter 7)
:r5+:r4—4:r3-—3a:2+3zv+1=0,
with
solutions
polynomials
my-+1
that
=
belong
Xf—X2~—2,
X1X2
~
X1
the
2cos
-
to
here
1,
2,
3,4,
0,
\xB04\x87
for j
=
are
some
BQ:
X§—X4—2,
X§-X3—2,
X4,
X2X3
of
polynomials
—
X2
'“
BK
X5,
rather
abstract
and
9.4
While
not
easily grasped, it must be made clear from the start that the set is
larger for less-complex equations, that is, those with particularly
even
set
may
seem
9.
The
Galois
Group
relations
many
of
the
among
Equation
an
solutions.
99
The
set
BK
is thus
of
sort
a
of the
measure
complexity of the underlying equation. A truly simple
characterization
of this complexity is obtained
with the help of the
Galois group,
which by definition contains
all permutations
of the 71
variables
X1,
a
,X,, that transform
polynomial in BK to another
polynomial in the set. This leads to the following definition.
.
.
.
Definition
9.2.
tions
coeflicientslie in
whose
field K) is the
that
a
that
for every
.
a
equation
without
field K, the
Galois
multiple
.
.
,a:,,)
.
.
symmetric
polynomial h(X1,
has
0, one
=
that
case
solu-
(over the
group
.
.
In the
polynomial
a
of all permutations
U in the
symmetric
group
3,,
indices
1,
,n of the solutions
in such
$1,.
, can
the
h(a:1,
on
set
permute
way
and
For
.
coefiicientsin K
,X,,) with
.
h(:cU(1), ,:v,,(,,)) O.
.
nontrivial
no
.
.
polynomials, exist,
.
=
.
relations,
that
the
group
Galois
is, those
based
not
consists
of all n!
permutations, and indeed, every polynomial in the set BK remains
unchanged under all permutations.
In contrast,
the example
earlier
of
given
of fifth
degree first solved by Vandermonde
leads to a drastically reduced
set of only five
permutations, whereby
in this case
the individual
polynomials in the set BQ are altered by the
permutations, with only the value 0 resulting from evaluation
at the
solutions
remaining unchanged. For example, the cyclic permutation
X1
equation
an
>
>
>
X2
>
X3
X4
X5
the
X1 transforms
2 into the polynomial
X2
X22 X3 2, which,
>
Xi?
—
—
—
belongs to
1 and
the
2 does
set
not
BQ. On
lead
to
other
element
an
immediately by investigating
see
the
result
X;
X1
——
~
—
of the
Galois
as
group,
the
one
can
2 and
polynomial Xf
X2
of the indicated
The resulting polynomial,
permutation.
2 does not belong to
BQ, since :33 2:1 2
233
ml % 0.
—
—
Another
the
the
polynomial
however, again
hand, simply permuting indices
example, one
biquadratic equation
to
which
we
shall
—
return
—
=
—
several
times,
is
£L‘4-—4:I1'3—4.’132-I-8£I3—2=0.
Without
going into detail,
the
identity
is
explained
of the
solution
$1223
+ 1102334
in the
section
procedure
note
We
=
first that
the
four
solutions
satisfy
0, Where the numbering of the solutions
on
computing
described
in
the
Galois
Chapter
In
View
should
keep
group.
3, one
in
such
that
mind
possesses
Since
tations
that
leave
Galois
group.
The
numbers
1,2, 3,4
Galois
in the
therefore
to
the
relationship
(—X2+ X3
Here
2X4)8
+ X3
253184(—X2
shall
we
+ X3
+ 16(—X2
—
that
permutations
—
to
belong
the
can
and
solutions
in the
above-
whether
check
a
single polynomial
a
2X4)7
-
~
~—
permu-
in
--
2X4) + 72256.
of the
explicit enumeration
The following taGalois
group.
with
ourselves
content
eight
the
among
for
candidates
as
remaining
group
belong
mentioned
section, where it is shown how one
permutation belongs to the Galois group using
BQ, namely
—
resolvent
is demonstrated
Galois
the
eliminated
are
That
group.
every
“respect”
in fact
tations
cubic
7é0 and $1902 + 1133584 % 0, only those permuthe polynomial X1X3 + X2X4 can
belong to the
of the twenty—four
result is that sixteen
(= 4!) per-
of the
membership
the
-1- (E2123
$1204
mutations
Equation
solution.
rational
a
an
of
Group
only when
exists
identity
an
Galois
The
9.
100
an
permutes the indices
eight permutations
is the identity,
denoted
here
first
The
00,
by
permutation,
1, 2, 3,4.
that
is, the permutation that leaves every index unchanged:
ble
As
each
how
shows
of the
it
already
stated,
without
reference
tion
is solvable
and
role,
where
alone,
can
to
the
1
2
3
4
1
2
3
4
3
2
1
4
1
4
3
2
3
4
1
2
2
1
4
3
4
1
2
3
2
3
4
1
4
3
2
1
be
determined
from
original equation,
the
group
the
equa-
whether
of the
roots
that
will
relations
that
exist
among
themselves,
that
is, relations
the
Galois
in the
degree
it is not only the size of the Galois
For such pronouncements
solution.
a certain
themselves
The
issue.
that
is
at
play
permutations
group
tations
what
of the
matters
Galois
group
is the
appear
the
permu-
in the
sense
9.
The
Galois
of
Group
of composition
an
Equation
of permutations,
discussed
in
procedure
for
101
Chapter
5. In
particular,
if one
performs Galois—group
or and
7' in
permutations
succession, the
result
is another
And this new
permutation.
permutation, denoted by
7- o 0,
has, like a and T, the property of changing all the polynomials
in the set B K into polynomials
in B K and thus is itself a member
of
the
Galois
9.5
A
lations
group.
universally
applicable
documenting all the regroup,
though not partic-
among
the
elements
of the
ularly elegant
due
to
its
explicitness,
is
how
this
in the
ation.
We
mentioned
shall
biquadratic
permutations
group.
The
the
index
01(1)
permutation
or5(a1(1))
take
the
permutation
to
05(3)
and
obtains
.
shows
nations
Each
from
that
at the
05
of two
o
01
of the
of the
context
of
example
an
01
and
permutes
the
solution
3 is
oper-
already
of
composition
permutations
index
group
05
permuted
from
the
index
by the
Ga1 to
second
=
the
table
2
PN\xAC
and
of the
07.
following result:
|
first 01
A look
the
table
a
4, the net result is a shift of index 1 to
does the analogous operations
for the other three
4. One
indices
As
01
Since
3.
=
05
=
works
equation.
shall
we
lois
solution
see
Galois
The
permutations,
then
05
1
2
3
2
4
3
4
[\3|-‘C/J
4
1
eight permutations
group
a
sort
table
consists
of the
Galois
group
of all such
combi-
glorifiedmultiplication table.
of first applying the permutation
of
entry in the table is the result
the top row
and then the permutation
in the
left
column:
relations
All
One
table.
group
the
among
of
Group
permutations
solutions
Equation
an
the
off from
be read
can
how the
longer tell
no
can
Galois
The
9.
102
and
their
indices
However, one can tell—and we
table
of its irnportance—-from
the group
shall say it again because
the underlying equation is solvable
in radicals and the
alone whether
for expressing the solutions.
degree of roots that are necessary
are
permuted by the permutations.
such
Why the Galois group contains
plausible by breaking the solution process
information
be
can
as
seen
of the
underlying equation
into individual
steps, each of which corresponds to the adjunction of a
single quantity, and then investigating how this adjunction alters the
Galois group.
Namely, by extending the set of possible coefiicients
to a larger field E, the set of
from a field K of “known quantities”
of the
used in the definition
“relational” polynomials B K, which was
is clearly enlarged to a set B E. The more
Galois
stringent regroup,
of the field
linked
with the extension
quirement on the permutations
of coefiicients may
then
tions
the
Galois
group.
It is thus
of the
Galois
group
is intimately
to
belonging
reduction
in size
lead
to
restriction
a
of permuta-
in the
set
seen
that
the
possible
bound
up
with
the
To put it more
concretely, under
properties of the adjoined values.
certain
conditions, the adjunction of a value that is the mth root of
of
an
already known quantity has the effect of reducing the number
permutations in the Galois group by a factor of m.
We shall
9.6
solving
of the
now
take
a
given equation
a
Galois
Its
equation.
solution,
can
group.
We
detailed
are
look
reflected
use
at how
in the
standard
our
the
steps of
individual
corresponding reductions
example of a biquadratic
possesses
solutions, since the cubic resolvent
be expressed solely in terms
of square
roots:
a
rational
a:1,3=1+x/iix/3+\/'2‘,
m2,4=1~\/ijzx/3—\/5.
ori
Beginning with the field of rational
known
quantities (given the rational
we
add
the
following
as
our
numbers
the
as
coefficients of the
first additional
known
quantity
the
two
in
solution
process
adjunctions
the
set
number
the
of
a
pri-
equation),
@?9 For
numbers
The
Galois
V 3 +
\,f2 and
V 3
root
of
now
observe
9.
the
square
Let
us
reduces
the
field K
Equation
an
\/§ present
—
known
a
Galois
how
themselves, in
the
each
involving
case
adjunction of these three quantities
initial
adjunction of \/§ to the base
The
the
to
103
quantity.
group.
Q leads
=
of
Group
result
that
others, the polynomial
among
X1—X2+X3—X4-*4\/3
belongs
to
the
permutations
four
the
set
since
BQ(\/5),
0
=
04,
931
05, 06,
:z:C,(2)+:rC,(3)—.’I30.(4)—4\/§,
they
=
after
group
the
extension
Versely, in analogy
Galois’s method
the
Galois
that
group.
ties” to include
the
07
can
Q to
=
01,
02, 03
Thus
the
of the
Value
\/5 reduces
size
in fact
of the
using
belong to
“known quanti-
of
set
show
can
Galois
by
group
one—half.
With
Since
Con—
Q \x90m\x98
=
Q, one
00,
the
E
=
permutations
the
4x/5.
=
:34
-
original field K
expansion
—
satisfy the condition
scam
no
longer belong to the Galois
field K
of the
the
to
11:2 + $3
—
the
subsequent adjunction of the second intermediate
value,
V 3 + \/3, the set of polynomials that refiect the
polynomial relations
the solutions
is enlarged, for example, on account
among
of the iden2V3 + x/5, to include
tity :01
$3
—
=
X1—X3—~2V3+\/§.
Since
for the
-2
x/Q‘
are
holds, these permutations
eliminated
from the Galois
when
V 3 + \/-2-is adjoined to the field Q
Conversely, one
\x80\xFD\x89
two
permutations
0
=
0'3 the
01,
equation
av(,(1)—a:C,(3)
:-
V 3 +
group
show
can
that
the
two
permutations
00,
02
in fact
belong to the
Galois
out, then
all four
group.
finallythe adjunction
If
solutions
the
931,
adjoined
.
.
.
need
result
be
can
numbers
definition of the
we
,a34
the
Figure
9.1 the
expressed
group
the
is that
x/5 is carried
—
in terms
using the basic
Galois
to consider
of V 3
Galois
in terms
of rational
arithmetic
of the
numbers
operations.
obtained
extension
four
and
By the
field,
polynomials X, —m, for 2' 1, 2, 3, 4. The
contains
group
only the identity permutation
=
00.
In
group
are
illustrated.
three
There
adjunctions
the
notation
and
their
‘effecton
K(a, b,
.
.
the
indicates
\xC0)}
Galois
the
The
9.
104
field formed
extension
Galois
of
Group
by the adjunction
of the
Equation
an
numbers
a,
b,
.
.
.
to
a
that
is, this field is defined as the totality of all numbers
be obtained
on
the numbers
using the four arithmetic
operations
field K.
can
b,
a,
.
.
That
with
together
.
Steps in
Equation
the
Solving
numbers
“Known
root
square
Current
Galois
I
Quantities”
the
\/E
Ecniation
0'0
00,02
p~y
s
of
Group
Q(\/§,\/3+\/5)
root
square
field K.
in the
Q(\/§,\/3+\/5m/3—\/5)
\/3+\/5
T
of
Fields
\/3—\/5
T
the
a0,a1,a'2,a3
root
uare
coefFiTcien(i'.s
equation
of
the
9.1.
Figure
by stepwise
how
the
Q
Solving
the
extension
of
associated
:70,
equation
:24
the
of
extension
set
~
49:3
—
“known
fields reduce
the
U1,
4372 +8a:
:74,
:73,
0'2,
~
2
:
0'5,
0'5,
07
0
quantities”and
Galois
group.
stepwise expansion of the set of known
root
of a previously
known
quantities
by the square
quantity does
more
than reduce
the size of the Galois
group
by a factor of 2. Each
of these
a decomposition
of an appropriately
adjunctions represents
table
into
four equal parts,
each of which
contains
arranged group
permutations
only from one or the other half of the Galois group.
For example, from the first adjunction
one
obtains
the following de—
Where the further
in the
composition,
decompositions can be seen
upper-left—hand
square:
It remains
to
0'7
that
the
00
U1
U2
03
04
05
(76
07
0'7
0'6
05
0'4
0'3
0'2
(71
0'0
between
correspondence that we have observed
adjunctions
and decompositions
of the group
table
holds generally for analogous
into m X m squares,
where
m
is a prime number,
decompositions
9.7
The
note
9.
The
Galois
under
mth
the
roots
The
that
of
been
unity have
the
Galois
of
described
one
table
to
Since
mth
Theorem
root, if it leads
X in
decomposition
root
whose
as
corresponds
group
table
into
in
X
equation
(In recognition
containing
to
in
each
called
soluable.)8
With
this
theorem,
it becomes
clear
which
be
to
reduces
m
squares
the
make
Chapter 7, can
to the
in
in
we
group
step by step
to
—
that
the
solutions
with
nested
allow
table
next
expressions
chapter
unnecessary.
we
shall
where
allowing such
groups
have
not
is
valuable
so
even
:3
begun to
in analyz-
Thus,
1
~
=
120
only once:
with
learn
rational
a
a
solving the original equation.
us
to
0
x
120 group
the
resulting 60 x 60
is precisely the reason
decomposition, and this
of the given
fifth—degree
equation
root
one
which
associated
squares
a
determine
further
no
Bln the
into
the
pre-
principle, purely combinatorial
table
in
progress
51:5
be shown
always
radicals
example, for the equation
permits
group
following result:
solvable
Galois
sure
why the Galois
ing the solvability of an equation:
considerations
regarding the group
group
X
in
ofg/rhirh.rrwtainajhamtthlrpatt
equivalence,
are
decomposition
table
group
decomposition of the (suitably ordered)
a
of this
process
can
these
reduction
only the identity permutation,
squares
stepwise
it
of the
into
in
be reduced
can
of the permutations.
for
actual
an
adjunction
described
we
group
group,
operations
to
such
is
Galois
the
squares.
irreducible
step
root
that
Under
steps.
decomposition
a
An
each
prove,
and
be shown:
leads
the
Galois
irreducible
radicals, this equivalence
9.3.
element
is
adjoined in previous
of unity,
roots
in
cisely when
equation
105
upper—left—hand
square.
the
be expressed
the
effects
every
mth
find an
can
Equation
mth
an
group,
into m
Conversely, for
0
an
following can
adjunction
in the
as
of
assumption
conditions,
0
Group
cannot
table
permits
a
square
that
be represented
radicands.
definition that
will
make
dealing
with
the
Of course,
the
9.8
We would
the
Galois
an
appearance
the
like
to
of
groups
next
the
use
equations,
will
we
in which
demonstration
the
Galois
of this
Equation
an
chapter
some
chapters.9 As with
equation,
of
group
chapter.
remainder
other
some
in earlier
biquadratic
of the
topic of the
will be the
Group
purely
table is not the most
elegant approach.
and why it works the way
be simplified
of investigating
possibility
combinatorially via the group
can
How such an investigation
it does
Galois
The
9.
106
our
that
it is shown
determine
made
of which
have
standard
example
concentrate
generally
to
on
that
of
part
certain
permutations
the remaining
The proof that
cannot
belong to the Galois group.
must
belong to the Ga.lois group can in principle always
permutations
below on computing
out using the technique of the section
be carried
In many
the Galois
cases,
however, a simpler argument can
group.
that are of use in this regard will
be adduced.
However, the theorems
be presented only in the following chapter.
definition of the
generally excluded
of the chapter we
In the remainder
equations with multiple roots.
our
considerations
to irreducible
shall further
reduce
equations. At
this is equivalent, as will be shown in the
the level of the Galois group
next
chapter, to the property that for every pair of solutions
11:, and
0
that permutes
the solution
exists
at least one
ask there
permutation
Galois
In
one
that
the
k.
such
a case,
to
that
says
wk,
is,
a(j)
acj
on
the solutions
of the equation.
acts transitively
group
In the
Galois
group
we
have
2
a Galois
Quadratic equations that are irreducible
always possess
In addition
to the identity
group
consisting of two permutations.
is additionally a permutation
01 that
permutes
permutation
00, there
The group
the two solutions.
table has the following form:
9.9
9The other
examples
Galois
McKay, Computing
groups
over
Their
article
also
(1985), pp.
higher
273-281.
degrees.
have
been
John
Soicher,
the
Journal
Theory, 20
rationals,
of Number
of the sixth
and
contains
examples of equations
taken,
in
part,
from
Leonhard
9.
The
Galois
of
Group
9.10
Regarding the
there
are
two
Galois
possibilities:
permutations
of the
that
the solutions
permute
derived
from
the
three
Equation
an
107
group
of
Either
the
Galois
or
it contains
solutions
irreducible
an
three
all six
permutations
of such
of seventh
equation
equation,
contains
group
cyclically. An example
cyclotomic
cubic
equation,
an
degree, is
:v3+a32—2a:—1=0,
whose
three
of the
solutions
are
zj
=
2 cos
identities
(
for j
1, 2, 3. On
=
account
;v2=:r§——2,
:c3=a3§~2,
9:1:=:z:§—2,
the
three
polynomials
X2—Xf+2, X3—X§+2, X1—X§+2
belong to
the
the
group.
Galois
set
BQ of polynomials
As
consists
group
solutions
The
group
has
course,
be determined
Aside
from
difference
on
the
index
only the three
0(1).
in
determining
permutation
a
by its
0
effect
on
the
Ga-
permute
the
these
three
per-
equation
can
also
Therefore
permutations
that
cyclically:
mutations
Of
of
be considered
we
have that
consequence,
is already determined
group
a
belonging to the Galois
a
single index, for example
lois
to
table
for the
Galois
the
following form:
the
Galois
directly,
group
group
of the
comprising
previous
that
is, without
Galois’s general
knowledge of the
procedure,
one
product
of the
solutions, whose
may
square
also
is the
solutions.
calculate
discriminant.
the
9.
The
with
the
108
easily done
Chapter 5:
This
is most
given
in
Galois
general formula
(931 ac2)(:1cg .’133)(CI31
:33) :i:6z'\/3
—
——
where
p and
in the
case
coefficients
of the
under
equation
Equation
an
for cubic
equations
(gr,
Pa\xB3
=
—
the
are
of the
leads
This
q
of
Group
reduced
investigation,
equation,
—g
and
2
p
for which
q
=
——7—
27
‘
to
933)(901 $03) -7,
(901 902)(~”02
—
‘
which
shows
change
the
at
the
that
once
=
—
odd
permutations,
product, do not
sign of the difference
that
belong
is, those
that
the
Galois
to
group.
“Most” irreducible
9.11
cubic
x3 + :10
solved
in
Chapter
the
1 with
6
—
three
2
-
such
equations,
:=
the
as
equation
0
solutions
61
2
61
W:¢’\3/3+§\/”27+<2’\3/3
for
j
=
0,1,2,
lead
to
table
of the
a
Galois
group
-
that
contains
all six
permuta-
tions:
The
Galois
following form, where
the decomposition
into four 3 X 3 squares
corresponding to the adroot
of the discriminant,
which contains
only
junction of the square
is clear:
three different
permutations,
group
group
has
the
9.
The
Galois
Group
The
direct
relationship
of the
the
equation
third
When
Equation
an
between
is complicated
of
root
of
unity C
the
in the
109
Galois
group
and
of the
cubic
equation
case
3 must
+ \xD0Oe
—%
2:
be assumed
the solution
in that
“known.”
as
the
equation has rational
this means
that the direct
coeflicients,
between
the solution
correspondence
steps of the equation
and the
steps in decomposition of the Galois group is ensured
only when the
field Q
to
unity
enlarged by the third
root
looks
general
of
\x90G1
To that
such
one,
coefiicients is
the
containing
extent
that
equation
the equation
as
an
simpler than
the
m3—3:v2——3:r—1=0,
which
j
0,1,2,
=
For
the
at
appears
end
of
ac,-+1
=
have
can
the
difference
reduced
equation
product
p
three
-6
obtains,
and
q
=
six
comprising
group
one
=
the
solutions
1+0’?/§+C2"«3/Z,
Galois
a
are
Chapter 1, with
coefiicients of the
the
since
permutations.
—6, the value
(171 £I72)(.'E2
E \xA0&%
.’I33)(£I31
(113) \xC0\xBFV
—
—
—
with
the
result
that
the
Galois
only when
the
third
root
of
lois group
over
the
though
even
solutions
are
a
numbers
In
is present
and
the
Ga-
all six permuta-
$3 + $2
—
therefore
first be
of unity must
roots
permutations
contrast,
contains
investigated equation
real, the three
three
to
unity C is adjoined.
irreducibilis
casus
reduces
group
field of rational
tions.” As in the earlier
T:
2:0
—~
the
1
=
0,
three
adjoined, with
1OMoreover,since
at?~3a:,~ .—2 :g9”1€/5,
the
field, that
splitting
Q(C, PL
in which
The
the
name
equation
is, the
field
“splittingfield”
to
be
solved
can
arising
from
adjunction
comes
from
the
be
factored
into
fact
that
linear
of
all
the
it is the
factors.
solutions,
smallest
is
field
the
a
The
9.
110
Galois
third
not
group
reduced.
being thereby
the
allows
root
Galois
solution
of the
Group
of
Then
the
Equation
an
of
adjunction
equation.
Irreducible
biquadratic equations can have Galois groups with
are
In some
of these
four cases there
4, 8, 12, or 24 permutations.
different
possibilities as to which permutations
belong to the Galois
group.
However, from a qualitative point of view, in particular with
will
two Galois
regard to the solvability of the Galois group,
groups
be considered
if one
can
be transformed
the same
into the other
by a
of the rows
and columns
table.
of the group
possible rearrangement
of
are
said to be isomorphic.
this notion
Such Galois
With
groups
there
are
then
isomorphism of groups,
only five possibilities for the
of an irreducible
Galois
biquadratic equation: two with four
group
For each
and one
each with 8, 12, or 24 permutations.
permutations
of these
cases
will look at an equation.
We
9.12
We
the
begin with
of
examples
biquadratic
equations
with
the
equation
iE4+{E3+{132+:L‘+l=0,
solved
in
solutions
Whose
tion,
0, 1, 2, 3, where
the
7, obtained
Chapter
are
from
$341
the
numbering
of the periods
construction
relations
follows, in the
among
the
fifth—degree
cyclotomic equa-
cos
=
+ isin
j
\xB0\xD70
for
=
already been done in
In what
for the cyclotomic equations.
the solutions
the following symmetry
is chosen
as
has
will be apparent:
CBj+1
Thus,
as
with
every
other
=
0\x9A+
equation
for
periods
of
a
cyclotomic
belonging to the Galois group 0 is determined
by its action on a single index, for example by the index 0(1).
This leads to the result
that
those
that
only four permutations,
permute
the solutions
The folcyclically, belong to the Galois group.
of the Galois
and the
lowing tables show the permutations
group
equation,
associated
a
permutation
group
table:
The
9.
9.13
Galois
of
Group
Equation
an
111
Similarly, the equation
&+1=0
leads
to
lutions
a
are
Galois
group
with
four
permutations.
four
of the
eighth
roots
of unity,
`
2'—1
2
arj =cos
equation’sso-
This
namely
2'—1
2
\x80\xA7\xA3
+isin
for
j:
1,2,3,4.
in
First,
analogy to how we dealt with the previous equation, a permutation
a
belonging to the Galois group is determined
solely by 0(1),
that is, by its action
on
the first index, on account
of the relation
2j—1
$3 —:v1
‘_
Therefore,
listed
in the
the
the
permutations
left—handtable
.
the
Galois
group
then
obtains
the
belonging to
below.
One
lead
table
right, where the four identical
permutations
that
no
can
renaming of the permutations
of the given equation:
9.14
An irreducible
Galois
sisting
of eight permutations
9.4 and
9.5.
on
show
A very
biquadratic
with
equation
those
group
table
the
on
a
are
diagonal
the
group
group
con-
to
has
already been investigated in Sections
simple equation that leads to an isomorphic Galois
is
group
#—2=a
Whose
:c2w4
=
solutions
are
0, we
may
investigated.
We
32,-
=
proceed
note
for j
2'j”1\"/§
analogously
only that
the
=
1,2,3,4.
to
the
Galois
Since
previous
group
reduces
:1:1w3 +
equations
to
four
112
The
9.
Group
field is extended
if the
cyclic permutations
Galois
to
of
Equation
an
root
of
permutations
is
Q by the fourth
1'.
unity
A Galois
9.15
obtained
for the
all
comprising
group
twelve
even
equation
:c4—|—8$—|—12=0.
The
is that
reason
using the
formulas
of
5 for
Chapter
cubic
the
resolvent
z3—12z+8=0,
for the
obtains
one
difference
the
product
value
2
Hm» 53k)
=
—
8
j>k
48i\/§
=
\x80\xEE
@\xE5\x9A
+
j>k
=
An
9.16
H(zj- zk)
-
maximal
-576.
irreducible
size
biquadratic
(24 permutations)
equation
is the
with
Galois
group
degree,
there
a
of
following:
a34+a:+1=0.
9.17
In
the
case
only five possibilities
groups
of
group
falls
into
while
those
for the
There
of irreducible
equations
for the
group
Galois
of fifth
(up to isomorphism), being
5, 10, 20, 60, 120 permutations.
are
one
of the
last
two
first
three
cases
are
only five permutations
Inonde’s equation
3:5 + 3:4
4m3
are
Equations
cases
solvable
are
whose
Galois
in
radicals,
not.
in the
Galois
32:2 + 3n: + 1
group
of Vander-
0, whose solutions
are
two—memberperiods of the cyclotomic equation of degree 11. As
described
in Sections
9.3 and 9.4, one
can
determine
the permutations
belonging
to
the
::c2+2
suchasatfi
Galois
—
group
—
with
andmg =a:3+2.
the
2
help of polynomial
relations
9.
The
Galois
of
Group
The
Equation
an
equation :35 5:3 + 12
Its solutions
permutations.
are
—
113
0 leads
=
to
a
Galois
with
group
ten
.’13j;1“€‘7§/
\/1 5x/5+3 5+125
0111-‘
+e3J\5/—1—g\/5
11.11
+e\/-1-5\/5+3
2
-
\x80\xD2=
p\xC9Vp\xC9V
125
2a'5__E
+e
1
i1_ a
2
.
11
+
125
3
4j5_1
)-—‘
25
5
where
e
=
+ isin \xF0:T
for j
\xC0#N
cos
from
calculating the
in Chapter 8, which
the identity
value
E
of the
O,1,2,3,4.
=
bicubic
corresponds,
,
This
resolvent
to
up
E
z
be
can
seen
5 investigated
=
sign to be determined,
a
to
331562 + 5132933+ 11331104+ 1134135 -|- $5$1
The
Galois
preface
to
The
a:j
1221134
of this
equation
group
this
book, in Figure
Galois
of the
group
for j
e7’1\5/2,
2:
£3125
931.133
1,
2
.
.
.
appears
110,determined
by its
all of the
5X 4
with
two
the
Galois
=
action
equation
,5, contains
20 ways
solutions
two
p=1,...,4andq=0,...,5,
a
—10.
table
group
in
twenty
0, whose
::
in the
solutions
twenty permutations.
of associating
these
2
~
Galois
solutions
$1, {U2 result
Indeed,
group.
$5
of the
the
on
as
—
0.1.
each permutation
ac‘§Tj:c§—1,
=
£41131 115mg
a
a
ml
group
and
permutations
Since
is
completely
(U2. And in fact,
pair of two different
permutation
are
solutions
belonging to the
are
defined,for
by
Upyq
(gs/g)
:
€pj+q\5/Z
forj=0,1,2,3,4.
are
The
equation
the
even
$5 + 20$ + 16
permutations,
=
those
0 yields
that
sixty permutations.
leave
unchanged
the
These
integer
value
of the
difference
5 in
Chapter
8, the
An
of 120
Galois
The
9.
114
product.
difference
Group
of
Equation
an
Using the formula given in footnote
product is either +32000 or —32000.
equation
example of a fifth—degree
is 305 :1: + 1
0.
permutations
with
Galois
maximal
group
=
~
of Galois
containing his
chapter with a theorem
findingsabout the solvability of equations in the form of a “traditionare
criterion
That
this
formulated
criterion.
equations
satisfying
ally”
solvable
was
hypothesized before Galois by Abel, in 1828, in a letter
to
this
We close
9.18
(1780—1855).11
Crelle
Theorem
An irreducible
9.4.
radicals
if and only if all the
in
arbitrary
two
of prime degree is solvable in
be expressed as polynomials
can
equation
solutions
solutions.
numbers) fifth—
particular, an irreducible
(over the rational
be
cannot
real
and
two
nonreal
solutions
three
with
degree equation
17 is
solved in radicals.
Thus, for example, the equation 12:5 17a:
immediately seen to be unsolvable, since by Eisenstein’s irreducibility
it has three real
criterion
the equation is irreducible, and furthermore,
solutions, and there is no Way that two of them can be used to express
in terms
of polynomials (with rational
all the solutions
coeflicients).
In
—
~
the
n
is
An additional
consequence
Galois
group
of
always
a
divisor
of Galois’s criterion
an
irreducible
of
n(n
—
Computing
As
we
of the
the
set
is far
too
have
mentioned,
Galois
group
the
is that
size
of
equation of prime degree
of n.
a multiple
solvable
1) and
the
Galois
Group
polynomials used
large to list explicitly.
B K
of
in
Even
definition
the
a
complete
of
and
editor
as
the
founder
remembered
primarily
und
The
Journal
German
mathematical
the first
Angewandte
fiir die Reine
journal.
referred
to
as
“Crelle’s journal.” Abel’s proof
is even
Mathematik
today frequently
in 1826
in volume
1 of Crelle’s journal
of unsolvability
was
(Beweis der
published
allals dem
Von
hoheren
vierten,
Graden,
Gleichungen
algebraische
Unmcjglichkeit,
Théoreme
is also
formulated
as
9.4
pp.
gemein aufzulosen,
65—84).Our Theorem
une
Classe
IV
particuliere
d’équations résolubles
(p. 143) in Abel’s Mémoire sur
11August Leopold
Crelle
is
algébriquement,Crelle,
4
Niels
solvable
131-156; see
Archive
equations,
Henrik
Abel
(1994), pp.
81-103.
and
(1829), pp.
Skau,
Gérding, Christian
48
Sciences,
for History
of Exact
also
Lars
9.
The
Galois
description
Group
of
Equation
an
simple.”A way
is not
115
of this
out
explicit
of the Galois
computation
group,
originally by Galois for defining
the Galois
Galois constructed, for the
of
this
in the
the
suitably chosen numbers
always find numbers
ml,
can
to
resulting
from
quantity
25 thus
that
erty
in particular
all
=
.
.
distinct solutions
ml,
Galois resolvent.
He
0
constructed
now
solutions
ml,
without
using
12For readers
.
from
.
.
.
.
,mn.
Galois
in the
field K such
of the
,a:,,
.
root
-
-
indices
1,.
.
as
hl,
form
.
.
,
,
sun
that
one
all n! values
can
be represented
distinct.” A
are
the
by polynomials
propin t,
operations:
ml
gl (t),
,:cn
g,,(t).14
BK used to define the Galois group
thus
=
set
.
.
that
Lagrange determined,
The
.,
(basis) polynomials
observed
,n
..
has,
who
.
.
-1- mnfEa(n)
-
already know (almost) everything:
ring K[Xl,
n]. From the Hilbert
basis
polynomial
then
.
the
.
+-~+mn:vn
,m,,
.
used
expressed
m11Ig(1) + m2H3g(2) +
permutations
Every polynomial
many
of the
ml,
method
n
so-called
+m2w2
by the
an
group.
form
t=m1:v1
with
is offered
hypothesized
given nth—degree
equation,
a
dilemma, allowing for
hm
such
that
the
set
.
=
.
BK is an
ideal
in the
there
exist
finitely
Bk comprises
the polynomials
theorem,
set
flh1+""l'fmhm
for
some
hl,
.
.
.
mutation
Galois
,
polynomials
hm, then one
be
can
If
could
compute
individually checked
the
could
Galois
with
these
one
determine
be
satisfied
values
ml,
.
.
to
,m,,
.
be
chosen,
using
polynomials
such
the
for
of the
equations
group
group.
13For the
may
fl, ...,f,,,.
none
ml
(wan) 9=r<1>)+'-‘+
mu
(='-‘o $r
)
for
two
basis
polynomials
fact that
each
per-
membership
in
the
—
—
=
0
distinct
0
permutations
and
7'.
Each
of these
%n|(nl
1)
thus
equations
limits
the
possible
selection
of values
a
ml,
, mn
in
by
hyperplane
K":
for 11, = 2 this
is a straight
line in K2, for n
3 a plane in
and so on.
K3,
Thus
in any
there
remain
case,
infinitelymany ways of choosing the values ml ,
,m,,.
14The proof of this theorem——the“modern”
variant
can
K(:1:l,...,a:n)
be found
K(t)
in books
on
abstract
algebra as a theorem
on
the existence
of a primitive
element-——was
only sketched
by Galois.
to the constructed
Corresponding
Galois
resolvent
formed
t, Galois
the polynomial
—
.
.
.
=
.
.
.
=
aESn
0(1):].
(n—1)I
Z
=
9k(X1:
:Xn)Tk:
k=CI
based
on
all
permutations
that
fix
polyno1nial’s(77. 1)l factors, where
the
coefficients gk(Xl,
in the
,Xn) appearing
sum
are
in the
polynomials
variables
that
Xl,
are
,Xn
in the variables
symmetric
X2,
X". If one
now
considers
the
polynomials
gl, (Xl,
as
,Xn)
in the
polynomials
variable
is dealing
Xl, then one
with
symmetric
in the
polynomials
variables
X2,
,X,,, which therefore
can
be ex»
as
pressed
elementary symmetric
in these
polynomials
variables.
Since
each
furthermore,
of these
elementary symmetric
can
be expressed
polynomials
as
a
polynomial
in the variable
Xl as well as the elementary symmetric
in the
polynomials
variables
.
.
.
.
0
the
—
.
.
.
.
.
.
.
.
.
.
.
,
Galois
The
9.
116
of
Group
Equation
an
reis satisfied by the
that
Galois
equation
polynomial
corresponds
in one
such polynomials
It. And
variable, as explained in Point 3
solvent
on
section
in the upcoming
computing with polynomials, are all multiples
Therefore
of an irreducible
polynomial over the field K with t as a zero.
with
can
be examined
respect to this single equation
permutation
every
Galois
And furthermore,
it belongs to the
whether
to determine
group.
this single irreducible
polynomial with t as a zero can generally be found
of
of
and
the
the
method
product
Lagrange
taking
b
y
employing
“easily”
factors
all n! linear
(T ta) and then decomposing this degree-n! equation
into irreducible
factors, whereby one then must seek the factor Q5(T) that
to
a
—
has
75 as
zero.
a
help in clarifying
equation should
use
the numercan
these
matters
as
well as in showing how in practice one
In fact, due to the inevitable
the Galois
to compute
ical solutions
group.
for proving
not
suitable
values
are
the numerical
due to rounding,
errors
often
which
to
an
can
be
used
an
prove
inequality,
equality, though they
the equation
suflices. As an example, let us consider
The
of
investigation
particular
a
a:4~4:v3—4m2+8a:—2=0,
X2 +
(for example,
X2,...,Xn
«
~
+Xn
-
(X1 -1-
=
-
-l-X") —X1), one
-
obtains
hj,k(X1y-~-:Xn)Xf{:
2
9k(X1:---:Xn)=
-
j=0,1,...
the
where
One
defines the
now
h_,-,;¢(X1,,Xn)
polynomials
.
.
.
Xn.
,
hjlk (331,
of the
original
show, the
equation,
algorithm
equation
F(X)
.
.
.
,
j
without
w,-L)can
values
coeflicients
That
.
.
EZhj,;,~($1,. ..,:z:,,)tkXj.
=
k
from
X1,
polynomial
F(X)
The
variables
the
in
symmetric
are
.
equation,
so
0 has
=
exception
that
they
only
as
:01
a
be
expressed
lie
must
in
in
solution
can
be
in
polynomials
as
the
shall
the
field K.
As
We
common
with
the
original
factor
linear
that
the
using the Euclidean
computed
ml)
(X
that
of the two equations,
is,
(see the following section) from the coefficients
arithmetic
with
in the field K together
operations.
values
75, using the four basic
of linear
can
be shown
division
in fact
do without
algebra
one
can
by the methods
—
so
(see Section
10.9).
the
still
to investigate
We have
with
associated
the identity,
factor
we
have
unchanged,
F(a21)
=
zeros
that
of the
is, the
polynomial
permutation
\x80\xB6Clearly,
leaves
that
the
from
everything
0.
..,£E,-L)
,:IJn)tk
G(t,I1I1,.
.,£I3n)tk£lI{
Zg;,~(.’.‘C1,..
Z::hj,k((I,‘1,.
=
.
k
=
=
.
k
PN\xC1
Furthermore,
.,:1:n)tkH2‘;
F(.’1)2) ZZhj‘k($1,$2, ,2:-,,)tkCB‘;
ZZhj‘k(:z:2,a:1,
=
.
k
=
.
=
.
.
k
3'
Zgk(w2,w1, -,wn)t'° H
-
=
-
Ic
(t
—
.7"
(m1a:2 +m2ma(2)
aesn
a(1)=2
#0!
with
corresponding
results
for
the
other
solutions
.
1:3,
.
.
.
,:c,,.
+
-
-
i
+
mnma(n)))
9.
The
Galois
already analyzed
in
whose
numerical
values
mation
algorithms:
One
now
-322
+ $3
one
can
seeks
a
Galois
show
that
..
9.4
and
9.5.
found
by
the
4!
the
It possesses
of
any one
:32
=
,
1:4
=
by trial
required
four
a
.
..
—1.67349368
error,
property:
solutions,
of approxi-
,
.
and
..
.
where, for example, t
By numerical
calculation
—:r,,(2)+ w,,(3)
24 linear
factors
~
:
24th—degree
equation with
coefficients, and therefore, with minimal
rounding of the numerical
one
can
determine
the nearest
integer values exactly.
one
In
the
the
Galois
search
efficients
that
numerical
values
for
has
resolvent
factor
a
the
Galois
be
multiplied
resolvent
t
to
as
K
over
a
with
We need
advantage.
to
results
rational
knowledge
zero,
integer
co-
about
check
the
which
factors
(T
can
a
Q5(T) irreducible
again be used
can
linear
twenty—four
of the
t
=
distinct.
2a:a(4) are
-
(—=Ba<2> 2))
+ macs)
—
real
number
0.84506656
24 values
=
by the
(T
117
,
resolvent
Then, by multiplication
finds for
Equation
..
.
2934 possesses
—
an
be
can
031321057.
=
$3
Sections
451521655
=
cvl
of
Group
"
together
(—93a<2> 3$a<4>))
+ M3)
to
yield
—
with
polynomial
integer coefficients.
of permutations
Clearly, every combination
can
be rejected for which
the
numerically calculated
product is not close to an integer polynomial. In the
converse
result
indeed
case, when the numerical
corresponds
approximately
to an integer polynomial, this
allegedly integer polynomial must be checked
to see
whether
it is indeed
a
divisor
of the degree-24 polynomial
to be
a
factored.
For
irreducible
Q5(t)
=
concrete
our
example,
the
obtains
for
the
Galois
resolvent
numbers) eighth—degree
polynomial
(over the
rational
em
T8 + 16T7
02
=
+
The
one
linear
factors
~
4OT6
—
1376T5
34048T3 + 222O8T2
that
following permutation
constitute
of indices:
the
—
—
if an
Q5(T) with
928T4
253184T
polynomial
+ 72256.
Q5(T) correspond
to
The
9.
118
Galois
Group
of
Equation
an
2
tx3p
n-I:
1-=
2
2
4
4
1
1
3
#-
3
The
of permutations
set
method
is the
desired
[O
one obtains with
group.
Then, on
that
Galois
03
this
generally applicable
the
one
the
from
hand,
‘
identity
(-2132+ $3
—
—
+ :c3g—
253184(—o:2
2534)+ 72256
—
obtains
+ 16(—(132
+ £63
2.’l24)8
2:I34)7
-
-
"
0
=
belonging to the set BQ. And this identity reof the underlying condition
mains
in the construction
valid, on account
of the Galois
resolvent, only for those permutations
corresponding to one
of the eight linear
factors
of the irreducible
polynomial ®(T),which are
in the above table.
can
On the other hand, one
precisely the permutations
0
satisfies all other
prove
conversely that every one of these permutations
valid for the solutions
That
polynomial identities
:31,
,mn.
is, one always
has h (m,,(1), ac,(,,)) 0 whenever
h(:1:1, ,:c,,) 0 holds.”
one
polynomial
a
.
.
15Such
of the
terms
.
.
.
Galois
might
begin
resolvent
t.
inserts
obtains
these
.
with
polynomial
That
is,
:91“):
ml
If
.
=
,
proof
a
.
representations
of the
in
solutions
51311 =g71(t)'
-~'1
into
=
.
the
beginning
equation
f (:12) O to be solved,
one
factor
f(gj \xC0] 0. From point 3 of the following section, the irreducible
be a divisor
must
Q5(T) belonging to the zero t of the constructed
nl—degree
polynomial
of the polynomial
For each
a
permutation
sought on the basis of the Galois
f(g_,-\xB0b\xCB
the associated
zero
O as well, so that
resolvent,
t, of ®(T) satisfies f(g,- (156))
every
value
of the
Solutions
If two
such
values
equal one
1131,
,a:,,.
gj (ta) must
gj (ta)
and
is a zero
of the
then
associated
t,
gk(t,,) are equal for some
permutation
0,
difference
therefore
must
be divisible
polynomial
gk (T), which
gj (T)
by Q3(T). It
follows
that
this
shows
that
the values
gk(t), that is, :1:_7- wk. Altogether,
gj (t)
to a permutation
of the solutions
g" (to) correspond
g1(tc,),
ml,
, :z:,,:
one
representations
=
=
=
.
.
.
—
=
.
.
.
=
,
.
5131(1)
That
the
the
permutation
'r
is
in
fact
the
‘
-
=
33-r(n)
-3
permutation
0
.
gn(ta)be
can
seen
from
the
fact
that
Q5(T) and
thus
also
polynomial
T
has
has
91(to):
:
.
T
the
=
t
as
value
a
zero,
ta
as
and
a
(m191(T) +
—
therefore
-
'
mn9n(T))
+
-
is divisible
by the
polynomial
zero:
to
=
+
7n1g1(t0‘)
'
For
'7' corresponding
to g1(t,,),
permutations
hold
However, this equality can
only for a
=
'
‘i’ mn9n(to)-
'
.
.
7',
.
,g,,
since
(to)
one
all
nl
therefore
has
possible
values
t,
=
of t,
t.,..
are
9.
The
Galois
Group
of
In summary,
to compute
and the associated
computed
irreducible
factors,
the
of the
Galois
description
solutions
(171,.
from
passing
factor
.
the
Equation
the
Galois
Q5(T). In this
In the
group.
resolvent
the
way,
to
=
finish,we
To
circumstances
resolvent
the
to
be factored
can
t
that
note
as
g1(t,,),
result
into
that
-
‘
permutation
of the
single value, namely, by
of the
t,
irreducible
+
7nn~’6a(n)-
=
g1
-
(t),
.
.
.
,
run
=
field K
lead
can
polynomial
irreducible
(’5(T),
factor.
factor
The
by
g,,
(t), one
in
certain
gn(t,,).
=
2200,)
of the
one
describes
every
implicitly determined
are
$1
...,
the
than
more
complete
zero
extension
an
a
t to
+ m2£Ua(2)
+
Among its
provides
a
cr
t is
zero
by changing
Beginning with polynomial representations
additionally has the formulas”
mag)
a
another
resolvent
constructed.
process,
permutations
m1$a(1)
=
first a Galois
group,
(’5(T)
having 15 as
factor
Galois
119
degree—n!
polynomial
is determined
,:v,,
.
an
that
has
K,
over
the
Galois
defined in terms
of the
extension
field. For this reason
Galois
himself
studied
the properties
of
such a decomposition——in
particular, all the factors have the same
degree——
in order
to discover
the precise behavior
of the Galois
in extension
group
fields.
A
The
a
zero
Quick
Course
determination
requires
allows
us
to
restrict
relatively
are
the
most
noted
at
It
easy
to
work
that
outset
now
the
0
:
some
to
with.
the
with
For
respect
in the
where
of a
polynomial
remains
used
four
is indeed
91(ta),
relation
.
.
16A proof
polynomials.
assembled
variable.
divisibility and
to
the
here
It should
be
related
properties,
one
sees
at
yields
the
desired
integers.
$a(n)
---y
h(:r1,
.
.
,a:n)
.
0 is
=
=
9n(ta)-
given, then
.
once
that
result
:
*
to
Galois’s procedure
have
to
is divisible
,g,, \x90\xFD
h(g1 (T),
by ('5(T). This
h(g1(tU)v' 197100)) h 030(1): -'17o'(n))
'
section,
single variable, which
we
one
previous
that
polynomial
It
That
Polynomials
that
reason,
of polynomials of
'
been
with
polynomials
display striking analogies
$a(1)=
If
Calculating
attention
our
properties
follows
group
as
described
groups,
investigation of polynomials,
important
polynomials
distinct.
in
Galois
of Galois
extensive
an
the
note
times.
the case,
of which
that
This
here
a
~
single
'
:
'
argument,
namely the divisibility
leads
one
to suspect
that
a
universal
principle
as
will be discussed
in the following chapter.
in Point
appears
3 of the
on
upcoming section
by G5, has
is in play.
calculating
with
The
9.
120
We
1.
Point
with
begin
9.5.
A polynomial
g(X) with
result
a
q(X) and
quotient
degree of the
remainder
of
Equation
an
integer division.
remainder
q(X)9(X)
=
Group
be divided
f (X) can
f(X)
the
of
analogue
an
Theorem
with
Galois
polynomial
by a nonzero
r(X)
such
that
7’(X),
+
polynomial r(X) of lower
degree than g(X
of the two polynomials q(X) and r(X) is
practice, the computation
of a procedure
accomplished by means
analogous to familiar long division,
is no coincidence, since integers written
which
in base 10 can
be understood
as
the values
of a polynomial
evaluated
to the
at 10. However, in contrast
long-division algorithm, there is no borrowing, so that the polynomial case
is actually
We will content
ourselves
with
an
simpler.
example, since a
general description of the algorithm would yield less clarity:
In
2X3
3X2
(X4
X2)
(X4—2X3—
:
X
2) : (X2
:
1)-X2+4
2X
4X2— X+2
4X2—8X~—4
7X+6
The
result
is therefore
X4—2X3+3X2—X+2
=(X2—2X~—1)(X2+4)+7X+6.
Point
2.
In
with
the
of
polynomial g(X) is
of a polynomial
called
a
divisor
f(X) if f(X) is divisible by g(X) with
zero
remainder.
A greatest
common
divisor
of two polynomials
f (X) and
by gcd (f(X), g(X)), is a polynomial of maximal
degree
g(X), denoted
that
divides
both
f (X) and g(X We note first that the greatest common
divisor
remains
unchanged when f (X) is replaced by f (X)
h(X)g(X);
analogy
case
integers,
the
—
that
is,
gcd (f(X),9(X))
The
for
reason
divides
f (X)
formation
—
this
is that
h(X)g(X),
=
gcd (f(X)
common
every
and
the
converse
h(X)9(X),9(X))-
*
divisor
of
holds
because
f (X)
and
the
g(X)
also
given trans-
is reversible.
Of special significance
is the special case
in which the polynomial h(X)
is equal to the quotient
of f (X)
q(X) resulting from division with remainder
the two polynomials
by g(X): in this special transformation
f1 (X)
g(X)
:
and
second
a
g1 (X)
f(X)
=
polynomial
remainder
of
a
—
q(X)g(X)
is smaller
division.
than
obtained,
are
that
of the
whereby the degree of the
polynomial g(X), since it is
The
9.
Galois
As in the
of
Group
Equation
an
121
of integers,
by repeating such steps one can compute the
common
greatest
divisor.
The procedure,
called
the Euclidean
algorithm,
begins with the pair of polynomials fo(X)
f (X) and g0(X)
g(X) and
reaches, at the jth step, the pair
case
=
fj(X)
where
the
9j—l(X): 9j(X)
=
of the
degree
fj—1(X) qj—1(X)9j—1(X)>
=
—
polynomial
gj (X) is
always
the
algorithm
must
polynomial
Therefore,
number
some
g,~_1 (X
of steps with
=
gm
(X)
One
0.
=
less
then
for
obtains,
9', for
has
the
obtained
for suitable
polynomials
fm(X)
71]‘
(X This
if working
backward
3'
is
one
satisfied,
trivially
finite
a
:
fm (X) thus
a
after
:"'
index
every
of the
(f1(X)>91(X))
gcd (fm(X)v0) p\x9D\xCA
=
now
that
terminate
gcd (f(X):9(X)) 7-gcd
One
than
representation
Uj (X) and
from the index
considers
=
2
the
greatest
uj (X)f,~ (X) +’l}j
(X)g,- (X),
is clear at once
inductively
for which
m,
divisor
common
equation
such
a
representation
corresponding
the
to
jth
step:
fm(X)
'“J'(X)fj(X)
+”J'(X)9j(X)
’uj(X)9'j—1(X)+
vj(X) (fj~1(X)
=
(X)9j—1(X))
+ (%‘(X)
Uj(X)fj—1(X)
vj(X)qj—1(X))9a'—1(X)-
=
=
equation, proved by induction, now
important property of the greatest
common
the
(H-1
—
This
an
r
reveals
to
divisor
g(X)
9.6.
can
nomials
Point
role
We
now
in his
Theorem
mon
9.7.
with
zero
first that
is
mial’s coelficients are
to
the
result
stated
in
of two polynomials f (X) and
—|—
v(X)g(X) for suitably chosen poly-
proved by Galois
that
played
If the polynomial f(X) is irreducible
and
the polynomial g(X), then
g(X) is divisible
irreducibility
respect
a
0
=
divisor
common
u(X)f(X)
to
come
j
case
an
impor-
investigations.
We observe
since
greatest
be expressed
as
u(X) and v(X).
3.
tant
The
in the
fm (X), as
following theorem.
Theorem
us
the
formulation
always in
reference
to
presumed
to
Here
theorem
may
belong.
field K.
Furthermore,
the
of the
be
the
possesses
a
com-
by f(X).
theorem
is not
quite exact,
in
which
the
set
irreducibility
is meant
polynowith
that the zeros
interpreted to mean
of an irreducible
polynomial are algebraically indistinguishable.
That
is,
with
to the
every
four basic operations
property
respect
enjoyed by one
zero
holds as well for the other
zeros
of an irreducible
polynomial.
Galois
The
9.
122
of
Group
Equation
an
first
determines, using the Euclidean
algorithm, the
which
divisor
of
and
we
common
shall denote
by d(X
f
g(X),
(X)
greatest
must
reside
in the field K.
Its coefficients
Moreover, executing the Euclidean
algorithm provides us with a representation
u(X ) f (X) +
d(X)
common
and
The
for
chosen
suitably
polynomials
u(X)
u(X
u(X)g(X)
of d(X),
of the two polynomials f (X) and g(X) is therefore
also a zero
zero
which
implies that the degree of d(X) is at least 1. This shows that the
polynomial f (X), must be
polynomial d(X), as a divisor of the irreducible
c in
K.
That
constant
cd(X) is a
is, f(X)
equal to f(X) up to some
of the polynomial g(X
divisor
For
a
proof,
one
=
=
Point
Theorem
is
For
ducible
gj (X
nomial
number
The
9.8.
unique
up
would
also
We
4.
the
to
like
make
to
factorization
arrangement
following theorem.
of the
note
factors
of a polynomial into irreducible
of the factors and multiplicative constants.
irregr (X) into
g1 (X)
f1 (X)
f3 (X)
with a factor
zero
factors, the factor f1 (X) must have a common
9.7, therefore, f1 (X) is a divisor of the polyAccording to Theorem
some
for
we
have
and
Therefore,
cgj
(X)
conversely.
f1
(X)
gj (X)
two
factorizations
-
=
-
-
-
-
»
=
a
field of coefficients
in the
by f1 (X), one
original equation
the proof.
Literature
on
H.—W. Alten
Galois
Groups
Helmut
Koch,
Edwards,
Galois
Einfuhrung
factor
and
New
York,
Their
die
klassische
sides
to
of the
complete
History
2003.
1960.
Theory, New York,
in
both
by factor,
Algebra, Berlin,
Edgar Dehn, Algebraic Equations,
M.
divides
now
proceed,
can
(et al.), 4000 Jahre
Harold
If one
K.
1984.
Mathematik
I, Berlin,
1986.
Gerhard
Kowol, Gleichungen,
Stuttgart,
principles
Radloif, Evariste Galois:
Mathematica, 29 (2002), pp. 114-137.
Ivo
Jean—Pierre Tignol,
2001.
Galois’ Theory
1990.
and
applications,
of Algebraic Equations,
Historia
Singapore,
Exercises
123
__________.j__________:_______
Exercises
(1) Determine
using the Euclidean
sion
integers) the greatest
for
145673
(2) Show
two
and
that
2134197.
for
nonreal
contains
and
leaves
(3) In what
an
equation
zeros,
the
permutation
a
all other
Galois
the
coefficients
over
group
that
of the
steps
2
—
solution
and
precisely
the
rational
numbers
two
nonreal
solutions
of the
equation
0
=
is, over what fields between
Q and Q (i, Æ
take place? Furthermore, provide for each chain
That
arranged?
can
rational
exchanges the
solutions
unchanged.
the
can
ways
with
:34
be
Veralgorithm (in the “original”
common
divisor
of the integers
solution
fields the
of extension
associated
decomposition
of the
different
possibilities
Galois
group.
Hint:
such
Altogether,
there
are
seven
of field extensions.
chains
One
for the
associated
decompositions
of the
effort.
That
number
over
tensions
will
Determine
this
be shown
the
Galois
carries
in the
and
show
cess
in the
sions.
that
there
Galois
—
1
group.
Give
one
as
to
the
group
chains
number
with
some
of field ex-
cyclotomic equation
0
=
is precisely
Galois
this
chapter.
of the
group
$17
next
establish
can
for
stepwise decomposition
well
the
associated
pro-
field exten-
10
Chapter
Structures
Algebraic
and
Galois
The
Theory
author
the following entry on
of this book associates
Galois
theory in the Brockhaus
Encyclopedia, sixteenth
edition, with his fruitless attempts as a fifteen-year-old
to understand
student
why a general equation of the fifth
in radicals:
degree should have no solution
to Galois
“According
theory, solving an equation is equivalent
to the construction
of the fieldE over the fieldK
of coeflicientsof the equation formed by adjoining the
solutions.
The set of permutations
sought—for
of the solutions
a group
of the equation induces
of maps of E to
itself (automorphisms) that leave unchanged the elements
of K. By determining all the subgroups of this group of
the fieldE step
automorphisms, it is possible to construct
by step using the subgroups that correspond to intermediate fields. The advantage of such a method
is that the
relations
between
fields, with their two operations, addition and multiplication,
is replaced by relations
among
which have a single operation.”
groups,
10.1
This
What
is the
theory
and
last
chapter
relation
the
between
material
is meant
in
to
this
description of Galois
the preuious chapters?
two
bridge between
points of View on Galois theory, namely, the “elementary”
point of
View of the previous chapters, focused
on
polynomials, and the “modern” point of View that became
dominant
at the beginning of the twenserve
as
a
tieth
of our
discussion
we
shall show that this
century. In the course
modern
point of view, based as it is on abstract
algebraic structures,
125
10.
126
be easier
to
turns
out
level
of abstraction.
duction
to
Structures
Algebraic
to
understand,
and
despite,
indeed
or
To understand
and
appreciate
theory, the
reader
should
the
modern
the
equivalent
abstract
algebra
Theory
Galois
of, its
because
this
have
simple
had
a
intro—
semester
algebra
and be familiar
with such concepts
as
normal
group,
subgroup, quotient
group,
field,vector space, basis, dimension, homomorphism, and
Since this book is directed
at readers
who lack such
automorphism.
preparation in Whole or in part, the conceptual apparatus required for
an
will be introduced
outline
in bare
understanding of the material
as
of the chapter.
required in the course
or
course
As in the
mial
on
chapter,
previous
begin with
we
linear
and/or
nth-degree
an
polyno-
equation
ac" +
a.,,_1ac"
-1
-2
a,,_2:c"
+
+
-
-
-
+ alaz
+ a0
0
=
complex coeflicients an_1,
(L1, 0.0 and without
multiple solutions.
That is, the solutions
In
to be distinct.
assumed
.731,
,:cn are
contrast
to the previous chapter, however, the field extension
obtained
of the equations
to the field of coefiicients
by adjoining the solutions
will be in the foreground:
Again, We will begin with a field K that
contains
the coefficients
of the equation.
As we did in the previous
chapter, we adjoin to this field K all the solutions
ml,
,:1cn, thereby
field was
defined
obtaining the field K (:01, ,w,,). This extension
with
.
.
.
.
.
.
,
.
.
as
the
of numbers
set
to
the
the
same
operations
shall
.
that
elements
.
.
.
arise
from
of K
and
of numbers
applying
the
the
solutions
basic
$1,
.
.
arithmetic
.
,w,,.
As
we
if we
forgo division, so that
number
of the field K ($1,
every
,:n,,) can be expressed as a polynomial
in 1101,
coefiicients in K. We note finallythat the
,:z:n with
field K(:1c1, ,m,,) is called
the splitting fieldof the given equation,
since it is the smallest
field in which the equation splits into linear
see,
set
arises
.
.
.
.
.
.
.
.
.
factors.
The
10.2
central
ter, Where
subsets
We
of the
subject.1
(1831-1916),
limited
a
complex numbers
A brief
a
of
historical
in a paper
field was
ourselves,
We would
operations.
1From
notion
look
point
that
since
it sufiiced for
closed
under
like to
now
at
the
defined in the
chap-
purposes,
to
our
the four basic
go somewhat
arithmetic
deeply into the
general definition of a field,one that
view, fields were
in 1871.
It
appeared
of
previous
more
first defined
was
only twenty
by
Richard
years
Dedekind
later
that
the
10.
Algebraic
goes
beyond
and
fields.
As
mial
K
Structures
the
and
in the
seen
K (931,
.
.
.
is
Theory
given in the
bound
with
up
,a:,,). Therefore,
follows
all such
fields,and
the
need
group.
But
first,we
definition of the
Galois
group.
10.3
thus
the
We
have
used
the
“Galois group.” Since
term
deeply into
more
far
the
formal
definition of
Theorem
10.1.
on
come
up
follows
the
attempt
to
of
so
in terms
with
an
alternative
a
of
component
will
we
of permutations,
stating
lie between
do
as
“group”
in what
after
group,
to
groups
of a polyno-
shall
we
shall
we
word
composition
a
section
fields that
the
in what
classify systematically
Galois
127
previous chapter, the solution
is intimately
equation
Galois
complex numbers,
have
we
and
need
shall
we
to
go
present
a
following theorem.
Every Galois group, when taken as a set of permutations
together with the binary operation of composition of permutations, forms a group.
That
is, the Galois
Definition
A group
10.2.
satisfies the
group
of
consists
a
following definition.
G
set
binary operation 0 (that is, for all 0, r E G, it follows
such that the following conditions
are
satisfied:
o
o
0
The
binary operation
has
(0‘O’7')OI/=(TO(7'O1/).
The
set
is
G possesses
associative;
ooe=ofor
allo€G.
Each
element
0
of G possesses
the
operation
0
such
We have
that
asserted
7'
of two
permutations
belongs to the
Galois
group,
a
of several
notion
of
o
group
field was
tables.
extended
to
0'1
in the
and
The
o
reason
o
we
is, for
5
that
(7 o r
o, r,
1/
such
=
a”1 o 0
=:
and
the
is
that
Galois
demonstrated
for this
G)
G, one
o
a
=
8.
chapter
have
5
E
by o"1, under
previous
7” of
E
that
inverse, denoted
an
o
already
position
element
identity
an
that
is defined a
which
on
simply
the
group
com-
again
this
by means
that
the
set
of
other
mathematical
with
systems
analogous
even
those
that
could
considered
as
properties,
subsets
of the complex
numbers,
for example
Weber
by Heinrich
and
references
can
(1842-1913) in 1893. More details
be found
in Erhard
der Ga1ois—Theorie,
Scholz, Die Entstehung
in: Erhard
Scholz
(ed.),
Geschichte
der Algebra,
Mannheim,
1990, pp. 365-398.
a
encompass
not
be
10.
128
Algebraic
definition of the
polynomials BK in the
of 7‘ and
action
by the
itself
(00 TXBK)
Galois
group
of
again by the action
a'(7'(BK))
=
C
Galois
and
Structures
Theory
is carried
to
for
ar—
0:
U(BK) C BK.
verify associativity easily. Indeed, it holds
for three
and mappings.
Thus
permutations
bitrary functions
has
of the Galois
and for a given index j, one
group
One
also
may
(0
0((’F 1/)(.7')) “(T0/(J'))) (0 ’T)(l/(3'))
(T I/))(j)
0
cr,7',1/
0
0
=
=
0
(0 7') \xD0Ӓ
o
0
We shall
make
of the
use
follows
in what
law
associative
mostly
to
implicitly, namely by leaving off parentheses entirely. It is thanks
notation.
the associative
law that We can
indulge in such “imprecise”
is the
of the Galois
group
identity element
that leaves every index
that is, the permutation
identity permutation,
in its place. It is clear that
this permutation
belongs to every Galois
It
is clear
the
that
group.
as
obvious
permutation
7‘ with
It is almost
inverse
that
the
for every
permutation
desired
properties. One
k for each index
j, where the index
7'(j)
by cr(k)
j. Then we have the identity
:
other
hand,
it is much
less
clear
er
in the
Galois
group
its
inverse
7'
as
well.
The
Galois
group
of the
fact
0
o
0‘, 03
=
that
0
o
09, with
representation
crp
=
Thus
If
operation
the
0‘ o
0,
p
>
.
a
group
and
simplest argument
that
contain
two
terms
that
G contains
one
01”“;
that
element
inverse
inverses,
2
sequence
it follows
a
7
in the
subset
calls
U
On
id.
=
7'
this
makes
is so
use
cf, 02
of powers
=
From
that
are
equal.
id, so
that
is a
a1"“‘1‘1
group
section
U that
a
=
every
cr‘1.
:
Galois
every
appear
an
permutation
a‘1 necessarily belongs to the
the
shown
following theorem.
for
0
U
=
a
whether
must
.
q,
of the
have
.
o
7'
finite group,
examples of groups
Further
10.4
we
in any
is
simply defines
k7 is uniquely determined
:2
the
there
0
on
is closed
is indeed
a
groups
and
fields.
under
the
group
group.
subgroup of G. We will require
10.
Structures
Algebraic
and
Galois
Theorem
10.3.
|U|, the
number
If a finite group G contains
of elements
of U, is a divisor
elements
in the
group
ln the
special
and
by Galois,
subgroup U, then
the
of ]G{, the number
of
G.
of
case
Galois
a
this
group,
fact
first
was
recognized
implicitly by Lagrange previously.2 To prove
one
forms
for
left cosets
of o,
defined by
orem,
129
Theory
element
every
of the
0
G the
group
the
the-
collection
of
aU={oor|TEU}.
If two
products
that
r1
coset
has
0
0
and
7'1
0
o“1 0007120”
=
1
0
T2
o
a
precisely ]U| elements,
Furthermore, the cosets constitute
of the
partition
This
group.
in
o
oU
:
T2
and
then
it
follows
shows
that
every
equal,
are
This
72.
thus
size.
they are all of the same
a disjoint
(that is, nonoverlapping)
is so
because
if two
had
cosets
nonempty
a
intersection,
mUfl@U%b
there
would
exist
two
0272,
with
the
result
overlapping
cosets
=
0171
is, the
two
entire
group
same
size
of
G can
elements
be
that
o1U
are
in fact
into
partitioned
|U|, it follows
that
[G]
the
in
71,72
=
a
0
02
0
T2
'rf1U
one
and
set
of, say,
n|U|, and
2
U such
subgroup
so
the
n
that
a2U. That
=
same.
Since
cosets
all of the
]U| must
be
theorem
that
a
the
divisor
|G
We
draw
two
direct
consequences
10.4.
For
every
element
such
that
o"
can
of the
we
have
just proved.
Corollary
positive
of the
integer
number
The
n
=
of a, finitegroup
called
5,
the
order
G the smallest
of o,
is
a
divisor
of G’.
|G| of elements
truth
a
of the
if for a given elecorollary can be seen at once
ment
0 we
consider
the subgroup {e, a, 0'2,
The first duplication
`{@
07’
04 that appears
in this listing, which must
occur
since the group
is finite,must
0 and q
hold, as we showed in Section 10.3, for p
n,
that is, for e
0". The subgroup then contains
precisely n elements,
.
.
.
:
=
=
=
and
this
number
must
2The
quotient
\xC0R<
in
what
characterization.
ever,
follows
we
be
shall
is
a
divisor
called
not
use
of
JG
the
indea:
of the
this
terminology,
subgroup
U
in
in
of
a.
favor
G’.
more
Howdirect
10.
130
A group
10.5.
Structures
Algebraic
that
Galois
Theory
of elements
be listed completely as {e, o, 02,
n can
,o”“1} for a suitably chosen
0.
Such a group
is called a cyclic group
element
of order n.
Corollary
contains
and
.
.
The
a
other
{e, a, 0'2,
Since
1 and
k: is
is in fact
the
subgroup
Simplypchoose for
corollary is easily seen.
than
the identity element.
subgroup
lc >
.
of this
validity
element
any
number
prime
a
.
.
,cr’°‘1
}, where
.
of n,
divisor
a
k; is the
entire
order
it follows
of the
that
the
consider
Then
k
=
element
0.
and
the
n,
group.
the
knowledge that we have gained
to expand our
investigations of the previous chapter. Let us begin
in which the field K that we have been considering
with the situation
10.5
like
We would
extended
to
polynomials
B K
is
the
Galois
We
may
thus
field L.
a
the
use
The
add
of
set
a
divisors, as
previous chapter.
of the
have
we
of
operation
of
a
extends
group
are
0
The
integers
Z under
addition.
0
The
rational
numbers
Q under
R and
0
The
the
complex
nonzero
rational
real
nonzero
of
set
group
in
a
shrink
to
one
examples
in
the
with
the
must
numerous
Fields
and
far
Here
composition.
Galois
through
seen
Groups
concept
B L results
polynomials
of elements
number
of its
The
of the
expansion
consequent
shrinking of
in the previous chapter.
as
we
have seen
However,
and
this extension
that
must
lead to a subgroup,
to
group,
now
to
now
beyond sets of permutations
some
examples of groups:
addition,
well
as
as
the
real
numbers
numbers
C, also under addition.
numbers
and also the
Q)‘ under multiplication,
Rx and the nonzero
(CX also
complex numbers
numbers
,
under
multiplication.
o
The
n—dimensional real
0
The
set
of
n
X in
0
The
set
of real
n
real
vector
matrices
R" with
space
under
addition.
coordinatewise
addition.
matrices
with
nonzero
determinant
under
matrix
3 X 3 matrices
with
nonzero
determinant
whose
associ—
X n
multiplication.
o
The
set
of real
ated
mappings
take
the
set
of vertices
of
a
Platonic
solid
with
center
10.
Algebraic
the
Structures
into
origin
tions
and
itself.
is matrix
The
Galois
Theory
for these
operation
symmetry transformais equivalent
to composition
of
which
multiplication,
131
mappings.
For a positive integer n 2 2, the set of n
remainders
“modulo n” forms a group;
together with addition
example for n
3, it makes no difference
which
o
this
=
three
,n—1 }
is because, for
{ 0, 1, 2,
.
.
numbers
.
from
the
cosets
{O,3,...,—3,—6,...},
{1,4,...,—2,—5,...},
{2,5,...,~1,—4,...},
chooses
one
the
on
0
and
two
then
cosets
adds.
chosen, not
the
numbers
chosen
always depends only
themselves.
order
Z/nZ.
is, the cosets,
residue
classes
modulo
We have
n.
(isomorphic) form, namely
as
For
residue
prime number
a
the
excluding
the
for Platonic
mentioned
here
i.e., for
are
ze_ro
field is defined as a
usually by + (addition) and
The
set
ment
denoted
by 0.
The
set
K
{0} is an
K
is
—
abelian
an
identity element
The
The
obtained
from
at
-
multiplication.
multiplicative
matrix
groups,
'r
in the
K
0
a
group,
together with
addition,
abelian
group
by 1.
under
=
m-y+a:~z
=
the
symme-
all the
groups
is
o
7'
commutative,
a.
operations, denoted
that the following con-
under
is, for
7'
by
two
group
(y+z)
division
under
being denoted
law holds; that
distributive
K, the identity
classes
(multiplication), such
-
satisfied:
are
groups
a
by
called
are
group
0,
set
denoted
in an equivalent
groups
of nth roots
of unity.
solids, and the permutation
groups,
abelian, that is, the group operation
pair of elements
every
and
n
The
these
seen
the
class, form
of the
A
ditions
the
n,
exception
try groups
0
coset
is called
the cyclic group
or
group
The elements
of the group
Z/nZ, that
With
o
on
resulting
resulting
n,
o
The
any
with
the
identity
ele~
multiplication,
with
the
three
a:,y,z
in
elements
holds.
examples of a field are the sets of rational
numbers Q, real numbers
C under
R, and complex numbers
the usual
addition
and multiplication.
Another
example is the field (@(a,b, c,
constructed
\x80\x8A(
by the adjunction of complex numbers
to the
field of rational
a, b, c,
familiar
most
.
.
.
..
numbers.
example of a field is that of the rational
functions in the
variables
is a fraction
X1,
,X,,. Such a function
whose
numerator
and
denominator
are
polynomials in the variables
X1,
,X,, with coefficients
in a field K. The denominator
may not be the zero
polynomial, of course.3
Another
.
.
.
.
3For clarity,
is historical
and
it
not
should
entirely
be
out
pointed
since
correct,
that
a
the
rational
.
.
nomenclature
function
“rational function”
is a formal
of
quotient
10.
132
One
obtains
the
variables,
the
the
in
.
variables
field of all rational
.
X1,
functions
in the
polynomial equation
.
.
,X,,.4
The
extension
the
basis
for
forms
elementary
of the
field to
the
general
studying
of fields (and their
context
which
for
replaced by the
.
.
Theory
functions
rational
allowing only those
by A0,
,A,,_1, are
denoted
Galois
and
Structures
subfield by
a
polynomials
Algebraic
automorphisms).
Among the examples of groups
given above one can findisome finite
fields
to the other
contrast
In
a
field.
is
fields.For a prime number
n, Z/nZ
fields
finite
have
looked
the
we
of
subfields
numbers,
complex
at, namely
of the multiplicative
have the following property:
The n—fold sum
identity
of
or
of a finiteCh(l’I‘(1Ct67"iSt’iC,
1 is equal to zero.
One speaks in such a case
which is the designation
characteristic
to characteristic
zero,
n, in contrast
for all subfields of the
With
numbers.
modifications
some
can
one
complex
in
not
contained
fields
other
and
for
finite
fields
a
Galois
develop
theory
Galois
be of help in computing
can
results
the complex numbers,
whose
groups
of extensions
10.6
The
field of rational
of the
next
that
concept
numbers.
wish
we
to
motivation.
phism. We begin with some
morphism will enable us to characterize
from
extensions
field K to K (:31,
the
.
discuss
is that
of automor-
The
concept
of
Galois
the
.
.
auto-
of
in terms
group
of as
,:n,,,)instead
an
previously
equation; in particular, two equations with coefficients
Galois groups.
splitting fields have identical
having identical
in terms
in K
Up
of the
to
the
now,
elements
of the
sively as permutations
him
had
Lagrange,
tations
can
made
be viewed
as
of the
Galois
solutions.
viewed
were
group
exclu—
Galois, and before
However,
intensive
use
of the
fact
that
these
functions
that
map
each
value
that
permucan
be
51:1,
,3)” to another
by a polynomial in the solutions
if a polynomial
Thus
such value.
,X,,) is given with coefh(X1,
cr
one
ficients in the base field K, then
defines,for a permutation
belonging to the Galois group,
represented
.
.
a(h(:r1,
.
For
example,
the
function
a
.
number
value
0(2)
.
2
=
,:1:n))
:
h
represented
.
.
.
.
(a:g(1), mC,(,,)).
.
as
2
=
.
.
,
3311821133 is
mg
—
assigned
the given
general,
1Ei(2)—.'l3U(1)1l7U(2)$U(3).
In
sense
of the term,
a
function
in the usual
and
not
although a function
polynomials
of a rational
function.
be easily defined on
the basis
For
functions
that
are
all rational
4This field contains
every
symmetric:
a priori
to be
do not
need
and
denominator
whose
numerator
rational
function,
But
the fraction.
can
be made
symmetric
by extending
metric, the denominator
must
also be symmetric.
fraction
of the
extended
the numerator
can
such
sym-
then
10.
Algebraic
definition of
Structures
and
cr(z) makes
sense
Galois
Theory
only because
it is
133
independent of the
:0
polynomial representation
h(a:1, ,m,,), which is obviously never
unique. Namely, if one has two polynomials hl and hg with identical
:
.
values
h1(:1:1,. .,:L',,)
.
h2(m1,
=
.
.
.
.
,:v,,), then
.
the
difference
of these
two
polynomials belongs to the set B K that was used in the previous
the Galois
chapter in defining
And indeed, the Galois
group.
group
defined in such
was
a
that
way
the
difference
belonging to B K results
equality hl (:lIU(1),...,:I3U(,.,))
hz (:L‘c,(1),..
with
,:CO.(,,)),
in the
=
result
that
it makes
in terms
of
With
hl
difference
no
whether
the
given definition we have thus
the
:31,
this
proof,
.
.
.
obtain
can
,:r,,} belonging
splitting field K (931, ,m,,) if one
in the polynomials
in the solutions
1:1,
vanishing denominators
arise, since as we
always the case that a(y) yé0.
.
.
.
.
the
to
,
of the
extension
an
entire
.
of definition to
realm
allows
.
.
These
.
the
extending
But even
splitting field K(:1:1,
without
\xA3\x90
unproved fact, for the permutations
belonging to the
one
group
have
we
.
in
by polynomials in the
already mentioned, though without
is the
to this
Galois
As
,:L',,.
.
set
reference
of 0(2) is defined
succeeded
.
solutions
value
hg.
or
definition of permutations
of the set {$1,
Galois group
to the set of values
representable
the
the
.
rational
functions
,.r,,.
No
problems
shall
see,
for
.
extended
0 it is
#
y
with
of def—
in reference
permutations,
inition, now satisfy the properties that we Wish
of a field automorphism.
First, we see at once
to invoke
in the
from
the
definition of
the
hold
for two
extended
values
3; and
identities
permutations
two
representable
polynomially
z
U(y +
2)
to their
that
in the
domain
solutions
notion
arbitrary
$1,
.
.
,acn:
.
C(21)+ 0(2),
=
U(3/2) 0(y)U(z>~
=
More
difficult
that
the
mapping
5Given
ciated
with
by
the
Galois.
We
operators
a
value
polynomial
And
shall
soon
onto
the
a(y) 75 0 for
a
proof that
is
a(y)
resolvent
t,
analogy
g(t)
therefore
see
or(y)
to
=
0 and
one
has
a
0
=
argumentation
is
divisible
by the
g(T) also has t as
that
the
mappings
finite—climensional K-vector
1/ ;£ 0, every
one
hold
0 can
=
only for
y
0 and
=
invertible.5
with
y
Galois
g(T). In
a
polynomial
that
is the
of the
mappings
polynomial
a(y)
=
representation
g(t,),
that
used
y =
is a
is, t,
previously,
g(t)
asso-
of the
zero
that
we
have
irreducible
polynomial
a zero,
that
is, 0 = g(t)
that
we
have
constructed
Q5(T) constructed
are
linear
K(X1,
Therefore,
since
space
0'
is invertible.
.
.
.
,Xn).
=
it follows
y.
here
10.
134
it follows
Finally,
determining
.
determine
a
that
the
four
permutation
in
in this
(more usual
0,
defined on
the
y for
field K(a:1,
the
.
all y E K
For
j
.
the
.
Galois
term
,:1:,,),one
can
.
.
.
it
,
also
use
+
is, o(a:j) is also
a
-
-
of
sides
O,
=
j 75 k: one
solution
of the
a(a:j) +
o(—1)cr(:rk)
o(a:]~ ask) aéO,
Since
equation.
by
obtains
a1o(a;,~)+ do
+
-
function
both
on
a
thereby
one
a
mapping), denoted
a,,_1o(:n,-)”—1
o(:r,-)" +
chooses
conversely
serve
Given
group.
be the
1,
=
this, for
see
simply
one
0,
given properties
.
To
,
=
.
would
context
original equation.
that
o(y)
=
value
It is important
to
that
Theory
Galois
and
o(y) of a permutation
polynomial h(X1,
g.
,X,,)
the
constant
the
Structures
Algebraic
for
has
o(:I:,-) o(mk)
—
=
are
distinct,
,0'(.’I)n)
the
values
with
a
with
coefiicients in the
o(m1),
.
permutation
.
.
=
of the
solutions.
field K
that
so
For
and
—
is acutally
one
polynomial h(X1,
a
h(w1,
.
.
.
,:r,,)
=
h(o(m1), ,o(:z:,,)) o'(h(m1,..,.'z:n)) 0. Thus
.
.
=
.
defined by the
third
function
in fact
0
to
belongs
the
proof of this equivalence complete,
the
With
=
.
of the
characterization
definition in terms
of the
Theorem
A Galois
Galois
group.
dealing
.
.
.
,
Xn)
0, one also has
the permutation
Galois
group.
we
have
In addition
to
obtained
the
a
original
polynomial set B K and the characterization
in terms
of the Galois
on
comin the section
resolvent, as described
puting the Galois group in Chapter 9, we have the following theorem.
can
10.6.
be obtained
tomorphisms
all elements
The
term
Definition
ping
the
0
that
as
of a given equation
group
Aut
over
the
fieldK
all
authe
set
of
.,a:n)
]
(K(:1:1,
K),
splitting fieldK (231, ,ac,,) that leave unchanged
the
set
.
.
of the
of the fieldK.
automorphism
.
An automorphism
assigns to each value y
are
U(y +
.
is defined as
10.7.
following conditions
.
follows.
of a field L is an
E
L
a
value
satisfied:
Z)
=
=
0(2),
0(y)0(z)-
invertible
o(y) E L such
map-
that
10.
Algebraic
In
this
equation
and
Galois
Theory
of the
Galois
only implicitly, namely,
in the
form
K(ac1,
of the
solutions
third
characterization
appears
of K
tension
As
Structures
to
.
.
.
,$,,) in
terms
the
original
of the
field ex~
group
51:1,
.
.,:tn.
.
have
we
stated, two equations with coefficientsin K and identical
fields automatically
have the same
Galois group.
Another
splitting
advantage
taken
of this
characterization,
definition of the
the
as
third
Galois
which
nowadays is generally
is its universality. Auto-
group,
morphisms can be studied for field extensions
to a splitting field,though in such
a case,
to
135
us
here
are
We
now
10.7
be found
to
would
that
the
do not
correspond
properties
of interest
only in part.
like
to
investigate the properties of the Galois
where from now
on
we
shall adopt the definition of the Galois
group,
as
a group
of automorphisms
group
G
Aut (K(a:1, ,mn) I
K),
and therefore
we
no
that the solutions
longer need to assume
of the
equation are all distinct.
=
.
first of three
The
known
were
in
important theorems
by Galois.
essence
that
we
.
.
going to derive
are
If one adjoins to a fieldK all the roots 3:1,
in K
of an equation with coefiicients
then the set of values
splitting fieldK(a:1,
,a:,,) that remain unchanged by every
morphism in the Galois group is the fieldK.
Theorem
10.8.
.
We thus
need
automorphisms
have
Galois
of
of
terms
resolvent
consider
we
our
previous
t, for
which
as
cr
discussions,
have
we
:
g1(t¢7)>
~
value
Galois
a
way,
2
group,
where
IGI z
=
h(a:1,
=
one
.
.
sums
1G] is the
2m)
or
=
.
,z11n) that
-
then
a(m1)=g1(tcr)>
For
the
this
proof
a
the
proved
automorphism,
an
need
we
for
which
belongs
group
that
2
to
for the
0(2)
the
=
the
for all
z
field K.
We
proof aleady in the
solving the cyclotomic equation.6
mo'(1)
If
value
a
Galois
argumentation
special situation
6In
in the
that
,:v,,
auto-
.
show
to
o
the
seen
.
.
in
,
.
.
»---3
remains
-
these
most
simply
given
using the
properties
7:001)
2
be
can
Z
identities
971 (to)can
be
reformulated
as
0(mn):9n(tU)‘
under
all automorphisms
the
different
of the value
z
and obtains
representations
number
of elements
in the Galois
group,
,
Zh(a
.
.
unchanged
-:‘7(mn))Zh(91(t0)1---:gn(ta))i
=
,
:7
of
in
10.
136
10.8
It is high time
some
concrete
Structures
and
we
solidified the
theoretical
For
the
Algebraic
that
examples.
Galois
Theory
with
discussion
quadratic equation
9c2—6m+1=0,
the
two
solutions
used
are
the
extend
to
field of the
rational
numbers
Q to the field
\xF0\xBE\xA1
={a+b\/§Ia,bEQ}.
The
Galois
identity permutation
be extended
the
to
of two
consists
group
where
permutations,
solutions
exchanges the two
mapping (a, b E Q)
01
3 ::
the
non-
2\/§. It can
=a~b\/5.
o1(a+b\/5)
of
interpretation
example to offer another
of the Galois
That
groups
automorphisms.
is, the automorphisms
a
field is viewed
as
are
linear
transformations, where the extension
We
vector
wish
these
remaining
Vector
this
concepts
will
readers
with
the
field K.
base
certainly understand
limited
knowledge
calculations
coordinatewise
the
use
defined over
space
with
to
with
vectors,
readers
Those
what
of
and
we
analytic
perhaps
mean.
familiar
For
the
geometry and
even
matrices,
representation
°1l(Z)l
(51) (Z) (—‘2)»
=
=
must
is
more
we
\/5
1,
However,
suggestive.
{
},
stress
that such coordinatewise
a.re
of use only for a bit
representations
of emphasis.
The advantage in using constructs
from linear
algebra
consists
in allowing one
to avoid
such explicit representations,
which
always depend on the particular choice of basis. It is important only
in
such representations
exist
and that
of elements
that
the number
is independent
of the particular
of a
a basis
choice.
This invariant
as
the
vector
known
associated
with
space,
dimension, is of course
related
to
the
basis
be
values
the
sum
on
the
t,,
right—handside can
in terms
of the polynomial
has these
This
that
values
expressed
it‘, as zeros.
polynomial
is the irreducible
factor
constructed
from
the
by T
t, as was
by Galois
Q5(T) divisible
resolvent
in Chapter
the
Galois
9 on
nth—degree
equation
computing
(see the section
of this
polynomial
group). Since the coefficients
Q§(T)are in the field K, the value z
because
of
the
symmetry
in
the
—
also
belongs
to
If.
10.
Algebraic
every
other
extension.
field
to
equal
vector
10.9.
the
in E
such
in the
then
can
is the
size
of
a
find exactly
m
values
field E
in the
basis
of E
e1,
be expressed
can
field K
.
.
as
a
,em
.
uniquely
form
‘i’
klel
with
field E of a base
extension
that
m
is, one
value
every
degree of the
of the
speaks
following definition.
number
K ; that
137
Theory
one
degree of an
natural
that
Galois
Where
the
The
over
space
and
extension,
We have
Definition
is
Structures
values
For
(coordinates) k1,
.
.
,
'
‘i’ kmem
'
km from
field K.7
the
of Q is
degree of the field extension
Q \xB0\xE5\xA0
example is the adjunction of the fifth root of
the
example,
.
‘
equal to 2. Another
to obtain
the field Q(().
+ isin \xF0F\x9E
To show
unity Q" cos
that
the degree of the extension
from Q to Q(C) is 4, we consider
the
equation, already investigated in Chapters 7 and 9,
=
:I;4+:z:3+m2+:n+1:O,
whose
four solutions
basis
of the
division
these
vector
in the
four
the complex fifth roots of unity C, C2,C3,(4. A
which also allows a simple multiplication
and
are
space,
field Q(§),is the
four
1, Q, (2, C3. The proof that
values
values
actually form a basis is most instructive, since the
arguments used can easily be generalized. First, we observe that every
number
that can be expressed as a polynomial in the rational
numbers
and the root of unity Q is of the form Ic01+k1§+-+k_,C5 with rational
coefficients
-1
kg, k1,
However, since (4
C (2 (3, one can
with k4
0. In this case, the
always find a representation
k5
coefficients kg, kl, kg, 14:3are uniquely determined.
For otherwise, that
two difierent coordinate
for one
and
is, if there were
representations
the same
value, one would have for the root of unity C an equation of
at most
the third
coefficients.
Since the above
degree with rational
-
.
.
=
..
=
7Those who wish
following degree formula
Theorem
10.10.
extension
of E
L
For
over
for
is
of extension
towers
nested
a
K
“experiment”with
to
equal
tower
to
the
—
=
-
-
-
~
=
this
definition
fields:
of fields K C
product
of the
L
—
may
E, the
degree ofE
C
wish
prove
the
of the
total
to
degree
L
over
and
that
K.
over
To
prove
the
theorem,
one
selects
from
forms
all products
of one
element
is a basis
of the total
extension.
This
is used
in
degree formula
construction
see
the section
problems;
two
bases
each
basis,
the
on
for the two
and
convinces
intermediate
oneself
proofs of the impossibility
this
topic at the end of the
extensions,
that
of
the
chapter.
the
result
classical
of
10.
138
equation
fourth—degree
of an equation
solution
It remains
Structures
Algebraic
and
Theory
Galois
of its solutions
irreducible, none
of lower degree (see Theorem
9.7).
is
only to show
the
that
can
be
a
set
{Ice+ k1<+k2<2 +k3<3 I Ieo,k1,k2,k3 e Q}
is closed
under
Indeed,
for two
division, so that
polynomials f(X)
if g(()
74 O, we
have
no
gm
.
this
is in
set
and
fact
the
coefficients,
rational
g(X) with
field Q(§).
my (:2) 9 (<3)9 (:4)
g<<2>g
g<<4>‘
g<<2>g
I
side of this identity has the desired
right—hand
of the form hol + I
I+ k2C2—1—
representation
h3g3, since the denomiTo see this, one
could multiply out the denominator
nator
is rational.
for a number
of the form g(()
kgl + lc1C+ k:2(2+ k3C3. Fortunately,
one
can
avoid such drudgery if one uses
the generally valid argument
that will be presented in the following section.
The
fraction
the
on
=
To
10.9
Q(C), We
generalize
like
would
the
to
result
prove
that
the
for the
established
have
we
field
following theorem.
If one adjoins to a fieldK all the solutions
of an
in K, that is, $1,
the degree
equation with coejficients
,:1:,,, then
is equal to |G], the number
in the
of this fieldextension
of elements
Theorem
10.11.
.
Galois
group
To prove
resolvent
G
Aut
=
k1t+-
-
.
theorem, we shall use
is completely analogous
whereby every value
a quotient
in which
to that
field K (:31,
numerator
and
.
.
.
used
,a:,,) can
denominator
It, belong to the
for the field Q(C),
be
represented
take
the
field K.
form
by
k0 +
Moreover,
is
highest power
[GI 1, since \xA0>|
the degree of the polynomial ®(T),which has t as a zero.
It remains
to show that every
such quotient can
be expressed as a polynomial
in t. To that
K —vector space
in the |GI—dimensional
end, we consider
the
associated
E
+
+
kg,
K[t]
1
l
{ho
l
.,k1G|_1 K}
in
=
~
-
can
be restricted
using the Galois
argument
an
in the
-+kmt’"and all coefficients
the
.
(K(a:1, .,a:,,) I \xA0\xC1Z
.
this
t that
.
to the
value
-
.
.
mapping
h(t) E K[t]
»—>
g(t)h(t)
—
10.
Algebraic
Structures
and
Galois
139
Theory
defined by multiplication
by an element g(t) E K. This mapping is
linear.
Furthermore, the interpretation of the mapping as multiplication shows that in the case
element
is mapped
g(t) 74 0, no nonzero
to zero.
Results
from linear
algebra on systems of linear equations
tell
that
us
element
every
of K [t] possesses
preimage,
a
where
in par-
1 is the number
ticular, the preimage of the number
1/ g(t). This
preimage allows us to represent expressions with g(t) in the denominator
as
polynomials in t, so that every value in the splitting field
K (.731,
kn) can be expressed polynomially in terms of the Galois
resolvent
t. The theorem
is proved.
.
.
.
,
Moreover,
number
every
in
t
considering
in the splitting
=
+
mlml
-
-
+ mnccn,
-
one
field K (1121, ,:cn) can
.
.
that
sees
be expressed
.
as
Thus in adjoining the solupolynomial in the solutions
ml,
,:v,,.
the operation
of division.
This fact, which
tions, one can do without
was
but not proved, has now
been established.
previously mentioned
a
.
10.10
The
and
group
close
its
relation
underlying
that
.
.
have
we
field extension
observed
between
will
be extended
now
the
Galois
to
the
following theorem.
Theorem
10.12.
If one adjoins the solutions 9:1,
,:z:,, of an equation with coefficients in a fieldK
then in the resulting splitting field
K (:31,
,mn), the set of values that remain unchanged under the
action
of every automorphism of a subgroup U of the Galois group
G
Aut (K(:v1, ,mn) I
is
to
K
equal
only when U is the full
K)
.
.
.
,
.
.
.
=
.
Galois
.
.
group.
That
all elements
tomorphisms
ever,
is the
that
are
not
in K.
of the
converse:
of the
field K remain
subgroup
U is clear.
If aside
from
the
unchanged under all auWhat
is important,
how-
field K there
are
no
elements
unchanged by every automorphism of the subgroup U, then
this subgroup must
be the full Galois
In other
group.
words, every
proper
elements
subgroup of the Galois group leaves unchanged some
To prove
this
theorem,
begins with
one
lois group
in which the elements
in the extension
field K(:z:1,
.
tomorphisms
of the
Galois
.
of the
.
,
group.
mu) that
With
a
subgroup
field K
remain
the
are
U of the
Ga-
the
only numbers
fixed by all the au-
Galois
resolvent
t
one
10.
140
the
forms
Algebraic
and
Structures
Galois
Theory
polynomial
H (X aw)—
(TEU
If one
'r of the
applies an automorphism
subgroup U to the coefficients
of this polynomial, this corresponds
to a permutation
of the linear
remain
factors, so that the coefficients
unchanged.8 According to
our
lie in the field K,
must
coefficients
assumption, the polynomial’s
since
under
of U. If U were
all automorphisms
they are invariant
a
be a solution
t would
proper
subgroup, then the Galois resolvent
of an
less than
in K whose
equation with coefficients
degree was
the
number
the
fact
of elements
that
|G| of the
Galois
resolvent
t is
Galois
the
Galois’s construction,
that
is, of an
of
zero
a
irreducible
contradicts
which
group,
Q5(T) according
of
polynomial
A quite similar
line of reasoning can
be used to prove
noted
in Section
9.8 that the Galois group
G of an irreducible
always acts
transitively
on
of solutions
50,-, am, there
is an
which
a(:r,-)
10.11
The
to
us
Galois
the
solutions, that is, that
automorphism cf of the Galois
theorems
a
few
Given
our
in
prove
this
theorem
of Sections
10.7, 10.9, and
quick strokes
journey thus
K(m1,
.
.
theory
has
it
far,
immediate
no
(K(:1:1, ,a:,,) [ K ) and
.
,m,,). The
of all of
out
8For
also
.
.
fundamental
relation
distinct.
namely, in the form
9If one applies
polynomial
7'
an
0
(7'”11/)
o
=
the
to
a
sur-
question
establishes
it
of the
Galois
K
and
thereby makes a coherent
and examples of the previous
and
of U, it follows
that
'r
u
of U is obtained
automorphism
automorphisms
Furthermore,
every
as
of
theorem
investigations
our
theorem
come
may
al—
now
fields lying between
the
.
distinct
two
for
group
10.10
fundamental
the
in radicals.
equation is solvable
Rather,
one-to—one correspondence
the subgroups
between
Aut
fact
equation
for every
pair
an
group
the
a:k.9
theory.
whether
are
=
three
prise that
a
degree
K.
|G| over
low
to
0'1
0'2
0
0'1
in
and
this
'r_ 0 (T2
way,
V.
automorphism
7‘
of
H (X
the
—
Galois
to
group
the
coefficients
of the
o‘(:I:j)),
UEG
the
result
changed.
ity of the
solution
is
permutation
By Theorem
10.8,
original polynomial
wk,
a
must
appear
in
of the
linear
the coefficients
and
Theorem
the
product.
uncoefficients remain
must
lie in K.
Because
of the irreducibiland in particular
the
9.7, all the solutions,
factors,
so
that
the
10.
Algebraic
Structures
and
Galois
141
Theory
See, for example, Figure 9.1 in Chapter 9. Since subgroups
of a finite group
are
case
relatively easy to findflindeed,
in the worst
chapter.
one
can
out
try
fundamental
finite number
a
in
the
fundamental
is then
possible
on
which
intermediate
state
now
the
from
complete classification of intermediate
fields.
to obtain
special cases
immediate
information
fields,if any, arise by the adjunction of roots.
theorem
It
We
possibilities—one
obtains
of
a
theorem
of Galois
theory.
K of the complezt numsubfield
bers all the solutions
sun of an
mi,
in K,
equation with coefiicients
then the Galois
G
Aut (K(a:1, ,mn) 1 K) of the field6:17group
tension
thereby obtained, that is, the group of all automorphisms
of
the field K(m1,...,mn) that
leave the base fieldK fired, possesses
the properties
enumerated
below. In detail, these properties
are
with
regard to the intermediate
fieldsL, that is, fieldssuch that K C
L C K (:1:i,...,9cn), and to each such field the associated
subgroup
Aut (K(:ci, .,mn) I L), which
comprises all automorphisms
of the
Galois
that
leave
element
group
every
of L fizzred.
(See also FigTheorem
10.13.
If one
to
adjoins
.
.
.
a
,
=
.
.
ure
.
.
.
10.1.)
(1) The mapping
L)
with
the
spondence
and
the
that
associates
the
fieldL
intermediate
the
subgroups of the Galois
number
of elements
Aut
establishes
one—to—one corre-
a
intermediate
fields
G.
group
.
is the
number
.
.
K011,
2
.
.
of automorphisms
.,ym)
is
joining to K the roots yi,
ym of an equation
in K, all lying in the fieldK(:I:i,
,a:,,), then
.
.
.
,
.
Aut(L | K)
.
(K(:1:1,...,:tn)] L)
fieldL
intermediate
the
.
from L to K(:t1,
,:1:n)is equal
the associated
subgroup
in
This
of the Galois group.
fir every element
of L.
(3) If an
]
(K(;r1, .,.’13n)
(that is, a bijection) between
(2) The degree of the fieldezctension
to
subgroup Aut
.
.
that
obtained
with
the
by adcoefiicients
Galois
group
|G|/|Aut (K(a:1,...,:1:n)| L)| automor-
contains
automorphisms
of this Ga—
lois group
Aut(L ] K) by restricting the domain of definition
K (:31, .,£l3n)
of the automorphisms
belonging to G to the inOne
phisms.
.
termediate
thus
can
obtain
all
.
fieldL
:
K(y1,
.
.
.
,ym).
10.
142
Algebraic
and
Structures
K'(w1:-Hymn)
U
n
U
n
~13» U=Aut(K(¢v1»---,$n)lL)
U
n
U
r1
K
The
The
.
.
ring to the theorems
We investigate the
Galois
fundamental
already
G
=
K
with
C
the
(K(a:1, ,rvn) [K
Aut
.
consists
proven
in Sections
of
intermediate
an
theory:
that
correspondence
theorem
extension
K ($1,
fields L such
is, the
group
Galois
of
theorem
in one-to-one
,a:n), are
U of the
A proof of the
,:I:,,)
.
fundamental
fields L, that
intermediate
subgroups
L(ac1,
G=Aut(I((:l:1,.--;5Bn)|I{)
10.1.
L C K (:01,
Theory
{id}
L
Figure
Galois
.
.
basically in refer10.7, 10.9, and 10.10.
field L to
field
the
It remains
(En)using those three theorems.
to note
that the fields L(a:1,
,m,,) and K(:z:1, ,.'1c,,)are identical,
since the field L is simultaneously an extension
field of K and a subfield of K(a:1,
,a:,,).
.
.
.
=
.
.
.
,
.
.
With
.
these
.
.
.
.
.
.
observations
of the
out
part
one
way,
of the
asser-
field L is uniquely determined
tion, namely, that an intermediate
by the associated
subgroup Aut (K (ml, .,:c,,) | L) (that is, that
the mapping is injective), is clear, since it is a consequence
of Theorem
as
the set of
10.8, according to which L can be characterized
all elements
of the field K (:31,
:3”)that are fixed by all automorphisms of Aut (K(w1, ,a:,,) ] L). To show that the mapping is
in fact a bijection, we need to show that
for every
subgroup U of
the Galois group
there
is always a corresponding
intermediate
field L
such a field
(thus that the mapping is surjective). We can construct
L with the desired
Aut (K (:51,
property U
,w,,) | L) by taking
all elements
of K(a:1,
337,)that are fixed by every automorphism
in U. That
is,
.
.
.
.
.
.
.
,
.
=
.
.
.
.
.
.
,
L={z€K(a:1,...,:1:n)|a(z)=zforalla€U}.
That
the
set
L thus
struction
it is referred
checking
that
for
every
defined is in fact
a
field—on account
fixedfield-can easily be
of L, their
pair of elements
sum,
to
as
a
of conshown
by
product,
10.
Algebraic
Structures
and
quotient,
difference
field L contains
now
the
comes
associated
and
with
Galois
again in L.
are
the
field K
and
is
main
point:
The
group
the
intermediate
the
group
Aut
(K(a:1,...,:1c,,)[ L).
It is clearer
subfield of
a
143
Theory
U
,aJ,,). And
(K(:n1, ,:z:,,) | L)
K(m1,
Aut
=
fixed
the
that
.
.
.
.
.
field L thus
.
contains
constructed
in U fixes
U, since every automorphism
element
of
every
L. Moreover, this subgroup has the property,
given the construction
of the field L, that
of L are
fixed by all the automoronly elements
phisms of the subgroup. According to Theorem
10.12, this has the
that the subgroup U must
consequence
equal the entire Galois group
The
second
part
Theorem
10.11
applied
The
third
part
that
we
have
tion
of the
fundamental
theorem
follows
the
extension
of L to
K(a:1,
of the
fundamental
theorem
refers
examined
by means
to
of
number
a
the
at
.
.
.
,w,,).
to
of
from
once
the
situa-
examples
in
chapter when all the solutions
yl,
,y,,, of a resolvent
of such intermeadjoined to the field K. The elements
diate
fields L are
of the special condimapped again to L because
tions
on
all automorphisms
0‘ of the
Galois
G. To see this,
group
one
need
0
of the Galois
only apply such an automorphism
group
G
Aut (K(:v1, ,a:n) I
to
K) both sides of the underlying equaprevious
equation are
.
.
.
=
.
tion
of the
Since
L
by a
Galois
K(y1,
Aut(L I K
a"1
o
with
in this
7
the
.
.
another
to
.
This
shows
solution
mapped
G, for
group
2
striction
field L.
field L is
the
definition from
its
.
intermediate
yj is mapped
of the
.
each
a(yj)
.
two
:
.
.
the
an
.
way
follows
field extensions.
completely
proven.
most
The
simply
fundamental
That
one
from
field
intermediate
automorphism of the group
6 G yield the
same
re0,7
identity on L, that is, when
precisely when a‘1 o 7' is the
belongs to the subgroup Aut (K(m1,
field L.
solutions
yk.
,:1:,,) to
automorphisms
intermediate
of the
by all the automorphisms
in G one
can
restrict
automorphism
,ym), thereby obtaining
Here
each
itself
to
field K (331,
the
that
the
theorem
.
obtains
.
,
537,)| L) associated
all
automorphisms
degree formula for
of Galois
theory
nested
is
now
10.
144
The
Algebraic
Fundamental
and
Structures
Theorem
An
of
Galois
Galois
Theory
Theory:
Example
fields between
example of computing all the intermediate
field and the splitting field,we return
to the biquadratic
equation
made
use
of in Chapter 9, namely,
As
the
an
base
that
We
w4~4a:3~4ac2+8a:—2=O.
The
four
solutions
of this
equation
are,
stated,
previously
as
UI31,3=1+\/§:i:\/3-l-\/3,
:v2,4=1—x/2d:\/3—\/2.
The
Galois
If the
eight elements.
the
solutions
determined
in Chapter
9 to be
elements
are
considered
permutations,
was
group
as
a
with
group
they
act
on
determined
In
ad-
group
by trial and error.
consisting of the identity
alone,
follows:
1234
1234
3214
1432
3412
2143
4123
2341
074321
The
dition
to
possible subgroups
the
whole
group
can
and
be
the
have
either
two
or
four
subgroup must
subgroups arise from each of the five elements
any
elements.
of order
The
two-element
two:
{(70501}: {O-0102}: {OM03}: {(70:04}: {00>07}easily find, using the
three
subgroups with four
In
addition, one
previous chapter,
can
{ao,a1,a2,a3
It
remains
four,
while
the
group
(Z/2Z)2.
to
other
According to the
to-one
correspondence
two
that
the
last
four-element
fundamental
between
this
{ag,a3,a5,05
of these
groups
subgroups
theorem
collection
in, the
computed
elements:
}, {ao,a3,U4,cr7},
observe
table
group
of Galois
are
\xA0\x9CI
cyclic of order
isomorphic to the
is
theory,
of subgroups
there
is
and
the
a
one-
fields
10.
Algebraic
Structures
that
lie between
the
and
Galois
145
Theory
fields Q and
two
(V3 .
Q(.7;1,a:2,m3,a34) Q
x/E,
+
=
Moreover, these fields can be determined
ing the fixed fields. To do so, we shall
from
the
use
the
subgroups by determin-
identities
1
\/-2-=
-332-i-033
-1104),
Z(w1
%(501—fl33),
V3—\/§=%(9»’2*
\/7:
»"03)(fC2
-334)$0101
\/3+\/5:
—
With
the
these
identities
automorphisms
tabulated
fixed fields can
be
determined.
the
inclusion
10.12
Now
that
theory,
we
together
with
relations
among
have
We
would
bit
a
under
As
permutations.
direct
a
In
the
fundamental
more
about
seen,
Aut
group
theorem
of Galois
As we
significance.
corresponds
precisely
its
field L there
for every intermediate
subgroup G of the Galois
numbers
Figure 10.2, the subthe corresponding
intermediate
fields
the various
objects.
proved
like to say
of these
images
the
showing
one
the
from
represented
are
groups
determine
can
directly
the
consequence,
have
we
(K(a:1, ..,:nn) ] L) and
.
conversely.
field L will not
A particular
intermediate
to itself
by an automorphism
image a(L),
as
one
another
intermediate
And
subgroup
group
this
U associated
can
of the
0
a
in fact
can
with
subgroup
G. But
group
mapped
since
the
readily check, is necessarily a field,and thus
field, it must have a corresponding subgroup.
L.
be
directly
the
must
from
the
be
a
conjugate
automorphisms
the automorphisms
leave
an
element
This
in
group
Its
U all the
subgroup U, then U is called
to the third
here, in reference
Galois theory, this implies that
determined
Namely,
subgroup of the form 0U (F1.
0(2) unchanged precisely when
U leave the value
2 unchanged.
If for
Galois
be
necessarily
is illustrated
conjugate
a
normal
sub-
of the
subgroup
Figure
10.3.
subgroups are simply the
subgroup. In the situation
item
in
the
every
automorphism
fundamental
of K
of
theorem
(231, ,mn)
.
.
.
10.
146
Algebraic
Galois
and
Structures
Theory
Q('\/3+«/'2—,«/3—\/E)
V
///I
,4,,./
/
"
\__’
///
~~~~~~
,4
_\_
Pm\x91
Q0/3—«/5)
one/3+J§)
L;
0295:»/5)
Pg\x90
1
/;;://i*
rx}"//
5
Q(J§)
;
~’\“‘>§;.:.
mil)
aw?)
Q
with
1,,=
ll)(\[2—(\l3+\/§’+\/3*
L2=
and
The
10.2.
Figure
to-one
L2
of the
subgroups
with
correspondence
the splitting
field.
the
Galois
are
group
fields between
the
in
base
one-
field
of L.1° We have already
yields an automorphism
0
mentioned
that two automorphisms
and 7' in Aut (K(a:1, ,m,,) ]
K ) are
cr“10 7' is the
the same
when restricted
to L precisely when
identity mapping on L and therefore
belongs to the subgroup U
restricted
to
L
.
.
.
=
Aut
Then
I
(K(:1:1,...,a3n) L).
once
recognizable
as
1°An example of
leading to two distinct
example
(see Figure
a
the
set
subgroup
of the
that
equal
to
be
found
is
not
fields, can
Q
Aut(L | K) is at
group
G / U of cosets
intermediate
10.2), namely
Galois
the
a
subgroup
thus
subgroup,
conjugate
in a previously
examined
(V \xE0\xA8k
(
\x80at
3 +
and
Q
U, where
3
—
10.
Structures
Algebraic
and
Galois
K(x1,...,x,, )
L
{id}
/U\
“""’
a(L)
/n\
\x90‫ױ‬
\“/
K
G=Aut(K(x1,...,x,,)lK)
10.3.
The
fundamental
Figure
showing the relationships among
images a(L) under automorphisms
their
associated
Due
subgroups.
dence
between
intermediate
holds
precisely
when
the
be derived
10.13
There
is
U
theorem
of the
0
from
the
Galois
theory,
fields L, their
Galois
and
group,
one—to—one correspon-
the
to
fields and
:2
of
intermediate
subgroups,
L
within
the
a(L)
=
aUa'_l.
of automorphisms
composition
K) can
GU04
U
\“/
of course
147
Theory
of the
composition
group
Aut(L |
group
G.
construction
principle at the heart of what we
have just described
that
can
be greatly generalized, even
beyond the
bounds
of Galois
theory, by defininga composition for the cosets
formed
from a normal
subgroup U of a group G:
a
(a1U)
The
fact
that
U is
definition is well
on
the
choice
of
result
o
normal
a
defined.
of the
(a2U)
(01 a2)U.
o
=
that this
subgroup is what guarantees
That
is, the definition does not depend
elements
01
and
02:
One
must
check
that
the
remains
unchanged when 01 or 02 is replaced
7' in
U. For 02, this is
by 01 o 7' or 02 o '7' for some
automorphism
clear.
o
of ((01 o 7‘)
However, for 01, it holds only on account
a2)U
(010020 (a2'1o*roa2))U, with a2"1oToa2 E U for7' E U.
a
composition
=
Much
less
thus
composition
in its
own
right.
made
use
of such
group
nZ
is
abelian
We
previous
than
surprising
defined leads
In the
section
notation
obviously
with
a
normal
such
a
to the
on
definition is the
set
groups
of cosets
and
fact
that
G / U being
fields we
have
the
a
the
group
already
example Z/nZ, where the
the group
subgroup because
subZ is
(commutative).
have
such
already encountered
quotient
chapter, in the form of decompositions
groups
G/ U in the
of the
group
table
10.
148
(see also
Figure
theorem
of
anisms,
when
the
Algebraic
10.4). Moreover,
Galois
theory
We
restricted
field K(a:1,
.
.
.
,a:,,) to
the
proof
of the
been
involved
with
in
have
we
automorphisms
of the
intermediate
field L
the
Galois
and
Structures
Theory
fundamental
Galois
group
K(y1,
=
mech—
such
.
.
.
of
,:vm):
subgroup U was a normal subgroup, the intermediate
field L was
mapped to itself by all the automorphisms of G, so that
the restriction
of the definition to L yields an automorphism
group
the
Whenever
Aut(L | K) with
4
Galois
from
9, indeed
for
figure. Above
is shown
the
from
the
results
as
of the
decomposition
4 squares,
of Chapter
X
9.1
The
The
10.4.
Figure
elements.
0n
table
group
into
four
the
in Figure
example shown
in that
the first adjunction
shown
quotient
group
G/ U constructed
cosets.
great
of Point
significance
3 of the
theory
(Theorem 10.13) is
above
be
extensions
can
tensions,
and
situation
is achieved
broken
indeed,
in
that
down
such
a
into
a
Way
that
satisfies the
fundamental
all
that
initial
the
of
field
complicated
of suitable
succession
with
theorem
subeX—
first extension
assumptions.
Here
a
the
10.
Structures
Algebraic
for
requirement
all the
the
an
intermediate
of
an
solutions
normal
and
Galois
field resulting
with
equation
of associated
overview, Figure
10.5
the
theorem
theory together
G
of Galois
Aut(K(x,,
=
10.5.
The
with
fundamental
field L and
intermediate
that
U is
in K
involved
in the
of its
some
to
clearer
a
fundamental
assertions.
determine
the
of
situation
in which
the
achieved
through
Artin’s
Version
the
an
subgroup U. In the
(or equivalently, if L is a
Aut(L | K) can also be dequotient
group
G/ U. We have yet
properties
the
theory:
group
which
acterize
of Galois
associated
subgroup
Galois
as
theorem
the
normal
a
splitting field),the
termined, namely,
to
adjunction
is equivalent
For
subgroups.”
objects
of
the
x,,)|K)
...,
Figure
case
shows
from
coefficients
property
group
149
Theory
a
of
adjunction
normal
subgroup
extension
of K
root
K‘/E.
a
of the
Fundamental
Galois
Theory
to
U charL
be
can
Theorem
of
theory, Emil Artin presented a proof of the fun—
damental
theorem
of Galois
from the one
theory quite different
presented
here.” In contrast
to earlier
proofs, Artin’s proof is accomplished without
use
of the Galois
resolvent
use
of the corresponding
theorem
of
(or without
the primitive
element). Instead, Artin built up Galois theory in a Way that
In
his
1942
“For
normal
a
3‘
1,
=
.
.
.
book
Galois
field
intermediate
of the
subgroup
the equation
,7n,
an
L
Galois
K(y1,
,ym) whose
group
Aut(L | K), one
=
H
0
whose
Then
L
:
coefficients
in
addition
K(y1,
12For
Weber
bis
.
.
.
in
are
to
yl,
the
.
.
,3/m) having
.
EG
.
(X
—
be
cr(z/1))
=
form,
may
subgroup
for
each
U
is
index
0:
of whose
U'(yj) can
solutions
belong to the field L.
be adjoined,
without
the field
enlarged.
see
B. L. van
der Waerden,
discussion,
Emil
Artin, Archive
for History
of Easact
a
associated
.
/ U
field K and
all
the solutions
, ym,
to
.
Die
Gal0is—Theorie
Sciences,
9
(1972),
von
pp,
Heinrich
240—248.
10.
150
Algebraic
and
Structures
Galois
Theory
possible to prove the fundamental
theorem, reformulated
only in
of its assumption,
some
using exclusively concepts from linear algebra. In
of linear
of systems
equaparticular, he investigates the set of solutions
makes
it
tions.
An
argumentation
involving polynomials and their solutions
for a proof of the fundamental
in this variant
theorem
unnecessary
it is, of course,
for its
is then
(though
subsequent interpretation).
The situation
investigated by Artin is the following: Given a field13E
and a finite group
G of automorphisms
of the field,a field K is constructed
of G.
fixed by all automorphisms
of E that
are
consisting of those elements
For this situation,
Artin first proves,
systems of
by analyzing various
that
equations
to
E
is
equal
.
degree of the field extension
constituting
[G[ of automorphisms
he constructs,
to the number
that
from
the
the
K
group
G.
With
this
theorem
of Galois
theory is
terminology, it goes
G of automorphisms
following assertions
hand, Artin’s version
relatively easy to prove.
like
In
this:
of this
the
In
of
a
a
theorem
simplifiedas
version
a
situation
field,and
fundamental
of the
in
field E,
to
finite group
a
fixed field K
for the
intermediate
field L with
group
G, the
hold:
o
The
o
in G that
are
the identity on L is a bijec~
automorphisms
the intermediate
fields
tion, that is, a one—to—one mapping, between
and the subgroups of G.
The degree of the field extension
of L to E is equal to the number
of
elements
{U}of the subgroup U.
The subgroup U is a normal
subgroup of G precisely when the field K
can
also be represented
as
a fixed field of a group
of automorphisms
o
group
U of
of the
intermediate
10.14
In
fill
some
To
this
Galois
have
that
mapping
the
each
associates
sections
of this
chapter
holes
left
in the
exposition
of the
we
will
consider,
in
light of the
fundamental
theory,
what
structure
end
in order
that
sub-
field L.
remaining
of the
the
the
a
equation
Galois
group
be solvable
of
we
would
previous
an
to
chapter.
theorem
equation
in radicals.
like
In
of
should
particular,
in Point
3 of the theorem
to the case
specialize the situation
of a successive
adjunction of roots, Where each radicand
belongs to
a field that
arises
from one
of the previous adjunction
steps. Since
root
so
that
expressions can always be reformulated
only roots with
we
must
13The theorem
complex numbers.
is
valid
for
all
fields, not
only
those
that
are
subfields
of
the
10.
Structures
Algebraic
and
begin with the
{/5 is adjoined to
root
3 of the
solutions
If
of the
one
all nth
that
,
be
to
this
ac"
of unity, to
a
we
case,
field K, which
for
field L
by adjoining
coeflicients in K, then using the
determine
Whether
of
root
this
extension
(‘/5of some
be obtained
can
number
in
a
field K,
some
K.
must
that
field K
0 lie in the
=
conditions
assume
in the
already
a
—
its
applicable,
is the
C"”1are
the
with
nth
.
extends
now
roots
.
equation
equation
an
a
ensure
.
attention
our
of a one—step
adjunction, in which an nth
field K, with the radicand
a in K.
For Point
theorem
unity §,(2,
of
roots
restrict
to
us
case
fundamental
be satisfied. To
151
Theory
it suflices for
prime degree are involved,
to prime degrees.
We
Galois
so
the
nth
that
all
( \x81
prime
n
contains
all the
solutions
Galois
group
of
one
an
can
through adjunction
Namely, we have the
following theorem.
Theorem
Given
10.14.
fieldK
a
that
(,(1,...,C""1
for a prime number n and
from the adjunction of all solutions
arises
all nth
contains
roots
of unity
field L that
equation with coef-
extension
an
of an
ficients in K there exists in L an nth root (‘/5such that a E K but
(/5 ¢ K and such that L
\xB9& if the Galois group Aut(L | K)
is cyclic of order n, that is, ifAut(L |
K)
{id,cr,02, ,o”‘1} for
a suitably
chosen
automorphism 0.
,
=
::
.
To prove
In this
case,
this
theorem, we begin with the situation
automorphism o of the Galois group
every
Aut
is uniquely
account
determined
of
(0 ( p5
C“3/5 for some
by 0 to
compositions
of
=
0(a)
effect
=
a,
on
the
the
element
lc.
exponent
=
.
L
K
=
( \xC0$
(K ({/6) I K)
element
Note
{‘/a.
{/5 must
Moreover, exponents
be
that
on
mapped
add
under
CI“
a({/E)
{/5 and r( (E)
("+1 \"/E. The Galois group
(TOG) u
If
automorphisms:
(j (/5, then (007)
therefore
by its
.
=
=
=
“corresponds
to,”that
is, is isomorphic to, a subgroup of
the cyclic group
Z/nZ. Since n is a prime number, the Galois group
must
consist
of one automorphism
or
of all n of them.
Since {/5 ¢ K,
the first possibility is excluded, so that the Galois group
can
be given
as
Aut
(K({‘/al)
p\x96 {id,cr,a2,...,a"_1}
=
10.
152
Algebraic
Galois
and
Structures
Theory
a({/5) =C\”/E.
with
of K to a field L
begin, conversely, with an extension
is equal to Aut(L l K)
Whose Galois
group
{id,a,a2, .,a"‘1 }
b of the field
for a suitably chosen
For every element
0.
automorphism
L one
can
form the Lagrange resolvent, which we met in Chapters 5
We
now
=
and
.
.
7:
(c, b)
From
b +
=
cow)
one
definition,
the
a((<,12)) aw)
=
and
thus
one
finds an
is not
equal
automorphisms
(C,b)”,so
=
that
-
-
<"“1a"~1(b>.
+
~
the
-
=<‘1-(ab),
<""1a”(b)
+
-
identity
field L whose
b in the
to
zero,
then
since
other
than
the
identity
From
the
Lagrange
=
leaves
field K.
be in the
(C, b)" must
element
((§,b)) unchanged.
field K
+
+
«
obtains
immediately
cam») c2a3(b>
+
((C,b)")
(I
<2a2(b>
+
+
resolvent
If
(C,b)
none
C‘j(C,b),
of the
all the
of the
elements
of Galois
theorem
fundamental
subfield of
not be a proper
theory, the field K ((C,b)) can therefore
L; that is, K ((C,b)) L. Therefore, the field L arises from the field
K by adjunction of an nth root
of an element
of the field K, namely,
=
0
(C,b)"-
=
still
We have
to
that
prove
one
always find in L an
can
b
element
Lagrange resolvent
, b) does not Vanish. We can extend the
for
selection
to Lagrange resolvents
with
k
1,
2,
,n
1,
((k, b),
an
from 1, since such Lagrange
arbitrary nth root of unity different
can
be used in accord with the previous
resolvents, if they are nonzero,
construction.
If one
forms the sum
now
of the Lagrange resolvents
whose
=
(ck,b)
for the
cw») <2ka2
1»+
=
exponents
"'+C("“1)j:0
k:
+
=
0,1,
.
.
.
,n—
+
-
-
+
~
.
.
,
<<"~1>’“aH
obtains,
1, one
—
.
since
1 +Cj +C2j +
(forj:1,...,n—1),
7?.
I
»—t
(ck,b)
=
nb.
a H )—‘
Were
1, 2,
since
the
.
.
.
,n
all
—
the
1 to
only the
value
Lagrange
vanish, then
first summand
(1,b) remains
(gk,1))
resolvents
one
would
have
in the
above
unchanged
under
for
the
the
equality (1, b)
sum
all
would
k
exponents
remain.
automorphisms,
::
2
nb,
Since
the
10.
Algebraic
number
Structures
(1, b)/ n must
b not
belonging
leads
Galois
field K.
lie in the
K
to
and
Theory
Thus
at
least
one
can
prove
an
to
10.15
From
making
use
this
theorem
we
again of Point
3 of the
This corollary answers
theory.”
in which
for the
be generated
solution
of
by adjunction
a
nonvanishing
Lagrange
immediate
question
equation,
an
of element
fundamental
the
of
choice
every
(§’°,b).
resolvent
153
Such
root.
theorem
to
as
an
a
corollary by
the
of Galois
circumstances
intermediate
field can
field is called
a
radical
extension.
If a fieldK containing all nth roots of unity for a
is extended
to a fieldL by adjoining all solutions
of
in K
then there is a normal
coefiicients
subgroup U
10.15.
Corollary
n
prime number
an
equation with
of the
Galois
G
group
the
emtension
but such
that
in
,
fieldL there
its nth
We have
ezctension
recall
only to
ready been
mentioned
9.7). Whether
the
is
an
that
in the
extension
as
the
unsatisfactory
solutions
of
of
roots
if we
are
b that
of K lying within
the
criterion
itself in K
it follows that the
L
just formulated
of group
tables
extension
then
be read
analogously
Because
unity, the
of the
criterion
seeking a direct
must,
answer
can
necessary
to
al-
be
from
cor-
the
assump-
however,
as
has
(see Section
first radical
off
when
is not
of the
=
about
n|U| precisely
2
(from which
terminology
respondingly decomposed can
subgroup U
Aut(L I K
tions
|G]
element
I)" is in K
power
fieldK (b) is a radical
with
Aut(L | K)
::
be
whether
seen
the
of radicals.
We
equation can be expressed in terms
need particularly to explain what change in the Galois group
is caused
when the required roots
of unity are
adjoined.
10.16
To
an
be
able
to
the
theorem, the roots of unity
of the relevant
degree must already belong to the base field K. To
be sure,
in the examples
of the previous chapter we often
had the
rational
numbers
Q as the base field,on the basis of which we initiated
our
of the Galois
To
investigations into determination
group.
use
the previous
first to adjoin appropriate
theorem, it is necessary
elements—elements that
do not
necessarily belong to the splitting
the
“Including
extension
use
discussed
previous
in
Section
10.13.
10.
154
field K ($1,
of the
.
.
,:1c,,)of the
.
field K, there
of the
instead
Algebraic
appears
field K (:31,
splitting
Galois
group
takes
.
,:r,,)
now
see
.
field K’(a:1,...
,:Bn). We
the
field K’, and
extension
an
shall
.
place, as illustrated
in
of the
Theory
Then
investigation.
under
equation
Galois
and
Structures
analogously,
current
how
the
Figure
10.5-
instead
equation,
the
in
change
.
K,(x1,...,x,1)
Aut(K’(x1,...,x,,)|K’)
U“wwm‘»QVVe~
K(x,,...,x,,)
K)
Aut(K(x,,...,x,,)l
KI
K
10.6.
Figure
the resulting
An
Extension
in the
change
automorphism
of the
of the
Aut
Galois
“new” Galois
the
field K ($1,
.
.
.
.
a
group
.
solutions
51:1,.
the
.
Galois
Thus
.,:cn.
of the
restriction
,:v,,). Then
it is
of def~
domain
group
(K'(a:1, ,:c,,) IK’)
subgroup of the original Galois group Aut (K(:v1, ,m,,) | K ) .15
Aut
is
K’ and
field
a
(K'(:r1, ,a:,,) I K)
.
by its images of the
uniquely determined
by the possible
to
to
group.
is determined
inition
field K
base
.
.
.
.
10.17
The
cyclotomic
equation
m"
1
.
.
0 with
prime degree n represents
a very
instructive
and at the same
time important
application
of these considerations.
Indeed, we have already mentioned, in Chapter 7, the stepwise path to solving such cyclotomic equations,
but it
is appropriate
now
to do this again in a more
In
systematic fashion.
particular, we wish to use the results of the previous section to prove
the
=
following theorem.
15It remains
sion
—
of the
original
field
observe
to
K
investigations.
to
the
that
this
field K’, is
subgroup
just the
relation,
relation
which
that
results
Galois
from
established
an
exten-
in
his
10.
Algebraic
Structures
Theorem
10.16.
The
degree n
is solvable
in
represented
by an
and
Galois
cyclotomic
Theory
equation
as"
155
1
——
0 with
=
prime
radicals; that is, each of its solutions
expression involving nested roots and rational
be
can
num-
bers.
If
(_fis
tension
nth
an
from
of
root
unity different
Q(§) is of degree n
Q to
from
1, as
—
1, then
the
field eX~
have
seen
already
we
10.8, where (,C2, .,C”‘1is a vector-space basis. Each
of the automorphisms
0' of the
Galois group
is uniquely determined
by the value of o((), which is always a power of C. Therefore, the
Section
in
Galois
.
group
determined
To
.
Aut(Q(€)
[ Q) consists precisely
by ok(()
(k for is 1,2,
,n
=
=
in radicals
expressions
.
automorphisms
1.
—
.
obtain
of the
.
for the
solutions
of the
fields of the extension
equation, we seek intermediate
Q(() that correspond to the steps of a solution in radicals.
tomic
the
fundamental
theorem
of Galois
theory,
there
exist
cyclofrom Q to
Based
on
corresponding
subgroups of the Galois group Aut(Q(() I Q). However, to determine
these
is by no means
simple, unless the powers
Ck are represented
again as in Chapter 7 in the form C9], where g is a primitive root
n.16 The
modulo
Galois
then
can
group
be written
{1d,og,og,...,og
that
so
for each
divisor
f
(and only one) subgroup
Us
One
obtains
from
the
the
field Z((),
a
rational
phism
og. This
this
1, setting
—
{id
025
>g7g)‘
element
is the
case
.
z
we
"-11,
f
:
obtain
one
.
\xA0\xE6
taking
it in coordinate
m1€+ m2C2+
ml,
e
..,a;
subfield by
representing
==
,
f elements, namely,
a6
coordinates
circumstances
n
with
associated
2
with
=
of
form
n—«1
2
-
in the
.
,
-
-
-
precisely
generic
element
2:
form
mn~1C”—1
and
m,,_1
remains
+
a
then
unchanged
checking under
under
the
what
automor-
for
’n’Lgo=7‘n,ge=’rng2e=---,mg1=mge+1=mg2e+1=---,
...,
-»
16From a purely group—theoreticpoint of view, a primitive
root
modulo
71, yields
an
between
the
isomorphism
cyclic group
Z/(n
1)Z and the multiplicative
group
This
makes
it much
easier
to find the subgroups
Z/nZ~—{0}.
isomorphism
of Z/nZ-{0}
—
10.
156
so
element
the
that
period
2
described
as
Algebraic
Z
in
be
can
in
expressed
Chapter
Galois
and
Structures
terms
Theory
f—member
of the
7:
+ ’I’)’Lge—1’I7e_]_.
+
77190770-1- ’IT7,g1’]’]1
=
-
-
-
periods
f—member
The
770:Pf(€)>77l:Pf(
now
the
possess
that
property
7]6—1:Pf<<‘g
--'7
77l:Pf(
they
are
phisms of the subgroup U5 on the one
not
changed by every automorphism
Q(no), @071),
.
.
.
the
automor-
and
the
other
hand,
in
all be the
,Q(77e-1) must
fixed by all
the
Therefore,
according
U6.
same
for
on
the
to
are
fields
fun~
of the
solution
cyclotomic
equation in radicals, it suffices to find,for every possible period
for a single f—member
period, say
length f, an expression in radicals
from 770 using
periods can be calculated
f—member
770, since the other
the four basic arithmetic
operations.
damental
of Galois
theorem
The
that
steps
make
equation in radicals
decomposition of the integer
n
1
-
=
-
plpg
~
-
of
the
possible
clotomic
factors:
Now
theory.
can
ps.
now
n—
1 into
One
may
cynth—degree
of the
solution
planned on the basis of a
(not necessarily distinct) prime
assume
inductively that all the
be
have
degree pj
a
been
solved
in radicals.
We
cyclotomic
equations
shall
denote
these
roots
by K’ the field that arises from the field Q by adjoining
of unity. Beginning with the increasing chain of fields
Q
Where
one
the
then
C
Q (P(n_1)/131
(O)
C
Q(P JO)
Galois
group
considers
the
C
C
'“
Q(€),
of each
chain
C
K’ (Pps(C)) C
step is cyclic of degree pj,
extension
of extension
K’ C K’ (P
/p1(C))
C
Q (P
/(€))
C
fields
K’ (P
/(C))
C
'
"
K10-
step has
10.6, each extension
for the
that
is a subgroup of the corresponding
a Galois
group
group
of eleoriginal chain of fields. However, groups with a prime number
ments
and the one—elementgroup
as
have only themselves
subgroups.
chain, the
step of the second
Therefore, for each actual extension
According
Galois
group
to
the
is the
results
same
of Section
as
that
of the
cyclic Galois
group
of the
10.
Algebraic
Structures
and
corresponding
step in the
first chain.
extension
can
step
therefore
Galois
Theory
157
Using Lagrange resolvents, the
generated by adjunction of a single
be
root.
Altogether,
solved
in radicals.
considerations
have
we
shown
It remains
that
cyclotomic
observe
to
that
in
equations
comparison
can
be
to
the
of
have been
Chapter 7, here complex calculations
avoided.
However, the price paid is a higher degree of
completely
abstraction.
We
10.18
allows
The
now
determine
to
one
basis
of the
follows.
Definition
10.17.
the
whether
theorem
is defined as
chain
finallyto
come
is the
criterion
(and its proof) that
is solvable
equation
an
of solvability of a group,
notion
A finite group
in radicals.
G’ is called
solvable
which
if there
is
next
group
a
of groups
{ld}==G0CG1CG2C"'CGk_1CGk=G
for
which
in the
the
chain
G, is a normal
subgroup
Gj+1, such
that
the
of the
subgroup
quotient
group
prime order.”
G]-+1/Gj is cyclic of
Calling a group thus defined solvable makes sense
only because
the solvability of an equation
and the solvability of a group
are
so
We have the following theorem, whose statement
in
closely related.
terms
of group
tables
was
given in the previous chapter.
Theorem
its solutions
be
can
To
can
prove
the
.
theorem,
the
,
.
four
basic
op-
a
solvable
The
equation.
stepwise through ad-
be extended
of prime
whereby the quotient
group
18It is by no means
.
begin with
we
the
is solvable.
group
coeflicients can
“Apparently
weaker,
K(a:1,
of the coefiicients
using
terms
the solutions
degree to a field that contains
a:,,.18 We shall look at only a single step, in which the field K
of roots
junction
.
in
if and only if its Galois
field containing
.
equation is solvable in radicals, that is, all of
be expressed in terms
of nested roots whose radicands
expressed
erations,
:31,
An
10.18.
.
,:En).
but
has
in
fact
only
excluded
equivalent,
to
that
be
is
a
modification
in
the
definition
abelian.
the
field
extensions
lead
outside
the
field
10.
158
Algebraic
and
Structures
Galois
Theory
is extended
arising from the previous radical extensions
by adjunction
of pth roots, p a prime, of an element
of K to a field L. We assume
such
a
of adjunction
sequence
tensions
for the
solution
the
steps in which
radical
necessary
ex-
pth cyclotomic equation have already
been carried
of unity. Now one
the pth roots
out, so that K contains
sees
that
the four fields K, L, K(:I:1,
,a:,,), and L(.'I31, ,:1:,,) are
in relation
to one
the
another
as
in Figure 10.7. In particular,
shown
of the field K(:v1,
degree of the extension
sun) to L(w1, ,m,,) is
either
1 or p, depending on Whether
of unity is
the adjoined pth root
of the
.
.
.
.
.
or
is not
an
element
of K(ac1,
.
.
.
.
.
.
.
.
,
.
.
,:c,,).
L(x1,...,x,,)
Ej\\clegree:
lor
de
la
ree:
or
p
K(x1,---,X,,)
m
p
degree:
L
N
In
Mm»
U
..
degree:
p
\.
K
10.7
Extension
Figure
adjunction of pth roots.
of the
.
These
0
two
In the
cases
the
field L is
lois
group
(K(:z:1,
to
Aut
.
.
,
second
.
by the
-awn);
field K (:31,
.
3.
.
normal
:12”)
I K ), where
order
cyclic with
In the
field L
(K(a:1, ,m,,) | L) is then,
.
theorem,
.
:L($1a'-
subfield of the
a
Aut
fundamental
0
a
case
K($1a-"a$n)
Aut
to
follows:
as
appear
field K
the
prime
in which
case,
.
.
.
.
.
.
Aut
Ga-
by Point
3 in
the
associated
Galois
quotient
group
group
is
p.
the
field extension
,:3,,) has degree p, we have
(L(:v1, ,a:,,) I L) is “equal”to the
L(a:1,
The
of the
subgroup
the
.,:r:,,).
.
that
Galois
(K(a:1,...,:1:n)| K);
of K ((121,
.
the
Galois
group
.
.
,:1:,,)
group
10.
Structures
Algebraic
that
and
Galois
is, all automorphisms
of the
former
when
its
of the
latter
159
from
result
group
those
of definition is restricted.
domain
Step by step, that is, with
field L, one
obtains
the desired
Theory
additional
of roots
adjunctions
chain
---CAut(L(a31,...,:v,,)|L) CAut(K(a:1,...,:z:,,)[K)
C
of
of the
subgroups
It
remains
Galois
to
the
to
\x80߁
group.
the
prove
That
converse.
is, we
begin
now
with
assumption that the Galois group G’ Aut (K (:31, ,m,,) I K )
is solvable, that
is, that one has the requisite chain of subgroups
G. We HOW begin to
Gknl C Gk
{id} G0 C G1 C G2 CC
the
=
.
=
a
unity.
In
field K’ from
elements
of G.
subgroup
of the
be
field K
the
roots
of
by adjunction of suitable
roots
of unity of every prime degree
particular, we adjoin the
or
equal to the greatest proper
than
can
.
=
create
less
.
The
Galois
solvable
Galois
the
by modifying
seen
H
group
G and
group
of the
number
|G| of
I K’)is
(K’(a:1,...,:r,,)
Aut
=
above
divisor
chain
of
is itself
solvable.
subgroups
a
This
follows:
as
{id}CG'1flHCG2flHC---CGk_1flHCGkflH=H.
Each
of these
chain.
Moreover,
is
of the
each
of the
associated
subgroup
of the
quotient
group
comprises precisely those cosets that
of H.
Therefore, either G,-+1 0 H
contain
at
least
one
Gj fl H
or
the
quotient
order.
This
we
a
View
may
(GJ-+1(1 H)/(Gj PI H) as
which
ment
normal
subgroup
groups
(G3-+1D H(Gj
group
the
subgroup
H itself
sume
without
loss
shortened
such
that
of
F]
a
H) is cyclic of prime
is solvable.
appear
The
are
for the
Thus
generality that the
all the subgroups
{ld}=H0CH1
that
=
C
chain
-'-CHk_1
rest
next
in the
group
quotient
groups
G’j+1/Gj,
shows
ele-
that
of the
proof we asof subgroups has been
CHk=H
distinct.
penultimate
H k_1,
is
normal
subgroup of the group
H.
is cyclic, of order
Furthermore, the associated
quotient group
a prime
number
p.
According to Section 10.15, there is thus a pth
’
root
of an element
a in the
field K whose
adjunction, as shown in
field. Then
in an analogous
Figure 10.8, yields the first intermediate
fashion
the remaining fields can
be constructed
stepwise in relation
group,
a
10.
160
H k._2,
to
the
the
field K’(:r1,
shows
radicals
subgroups
.
that
the
with
Algebraic
.
.
,
.
.
,H1
237,),whose
solutions
the
.
21:1,
radicands
.
.
as
and
continuing
extensions
Galois
.
Galois
Structures
,:13,, can
is
group
Hk_1.
Theory
through
All in all, this
up
expressed in terms
be
of nested
field K.
in the
)
K’(x1,...,x,,
E \xF3
H
KW?)
{
G
K’
\xE5
adjunction of
of unity
10.8.
Figure
10.19
the
tion
for solvable
With
the
How
to
find
a
\xA0zT
K
solution
of the
underlying
equa-
groups.
completion
of the
proof above we have accomplished
in the introduction,
using Galois
goal of this book. As announced
whether
a
theory alone, one can determine
given equation is solvable in radicals.
Properties of an equation and of the field generated
can
be obtained
by the solutions
through the investigation of much
Of course,
in a specificcase,
it is
simpler objects, namely groups.
not recommended
to test
the solvability of a group
by a step—by-step
decomposition of the group table, as we have done in the previous
obtained
from the general
chapter. Much simpler is to use methods
from
theory of finite groups.
However, we have deliberately refrained
in this book
in order
to limit
the scope
of
explaining such methods
this introductory
excursion
into the theory of polynomial equations.
We note here only that
the symmetric
group
Sn, that is, the group
of all permutations
on
n
elements, is not solvable for every n > 5.
A proof of this fact is given in the epilogue. Equations
of the fifth
not solvable
in radicals.
An
degree with Galois group 85 are therefore
10.
Structures
Algebraic
Galois
Theory
161
equation—andindeed, such equations
exception~was
given in Section 9.17.
example of such
rather
and
an
than
the
The
Unsolvability
of the
Classical
the rule
are
Construction
Problems
We
stated
in
doubling
the
that
circle
the
Chapter 7 that the
cube, and trisecting
transcendence
cannot
be
doubled
of the
number
7r.
classical
problems of squaring the
The
angle have no solution.
an
lies
The
proofs
of the
impossibility
not
rest
on
the
deep
results
we
shall
now
see,
to
consider
complete
the
proofs, then,
Galois
theory,
other
two
constructions
theory.
In
of the
of Galois
the
in
not
formula
circle,
reason
but
in
the
also
do
general, it sufiices,as
degrees of towers of field
for the
extensions.
To
ter
of fields: If
7 in terms
a
shall
we
construction
reformulate
be
the
carried
results
of Chap—
with
out
straightthat
leads to the construction
of a point z in the complex
edge and compass
plane, this is algebraically equivalent to the point z lying in a field obtainable from the field Q of rational
numbers
of degree
by stepwise extensions
2. According to the degree formula
for nested
field extensions, the number
z must
therefore
belong to a field whose degree over the field Q of rational
numbers
is
Since
a
of 2.
power
field
the
can
Q (
has
degree
three
the
over
rational
numbers,
formula, lie in a field whose degree
a power
of 2. Therefore,
a segment
of length 3/2 cannot
be constructed
with straightedge
and compass.
Thus
the problem of doubling the cube
3/2 cannot,
based
the
on
degree
is
is
19
unsolvable.
In the
of
problem
angle, again the key is in the construc~
tion of a cubic
extension
field. The problem is equivalent to constructing
a
number
a on
the unit circle, satisfying 23
a.
z, given a number
Beginning
with
the field Q((, a), where
C is a cube root of unity not equal to 1, the
2 yields one
of the following two scenarios:
adjunction of the number
trisecting
an
=
(1) The
number
z
lies
in the
number
2
does
not
degree
3.
field Q((, a), and
no
field extension
genuine
arises.
(2) The
extension
19A way
of
lie in the
field (@(C,
a), and
the
of
as
Well as the other
classical
solving this problem,
without
lems, in an elementary
fashion,
using the degree formula
Galois
in Eine
elernentare
theory, is described
by Detlef
Laugwitz
bei Konstruktionen
mit
Zirkel
und
Unmoglichkeit
Lineal, Elements
17
(1962),
necessary
pp.
to
54—58.
represent
In
addition
the
roots
there
of cubic
is
a
discussion
equations.
of
how
result
is
construction
or
other
field
a
probresult
fiir
Methode
der Mathematik,
few
square
roots
of
die
are
10.
162
In
(2), which
case
Algebraic
Galois
and
Structures
for example, in the
occurs,
the construction
of the point
problem
Theory
of
trisecting an
straightedge and
z with
angle of 120 degrees,
is impossible.
compass
We note
finally that one can prove the constructibility
for
structibility of regular polygons using the degree formula
or
noncon-
field
nested
extensions.
Literature
on
Galois
Theory
Theory,
Notre
and
Algebraic
Structures
Emil
Galois
Artin,
Dame, 1942.
2001.
Siegfried Bosch, Algebra, Berlin,
Jean
Escofier,Galois
Pierre
Theory,
2002.
Serge Lang, Algebra, Springer,
Bartel
Leendert
der
van
York, 2001.
New
Algebra I, Berlin,
Waerden,
1971.
Tignol, Galois’ Theory of Algebraic Equations,
Jean-Pierre
Singapore,
2001.
Exercises
(1) Prove
that
the
K C L C E,
degree formula for nested field extensions
E over
K is equal to
is, that the degree of the extension
products of the degrees of E over L and L over K.
the
(2) Show
that
power
of
(3) If for
ates
the
a
number
prime number
transitively on the
H
G with
C
one
must
Why
Conclude,
ducible
into
these
the
.
finite field must
be
a
is the
other
of prime
solution
sub-
transitively
on
into
so-
can
be
set
set
of all elements
{1,2, ..,n}
.
suitable
all be of the
that
operates
the
by a
oper-
.
gé {id} also
moreover,
in the
.
Decompose
orbits
equation
group
set
of which
orbits, each
translated
a
G C 3,,
permutation
group
,n }, then
any normal
{ 1, 2,
a
n,
H
Hint:
{1,2,...,n}.
called
of
prime.
a
group
mate
of elements
if the
same
that
in H.
permutation
size?
Galois
group
of
degree n is solvable, then
is cyclic of order
n.
the
an
irre~
penulti-
Exercises
163
(4) Show
that
for
mations,
that
fa,1,(a:)
am
a
the
n,
set
is, functions
fa,g, 2 Z/nZ
+ b for given residue
classes
of linear
transfor-
Z/nZ such
(2,1) 6 Z/nZ,
that
——>
form
group.
In
o
0
0
the
addition,
prove
No
linear
transformation
that
is, for each
residue
class
a;
following:
has
two
transformation
with
or
fixed points,
more
fag, there
fa,g,(a:)
=
is at
most
one
ac.
Every element of order n has the form fm, with b 75 0.
Every linear group, that is, a group of linear transformations
including the transformation
f1,1, is solvable. Hint
for the last part:
The subgroup generated
by f1,1 is a
component
of the
Among the examples
in Section
the
number
prime
a
9.17, it was
field of rational
solution.
of Galois
stated
numbers
groups
of equations
that
the
equation
leads
to
a
Galois
of fifth degree
935
group
—
2
with
=
0 over
twenty
elements, while the equation $5 53: + 12
0, despite the more
complicated root structure, results in a Galois group of only ten
elements.
How is this phenomenon
to be explained in the light
of our considerations
in this chapter on radical
extensions
of the
underlying field and subgroups of the Galois group?
——
=
Epilogue
In the
end
was
the
beginning.
groups
and
finite fields;
and
to the
historically and in relation
thematic
framework
of this introduction,
the end result
creates
a new
beginning: Although the problem of solving polynomial equations in
radicals
and Ferrari
was
able to be answered, the
posed by Cardano
in the solution, groups
and fields,raise many
new
objects involved
of
questions about their general properties, and not only in the sense
“art for art’s sake.” The knowledge that
these
objects, and the associated
applications and techniques, are applicable in many fields of
that
inquiry has allowed algebra, that is, the subfield of mathematics
deals with basic arithmetic
itself as a major
operations, to establish
mathematical
discipline. In the field of abstract
algebra, the objects
of consideration
are
defined and
“classified” in the broadest
possible generality and categorized according to their basic structure.
To
do this with maximum
efiiciency,
general classifications are refined
as
and fields with their
needed, for example, groups
subcategories
abelian
Both
such
classifications are
also
gen-
example, with the definition of a commutative
ring, which
satisfies all the requirements
of a field except for the invertibility
of
eralized,
for
multiplication.1
1The best—known examples of rings that are not fields are the integers,
the set
in one
or
several
classes
and the set of residue
polynomials
variables,
Z/TLZ for 71. not
of
a
prime.
165
166
Epilogue
There
several
are
by such
axiomatic
an
Mathematics
0
0
advantages to developing
mathematical
objects
In
one
method:
becomes
transparent.
more
recognize fundamental
ematical
objects that
exhibit
Mathematics
liberated
properties
becomes
a
in
a
of various
collection
of properties
number
from
particular,
in
can
math-
common.
“truths” taken
fundamental
for granted once
it has been freed from particular
and applications.
Thus, for example, it was with
interpretations
Such
with
the
generalization of the parallel postulate
to non—Euclidean geometries that
it became
the unprovability of the parallel
possible to establish
axiom, a problem that had been festering since antiquity.
0
economical, at least
whole, since important facts
is
approach
an
mathematics
as
a
more
do
respect
to
have
to
not
be
situations.
proved over and over in different
Moreover, these
in mathegeneral principles, which in fact are of central interest
from more
matics, can often be derived as special cases
generally
valid
theorems.
Although such
constituted
axiomatically
mathematics
diverges
from the descriptive natural
sciences
in being only indirectly
connected
with our
physical perception of the world, one should note
that
classification plays an important
role in those
sciences
as
well,
from the Linnaean
taxonomic
system of biological classification to the
to the classification of symmetries
of
periodic table of the elements
fundamental
particles.
If this
before
book
and
chapter,
that,
terested
employed such a structural
approach only in the last
but pragmatically
in the chapter
perhaps half—heartedly
the
reason
to
was
nonmathematician.
cepts that
seem
barrier
to
the
chapter
will
intention
an
opaque
minimize
The
on
multiplicity
first contact
nonmathematician.
have
of the
received
such
the
an
presents
difficulties
for
the
of definitions and
an
almost
in—
con-
insuperable
readers
of the last
Perhaps some
impression, despite the contrary
author.
deliber—
things were
of which are related
to polynomials.
We tacitly
ately excluded, some
a formal
a polynomial
as
a formal
sum
accepted, without
definition,
To
avoid
unnecessary
complications,
some
Epilogue
of
167
of
products
values
in
also
have
variables
more
X, Y,
Generally, this
taken
in additional
polynomials
Such
or
fixed set.
some
could
one
one
the
.
set
.
.
,
was
ring of integers
coeflicients taking
and
a
or
particular
field,but
indeed
set
the
of all
variables.
formal
polynomials are to be distinguished from the functions
that
such polynomials
define when
the variables
are
replaced
values
set
of numbers.
Now one
by concrete
CL,b,... from some
can
calculate
both
with polynomials
themselves, taking their sums
and products,
and with their
functional
values.
It is clear that
the
two
forms
of calculation
are
compatible, for example that one has
f (a) g(a). However, one should prove that this is the
(f g)(a)
=
-
-
case.
of our
also serves
the purpose
of
simplification
presentation
to subfields of the complex numbers.
It
specializing the discussion
was
of the fundamental
theorem
of algebra, that a
clear, on account
splitting field exists for every polynomial with complex coefficients.
Despite the practicality of such an approach, and despite the impor~
tance
of the fundamental
theorem
of algebra, the form of argumenta—
tion has little to do with algebra. It is not only that the fundamental
theorem
is proved using mathematical
analysis (calculus), an argu~
ment
of distance
and intermediate
involving estimates
values, which
renders
the theorem’s appellation a historical
It is also that a
artifact.
generalization to other cases, for example that of finite fields,cannot
be carried
out by such methods.
The
For
these
it
is understandable
that
in
algebra a comtack is generally taken
for constructing
the splitting
pletely diiferent
fields that
are
crucial to Galois theory. Beginning with a field K
and a polynomial irreducible
over
K, a field extension
containing the
elements
that
solve the corresponding
equation
reasons
_1
ar"+a,,_1m“ +a,,_2:r"
is constructed
adjunction
of the
in
of
a
a
formal
completely
value
_
=0
2+-~-—|—a1$+ag
formal
oz, where
One
does
this
calculating
with
expressions
way.
in
form
k0+k1oz+k2a2 +---+k,,,o/",
by the
168
Epilogue
with
kg,
.
.
.
,
km E K,
so
that
that
it
the
employ
the
simplification
—a,,_1o/‘*1an_2oz”
_2
TL
a
We
=
—
always
be achieved
can
that
m
-
—
3
-
~
n
—a1a
~
1. One
—
—
as,
can
then
show
set
K[a]={kg+k1a+k2a2+---+k,,_1oz""‘1I/cj€K
forms
field that
a
equation?
a solution,
namely a, of the given
clearly contains
is tricky here is the proof that
the set K [a] is closed
What
division.3
under
If the
ducible
polynomial
is then
factored
over
with
additional
the
field K[a] into
irre—
adjunction steps. In
this way,
one
a
finallyobtains
completely algebraically constructed
splitting field/1 It is uniquely determined, as can be shown, in that
every other splitting field is isomorphic to this one, that is, that the elements
are
related
by a one—to-one correspondence that is compatible
with
now
factors,
one
can
proceed
the
basic
arithmetic
operations.5
With
the
formulation
thus
be
made
amenable
to
purely algebraic methods.
Chapter 5 as the equation
2From
in
which
formal
variables
:51,
.
.
.
,m,,
in
is similar
to that
of a quotient
point of view, this approach
group
from
a
normal
It is an
of a ring of residue
can
be
subgroup.
example
classes, which
constructed
from
a
of a ring called
an
ideal.
It is such
methods
ring and a subset
of constructing
new
the
axiomatic
definition of such
as
objects that
requires
objects
and
of the
as
we
were
groups
fields, not just as subgroups
symmetric
group,
always
able
to do in the
case
of finite
and
subfields of the complex
numbers.
groups,
from
Section
10.9
can
be easily extended.
That
3Essentially, the arguments
is,
one
the
linear
of equations
that
to multiplication
investigates
system
corresponds
by
the inverse
of an
element
of K[oz]. However,
it is also
for the considerations
necessary
of Section
10.9 to prove
that
the product
of two
nonzero
elements
is again
nonzero.
4This purely algebraic construction
can
in fact
be used
to prove
the fundamental
theorem
of algebra
is over
the
using complete induction.
(The induction
highest power
of 2 that
divides
the degree of the equation.)
enter
the picture
Analytic arguments
only
in the
form
of that
value
that
oddfact, provable
by the intermediate
theorem,
every
degree polynomial with real coeflicients has a real zero.
See Jean—Pierre Tignol,
Galois’
theory of Algebraic Equations,
5 at
Singapore,
2001, pp. 119, 121—122,and Exercise
the end of this
chapter.
5A field automorphism
is simply
an
of a field with
itself.
isomorphism
a
formal
described, the general equation can
treatment
of
by Galois theory in terms
We have seen
the general equation
in
Epilogue
the
169
associated
elementary
symmetric
polynomials
S1(.’B1,...,.I1n)=(I}1+{I12+'--‘l-(I:.,,,
82($1,
fin) $1552 +!I31iB3 +
+93n—1$n,
-
3n(371:
-
-
~
-
=
amn)
‘
'
'
-'1315172"'«Tn:
:
be determined.
to
are
-
language of field extensions, this corresponds to the situation in which
beginning with a field K of polynomial coefficients, one
is to investigate
the extension
of the field K (31,
Sn) to the field
for symmetric
K(:1:1, ,cc,,). Due to the uniqueness theorem
polynomials
(see the section on this topic in Chapter 5), one may treat
the symmetric polynomials in K (s1,
sn) as though they were formal variables
with no formal
them
polynomial relations
among
(one
speaks of algebraically independent quantities). One thereby obtains
an
of the general equation,
additional, fully equivalent, interpretation
in which
now
the equation’s
coefficients
are
variables
a,,_1
a0, a1,
for which, as described, a splitting field can
be constructed.
Since
the solutions
have no relations
themselves—in the first place
among
by definition and in the second place by the equivalence6-the Galois
of the general equation is the full symmetric
group
group.
In the
.
.
.
.
.
Theorem
E.1.
The
symmetric
group
solutions
$1,.
As
consequence,
a
covered
.
.
.
course,
a
,X,,)
with
.
.
,
Sn; that
of the general nth—degree
equation
all permutations
is, it contains
of the
the
for the
Galois
group
,:1c,,.
.
by Lagrange,
6Of
h(X1,
.
results
general equation,
first dis-
as
special case of Galois theory. Here
field is generated by polynomials in the variables
intermediate
every
,
,
.
n
.
.
.
is the
.
as
appear
direct
is
proof
h(a:1,
.
.
.
g(X12--
~
also
urn)
,
a
yX7l)
0,
=
with
Beginning
the product
possible:
forms
one
H
2
h
(Xa(1)7 ':X0(n))
-
'
a
given
polynomial
-
UES11
Since
the
in the variables
as
polynomial
g is symmetric
X1,
,Xn, it can be expressed
a polynomial
in the elementary
in these
There
is thus
variables.
symmetric
polynomials
a polynomial
u(Y1,
,Y,,) such that the polynomial
g(X1,
,X,,) can be expressed
in
the
tutes
form
the
'u.(a,,_1,
tities
.
g(X1,
solutions
.
.
.
,
.
.
.
.
.
.
,
3:1,
0.0). This
finally show
.
.
.
X”)
.
.
=
,:v,,
.
shows
u
(s1(X1,
into
.
this
.
.
,Xn),
that
.
.
,
then
identity,
immediately
.
u
=
s,,(X1,
.
obtains
one
0.
.
.
The
.
.
,X,,)).
If
O =
g(:z:1,
previous
substi-
one
.
.
polynomial
.
,:z:,,)
ideni
that
g
=
O and
h
=
0.
=
170
Epilogue
that
remain
E.2.
The
of the
automorphisms
of permutations.
it naturally follows
associated
group
Furthermore,
that the solvability of the general equation of a particular
degree n is
equivalent to the solvability of the symmetric group S,,. Abel’s imthus corresponds to the following gro.up~theoretic
possibility theorem
a,-1,
.,a:,,
.
.
the
under
unchanged
theorem.
Theorem
symmetric
group
is unsoluable
Sn
for n 2 5.
theorem
usually is
proof of a group—theoretic
used to derive
A proof is possible with arguments
Abel’s theorem.
similar
on
this topic at the
to those
used by Ruffini (see the section
end of Chapter 5). To this end, one first proves
the following theorem.
In
this
textbooks,
Theorem
If G
subgroup of the symmetric group 8,, for
n
all three—cycles,
that is, all cyclic permutations
2 5 containing
of
the form a —-> b —> c —> a of three distinct
elements
a, b, c, and ifN is
a normal
subgroup of G with commutative
quotient group G/N, then
this normal
all the three—cycles.
subgroup also contains
E.3.
To
this
prove
three-cycle
a
theorem
preparatory
b —>
—+
a
is
—>
c
a
the
as
one
represents
arbitrary
an
product
(cl—+b—>a—>d)"1o(a—+e—>c—>a)_1
o(d—>b—>a—>d)o(a—->e—>c—>a),
where
d and
e
from
a,
must
lie in the
Since
b, c.
asserted,
then,
On the
solution
The
chain
quotient
elements
is
group
that
every
three—cycle
belongs
basis
of the
can
the
distinct
coset
step by step that
a
arbitrary
are
of the
theorem
every
group
symmetric
therefore
single element,
and
Moreover,
the
so
not
the
the
represents
that
are
also
distinct
commutative,
the
product
is, in
N.
As
subgroup
N.
that
identity,
to
the
normal
just proved, it can
in an ascending chain
Sn must contain
group
end
in
symmetric
the
group
cannot
be deduced
corresponding to
all three—cycles.
group
containing a
be solvable.
applied to the alternating
With refergroup
An, defined as the group of all even permutations.
ence
to the alternating
group
An, we note that it is a normal subgroup
of the symmetric group
Sn, since the quotient group is a commutative
same
argument
can
be
trivial
now
Epilogue
171
two—element group.
In the
case
of the
to
the
intermediate
ing group
corresponds
adjunction of the square
To the
that
extent
of the
root
the
alternat-
field that
arises
through
discriminant.
field K
base
the
general equation,
for the
general equation is a
subgroup of the complex numbers, the implicitly assumed
possibility
of extending Galois theory and its applications to radical
extensions
is unproblematic.
In an extension
of Galois theory to arbitrary fields,
however, two additional
complicating factors need to be considered:
only if every irreducible
polynomial
distinct
zeros.
of
possesses
Otherwise, not every automorphism
the splitting field is associated
of
uniquely with a permutation
0
0
The
generalization
the
zeros,
can
be
and
finite fields cause
fields of characteristic
Nevertheless,
no
problems in this respect.
The
characterization
of radical
and
works
the
moreover,
problematic.
resolvents
assumes
that
one
extension
(see the
end
of the
finite characteristic
with
of Galois
construction
extensions
this
zero
of Lagrange
in terms
by the degree of the
divide
can
resolvents
field
proof in Section 10.14). In fields
is not necessarily possible.7
to finite fields,
preceding chapter relates
examthan
which
We
have used only indirectly, other
giving some
classes
ples in Chapter 10, namely, in the form of fields of residue
of the existence
of a
In particular,
made
use
modulo
a prime.
we
the cyclotomic equation could be
primitive root modulo
n, so that
solved using suitable
sums
of roots
of unity, that is, the periods. Thus
of an integer g such
for prime numbers
assumed
the existence
n we
Another
that
the
classes
hole
in
g1, g2,
numbers
1, 2,
.
.
.,n
the
.
.
,g"“1represent
.
E.4.
residue
nonzero
1.
—
this
Using algebraic structures,
greater generality.
Theorem
distinct
fact
be formulated
can
in
Every finitesubgroup of the multiplicative
slightly
group
of a
of
finite
fieldis cyclic.
The
application
field Z/pZ,
was
first proved,
7Indeed, the general
ample,
is
not
solvable
of interest
in
quadratic
radicals.
here, relating
in
formulation
a
equation
See
B.
L.
the
over
van
der
subgroups
to
as
two-element
Waerden,
a
a
statement
about
field
for
Algebra
Z/ZZ,
I, Section
ex-
62.
172
Epilogue
residue
classes, by Legendre (1752#1833). Earlier
proofs by Euler
be considered
incomplete. Although a proof can be given based
must
extensive
computations
proof of the generalized
on
a
classes,8 we
in residue
is shorter
which
theorem,
would
like to offer
and
easily
more
understood.
We
begin with
associates
{1,2,
and
with
.
.
5 are
go(8)
=
natural
a
of Euler’s
investigation
an
d the
number
.,d} relatively prime
d.
to
of integers in the
number
For
set
example, 90(6) 2, since 1
1 and 6 relatively prime to 6; and
prime to 8. Euler’s phi function
the
only integers between
4, with 1,3,5, 7 relatively
satisfies the
which
phi function,
=
relation
2
n.
=
dln
We will first justify this
where
formula,
the
is over
sum
all divisors
d of
To this
for each residue
class j modulo
end, consider
77,, represented
for example by the n integers 0, 1,
,n
1, its order (1 as an element
77,.
—
.
of the
the
Z/nZ
group
residue
by
an
the
residue
class
integer
d, and
m
thus
j must
lie in
that
so
associated
obtained
corresponds
with
a
the
precisely to
preparations
we
natural
number
divisor
when
it is
contains
and
the
U of the
that
Z/nZ thus
formula.
summation
such
of n,
precisely
n—elementgroup
address
can
finite subgroup
a
therefore
of the
partition
a
represented
the subgroup generated
by
subgroup is cyclic of order
3-. This
Z/dZ, and
to
d. The
theorem, namely,
field. If d is
be
§,
of order
a
d must
d precisely
isomorphic
these
order
order
elements
After
such
.
j will have
-
class
Each
.
.
actual
multiplicative
there
is
of the
content
group
element
an
as
of
in U
acd‘1
by m is the group { 1, w, x2,
}
of d elements, then according to Section
10.4, d is a divisor of |U], the
number
of elements
in U. Since rrd
of this group
1, every element
is a zero
of the polynomial X d
1. Since we know from Section
4.2
for which
the
generated
group
.
.
.
,
=
—
that
for each
and
thus
an
element
this
zero
of U outside
d~element
a
either
element
8See, for example,
Chapter
10.
polynomial
a
polynomial
generates
no
of
that
Jay
can
have
we
at
may
most
split
d zeros,
ofi? a linear
there
factor,
cannot
exist
that
subgroup {l,.’12,£L‘2,
,md“1}
subgroup. Therefore, in the group U there is
a d—element
generates
subgroup or there are
R.
of the
Goldman,
.
The
Queen
of Mathematics,
.
.
Wellesley,
1998,
Exercises
173
If
of them.
the
composed
each
element
one
again decomposes
to
Z/nZ according
group
for
then
generates,
group
U
the
size
of the
earlier
de-
subgroup
that
as
we
summation
the
obtain
IUI we
=
n
the
formula
n
where
of the
each
are
the
-
dln
6,1 is either
numbers
1. One
0 or
to
the
previously
must
always
derived
then
immedi-
sees
formula, that
1. In particular,
have 64
there
divisors
d of n we
cp('n)elements of U that generate an n—elementsubgroup, that is,
U is therefore
entire
U. The group
cyclic.
group
ately a similarity
for
Z <,o(d) 5,1,
=
summation
=
Exercises
Fermat’s little
(1) Prove
positive integer
divisible
by n.
Wz'ls0n’stheorem:
(2) Prove
For
a
a
to
prime
relatively
a
For
theorem:
n, the
prime
number
prime
a"‘1
number
number
n,
and
n
the
a
1 is
—
number
from this that
a
by n. Then conclude
2 2 is prime if and only if (77, 1)! + 1 is
(71 1)! + 1 is divisible
—
natural
number
divisible
by n.
the
(3) Prove
of Fermat’s little
generalization
number
natural
——
n
and
n
number
natural
a
a
theorem
that
relatively prime
for
a
to
n,
a‘*’(") 1 is divisible
by n. Hint: First show that the
residue
classes in Z/nZ represented by integers relatively prime
under
to n form a group
multiplication.
the
number
that
(4) Prove
p and
q and
is divisible
—
if n
=
pq
if u and
by (p
am’
is
two
‘U are
1)(q
—
product
a
-
of two
numbers
natural
1), then
distinct
for
prime numbers
such
everynatural
that
M)
number
—
1
a,
by n.9 In such a case, the pairs
is
one
where
as
can
serve
and
keys,
cryptographic
(21,n)
(u, n)
One speaks of an
for decryption.
the other
used for encryption,
to
In
contrast
symmetric
algorithm.
cryptographic
asymmetric
the
number
—
a
is divisible
m l~—> m”
residue
class
exercise
is
that
two
of
this
the
mappings
significance
is used
a construction
Such
inverses
of each
other.
itself
are
and
:1: I—> m” of Z/'n.Z into
is,
Here
in the
in cryptography,
RSA
large primes, that
very
encryption
procedure,
such
two
hundred
of several
prime numbers
digits each, are used, so that determining
Of
of
=
millions
after
even
'n.
would
be
years
their
impossible
product
pq
q given
9The
12,
computation
on
today’sfastest
computers.
174
Epilogue
in which
algorithms,
with
the
RSA
algorithm
be published without
can
able
and decryption use a single key,
of the keys, that for encoding, say,
encryption
one
fear
unauthorized
that
will
persons
be
decode
to
One therefore
refers to the
encrypted messages.
RSA algorithm as public key encryption. Hint: The assertion
can
be reduced
to that
is divisible
by p or
am’
—
a
Prove
3. To include
of Exercise
by p and
the
the
might demonstrate
by q separately.
one
q,
the fundamental
theorem
proving that for a nonconstant
if one
factors
coefiicients,
f
f(X)
in which
a
divisibility
of
case
of algebra in
an
polynomial
f (X) with
into
linear
algebraic way by
factors
complex
as
(X-=v1)°--(X-wn)
=
field of (C (which is always possible
algebraic extension
via an algebraic construction), one
in fact has that 331,
mm 6
C. First show the following:
o
The theorem
holds for quadratic
polynomials f (X) (see
also Exercise
1 in Chapter 2).
0
It suflices to prove
the existence
of a single complex solution acj. Moreover, one
can
restrict
attention
to polynoin
some
.
mials
Now
with
.
.
,
coefficients.
real
the
inproof can be carried out using mathematical
duction
on
the highest power
of 2 that
divides
the degree of
the polynomial.
For the base step of the induction, one
uses
the
of the
version
theorem
for real
polynomials of odd degree,
which is proved using mathematical
analysis (calculus). For the
induction
step, one investigates polynomials of the form
gc(X)
H (X
=
+ 10,-+ ca:,~mj))
\xA0\x974
—
¢
for
For
to
suitably
a
chosen
parameter
c.
m
and a natural
number
a relatively
prime number
prime
the Legendre symbol is defined as follows:
a
m,
if a
(%> {+1
=
.92 + km for suitable
=
-1
if a does
not
have
such
integers
representation.
.3
and
k,
Exercises
175
The
Legendre symbol therefore
represented by a is a square
class
of residue
classes
Z/mZ
tells
in
the
residue
multiplicative
group
whether
the
the
value
of the
Le-
by finite trial
and
error,
one
Therefore,
show
Even though
0]\xA1
—
gendre symbol can be determined
in a direct
is naturally
interested
calculation.
first that
a/2
(E)
_
m
is divisible
by 777,.
properties
Other
of the
symbol can be obtained
of unity. For a second
n 2 3
with the use of roots
prime number
of unity §
the nth root
we
+
again let Q denote
cos
isin \xD0\x9B
while the periods of length (71 1)/ 2 (see Section
7.2)
and 771
where
denoted
are
by 770
P(n_1)/2((9),
P(n_1)/2(()
modulo
a primitive
root
n.
Show
the integer g again represents
Legendre
=
—~
=
=
that
0«1C+ + Cln—2Cn_2)
(G0
\xB0\xD5r
(770 771)
(770 771)m
with
integers
done
so
a0, a1,
.
.
+
'
'
also
Show
,an_2.
.
'
have
(if you
not
already
Chapter 7) that
2 of
in Exercise
777»
=
—
—
~
(770 771)2 (—1)("_1)/27%
=
—
Finally,
show
the
how
identity
resulting
l<“1>"%”‘1
(%> Gil
*
with
integers
(be
+
+ 171C
m
=
b0,b1,
.
.
.
-
-
("O W2
M
bn~2C"—2)»
+
-
the
,b,,_2, yields
law
reci-
of quadratic
p7"ocz'ty1°
(n) —1——<—>i
n—1m—l
Tn
(7) For
a
prime
group
number
p,
of
quadratic
m
-
-
=
-
»
-
,gp,g1)
first proved
in his diary.
8, 1796, as documented
ramifications.
of number
Furthermore,
theory with many
other
elementary
properties
together with some
reciprocity
of arbitrary
the values
quickly.
Legendre
symbols rather
April
by a
mapping
:91?) (92,
was
reciprocity
by an entry
[G] is divisible
of elements
defines the
one
s0(91,92,
law
’
>2
number
G whose
.
10The
<
=
—
TL
Friedrich
It is a fundamental
using the law of
can
of integers,
one
by
Carl
Gauss
on
result
quadratic
compute
176
Epilogue
for g1,g2,...,gp
E
G, as
well
as
the
set
X={(91,g2,-.-,9p)€G"lg1g2~~9p=6}»
Where
again denotes
6
Prove
the
°
IX!
=
0
The
0
following:
lGl”'1=
are
equal.
Every
orbit
of either
0
The
.
element
that
of
element
a
theorem
:11Cauchy’s
exists
that
one-element
after Ludwig Sylow
(1832-1918).
of a group
of order
the power
of
The
a
element
an
there
exists
E
:2:
X
with
by 19.
a
one-
ele—
least
one
other
orbit, and thereby
prove
the
existence
p
formulated
usually
in X is divisible
at
of G of order
is
G1’,consists
6
m
.
p elements.
or
there
orbit, conclude
with
.
of one—elementorbits
number
ment
an
where
{.'I:,(p(.'17),<,02(:l2),
},
one
Assuming
element
identity.
group
mapping (,0 maps the set X into itself.
If the identity (pk\xB0'_{It holds for an element
zv E G?’ and
an
of
integer k not divisible by p, then all the coordinates
:3
0
the
(Cauchy’s
theorem).11
in
a
Sylow theorems
prime.
general
more
make
form,
assertions
which
about
is named
subgroups
Index
Abel,
Niels
Henrik,
xii, xix, 49, 50, 53,
55, 95, 114
absolute
value, 14, 32, 34
Acampora,
Renato,
1, 4, 9
adjunction,
95~97, 102—105,109
171
algebraic structure,
169
algebraically
independent,
al-Khwarizmi,
2, 3, 5
170
alternating
group,
136
analytic
geometry,
4
Archimedes,
Argand, Jean Robert, 34
Ars
Magna, x, xi, 5, 6, 8-10, 18, 23,
24, 29, 30
Artin, Emil, xix, 95, 149, 150, 162
associative
law, 128
automorphism,
126, 132-135,
139—143,
151, 152, 154-156
group,
125, 135, 148
foundations
of mathematics,
145~147,
automorphism
axiomatic
166, 168
Ayoub, Raymond
G.,
51
Bachmann,
Paul, 79
basis, 6, 115, 126, 136, 137, 155
Bézout, Etienne,
46, 83
bicubic
resolvent,
85, 86, 88, 113
141, 142, 150
bijection,
biquadratic
equation,
xi, xii, xviii, 2428, 37, 38, 40-42, 44, 53, 76, 99,
101, 102, 106, 110~112
Bombelli,
Rafael, 11, 18
Bos, Henk, 28, 70
Bosch, Siegfried, 162
Breuer,
Samson, 90
Bring, Erland
Samuel, 83, 89, 90
Bring—Jerrardtransformation,
83, 90
xvi
Buchmann,
Johannes,
Cantor,
Moritz,
11
Cardano
formula, 6, 7, 11
4
suspension,
wave,
4
Geronimo,
xi, xvi, 23, 24
Cartesian
coordinates,
69, 70
casus
irreducibilis,
10, 16, 19, 20, 75
176
Cauchy’stheorem,
34
Cauchy, Augustin-Louis,
characteristic
of a field, 132, 171
closed, xiv, xix, 96, 126, 128, 138, 168
complex
Cardano,
analysis, 14
number,
10v13, 31, 32, 34, 64, 74,
76, 96
plane, 14, 15, 64, 161
of permutations,
composition
43, 94,
101, 127
58
computer
algebra system,
subgroup,
145, 146
conjugate
of a function,
14, 31—33
continuity
coset, 129, 131, 146, 147, 159, 170
Crelle, August Leopold, 114
xvi, 173
cryptography,
cubic
equation,
x, xi, 1, 4-10,
17-20,
24, 25, 38, 40, 41, 53, 55, 75, 97,
107—109,161
cubic
resolvent,
100, 102, 112
cycle, 54
cyclotomic
xviii, 17, 39, 59,
equation,
60, 65, 67, 70, 72, 73, 75, 77, 79,
80, 94, 107, 110, 112, 123, 135,
154-158, 171
~
Davies, Philip J., xi
126
Dedekind,
Richard,
177
178
Index
formula
degree
for
161
tower
a
of
fields,
137, 143,
Dehn, Edgar, 122
75
Delahaye,
Jean-Paul,
Descartes,
René, 28~30, 69, 70
difference
product,
108, 109, 112, 114
dimension,
126, 136
discriminant,
40, 41, 46, 88, 107, 108,
171
disjoint
129
partition,
distributive
law, 13, 131
Difirrie,Heinrich, 26
doubling the cube, 161
Edwards,
Harold
Eisenstein
irreducibility
M., 95,
122
criterion,
59,
60
Eisenstein,
Ferdinand
Gotthold
encryption
procedure,
Max,
59
ElGamal
elliptic
curve,
encryption
asymmetric,
symmetric,
xvi
xvi
abelian,
xvi
173
162
73, 97, 116, 122,
123
Euler’s phi
function, 172
Euler, Leonhard,
16, 46, 81-83, 172
extension
field, 96, 97, 104, 119, 126,
136, 137, 139, 142, 151, 153, 154,
161, 174
function,
Fermat
prime,
Fermat’s little
group,
group
Escofier,Jean Pierre,
Euclidean
algorithm,
factorial
xiii—xv, xix, 93—95,97~
116, 118, 119, 122, 123, 127-130,
132~136, l38—151,l53—l60, 162,
163, 169
solvable, 105
Galois
133~
resolvent, 98, 115, 117—119,
135, 139, 140, 149, 171
Galois
theory, xii, xiv, xv, xviii, xix,
53, 75, 80, 82, 95, 125, 132, 147,
149, 160—162,
167-169, 171
Galois, Evariste, xii, xv, 93, 122
Giirding, Lars, 114
Gauss, Carl Friedrich,
12, 31, 56, 6365, 67, 69, 70, 75, 77, 94, 175
Gaussian
plane, 65
general equation,
38, 39, 42, 45, 49, 50,
52, 53, 55, 77, 80, 93~95, 168~171
Girard, Albert, 30
Goldman, Jay R., 172
Gottlieb,
Christian, 73
common
greatest
divisor, 120
Galois
43
72, 73
173
theorem,
Ferrari, Ludovico, 1, 23, 165
Ferro, Scipione del, 4
field, xiv, xvi, xix, 13, 16, 131, 165
finite, xvi, 132, 162, 165, 167, 171
field extension,
135, 137, 139, 141, 148,
150, 158, 161, 162, 169
degree of, 137, 150
see
fifth-degree
equation,
quintic equation
Fior, Antonio, 1, 3, 4
fixed field, 142, 143, 145, 150
Fontana, Niccolo, see Tartaglia
Fiihrer, Lutz, 21
function
theory, 14
fundamental
166
particle,
fundamental
theorem
of algebra,
30—
32, 34, 37, 167, 168, 174
fundamental
theorem
of Galois
theory,
140-145, 147~150, 152, 153, 155
fundamental
theorem
of symmetric
polynomials, 45, 46
131,
165
170
alternating,
cyclic, 130, 131, 151, 155
of integers
mod
n,
131,
132,
151,
163, 165, 172, 173
linear, 163
solvable, 160
symmetric,
43, 99, 160, 169, 170
group
table, xiii, xiv, 101, 102, 104108, 110, 111, 113, 127, 144, 147,
153, 160
Henn,
73
Hans-Wolfgang,
heptadecagon, xi, xviii, 63, 68, 70, 72
Hermes, Johann
Gustav, 73
Hilbert
basis
theorem, 115
126
homomorphism,
Hudde,
Jan,
97
ideal, 115, 168
identity element, 43, 127, 128, 130, 131
imaginary
part,
14, 19, 21
imaginary unit, 14
injective, 142
integers,
14, 65, 73, 130, 165
intermediate
value
theorem, 31, 168
inverse
element, 13, 128
irreducible,
59-61, 76, 82, 105~108, 110—
112, 114, 116-119, 121, 122, 133,
136, 138, 140, 162, 167, 168, 171
isomorphic
groups,
110, 111, 144, 151,
172
isomorphism,
110, 155,
Jerrard, George Birch,
Jfjrgensen, Dieter, 4
168
90
Index
179
Kabayashi,
Sigeru, 90
Katscher,
Friedrich, 1
B. Melvin,
95
Kiernan,
King, R. Bruce, 90
Klein, Felix, 73
known
quantity,
97, 102-104
Koch, Helmut, 122
122
Kowol, Gerhard,
35
Lagrange
interpolation,
Lagrange
resolvent,
46, 78, 152, 157
Lagrange,
Joseph Louis, 43-47, 49, 52,
77, 82, 83, 94, 95, 115, 116, 129,
132, 169
Lang, Serge, 162
Laugwitz,
Detlef, 161
Lazard, Daniel, 91
Legendre symbol, 174, 175
172
Legendre,
Adrien-Marie,
12
Leibniz, Gottfried
Wilhelm,
lexicographic
order, 47, 48
Carl Louis
Ferdinand
Lindemann,
von,
75
algebra,
116, 150
linear
equation,
139, 150
linear
factor, 29, 30, 38, 45, 54-56, 5860, 76, 83, 97, 109, 116-118,
126,
140,172,174
linear
transformation,
136, 163
Liouville,
Joseph, 94
period, 3, 37, 66-68, 70-74, 77-80, 110,
112,156,
171, 175
periodic table, 166
permutation
cyclic, 43, 170
even,
88, 113, 170
identity, 43, 103, 105, 106
odd, 88, 108
Pertsinis,
Tom, 93
Pesic, Peter, 53
12
Pieper, Herbert,
Pierpont,
James, 82, 90
Platonic
solid, 130, 131
19
polar coordinates,
polynomial
elementary
symmetric,
38, 42, 44,
45, 47, 48, 50, 88, 115, 169
monic, 56-59, 61, 86
polynomial
ring, 115
primitive
element,
115, 149
primitive
root, 65, 66, 71, 73, 77, 80,
155, 171, 175
public key encryption,
xvi, 174
linear
Giovanni
Malfatti,
Francesco,
82, 83,
86, 87
134
mapping,
matrix,
130, 131
Matthiessen,
Ludwig, 26, 37
McKay, John, 106
modular
arithmetic,
66, 71, 77, 80, 131,
155
de
de
16, 76
16, 32,
Abraham,
Moivre’s formula,
Moivre,
34
monomial,
47, 48
multiple root, 39, 40, 97, 106
Nahin, Paul J., 21
Nakagawa,
Hiroshi,
Neuwirth,
Lee, 94
Newton,
Isaac, 45
non-Euclidean
90
geometry,
166
98
nonsymmetric
polynomial,
normal
subgroup,
126, 145, 147-150,
157-159,
168, 170
orbit,
order
parallel
176
of a group
postulate,
63, 73
pentagon,
element,
166
130
quadratic
equation,
ix, 1-3, 5, 27, 38,
40, 55, 67, 68, 71-74, 89, 171
175
quadratic
reciprocity,
see
quartic equation,
biquadratic
equation
xii, 37, 49, 81, 82,
quintic
equation,
86, 99, 163
126, 147, 149, 157, 159,
quotient group,
168, 170
radical
153, 158, 163, 171
extension,
Radlofi‘,Ivo, 51, 122
xi
Ramanujan,
Srinivasa,
rational
131-133
function,
rational
numbers,
31, 59-61, 69, 74,
75, 82, 96, 102,109,
114,117,123,
130-132,
136, 153, 155, 161, 163
real numbers,
13, 14, 16, 31, 32, 130
regular polygon, 63, 64, 70, 162
Reich, Karin, 28, 70
102
relational
polynomials,
residue
class, xvi, 131, 163, 165, 168,
175
171-173,
54
resultant,
Ribenboim,
Paulo, 72
F. J., 73
Richelot,
Toti, 93
Rigatelli, Laura
of unity,
root
46, 68, 70, 73, 78, 80,
109, 137, 152, 155, 158, 161, 175
Rothman,
A., 93
RSA encryption,
xvi, 173, 174
Ruffini, Paolo, 49-53, 82, 95, 170
Runge, C., 88
28, 70, 95, 127
Scholz, Erhard,
Schultz, Phillip, 4
180
Index
Schultze, Reinhard
Siegmund, 90
see
seventeen—gon,
heptadecagon
Skau, Christian,
51, 114
106
Soicher, Leonhard,
solution
formula, xii, xviii, xix, 2, 4,
37, 42, 44, 46, 50, 52, 55, '76, 77,
95
solvability
in
radicals,
xii-xiv,
81, 94, 157
xix, 50,
solvability of a group,
157, 160
Sossinsky, Alexei, 94
Spearman, Blair K., 91
splitting field, 109, 132-135, 139, 144,
146, 149, 153, 154, 167
the circle, 161
squaring
Stewart, Ian, 69
Stillwell, John, 53
straightedge and compass,
xi, 64, 69,
70,74,75,161,162
Stubhaug, Arild, 50
subgroup, xv, 125, 128-130, 139-147,
151, 153-157, 159, 160, 163, 168,
170-173, 176
normal,
153
subset,14,94,96,126-128,168
142
surjectivity,
Sylow, Ludwig, 176
symmetric
polynomial,
42, 44, 45, 47,
48, 50, 52, 88, 95, 98, 99, 115
Tartaglia,
1, 4,
9
52, 170
69
Tietze, Heinrich,
Tignol, Jean—Pierre,122, 162, 168
161
transcendence,
transitive
operation,
106, 140, 162
54
transposition,
32
triangle inequality,
an
trisecting
angle, 161
three-cycle,
Tschirnhaus
transformation,
Tschirnhaus,
Ehrenfried
Von,
83, 89,
90
Walther~«Graf
90
theorem
for symmetric
uniqueness
nomials, 48, 169
poly-
der
Waerden, Bartel
Leendert, xix,
19, 95, 149, 162, 171
Alexandre
Vandermonde,
Théophile,
47,
77, 78, 82, 83, 88, 94, 99
vector
space,
126, 130, 133, 136-138
Viete, Francois, 27
Viete’s root
theorem,
28, 29, 38, 40,
67,86
van
Weber, Heinrich, 88, 95, 127,
Wessel, Caspar, 12
Williams, Kenneth, S., 91
Wilson’s theorem,
173
zero
of
172
a
function,
149
118, 122, 133, 140,