Math $$$
HW
1
February 5, 2018
Due Wed, Feb 21. Be sure to justify all of your answers to receive full credit.
In this assignment we will show that a group of order 6 is either commutative, or is isomorphic
to S3.
1. Let G be a group. Suppose that for any element x ∈ G, x2 = e. Show that G is abelian.
2. Suppose G has even order, show that there is at least one element of G which has order 2.
3. Now let G = n. Suppose x ∈ G has order o(x) > n
2
. Show that o(x) = n.
4. Now let G be a group of order 6 such that G is not abelian.
(a) Show that G has an element σ of order 3, and an element τ of order 2.
(b) Show that στ = τσ2.
(c) Conclude that G is isomorphic to S3.
5. Saracino: 2.10, 3.1, 3.4, 3.6, 4.6, 4.13, 5.4, 5.6, 5.25
1
ABSTRACT
ALGEBRA
Second Edition
ABSTRACT
ALGEBRA
A First Course
Second Edition
Dan Saracino
Colgate University
WAVEIAND
PRESS, INC.
Long Grove, illinois
For information about this book, contact:
Waveland Press, Inc.
4180 IL Route 83, Suite 101
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www.waveland.com
Copyright © 2008, 1980 by Dan Saracino
I Odigit ISBN 1577665368
13digit ISBN 9781577665366
All rights reserved. No part of this book may be reproduced, stored in a retrieval
system, or transmitted in any form or by any means without permission in writing
from the publisher.
Printed in the United States of America
7 6 543 2
CONTENTS
o Sets and Induction ………… ……… …. …. ……………. ……… …. …….. … …………… …… 1
1 Binary Operations ….. …. … ……….. .. … ……… … ………. …. .. .. ……. ….. .. … .. ……….. 10
2 Groups .. … …………. ….. ……………… … ……….. …….. …………. ……. …. …. .. … ………. 16
3 Fundamental Theorems about Groups …… ……. .. … ………. …… …….. ………. …… 25
4 Powers of an Element; Cyclic Groups …………………………………………………. 33
5 Subgroups ………. ………. ……………. .. …….. ……. …….. …………… … ……. …….. …….. 43
6 Direct Products ………………………………………………………………………………… 55
7 Functions ………………………………………………………………………………………… 59
8 Symmetric Groups ………………………………………… ………………… ……….. ……. 66
9 Equivalence Relations; Cosets ……. … …. … …………… …………. …………… ……… 80
10 Counting the Elements of a Finite Group …………………………………………….. 88
11 Normal Subgroups …………………………………………………………………………… 99
12 Homomorphisms ……………………………………………………………………………. 109
13 Homomorphisms and Normal Subgroups ……………………………… ….. .. … …. 121
14 Direct Products and Finite Abelian Groups ……. … ………. ……… …. … …… …… 133
15 Sylow Theorems ………………………………………………………………… ….. … .. …. 143
16 Rings ……………………………………………………………………………………………. 153
17 Subrings, Ideals, and Quotient Rings ………………….. …. ………… .. ……. ……… 164
18 Ring Homomorphisms ……………………………………………………………………. 177
19 Polynomials ……………… ….. ……. ….. ………….. ………… ……….. …………….. …. … 191
20 From Polynomials to Fields …. ……… …….. …… .. ……. ……………… ……….. … … 205
21 Unique Factorization Domains …… …. ….. …….. ………… ………. ……… … ………. 211
22 Extensions of Fields …………….. ….. .. …. ………………….. …………… …….. …… .. . 227
23 Constructions with Straightedge and Compass ………. .. ………… ……….. … .. .. 240
24 Normal and Separable Extensions … ….. …. ………………….. ……… ……. ….. … … 249
25 Galois Theory ………………………………………………………………………………… 265
26 Solvability ……….. … ……….. ………………. …… …. …………………………………….. 279
Suggestions for Further Reading …. .. ………. …………… ………… ….. …. …….. … 297
Answers to Selected Exercises ………………….. ……. …….. …………. … …. ……… 301
Index ……………………………………………………………………………… …….. ……. .. 307
PREFACE
This book is intended for use in a juniorsenior level course in abstract
algebra. The main change from the first edition is the addition of five new sections
on field extensions and Galois theory, providing enough material for a two
semester course. More minor changes include the simplification of some points in
the presentation, the addition of some new exercises, and the updating of some
historical material.
In the earlier sections of the book I have preserved the emphasis on providing
a large number of examples and on helping students learn how to write proofs. In
the new sections the presentation is at a somewhat higher level. Unusual features,
for a book that is still relatively short, are the inclusion of full proofs of both
directions of Gauss’ theorem on constructible regular polygons and Galois ‘
theorem on solvability by radicals, a Galoistheoretic proof of the Fundamental
Theorem of Algebra, and a proof of the Primitive Element Theorem.
A onesemester course should probably include the material of Sections 013 ,
and some of the material on rings in Section 16 and the following sections.
Sections 14 and 15 allow the inclusion of some deeper results on groups. The
results of Section 14 are used in Section 15, and the First Sylow Theorem from
Section 15 is used in Sections 25 and 26.
In two semesters it should be possible to cover the whole book, possibly
omitting Section 21.
I want to express my appreciation to my students who used the manuscript for
the five new sections as a text and pointed out to me parts of the presentation that
needed clarification. I also want to thank all those who have sent me comments
about the book over the years, and those who suggested that a new edition would
be a good idea. I hope this second edition will be useful.
Dan Saracino
SECTION 0
SETS AND INDUCTION
One of the most fundamental notions in any part of mathematics is that of a
set. You are probably already familiar with the basics about sets, but we will
start out by running through them quickly, if for no other reason than to
establish some notational conventions. After these generalities, we will make
some remarks about the set of positive integers, and in particular about the
method of mathematical induction, which will be useful to us in later proofs.
For us, a set will be just a collection of entities, called the elements or
members of the set. We indicate that some object x is an element of a set S by
writing xES. If x is not an element of S, we write x f£ S.
In order to specify a set S, we must indicate which objects are elements of
S. If S is finite, we can do this by writing down all the elements inside braces.
For example, we write
S={1,2,3,4}
to signify that S consists of the positive integers 1,2,3, and 4. If S is infinite,
then we cannot list all its elements, but sometimes we can give enough of
them to make it clear what set S is. For instance,
S= {1,4, 7,10,13,16, … }
indicates the set of all positive integers that are of the form 1 + 3k for some
nonnegative integer k.
We can also specify a set by giving a criterion that determines which
objects are in the set. Using this method, the set {l,2,3,4} could be denoted
by
{xix is a positive integer ~4},
where the vertical bar stands for the words “such that.” Likewise, the set
{1,4,7, 10, 13, 16, … } could be written as
{xix = 1 + 3k for some nonnegative integer k}.
1
2 Section O. Sets and Induction
Some sets occur so frequently that it IS worthwhile to adopt special
notations for them. For example, we use
Z to denote the set of all integers,
Q to denote the set of all rational numbers,
R to denote the set of all real numbers, and
C to denote the set of all complex numbers.
The symbol 0 denotes the empty set or null set, i.e., the set with no elements.
Sometimes we wish to express the fact that one set is included in another,
i.e., that every element of the first set is also an element of the second set. We
do so by saying that the first set is a subset of the second.
DEFINITION If Sand T are sets, then we say that S is a subset of T, and write
S k T, if every element of S is an element of T.
Examples If S={1,2,3} and T={l,2,3,4,5}, then SkT.
If S= {7T, V2} and T= {7T,5, V2 }, then S k T. We write S g{5, V2}
because 71′ E S but 71′ f1. {5, V2 }.
If we let
z+ = {xix is a positive integer},
then Z+ kZ; similarly, we have Q+ kQ and R+ kR
Observe that for any set S, S k S, that is, S is a subset of itself. Also
observe that 0 k S, no matter what set S is. Perhaps the best way to see this is
to ask yourself how it could be false. If 0 g S, then there is some x E0 which
is not in S; but this is nonsense, because there is no x E0, period.
We say that two sets Sand T are equal, and we write S= T, if S and T
have the same elements. Clearly, then, saying that S = T is equivalent to
saying that both S k T and T k S. If S k T but S=I=T, we say that S is a
proper subset of T. If we wish to emphasize that S is a proper subset, we write
S 1 T.
Very often we consider sets that are obtained by performing some
operation on one or more given sets. For example, if Sand T are sets, then
their intersection, denoted by S n T, is defined by
S n T= {xlxES and xE T}.
The union of Sand T, denoted by S U T, is given by
S U T = { x I xES or x E T or both}.
Section O. Sets and Induction 3
The union and intersection of more than two sets are defined in an analogous
way; for instance,
S n Tn U = {xix E S and x E T and x E U}.
Examples Let S={1,2,3,4,5} and T={2,4,6}. Then SnT={2,4} and
S U T= {l,2,3,4,5,6}.
Again let S= {l,2,3,4,5}. Then S n ~= Sand S u ~= IR.
We can illustrate many of the notions we have introduced by generalizing
this last example.
THEOREM 0.1 Let Sand T be sets. Then S ~ T if and only if S n T= S.
PROOF. We must show that S ~ T implies S n T= S, and that conversely
S n T= S implies S ~ T.
Assume S ~ T. To show that S n T= S we have to show that S n T ~ S
and S ~ S n T. The first is clear; for the second we must show that every
element of S is an element of S and of T. Clearly any element of S is an
element of S; and since we are assuming S ~ T, any element of S is an
element of T, too, so we are done with the first half of the proof.
Now assume S n T= S; we show that S ~ T. Why is it true that any
element of S is an element of T? Because any element of S is an element of
S n T by our assumption, and any element of S nTis clearly an element of
T·O
It is also true that S ~ T if and only if S u T= T. The proof of this is left
as an exercise.
As a matter of notation, we adopt the abbreviation “iff” for “if and only
if.” Thus we say that S ~ Tiff S u T= T. Sometimes the symbol ~ is used in
place of iff; using ~, we would write S ~ T~S U T= T.
One set that is particularly important in mathematics is the set 1.. + of
positive integers. We will see that, in abstract algebra, concepts defined in
terms of positive integers can often help to clarify what is going on. For this
reason, methods for working with integers can be very valuable tools. Perhaps
the most useful strategy for proving things about 1..+ is the method of
mathematical induction.
Suppose we have in mind a statement P(n) about the integer n. For
example, P(n) might say “n is even,” or “n is the square of some integer,” or
“If p is a prime, then every group of order p n has nontrivial center” (whatever
that means). Mathematical induction provides us with a way of trying to
prove that P(n) is true for every positive n.
4 Section O. Sets and Induction
The technique rests on an intuitively acceptable axiom called the
WellOrdering Principle: Every nonempty subset of 7L.+ has
a smallest element.
The wellordering principle yields two slightly different forms of induction,
both of which are good to know.
nIEOREM 0.2 (Mathematical Induction, first form) Suppose pen) is a state
ment about positive integers, and we know two things:
i) pel) is true;
ii) for every positive m, if P( m) is true, then P( m + 1) is true.
Under these circumstances, we can conclude that pen) is true for all positive
n.
PROOF. Suppose pen) is false for some positive n. Then S={nlnE7L.+ and
P( n) is false} is a nonempty subset of 7L. +. By the wellordering principle, S
has a smallest element, say no’ Clearly no*” 1, because P(1) is true by (i).
Therefore, no – 1 is a positive integer, and P( no – I) is true because no – I is
smaller than no’ By (ii), this means that P(no I + I) is true; that is, P(no) is
true, and this contradicts the fact that P(no) is false!
Since the supposition that Pen) is false for some n has led us to a
contradiction, we conclude that P( n) holds for all n E 7L. +. 0
What you do to prove something by induction, then, is this. You first
show that P(I) is true (this is usually trivial). You then show that for an
arbitrary positive m, if P(m) is true, then P(m + I) is true. You do this by
assuming that P(m) is true and using this assumption to establish that
P(m + 1) is true.
Sometimes people are bothered by the word “assuming” in the last
sentence. They get worried that “assuming that P(m) is true” amounts to
assuming what we are trying to prove. But it does not, for we are not assuming
that P(m) is true for all m. Rather we are arguing, for an arbitrary fixed m,
that if P( m) is true for that m, then so is P( m + I). The only way of doing this
is to show that P( m + 1) is true on the basis of the assumption that P( m) is.
Examples
1. You may recall that, in calculus, when you are evaluating definite
integrals from the definition as a limit of Riemann sums, you run into sums
such as 1 + 2 + 3 + … + n, and you need formulas for these sums in terms of
n. The formula for 1 +2+··· + n, for example, is n(n+ 1)/2. Let’s prove it,
by induction.
We take for pen) the statement that
1+2+··· +n=n(n+l)/2;
Section O. Sets and Induction 5
we hope to show that P(n) is true for all positive n. First we check P(l): It
says that 1 = 1(1 + 1)/2, which is certainly true. Second, we assume that P(m)
is true for some arbitrary positive m, and we use this assumption to show that
P(m + 1) is true. That is, we assume that
1 +2+··· + m=m(m+ 1)/2 [0.1]
and we try to show that
1 + 2 + … + m + m + 1 = (m + 1)( m + 1 + 1)/2.
The natural thing to do is to add m + 1 to both sides of Eq. [0.1]:
1 +2+··· +m+m+ 1 = [m(m+ 1)/2] +m+ 1.
Now
[U]
[m(m+ 1)/2] +m+ I = [m(m+ 1)+2(m+ 1)]/2= [(m+ 1)(m+2)]/2,
so we have Eq. [0.2].
Therefore, by induction, P(n) holds for all positive n, and we are done.
2. A similar formula, also used in calculus, is
12+22+ … + n2= n(n + 1~(2n+ 1) .
If we take this equation as P(n), then P(I) says
12= 1(1+1)(2+1)
6 ‘
which is true. To show that P(m) implies P(m+ I), we assume that
12+22+ … +m2= m(m+ 1)(2m+ I)
6 ‘
and we try to show that
2 2 2 (m+ 1)(m+2)[2(m+ 1)+ 1]
1 +2 + … +(m+ 1) = 6 .
If we add (m+ 1)2 to both sides of Eq. [0.3], we conclude that
12+22+ … +m2+(m+ 1)2= m(m+ 1!(2m+ 1) +(m+ 1)2,
and the righthand side is
m(m+ 1)(2m+ 1) +6(m+ Ii (m+ 1)[ m(2m+ 1)+6(m+ 1)]
6 = 6
[0.3]
[004]
(m+ 1)(2m2+7m+6) (m+ 1)(m+2)(2m+3)
= 6 = 6 ‘
so we have Eq. [0.4].
Thus our formula is established by induction.
6 Section O. Sets and Induction
3. The following popular example illustrates the fact that some care is
necessary in trying to prove something by induction. We shall “prove” that
all horses are the same color.
Let P(n) be the statement: “For every set of n horses, all the horses in the
set are the same color.” We “prove” P(n) for all n by induction. Clearly, P(l)
is true since any horse is the same color as itself. Now assume that P(m) is
true and let us show that P(m + 1) holds. Let S be a set of m+ 1 horses; say
the horses in S are h l ,h2, ••• ,hm + 1 (h for horse). Now h l ,h2, ••• ,hm comprise a
set of mhorses, so since P(m) holds, hl ,h2, ••• ,hm are all the same color.
Likewise, h2, h3, • •• , hm + I make up a set of m horses, so h2, h3, ••• , hm + I are all
the same color. Combining these statements, we see that all m + I horses are
the same color (for instance, they are all the same color as h0.
There must be something wrong with this, but what?
The second form of induction is similar to the first, except that (ii) is
modified somewhat.
THEOREM 0.3 (Mathematical Induction, second form) Suppose P(n) is a state
ment about positive integers and we know two things:
i) P(l) is true;
ii) for every positive m, if P(k) is true for all positive k
the only positive integers that divide p are 1 and p. The first few primes are 2,
3,5,7, 11, 13, 17, ….
Section O. Sets and Induction 7
The Fundamental Theorem of Arithmetic asserts that every positive n> 1
can be written as the product of finitely many primes, and that except for a
possible rearrangement of the factors, there is only one such factorization.
TIlEOREM 0.4 (Fundamental Theorem of Arithmetic) Let n > 1. Then there are
primes PI’ P2′ P3″” ,Pr (not necessarily distinct) such that n = PIP2′ .. Pr’
Moreover, if n = ql q2′ .. qs is another such factorization, then r = s and the p;’s
are the q/s, possibly rearranged.
PROOF. We prove only the existence of a factorization here. The proof of the
uniqueness assertion requires some more groundwork and is left to the
exercises in Section 4.
We wish to prove that pen) holds for every n ~ 2, where pen) says that n
can be written as a product of primes. Accordingly, we start our induction at
2 rather than at 1; we show (see Exercise 0.13) that P(2) is true and that for
any m > 2, if P(k) holds for all 2 <. k
holds. If m is not a prime, then we can write m = ab, where neither a nor b is
m. Thus 2 <. a, b
0.9 Prove that
for all n> l.
0.10 Prove that
for all n> l.
1 +3+5+ … +(2n+ I)=(n + 1)2
2 + 4 + 6 + … + 2n = n( n + 1)
0.11 Prove Theorem 0.3.
0.12 Prove the following more general form of Theorem 0.2:
THEOREM. Suppose P(n) is a statement about positive integers and c is some
fixed positive integer. Assume
i) P(c) is true; and
ii) for every m >c, if P(m) is true, then P(m+ 1) is true.
Then P(n) is true for all n >c.
0.13 Prove the following more general version of Theorem 0.3:
THEOREM. Suppose P(n) is a statement about positive integers and c is some
fixed positive integer. Assume
i) P(c) is true; and
ii) for every m >c, if P(k) is true for all k such that c
1· 2 + 2·3 +3 ·4+ … + (n – I)n = ‘.,(n_—–“I).”,(n.<)'.,(n_+_1....<..,)
3
0.15 By trying a few cases, guess at a formula for
1 1 1 1
+++ ... + n>2
1·2 2·3 3·4 (nI)n’ .
Try to prove that your guess is correct.
0.16 Prove that for every n> 1, 3 divides n3 – n.
0.17 Prove that if a set S has n elements (where nEZ+), then S has 2n subsets.
Section O. Sets and Induction 9
0.18 The Fibonacci sequence 11.J2.J3,'” is defined as follows:
11 = 12= 1, 13=2, 14=3, Is=5, 16=8, … ,
and in general,
In=lnl+ln2 foralln~3.
Prove thatlsk is divisible by 5 for every k;;;. 1, that is, 5 divides every 5th member
of the sequence.
0.19 As in the preceding exercise, letfi denote the kth Fibonacci number. Prove that for
every n ~ 1,
g+l /,,[,,+2 = (1)”.
0.20 Again let//; denote the kth Fibonacci number. Let
l+JS IJS
a=– and p=–.
2 2
Prove that for every n ~ 1
a” _po
1.= JS .
0.21 The nth Fermat number is Fn = 2(2″) + 1. Prove that for every n ~ I
F’oF;Fz … Fn_1 = F” – 2.
0.22 Prove that the following statement is true for every integer n ~ 1: If the number of
squares in a “checkerboard” is 2″ x 2″ and we remove anyone square, then the
remaining part of the board can be broken up into Lshaped regions each consisting of
3 squares.
SECTION 1
BINARY OPERATIONS
In high school algebra, we spend a great deal of time solving equations
involving real or complex numbers. At the heart of it, what we are really
doing is answering questions about the addition and multiplication of these
numbers.
In abstract algebra, we take a more general view, starting from the
observation that addition and multiplication are both just ways of taking two
elements and producing a third, in such a way that certain laws are obeyed.
We study the situation where we have a set and one or more “operations” for
producing outputs from given inputs, subject to some specified rules.
From this description, it i~ probably not clear to you why abstract algebra
is any more profitable than going off and counting the grains of sand on the
nearest beach. But it can be very profitable, for a number of reasons. For one
thing, the abstract approach may clarify our thinking about familiar situa
tions ~y stripping away irrelevant aspects of what is happening. For another,
it may lead us to consider new systems that are valuable because they shed
light on old problems. Yet again, a general approach can save us effort by
dealing with a number of specific situations all at once. And finally, although
it can take some time to appreciate this, abstraction can be just plain
beautiful.
DEFINITION If S is a set, then a binary operation * on S is a function that
associates to each ordered pair (S),S2) of elements of S an element of S, which
we denote by s) * S2′
Observe that the definition says ordered pair. Thus (S),S2) is not neces
sarily the same thing as (S2’S)), and S) * S2 is not necessarily the same thing as
S2 * s). Notice too that * must assign an element of S to each and every pair
(S),S2)’ including those pairs in which s) and s2 are the same element of S.
10
Section 1. Binary Operations 11
Examples
1. Addition is a binary operation on Z+: a * b = a + b. Subtrac
tion (a *b = a – b) is not; but subtraction is· a binary operation on Z.
2. Multiplication (a*b=a·b) is a binary operation on Z+ or Z or ~ or Ill.
Division (a * b = a / b) is a binary operation on III + or ~ + but not Z or Z + or
~ or Ill.
3. a*b=a3 +b2 +1 is a binary operation on Z, 1Il,~, Z+, 1Il+, or ~+.
4. Let X be some set and let S be the set of all subsets of X. For example,
if X={1}, then S={0,{I}}, and if X={1,2}, then
S = {0, { I }, {2 }, { 1,2} }.
The operation of intersection is a binary operation on S, since if A,B are
elements of S, then A * B = An B is an element of S. Similarly, union gives us
another binary operation on S.
S. Let S be the set of all 2 X 2 matrices with real entries. Thus an element
of S looks like (: !), where a,b,c,deR. Define x*y to be the matrix
product of x and y, that is,
f)=(ae+bg
h ce+dg
Then * is a binary operation on S. For instance,
af+bh).
cf+dh
I ) _(.”
1 I
3.,,) o .
The definition of binary operation doesn’t impose any restrictions on *,
and in general a binary operation can be wildly misbehaved. Ordinarily, we
want to consider operations that have at least something in common with
familiar, concrete examples.
DEFINITIONS If * is a binary operation on S, then * is called commutative if
3 1 *32 =32 *31 for every 3 1,32 E S. On the other hand, * is called associative if
(31 *32)*33 =SI * (S2 *33) for every 31,32,S3ES.
Examples
1. Subtraction on Z is neither commutative nor associative. For example,
12::;=21 and (12)3*1(23).
2. Let a*b=2(a+b) on Z; then clearly * is commutative. Now
(a * b) *c =2(a + b) *c=2(2(a + b) + c)=4a+4b+2c,
12 Section 1. Binary Operations
and
a. (b.c) = a .2(b+ c)=2(a+2(b+ c»=2a +4b+4c.
Since these are not always the same, • is not associative.
3. Let a. b = 2ab on lL. Commutativity is again clear, and since
(a.b).c =2ab .c=4abc and a. (b*c)= a*2bc=4abc,
* is also associative in this case.
4. Multiplication of 2 X 2 matrices is associative:
as may be verified by multiplying out both sides. This operation is not
commutative, however. For example,
(01 01)*(31 42)=(31 4) (1 2) (0 1) (2 1) 2 but 3 4 * 1 0 = 4 3·
Before we consider a slightly more complicated example, observe that
these first four examples say something about the relationship between
commutativity and associativitynamely, that there isn’t any! A binary
operation can be commutative with or without being associative, and it can
be noncommutative with or without being associative.
5. We introduce another operation on sets: If A,B are sets, then A t:;.B
denotes the symmetric difference of A and B. This is by definition the set of
all elements that belong to either A or B, but not to both. Thus
A t:;.B=(A – B)U(BA),
where A – B denotes the set of elements of A that are not elements of B, and
BA denotes the set of elements of B that are not elements of A. We can
also write
At:;.B=(AUB)(AnB).
In Fig. 1.1, if A is the set of points inside the square, and B is the set of
points inside the circle, then At:;. B consists of the points in the shaded region:
FIgure 1.1
Section 1. Binary Operations 13
For example, if A = {1,2,3,4,5} and B= {2,4,6}, then A .::.B= {l,3,5,6}.
If
A={xlxElandx>O} and B={xlxElandx
rn=a(q+ l)n > 0,
so (q+ l)n is a multiple of n that is <; a, contradicting the choice of qn as the
greatest such multiple.
so rzrl is a multiple of n; but n< rzrl < n, so rzrl=O, so rz=rl. Thus
q1n=qzn, so ql =qz· 0
We denote the unique r guaranteed by the lemma by ii, and call it the
remainder of a mod n.
Observe that two integers a l and az have the same remainder mod n iff
a l  az is a multiple of n. When this happens, we say that a l and az are
congruent modulo n, and we write a l == az(mod n).
Examples Let n=7. Then if a= 16, we get 16=2·7+2, so q=2, r=2, and
16=r=2. If a=35, we get 35=5'7+0, so 35=0.
By the lemma we can unambiguously define a binary operation E9 on In
by setting xE9y =x+ y. For example, if n=7, then In = {O, 1,2,3,4,5,6}, and
3E92=3+2=5=5, 3E94=3+4=7=0, and 6E93=6+3=9.=2.
We claim that this operation turns In into a group (In, e).
Associativity: The question is whether (xE9y)E9z = xE9(yez), that is,
 ? 
x+y E9z = xey+z,
? ===
x+y+z=x+y+z.
Well,
x+y +z = (x+y)+z
since x+y and (x+y) differ by a multiple of n; similarly,
x+ y+z = x+(y+z).
So all we have to check is whether (x+y)+z equals x+(y+z). But this is
obviously so, since (x+y)+z=x+(y+z).
22 Section 2. Groups
Identity: The identity element is O. Why? Qffix=O+x=x=x if O<;x<;nl,
and, similarly, xffiO= x.
Inverses: We know that 0 is an inverse for 0, since OffiO=O+O=O=O, the
identity element. Now if x * 0, then x E {I, 2, ... , n  I}, so n  x E
{1,2, ... ,nl}, and we see that nx is an inverse of x:
xffi(nx)= x+nx =n=O,
and, similarly, (nx)ffix=O.
For example, in (1:.7' ffi), 1 is an inverse for 6, and 3 is an inverse for 4. In
(1:.)3' ffi), 5 is an inverse for 8.
The group (1:.n' ffi) is called the additive group of integers mod n. Notice
that, for n = 1, we have 1:.) = {OJ, so (1:.), ffi) is a group with only one element
in it. In general, any group having only one element is called trivial. If x is
any object whatsoever (e.g., x = Whistler's Mother), then we get a trivial
group ({x},*) by defining x*x=x.
EXEROSES
2.1 Which of the following are groups? Why?
a) R+ under addition
b) The set 3l of integers that are multiples of 3, under addition
c) R  {O} under the operation a. b = labl
d) The set p, I} under multiplication
e) The subset of Q consisting of all positive rationals that have rational square
roots, under multiplication
f) The set of all pairs (x,y) of real numbers, under the operation (x,y).(z, w)
"'(x+z,y w)
g) The set of all pairs (x,y) of real numbers such thaty #0, under the operation
(x,y).(z, w)=(x+ z,yw)
h) R {I}, under the operation a*b=a+ bab
i) l, under the operation a. b = a + b  1
2.2 a) Of those examples in Exercise 2.1 that are groups, which are abelian?
b) Which of the groups in Examples 18 on pp.1720are abelian?
2.3 Let X be a set and let P(X) be the power set of X. Does P(X) with the binary
operation A • B = A n B form a group? How about P(X) with the binary
operation A .B=A UB?
Section 2. Groups 23
1.4 The operation in a finite group can be specified by writing down a table (see
Exercise 1.9). Write down the tables for the following.
a) (l4' $)
b) (ls, $)
c) (l6, $)
1.5 The following table defines a binary o~ration on the set S= {a,b,c}.
Is (S,.) a group?
• abc
a
b
c
a
b
c
b
b
c
c
c
c
1.6 The following table defines a binary operation on the set S={a,b,c}.
• a b c
a a b c
b b a c
c c b a
Is (S,.) a group?
1.7 Let S = {a,b}. Write down a table that defines a binary operation. on S such
that (S,.) is a group. Show that your table works.
1.8 Let G be the set of all realvalued functions f on the real line which have the
property that f( x) * 0 for all x E R. Define the product f X g of two functions
f,ginGby
(fXg)(x)= f(x)g(x) for all xEIIl.
With this operation, does G form a group? Prove or disprove.
1.9 a) Show that for 2 X 2 matrices A and B,
determinant of AB=(determinant of A)(determinant of B).
b) Show that a 2 X 2 matrix A is in GL(2, R) iff the determinant of A is not O.
c) Use the results of (a) and (b) to give another proof that GL(2, Ill) is a group
under matrix multiplication.
1.10 Let G be the set of all 2 X 2 matrices (a b), where a, b E III and a2 + b2 * O.
b a
Show that G forms a group under matrix multiplication.
1.11 Let G be the set of a1l2x2 matrices (~ ~), where a and b are nonzero real
numbers. Show that G forms a group under matrix multiplication.
l.ll Let G be the set of all triples (a,b,c) such that a,b,c are elements of~. Define * by
(aJ, bJ, CI) * (a20 b2, C2) = (al $ a2, bl $ b" CI $ C2 E9 bla~,
where all additions and multiplications are performed mod 3. Prove that (G, *) is a
group.
24 Section 2. Groups
2.13 Let al,a2, ... ,an be elements of a group G. Show that
has an unambiguous meaning in the sense that no matter how we insert
parentheses into the expression to indicate the order in which the multiplications
are carried out, we always get the same result. [Suggestion: Show that any
insertion of parentheses gives the same answer as
al. (a2. (a3. (a4 • ...• (anl .an)·.· »).
To do this, use induction on n.]
2.14 If Xis a set and At, ... ,An are elements of (P(X),6.) then by Exercise 2.13 A lM 26.· . 'Mn
has an unambiguous meaning. Prove that for every n ;::: 1 the elements of X that are in
AlM 26.· . 'M" are exactly those elements that are inAj for an odd number ofj's in {l,
2 •...• n}.
SECTION 3
FUNDAMENTAL THEOREMS
ABOUT GROUPS
If we hope to get anywhere in working with groups, there are certain
fundamental facts about their behavior that we must master at the outset. The
operation * in a group comes to us endowed only with the properties given to
it by the group axioms. Everything else must follow from these, and our first
task is to use the axioms to set down some basic rules of operation that
enable us to carry out with ease at least some elementary calculations.
First, a convention: We usually call the operation in a group "multiplica
tion"; but very often the operation is called "addition" if the group happens
to be abelian.
1HEOREM 3.1 (Uniqueness of the identity element) If (G, *) is a group, then
there is only one identity element in G.
PROOF. We must establish that if e and e} are two elements of G both of
which satisfy the defining property of an identity element in G, then in fact
e= e}. That is, we assume that x *e= e*x= x for all x in G and x *e} = e} *x
= x for all x in G, and we then proceed to show that e must equal e}.
By the assumption on e, we have in particular (taking x to be e})
e}*e=e*e}=e}.
By the assumption on e}, we have (taking x to be e)
e*e}=e}*e=e.
Thus we have e} = e * e} = e, and the proof is complete. 0
1HEOREM 3.2 (Uniqueness of inverses) If (G, *) is a group and x is any
element of G, then x has only one inverse in G.
PROOF. We must show that if both y} and Y2 satisfy the definition of an
inverse of x, that is, if x*y}=y}*x=e and X*Y2=Y2*x=e, then in fact
Y}=Y2·
25
26 Section 3. Fundamental Theorems about Groups
We will take some of the given information and use it to derive an
equation which has YI on one side and Y2 on the other. It may leap to your
eye, for example, that
because both sides are e. Once this is seen, all that remains is to get rid of the
x's. We can do this by using more of the given information, namely the fact
thatYI*x=e. We multiply both sides by YI:
YI*(x*YI)=YI*(X*h)·
By using associativity, we get from this
as desired. 0
(YI*X)*YI =(YI*X) *Y2'
e*YI = e *Y2'
YI=Y2'
The proof is finished, but we are going to do it again in a slightly
different way to illustrate the fact that there are often several different ways
to prove something. Suppose, for example, that the equation x * Y I = X * Y2 did
not leap to your eye, and that you just took one of the equations given, say
to start with. If you want to get from this an equation withYI on one side and
h on the other, then certainly you observe that YI is already on the left, and
we can get Y2 on the right by multiplying both sides by h:
(YI *x)*h=e*Y2
=Y2·
Now if only we could get rid of the x and Y2 on the lefthand side, we'd be
done; but we know from the equations we had to start with that x * Y2 = e, so
we rewrite the left side by using associativity, so as to bring x * Y2 into play:
and we are done. 0
Y I * (x * h) = h,
Yl*e=Y2'
YI=Y2'
We emphasize the significance of what we have just proved twice: The
group axioms simply assert that for any x, there must exist an inverse; our
theorem says that once you know you've got a group, any x has precisely one
inverse.
Section 3. Fundamental Theorems about Groups 27
Example Let (G, *) be G L(2, R). Then by applying our general result in this
specific case, we conclude that if (: !) is an invertible matrix, then there is
only one matrix (; ~) such that
0) I .
This fact can of course be established directly, by working with systems
of linear equations in two variables. But one is struck by the economy and
elegancethe "cleanness" of the proof we have obtained by viewing the
collection of all invertible 2 X 2 matrices as a group.
Henceforth we will usually denote the unique inverse of x by x I. When
we are dealing with an abelian group and referring to the group operation as
addition, however, we will sometimes denote the inverse of x by  x. For
example, in GL(2, IR) we write
5)1={ 7
7 4
5)
3 '
and in (Z7' $) we write  3 =4.
The next result will enable us to conclude that no two distinct elements of
a group G can have the same inverse.
TIIEOREM3.3 If (G,*) is a group, then for any xEG we have (XI)I=X.
PROOF. Since XI is the inverse of x, we know that xI*x=x*xI=e. By
these equations, x satisfies the definition of (XI)I, so x=(XI)1 by the
uniqueness of inverses. 0
That was slick, but let's do it again in a slightly different way. We know
that x  I * X = e. We want to get from this to an equation with x on one side
and (x  I)  I on the other . We need to get rid of the x  I on the left side, so
let's multiply both sides by (XI)I:
(XI) I * (XI *x)= (XI) I *e
(x I) I *x I) *x = (XI) I (using associativity on the left)
e*x=(xI)I
x=(xI)I. 0
28 Section 3. Fundamental Theorems about Groups
Example Let G = (Z, +). Then for this example the theorem says that  (  n)
= n for any integer n.
Now suppose x,y are elements of a group (G,*) and XI=yI. Then by
taking inverses on both sides we get (X I)I=(yI)I, so, by the theorem,
x = y. Thus, as promised, if two elements have the same inverse then they
must in fact be the same element. It is possible to prove this without reference
to the preceding theorem; see Exercise 3.8(b).
Next we examine the inverse of a product.
THEOREM 3.4 If (G, *) is a group and x, y E G, then
PROOF.
= x * ((y *y I) *x I) = x * (e*x I) = x*x I = e,
and similarly we can show that (yI*xI)*(x*y)=e. (Do it!) Thus the
elementyI*x I satisfies the conditions that define (x*y)\ and since we
already know that inverses are unique, this implies that y I * X  I and
(x * y) I must be the same element. 0
It is worth emphasizing the reversal of order in the above result. In
general, it is not true that (x * y)  1 is X I * Y  I. This does, of course, hold true
in abelian groups, for then x I * Y I is the same thing as y I * XI; in fact,
the equation (x * y)I = XI * Y I holds for all x, y in a group G if and only if
G is abelian (Exercise 3.9).
It is somewhat bothersome to have to check both the conditions
(x*y)*(yI*xI)=e and (yI*xI)*(x*y)=e in the above theorem and,
in fact, it is not hard to show that it is really sufficient to check either one of
them.
THEOREM 3.5 Let (G, *) be a group and let x, y E G. Suppose that either
x * y = e or y * x = e. Then y is x  I.
PROOF. Suppose that x*y=e. We wish to solve this equation for y, so let's
mUltiply both sides by x  I :
Section 3. Fundamental Theorems about Groups 29
Thus
(X1*X)*y=X 1,
e*y=xt,
y=x 1.
A similar argument shows that y * x = e is also by itself sufficient to guarantee
thaty=x 1. (Do it!) 0
The same solving of an equation proves the more general
THEOREM 3.6 (Cancellation laws) Let (G,*) be a group and let x, y, zEG.
Then:
i) if x * y = x * z, then y = z; and
ii) ify*x=z*x, theny=z.
The proof is left as an exercise. Part (i) is called the left cancellation law,
and Part (ii) the right cancellation law.
We close this section by giving another formulation of the axioms for a
group which is equivalent to our original definition, but somewhat simpler to
work with in establishing that some system is in fact a group.
THEOREM 3.7 Let G be a set and * an associative binary operation on G.
Assume that there is an element eEG such that x*e=x for all xEG, and
assume that for every x E G there exists an element y in G such that x * y =
e. Then (G,*) is a group.
The element e is called a right identity, and the element y associated to x
is called a right inverse of x. In order to prove the theorem we have to show
that G satisfies all the axioms for a group, and since we have assumed that *
is a binary operation on G and * is associative, we have only to verify that e
is, in fact, also a left identity, that is, e * x = x for all x E G; and that a right
inversey of x is also, in fact, a left inverse of x, that is,y*x=e.
PROOF OF THE THEOREM. First we show that e *x = x for all x E G. Let us do
the proof backwards by trying to obtain some equations that we know would
yield e * x = x. Let x' denote a right inverse of x. Certainly it would be enough
to have
(e*x) *x' = x *x', [3.1)
for then we could mUltiply both sides by a right inverse of x'. But having [3.1]
is the same thing as having
e * (x * x') = x * x',
30 Section 3. Fundamental Theorems about Groups
by associativity, and this is the same thing as having
by the definition of the right inverse x'. Certainly we know that e * e = e,
because x*e=x for any xEG.
We have proved e*x=x, because each successive equation in our proof
implied the one before it, so that when we finally arrived at a true equation,
its truth implied the truth of all the previous equations. It is crucial to realize
that in this situation it would not have sufficed to have each equation
implying the one after it. In other words, if we want to prove e * x = x, it
suffices to show that e * x = x is implied by the true statement e * e = e, but it
would not be enough to show that e*x=x implies e*e=e. (A simple
example: The false statement" 1 = 1" implies the true statement "I = 1," as
we see by squaring both sides; but that doesn't prove  1 = 1.)
Now let's finish the proof of Theorem 3.7 by showing that a right inverse
x' of x is also a left inverse of x, that is, x' * x = e. We know that there is some
(x')' such that x' * (x')' = e, and it will suffice to show that x = (x')'. Now from
x*x'=e
we get
(x *x') * (x')' = e * (x')',
and since we know that e is a left identity, this yields x = (x')'. D
An analogous proof shows that assuming associativity and the existence
of a left identity and left inverses is also sufficient to guarantee a group. It
should be observed, however, that associativity plus the existence of a right
identity and left inverses (or a left identity and right inverses) is not enough.
Example Consider the set Z with the binary operation given by
It is easy to check that * is an associative binary operation on Z and that 1 is
a right identity element and also a left inverse for every element of Z.
However (Z, *) is not a group since, for example, there is no twosided
identity element in (Z, *).
Since the axioms in Theorem 3.7 are manifestly simpler than those in our
definition of a group, you may wonder why we didn't use the simpler version
as the definition and then show that the stronger axioms follow. We could
have done so, but we decided against it in favor of emphasizing the fact that
in a group the identity element and inverses work from both sides.
Section 3. Fundamental Theorems about Groups 31
EXERCISES
3.1 In (Z12,EB), solve the equation 2EBxEB7= I for x.
3.2 Let X= {l,2,3,4,5,6, 7,8,9, IO}. In (P(X), 6), consider the elements A =
{l,4,5,7,8} and B= {2,4,6}, and solve A u= B for x.
3.3 Find elements A,B, C of GL(2, IR) such that AB = BC but A =foe.
3.4 Let g be an element of a group (G, • ) such that for some one element x E G,
x.g=x. Show that g=e.
3.5 If (G,.) is a group and x, y, z E G, then we can unambiguously write x • y *Z to
denote either (x .y) *Z or X. (y .z), since, by associativity, these are the same
element. Show that
3.6 Prove the cancellation laws (Theorem 3.6).
3.7 Let G be a finite group, and consider the multiplication table for G, i.e., the table
that gives the binary operation of G (see Exercise 1.9). Show that every element
of G occurs precisely once in each row of the table and precisely once in each
column.
3.8 Use the cancellation laws to give alternative proofs of:
a) Theorem 3.1;
b) the fact that if X1=yl then x=y.
3.9 Let (G,.) be a group. Show that (G, .) is abelian iff
(X.y)I=X1.yl for all x,yEG.
3.10 Let (G,.) be a group and let g be some fixed element of G. Show that
G={gulxEG}.
3.11 Let (G,.) be a group such that x 2 = e for all x E G. Show that (G,.) is abelian.
(Here x 2 means x.x.)
3.12 Let (G,.) be a group. Show that (G,.) is abelian iff (x .y)2= x 2 .y2 for all x, y
in G.
3.13 Let G be a set and let. be an associative binary operation on G. Assume that
there exists a left identity element in G and that every element in G has a left
inverse. Prove that (G, .) is a group.
3.14 Let G be a nonempty set and let. be an associative binary operation on G.
Assume that for any elements a, b in G, we can find x E G such that a • x = b,
and we can find y such that y • a = b. Show that (G, .) is a group.
3.1S Let G be a nonempty set and let. be an associative binary operation on G.
Assume that both the left and right cancellation laws hold in (G,.). Assume
moreover that G isfinite. Show that (G,.) is a group.
32 Section 3. Fundamental Theorems about Groups
3.16 Give an example to show that if the assumption that G is finite is omitted from
Exercise 3.15, then the conclusion need no longer follow, i.e., (G,.) need not
be a group.
3.17 Suppose G is a set and * is an associative binary operation on G such that there is a
unique right identity element and every element has a left inverse. Prove that (G, * ) is
a group.
SECTION 4
POWERS OF AN ELEMENT;
CYCLIC GROUPS
Before going any further in our investigation of groups, we pause to stream
line our notation. You have probably already started to get tired of writing *
every time you want to indicate the operation in a group (G, *). It is common
practice to avoid this encumbrance by writing xy in place of x * y, so long as
no confusion can arise. For example, the equation
(X*y)I=yI*X 1
is usually written
and the associative law
is written
(XY)Z=X(yz).
In keeping with this simplification, we will usually refer to an abstract group
as G, rather than (G,*).
In discussing concrete examples, we continue to use whatever notation is
appropriate. For example, if we are talking about (1'., +) we write x+y. You
are also reminded that the additive notation is very commonly used in
discussing any abelian group.
Another economy in notation is achieved by taking advantage of the
associative law in order to eliminate parentheses. For example, we can
unambiguously write xyz to denote either (xy)z or x(yz), since these two
elements are the same. Similarly, if Xl'''' 'Xn are elements of a group, then
XIX2X3' .. xn has an unambiguous meaning; no matter how we insert
parentheses into the expression, the resulting product always equals
x l(xix 3('" (XnIXn)'" ))).
33
34 Section 4. Powers of an Element; Cyclic Groups
Verifying this for yourself is a good way to check your understanding of the
associative law. (See Exercise 2.13.)
A word of caution: There are times when parentheses cannot be omitted
without changing the meaning of an expression. For example, (xy)I is not in
general the same thing as xy  I.
Now let x be an element of a group G. We define the powers xn of x (for
n E l) as follows:
xO=e;
xn = xxx' .. x (n factors), if n > 0;
n (I)n I I I I x = X =x x x “‘x , if n >0.
Here are the rules for working with exponents.
1HEOREM 4.1 Let G be a group and let x E G. Let m, n be integers. Then:
i) xmxn = xm+n;
ii) (xn)I=x n;
iii) (xmr=xnm=(xn)m.
PROOF. i) First suppose that m and n are both positive. Then
xmxn = xx'” x . xx’ .. x = xx” . x (by associativity),
‘—y—‘ ‘—r—‘ ‘—y—‘
m factors n factors m + n factors
and this is xm+n. Next, if m and n are both negative, say m= r and n= s,
then
xmxn=xrxs=(XIY(XI),=(XIy+s (by the first case),
and this is x(r+s>, that is, xm+n. If m= r
have
The remaining cases can be treated similarly, and are left to the reader.
(ii) and (iii) are exercises. 0
We observe that if we are writing the group operation additively, then x 2
means x + x, x 3 means x + x + x, and so on. In this context, we usually write
nx in place of xn; then (i) above becomes mx+nx=(m+n)x, (ii) becomes
(nx)=( n)x, and (iii) becomes n(mx)=(nm)x=m(nx).
DEFINITIONS If G is a group and x E G, then x is said to be of finite order if
there exists a positive integer n such that xn = e. If such an integer exists, then
the smallest positive n such that xn = e is called the order of x and denoted by
o(x). If x is not of finite order, then we say that x is of infinite order and write
o(x)= 00.
Section 4. Powers of an Element; Cyclic Groups 35
Examples
1. Let G be (Z3; EEl). Then 0(1) = 3, since
17e O,lEElh”‘O, and 1 EEl 1 EEl 1 =0.
2. Let G be (Z, +). Then 0(1)= 00, since
he 0, 1+ heO, 1 + 1 + 1″,,0, etc.
3. Let G be (10+, .). Then 0(2)=00, since
2″,,1, 22 “,,1, 23”,,1, 24 “,,1, etc.
4. Let G be G L(2, R). Then o( ( ;1 ~ I ) ) = 2 since
° )2 = ( 1 0)
1 ° 1·
Since the notion of “order of an element” is defined in terms of integers,
it is not surprising that one needs some information on integers in order to
investigate its properties. We include this material at this point as a change of
pace.
If m,n are integers, not both zero, then (m,n) denotes the greatest
common divisor (g.c.d.) of m and n. This is by definition the largest integer d
that divides both m and n. [If m = n = 0, then (m, n) doesn’t exist because
every integer divides 0: O=k·O for any k.] It is clear that (m,n)=(lml,lni), so
that in what follows we can assume that m and n are nonnegative integers, at
least one of which is not zero.
There is a process called the Euclidean algorithm which enables us to find
(m,n) by doing some arithmetic. Say n
ordering principle, such a decreasing sequence of nonnegative integers cannot
go on forever, so some r; must eventually be 0. If so, then
Thus (m,n) is the last nonzero remainder arising from our repeated divisions.
Example Let’s find (1251, 1976):
1976= 1251·1 + 725,
1251 =725·1 +526,
725=526·1+199,
526= 199·2+ 128,
199= 128·1 +71,
128=71’1+57,
71=57·1+14,
57= 14·4+ I,
14= 1·14+0.
Here r= 725, rl = 526, r2 = 199, r3 = 128, r4 = 71, rs = 57, r6= 14, r7= I, rs =0.
Thus (1251, 1976) = r7 = l. We indicate the fact that the g.c.d. is I by saying
that 1251 and 1976 are relatively prime.
Actually our interest in this process is not so much that it enables us to
find (m,n), but that it allows us to establish the following fact.
TIlEOREM 4.2 If m and n are integers, not both zero, then there exist integers x
and y such that
mx+ny=(m,n).
Thus the g.c.d. of m and n can be written as a “linear combination” of m and
n, with integer coefficients.
The utility of this information will become clear in a moment.
Section 4. Powers of an Element; Cyclic Groups 37
PROOF OF THE THEOREM. We write down the steps in the calculation of (m, n)
by the Euclidean algorithm, and then use them in reverse order. We have
m=qn+r,
n=q1r+rl,
r=q2rl+ r2′
r l = q3r2 + r3,
ri – 4 = qi2ri3 + ri 2,
ri 3 = qil ri2 + ri_ i’ (ri _ I 7’=O)
ri 2 = qiriI +0,
so ri – I =(m,n). Now the nexttolast step can be written as
ri _ 1 = l·ri_ 3 – qi (ri 2, [4.1]
so (m,n) is written as a linear combination of ri – 3 and ri – 2• The preceding
step (ri – 4 = qi2ri3 + ri 2) can be used to replace ri 2 by rj_ 4 – qj2rj3 in
Eq. [4.1], resulting in an expression for (m,n) as a linear combination of rj _4
and rj _ 3• Using all the equations from the Euclidean algorithm in reverse
order, we eventually arrive at an expression for (m, n) as a linear combination
of m and n. D
Example We find x and y such that (1251, 1976)= 1251x+ 1976y. Referring
back to our calculation of (1251, 1976)= 1, we get:
1=1·574·14
= 1· 57 4(71 57)= 4· 71 + 5· 57
= 4·71 +5(12871)= 9·71 +5·128
=5·1289(199128)= 14·1289·199
= 9’199+ 14(526199·2)= 37·199+ 14·526
= 14·52637(725 526) = 51·52637 ·725
= 37(725)+51(1251725)= 88·725+51·1251
=51·125188(19761251)= 139,125188·1976.
Thus we can take x = 139 and y = – 88.
After that, one should be in a good frame of mind to appreciate some
good clean abstraction, but before returning to groups we derive a con
sequence of the last result that will be useful.
THEOREM 4.3 (Euclid) If r,s, t are integers, r divides st, and (r,s) = 1, then r
divides t.
PROOF. Since (r,s)= I, there exist integers x andy such that
rx+sy= 1.
38 Section 4. Powers of an Element; Cyclic Groups
Multiplying both sides of this equation by t yields
rxt+syt= t, or
r(xt) + st(y) = t.
Manifestly, r divides r(xt); and r divides st(y) since it divides st by assump
tion. Thus r divides the sum of r(xt) and st(y), i.e., r divides t, as claimed. 0
There are many simple results in number theory which never lose their
charm, and that’s one of them.
Back to groups.
THEOREM 4.4 Let G be a group and x E G.
i) o(x) = o(x 1).
ii) If o(x)=n and xm=e, then n divides m.
iii) If o(x)=n and (m,n)=d, then o(xm)=n/d.
PROOF. The proof of part i) is left as an exercise.
ii) We have xm = e and we seek to make something of the fact that n is
the smallest positive integer such that xn=e. Write m=qn+r, where O
and therefore, since nld and k are positive integers, nld:s; k. We have xnk = e, so by
part (ii) we know that n divides mk, which implies that nld divides (mid) . k. Since
(mid, nidi = 1 (why?), this implies that nld divides k (by Theorem 4.3). 0
We will use these results on the order of an element in Section 5, to help
us obtain some results about what are known as cyclic groups. For now, we
will just introduce cyclic groups.
We remarked that the study of abstract group theory evolved from the
study of specific examples. The abstract concept was formulated in an effort
Section 4. Powers of an Element; Cyclic Groups 39
to bring together certain concrete cases. Once this was done, however, there
was, of course, a new problem. How farreaching was the abstract concept?
What kinds of groups were there other than those that motivated the abstrac
tion?
A central goal of group theory is to classify all groups, i.e., to see what
kinds of groups there are. One would like to start with the easiest groups. It
turns out that these are the cyclic groupsthose groups that are just the set
of powers of some one element.
DEFINITIONS A group G is called cyclic if there is an element x E G such that
G= {xnln EZ}; x is then called a generator for G.
It will be convenient to have a more compact notation for the set
{xnlnEl}. We will denote it by (x). Thus G is cyclic with x as a generator
iff G=(x).
In additive notation, (x)={nxlnEI}.
Examples
1. (ll’ €B) is the trivial group {O} consisting of just an identity element.
Clearly, then, (ll’ €B) = (0).
2. If n > 2, then (In’ €B) = (1 ), for the powers
\ 1, 1 €B 1,1 €B 1 ElH, … ,.1 €B 1 €B 1 ~ … €B 1,
n terms
exhaust (Zn’ €B).
3. (l, + ) is cyclic with generator 1, that is, (l, + ) = (1). In this case
we have to use all the powers of the generator to get all of the group:
0,1, – 1, 1 + 1, – 1 1, 1 + 1 + 1, – 1 1 1, and so on.
4. (0, +) is not cyclic. For clearly 0 is not a generator, and if q*O
then we can easily exhibit rational numbers that are not in (q) =
{nqlnEZ}. An example is q/2.
It should be made explicit that the powers of an element need not all be
distinct. In fact, we have the following result:
TIIEOREM 4.5 Let G=(x). If o(x)=oo, then xj*xk for J*k, and conse
quently G is infinite. If o(x) = n, then x j = Xk iff} == k (mod n), and conse
quently the distinct elements of G are e, x, x 2, ••• ,xn I •
PROOF. Suppose thatJ*k and xj=Xk. If, saY,J>k, then we obtain xjk=e,
and J – k > 0, so x has finite order. This proves the first statement.
For the second, suppose that o(x)=n. Then xj=Xk iff xjk=e iff (by
Theorem 4.4 ii) n divides j – k iff j == k (mod n). 0
40 Section 4. Powers of an Element; Cyclic Groups
DEFINITION The order of a group G, denoted by jGj, is the number of
elements in G.
Theorem 4.5 has the following immediate consequence, or corollary.
COROLLARY 4.6 If G=
m or n or both. Say m=p{tp~2 .. ·Pr;’ and n=p{tJ1i2 … pt,. Sho:w that
(m,n)=pf tpf2 .. ·Prlc” where kt is the smaller of i, andj” for each t.
4.30 Let nt, … , nk be integers, not all O. The greatest common divisor of n t ,.·., nk,
denoted by (n t , ••• ,nk), is the largest integer that divides all of nt,n2, … ,nk. Show
that there exist integers at, … ,ak such that
atn t+ a2n2+··· +aknk=(nl> … ,nk).
[Suggestion: Use induction on k. Use the inductive hypothesis to show that
(nt,·.· ,nk) = «n!> … ,nkt),nk),
and apply Theorem 4.2.]
4.31 If m and n are integers, we define their least common multiple, [m,n], as follows.
If m=O or n=O, we set [m,n]=O; otherwise we let [m,n] be the smallest positive
integer that is divisible by both m and n.
a) Show that if m and n are both positive and m=p:tp~2 .. ‘p/’, n=p{tJ1iz … pt”
as in Exercise 4.29, then [m,n]=pftpi2 .. ‘P:’, where I, is the larger of i, and
j” for each t.
b) Show that if m and n are both positive, then
mn=(m,n)[m,n].
4.32 Let G be an abelian group, and let x andy be elements of G such that o(x) = m
and o(y)=n. Show that G has an element z such that o(z) is the least common
multiple of m and n ..
4.33 a) Show that in the group of Exercise 2.12 we have (xy)3 = xV and
(xyt=xY for all x andy, but the group is not abelian. (Compare Exercise 3.12.)
b) Prove that if G is a group and there exists a positive integer n such that for all
x, y in Gwe have
(xyr = X’j’ and (xy)n+l = rly”‘l and (xyr2 = r2y”‘2
then G is abelian.
SECTION 5
SUBGROUPS
Up to this point we have been considering groups as separate entities,
unrelated to each other. Even so, you have probably observed that some
groups sit inside others. For example, in (Z, +), the set 2Z of even integers is
itself a group under +: Addition is a binary operation on 2Z since the sum of
two even integers is even; addition is associative on 2Z since it is associative
on all of Z; 2Z contains the identity element 0 of (Z, +); and if xE2Z then
– x E2Z, so 2Z contains the inverse of each of its elements.
Indeed, one of the most natural questions one can ask about a group G is
“What groups sit inside G?” Those that do are called subgroups of G.
DEFINI110N A subset H of a group (G,*) is called a subgroup of G if the
elements of H form a group under *.
It is worth emphasizing the “under *.” For example, (0+,·) is a group
and (0, + ) is a group, and ° + k 0, but (0 +, .) is not a subgroup of (0, +)
because the operation on (0+, .) is not the operation on (0, +).
Observe that if H is a subgroup of G, then H cannot be empty because H
must contain an identity element. In fact, the identity element of H must be
e, the identity element of G. For suppose e’ is the identity element of H; then
in particular e’ * e’ = e’, a relationship between elements of the group G. Thus,
multiplying by (e’) 1 in G, we get e’ = e.
It is convenient to have a more compact criterion for a subset of a group
to constitute a subgroup.
TIIEOREM S.l Let H be a nonempty subset of a group G. Then H is a
subgroup of G if and only if the following two conditions are satisfied:
i) for all a,bEH, abEH, and
ii) for all aEH, aI EH.
43
44 Section 5. Subgroups
Condition (i) is expressed by saying that H is closed under the operation
in G, and condition (ii) is expressed by saying that H is closed under inverses.
PROOF OF THE THEOREM. If H is, in fact, a subgroup of G then it is clear that
(i) is satisfied. As for (ii), if we let a _ I denote the inverse of a in H, then
aa_ 1 = e (we have remarked that the identity element of H is the same as the
identity element of G) in G, which implies that a_I is in fact a I, the inverse
of a in G. Thus a I E H for any a E H, so (ii) is satisfied.
Conversely, assume that the nonempty subset H is closed under the
operation. in G and that H is closed under inverses. To show that H is then
a group under. it suffices to check that e E H and that associativity holds in
H. Since H is nonempty by assumption, we can let x denote some element
of H. Then by (ii) xIEH, so by (i) xxIEH. But xxI=e, so eEH.
Finally, H inherits associativity from G: if a,h,cEH then a,h,cEG, so
(ah)c = a(hc) by associativity in G. 0
Examples
1. (Q + , .) is a subgroup of (iii + , .), which is, in turn, a subgroup of
(R {O},·).
2. Let G be any group and let aEG .. Then
~I/
A line going upward from one group to another indicates that the bottom
group is a subgroup of the top one.
4. If G = (I, + ) and n is any integer, then the set nl = < n > of multiples of
n forms a subgroup of G. In particular for n = 0 we get the subgroup {O}
consisting of just the identity element, and for n = I we get G itself.
In fact the nl are all the subgroups of (I, +). For if H is a subgroup
other than {O} there exist positive integers in H. (Why?) If we let n be the
smallest positive integer in H, then we claim that H is nl. Oearly, nl ~ H
since n E Hand H is a subgroup. But also H ~ nl, for if hE H we can write
h=qn+r, with O
(not all of which are distinct if n > 2: for example, (1 > =
the inverse of 1), and an argument like that in Example 4 would show that
these are, in fact, all the subgroups of (In’ EB). However, rather than go
through the argument in the context of (In’ EB), we will prove a general result
that covers Example 4 and (In’ EB) simultaneously.
THEOREM 5.2 Let G be a cyclic group. Then every subgroup of G is cyclic.
PROOR Suppose G=
clearly H is cyclic, so assume H *’ {e}. Let n be the smallest positive integer
such that xn EH. (Why does n exist?) We assert that H=
/”” <2> <3>
/”‘ /
<4> <6>
“” / <0>
52 Section 5. Subgroups
Here is the situation for infinite cyclic groups.
THEOREM 5.7 Let G=
i) G=(P(X), £» ; H=P(Y), where Y(;:X.
5.2 Let G be the group of realvalued functi!Jns on the real line, under addition of
functions. Let H be the set of differentiable functions in G. Show that H is a
subgroup of G.
5.3 Let H be the set of elements ( : !) of GL(2, R) such that ad – be = 1. Show that
H is a subgroup of GL(2, R). H is called the special linear group of degree 2 over
R and is denoted by SL(2, R).
5.4 a) How many subgroups does (liS’ EB) have? What are they?
b) How many subgroups does (l3S’ EB) have? What are they?
c) How many subgroups does (l36′ EB) have? What are they?
5.5 Find all the subgroups of Qs. Show that Qs is an example of a nonabelian group
with the property that all its proper subgroups are cyclic ..
5.6 a) Let G be a cyclic group of order n. Show that if m is a positive integer, then
G has an element of order m iff m divides n.
b) Let G be a cyclic group of order 40. List all the possibilities for the orders
of elements of G.
5.7 Let G =
and only if (m,n)= I. Thus the number of generators of a cyclic group of order
n is the number of integers m in the set {O, 1, .. . ,n l} such that (m,n)= I. This
number is denoted by
role in number theory.
Section 5. Subgroups 53
5.8 Let G=(x) be a cyclic group of order 144. How many elements are there in
the subgroup (X 26 )?
5.9 Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent
to mZ r:;;, nZ? What condition on m and n is equivalent to mZ U nZ being a
subgroup of (Z, + )?
5.10 Prove that every subgroup of an abelian group is abelian.
5.11 Let G be an abelian group, and let n be a positive integer. Let H be the subset
of G consisting of all x E G such that x n = e. Show that H is a subgroup of G.
5.12 Find the center of:
a) V;
b) Qs.
5.13 Let H be the group introduced in Example 7 on p. 46. Find Z(H).
5.14 Prove that the intersection of two subgroups of a group G is itseH a subgroup
of G.
5.15 Show that if Hand K are subgroups of the group G, then H u K is closed under
inverses.
5.16 Give an example of a group G and a subset H of G such that H is closed under
multiplication but H is not a subgroup of G.
5.17 Suppose H is a nonempty finite subset of a group G and H is closed under
inverses. Must H be a subgroup of G? Either prove that it must, or give a
counterexample.
5.18 a) Show that it is impossible for a group G to be the union of two proper
subgroups.
b) Give an example of a group that is the union of three proper subgroups.
5.19 Let G=(x) be an infinite cyclic group. Show that all the distinct subgroups of
G are (e),(x),(x2),(x3),(x4),(x5), ….
5.20 Let G be a finite group with no subgroups other than {e} or G itseH. Prove that
G is either the trivial group {e} or a cyclic group of prime order.
5.21 Let G = (x) be a cyclic group of order n. Find a condition on the integers rand
s that is equivalent to (x’) r:;;,(x’).
5.22 Let G be a group. Prove that Z(G) is a subgroup of G.
5.23 Let G be a group, and let g E G. Define the centralizer, Z( g), of gin G to be the
subset
Z(g)= {xEGlxg=gx}.
Prove that Z(g) is a subgroup of G.
5.24 Let G be a group and let H be a nonempty subset of G such that whenever
x,y E H we have xy I E H. Prove that H is a subgroup of G.
54 Section 5. Subgroups
5.25 Let G be a group and let a be some fixed element of G. Let H be a subgroup
of G and let aHa – 1 be the subset of G consisting of all elements that are of the
form aha – 1, with h E H. Show that aHa – 1 is a subgroup of G. It is called the
conjugate subgroup of H by a.
5.26 Let H be a subgroup of the group G and let N( H) = { a E G I aHa ) = H}. (See
Exercise 5.25 for the definition of aHa 1.) Prove that N(H) is a subgroup of
G.
5.27 Let G be a finite abelian group. Show that G is cyclic iff G has the property that
for every positive integer n, there are at most n elements x in G such that xn = e.
5.28 Prove that every inimite group has inimitely many subgroups.
5.29 Suppose G is a group and there is an element g E G such that g ¢ e and g is in every
subgroup of G other than the trivial subgroup {e}. Prove that every element of G has
finite order.
5.30 Prove that (il,. $) has an element g as described in Exercise 5.29 if and only if n is a
power of a prime.
SECTION 6
DIRECT PRODUCTS
In dealing with an abstract concept, it is very useful to have a good supply of
concrete examples on hand. The examples make the abstraction come to life,
and they also provide us with a means of testing out general ideas on specific
cases. In the preceding section, we saw that new examples of groups can
sometimes be found sitting inside old ones. Now we take the opposite tack by
considering how we can “patch together” given groups to make new ones.
Actually there are a number of different ways of doing this. We will
consider the simplest and most frequently used method, called the direct
product construction.
Suppose G and H are groups (not necessarily distinct). To form the direct
product of G and H, we consider the set of all ordered pairs (g,h), where
gEG and hEH. We introduce an operation on this set by multiplying
componentwise:
(g I’ h l)( g2′ h2) = (g I g2′ hi h2),
where gl g2 is computed in G and hlh2 is computed in H. It is clear that this
definition gives us a binary operation, and we have associativity as a con
sequence of associativity in G and H:
[(gl,h l)(g2,h2) ](g3,h3) = (gl g2,hlh2)(g3,h3) = [(gl g2)g3,(h lh2)h3]
= [ gl( g2g3),hl(h2h3)] = (gl,hl)( g2g3,h2h3) = (gl,hl)[ (g2,h2)( g3,h3)]·
The identity element is (eG,eH), where eG and eH are the respective identity
elements of G and H, and the inverse of (g,h) is (gI,h I).
Thus we have a new group, which we denote by G X H and call the direct
product of G and H. The groups G and H are called factors of the product.
In a completely analogous fashion, we can form the direct product
G I XG2 X'” XGn
55
56 Section 6. Direct Products
of n groups. The elements of this group are ntuples (gl,g2′ … ,gn) with gj E Gj,
and the multiplication is defined componentwise. As a matter of fact, there is
no reason why we have to restrict ourselves to finitely many factors, but we
will rarely use infinitely many.
Examples
1. Let G1 = G2 =··· = Gn=(IR, +). Then
is ordinary nspace IRn under addition of ntuples.
2. Consider Z2 X Z2′ where the operation on each factor is addition mod
2. This is a group of order 4, and already reveals some interesting things
about direct products.
First of all, Z.2 X Z2 is not cyclic, although both factors are cyclic.
Denoting the operation on Z2 X Z2 by +, for simplicity, we have
(0,1)+(0,1)=(0,0),
(1,0) + (1,0)= (0,0),
and (1,1) + (1,1) = (0,0),
so that every nonidentity element has order 2, and there is no element of
order 4 to generate the group.
Notice something else. In general, a direct product G X H has certain
“obvious” subgroups, because if A is a subgroup of G and B is a subgroup of
H, then A X B is a subgroup of G X H (Exercise 6.5). Z2 X Z2 points out that
G X H may have subgroups other than those of the form A X B. For instance,
if the cyclic subgroup
«1,1» = {(I, 1), (O,O)}
were of this form, then A and B would both have to be Z2; but then A X B
would be the whole group Z2 X Z2′ not «1,1».
3. For Z2 X Z3′ things are different. The group has order 6, and is cyclic,
because (1,1) has order 6 [its first six powers are (1,1), (0,2), (1,0), (0,1), (1,2),
(0,0)]. Also, by Theorem 5.5, we know that there is a unique subgroup of
order m, for each m dividing 6. Since
are subgroups of order 1, 2, 3, and 6, these are the only subgroups, so every
subgroup has the form A X B.
Section 6. Direct Products 57
The following result goes a long way toward explaining the difference
between Examples 2 and 3.
THEOREM 6.1 Let G = G1 X G2 X … x Gn•
i) If g; E G; for 1 <.i <.n, and each g; has finite order, then O«gl,g2' ... ,gn» is
the least common multiple of o( g I)' o( g2)' ... , o( gn)'
ii) If each G; is a cyclic group of finite order, then G is cyclic iff I G;I and 101
are relatively prime for i:l=i.
PROOF. i) If m is a positive integer, then
(gl,g2"'. ,gn}m = (gj,gi, ... ,gnm).
It follows, by Theorem 4.4, that (gl,g2, ... ,gn)m=(eG"eG2, ••• ,eG) iff m is
divisible by each o(g;). Thus 0«gl,g2' ... ,gn» is the smallest positive integer
that is divisible by each o( g;).
ii) If G is cyclic, let g=(gl,g2, ... ,gn) be a generator. Then for 1 <.i <.n, g;
generates G; (why?), so by Corollary 4.6 we have o( g;) = I GJ Thus, by part (i),
o(g) is the least common multiple of IGII,IG21, ... ,IGnl. But since g generates
G,
o(g}= IGI = IGtI·IG21· ... ·IGnl
(see Exercise 6.4). We conclude that the least common multiple of
IGII,IG2 1, ... ,IGnl is IGtI·IG21· ..• 'IGnl, and this means that IG;I and IGjl are
relatively prime if i:l=i.
Conversely, if IG;I and IGjl are relatively prime for i:l=i, then the least
common multiple of IGII, ... ,IGnl is IGII· '" ·IGnl. If we let g; be a generator
for G;, then o(g;)=IG;I, so by part (i), (gl, ... ,gn) has order IGII···· 'IGnl in
G. Thus (gl, ... ,gn) generates G, and G is cyclic. 0
Examples In l12' 0(8) = 3, and in llg, 0(15) = 6. Thus in ll2 X Zig, the order of
the element (8,15) is the least common multiple of 3 and 6, namely 6.
The groups ll4X.llS and .lgx.l9XlS are cyclic; ll4x.ll6 and .lgx.l9x
l6 are not.
EXERCISES
6.1 Calculate the order of the element
a) (4,9) in llSXl lS'
b) (7,5) in l12XlS.
c) (8,6,4) in llsXl9 Xls.
d) (8,6,4) in 19x117XllO.
58 Section 6. Direct Products
6.2 Which of the following groups are cyclic?
a) lllXZ,
b) llOXl85
c) l4Xl2Sxl6
d) l22 X l2\ X l6S
6.3 Is l X l cyclic? [Here l means (l, + ).]
6.4 Show that for finite groups GI , G2, • •• , Gn ,
IG) xG2 x··· X Gnl=IG)I·IG21· ... ·IGnl.
6.5 Let A be a subgroup of G, and let B be a subgroup of H. Show that A X B is a
subgroup of G X H.
6.6 Show that G) X G2 X ••. X Gn is abelian iff each G; is abelian.
6.7 Construct a nonabelian group of order 16, and one of order 24.
6.8 Construct a group of order 81 with the property that every element excep. the
identity has order 3.
6.9 Show that Z(G) X G2 x··· X Gn)=Z(G)XZ(G0x··· XZ(Gn).
6.10 Find all subgroups of l2 X l4.
6.11 Find all subgroups of l2 X l2 X l2.
6.12 Let G and H be finite groups. Show that if G x H is cyclic, then (i) G and H are
cyclic, and (ii) every subgroup of G X H is of the form A X B for some subgroups
A and B of G and H, respectively.
6.13 Prove the converse of the result in Exercise 6.12; that is, show that for finite
groups G and H, (i) and (ii) of Exercise 6.12 (taken together) imply that G X H
is cyclic.
6.14 Use Theor:em 6.1 to prove the Chinese Remainder 1beorem: If m), ... ,"", are
positive integers such that "'t and mj are relatively prime for i +j, and k 1, ••• ,kn
are any integers, then there is an integer x such that x == k,(mod "'t) for I <.i <.n.
(Hint: Consider the generator (1,1, ... ,1) for lift, X ••• Xl ..... )
6.15 Prove that if G is an infinite group and H is a group then G x H is cyclic if and only if
G is cyclic and H = {eH}.
SECTION 7
FUNCTIONS
In this section we will present some elementary results about functions. These
results will be useful in Section 8, when we investigate what are called
symmetric groups, and later on, when we discuss homomorphisms and isomor
phisms.
DEFINITION If Sand T are sets, then a function f from S to T assigns to each
s E S a unique element f(s) E T.
As a definition this is somewhat strange, in that it tells you what a
function does rather than what it is. Sometimes this difficulty is avoided by
saying that a function is a "rule" that assigns elements of T to the elements of
S, but this isn't any better because "rule" isn't defined. Besides, for some
functions the "rule" is obscure at best, and it may be so hard to state that
most people wouldn't call it a rule at all.
The above definition is fine as a working definition, and is how we
usually think of functions. A more precise definition is as follows.
DEFINITION (precise) A function from S to T is a set of ordered pairs (s,t),
where each s E S and each t E T, such that each s E S occurs as the first
element of one and only one pair (s,t).
Obviously this formulation captures the intent of our working definition;
if s E S, then there is only one pair (s, t) with s as its first element, and the
function assigns the second element, t, of that pair to s. We write j(s} = t, and
sometimes we say thatfsends s to t, or m~ps s onto t. (The word "mapping"
is sometimes used for "function.")
We write f: S~ T to indicate that f is a function from S to T.
59
60 Section 7. Functions
Examples
1. S= T=R; J: S~T is given by J(x)=x2.
2. S={1,2}, T={3,4,5};J:S~Tisgiven byJ(I)=3,J(2)=5. In the
precise formulation,J is {(l, 3), (2, 5)}.
3. S = R, T= [  I, I], the closed interval  I "x" I; J: S~T is given
by J(x) = sinx.
4. S = GL(2, R), T= R; J: S~ T is given by
~(~ :))=determinant of (~ :)=adbc.
S. S = set of continuous functions from R to R, T= R; F: S~ T is given
by F(g) = folg(x)dx.
6. S=l2X l4' T=l4; J: S~T is given by J«x,y»=y.
7. Let G be a group, and let a E G. Define J: G~ G by J(x) = ax. In the
precise formulation, J is {(x,ax)lx E G}.
8. If S={1,2}, T= {3,4,5}, thenJ={(1,3)} is a function from {l} to
T, but it is not a function from S to T because J(2) isn't defined. Also,
{(1,3),(2,5),(l,4)} isn't a function; because 1 appears as the first
element of more than one pair.
Certain kinds of functions are particularly relevant to group theory.
DEFINITIONS Let J: S ~ T be a function. J is onto if for each t E T there is at
least one s E S such that J(s) = t. J is onetoone if whenever Sl and S2 are two
different elements in S, we have J(s\)=I=J(S2).
Thus J is onto iff every t E T comes from at least one s E S; saying it yet
another way, everything in T gets hit by J. You can think of J as a cannon
firing shells (elements of S) at T (the target):
J is onto just if it doesn't miss anything in T.
On the other hand, J is onetoone just if nothing in T gets hit twice. That
is, anything in T is hit either just once or not at all.
Section 7. Functions 61
Observe that if f is both onetoone and onto, then every element of T is
hit once and only once, so that f establishes a onetoone correspondence
between the elements of S and those of T: t E T is paired with the unique
sES such thatf(s)=t.
There is some standard terminology that is good to know in connection with
functions. If f ST is a function, then S is called the domain of j, and
the set of elements of T which are hit by f (more precisely {t E TI for some
s E S, f(s) = t}) is called the image (or range) of f:Thus f is onto iff the
image of j is T:
Examples Consider again Examples (1)(7) on p. 60. The functions in Exam
ples (3), (5), (6), and (7) are onto, and those in Examples (2) and (7) are
onetoone.
The functionf(x) = x 2 in (I) is not onto since, for instance,  1 flimage of
f. It is not onetoone since f( 1) = f(I).
The function 10 (4) is not onto since Oflimage of f, and it is not
onetoone since
The function F in (5) is onto since for any rET we can find some g E S
such that F(g) = r. For instance, the constant function g such that g(x) = r for
all x will do. F is not onetoone since if gl and g2 are defined by
gl(x)=O for all x, g2(x)=xi,
then gl and g2 are different elements of S, but
F(gl)= foIOdX=O= fol( x ~ )dX=F(g2)'
Finally, consider the functionf: G~G in (7), given by f(x) = ax.fis onto
since for every yEG there is some x such that f(x)=y. In fact, x=ay
works, because then
f(x) = f(a y) = a(ay) = y.
This function is onetoone by the left cancellation law: if f(x I) = f(x0, that is,
ifax i = ax2, then XI = x 2. (Here we have used the definition of "onetoone"
in the following form: If f: S~ T, then f is onetoone iff whenever SI,S2 E S
and f(sl) = f(s0, then Sl =S2')
The term "injective" is sometimes used in place of "onetoone." This
terminology expresses the fact that the domain is carried intact into the
image, without any collapsing taking place. People who use "injective" for
"onetoone" often use "surjective" for "onto"; this word indicates that the
function throws the domain onto the set T.
62 Section 7. Functions
Assume now that f: S~ T is onetoone and onto. As we have seen, f
accomplishes a onetoone pairing off of the elements of S with those of T.
Therefore, f provides us with a function fI : T ~S, which maps any t E T
onto the s E S with which t is paired by f. fI is called the inverse function
off·
Let us examine fI in a little more detail. Take any t E T. Since f is onto,
there exists at least one s E S such that f(s) = t. Since f is onetoone, there is
only one such s. Thus we can unambiguously define f ie t) by setting
fl(t)=S.
Observe that if f were not onetoone, then there might be two different
elements Sl and S2 of S such that bothf(sl)=t andf(s2)=t. We would then
face a quandary in trying to define f I ; the idea of the inverse function is
that it is supposed to undo everythingfdid, and iff(sl)=f(S2)=t, then we
cannot define fI(t) so that 11 undoes what f did to both SI and S2. For
example, if we were to define fI(t) = sl> then f would send S2 to t and fI
would send t to SI’· rather than back to S2. Similarly, if we were to define
fI(I) = S2′ then fI would not undo what f did to SI.
If f is onetoone but not onto, then f has an inverse function fI with
domain equal to the image of f. In order that f have an inverse function with
domain T, it is thus necessary and sufficient that f be onetoone and onto.
Observe that if f: S~ T is onetoone and onto and we view f as a set of
ordered pairs (s, t), then fI : T ~S is the function obtained by switching the
entries in each pair inf: (s,t) is replaced by (t,s). It is not difficult to see that
fI is itself both onetoone and onto, so that it has an inverse UI)l : S~
T. In fact U I ) I is f, because we obtain it by switching all the pairs (t,s)
back to (s,t).
Examples
1. Let S= T=71.. and letf: S~T be given by f(n)=n+ 1. Thenf is
onetoone and onto, so it has an inverse function fI. In fact
fI(n)=nl forallnE71…
2. Let S= iij and T=[ 1,1], and letf: S~Tbe given by f(x) = sinx.
Thenfis onto but not onetoone. (For example,j(O)= f(‘IT)=O.} We
can restrict the domain of f so that f becomes onetoone and onto,
however. For example, if we restrict the domain to [ – ‘IT /2, ‘IT /2], then
f has an inverse function which is called sin – I or arcsin. This
function is probably familiar to you from calculus.
3. Let S= iij and T= iij+, and letf: S~Tbe given by f(x) = eX. Then f
is onetoone and onto, and its inversef I is given by fI(x)=ln x
for all x E iij+
Section 7. Functions 63
The use of the term “inverse” is no accident visa.vis group theory. If X is
a nonempty set, then we are going to define a binary operation on the set of
all onetoone onto functions X ~X, which turns this set into a group in such
a way that the inverse of f in the above sense is the inverse of f in the group.
It will be useful to introduce this operation in a more general context, so
let f: S ~ T and g: T ~ U be functions. We define the composite function
go f: S~ U by setting
(g 0 f)(s) = g(j(s))
for all s E S. Observe that f(s) E T, so g(f(s» makes sense and is an element
of U.
Examples
1. Let S=T=U=R, and letf(x)=x5+1 and g(X)=X3 for all xER
Then
for all x E R What is fog?
(Jog)(x) = f(x 3)=(X3)5+ I =X I5 + 1.
Thus we see that go f and fog are not necessarily the same function.
In general we say that two functions fl and f2 are equal and write fl = f2 if
fl and f2 have the same domain and fl(x) = fix) for all x in that domain.
Thus in the present example we write gof=l=fog. Observe that in terms of the
precise definition of a function, two functions are equal iff they are the same
set of ordered pairs.
2. Letf: S~T be onetoone and onto. Let is be the identity function on
S, that is, is(s) = s for all s E S. Similarly, let iT be the identity mapping on T.
Then
fIof:S~S and fofI: T~T.
We havef I of= is andfofl=iT .
3. Let S= T= {(a,b)la,b ER}. Let f«a,b»=(b,a), and let g«a,b»
=(a,b+ I). Then
and
Here again, go f=l=f 0 g.
(g of)«a,b))=(b,a + 1)
(Jog)«a,b»=(b+ I,a).
Now let X be a nonempty set, and let Sx denote the set of all onetoone
onto mappingsf:X~X.
64 Section 7. Functions
nmOREM 7.1 (Sx, 0) is a group.
PROOF. If j and g are both in S x’ then certainly jog is a function from X to
X. To check that 0 is a binary operation on Sx’ we must verify that if j,g are
both onetoone and onto, then jog is onetoone and onto. This is left as an
exercise.
Associativity of 0 requires that ifj,g,hESx then (fog)oh=jo(goh). In
other words, we must show that
[(fog) oh ](x)= [Jo(goh) ](x)
for every x EX. But
[(fo g) oh ](x) = (f 0 g)(h(x» = j(g(h(x»),
and
[J 0 (g 0 h) ](x) = j(g oh)(x» = j(g(h(x»).
The identity element of (Sx, 0) is ix ‘ That is, j 0 ix = ix 0 j= j, for every
jESx’
Finally, ifjESx andjI denotes its inverse function, thenjIESx and
jojI=jIoj=ix ,
so jI is the grouptheoretic inverse of j. 0
EXEROSES
7.1 In each example below,jis given either as a rule or as a set of ordered pairs. In
each case, determine whether or notjis a function from S to T. For those cases
in which it is, determine whether it is onetoone, and whether it is onto.
a) S= {l,2,3,4,5}, T= {6, 7,8,9, lO}, j= {(I, 8), (3, 9), (4, lO), (2, 6),(5, 9)}
b) Sand T are as in (a),j= {(l,8),(3, 10),(2,6), (4,9)}
c) Sand T are as in (a),j= {(I,7), (2, 6), (4, 5),(1,9),(5, 1O)}
d) S= T=R,j{x)=x2 x
e) S= T=R,j{x)=x3
f) S=T=R,j(x)=Vx
g) S= T=R,j{x)= l/x
h) S= T=Z+,j(x)=x+ I
i) STZ+,j{X)={ I
xI
j) S= T=R+,j(x) .. _x_
x 2 +1
7.2 Letj: R_R.
ifx=1
ifx>1
a) Give a condition on the graph of y = j{x), in terms of its intersections with
horizOntal lines, that is equivalent to j being onetoone.
b) If g.: R_ R and f and g are both onetoone, must f + g be onetoone?
Section 7. Functions 65
73 Letf: IR~IR be given by f(x) = ax+ b, where a and b are fixed constants.
a) Show that if a * 0 then f is onetoone and onto, so that f I exists.
b) Assuming that a * 0, find an explicit formula for the inverse function f I.
7.4 Let S = T= the set of polynomials with real coefficients, and define a function
from S to T by mapping each polynomial to its derivative. Is this function
onetoone? Is it onto?
7.5 Let X be a set, and let A kX. Define a functionf: P(X)~P(X) by f(B)=A n B,
for B E P(X). Under what conditions is f onetoone and onto?
7.6 Let X be a set, and let A kX, Definef:P(X)~P(X) by f(B)=A Lo.B. Isf
onetoone? Is f onto?
7.7 Let G be a group and let aE G. Define a functionf: G~G by f(x)=axa I for
all x E G. Is f onetoone? Is f onto?
7.8 Let G be a group, and letf(x) = x I for all x E G. Isf a function from G to G?
If so, is it onetoone? Is it onto?
7.9 Show that 0 is a binary operation in Theorem 7.1.
7.10 Letf: S~T.
a) Show thatf is onetoone if and only if there exists a function g: T ~S such
that go f = is.
b) Show thatf is onto if and only if there exists a function g: T ~S such that
fog= iT’
c) Show thatf is onetoone and onto if and only if there exists a function
g: T~S such that gof=is andfog=iT.
7.11 Letf:S~T and g: T~U.
a) If g of is onetoone, must both f and g be onetoone?
b) If go f is onto, must both f and g be onto?
7.12 Letf: S~T. For any subset A of S, define
f(A)= {J(s)ISEA}.
a) Show that if A,B are subsets of S, thenf(A U B)= f(A)uf(B).
b) Show thatf(A n B)kf(A)nf(B). Construct an example where the inclusion
is proper, i.e., f(A n B)$f(A) n feB).
SECTION 8
SYMMETRIC GROUPS
If X is a nonempty set, then a onetoone onto mapping X ~X is called a
permutation of X. We have seen that the set of all such permutations forms a
group (Sx, 0) under composition of functions.
DEFINmON (Sx, 0) is called the symmetric group on X.
Symmetric groups were used in mathematics before the abstract concept
of “group” had been formulated. In particular, they were used to obtain deep
and incisive results about the solutions of polynomial equations, and the
success of these efforts gave an impetus to the development of the abstract
theory. After the abstract notion was established, the English mathematician
Arthur Cayley (18211895) again demonstrated the importance of symmetric
groups by showing that every group can be thought of as a subgroup of some
symmetric group. We will state Cayley’S result precisely, and prove it, in
Section 12.
For now, we will try to get familiar with symmetric groups by investigat
ing symmetric groups on finite sets. If X is finite and has, say, n elements,
then we can represent X by {1,2, … ,n}, and we accordingly denote (Sx, 0) by
Sn in this case. Sn is called the symmetric group of degree n.
Let f E Sn’ Then f shuffles the elements 1,2,3, … , n, and we can represent
f explicitly by writing
2 3
f(2) f(3)
where f(k) is placed under k for each k between 1 and n.
66
It is easy to calculate products in Sn’ For example, consider
2
4
3
1
2
2
3
4
Section 8. Symmetric Groups 67
in S4. To see what goes under 1 in the product we just recall that the product
is the composition of the two given permutations, the one on the right being
performed first:
U 2 3 j)o(; 2 3 1 )(l)=(~ 2 3 j )(3)= 1. 4 1 2 4 4 1
Similarly,
U 2 3 j)o(; 2 3 1)(2)=4, 4 1 2 4
because 2 goes to 2, which then goes to 4. The product is
(~ 2 3 j)o(; 2 3 1) = (~ 2 3 i)· 4 1 2 4 4 3
Observe that the notation for the product
( ~ 2 3 i) 4 3
is rather uneconomical. Nothing happens to 1 or 3 (they are left fixed), and
the whole permutation does nothing more than interchange 2 and 4. To
achieve a more efficient notation for permutationsand to introduce an
important subgroup of Snwe consider special permutations called cycles.
Let X 1,X2′ ••• ‘X” 1 <.r<.n, be r distinct elements of {l,2, ... ,n}. The
rcycle (X\lX2' ••• 'X,) is the element of Sn that maps Xl~X2' x2~X3,.·.,X'_I~
X" X,~Xl' and leaves all elements of {l,2,3, ... ,n} other than X\lX2' •.• 'X,
fixed:
This cycle could just as well be written
For example,
( ~ 2 3 4 3
can be written more simply as (2,4), (or (4,2), which is the same thing). The
68 Section 8. Symmetric Groups
identity permutation
( ~ 2 2 3 4 3 4 ~)
in Ss can be written as (I), or as (2), or (3), or (4), or (5). It can also be written
as (2,5) 0(2,5), since the righthand factor just switches 2 and 5, and the
lefthand factor switches them back.
Two cycles (XI,X2""'x,) and (YI'Yl, ... ,Ys) in Sn are called disjoint if no
element of {I, 2, ... , n} is moved by both cycles. If r;;> 2 and s ;;> 2 this can be
expressed by saying that
{X I,X2,· .. ,x,} n {YI’Y2′” .,Ys} =0.
It is not difficult to see that every permutation can be. written as the product
of a finite number of cycles, any two of which are disjoint.
THEOREM 8.1 Let 1 E Sn’ Then there exist disjoint cycles 11.J2, … .Jm in Sn such
that 1 = 11 012 ° . . . ° 1m’
PROOF. Choose some XI E{l,2, … ,n}. Let x2= l(x l ),x3= l(x0, and so on.
Since {l, 2, 3, … , n} is a finite set, there must be a first element in the
sequence X I ,x2’X3′ … which is the same as a previous element. Say this
element is Xk and xk=xj>j
Xk _ 1 = xj _ 1 (since 1 is onetoone), contradicting the minimalityof k. Thus the
first kI elements of the sequence XI’X2’X3′ … are distinct and Xk=x l ‘ Thus
1 includes the cycle 11 =(XI,X2,” “xk_I)’ and
where hi permutes the elements of {1,2, … ,n} other than XI’ … ‘Xk_ l • Repeat
ing our argument on hi’ we write
hi =12 oh2,
where 12 is a cycle disjoint from 11′ and h2 permutes the elements of
{I, 2, … , n} not contained in either 11 or 12, If we continue this process long
enough, we must come to a point where hm has nothing left to permute, that
is, hm is the identity permutation. Then
1=110120130′” 0Imo hm=11012 °'” °lm’ 0
Example Consider the element
I=U 2 3 5 7 4 5 6 428 7 1
Section 8. Symmetric Groups 69
in Ss. We have
f=(l,3,7)0(~ ; ~
=(1,3,7)0(2,5)0(12
I 2
= (1,3, 7) 0 (2,5) 0 (6,8).
4
4
5
2
3
3
4
4
6
8
7
7
5
5
!)
6 7
8 7 ~)
We might also have written this as (1,3,7) 0 (2, 5) 0 (4) 0 (6, 8), but (4) is just
the identity element, and it is more economical to omit it.
The factorization of permutations into disjoint cycles is very much like
the factorization of integers into primes in the Fundamental Theorem of
Arithmetic. It is easy to see that disjoint cycles commute (Exercise 8.8), so
that once we have
we can also write
and so on. However, if we omit all factors which are the identity, then the
factorization is unique, except for this ability to rearrange the factors.
A 2cycle, i.e., a cycle that just interchanges two elements, is called a
transposition. We continue our decomposition of permutations by proving
THEOREM 8.2 If n > 2, then any cycle in SrI can be written as a product of
transpositions.
PROOF. A Icycle is the identity, hence can be written as (1,2)0(1,2). For an
rcycle with r> 2, we have
(x\,x2, … ,xr)=(x\,xr) 0 (x\,xr_\)o (X\,Xr_2) 0 .,. 0 (x\,x3) 0 (x\,x2). 0
Example Referring to the previous example, we have (1,3,7)= (I, 7) 0 (1, 3).
We also have (1,3,7)=(4,7)0(1,7)0(1,4)0(1,3), and in general there are
many ways in which a cycle can be written as a product of transpositions. We
have now lost the uniqueness of factorization.
Combining the last two theorems proves
THEOREM 8.3 If n > 2, then any element of SrI can be written as a product of
transpositions.
70 Section 8. Symmetric Groups
Example Referring again to the example of
we have
and also
2
5
3 4 5 6 7
7 4 2 8 I
1 = ( I, 7) 0 ( I , 3) 0 (2, 5) 0 (6, 8),
1 = (4,7) 0 (1,7) 0 (I ,4) 0 (1,3) 0 (2,5) 0 (6,8).
DEFINITION A permutation is even if it can be written as the product of an
even number of transpositions. It is odd if it can be written as the product of
an odd number of transpositions.
We would not want it to be possible for a permutation to be both even
and odd, and it turns out that it isn’l possible. Although a given permutation
may have many representations as products of transpositions, it will always
be the case that either these products all have an even number of factors or
they all have an odd number of factors. The neatest, most natural proof of
this fact that we have seen is a proof published in 1971 by William Miller (“Even
and Odd Permutations,” Mathematics Association of TwoYear Colleges Journal,
vol. 5, p. 32). Here it is:
TIlEOREM 8.4 No permutation is both even and odd.
PROOF. Suppose that 1 is both even and odd. Then we have
1=1112.” tk =SIS2″ ‘Sf’
where k is even, e is odd, and the I’s and s’s are transpositions. Thus
that is,
t l t2··· tk·seI·se=\··· S;ISI1 = identity,
t l t2··· Iksese_I'” S2S1 = identity,
so since k + e is odd, the identity permutation is written as the product of an
odd number of transpositions. We are going to show that this is impossible,
hence 1 could not have been both even and odd.
Consider for a moment an arbitrary product P = t I t2 • •• tm of transposi
tions. We assert that either
(I) P is not the identity permutation, or
(II) P equals a product of m – 2 transpositions.
To see this, let a be some element of {I, 2, … , n} that occurs in some t, and let
~ be the first t 1rom the right in which a appears. Say ~ is (a,b).
Section 8. Symmetric Groups 71
Now if ~ is t., we have (I), since then P does not fix a. If ~ is not t l , then
consider ~_I’ It must be either:
i) (a,b), in which case ~_I~=identity;
ii) (a,c), for some c=l=b, in which case ~_,0=(a,c)(a,b)=(a,b)(b,c);
iii) (b,c), for some c=l=a, in which case ~_,0=(b,c)(a,b)=(a,c)(b,c);
iv) (c,d) for some c=l=a,b and d=l=a,b, in which case ~_I~=(c,d)(a,b)=
(a,b)(c, d).
If Case (i) occurs, then we can delete ~_I~ from P, and we have (II). If any of
the other cases occurs, then by using the indicated equalities we can find an
expression equal to P in which the rightmost occurrence of a is one factor
further to the left than it was when we started.
If we now keep repeating our argument on a, then either Case (i)
eventually occurs, and we have (II), or else we eventually replace P by an
equivalent expression in which the rightmost occurrence of a is in the leftmost
factor, and we have (I) as above.
Thus we have proved our assertion about P. It now follows that if m is
odd, then P = t I t2 • •• tm cannot be the identity permutation. For if it is, then
(II) must hold for P, so we can write the identity as a product of m – 2
transpositions. Then (II) holds for this shorter factorization, so we can find
another factorization with m – 4 factors. Continuing in this way, we eventu
ally reduce the number of factors to 1 (since m was odd to start with), and
then we are done because it is c1ea~ that a single transposition is not the
identity. 0
You might enjoy wntmg down a few products of transposItIons for
yourself, and seeing how either (I) or (II) comes true for them.
Now let n > 2, and let An denote the subset of Sn consisting of all the even
permutations. Recall that if n is a positive integer, then n! (read “n factorial”)
denotes the product n(nI)(n2)··· (3)(2)(1).
THEOREM 8.5 Let n > 2. Then An is a subgroup of Sn’ ISnl = n!, and IAnl =
n!/2.
PROOF. To see that ISnl = n!, we consider what it takes to determine an
elementfESn • We must choosef(l) from among {l,2, … ,n}; for any choice
we make, there are n – 1 ways of choosing f(2), because we must choose f(2)
from the set {I, 2, … , n} – {f( I)}. Thus there are n( n – 1 ) ways of determining
what f(1) and f(2) are going to be. For anyone of these ways, there are n – 2
ways of choosing f(3), so there are n(n – l)(n – 2) ways of choosing the first
three values of f. Continuing in this way, we see that there are n! ways of
specifying an element of Sn’
72 Section 8. Symmetric Groups
Now to the claims about An’ An is a nonempty subset of Sn and therefore
a nonempty finite subset of Sn’ (Subsets of finite sets are finite.) Since the
product of two even permutations is even (why?), An is closed under the
group operation in Sn’ Therefore, An is a subgroup of Sn by Theorem 5.3.
To see that IAnl = n!/2, we want to show that An contains exactly one half
of the elements of Sn’ Observe that if fl’ f2″‘” fk are all the distinct elements
of An’ then it suffices to show that there are exactly k distinct odd permuta
tions in Sn’ If we let g be (1,2), then
gfl’ gf2′ … , gfk
are all distinct and all odd. (Why?) Furthermore, these are all the odd
permutations in Sn’ For if h is odd, then gh is even, so gh is one of
fi> f2, .. ·,Jk and
g(gh) E {gfl,gf2′” ·,gfk}·
But g(gh) is h, since g2 is the identity. Therefore, every odd permutation is
one of gfl,gf2, … ,gfk’ and we have shown that there are exactly k odd
permutations in Sn’ as desired. 0
An is called the alternating group of degree n. Alternating groups (espe
cially A4) will be useful to us a little later on.
We conclude this section by working out some examples in detail.
First let’s look at S3′ By Theorem 8.5, S3 has 3! =6 elements. They are
n 2 3 )=e 2 3 ‘
(~ 2 i) = (1,2,3), 3
(j 2 3 )_ I 2 (1,3,2),
( ~ 2 3 )_ 3 2 (2,3),
(~ 2 ~) =(1,2), I
(~ 2 i)=(1,3). 2
If we denote the second element in the list by f, then
f2=(~ 2 i)(;
2
;)=(j
2 D, 3 3 I
and
f3={j 2 ;)U 2 i)=e. I 3
Section 8. Symmetric Groups 73
Thus 0(1) = 3, and f generates the cyclic subgroup
a) Prove that for every h E S”
Section 8. Symmetric Groups 79
h 0 (Xh X2, … , x,) 0 hI = (h(xl), h(X2)’ … , h(x,)).
b) Prove that if.fi. 12 E S” then there exists h E S” such that 12 =h 0 .fi 0 h I iff for every
r ~ 2 the factorizations of.fi and12 into disjoint cycles contain the same number of
rcycles.
c) How many elements of S” are conjugate to (1.2 ….. n) in S,,1
8.27 a) Let Hbe a subgroup of S” that contains the transposition (1,2) and the ncycle
(1,2,oo.,n). Show thatH= Sn,
b) Let p be a prime, and let H be a subgroup of Sp that contains a tr~position and a
pcycle. Show that H = Sp.
SECTION 9
EQUIVALENCE RELATIONS;
CO SETS
Many times in mathematics we run into the following kind of situation. We
have a set S and we wish to identify certain elements of S with each other,
i.e., to regard certain elements as being “essentially the same” even though
they are different elements. This comes about when we are considering some
relationship that mayor may not hold between two elements of S, and we
wish to “lump together” any two elements between which the relationship
holds. For example, if we were considering the set of all triangles in the plane,
we might want to regard as “the same” any two triangles that were congruent
to each other.
Let’s examine this general situation a bit more precisely. First of all, what
do we mean by a “relationship” that mayor may not hold between two
elements of S? By a relation R on S we mean a set of ordered pairs of
elements of S. If SI,S2 E S, then SI is in the relationship R to S2 if and only if
the ordered pair (SI’S~ is one of the pairs in R. For example, if S = Z and R is
the set of all pairs (m,n) where m and n are both even or both odd, then 2 is
related to 6 by R, since 2 and 6 are both even [(2,6) E R], but 2 is not related
to 3, since 2 is even and 3 is odd [(2,3)E;eR]. For converiience we usually
express the fact that (SI’S~ E R by writing SI Rs2; thus in our example we
have 2R6 but not 2R3. Intuitively, sIRs2just means that SI is related to S2
by R.
Not every relation R on S is suitable for use in performing identifica
tions. For example, we certainly want to identify any element S with itself, so
if we aim to identify SI with S2 if and only if SI Rs2, then we want R to have
the property that S Rs, for every S E S. Similarly, if we are going to identify SI
with S2 then we want to identify S2 with SI’ so we want R to have the property
that SI RS2 implies S2RsI. Finally, if we identify SI and S2 and also S2 and s3
then we want to identify SI and S3′ so we want it to be the case that whenever
80
Section 9. Equivalence Relations; Cosets 81
SIRs2 and s2Rs3 then SIRs3′ These considerations lead us to the definition of
a special kind of relation on S called an equivalence relation.
DEFINI110N A relation R on S is called an equivalence relation on S if R has
the following three properties:
Reflexivity: For every sES, sRs;
Symmetry: For every Sl and S2 in S, if Sl RS2 then S2RsI;
Examples
1. Let S be any set and let R be the relation of equality on S, that is,
sIRs2 iff SI=S2′ Clearly R is an equivalence relation on S, and it is the
smallest one in the sense that it relates two elements iff they are related by
every equivalence relation on S.
2. Let S = Z and let aRb iff a < b. R is not an equivalence relation, since
the fact that 1 ={e,g}. Then we have:
H= He={e,g} =Hg =e =g,
Hf={j,gf} =Hgf =] = gf,
Hf2={j2,gf2}=Hgf2= f2 = gf2.
5. Let G = (0, +) and let H = Z. We assert that the distinct right cosets of
Z in (0, + ) are precisely the cosets Z + q, for 0..;; q < 1.
First of all, if q\,q2EO, O";;q\ < 1, O";;q2< 1, and q l =l=q2' then Z+q\=I=
Z + q2' because ql and q2 do not differ by something in Z, that is, q\  q2 is
not an integer. Secondly, for any rEO, the coset Z + r is the same as Z + q for
some q, 0..;; q < 1, because there is such a q which differs from r by an integer.
Observe that, in this example, there are infinitely many cosets.
Having considered right cosets, it is natural to wonder about left ones. If
H is a subgroup of G, then a left coset of H is of course a subset of G of the
form aH = { ah I h E H}, where a E G. Arguments just like the ones we have
gone through show that the left cosets of H are the equivalence classes under
the equivalence relation H == defined by
x H== y iff xIy EH
(Exercise 9.16). The left cosets of H need not be the same as the right cosets.
Example Let's find the left cosets of the subgroup H = {e,g} in S3' We have:
H = eH = {e,g} = gH, (Check it!)
fH = {j,jg} = fgH,
f 2H= {j2,j2g} = f2gH.
Observe that H is both a left coset and a right coset, but no other left coset is
a right coset. We will later single out for special attention those subgroups
with the property that right cosets and left cosets are the same thing.
EXERCISES
9.1.Determine which of the following relations R on Z are equivalence relations.
a) aRb iff a  b ? 0
b) aRb iff lal = Ibl
c) aRb iff ab? 0
d) aRb iff la  bl ~ 1
86 Section 9. Equivalence Relations; eo sets
9.2 LetS= {1, 2, 3} andletR= {(I, 1), (2, 2), (3, 3), (1, 2), (2,1), (3, 2)}.
Which of the properties of an equivalence relation hold for R?
9.3 Define an equivalence relation R on the set of points in the xyplane by:
(XI, YI) R (x:z, Y2) iffy I  XI = Y2  X2'
Describe the equivalence classes of R geometrically.
9.4 Let R be an equivalence relation on S. Show that for all SI, Sa E S we have
~ = S2 iff s1Rs2.
9.5 Let G = Qg. Find the right cosets of H in G for:
a) H=
b)H=<I>
9.6 Let G= D4 and letH= {e,f2g}. Find the right cosets of Hin G, and the left cosets.
9.7 Find the right cosets of the subgroup H = {(O, 0), (1, 0), (2, O)} in Z3 x Z2.
9.8 Find the right cosets of the subgroup H = {(O, 0), (0, 2)} in Z4 X Z4.
9.9 Find the right cosets of the subgroup H = < (I, 1) > in Z2 X Z 4.
9.10 LetX= {1,2,3,4} and let Y= {1,2}. Let Gbe the group (P(X), ~), Hthe subgroup
(P(y), ~). Find the right cosets of H in G.
9.11 For sets A and B, let ARB mean that there exists a onetoone mapping from A onto
B. Show that R is an equivalence relation on the class of all sets.
9.12 Let G be a group and for elements a, bEG let aRb mean that there exists an element
X E G such that a = xbx I. Show that R is an equivalence relation on G.
9.13 Suppose G is a group and A and B are subgroups of G. Define a relation R on G by:
X R Y iff there exist a E A and b E B such that X = ayb.
Prove that R is an equivalence relation on G.
9.14 Let G be a group and for a, bEG let aRb mean that ab = ba. Must R be an
equivalence relation on G? If so, prove it; if not, indicate for which groups R is an
equivalence relation.
9.15 Suppose G is a group and define a relation R on G x G by:
(a, b) R (c, d) iff ad = cb.
What condition on G is equivalent to R being an equivalence relation on G?
Prove your answer.
9.16 Let Hbe a subgroup of a group G and define fl= on G by letting
X IF Y iffxIy E H.
a) Show that fl= is an equivalence relation on G.
b) Show that the equivalence classes under n= are the left cosets of H in G.
Section 9. Equivalence Relations; eosets 87
c) Show that for a, bEG, aH= bH iff a1b E H.
9.17 Prove the second half of Theorem 9.1.
9.18 Suppose Hand K are subgrol1ps of a group G and x, y E G. Prove that if Hx = Ky then
H=K
9.19 Let G be the subgroup of SQ consisting of all permutations that are of the formj{x) =
ax + b with a, b E Q and a :f. O. Let H be the subgroup of G consisting of all
permutations of the formj{x) = x + n, with n E Z. LetJi andJi be the elements of G
given by Ji(x) = 2x +1 andJi(x) = 2x. Show thatJiH :f. JiH and HJi = JiHu JiR In
particular, it is impossible to choose exactly one representative from each left coset of
H in G in such a way that we have also chosen exactly one representative from each
right coset.
9.20 SupposeK is a subgroup ofH andH is a subgroup ofG. Suppose ~, … ,hm are
elements of H that determine distinct right cosets of K and gl”‘” g n are elements of G
that determine distinct right cosets of H. Prove that if either i :f. s or j :f. t then h;& and
hsl?t determine distinct right cosets of K.
SECTION 10
COUNTING THE ELEMENTS
OF A FINITE GROUP
In this section we shall consider two different ways of counting the elements
of a finite group G. One way will lead us to a classic result known as
Lagrange’s Theorem; the other will yield a powerful tool called the class
equation of G. In both cases, the plan will be to use an equivalence relation to
split G up into disjoint subsets, and then to count the elements of G by
counting those in each subset separately and adding the answers.
We begin with
1HEOREM 10.1 (Lagrange’s 1beorem) Let G be a finite group and let H be a
subgroup of G. Then 1 H 1 divides 1 G I.
Lagrange’s Theorem has an obvious significance in that it says a lot
about what possibilities there are for subgroups of a given finite group. This
information can be used to derive a variety of interesting results with little or
no effort; but first let’s prove the theorem itself. This too is not hard, because
the groundwork was laid in the previous section. Notice, incidentally, that we
already know the theorem is true if G happens to be cyclic.
We do need the following easy preliminary result:
LEMMA 10.2 Let G be any group (not necessarily finite), and let H be a
subgroup of G. Let Ha and Hb be right cosets of H in G. Then there is a
onetoone correspondence between the elements of Ha and those of Hb.
PROOF. Define a functionf:Ha …. Hb by declaring.f(ha)=hb for every hEH.
Thenfis onto because every element of Hb has the form hb for some hEH;
andfis onetoone since iff(h)a)=f(h2a)that is, if h)b=h2bthen h)=h2
by right cancellation, so h)a = h2a. 0
The significance of this result is that any two right cosets of H in G have
the same number of elements. Thus, for example, if one right coset has
88
Section 10. Counting the Elements of a Finite Group 89
sixteen elements, then every right coset has sixteen elements. If one right
coset is infinite, then every right coset is infinite, and has the “same infinite
number” of elements, because of the onetoone correspondence exhibited
above. In general, for any two sets Sand T, we say that S and T have the
same cardinality, and we write lSI = I TI, if there exists a onetoone corre
spondence between the elements of S and those of T. Thus we can express
the result of the lemma by saying that any two right cosets of H have the
same cardinality.
PROOF OF LAGRANGE’S THEOREM. Let G, H be as in the statement of the
theorem. The idea of the proof is that we can split G up into a finite number
of mutually disjoint subsets, each having IHI elements. Thus IGI is IHI times
the number of subsets.
Let =H denote the equivalence relation given by a =Hb iff ab I E H. By
Theorem 9.1, the equivalence classes under =H partition G into a collection
of mutually disjoint nonempty subsets, and by Theorem 9.3, these equiva
lence classes are just the right cosets of H. Since G is finite, only finitely many
distinct cosets can fit into G, so we have
G=Ha l uHa2 u··· uHak
for some integer k and elements al ,a2, ••• ,ak in G. Now, by Lemma 10.2, all
the cosets have the same number of elements, namely IHI, since H is one of
the cosets. Thus, counting the elements on both sides of the last equation, we
get
IGI=IHI+IHI+··· +IHI,
where there are k terms on the right. Thus IGI=k·IHI, and we have shown
that IHI divides IGI· D
If G is any group (not necessarily finite) and H is any subgroup, then the
number of distinct right cosets of H in G is called the index of H in G. We
denote this number by [G: H]. Thus Lagrange’s Theorem tells us that if G is
finite, we have
IGI=[G:H]IHI·
Examples
1. Let G be Klein’s 4group, that is, G={e,a,b,c}, with a2 =b2 =c2 =e,
ab=ba=c, ac=ca=b, and bc=cb=a. Then if H=, IHI=2, and since
IGI=4, we get 4=[G:H]·2, so [G:H]=2. In fact the right cosets of Hare
H={e,a} and Hb={b,c}.
2. Let G = (:l12′ E9) and let H = <4>. Then I H I = 3, and the right cosets of
H in G are H={4,8,O}, HE91={5,9,1}, HE92={6,lO,2}, and HE93=
{7,1l,3}. We have IGI=[G:H]·IHI, that is, 12=4·3.
90 Section 10. Counting the Elements of a Finite Group
3. Let G=S3 and let H=(g>. Then IHI=2, so IGI=[G:H]·IHI says
that 6=[G:H]·2, or [G:H]=3. As we saw in Section 9, the distinct right
cosets of H in G are H, Hj, and Hj2.
4. For an easy example in which G is infinite, let G=(Z, +) and let
H=(2)=2Z. The right cosets of Hare Hand H+I, so [G:H]=2.
5. It is possible for [G: H] to be infinite. For instance, let G = (0, + ) and
HZ. We saw in Section 9 that there are infinitely many right cosets of H
in G.
We have defined [G:H] to be the number of right cosets of H in fl. It is
natural to wonder what the number of left cosets is. Since the left cosets need
not be the same as the right ones, it comes as a pleasant surprise that the
number of left cosets is always the same as the number of right ones (even if G
is infinite).
THEOREM 10.3 Let H be a subgroup of G. Then the number of left cosets of H
in G is [G:H].
PROOF. Let S be the set of all right cosets of H, and let T be the set of all left
cosets. To prove the theorem, we will show that there is a onetoone function
from S onto T.
We want to definej:S+T by j(Ha)=aIH,jor all ae G. The one thing
we have to worry about is whether this assigns a unique value to each right
coset. In other words, if He – Hd, then we have mapped this coset to e IH,
but also to d IH. Does e IH – d IH? It does, because if He – Hd, then
ed””,leH, that is, (eI)ldleH. By Exercise 9.16, this says eIHdIH.
Thus we have assigned a unique value in T to each element of S. Our
function j is onetoone by the reverse of the argument we just gave. That is,
if j(He)j(Hd), i.e., eIHdIH, then (elrldIeH, edIeH, and
HeHd. Finally,jis onto since if eH is any element of T, we have He I eS,
andj(HeI)=(eI)IHeH. 0
A word more about this proof. Instead of attempting to achievej(Ha)
aIH for every ae G, we might have tried choosing one particular representa
tive x for each right coset Hx, and definingj(Hx)=xIH justjor that special
representative. This approach is not as satisfactory as the one we adopted. It
involves arbitrary choices, hence is not as “natural,” and it also binds us to
the particular representatives chosen. For instance, in checking that j is onto,
we took an arbitrary eH e Tand said thatj(HeI)=eH; if we had used fixed
representatives for all the cosets, we would have had to worry about whether
e I was the fixed representative for HeI, in order to getj(HeI)=eH.
Section 10. Counting the Elements of a Finite Group 91
We will now look at several results that demonstrate the importance of
Lagrange’s Theorem.
TIIEOREM 10.4 Let G be a finite group and let xEG. Then o(x) divides IGI.
Consequently, x lGI = e for every x E G.
PROOF. If we are going to use Lagrange’s Theorem to show that o(x) divides
IGI, then obviously what we want to do is to find ourselves a subgroup H of
G such that IHI=o(x). Take H=
element x”‘l=e in G. Then by Lagrange’s Theorem, l
negative integers less than m that are relatively prime to m. This generaliza
tion can be proved by essentially the same technique we used above for the
special case. One considers the group of integers in the set {O,I,2, … ,ml}
that are relatively prime to m, under multiplication mod m. Since the order of
this group is cp(m), the result follows. If you know about this theorem, you
might be interested in checking through the details of the proof.
Section 10. Counting the Elements of a Finite Group 93
We conclude our discussion of Lagrange’s Theorem by mentioning that
the theorem does not have a complete converse. That is, if G is a finite group
and n is an integer which divides I G I, then it does not follow that G has a
subgroup of order n. (You will recall that if G is cyclic, then it does follow.)
The usual counterexample is the group G = A 4 , the alternating group of
degree 4. We know that IGI =4!/2= 12, but it turns out that although 6
divides 12, A4 has no subgroup of order 6. (See Exercise 11.24).
We now tum to our second major objective for this section: the class
equation.t We need some preliminary ideas.
In Exercise 5.23 we introduced the notion of the centralizer of an
element. If G is a group and y E G, then the centralizer of y, denoted by Z(y),
is the set of all elements that commute with y:
Z(y)= {x E Glxy=yx}.
The point of Exercise 5.23 was that Z(y) is always a subgroup of G. It is clear
that Z(G)<;;;;Z(y) (an element that commutes with everything certainly com
mutes withy!), but in general Z(y) may be larger than Z(G).
Examples
1. If G is abelian, then Z(y) = G for every y E G.
2. Let G=S3={e,j,j2,g,jg,Pg}. Then Z(f)={e,j,p}, since we can
verify by direct calculation that g, jg, and j2g do not commute with j.
Likewise, Z(g)= {e,g}, sincej,j2,jg, andj2g do not commute with g. In this
case, Z( G) = { e}, the trivial subgroup, so both Z(f) and Z( g) are larger than
Z(G).
3. Let G= D4 = {e,j,j2,j3,g,jg,j2g,pg}, where 0(f)=4, 0(g)=2, and gf
= jig for all i. Here Z(g)= {e,j2,g,j2g} (verify!), so Z(g) is larger than
[G:Z(a)J=l ~ Z(a)=G ~ aEZ(G), and therefore ak+I, … ,ak+s are pre
cisely the elements of Z( G). Hence 3 = I Z( G)I, and we are done. 0
Perhaps the most noticeable difference between the situation here and
that in the proof of Lagrange’s Theorem is that here it need not be the case
that all the equivalence classes have the same number of elements. For
instance, we have already seen that
S3=iujUg,
so the class equation of S3 is
6= 1 +2+3.
The class equation has a number of striking applications. Two are presented in
Exercises 10.28 and 10.29, and we will see another one in Section 15.
EXERCISES
10.1 Let G Qa. Find [G:H] for H=< I), H=
in the list and take the elements h\a\.h2a\ ….. hna\. Show that the 2n elements
written down so far are all distinct. If they do not exhaust G, pick an element
a2 not listed so far and write down h\a2.h2a2″ … hna2′ Show that the 3n
elements written down so far are all distinct. Continuing in this way. finish the
proof.
10.11 a) Show that if H is a subgroup of G. then all the left cosets of H in G have
the same number of elements.
b) Show that any right coset has the same number of elements as any left coset.
10.12 Show that if H is a subgroup of afinite group G. then the result of Theorem
10.3 can be established by redoing the proof of Lagrange’s Theorem using left
cosets instead of right ones.
10.13 In the proof of Theorem 10.3. try definingJ(Ha) = aH. for all aE G. Show that
this may not give a welldefined function. because Ha may equal Hb without
aH equaling bH. (Suggestion: Let G= S3′)
10.14 Show that there are essentially only two groups of order 6. as follows.
a) If I G I = 6 and G contains an element of order 6. then G is cyclic. Why?
b) If G is not cyclic. then all elements of G have order 1.2. or 3. Why? Show
in fact that there must be an element of order 3. Call it a.
c) Let b be an element of G that is not in (a). Show that e.a.a2.b.ab.a2b are
all the distinct elements of G.
d) Show that o(b)=2. Since b was chosen arbitrarily. it follows that also
o(ab) = 2 and o(a2b)=2.
e) Show that ba=a2b and ba2 =ab.
The above steps show that either G is cyclic. in which case the multiplication
in G is like that in (Zt;. EEl). or G is not cyclic, in which case the multiplication
in G is like that in S3. with a and b playing the roles of the elementsJand g.
10.15 Let G be a finite grouP. and let H be a subgroup of G. Let K be a subgroup
of H. Prove that [G:K]=[G:H][H:K].
10.16 Let G be an abelian group such that I G I is an odd integer. Show that the
product of all the elements in Gis e.
Section 10. Counting the Elements of a Finite Group 97
10.17 Show that the multiplication in l.p – to} is associative.
10.18 Carry out the proof of Euler’s Theorem suggested in .the text.
10.19 Wilson’s Theorem (Another application of the group (Zp {O}, 0 ) to number
theory):
a) Show that ifp is prime then (Pl)! == I (modp). [Hint: Consider which elements
of(Zp{O}, 0 ) are their own inverses.] This result was stated by John Wilson
(17411793). It also seems to have been known to Leibniz in the late 1600s. The
first published proof of the theorem was given by Lagrange in 1770.
b) Prove the converse of the result in (a): Show that if n 2:: 2 and (n – I)! == 1 (mod
n) then n is prime.
10.20 If G is a group, Hand K are subgroups of G, and g e G then the set
HgK= {hgkl h e Handk eK}
is called a double coset. Prove thatHgK is the union of exactly [H: Hn gKg1] of the :
left cosets of Kin G, and that HgK is the union of exactly [K: K n gIHg] of the right
cosets of H in G.
10.21 Prove that every group of order 77 must have an element of order 7 and an element of
order 11.
10.22 a) Suppose G is a finite group such that (xy i = xV for all x, y E G and 3 fiG I.
Prove that G is abelian.
b) Does the result of part (a) remain true if we replace “3” by “5” throughout?
10.23 In each of the following cases, show that the given sets Sand Thave the same
cardinality .
a) S = the set of even integers, T= the set of odd integers
b) S=Z, T=Z+
c) S= {x E lR 10
d) S= lR, T= lR+
e) S={xElR l1
[Sn :Anl=2.
Our last criterion for normality will follow as an immediate corollary of
lHEOREM 11.4 Let G be a group, H a subgroup of G, and g E G. Then gHg\
is a subgroup of G, with the same number of elements as H.
PROOF. To see that gHg J is a subgroup, note that gHg J is nonempty, and
it is closed under multiplication since if gh J gJ and gh2g’J are elements of
102 Section 11. Normal Subgroups
(gh l gI)( gh2g l) = ghlh2gl,
which is in gHg – I since hI h2 E H. To check closure under inverses, note that
(ghgI)I=ghIgI, which is ingHg 1 since hIEH. The verification that
gHg – I has the same number of elements as H is left as an exercise. 0
COROLLARY 11.5 If H is a subgroup of G, and no other subgroup has the same
number of elements as H, then H is normal in G.
PROOF. For any g E G, gHg I is a subgroup of G with the same number of
elements as H, so gHg 1 must be H by hypothesis. Thus H is normal. 0
Examples
1. In S3′ A3 (= <1» is the only subgroup consisting of three elements.
Hence we see again that A3 is a normal subgroup of S3'
2. For an example where neither of Theorems 11.2 or 11.3 is applicable
but Corollary 11.5 is, let H be the subgroup
H = {e, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)}
m
G = A4 = H U { (1,2,3),(1,3,2),(1,2,4),(1,4,2),(1 ,3,4),(1 ,4,3),(2,3,4),(2,4,3)}.
Since every element of G  H has order 3, none of these elements can be
contained in a subgroup of order 4. Thus H is the only subgroup of order 4,
so H is normal in A 4, although H is not contained in Z(A4) and does not have
index 2.
We will now proceed to the business that motivated the introduction of
normal subgroups in the first place: we will show that if H is a normal
subgroup of G, then there is a natural way to turn the set of right (= left)
co sets of H into a group.
First a standard piece of notation: We write H
H is a nontrivial normal subgroup of G, and by Lagrange’s Theorem we have
IGI = IG/HI· IHI,
where IGIHI < m. Now since p divides I G I and p is prime, p divides either
IHlor IGIHI. If P divides IHI then, since H is cyclic, H has a subgroup of
order p, and this is the desired subgroup of G. Otherwise, p divides IGIHI, and
since GIH is abelian (see Exercise 11.17) the inductive hypothesis implies that
GIHhas a subgroup of order p. This subgroup must be cyclic, so it is < Hg > for
some element Hg of order p. By Exercise 11.10, o(g) = kp for some positive
integer k, and thus o(/J = p. Thus < t > is the subgroup we seek in G. 0
We know that Theorem 11.7 is no longer true if we drop the assumptions
that G is abelian and p is prime. One outcome of Sections 14 and 15 will be
that it is still true if we drop either one of these assumptions and keep the
other in force.
106 Section 1l. Normal Subgroups
EXEROSES
11.1 Show that SL(2,JR)
a) Find the order of the elementH + (5, 8) in GIH and justify your answer.
b) Is GIH cyclic? Justify your answer.
11.15 D4/ Z (D4) is “just like~’ one of the groups with which you are familiar. Which one?
11.16 Show that (Q +) / (Z, +) is an infinite group every element of which has finite order.
11.17 Let G be abelian and let H be a subgroup of G. Show that G/H is abelian.
11.18 Let G be cyclic and let Hbe a subgroup of G. Show that GIH is cyclic.
Section 11. Normal Subgroups 107 , which is not a normal sub
11.19 Give an example of a nonabelian group G such that GIZ (G) is:
a) abelian;
b) nonabelian.
11.20 Let G be a group and let H be a normal subgroup of G such that [G : If] = 20 and 1111
= 7. Suppose x e G and x7 = e. Show that x e H.
11.21 Let G be an abelian group and let Hbe the subgroup of G consisting of all the
elements that have finite order. Prove that every element of GIH other than the
identity element has infinite order.
11.22 Let G be an abelian group of order pq, where p and q are distinct primes. Show that G
is cyclic.
11.23 Let G be a group and let Hbe a subgroup of index 2. Show that for every a e G,
cleH.
11.24 Use Exercise 11.23 to verify the remark made in Section 10 to the effect that A4 has no
subgroup of order 6.
11.25 Let H
indicated elements being distinct. Define cp: In~G by cp(j)= gj,O <:.j <:.n  1.
Then cp is an isomorphism from In onto G. For clearly cp is onetoone and
onto; and cp is a homomorphism since for j, k E In we have
cp(j$k) = ~GJk = ~+k = ~gk = cp(j)cp(k). D
The corresponding result for infinite cyclic groups is also true:
THEOREM 12.3 Let G be an infinite cyclic group. Then G;:;; (I, +). Conse
quently, any two infinite cyclic groups are isomorphic to each other.
PROOF. Exercise.
The next three theorems provide some information about the behavior of
homomorphisms with respect to elements and subgroups.
114 Section 12. Homomorphisms
TIlEOREM 12.4 Let
group of K.
We would like to use Theorem 12.6 to help us deliver on a promise we
made in Section 8. In discussing symmetric groups, we remarked that one
reason why they are important is that every group can be “thought of” as a
subgroup of some symmetric group. The precise statement is that every group
is isomorphic to a subgroup of some symmetric group.
THEOREM 11.7 (Cayley’s Theorem) If G is a group, then G is isomorphic to a
subgroup of SG’ the symmetric group on the set G.
PROOF. To define a mapping cp: G~SG’ we must assign a permutation of G to
each g E G. Given g, define ~ E SG by
fg(x)=gx,
for all x E G. We saw in Section 7 that ~ is a onetoone mapping of G onto
itself.
Section 12. Homomorphisms 117
Now define cp: G–+SG by
cp( g) =.ig.
We assert that cp is a homomorphism, that is, CP(glgJ=CP(gl)CP(g2)’ for all
gl,g2 E G. This equation says that
.ig. g2 = .ig. o.ig2′
in other words, that.ig.g2 andfg. o.ig, are the same element of SG’ To verify this,
we show that .ig.g,(x) = (.ig. 0 fg2)(x), for every x E G:
fg•g2(x) = (gl g2)X = gl(g2X) = fg.(g2X) = fg.(fg,(x» = (fg. 0 fg,)(x).
Since cp is a homomorphism, Theorem 12.6 tells us that cp(G) is a
subgroup of SG (consisting of all the .ig’s). G is isomorphic to this subgroup,
since cp is onetoone: if cp( g I) = cp( gJ, that is, if .ig. = .ig” then in particular
.ig.(e) = .ig,(e)
gle=g2e
gl =g2′ D
The impact of Cayley’s Theorem is lessened somewhat by the fact that Sa
is usually huge in relation to G. (For instance, if IGI = 10, then ISal = 10! =
3,628,800.) This makes it difficult to derive information about G from the fact
that G is isomorphic to a subgroup of Sa. In Exercise 13.28 we will develop a
generalization of Cayley’s Theorem which sometimes enables us to show that
G is isomorphic to a subgroup of a symmetric group smaller than Sa.
EXERCISES
12.1 Which of the following mappings are homomorphisms? Monomorphisms?
Epimorphisms? Isomorphisms?
a) G=(R {O},’), H=(R+, .); cp: G~H is given by cp(x) = Ixi
b) G=(R+, .); cp:G~G is given by cp(x)=Yx.
c) G … group of polynomials p(x) with real coefficients, under addition of
polynomials; cp: G~(R, +) is given by cp[p(x)]=p(l).
d) G is as in (c); cp: G~G is given by cp[p(x)]=p'(x), the derivative ofp(x).
e) Gthe grQUP of subsets of {l,2,3,4,5} under symmetric difference;
A={l,3,4}, and cp:G~G is given by cp(B)=ALl.B, for every B~
{1,2,3,4,5}.
12.2 Let G be an abelian group. Show that the mapping cp: G–+G given by
cp(x) … X I is an automorphism of G. Show that if G were not abelian, then cp
would not be an automorphism.
118 Section 12. Homomorphisms
11.3 Let G be an abelian group, let n be a positive integer, and 1et!p: G+G be given
by !p(x) = x”. Show that!p is a homomorphism. Need it be a monomorphism?
An epimorphism?
11.4 In each case, determine whether or not the two given groups are isomorphic.
a) (ZI2, EB) and (0+,·)
b) (2Z, +) and (3Z, +)
c) (R – {O}, .) and (R, + )
d) V and Z2 XZ2
e) Z3XZ3 and Z9
f) (R{O},·) and (R+”)X(Z2,EB)
g) (Z, + ) and (Z, .), where
a.b=a+bl
h) G and G X G, where G=Z2XZ2XZ2XZ2X'”
(one copy of Z2 for each positive integer)
i) (lR – to}, .) and the group of Exercise 2.1(h)
j) (Q, +) x (Q+) and (Q+) x (1£, +)
k) D3 x 1£4 and D4 xl£ 3
12.5 Let G and H be groups. Show that G X H ~H X G.
12.6 Let G, H and K be groups. Show that (G X H) X K:;;;.G .x H X K.
11.7 Show that if A:;;;.G and B:;;;.H, then A xB:;;;.G XH.
11.8 Is (ZI4′ EB) isomorphic to a subgroup of (Z35′ EB)? Of (Z56, EB)?
11.9 Is V isomorphic to a subgroup of Qs?
11.10 Let X be a set containing at least two elements. Show that V is isomorphic to
a subgroup of (P(X), 1:>.).
11.11 Let G = Z2 X Z4′
a) Find subgroups H and K of G such that H ::0: K but G / H *’ G / K.
b) Find subgroups A and B of G such that G / A ::0: G / B but A “” B.
11.11 Show that there exist five groups of order 8, no two of which are isomorphic
to each other.
11.13 Let!p: G+H be a homomorphism.
a) Show that if H is abelian and !p is onetoone, then G is abelian.
b) Show that if G is abelian and !p is onto, then H is abelian.
c) Show that if !p.is an isomorphism. then G is abelian iff H is.
12.14 Let!p: G+H be an isomorphism. Show that Z(G)SilZ(H).
11.15 Let!p: G+H be an onto homomorphism. Show that if G is cyclic, so is H.
Section 12. Homomorphisms 119
12.16 Consider the mapping 11′: S3~S3 given by IP(Jigi) = j2igi. Show that 11′ is an
automorphism of S3′
12.17 How many automorphisms does Klein’s 4group have?
12.18 (Assumes familiarity with n X n matrices.) Let GL(n, R) be the group of all
invertible n X n real matrices under matrix multiplication. Let H be the subset
of GL(n, R) consisting of all matrices such that each column consists of one
1 and (n 1) zeros, and each row consists of one 1 and (n 1) zeros. Show that
H is a subgroup of GL(n, R) and H >:;;;S”.
12.19 Prove Theorem 12.6(ii) and (iii).
12.20 (Exercise 12.3, revisited). Let G be a finite abelian group and let n be a positive
integer relatively prime to IGI.
a) Show that the mapping lJ'(x) = x” is an automorphism of G.
b) Show that every x E G has an nth root, i.e., for every x there exists some
yEG such thaty”=x.
12.21 Let G be the group of nonzero complex numbers under multiplication and let
H be the subgroup of GL(2, R) consisting of all matrices of the form ( a b),
b a
where not both a and bare O. Show that G =r.H.
12.22 Let G be a group and let g E G. Show that the mapping cp : G ……. G given by cp(x) =
gxgl is an automorphism of G. Any such automorphism, obtained by conjugating by
a fixed element g E G, is called an inner automorphism of G.
12.23 A subgroup H of a group G is characteristic if 11′( H) ~ H for every
automorphism 11′ of G.
a) Show that every characteristic subgroup is normal.
b) Show that the converse of (a) is false.
12.24 Suppose that H <] G and K is a characteristic subgroup of H. Prove that K <] G.
(See Exercises 12.22 and 12.23.)
12.25 Show that the center of a group is a characteristic subgroup. (See Exercise
12.23.)
12.26 Show that the commutator subgroup of a group is a characteristic subgroup. (See
Exercise 11.30.)
12.27 If G is a group, Aut( G) denotes the set of automorphisms of G. Show that
Aut(G) is a subgroup of (SG' 0).
12.28 Let G=(Z3, $). Show that Aut(G) is not a normal subgroup of SG'
12.29 Let p be a prime. Show that a cyclic group of order p has exactly p  1 distinct
automorphisms.
12.30 Let G be an infinite cyclic group. Prove that Aut(G);;:,,(Z2' $).
12.31 Let H be a proper subgroup of G and let 1/1 be an automorphism of H other
than the identity mapping. Define a mapping 11': G~G by
120 Section 12. Homomorphisms
'P(x) = {�(x)
Is 'P an automorphism of G'l Explain.
ifxEH
if xflH.
12.32 Let G and H be two isomorphic groups. Exhibit a onetoone correspondence
between the set of automorphisms of G and the set of isomorphisms from G
onto H.
12.33 If we label the elements of V={e,a,b,c} with the integers 1,2,3,4, respectively,
then the proof of Cayley's Theorem shows us how to find a subgroup of S4
isomorphic to V. Write down the elements of this subgroup.
12.34 (See Exercise 12.22.) Show that the set of all inner automorphisms of a group
G is a subgroup of Aut( G). Is it a normal subgroup?
12.35 Prove that if G is a finite cyclic group then (Aut( G), 0 ) is an abelian group whose
order equals the number of generators of G.
12.36 To what familiar group is Aut (Zd isomorphic? Prove your answer.
12.37 Prove that ifp is an odd prime then there is no fmite group G such that Aut (G) � 'lLp.
12.38 a) Suppose G is a finite group and cp is an automorphism of G such that the set
{g E G I cp(g) = g I} contains more than threefourths of the elements of G.
Prove that G is abelian and that cp(g) = g 1 for all g E G.
b) Give an example of a nonabelian fmite group G and an automorphism cp of G
such that cp(g) = g 1 for exactly threefourths of the elements g E G.
12.39 Give an example of two fInite groups G and H such that G and Hhave the same
number of elements of order n for every n E Z+ but G and H are not isomorphic.
SECTION 13
HOMOMORPHISMS AND
NORMAL SUBGROUPS
In this section we will establish a connection between the seemingly unrelated
concepts examined in the preceding two sections. The main idea is that all
normal subgroups can be obtained from homomorphisms, and all homomor
phisms can, in a sense, be obtained from normal subgroups.
First, let H
2. If p:G~G/H, then p(e) =P(f) =p(p) = H= eG/ H ; and p(g)=p(fg)=
p(f2g) = Hg, because the elements g,fg,Pg differ by elements of H. The
canonical homomorphism sends everything in H to eG/ H , and thus “wipes
out” differences that lie in H.
2. Let G= Qs and H={I, I}. If p: G~G/ H, then p(/)=p(/)=H=
eG/ H’ and p(J) = p( – J) = H·J. G / H has order 4, and as we saw in Section
11, every nonidentity element of G / H has order 2, so that G / H is isomor
phic to Klein’s 4group. Thus there is a homomorphism from Qs onto V, and
we say that V is a homomorpbic image of Qs’
Our next move is to try and recover H from p, and in fact this is very
easy. If we are given p, then we get H by taking the set of elements in G that
are mapped by p to the identity element in G / H. That is,
H= {aEGlp(a)=eG/ H }.
The success of the preceding paragraph leads us to consider the general
notion of the kernel of a homomorphism.
DEFINITION If cp: G~K is a homomorphism, then the kernel of cp is
ker(cp) = cp I( {eK }) = {gE Glcp(g) = eK }.
THEOREM 13.1 For any homomorphism cp: G~K, ker(cp)
Gjker(q:».
THEOREM 13.2 (Fundamental theorem on group homomorphisms) Let q:>: G~K
be a homomorphism from G onto K. Then K ~ G jker( q:».
Remark. If q:>: G~K is not necessarily onto, we get q:>(G)~G jker(q:».
PROOF OF THE THEOREM. We have a map q:> from G to K and we wish to
~onstruct a map q5 from G jker( q:» to K. Let us write ker( q:» = N for simplicity.
The elements of G j N are right cosets Na, and we have to decide where to
send each such coset. What other try is there but q5(Na) = q:>(a)?
Let us reiterate. We have a mapping q:> from G to K, and we wish to find
one from G j N to K:
Our strategy in deciding where to send Na is to take a representative, a, for
Na and see what q:> does to it. q:>(a) will be an element of K, and this is where
we send Na.
We have to check that this gives us a welldefined mapping, i.e., that if
Na is also Nb then q:>(a) = q:>(b), so that our definition of q5(Na) is independent
of which representative of Na we pick to make the definition. Now if
Na= Nb, then ab I EN =ker(q:», so
q:>( ab I) = eK •
But
q:>( ab I) = q:>( a)q:>(b I) = q:>(a) [ q:>( b)] I,
so q:>( a) = q:>( b) and all is well.
q5 is a homomorphism, since
q5(NaNb) = q5(Nab) = q:>(ab) = q:>(a)q:>(b) = q5(Na)ip(Nb).
124 Section 13. Homomorphisms and Normal Subgroups
The crucial step in this chain of equalities is the fact that cp( ab) = cp( a )cp( b).
Thus the fact that ij5 is a homomorphism is thrown back on the fact that cp
was one to begin with.
ij5 is onetoone since if ij5(Na)=ij5(Nb) then cp(a)=cp(b), so cp(ab1)=eK
and ab 1 Ekercp=N, yielding Na=Nb. Finally, ij5 is onto, since if kEK then
there exists a E G such that cp( a) = k (because cp was assumed to be onto), and
this means that ij5(Na) = k. 0
Examples
1. Again let cp: G L(2, IR)~(IR – {O}, . ) be given by cp( = ~) = ad – be. Then
cp is onto, since for any r*O we have (~ ~) E GL(2, IR), and cp((~ n) = r. By
the Fundamental Theorem we conclude that
GL(2, IR) Iker( cp) a.; (IR – {O}, . ),
in other words,
GL(2, IR) 1 SL(2, IR) a.; (IR – {O}, . ).
2. Let cp: (1, + )~(ln’ E9) be given by cp(m) = iii, the remainder of m (mod
n). Then cp is onto, so the Fundamental Theorem says that
that is,
This makes precise our observation in Section 11 that the addition of the
cosets of n1 “corresponds” to the addition of their representatives, mod n.
3. Let cp: Sn~({l, I}, .) be given by
cpU) = { _ 11 if J is even
ifJis odd.
Then cp is onto, and as we have seen, ker(cp)=An• Thus
4. Think of the complex numbers as the points in a plane by identifying
the number x + yi with the point (x,y). Let U be the set of points on the circle
of radius 1 about the origin. Thus U consists of all points x + yi such that
x 2+ y2= 1, and the points in U are precisely those that can be represented in
the form cosO+ i sin 0, for some real O. We assert that U is a subgroup of
(C – {O}, .), and
U~(IR, + )11.
Section 13. Homomorphisms and Normal Subgroups 125
To see this, define a mapping cp: IR~(C – {O}, .) by
cp(x) =cos2’ITX + isin2’ITx.
cp maps IR onto U, and cp is a homomorphism since for x, y E IR we have
cp(x + y) = cos(2’ITx + 2 ‘IT)’ ) + i sin(2’ITx + 2 ‘IT)’ )
= (cos2’ITxcos2’1T)’ – sin2’ITx sin 2 ‘IT)’ )
+ i(sin2’ITxcos2’1T)’ +cos2’ITx sin2’1T)’)
= (cos2’ITx + i sin 2’ITx)(cos 2 ‘IT)’ + isin2’1T)’) = cp(x)cp(y).
Thus U is a subgroup of (C – {O}, .) by Theorem 12.6. Its identity element is
I +Oi, so
ker( cp) = {x E IRlcos2’ITx = 1 and sin2’ITx = O} = lL.
By the Fundamental Theorem, (IR, + )/lLr;;r.U.
S. Let G and H be groups, and consider the normal subgroup G X { eH }
in G X H. It would seem that if we factor G X H by G X {eH }, we should get
essentially H. Indeed we do, since there is an onto homomorphism cp: G X H
~H given by cp[(g, h)] = h, and the kernel of this map is G X {eH }. Thus
(G X H)/( G X {eH})r;;r.H.
These examples demonstrate that the Fundamental Theorem is a useful
tool for obtaining isomorphisms. On the theoretical side, the theorem tells us
that the image of any homomorphism can essentially be recovered from the
kernel. Moreover, the proof of the theorem shows us how close we can come
to recovering the homomorphism itself.
In the proof, the isomorphism q5 was defined so that q5(Na)=cp(a) for
every aE G. Thus
q5(p(a» = cp(a)
for every a, or, in other words,
cp 0 p= cpo
This equation is often expressed by saying that the diagram
cp
G ) K
~y
G /ker(cp)
“commutes,” because going directly from G to K accomplishes the same thing
as taking the detour through G /ker( cp). The equation q5 0 p = cp tells us that cp
and p come as close as could be hoped to being the “same” mapping. They
126 Section 13. Homomorphisms and Normal Subgroups
map G onto groups that are isomorphic, and except for the isomorphism ip
they are the same mapping.
In the situation of the Fundamental Theorem, there is a onetoone
correspondence between subgroups of K and subgroups of G that contain
ker(cp), normal subgroups corresponding to normal subgroups. To discuss this
correspondence, we shall use the fact that if H is a subgroup of G containing
ker(cp) then cpl[cp(H)] = H. (For the notation, see the statement of Theorem
12.6.) To verify this fact, note that H ~cpl[cp(H)] automatically, that is, if
hE H then cp maps h to an element of cp(H). For the reverse inclusion, let
x E cp 1[cp(H)], i.e., suppose cp(x) Ecp(H), with the aim of showing that x E H.
Since cp(x) E cp(H), we have cp(x) = cp(h) for some hE H. Thus cp(xh I) = eK ,
so xh – I E ker( cp), and therefore xh – I E H by our assumption that ker( cp) ~ H.
Thus x = (xh I)h is the product of two elements of H, so x E H as desired.
Now to any subgroup H of G containing ker(cp) we associate the
subgroup cp(H) of K; cp(H) is a subgroup by Theorem 12.6. This association
is onetoone, since if HI is another subgroup of G containing ker( cp) and if
cp(HI)=cp(H), then
hence HI = H by the fact established above. The association is onto since if J
is any subgroup of K, then cpI(J) is a subgroup of G containing ker(cp), and
cpr cp I(J)] is J, because cp is onto (see Exercise 13.18). Notice that if H d
ker( cp) is normal, then so is cp(H); and if cp(H) is normal in K, then H
This completes the proof. 0
Example Let G = (l, + ), H = 4l, K = 6l. The theorem says that
4l 41+6l
=
4ln6l 6l
that is,
128 Section 13. Homomorphisms and Normal Subgroups
This conclusion can be verified by other means, because both 41:/121: and
21:/61: are cyclic groups of order 3.
An application
Let G be a finite group, let H be a subgroup of G, and let K
Qa/<I>
Observe that the order of the quotient group on the left side is 4/2=2
and the order of Qa/
that IHl divides IGI.
13.1S Let m and n be positive integers. Show that there exists an onto homomorphism q> ;
(Z”, EB) + (Z”” EB) if and only if m divides n.
13.16 Let A ; G + Kbe an onto homomorphism and letJbe a subgroup of K. Show that
q>[ q> – 1(.1)] = J.
13.19 Let q> ; G + K be a homomorphism. Prove that q> is onetoone if and only if ker (q»
= {eG}.
13.20 Let q> ; G + K be an epimorphism. Let J ; G + Kbe an epimorphism and letNbe a subgroup ofker(q» such thatN pl,.
The uniquely determined integers p It ;;. p 12 “> •.• “> P I” taken for all primes
that divide \G\, are called the invariants of the nontrivial group G. We adopt
the convention that the invariants of a trivial group are {I}.
Before we consider the proof of Theorem 14.2, we will list some
corollaries which indicate the kind of control the theorem gives us over finite
abelian groups. The first corollary tells us when two finite abelian groups are
isomorphic.
COROLLARY 14.3 Let A and B be finite abelian groups. Then A r;;;oB iff A and B
have the same invariants.
PROOF. If A and B have the same invariants, then A is trivial iff B is. If they
are both trivial, they are isomorphic. If neither is trivial, then when we write
them both as direct products of (nontrivial) cyclic groups of primepower
order, we get the same number of factors of each order in both cases, and
from this it follows that A r;;;oB.
Conversely, suppose A r;;;oB. If A is trivial, then so is B, so A and B have
the same invariants. If A is not trivial and we write A r;;;oA \ X A2 X … X Ar ,
then B;;;;,:A\ xA2x··· XA r , so again A and B have the same invariants. 0
If n is a positive integer, then by a partition of n we mean a sequence of
positive integers 1\ ;;. 12 ;;’ ••• ;;. Ir such that 1\+ 12 + … + Ir = n. The number of
distinct partitions of n is denoted by p(n). For example,p(4)=5, since we can
write 4 as 4, 3 + 1, 2 + 2, 2 + 1 + 1, or 1 + 1 + 1 + 1.
COROLLARY 14.4 If q is a prime and n is a positive integer, then the number of
non isomorphic abelian groups of order qn is p(n). If m = qftq22 … q:” for
distinct primes qi’ then the number of nonisomorphic abelian groups of order
m is p(n\)p(n2)·· .p(nk ).
136 Section 14. Direct Products and Finite Abelian Groups
PROOF OF THE FIRST STATEMENT. To any partition t, ;;;.t2 ;;;’ ••• ;;;’tr of n, we
associate the group
llql’ X llql2 X … X llql”
an abelian group of order qn with invariants q’l ;;;’q’2;;;’ … ;;;’q”. The groups
associated to different partitions are nonisomorphic because they have dif
ferent invariants. Thus we get pen) nonisomorphic groups. But any abelian
group of order qn must be isomorphic to one of these because it has the same
invariants as one of them. Hence there are precisely pen) nonisomorphic
abelian groups of order qn.
The proof of the second statement is left to the reader, but the idea
should be clear from the following examples. 0
Examples
1. Let’s find all abelian groups G of order 36. Since 36 = 22. 32, the
possibilities for the 2groups in the primepower decomposition of G are 1122
and 1121 X 1121, that is, 114 and 112 X 112′ The possibilities for the 3groups are 119
and 113 X 113′ Thus we have
114 X 119′
114 X 113 X 113′
112 X 112 X 119′
112 X 112 X 113 X 113′
a total of four different abelian groups of order 36.
2. To find all abelian groups of order 600, we write 600=23.3′.52• The
possibilities for the 2groups are llg, 114 X 112′ and 112 X 112 X 112′ There must be
one 3group, 113′ The 5groups can be 1125 or 115 X lls. So we have
llg X 113 X 112s’
llg X 113 X lls X 115′
114 X 112 X 113 X 1125′
1140 112 X 113 X lls X lls’
112 X 112 X 112 X 113 X 112s’
112 X 112 X 112 X 113 X lls X lls.
One of these groups must be isomorphic to 11600′ In fact, it is llg X 113 X
112s, because this is the only cyclic group in the list, by Theorem 6.1. We can
also write 11600 ~ 1124 X 1125 ~ llg X 117s ~ 113 X 112oo’ These examples serve to em
Section 14. Direct Products and Finite Abelian Groups 137
phasize the fact that two products of nontrivial cyclic groups can be isomor
phic without the number of factors being the same, and without the orders of
the factors being the same. By restricting ourselves to factors of primepower
order, however, we achieve uniqueness.
The information provided by Theorem 14.2 can sometimes be used to
handle the “abelian case” in more general contexts. For example, let us
sketch a proof that there are essentially only five groups of order 8 (non
abelian ones included).
We assert that any group G of order 8 must be isomorphic to one of
Zg,Z4 X Z2,Z2 X Z2 X Z2′ Qg,D4 •
If G is abelian, then we know that G is isomorphic to Zg, Z4 X Z2′ or
Z2 X Z2 X Z2. If G is not abelian, then G can have no element of order 8 (else
it would be cyclic), and G must have an element of order 4, since otherwise
we would have x 2 =e for every xEG, making G abelian (Exercise 3.11). Say
aEG and o(a)=4. Then 2. The group
7..2 X 7..2 X … X 7..2 has a nontrivial automorphism ((!, obtained by interchang
ing the first and second components of all the elements. If ‘” denotes an
isomorphism from G onto 7..2 X 7..2 X … X 7..2’ then ‘” I 0 ((! 0 ‘” is a nontrivial
automorphism of G. 0
The only place where we used the finiteness of G in this proof was in
obtaining an isomorphism from G onto a product of 7..2’s. It can be shown
that if G is an infinite abelian group such that x 2 = e for all x E G, then G is
isomorphic to a subgroup of the product of infinitely many 7..2’s (the subgroup
consisting of all the elements with only finitely many nOnzero components).
Our proof then shows that every group with more than two elements has a
nontrivial automorphism.
The preceding corollaries demonstrate that Theorem 14.2 is a powerful
tool. We will now take up its proof, which is essentially a reprise of the
“quotient groups and induction” theme we introduced in Section 11.
The proof is somewhat longer than any we have done before, so we will
split it into three steps:
Step 1. We show that every finite abelian group is isomorphic to a product of
abelian pgroups.
Step 2. We show that every finite abelianpgroup is isomorphic to a product
of cyclic groups of ppower order.
Steps I and 2 establish the existence of the primepower decomposition.
Step 3. We show that the primepower decomposition is unique.
Suppose G is abelian and IGI=p[l’~z … p!/, where the p;’s are distinct
primes, and each rj ;> 1. For each i, the set
G(pJ= {xEGlxP(I.=e}
Section 14. Direct Products and Finite Abelian Groups 139
is a subgroup of G, and is a Pjgroup. Step 1 of our program will be
accomplished if we can show that
G;;;G(PI) X G(P2) X· .. X G(Pk)’
and to do this it will suffice, by induction, to prove the following lemma.
LEMMA 14.7 Let G be an abelian group, and let IGI = mn, with (m,n)= 1. Let
A = { x E G Ix m = e} and let B = { x E G Ix n = e}. Then G;;;A X B.
PROOF. A and B are subgroups of G, and they are both normal, because G is
abelian. We must show thatAB=G andAnB={e}.
Since (m,n)= 1, there are integers rand s such that rn+sm= 1. If xEG,
then
x = Xl =x rn + sm = X Tnxsm,
and xTnEA, xsmEB, since mn=IGI. Thus G=AB.
If xEAnB, then x=xTnxsm=ee=e. 0
Step 2 of the proof takes a little more doing, but the plan of attack is
straightforward. Suppose G is a finite abelian pgroup; we wish to show by
induction on I G I that G is isomorphic to a product of cyclic groups of
ppower order. If I G I = 1, then G is already cyclic of order pO. Now assume
the result is true for all pgroups of order less than I G I.
We want to choose x =Fe in G, let A =(x), and find a subgroup B of G
such that we can apply Theorem 14.1 to A and B. If we can do this, then since
IBI < I GI, the inductive hypothesis will finish the proof.
Now if B exists at all, it has to be isomorphic to G / A, so it is natural to
look at G / A. The inductive hypothesis also applies to this group, so we have
G/A;;;(YI) X
o( x) 😉 o(y). As things stand this need not be true, but we can make it true by
Section 14. Direct Products and Finite Abelian Groups 141
making a special choice of x at the outset. We choose x so that o(x) >o(y) for
all y E G to begin with, and then Step 2 of our proof is complete.
For Step 3, observe that if G~GI x··· X Gk , where Gp ••• ,Gk are nontri
vial cyclic groups of primepower order, then I G I = I Gil’ … ·1 Gk I, so the
primes that occur in the orders of the G;’s are precisely the primes that divide
I G I. Also note that, for any such prime p, the product of those G;’s that are
pgroups is isomorphic to the subgroup of G consisting of those elements with
order a power of p. Thus, to establish the uniqueness of the primepower
decomposition, it will suffice to handle the case of a nontrivial finite pgroup.
So suppose that
proceed by induction on the (equal) orders of the two products involved, the
case of order p=pl being trivial. For the induction step, notice that any
isomorphism which gives us [14.3J must map the set of pth powers of all the
elements on the left onto the set of pth powers of all the elements on the right.
That is, [14.3J entails
products have smaller order than the ones we started with, since
14.9 Find a direct product of cyclic groups of primepower order that is isomorphic to
Aut (~o), and show that your answer is correct.
14.10 Let Gl> … ,G,. be subgroups ofG such that:
i) Gt. … ,G” are all normal;
ii) G = GIG2 •• ·G”, that is, every element of G can be written as g~2 … g,. with g/ E
G/;
iii) for 1 ~ i~ n, G j n GIG2 ••• GtI = {e}.
Show that G == GI X G2 X ••• x G,..
14.11 Show, by an example, that ifwe replace (iii) in Exercise 14.10 by the weaker
condition G1 n Gj = {e} for i “¢ j, then G does not have to be isomorphic to
GI xG2 x … xG”.
14.12 Let G, H, and K be finite abelian groups. Show that if G x H == G x K., then H == K
14.13 Show, by example, that if we allow the group G in Exercise 14.12 to be infinite, then
H need not be isomorphic to K.
14.14 Let G be an abelian group of order p”, where p is prime. An element x EGis said to
be of maximal order if o(x) ~ o(y) for ally E G. Show that the only subgroup ofGthat
contains all the elements of maximal order is G itself.
14.15 Let G andHbe finite abelian groups such that for every positive integer n, G andH
have the same number of elements of order n. Prove that G == H.
14.16 Let G be an abelian group of order P’ with invariants p~ ~ /2 ~ … ~ i r • Let H
be a subgroup of G with invariants p Z’J ~ P 1’2 ~ •.. ~ pU, . Show that s ~ r and UI ~
tl for 1 ~ i ~ s.
SECTION 15
SYLOW THEOREMS
We have seen that for finite abelian groups, Lagrange’s Theorem has a
converse: if m divides IGI and G is abelian, then G must have a subgroup of
order m. Of course for general groups G this falls apart; A4 is a group of
order 12 with no subgroup of order 6. We are left wondering what can be
salvaged in general; if G is a group and m divides 1 G I, then under what
conditions can we assert that G must have a subgroup of order m?
Well, look at A 4• It has subgroups of order 2, 3, and 4, namely
«1,2)(3,4», «1,2,3», and {e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)}. At least on
the basis of this (admittedly flimsy) evidence, we might suspect that the
trouble comes when we “mix primes,” i.e., when we try an m that is not just a
power of some one prime. This suspicion can be borne out. For we shall
prove in this section that if pk divides 1 G 1 for any finite group G, then G must
have a subgroup of order pk.
This assertion, together with some related facts, comprises what are
known as the three Sylow Theorems, after the Norwegian mathematician
Ludwig Sylow (18321918). We will state the three theorems together, then
look at some examples and applications, and finally present the proofs.
We need to recall a couple of old notions before we can get started. If H
is a subgroup of G and g E G, then the set gHg I is called the conjugate of H
by g. By Theorem 11.4, gHg – 1 is a subgroup of G, with the same number of
elements as H. If K is also a subgroup of G, we say that H and K are
conjugate if K = gHg – 1 for some g E G. Conjugacy is an equivalence relation
on the set of all subgroups of G by the same proof as for conjugacy of
elements. Finally, the normalizer of H in G is the subset
N(H)= {gE GlgHg1=H}.
By Exercise 11.27, N(ll) is a subgroup ofG. We have H!;; N (ll).
143
144 Section 15. Sylow Theorems
Example Let G=S3′ and let H=
We assert that no group G of order 28 is simple. For let H be a 7Sylow
subgroup of G, and consider the number of 7Sylow subgroups of G. This
number is [G: N(H»), hence must divide 4; but it is also of the form 1+ 7j.
Clearly, then, there is only one 7Sylow subgroup, so H
Section 15. Sylow Theorems 147
Thus, either there is only one 3Sylow subgroup, or there is only one
5Sylow subgroup. In other words, one of B, C must be normal. Thus BC is a
subgroup of G, of order (IBI·ICI)/(IBnCl)=15. By Theorem 15.5, BC is
cyclic; say BC=
IKnH*1 pi p P
This contradicts the fact that H is a pSylow subgroup of G. Thus Lemma
15.7 is proved and, with it, the First Sylow Theorem. 0
PROOF OF THE SECOND SYLOW THEOREM. Easy: If K and Hare pSylow
subgroups, then by part (ii) of the First Sylow Theorem, we have KJf gHg – 1
for some g E G. But K and gHg I have the same order, so K = gHg ,and K
and H are conjugate.
Next, if a pSylow subgroup H is normal and K is any pSylow subgroup,
then the fact that K=gHg 1 for some g means that K=H. Thus H is the
only pSylow subgroup. Conversely, if H is the only pSylow subgroup, then
H is normal by Corollary 11.5. 0
PROOF OF THE THIRD SYLOW THEOREM Let H be a pSylow subgroup. By the
Second Sylow Theorem, the number of pSylow subgroups is the number of
conjugates of H in G, and this is [G:N(H)] by Lemma 15.8. Finally, if
Hb … ,Hm are all the pSylow subgroups of G, we want to show that m = 1 + jp
for some j ~ O. If K is anyone of the pSylow subgroups, then as in the proof of
Theorem 15. 1 (ii), m is a sum of terms of the form [K: K (l N (Hi)], where each
term in the sum is among l,p,p2, … ,p” = IKI, and at least one term is 1. To
fmish the proof, it will suffice to show that exactly one term in the sum is 1.
But what does [K:K (“\ N (Hi)] = 1 mean? It means that K ~ N(Hi)’ which, by
Lemma 15.7, means that K ~ Hi’ Since IK I = IN; I, this means that K = Hi’
Thus [K: K (“\ N (Hi)] = 1 iff Hi = K, and the proof is complete.D
EXERCISES
15.1 Let H be a normal subgroup of a finite group G and suppose IHI = pk, where
p is a prime. Show that H is contained in every pSylow subgroup of G.
15.2 Let H be apSylow subgroup of G. Show that H is the only pSylow subgroup
of G contained in N(H).
15.3 Suppose that K
is not abelian.
15.12 Let P. q be primes such that p divides ql. Assume the following fact (which follows
from Corollary 19.4): 1ft is an integer that satisfies the conditions t _ 1 (mod q) and
t P == 1 (mod q), then any integer that satisfies these conditions must be congruent
modqto one oft, r ….. trl. Use this to prove that all nonabeliangroups oforderpq
are isomorphic to each other. [Suggestions: Note that the qSylow subgroup Q of any
such group must be normal. Let Q =< a > and take bEG – Q. Then < a >< b >= G
and babI = d for some t. Show that bPabP= a9′ ,and therefore t P == 1 (modq),
although t Ii! 1 (mod q). Note that if we choo!je a different generator for < b >for
example bi , 2 ~j ~plthen lIa(llrl = at~]
15.23 Let I GI = pn and let H be a subgroup of G such that IHI = p”‘, where m ~ n. Show that
if m ~ k ~ n then there exists a subgroup K of G such that IKJ = pk and H ~ K.
[Suggestion: Use the fact that, by the class equation, Z(G) has a subgroup of order p.]
15.24 a) In S3 x S3, let H = x and K = < (g, g) > . Show that HK is a subgroup of
order 18 in S3 x S3.
b) Prove that there are exactly five pairwise nonisomorphic groups of order 18, and
find them all.
15.25 a) Show that < if, 0) >< (g, 1) > is a subgroup of order 12 in S3 x 2:4.
b) Prove that there are exactly five pairwise nonisomorphic groups of order 12, and
find them all.
SECTION 16
RINGS
Up to now we have been studying sets with a single binary operation defined
on them. For example, we have encountered groups such as (0, +) and
(0 – {O}, .). But of course there are times in real life when one considers both
addition and multiplication simultaneously, for instance on 0, or more
basically on lL. We will now consider an abstract notion designed to capture
the essence of such situations where two operations interact with each other.
What is the essence of the situation for addition and multiplication on lL,
for example? If we just look at addition, then we have an abelian group
(lL, + ). If we concentrate on multiplication, we have an associative commuta
tive binary operation. There happens to be an identity element for ” but most
elements fail to have inverses. Finally, if we consider both operations at once,
then the most salient point is that they are connected by the distributive laws:
a(b + c)= ab + ac and (b + c)a= ba + ca.
During the nineteenth century, number theorists worked with systems
more inclusive than lL which satisfied the same properties with respect to +
and ‘. One motivation for their efforts was the hope that by considering such
systems, one might answer questions about lL that could not be answered by
thinking in terms of lL alone. Although this hope was only partially realized
(questions about lL can be hard!), a great deal was accomplished, and
moreover, the groundwork was laid for the development of an abstract theory
in the twentieth century.
The abstract concept which emerged is that of a ring. A ring has all the
properties of lL indicated above, except that in order to achieve greater
generality, one does not require that there exist an identity element for
multiplication, nor that multiplication be commutative. (A good deal of work
has also been done on systems for which multiplication fails even to be
associative, but we will not consider such nonassociative rings.)
153
154 Section 16. Rings
In writing down the axioms for a ring, we should perhaps use symbols
such as * and 0 to denote the two operations involved, to emphasize that
they do not have to be ordinary addition and multiplication of numbers.
However, we shall just use + and ” for simplicity. You are experienced
enough by this time to keep in mind that we are just talking about two binary
operations, even though we denote them by + and . and call them addition
and multiplication.
DEFINmON Suppose that R is a set and + and . are two binary operations
on R. Suppose further that:
i) (R, +) is an abelian group,
ii) . is associative, and
iii) the distributive laws hold, i.e.,
for all ‘,,’2,’3 in R.
Then R, together with the binary operations + and ” is called a ring. We
denote it by (R, +, .), or R for short.
The two distributive laws are referred to as the left and ,ight distributive
laws, respectively. Of course, if . happens to be commutative, then these two
laws say the same thing.
A ring for which . is commutative is called a commutative ring. In
general, the addition on a ring has already been assumed to be pretty nice,
and one gets more special rings by imposing more assumptions on the
multiplication.
The additive identity element of R, i.e., the identity element for (R, +), is
denoted by 0, or OR’ If there happens to be an identity element for’, then it is
an easy matter to see that there is only one such; it is called the multiplicative
identity element or the unity of R, and is denoted by 1, or 1 R’ A ring that
possesses a unity is called (what else?) a ring with unity.
Examples
1. (I, +, .) is a commutative ring with unity, as are (10, + , .) and
(IR, + , .). Here + and . denote ordinary addition and multiplication.
2. (21, +, .) is a commutative ring, but not a ring with unity.
3. Let R be the set of all real numbers that can be written in the form
a + b v’2 , where a, bEl. It is clear that the sum of two elements of R is in R,
Section 16. Rings 155
and if a + b Y2 and c + dY2 are in R, then their product
(ac + 2bd)+ (ad+ bc)Y2
is in R too. R is a commutative ring with unity under ordinary addition and
multiplication.
4. Let R = {O, 1,2, … ,n – I}, and let EB and 8 denote addition and
multiplication modulo n on R, that is,
aEBb= a+b and a8b= a·b,
where denotes remainders modulo n. Then (R, EB) is the familiar group
(In’ EB), and we claim that (R, EB, 8) is a commutative ring with unity, which
we will denote by (In’ EB, 8). In fact, the proofs of associativity for 8 and of
the distributive laws are very similar to the proof we gave for the associativity
of EB on In in Section 2 (Exercise 16.8). (In’ EB, 8) is commutative because
a8b= a·b = b’a =b8a.
The multiplicative identity element is 1.
The rings (In’ EB, 8) are interesting at this point because they already
begin to display behavior quite different from that of the prototype example
(l,+,·). For instance, look at (l6,EB,8). Here 283=0 although neither 2
nor 3 is 0. Things like that certainly don’t happen in l. To see something even
stranger, look at (ls, EB, 8). Here 23(=28282)=0, so a power of a nonzero
element can be 0.
We introduce some terminology in order to deal with such situations. An
element a E R is called a zerodivisor if there exists an element b =1= ° such that
either
ab=O or ba=O.
a is called nilpotent if there exists some positive integer n such that an = 0.
(Here an means a multiplied by itself n times.) Thus in l6′ 2 is a zerodivisor,
and so is 3 (and ° and 4). In ls’ 2 is nilpotent, as are 0, 4, and 6.
At the opposite extreme from these badly behaved elements are those
called units. Suppose R is a ring with unity. Then a E R is called a unit if there
exists an element b E R such that
ab=ba= 1.
Note right away that “unit” and “unity” are not the same thing! The unity is
a unit because 1·1 = 1, but there may be many units other than 1. For
instance, in ls, 1,3,5, and 7 are all units, since 12=32=52=72= 1. In lJO’ 1,
3, 7, and 9 are units because 12=92= I and 387=783= 1.
We shall see below that a unit can never be a zerodivisor. The general
situation for (In’ EB, 8) is that a Eln is a unit iff (a,n)= 1, and every element
156 Section 16. Rings
is either a unit or a zerodivisor (see Exercise 16.9). On the other hand, some
rings contain elements that are neither units nor zerodivisors. Can you give
an example?
It is easy to see that if a is a unit, then there is only one b such that
ab = ba = 1. We call b the multiplicative inverse of a, and denote it by a I.
Examples (continued)
5. Let ~ ffi ~ be the set of all ordered pairs (a, b) of real numbers, with
addition and multiplication defined componentwise:
(a,b)+(c,d)=(a+c,b+d) and (a,b)·(c,d)=(ac,bd).
Then ~ffi~ is a commutative ring with unity. An element (a,b) in ~ffi~ is a
zerodivisor iff at least one of a, b is 0, and it is a unit iff neither of a, b is O.
For example, the equation
(0,3) ·(1,0)=(0,0)
shows that both (0,3) and (1,0) are zerodivisors, while
(5,6) . ( ~, ~) = (1,1)
shows that both (5,6) and (L ~) are units. In this ring it is true that an
element is a zerodivisor iff it is not a unit.
The ring IR ffi IR is called a direct sum. It is common practice, in ring
theory, to speak of “direct sums” rather than “direct products.” The additive
terminology is also commonly used in discussing abelian groups.
6. Generalizing the previous example, let R I’ R 2, ••• , Rn be rings. Then
their direct sum R 1 ffi R2 ffi … ffi ~ is the ring whose elements are all ntuples
(‘I,’2″””n)’ with riERi, under componentwise addition and multiplication.
The direct sum is commutative iff each summand Ri is, and it has a
multiplicative identity iff each Ri has one.
7. Let R be the set of all realvalued functions defined on IR, under
addition and multiplication of functions. R is a commutative ring with unity.
An element fin R is a zerodivisor iff f(x) = 0 for at least one x E IR, and it is a
unit iff f(x) =1= 0 for all x E IR. Here, too, an element is a zerodivisor iff it is not
a unit.
Which elements of R are nilpotent?
8. Let MilR) denote the set of all 2 X 2 matrices with real entries, under
addition and multiplication of matrices. (Addition means adding correspond
ing entries.) Then M2(1R) is a noncommutative ring with unity. For example,
Section 16. Rings 157
the left distributive law requires that
and this is easily checked.
Note that there exist elements A, B in MilR) such that AB = (g g) but
BA *(g g). For instance, we can take
A =(: g) and B=(g 0) I .
This example explains why, in the definition of zerodivisor, we said” ab = 0
or ba=O.”
Note also that this ring contains nonzero nilpotent elements. For exam
ple,
0) o .
9. Let (G, + ) be any abelian group, and denote the identity element of G
by O. Define a multiplication on G by declaring a . b = 0 for all a, bEG. It is
then easy to check that (G, +, .) is a ring. We call this the ring on G with
trivial multiplication.
If, in particular, we start with G = {O}, the trivial group, then we get a ring
with one element. It is called the trivial ring, and is rather annoying. For
example, 0 satisfies the definition of a multiplicative identity element (O’x=
x·O=x for all x, right?), so 0 is 1. That should make you cringe, but there is
some comfort in noting that this anomaly can only occur in the trivial ring: If
R is a ring with unity and R has more than one element, then 0* I in R. In
order to prove this, we need some basic information.
We frequently denote multiplication in a ring by juxtaposition, writing ab
rather than a . b.
THEOREM 16.1 Let R be a ring, and let a,b be elements of R. Then:
i) a ·O=O·a=O;
ii) a( – b)=( – a)b= (ab);
iii) ( a)( – b)= ab;
iv) m(ab)=(ma)b=a(mb) for any integer m;
v) mn(ab) = (ma)(nb) for any integers m and n.
PROOF. i) To show that a ·0=0 it is enough to show that
a ·0 + a ·0 = a ·0,
158 Section 16. Rings
for then we can add (a ·0) to both sides. But
a·O + a ·O = a ·(0+0) = a ·0.
Similar reasoning works for O·a.
ii) To show that a(b)= (ab), it suffices to show that a(b)+ab=O.
But
a( – b)+ ab= a( – b+ b)= a ·0=0
by (i). A similar argument, using right distributivity instead of left, shows that
(a)b= (ab).
iii) Replacing a by – a in the first equality of (ii), we obtain
( a)( – b)= [ ( – a).]b= abo
iv) Exercise.
v) Exercise. 0
COROLLARY 16.1 Let R be a nontrivial ring with unity. Then 0+ 1 in R.
PROOF. Since R is nontrivial, we can pick a+O in R. Then if 0= 1 we get
a ·0= a . 1, that is, 0 = a, a contradiction. 0
COROLLARY 16.3 Let R be a ring with unity, and let u E R be a unit. Then u is
not a zerodivisor in R .
PROOF. We must show that if r is an element of R such that ur=O or ru=O,
then r=O. Now if ur=O, then
u1(ur) = u – 1(O)=0,
that is,
r=O.
A similar argument works if ru = O. 0
The next corollary is technical, but useful.
COROLLARY 16.4 If band c are elements of a ring R , define b – c to mean
b+(c). Then for any aER, we have
a(bc)=”abac, and (bc)a=baca.
PROOF.
a( b – c) = a( b + ( – c» = ab + a( – c) = ab + [ – (ac) ] = ab – ac.
Likewise for the second equality. 0
We mentioned above that one gets nicer and nicer rings by imposing
more and more assumptions on the multiplication. For example, we get rings
Section 16. Rings 159
that behave somewhat like (l, ED, 0) by adding the assumptions indicated in
the following
DEFINmON An integral domain (or just domain, for short) is a commutative
ring with unity in which 1 *0 and there are no nonzero zerodivisors.
Thus l, 0, and ~ are all domains. Another example is the ring in
troduced in Example 3 above, consisting of all real numbers of the form
a+bV2, with a,bEl.
The following simple observation yields an alternative characterization of
integral domains.
11IEOREM 16.5 Let R be a ring and let a, b, c E R. Assume that a is not a
zerodivisor. Then if ab=ac, we have b=c.
PROOF. From ab=ac we get abac=O, so a(bc)=O. Since a is not a
zerodivisor, this means b – c = 0, so b = c. D
COROLLARY 16.6 Let R be a commutative ring with unity 1 *0. Then R is an
integral domain iff whenever a,b,cER satisfy ab=ac and a*O, we have
b=c.
PROOF. Assume that R is a domain. Let a,b,cER, a*O, and suppose that
ab=ac. Then, since a is not a zerodivisor, the theorem implies that b=c.
Conversely, suppose that R is not an integral domain. We will find
a,b,cER, a*O, such that
ab=ac but b*c.
In fact, we know that there is a nonzero zerodivisor a E R. Let bE R be such
that b*O and ab=O. Then we have
a ·b=a ·0, but b*O. D
Finally, we consider rings with unity in which every nonzero element has
a multiplicative inverse.
DEFINmONS R is called a division ring if R has a unity 1 *0 and every
nonzero element of R is a unit. A commutative division ring is called a field.
Another way of saying that R is a division ring is to say that the set
R – {O} forms a group under multiplication. Saying that R is a field amounts
to saying that this group is abelian.
Familiar examples of fields include 0, ~, and the complex numbers C. A
less familiar example is (lp, ED, 0), where p is a prime number. lp is a field
because each r E {l, 2, … ,p l} satisfies (r,p) = I and is therefore a unit in lp
(see Exercise 16.9).
160 Section 16. Rings
Examples of division rings that are not fields are a little harder to come
by. In fact, a celebrated theorem of J. H. M. Wedderburn asserts that there
aren’t any finite examples: Every finite division ring is necessarily a field.
Proving this here would take us too far afield (so to speak), so we shall
content ourselves with a much simpler result. We shall see an infinite division
ring that is not a field in Section 17.
We obtain our easy substitute for Wedderburn’s Theorem by replacing
“division ring” by “integral domain.”
THEOREM 16.7 Every finite integral domain is a field.
PROOF. Let R be a finite domain. Then R is commutative, and 1 *0 in R. We
must show that if rER, r*O, then r has a multiplicative inverse in R.
Since R is finite we can list its elements as rl>r2, ••• ,’n’ Consider the
elements
Since r*O, these are all distinct by Corollary 16.6. Since they are all in Rand
R has only n elements altogether, they must account for all the elements in R.
In particular, one of them is 1, so
for some i, and r is a unit. 0
This proof will look familiar to you if you worked Exercise 3.15.
Of course, there exist infinite domains that are not fieldsoZ, for in
stance. On the other hand, every field, finite or infinite, is a domain, because
units are not zerodivisors. Thus the notions of “domain” and “field” coincide
for finite rings, but, in general, “field” is stronger.
EXERCISES
16.1 Let R be a ring with unity 1 R’ Show that ( – 1 R)a = – a for all a E R.
16.2 a) If r.s=2(r+s) and rDs=rs, is (IR,.,D) a ring?
b) If r.s=2rs and rDs=rs, is (IR{O},.,D) a ring?
c) If r.s=rs and rDs=rs, is (IR+,.,D) a ring?
16.3 Show that the set of all real numbers of the form a + b V2 , where a, bE Q,
forms a field under ordinary addition and multiplication.
16.4 Consider (0,., D), where. is the addition given by a. b = a + b 1, and ° is
the multiplication given by aDb = a + b – abo Is (0,., D) a field?
Section 16. Rings 161
16.5 (A construction of the complex numbers.) Let F be the set of all 2 x 2 matrices
of the form
(~ ~),
where a, bE IR. Show that F forms a field under addition and multiplication
of matrices.
Remarks. Note that if we think of
(~ ~)
as representing the complex number a + bi, then addition and multiplication
in F correspond to the usual operations on complex numbers. For example,
(~ ~)( ~ ~)=( :~=~~ :~t~),
which corresponds to (ac bd)+(ad+ bc)i. Thus this exercise shows you how
to construct a field having all the desired properties of C, starting only with IR.
16.6 Let Fbe a field. For a,bEF, b’l=O, define alb to mean ab I. Show that:
(.)(a)(c) ac ( .. )a c_(ad+bc) Ib’li = bd; 11 b+li bd
16.7 Let F be a field, let a,b E F, and assume a’l=O. Show that the equation
ax+b=O
can be solved for x in F; that is, there is x E F which makes the equation true.
16.8 Prove the following for (Zn’ Ell, 8):
a) associativity for 8;
b) the distributive laws.
16.9 a) Show that an element a E(Zn’ Ell, 8) is a unit iff (a,n)= 1.
b) Show that every element of Zn is either a unit or a zerodivisor.
c) Which elements of Zn are nilpotent?
16.10 Prove parts (iv) and (v) of Theorem 16.1.
16.11 Find all units, zerodivisors, and nilpotent elements in the following rings:
a) ZEllZ;
b) Z3EllZ3;
c) Z4 Ell Z6′
16.12 a) Show that the trivial ring is the only ring in which 0 is not a zerodivisor.
b) Show that in any ring except the trivial ring, every nilpotent element is a
zerodivisor.
16.13 a) Show that if R is a ring with unity, then the multiplicative identity element
in R is unique.
162 Section 16. Rings
b) Show that if R is a ring with unity and a E R is a unit, then the multiplicative
inverse of a is unique.
16.14 An element r in a ring R is called idempotent if r2 = r. Find all the idempotent
elements in the ring of realvalued functions on IR under addition and
multiplication of functions.
16.15 (See Exercise 16.14.) Let R be a nontrivial ring with unity. Let rER be
idempotent. Show that:
a) 1 – r is also idempotent, and
b) either r or 1 – r is a zerodivisor.
16.16 Let R be a ring with unity. R is called Boolean if every element of R is
idempotent. Show that if R is Boolean then:
a) 2r = 0 for every r E R (in other words, r = – r);
b) R is commutative.
[Hint jor (a): Consider (r + r)2.J
16.17 Let X be a set and let R be the set of all subsets of X.
a) Show that (R, Co. , n) is a commutative ring with unity, where Co. denotes the
operation of symmetric difference.
b) Show that (R, Co., n) is Boolean (see Exercise 16.16).
16.18 Let R be a ring with unity, and assume that R has no nonzero zerodivisors.
Let a, b E R, and assume that ab = 1. Show that ba = 1 too, and therefore a and
b are units.
16.19 Let R be a ring with unity, and let a E R. Assume that there is a unique bE R
such that ab = 1. Show that ba = 1, and therefore a is a unit.
16.20 a) Let S be the set of all realvalued functions on IR. Is (S, +, 0) a ring? (Here
o denotes composition of functions.)
b) Let R be the set of all realvalued functions on IR that are homomorphisms
of the additive group (IR, +). Is (R, +, 0) a ring?
16.21 Let G be the infinite direct product Z X Z X Z X Z X •.. , where there is one copy
of Z for each positive integer, and the operation on each copy is ordinary
addition. If ‘PI and ‘P2 are homomorphisms from G into itself, define ‘PI + ‘P2 by
(‘PI + ‘P2)(g) = ‘PI(g) + ‘P2(g)
for all gE G.
a) Show that ‘PI + ‘P2 is a homomorphism, and that the set of all homomor
phisms from G into itself forms a ring R with unity under the operations
+ and o.
b) Show that there exist elements ‘P and I/; in R such that ‘PI/; = 1 but I/;’P=F 1
(here 1 denotes the unity of R).
16.22 Let (R, +, . ) be a ring, and let S be a set. Let R S denote the set of all functions
from S to R. Show that R S forms a ring under addition and multiplication of
functions.
Section 16. Rings 163
16.23 Let R be a ring with unity, and let U denote the set of units in R. Show that
U is a group under the multiplication in R.
16.24 a) Let 1[ i] denote the set of all comple~ numbers of the form a + bi, where
a,b E1. Show that l[i] is a commutative ring with unity under ordinary
addition and multiplication of complex numbers. l[i] is called the ring of
Gaussian integers.
b) For r= a + bi E l[i], define the norm N(r) of r by N(r) = a2 + b2• Show that
if r,sE1[i], then N(rs)=N(r)N(s).
c) Show that r=a+bi is a unit in l[i] iff N(r) = l. Using this information, find
all the units in 1[ i].
d) (See Exercise 16.23.) The group of units of l[i] is isomorphic to a familiar
group. Which one?
16.25 Let R denote the ring of all real numbers of the form a + b VI , where a, bEl.
For r=a+bVI ER, define N(r) by N(r)=a2 2b2•
a) Show that if r,sER then N(rs)=N(r)N(s).
b) Show that r is a unit in R iff N(r) = ± l.
c) Show that there are infinitely many units in R.
16.26 Give an example of a finite noncommutative ring.
16.27 Give an example of a noncommutative ring with no multiplicative identity.
16.28 Let R be an integral domain. If there exists a positive integer n such that
n . I = 0, then the smallest such integer is called the characteristic of R. If no
such n exists, then we say that R has characteristic O.
a) Show that if R has characteristic n, then n ·r=O for every rER.
b) Show that if R has characteristic n > 0, then n is a prime number.
c) For each prime number p, give an example of a field of characteristic p. Give
an example of a field of characteristic O.
16.29 Must every ring with a prime number of elements be commutative? Either prove
that it must, or give a counterexample.
16.30 Let R be a finite nontrivial ring with no nonzero zerodivisors. Show that R
is a division ring.
16.31 Prove that if F is a finite field then there exist a prime number p and a positive integer
j such that IFI = pi. [Suggestion: Use the result of ExerC:lse 16.28(b).]
16.32 a) Suppose R is a ring with unity and for all x andy inR we have (xyi =~y. Prove
that R is commutative.
b) Give an example to show that the result of part (a) may fail if R does not have a
unity.
SECTION 17
SUBRINGS, IDEALS,
AND QUOTIENT RINGS
In this section we shall develop analogues, for rings, of some of the concepts
we encountered in dealing with groups. As for groups, the purpose of doing
this is to develop ways of talking about the internal structure of a given ring
and the relationships between different rings.
We begin with the analogue of “subgroup.”
DEFINmON Let (R, +, . ) be a ring. A subset S of R is called a subring of R if
the elements of S form a ring under + and ‘.
In particular, the definition requires that (S, +) be a subgroup of (R, +).
Thus if S is a subring of R, then we know that the additive identity element 0
of R is in S, and that S is closed under addition and under additive inverses.
The relationship between Rand S with respect to multiplication need not be
so clean. For example, if R has a multiplicative identity 1, then 1 need not be
in S, and it is even possible that S may have an identity element different
from that of R.
Examples
1. Let (l., +, .) be the integers under ordinary addition and multiplica
tion, and let 2l. be the set of even integers. Then it is easy to see that 2l. is a
subring of l.. Although l. is a ring with unity, 2l. has no unity.
2. Consider the ring REB R. Let S denote the set of all pairs of the form
(r,O), where rER Then S is a subring of REBR, and REBR has unity (1,1),
164
Section 17. Subrings, Ideals, and Quotient Rings 165
which is not in S. In this case, S has its own unity, namely (1,0). Note that
(1,0) is not a unity for IREBIR, but only for S.
As for subgroups, the definition of “subring” can be recast III more
compact form.
THEOREM 17.1 Let (R, +, .) be a ring, and let S be a subset of R. Then S is a
subring of R iff the following two conditions are satisfied:
i) (S, +) is a subgroup of (R, +); and
ii) S is closed under multiplication, that is, if rpr2ES then r(r2ES.
PROOF. It is clear that if S is a subring of R, then (i) and (ii) hold. Conversely,
assume that (i) and (ii) hold for S. Then + and . are both binary operations
on S, and (S, +) is an abelian group, so we need only check that . is
associative on S and that the distributive laws hold in (S, +, .). But associa
tivity and distributivity hold for all elements of R, hence for those of S. 0
Condition (i) can be reduced further by using your favorite subgroup
criteria. For example, by using the result of Exercise 5.24 we obtain the
following.
COROLLARY 17.2 Let (R, +,.) be a ring and let S be a nonempty subset of R.
Then S is a subring of R iff the following two conditions hold:
i) for every r(,r2 E S, we have r( – r 2 E S;
ii) for every r(,r2 ES, we have r(r2 ES.
Examples (continued)
3. Let’s find all the subrings of (Z, +, .). We know that any subring must
be a subgroup of (Z, +) and hence must be additively a cyclic subgroup of
the form mZ, for some m. We have only to figure out which of these
subgroups constitute subrings. The extra requirement that mZ must satisfy to
be a subring is closure under multiplication. But clearly if i, j are integers,
then (im)(jm) = (ijm)m E mZ. Thus every subgroup of (Z, + ) is a subring of
(Z, +, .).
Similar reasoning shows that every subgroup of (Zn’ EB) is a subring of
(In’ EB, 0).
4. Let R be the set of all realvalued functions defined on IR under
pointwise addition and multiplication of functions. Let S be the subset of R
consisting of all the continuous functions. Then S is a subring of R, for if j,g
are continuous functions so is j g and so is jg. (We have used Corollary
17.2.)
166 Section 17. Subrings, Ideals, and Quotient Rings
5. Let R be as in the previous example and let
S={jERIJ(O)=O},
so that S consists of those functions that take the value 0 at x = O. Then S is a
subring of R, for if J,g E S then J(O) = 0 and g(O) = 0, so
(1 g)(O) = J(O) – g(O) =0,
which shows that J – g E S, and
Jg(O) = J(O)g(O) = 0,
which shows that Jg E S.
Observe that if T= {f E R IJ(O) = I}, then T is not a subring of R, because
if J(O) = g(O) = 1 then (1 g)(O) = 11 =0.
6. Let MilR) be the ring of all 2 X 2 matrices with real entries. Let S
consist of all matrices of the form (~ ~). Then S is a subring of MilR), for if
(a b) and (e f) are in S then so is (a b)_(e f) and so is (a b)(e f).
Od Oh ‘ Od Oh’ OdOh
On the other hand, if T consists of all matrices of the form (: ~), then T
is not closed under multiplication, so T is not a subring of MilR).
7. Let M 2(C) be the ring of all 2 X 2 matrices with comp~ex entries. Let
1=(6 0) J=(i 1 ‘ 0
0) K= ( 0
– i ‘ 1 6)’ and L=JK=(~ i) O’
Let IHI be the following subset of MiC):
IHI = {a1 + bJ + cK + dLla,b,c,d EIR}.
(Note: The product of a constant and a matrix is obtained by multiplying
each entry of the matrix by the constant.) IHI is a subring of M 2(C), for it is
clear that IHI is an additive subgroup, and closure of IHI under multiplication
follows from distributivity in MiC) and the fact that {± 1, ±J, ± K, ± L} is
closed under multiplication. [{ ± 1, ± J, ± K, ± L}, under multiplication, is
Q8′]
IHI is called the ring of quatemions, and “IHI” is for Hamilton, the man who
discovered this ring. IHI is a noncommutative ring, for the same reason that Qs
is a nonabelian group: JK:I=KJ. IHI has an identity 1*(g g), and it is easy to
see that H is in fact a division ring. For if
a1+bJ+cK+dL=( a+bi
c+di
C+di)EIHI
abi
Section 17. Subrings, Ideals, and Quotient Rings 167
and at least one of a,b,c,d is not zero, we have
( a+bi
c+di
c + di ) – I I ( a – bi – c – di )
abi = a2 +b2 +c2 +d2 cdi a+bi
I
—– (aI – bJ – cK – dL) E IHI.
a 2 + b2 + c2 + d 2
IHI is the example of a noncommutative division ring that we promised you in
Section 16.
Hamilton discovered the quaternions in 1843, after he had spent ten or
fifteen years seeking a generalization of C that could be used in connection
with geometric and physical problems in 3space. One reason why it took him
so long was that he started out looking for a commutative generalization;
coming up with a noncommutative one was, at the time, a revolutionary step.
The definition of IHI via MiC) was not possible until 1858, when Cayley
introduced matrices. Hamilton thought of IHI while taking a stroll on the
evening of October 16, 1843; it occurred to him as a set of elements of the
form
a+ bi+ cj+ dk,
where a,b,c,dEIR and
i2 =f=k2 =ijk= I.
Although he made other distinguished contributions to science, Hamilton
considered the discovery of the quaternions to be the crowning achievement
of his life. He spent twenty years studying them, and wrote several huge
volumes about them.
In dealing with groups, we found that some subgroups were better than
others. For example, in attempting to construct the quotient group modulo a
subgroup H, we saw that it was crucial for H to be normal. We encounter a
similar situation when we try to construct quotient rings.
Let (R, +,.) be a ring and let S be a subring of R. We know that (S, +)
is a subgroup of (R, + ), and in fact there is no problem about (S, + ) being a
normal subgroup of (R, +), because (R, +) is abelian. Thus we already know
how to construct a quotient group (R/ S, +), the elements of which are the
co sets of Sin R, with addition defined by
(S+ a)+(S+b)= S+(a+b).
We would like to endow this quotient group with a multiplication, arising
naturally from the given multiplication in R, in such a way that the quotient
becomes a ring. As usual with these things, there is only one reasonable
168 Section 17. Subrings, Ideals, and Quotient Rings
attempt; we want to define
(S+a)(S+b)=S+ab.
What we have to check is that this is a welldefined operation, that is, if
S+a= S+a’ and S+b= S+b’,
then
S + ab = S + a’ b’,
so that the product doesn’t depend on which representatives we use to define
it.
So we assume a – a’ E Sand b – b’ E S, and we wish to show that
ab – a’ b’ E S. It is clear that if this is to work for all possible choices of
a,a’,b,b’, then S has to be rather special. For instance, if we take aES, a’=O,
b arbitrary, and b’ = b, then
aa’ES and bb’ES,
so we want
abObES, that is, abES.
In other words, we require that if a E Sand b is any element of R, then
ab E S. Similar reasoning shows that we also require that if bE S and a E R is
arbitrary, then ab E S.
Now we claim that these two conditions are enough to make our
multiplication work out. For suppose S satisfies both conditions. Assume
aa’ES and bb’ES. Then
ab a’b’ =(a a’)b+ a'(b b’).
By the conditions on S, (a – a’)b E Sand a'( b – b’) E S, so
(a a’)b+ a'(b b’) E S
since S is an additive subgroup. Thus ab – a’ b’ E S, and our multiplication is
well defined.
Subrings that have the special properties required to make multiplication
of the additive cosets well defined are called ideals.
DEFINl110N A subring S of a ring R is called an ideal of R if for every s E S
and rER we have rsES and srES.
We have seen that if S is a subring of R, then the natUral attempt at
introducing a multiplication on R/ S will succeed iff S is an ideal. It is now
easy to complete the proof of
Section 17. Subrings, Ideals, and Quotient Rings 169
THEOREM 17.3 Let (R, +, .) be a ring, and let S be an ideal of R. Then the set
R/ S of cosets of the additive subgroup (S, +) is a ring under the operations
(S+ a)+(S+ b)= S+(a+b),
(S+a)(S+ b)= S+ab.
PROOF. We know that R/ S is a group under the indicated addition, and this
group is abelian because (R, +) is. We have seen above that the indicated
multiplication yields a binary operation on R/ S, so all that we have to check
is that this multiplication is associative and that the distributive laws hold.
Associativity requires that
[(S+ a)(S+ b) J(S+ e) =(S+ a)[ (S+ b)(S+ e)]
for all a,b,e E R. This amounts to
(S+ab)(S+e)=(S+a)(S+be), that is,
S+ (ab)e = S+ a(be),
which is true by associativity of multiplication in R. Similarly, R/ S inherits
distributivity from R; we leave the details of this to the reader. 0
R/ S is called the quotient ring (or factor ring) of R by S.
There is a certain redundancy in our definition of “ideal,” in that the
condition that rs E Sand sr E S for every r E Rand s E S already implies part
of the condition that S be a sUbring. The next result provides a neater
characterization of ideals.
THEOREM 17.4 Let R be a ring and let S be a subset of R. Then S is an ideal
of R iff the following two conditions hold:
i) S is an additive subgroup of R (equivalently, S is nonempty and closed under
subtraction);
ii) For every r E Rand s E S, we have rs E Sand sr E S.
PROOF. Exercise.
We will usually denote an ideal by I, rather than S, from now on.
Examples (continued)
8. As we have seen, the subrings of (l, +, .) are precisely the additive
subgroups ml. In this case [likewise for (In’ EB, 0)], every subring is also an
ideal, because if we multiply an element of ml by any integer we get an
element of ml.
9. l is a subring of (Q, +, .), but not an ideal. For example, 1 Eland
4 E Q, but 1· 4 EtC l.
170 Section 17. Subrings, Ideals, and Quotient Rings
Actually, there is something more general going on in this example. If R
is a ring with unity 1, then the only ideal of R that contains 1 is R itself.
(Prove it!)
Incidentally, R will always be an ideal of R, no matter what ring R is. We
will call R the improper ideal; all other ideals are called proper. The ideal {O}
is called trivial.
10. Refer back to Examples 4 and 5, which present two examples of
subrings of the ring R of all realvalued functions on R. The subring S in
Example 4 is not an ideal, because if j E Sand g E R thenjg need not be in S,
since it need not be continuous. (Can you give an example?) The subring S in
Example 5 is an ideal, however, because if j(O) = 0 and g is any element of R,
then
jg(O) = j(O)g(O) =O·g(O) =0,
sojgES. Likewise gfES.
11. Let I be the subring of REBR consisting of all pairs of the form (r,O).
Then I is an ideal, because (r,O)(a,b)=(ra,O)EI for any r,a,bER, and
likewise (a,b)(r,O)EI. Note that I has a unity, and yet I is a proper ideal of
R. Why does this not contradict our observation in Example 9 about ideals
that contain the unity?
12. Let R be a commutative ring with unity, and let a E R. Let
aR={arlrER},
so that aR is the set of all multiples of a in R. Then aR is an ideal of R. First
of all, if ar1,ar2EaR then arlar2=a(rlr~EaR, so aR is an additive
subgroup of R. Secondly, if ar E aR and t E R, then
t(ar) = (ar)t = a(rt) EaR,
since R is commutative, and this verifies that aR is an ideal.
We call aR the principal ideal generated by a. Note that a EaR, since
a = a . 1. Also observe that aR.= Ra = {ralr E R}.
13. Let R = (Z, + , .) and let I be the ideal nZ for some positive integer n.
Then (R / I, +, .) has n elements, namely, 1+ 0,1 + 1,1 + 2, … ,1 + (n – 1). We
saw in Section 13 that (R/ I, +) is isomorphic, as a group, to (ZII’ $). In
Section 18 we shall introduce a notion of isomorphism for rings, and it will
tum out that (Z/ nZ, +, .) is isomorphic, as a ring, to (ZII’ $, 0).
Suppose now that R is a ring and I is an ideal of R. Then R/ I is a ring,
and we can sensibly (and profitably) ask which properties in R translate into
familiar properties for the elements of R/ I. For example, if a E R, when is
I + a a zerodivisor in R / I?
Section 17. Subrings, Ideals, and Quotient Rings 171
I + a is a zerodivisor iff there is some 1+ b ‘f= 1+ 0 such that either
(I+a)(I+b)=I+O or (I+b)(I+a)=I+O. This boils down to there being
some b f1. I such that either ab E I or ba E I. In particular, I + a is a nontrivial
(i.e., nonzero) zerodivisor in R/ I iff a f1. I and there is some b f1. I such that
abEl or baEI.
From this it is clear what conditions we need on I to rule out nontrivial
zerodivisors in R/ I.
DEFINmON Let R be a ring, I an ideal in R. Then I is prime if whenever
a, bE Rand ab E I, then at least one of a or b is in I.
Example Let p be a prime in 71.. Then p7L is a prime ideal, because if ab is
divisible by p, then one of a or b must be divisible by p.
TIlEOREM 17.5 R/ I has no nontrivial zerodivisors iff I is prime.
The proof is immediate from the above discussion. Specializing to the
case where R is a commutative ring with unity, we get:
COROLLARY 17.6 Let R be a commutative ring with unity, I an ideal in R.
Then R/ I is an integral domain iff I is a proper prime ideal.
PROOF. R/ I is a commutative ring with unity (see Exercise 17.14). Thus R/ I
is an integral domain iff it is nontrivial and has no nontrivial zerodivisors,
that is, iff I is proper and prime. 0
When is R/ I a field?
DEFINmON. An ideal I of R is called maximal if I is proper and there is no
proper ideal J 7. I.
Thus I is maximal iff it is proper and cannot be extended to a larger
proper ideal.
TIlEOREM 17.7 Let R be a commutative ring with unity. If I is an ideal in R,
then R / I is a field iff I is maximal.
PROOF. R/ I is a commutative ring with unity. Thus it is a field iff it is
nontrivial and each of its nonzero elements is a unit.
Now R/ I is nontrivial iff I is proper. And every nonzero element of R/ I
is a unit ~ for every a E R – I, there is bE R such that (I + a)(I + b) = 1+1,
in other words,
for every a f1. I, there is b such that ab – I E I. [17.1]
Thus R/ I is a field iff I is proper and [17.1] holds. To conclude the proof, we
will show that a proper ideal I is maximal iff [17.1] holds.
172 Section 17. Subrings, Ideals, and Quotient Rings
First suppose that I is maximal, and take a f/. I. Then the set
l={ar+xlrER andxEI}
is an ideal that properly includes I, hence IS not proper. Therefore, aro+ xo= 1
for some ro and x o’ and this yields
arol = xoE/,
which means that [17.1] holds. Conversely, if [17.1] holds, then let I’ be an
ideal such that I’ ~ I, with the aim of showing that I’ = R. Take a E I’ – I,
and take b such that ab – 1= y E I. Then a E I’ andy E I’, so ab – y E I’. Thus
IEI’, so I’ = R, and I is maximal. 0
COROLLARY 17.8 Let R be a commutative ring with unity. Then every maxi
mal ideal of R is prime.
PROOF. If I is maximal, then Rj I is a field, hence an integral domain. Thus I
is prime by Corollary 17.6. 0
This last result can fail when we try to weaken the assumptions. For
instance, let R be the ring on (Z2′ EEl) with trivial multiplication. Then R is a
commutative ring without unity, and {O} is a maximal ideal which is not
prime.
On the other hand, prime ideals need not be maximal, even if R is a
commutative ring with unity. According to our definitions, in fact, any ring is
a prime ideal of itself which is not maximal. If this seems like cheating, note
that {O} is a prime ideal in Z which is not maximal. And for an example of a
nontrivial proper prime ideal which is not maximal, take the subset [of zez
consisting of all pairs (a,O). Note that
{(a,2b)la,bEZ}
is a proper ideal which is strictly larger than [.
Some conditions under which proper prime ideals are maximal are
indicated in Exercises 17.2917.31.
Here are a few more examples of maximal ideals.
Examples
1. We have observed that every ideal in Z has the form nl for some n,
and that if n is a prime p, then pI is a prime ideal. If n > 2 is not prime, then
nl is clearly not a prime ideal, so the only other prime ideals in I are {O}
and I.
Section 17. Subrings, Ideals, and Quotient Rings 173
What are the maximal ideals? Since I is a commutative ring with unity,
any maximal ideal must be prime; and since {o} and I are obviously not
maximal, the only possible maximal ideals are of the form pl. In fact every
p I is maximal. For suppose J ;;. p I and J is an ideal. If x E J – p I then
(x,p) = I, so there are a,b such that ax+ bp = 1. Since ax EJ and bp EJ, I EJ,
so J=l.
Thus for I the maximal ideals are the nontrivial proper prime ideals.
2. The ideal I={(a ,2b)la,bEI} is maximal in lEal. For if J is an ideal
properly including I, we have (m ,2n + I)EJ for some m,n EZ. Since (m,2n)
EI, subtracting shows that (0, I)EJ. Also, (1,O)EI r:;;;,J. Adding, we see that
(1, I)EJ, so J=lfBI .
3. Let us return to Example 5 above. R is the ring of all realvalued
functions on JR, under pointwise addition and multiplication, and I is the
ideal consisting of all 1 such that 1(0) = 0. We want to show that I is maximal.
Suppose J is an ideal that is strictly larger than I, and take g E J – I. Then
g(O)*O, so there is hER such that
I
h(O) = g(O) .
We have g(O)h(O) = I, so if 11 denotes the multiplicative identity in R, the
value of the function gh – 11 at ° is 0. Hence
gh11 EI r:;;;,J,
so since gh EJ, 11 EJ. Thus J = R, and I is maximal.
The same argument demonstrates that for any r E IR, I, = U E R 11(r) = O}
is a maximal ideal in R.
4. Let S be the subring of MiR) introduced in Example 6 above, that is,
S={(~ !)la,b,dER}. We contend that I={(~ ~)la,bER} is a maximal
ideal of S. First of all, it is clear that I is closed under subtraction in S, and if
(~ ~)ES and (~ ~)EI, we have
(~ {)( ~
(~ ~)(~
~)=( ~
{)=( ~
~)EI and
a1+bh)EI
° ‘
so I is an ideal. To see that I is maximal, let J be an ideal of S that properly
includes I. Then we have (~ !) EJ for some a,b,d, with d=foO, and therefore
(~ l~d)(~ !)=(~ nEJ.
17 4 Section 17. Subrings, Ideals, and Quotient Rings
Since (~ ~) E /, we get
so J = S, as desired.
EXERCISES
17.1 Which of the following subsets of M2(1R) are subrings?
a) S=all matrices of the form (~ !)
b) S = all matrices of the form (: ~)
c) S = GL(2, IR)
d) S = all matrices of the form (: !).
17.2 Let R be the ring of realvalued functions on the real line, under pointwise
operations. Which of the following subsets S of Rare subrings? Which are
ideals?
a) S={jERlf(l)=O}
b) S= {jE R If(1)=O or f(2)=0}
c) S={jERlf(3)=f(4)}.
17.3 Let R be the ring with trivial multiplication on some abelian group G. What
are the ideals of R?
17.4 Let S, T be subrings of R. Under what conditions is S uTa subring of R?
17.5 Let R be a finite ring with, say, n elements. Let S be a subring of R that has
m elements. Show that m divides n.
17.6 Find a maximal ideal in
17.7 Find all the maximal ideals in lln.
17.8 Let R =2ll be the ring of even integers, under ordinary addition and
multiplication, and consider the subset S=4ll. Show that S is an ideal of R.
Is S maximal? Is it prime?
17.9 Let R={qEQlq=ajb, a,bEll and b is odd}. Show that R has a unique
maximal ideal.
17.10 Let X be a nonempty set and let R be the ring (P(X), 6, n).
a) Show that if Y s: X, then P(Y) is an ideal in R and has a unity different from
that of R.
b) Find a maximal ideal in R.
Section 17. Subrings, Ideals, and Quotient Rings 175
17.11 Prove Theorem 17.4.
17.12 Let R be a ring with unity. Show that a nonempty subset S of R is an ideal iff
the following two conditions are satisfied:
i) SI +S2ES for every Sl>s2ES;
ii) rs and sr are in S for every sE S, rER.
17.13 Show that if I is an ideal of R, then the distributive laws hold in Rj I.
17.14 Let R be a ring and I an ideal of R. Show that
a) if R is commutative, so is Rj I;
b) if R has a unity, so does Rj I.
17.15 Show that if S is a subring of R and I is an ideal such that I !;;;; S, then S j I is
a subring of Rj I. Show that if S is an ideal of R, then Sj I is an ideal of Rj I.
17.16 Show that M 2(R) has no ideals other than the trivial ideal and the improper
ideal.
17.17 Let S be the ring of all matrices of the form (~ :), with a,b,dER. Find a
maximal ideal of S other than the one found in the text, and show that every
proper ideal of S is contained in one of these two maximal ideals.
17.18 Let I(i] be the ring of Gaussian integers (see Exercise 16.24). Let I be the
principal ideal generated by the element 2 + 2i. How many elements are there
in I[i]j I?
17.19 Let R be a commutative ring with unity I ~O.
a) Show that R is a domain iff {OJ is a prime ideal in R.
b) Show that R is a field iff {OJ is a maximal ideal in R.
17.20 Let R be a commutative ring with unity, and let a E R. Show that aR = Riff
a is a unit.
17.21 a) Let R be a ring. Define the center of R to consist of all r E R such that rx = xr
for every x E R. Show that the center of R is a subring of R.
b)Must the center of a ring be an ideal?
17.22 a) Let R be a commutative ring and X a subset of R. Define
Ann(X) = {rERlrx=O for every xEX}.
Ann(X) is called the annihilator of X. Show that Ann(X) is an ideal.
b) In (112′ EB, 0), find Ann({2}).
c) If R were not commutative, then the set Ann(X) defined above would be
called the left annihilator of X. Show that in this case, if X is itself an ideal,
then Ann(X) is still an ideal.
17.23 a) Let R I,R2, … ,Rn be rings with unity. Show that every ideal of
RI EBR2 EB’ .. EBRn is of the form II EBI2 EB· .. EBim where I; is an ideal of
R; for each i.
176 Section 17. Subrings, Ideals, and Quotient Rings
b) Show that this result may fail if the R;’s are not rings with unity.
17.24 Let R b R2, ••• ,Rn be rings with unity. What do the maximal ideals of
Rl EBR2 EB· .. EBRn look like?
17.15 a) Let R be a ring and let I andJ be ideals of R. Show that I nJ is an ideal
of R.
b) Suppose that I and J are prime. Must I nJ be prime?
17.16 a) Give an example of a ring R, an ideal I of R, and an ideal J of I such that
J is not an ideal of R.
b) Show that J must be an ideal of R if it is a prime ideal of l.
17.27 Let R be a commutative ring.
a) Show that the set of nilpotent elements in R forms an ideal.
b) . Show that the quotient of R by this ideal has no nonzero nilpotent elements.
17.28 Let R be a commutative ring, and let I be an ideal of R. Define the radical, VI ,
of I to consist of all r E R such that some power of r is in l.
a) Show that VI is an ideal of R.
b) I is called semiprime if I = VI . Show that I is semiprime iff R/ I has no
nontrivial nilpotent elements.
17.29 Let R be a finite commutative ring with unity. Show that every proper prime
ideal of R is maximal.
17.30 Let R be an integral domain. We call R a principal ideal domain (PID) if every
ideal of R is of the form aR for some a E R. Show that in a PID every nontrivial
proper prime ideal is maximal.
17.31 Let R be a Boolean ring (see Exercise 16.16). Show that every proper prime
ideal of R is maximal.
17.32 Prove that every nontrivial finite subring of a division ring is a division ring.
17.33 a) Let I and J be ideals of R. Define their sum 1+ J by
I+J= {x+ylxEI,YEJ}.
Show that 1+ J is an ideal.
b) Find 6l+ 14l in (l, +, .).
17.34 Let I and J be ideals of R. Define their product IJ to be the set of all finite sums
of the form XIYl + … +xnYn, where each x;EI and eachy;EJ. Show that IJ
is an ideal, and that IJ ~ In J.
17.35 Let I, J, K be ideals of R. Assume that IJ ~ K and K is prime. Show that at
least one of I or J is contained in K.
17.36 Let R be a commutative ring with unity, and let I and J be ideals of R such
that I+J=R. Show that IJ=InJ.
17.37 Let I, J, K be ideals of R. Show that I(J+K)=IJ+IK.
SECTION 18
RING HOMOMORPHISMS
We have seen that group homomorphisms enable us to relate different groups
to each other. Ring homomorphisms do the same thing for rings.
In the context of rings, a homomorphism must be “sensible” with respect
to both operations:
DEFINITION Let Rand S be rings, and let cp: R~S be a function. Then cp is
called a (ring) homomorphism if for every a, b E R we have
i) cp(a+ b)=cp(a)+cp(b) and
ii) cp(ab)=cp(a)cp(b).
Thus a ring homomorphism is in particular a group homomorphism from
(R, +) into (S, +). As such, it has many familiar properties. For instance,
CP(OR)=OS’ and cp(na)=ncp(a) for every aER and nEZ.
A ring homomorphism cp must also preserve products, but in general this
doesn’t have as many consequences as the corresponding fact for sums,
because (R – {OJ, .) and (S – {OJ, .) need not be groups. It is entirely possi
ble, for instance, that Rand S are both rings with unity, and yet cp(l R) * Is.
Because of this, it is possible that u E R is a unit but cp( u) E S is not.
Examples Let cp: R~R be given by cp(r) = 0 for all r E R. Then cp is a ring
homomorphism, cp(IR)#’ IR’ and cp(r) is never a unit, although every r#’O is a
unit in R.
For a slightly less trivial example, let
cp: R~REBR
be given by cp(r)=(r,O) for every rER. Clearly cp is a ring homomorphism,
but cp(lR)=(l,O), which is not the multiplicative identity in REBR. Again cp
maps all the units in R to nonunits.
177
178 Section 18. Ring Homomorphisms
Many times people who are working solely in the context of rings with
unity tighten up their definition of homomorphism so as to insist that
‘P(lR) = Is· This is nice in that it avoids maps like the ‘P’s in our examples, but
in our general context our definition is probably better, and we will stick to it.
We record some basic properties of ring homomorphisms for reference.
11fEOREM 18.1 Let
remainder of x (mod n). Then cp is an onto homomorphism, and ker(cp)=nZ,
so
(Z, +,. )/nZ~(Zn’ $, 0).
The distinct subrings of (Zn’ $, 0) are (dl), … ,(dk ), where dl, … ,dk are
the positive divisors of n (Corollary 5.6). The distinct subrings of (Z, +, .)
that contain nZ are dI Z,d2Z, … ,dk Z, so the correspondence between the
subrings is very transparent in this case.
182 Section 18. Ring Homomorphisms
2. Let cP: lLffilL~lL be given by cp[(m,n)] = n. Then cP is onto and ker(cp) is
the ideal {(m,O)lm ElL). We have
(lLffilL) /ker( cp) ~ lL.
The ideals of lLffilL that contain ker(cp) correspond in an obvious way to the
ideals of lL.
3. Let R be the ring of realvalued functions on Ill, under pointwise
operations. Fix some real number r in mind, and define a mapping CPr: R~1Il
by
CPr(J)=f(r).
CPr is an onto homomorphism, and ker(cpr)=Ir={jERlf(r)=O}. Since
R/ker( CPr) ~ III
and R is a field, Theorem 17.7 tells us that Ir is maximal in R for every r. We
verified this by more heavyhanded means in Example 3 on p. 173.
The proofs of the second and third isomorphism theorems offer no
surprises, so we leave them as easy exercises.
THEOREM 18.6 (Second isomorphism theorem for rings) Let S be a suoring of
R and let I be an ideal. Then S n I is an ideal of S, and
S/(S n 1)~(S+ 1)/ I.
Here S+I is the subring {s+xlsES, xEI} of R.
THEOREM 18.7 (Third isomorphism theorem for rings) Let I and J be ideals of
R and suppose I r:;,J. Then J / I is an ideal of R/ I and
R/I
J/I~R/J.
With these theorems stated, we have now remodeled all our basic
machinery so that it is suitable for use in ring theory. We will proceed to
some new results, each of which involves ring homomorphisms.
We first want to show that every field F has a subfield that is isomorphic
either to Q or to (lLp, ffi, 0) for some prime p. In fact, more is true: F has
exactly one such subfield. We are interested in this result because it provides
a useful classification of fields. The nature of the solutions of equations
involving elements of F can be heavily influenced by whether it is Q, on the
one hand, or some lLp, on the other, which is isomorphic to a subfield of F.
THEOREM 18.8 Let F be a field. Then the intersection of all the sub fields of F is
itself a sub field of F, and is isomorphic either to Ql or to Zp for some prime p. This
is the only subfield of F that is isomorphic either to Ql or to some Zp.
Section 18. Ring Homomorphisms 183
PROOF. Let K be the intersection of all the subfields of F. As in Exercise 5.14 we
see that K is a subfield of F. K is contained in every subfield of F, and the
multiplicative identity element e of Fis in K. (We use “e” instead of “1” or “II’
for clarity in what follows.) We consider two cases depending on the order o(e)
of e in the group (F, +).
Case 1: If o(e) is infinite then S = {me(netl I m, nEZ, n:f:. O} !;; K. It is easy to
verify that S is a subfield of F, so we have K !;;; S and therefore K = S. The
mapping rp: Q +K given by rp (mIn) = me (netl is a welldefined isomorphism.
Case 2: If o(e) = n E Z+ then it follows from Exercise 16.28 that n is a prime
number p. If S = {OF> e, 2e, … , (p – 1) e} then S!;;; K. S is a subring of F isomorphic
to the field Zp , so S is a subfield of F and thus K!;;; S. So K = S and K == Zp.
To prove the uniqueness assertion, suppose L is a subfield of F and If/: L + Q
is an isomorphism. Then If/(e) = 1, so since Q = {m I n I m, nEZ, n:f:. O},
L = {me (netll m, nEZ, n:f:. O} = K
(because o(e) is infinite in (F, +». Likewise, if L == Zp for some prime p then L =
{OF> e, 2e, … , (Pl) e} = K 0
The subfield of F that is isomorphic to Q or some l.p is called the prime
subfield of F. If F is itself either Q or some l.p’ then F is its own prime
sub field, and Q and the l./s are accordingly called prime fields.
If the prime subfield of F is isomorphic to Q, equivalently if 1 F has
infinite order in (F, +), we say that F is of characteristic O. Thus Q, R, and C
are all of characteristic O. If the prime subfield of F is isomorphic to l.p’ we
say that F is of characteristic p. It is clear, for example, that every finite field
must be of characteristic p for some prime p.
Our next two results are examples of what are called embedding theorems,
in that they state that a ring of some kind can be embedded in a ring of some
other kind. To say that R can be embedded into S is simply to say that there
is an embedding (monomorphism) from R into S, that is, R is isomorphic to a
subring of S. The usual purpose of an embedding theorem is to remedy some
defect of a given ring by embedding the ring in a bigger ring, where things are
better. For example, we know that some rings lack multiplicative identities.
But:
TIIEOREM 18.9 Let R be a ring. Then R can be embedded in a ring with unity.
PROOF. Let S={(r,n)lrER, nEl.}, with operations defined by
(r,n) +(s,m) =(r+s,n+ m)
and
(r,n)(s,m)= (rs+ ns+ mr,nm).
184 Section 18. Ring Homomorphisms
It is clear that S forms an abelian group under addition, and a routine
calculation shows that multiplication is associative and the distributive laws
hold. Thus S is a ring.
We map R~S by
x) to mean thaty – x E F+. Show that, for all x,y,z E F:
190 Section 18. Ring Homomorphisms
a) exactly one of the following holds:
x
b) if x
d) if x
SECTION 19
POL YNOMIALS
We have all been familiar with polynomials since our days in high school
algebra. We are by now accustomed to thinking of them as functions of the
formf(x)=ao+a\x+'” +anxn, with x being a variable and the a;’s being
real constants. In this section we will take another look at them, from the
point of view of abstract algebra.
We ‘Shall denote variables by uppercase letters: X, Y, … . If R is a ring,
then by a polynomial in X with coefficients from R we mean an infinite formal
symbol
Qo+a\X+a2X 2 +a3X 3 + … ,
where each ajER and there is some n such that aj=O for all i>n. The a/s are
called the coefficients of the polynomial, and aj is called the coefficient of xj.
If an*O and aj=O for all i>n, then we usually write the above polynomial
more simply as
but we still regard it as having a coefficient aj for every j > O. This approach is
often very convenient. For instance, it makes it easy for us to say what we
mean by two polynomials being equal. If
and
are polynomials with coefficients from R, then we say that they are equal, and
we write f(X) = g(X), if aj = hj for every i. Notice that we don’t have to worry
about one of f(X) or g(X) having more coefficients than the other (for
instance, extra Ocoefficients) because we regard them both as having a
coefficient for each power of X.
191
192 Section 19. Polynomials
Observe that a polynomial f(X) with coefficients from R gives us a
function on R, obtained by plugging elements of R in for X and interpreting
addition and multiplication as the operations in R. However, our point of
view at the moment is that f(X) is a formal expression, and not the function
this expression induces on R. This distinction is a significant one, because it is
possible for two different polynomials to induce the same function on R. For
example, if R is (.l2,E9,O), then both 0+OX+OX 2+”. and 0+lX+IX2
give us the function that maps every element of R to O.
In denoting polynomials, we usually omit terms with coefficient 0,
wherever they occur. For instance, ao+OX + a2X2 can be written as ao+
a2X 2• On the other hand, if R is a ring with unity, then we usually do not
bother to write I R when it occurs as a coefficient. Thus if we are considering
polynomials with coefficients from .l2′ then we write X + X2 in place of
IX+ lX2.
We denote the set of all polynomials in X with coefficients from R by
R[X]. The elements of R[X] are also sometimes called polynomials over R.
We turn R[X] into a ring by introducing the natural addition and
multiplication. If
and
then we addf(X) and g(X) by adding corresponding coefficients:
f(X) + g(X) = (ao+ bo)+ (a l + bl)X + (a2+b2)X2+ ….
We multiply just as we did in high school algebra: f(X)g(X) = co+ c\X +
C2X 2+ … , where for each n,
cn = aobn + a\bn_\ + a2bn2 + … + an_\b\ + anbO’
Thus the product is obtained by multiplying everything out, using the rule
aXj·bXj = abXi+j, and then collecting terms that involve the same power of
X. The product is indeed a polynomial over R, because there exists n such
that cj=O for every i>n. Specifically, if aj=O for all i>m and bj=O for all
i > I, then we can take n = m + I. For with this choice of n, each term ajbj in
the expression for ck , k>n, must have i+ j>m+ I, so either i>m or j>l, and
therefore either aj = 0 or bj = 0, whence ajbj = O.
Observe that, under these definitions, a polynomial such as ao+a\X +
a2X 2 is actually the sum of ao’ a1X, and a2X 2 in R[X]. By the same token, if
R is a ring with unity, so that X ER[X], then a2X 2 is the product of a2X and
X.
It is not very difficult (but neither is it very exciting) to verify that R[X]
does form a ring under the given operations. We shall forego getting into this,
Section 19. Polynomials 193
and instead take up the more interesting question of which properties of R
carry over to the new ring R[X].
Some things can be seen at once. For instance, if R has unity 1, then
R[X] has unity 1+OX+OX2+ … , which we also denote by l. If R is
commutative, then so is R[X], for if
j(X)=ao+a\X+a2X 2 +… and
then the coefficient of Xn inj(X)g(X) is
aobn+a\bn_\+··· +an_\b\+anbO’
and the coefficient of xn in g(X)j(X) is
boan+b\an_\+'” +bn_\a\+bnao’
These are the same, by the commutativity of multiplication in R.
What if R is a domain? Must R[X] be one too? The answer is yes,
because if j(X) and g(X) are as in the last paragraph, and neither one is the
zero polynomial
0+OX+OX 2 + … ,
then we can let n be such that an*O and a;=O for all i>n, and let m be such
that bm*O and b;=O for all i>m. It then follows that the coefficient of xn+m
in j(X)g(X) is anbm, which is not ° since an,bm are nonzero elements of a
domain. Thusj(X)g(X) is not the zero polynomial, and it follows that R[X]
is a domain.
Arguments such as the one we have just given can be expressed more
succinctly if we allow ourselves some additional terminology. If j(X)ER[X]
is not the zero polynomial, then we can write
j(X)=ao+a\X+··· +anXn, where an *0.
The integer n is called the degree of f(x), and an is called the leading coeffi
cient. We denote the degree off(x) by deg (j), or deg (f(X). Notice that deg (j)
= ° ifff(x) is a nonzero constant polynominal, that is, ifff(x) is the polynomi
nal ao. for some nonzero ao E R. The zero polynominal is not assigned a degree.t
The argument we gave above can now be expressed more precisely by
saying that if R is a domain, andj(X),g(X)ER[X] have leading coefficients
an and bm, respectively, then j(X)g(X) has leading coefficient anbm. As a
consequence, we’ have
“tNo confusion should result from using the same symbol to denote both “0 E R and the constant
polynomial “0. In fact, R is isomorphic to the subring of R[X] consisting of all the constant
polynomials. and we often think of R itself as a subring of R[X].
194 Section 19. Polynomials
THEOREM 19.1 (Degree rule) If R is a domain and J(X),g(X) are nonzero
elements of R[X], then
deg(f(X)g(X) = deg(f(X) + deg(g(X).
The assumption that R is a domain is crucial here. For instance, in the
nondomain 1.6, we have (2X)(3X + 1)=2X, or, even worse, (2X)(3X +3)=0.
We have seen above that a number of properties will always be passed on
from R to R[X). One property that will obviously not be passed on is that of
being a field. For if F is a field, then the element X is clearly not a unit in
F[X].
Nevertheless, it does seem reasonable to expect that assuming F to be a
field will have a beneficial impact on F[X], beyond that of making it a
domain. Much of what can be said about F[X] depends on the following
analogue of the division algorithm for 1..
THEOREM 19.2 (Division algorithm for F[ X]) Let F be a field, and let
J(X),g(X)EF[X]. If g(X) *0, then there exist q(X),r(X)EF[X] such that
J(X) = q(X)g(X) + r(X)
and either r(X)=O or deg(r)< deg(g).
PROOF. If J(X) = 0, or if J(X) '1= ° and deg(f) < deg( g), we write
J(X)=O·g(X)+ J(X),
and we are done.
We now proceed by induction on deg(f). If deg(f) = 0, then by the above
we are done unless deg(g) = 0. But in this case both J(X) and g(X) are
constant polynomialssay, J(X) = ao, g(X) = boand thus we can write
J(X) = (aobol)g(X) + 0.
Note that bO 1 exists because F is a field and bo'l=O.
Now assume the result has been proved for deg(f)
roots and cannot have all the elements of S as roots. 0
COROLLARY 19.6 Let F be an infinite field, S an infinite subset of F. Suppose
J(X),g(X) E F[X] and J(s) = g(s) for every s E S. Then J(X) = g(X), that is,
J(X) and g(X) are the same polynomial.
PROOF. Apply Corollary 19.5 to the polynomiaIJ(X) g(X). 0
Examples
1. Two polynomials in R[X] induce the same function on R iff they are
the same polynomial.
2. Corollary 19.4 may fail if F is not a field. For instance, let R = Z2 Ea Z2′
and letJ(X)=X2+XER[X]. Then deg(f)=2, but each of the four elements
of R is a root of J(X).
Corollary 19.4 (and, with it, 19.5 and 19.6) does hold true for any domain
R, however. See Exercise 19.16.
3. Corollary 19.4 says that if deg(f)=n, thenJ(X) has at most n roots in
F. It may have fewer; indeed, it may have none. For example, the polynomial
X2+ 1 in R[X] has no roots in R
The problem of trying to determine the roots of a polynomial in R[X] is
one that we all encountered in our earliest experience with polynomials. One
of the standard methods for dealing with this problem is, of course, to write
the given polynomial as a product of simpler factors, and then to find the
roots of all the factors. You may remember that in R[X], every nonconstant
polynomial can be written as a product of factors that either have degree 1, or
have degree 2 and are irreducible in the sense that they cannot be factored
any further (Exercise 19.5).
DEFINITION Let F be a field. A nonconstant polynomialJ(X)EF[X] is called
irreducible in F[ X] (or irreducible over F) if J cannot be written as the product
of two nonconstant polynomials in F[X].
In general, F[X] may contain irreducible polynomials of degree higher
than 2. It is still true, however, that every nonconstant polynomial can be
written as a product of irreducible factors.
Section 19. Polynomials 197
THEOREM 19.7 Let F be a field, and let j(X) be a nonconstant polynomial in
F[X]. Then there exist irreducible polynomials jJ(X)”” ,Jk(X) in F[X] such
that j(X) = jJ(X)jz
Since p does not divide akbl but does divide all the other terms, p does not
divide Ck+I’ and Gauss’ Lemma is proved. 0
We put it to work:
LEMMA 19.10 Let f(X)EZ[X]. If f(X) can be written as the product of two
nonconstant polynomials in O[X], thenf(X) can be written as the product of
two nonconstant polynomials in Z[X].
PROOF. As in the discussion preceding the statement of Gauss’ Lemma,
suppose g,h EO[X] are nonconstant polynomials such that
f(X) = g(X)h(X),
and write g(X)=(I/ c)g·(X), h(X)=(l/ d)h·(X), with g·,h· EZ[X]. We can
now write
g·(X) = content (g.) ‘g\(X) and h·(X)=content(h·) ·h\(X),
with g\ and h\ primitive. Therefore,
f(X)= content(g.~~ontent(h.) g\(X)h\(X),
and since g\ and h\ have the same degrees as g and h, we will be done if we
can show that the constant factor on the righthand side is in Z. Now
cd·f(X) = content( g·)content( h·)g \(X)h\(X),
so, taking contents on both sides,
Icdl’ content(j) =content( g·)content(h·)content(g\h\).
By Gauss’ Lemma, content (g\h\)= I, so we obtain
(f)
– content( g·)content( h·)
content – Icdl
Since content(f) E Z, we have what we want. 0
We now reap the first benefit of the preceding discussion by stating a
criterion for irreducibility in O[X] due to Ferdinand Eisenstein (18231852).
200 Section 19. Polynomials
Eisenstein was a prize pupil of Gauss, and his result is very much in the spirit
of Gauss’ Lemma.
lHEOREM 19.11 (The Eisenstein Criterion) Let f(X) = ao + a\X + … + anxn E
l[X), and suppose p is a prime such that
plao,pla\, … ,plan_\,p.{an, and p2.{ao.
Thenf(X) is irreducible in Q[X).
PROOF. If not, then by Lemma 19.10 there are nonconstant g(X),h(X) El[X)
such that f(X) = g(X)h(X). Say
with 1
j(X) = g(X)h(X) (see Exercise 19.13), and since deg(h=deg(f) we must have
deg(g)= deg(g) >1, deg(h)=deg(h);;> 1. Thus AX) is reducible in 7l.p [X],
contrary to our assumption. 0
Example The polynomial X 3 + 2X + 20 is irreducible in Q[ X], because if we
reduce its coefficients mod 3, we get X 3 + 2X + 2 E 7l.3[X], which has the same
degree and is irreducible by Theorem 19.8. Note that if we reduce the
coefficients mod 2, we get X 3 E 7l.2[X], which is reducible. Thus if we reduce
mod some prime and get a reducible polynomial, it does not follow that the
original polynomial is reducible.
EXERCISES
19.1 Letf(X)=ao+ a1X + … +a,X’ EZ[X]. Suppose mlnEQ, with (m,n)= 1.
Show that if min is a root of f(X), then mlao and nla,.
19.2 Determine which of the following are irreducible in Q[X].
a) X 3+X+36
b) 2X3_8X26X+20
202 Section 19. Polynomials
c) 2X4+3X3+ 15X +6
d) X4+2X3+X2+X+l
e) XS+14X 2+4X+6
f) X4+X3+X2+X + 1 [This is (X sl)/(Xl). Recall how we dealt with
X4+ 1.]
g)x”3XZ+6X+l
h) X XZ+ 1
19.3 Write each polynomial as a product of irreducible polynomials over the given
field.
a) 2X3+X2+2, over I3
b) X 3+3X 2+X+4, over Is
c) X2+5, over I7
d) X4+X3+2X2+X+2, over I3
e) X5+X2Xl, over I2
19.4 a) Suppose thatf(x) E JR[X] and c = a + bi E C is a root ofj{X). Show that the
complex conjugate c = a – bi of c is also a root ofj{X).
b) The Fundamental Theorem of Algebra asserts that every nonconstantj{X) E C[X]
has a factorization in C[X] of the form
Assuming this, show that every nonconstantf(X) E JR[X] can be factored in JR[X]
as a product of irreducible polynomials of degree at most 2.
19.5 Suppose F is a field and a(), ai, … , an E F. Let
j{X) = a”x’ + … + atX + ao
and
Prove thatj{X) is irreducible over F iff g(X) is irreducible over F.
19.6 Letj(X) = aX2+ bX + c E IR[X], a,..,O. Show thatj(X) is irreducible in R[X]
iff b2 – 4ac < O.
19.7 Let a E I +. Show that X4 + a is reducible in Q[X] iff a = 4b4 for some integer
b.
19.8 Letj(X),g(X) be nonzero polynomials in I[X]. Show that
content(fg) = content(f)· content( g).
19.9 Show that the polynomials q(X) and r(X) in Theorem 19.2 are uniquely
determined by j(X) and g(X).
Section 19. Polynomials 203
19.10 Let F be a field.
a) What are the units in F[Xl?
b) Show that if cEF, c~O, then for anyj(X)EF[Xl,j(X) and cj(X) generate
the same principal ideal in F[Xl.
19.11 Let R be a ring. Verify the following for R[Xl:
a) the left distributive law;
b) associativity of multiplication.
19.12 Let R be a commutative ring, rER,j(X),g(X)ER[Xl. Let h(X)= j(X) + g(X)
and k(X)= j(X)g(X). Show that
h(r) = j(r) + g(r) and k(r) = j(r)g(r).
Thus the mapping (fir: R[Xl~R given by (fIr(J(X» = j(r) is a homomorphism.
It is called an evaluation homomorphism.
19.13 LetR and Sbe rings and let cp : R ~ Sbe a homomorphism. Forf(X) E R[X], letf"(X)
E S [X] denote the polynomial obtained by replacing each coefficient aj off(X) by
cp(aj). Show that the mapping R[X] + S[X] given by f ~ f" is a homomorphism.
19.14 Give another proof of Eisenstein's Criterion, by considering the homomorphism
Z[Xl~Zp[Xl obtained by reducing all the coefficients of polynomials in Z[Xl
modp.
19.15 Let F be a field, let bb ... ,bn +) be n+ 1 distinct elements of F, and let
c), ... ,cn +) be n + 1 elements of F (not necessarily distinct).
a) Find a polynomialj(X)EF[Xl such thatj(X)=O or deg(f)
a) Show that m is uniquely determined by f and a. That is, if we also have
f(X)=(X – a)’h(X), where Xa does not divide h(X), then r= m.
b) If m:> 2, we say that a is a multiple root of f . Show that a is a multiple root
of f iff 1′(a)=O. (See Exercise 19.17.)
19.19 Let F be a finite field with q elements.
a) Show that a q – I = I for every a*O in F.
b) Letf(X)EF[X]. Show that there exists a polynomialr(X)EF[X] such that
either r =0 or deg(r) deg(f).
So let S={deg(g)lgEI and g*O}. Since S is a nonempty set of non
negative integers, it has a smallest element, n. Choose J(X) in I such that
deg(f) = n. It is clear that (f(X»!: I, and to establish the reverse inclusion we
use the division algorithm for F[X].
Let g(X)EI, and write g(X)= J(X)q(X) + r(X), where either r(X) =0 or
deg(r)
PROOF. We can assume thatf(X) is irreducible in F[X), because if it isn’t we
can work with one of its irreducible factors.
Under this assumption, F[X]/U(X» is a field, in which f(X)=O. If
f(X)=ao+aIX+··· +anXn, this means that
iio+iiIX+·.· + iinxn =0. [20.1)
We have an embedding ‘P:F~F[X]/(f(X», given by ‘P(a) = ii, for every
a E F. ‘P is clearly a homomorphism, and it is onetoone since if aI’ a2 E F and
208 Section 20. From Polynomials to Fields
al=a2, thenf(X) divides al a2 in F[X]. This means that ala2=0, since
deg(f) > 1, so a l = a2•
Thus { ala E F} is a subfield of F[X]/(f(X», isomorphic to F. If we
replace each element a by a, we obtain a field K that extends F, in which
ao+aIX+· · · +anXn=O, by [20.1]. 0
Theorem 20.3 is the key to an indepth analysis of the roots of polynomi
als, because it tells us we can always get our hands on a complete set of roots
to work with:
COROLLARY 20.4 Let f(X) E F[X] have degree n> 1. Then there is a field
K d F such that in K[X] we can write f(X) = a(X – c1)(X – c~· .. (X – cn)’
PROOF. By induction on n.For n= 1 we have f(X)=aX+b, with a,bEF,
a*O, so we can writef(X)=a(X(b/a» in F[X]. Now suppose deg(f)=
m and the result is proved for polynomials of degree m – 1. By Theorem 20.3,
we can let KI d F be a field containing a root c1 of f(X). In K1[X], we can
write f(X) = (X – cl)g(X), where g(X) has degree m – 1 and is therefore
subject to the inductive hypothesis. Let K d KI be such that in K[X] we have
g{X) = a{X – C2)’ .. (X – cn)’
Then K is an extension of F, and we have
f{X) = a{X – cl)·· . (X – cn)
in K[X]. 0
Of course, some of c1,,, , ,cn may already lie in F, and c1″ “,cn need not
be distinct.
Example Let f(X) = X4 X 3_ X + 1 EIs[X]. Then 1 is a root of f(X), and
long division yields
f{X)={X 1){X3 1)
in Is[X]. Since 1 is again a root of X 3 – 1, we obtain
f{X) = (X 1){XI){X2+ X + I).
The factor X2+ X + 1 has no roots in Is, but we know there is a root C in
some extension K of Is. We divide X2+ X + 1 by X – c in K[X]:
X +(c+l)
X+l
cX
(c+ l)X + 1
(c + l)X – c( C + 1)
c( C + I) + 1 = c 2 + C + 1 = 0.
Section 20. From Polynomials to Fields 209
Thus, in K[X),
f(X)= (X 1)(X 1)(X – c)(X – ( cl».
EXEROSES
20.1 Letp be a prime. Show that lp[Xl!(X 2 + 1) is a field iff the equation x 2=1
has no solution (modp).
20.2 Is O[X]/«X – 1)2) a domain?
20.3 Is O[XJ!(X 3 +2X +2) a field? How about IR[XJ!(X 3 +2X +2)?
20.4 LetK= {O, 1,X,X + I} be the fourelement field constructed in Example 1 on
pp. 206207. Write ~ + X + 1 as a product of factors of degree 1 in K[X].
2O.S The elements of the field K in Exercise 20.4 can each be written uniquely in
the form a+bX, with a,bEl2• Find a general rule for writing the product
(a + bX)( c + dX) in this form.
20.6 a) Let f(X) be irreducible in F[X], and let K be the field obtained from
F[XJ!(f(X» by replacing a by a, for each a EF. Show that if deg(f) = n,
then every element of K has a unique representation in the form
ao+alX + … +an_IXnt,
with ajEF.
b) Show that if we start with F = lp and deg(f) = n, then the field K in part
(a) has p n elements.
20.7 Use the result of Exercise 20.6 to construct a field with m elements, for m=
a) 8;
b) 9;
c) 27;
d) 25;
e) 125.
20.8 Let F be a field. What are the prime ideals in F[X]?
20.9 Let n;;. 1, letp be a prime, and letf(X)=XP· – X Elp[X]. By Corollary 20.4,
let K be an extension of lp such that in K[X] we have
f(X)=(X – c l )”, (X cp.).
a) Show that in this case CI>””cp • are all distinct. (Use Exercise 19.18.)
b) Show that {CI,’ .. ,cp .} is a subfield of K, and hence that there exists a field
with p n elements.
20.10 IfF is a field and.f{X), g(X) are elements of F[X] then a polynomial h(X) E F[X] is
called a greatest common divisor (g.c.d) of.f{X) and g(X) if
i) h(X) divides both.f{X) and g(X), and
ii) every k(X) in F[X] that divides both.f{X) and g(X) divides h(X).
210 Section 20. From Polynomials to Fields
a) By applying Theorem 20.1 to the ideal
/= {a(X)f{X) + b(X)g(X) I a(X), b(X) E FIXJ},
prove that any two polynomialsf{.x), g(.x) in FIX] have a g.c.d. that can be written
as a(X}f{.x)+b(.x)g(.x) for some a(.x), b(.x) E FIX].
b) Prove that if at least one of f{.x) or g(X) is not the zero polynomial then any two
g.c.d.’s off{.x) andg(.x) differ by a constant factor.
20.11 Let F be a field, let a E F, and let p be a prime. Let fiX) = ~ a. Prove that f (.x) is
irreducible in FIX] if and only iff{X) has no roots in F.
20.12 Let F be a field of prime characteristic p, let a E F, and let n E Z+. Prove that X pn –
a is irreducible in FIX] if and only if ~ – a has no roots in F.
SECTION 21
UNIQUE FACTORIZATION
DOMAINS
We have called a positive integer p a prime if p~ I and the only positive
divisors of p are I and p itself. This notion of “prime” generalizes in a natural
way to allow negative primes: an integer n is called prime if n~O, n~ ± I,
and the only divisors of n are ± I, ± n. Thus 2, – 2, 3, – 3,5, – 5,7,
– 7, II, – II… are all primes in lL..
In these terms, the Fundamental Theorem of Arithmetic may be taken as
the statement that every integer n which is neither zero nor ± I can be
factored into a product of primes, and that if
n=PIP2″‘Pr and n=QIQ2″‘qs
are two such factorizations, then r = s and, after rearranging the q;’s if
necessary, we have Pj = ± qj. For example, we have – 60 = ( 2)(2)(3)(5) =
( – 3)(2)( – 5)( – 2), and if we rearrange the second factorization it becomes
( – 2)(2)( – 3)( – 5).
The Fundamental Theorem of Arithmetic seems obvious to most of us,
because we all grew up with it in school. For this reason, it is easy to get
lulled into thinking that the theorem doesn’t require any proof. One purpose
of this section is to convince you that it does require proof by showing you
some integral domains for which the analogue of the Fundamental Theorem
is false. With such examples in hand, it becomes interesting to try to
enunciate some conditions on a domain which will guarantee that the
analogue of the Fundamental Theorem does hold.
The necessity for such an investigation is underscored by the fact that
some notable mathematicians have fallen into the trap of assuming that the
Fundamental Theorem holds in cases where in fact it does not. Efforts to
salvage the results which they based on these faulty assumptions have been
largely responsible for the development of ring theory as we know it today.
t This section can be omitted without loss of continuity.
211
212 Section 21. Unique Factorization Domains
Before we recount a specific instance, let us establish some vocabulary, so
that we can say things a bit more precisely.
Let D be a domain. (We concentrate on domains because domains are
the natural generalization of 7L.) An element dE D is called irreducible if d is
neither 0 D nor a unit, and whenever a, bED and d = ab, then either a or b is a
unit.
Examples
1. Since the units in 7L are ± 1, the irreducible elements of 7L are precisely
the primes.
2. If F is a field, the units in F[X) are the nonzero elements of F. Thus
the irreducible elements of F[X) are the nonconstant polynomials that are
irreducible according to the meaning of the word in Section 19.
3. We claim that a nonconstant polynomial J(X) E7L[X) is irreducible in
7L[X) iff J(X) is primitive and cannot be written as the product of two
nonconstant polynomials in 7L[X).
First of all, if J is irreducible then J must be primitive, for otherwise we
could write J(X)=p·g(X) for some prime p and nonconstant g(X), and
neither p nor g(X) is a unit in 7L[X]. Likewise, J cannot be written as the
product of two nonconstant polynomials.
Conversely, suppose that J has the two indicated properties. Then if
J(X)=g(X)h(X) in 7L[X), either g or h must be a constant; since J is
primitive, this constant must be ± 1, so it is a unit in 7L[X). Thus J is
irreducible.
It is easy to see that a constant is irreducible in 7L[X) iff it is a prime in 7L,
and therefore the irreducible elements of 7L[X) are the primes of 7L and the
nonconstant irreducibles described above.
In light of Example 2, our use of the term “irreducible element” seems
very natural; but in terms of Example 1, “prime element” might seem more
appropriate. In general, the word “prime” is used in connection with a
concept that is sometimes stronger than irreducibility. If D is a domain, and
x,y ED, then we say that x dividesy, and we write xly, if there is some zED
such that y = xz. An element d ED is called prime if d is neither 0 D nor a unit
and whenever a,bED and dlab, then dla or dlb.
Prime elements are always irreducible, in any integral domain:
THEOREM 21.1 Let D be a domain, and let d E D be prime. Then d is
irreducible.
Section 21. Unique Factorization Domains 213
PROOF. Suppose d=ab; we must show that either a or b is a unit. Since d is
prime, we know that d divides either a or b. If dla, then we have a = de for
some e, and so
d=(de)b.
Since d=f’O and D is a domain, we conclude that 1= eb and b is a unit.
Similarly, if dlb then a is a unit. 0
For (Z, +, .), the notions of “prime” and “irreducible” coincide (Theo
rem 4.3) and in general the truth of the implication “irreducible ~ prime” for
a domain D is very intimately connected with the truth of an analogue of the
Fundamental Theorem for D (see Exercise 21.9). We will soon see examples
of irreducibles that are not prime.
Two elements a and b of a domain D are said to be associates if a = bu
for some unit u. For example, any element a is an associate of itself, because
a = a . 1. Similarly, if a and b are associates, then b and a are associates,
because a = bu implies b = au – ‘. The relation
aRb iff a and b are associates
is in fact an equivalence relation; transitivity is easily verified.
We now have the words we need in order to say precisely what it means
for an analogue of the Fundamental Theorem to hold for D.
DEFINITION Let D be an integral domain. Then D is called a unique factoriza
tion domain (UFD) if:
i) Every element d ED that is neither 0 nor a unit can be factored as the
product of a finite number of irreducible elements; and
ii) If d=P,P2·· ·Pr and d=Q,q2··· Qs are two such factorizations, then r=s,
and there is a permutation f of {1, 2, … , s} such that Pi and Qf(i) are
associates, for each iE{1,2, … ,s}.
Note that this is a direct generalization of the statement of the Funda
mental Theorem that we gave on p. 211, because in Z, two elements a and b
are associates iff a = ± b.
Now for a bit of history. We all know that there are triples x, y, z of
nonzero integers such that x2 + y2 = z2; for instance, 32 + 42 = 52. In 1637,
Pierre de Fermat claimed to have discovered a “truly remarkable proof’
that if n > 2 then there do not exist nonzero integers such that xn + yn = zn.
This result is referred to as Fermat’s Last Theorem, although Fermat
never revealed a proof and there is a great deal of doubt as to whether he
ever had one. Mathematicians sought a proof for over 350 years, until a
proof was finally published by Andrew Wiles in 1995.
214 Section 21. Unique Factorization Domains
The problem is very easily reduced to showing that x n + y n = Z n is
impossible (for nonzero x,y,z) if n is 4 or an odd prime. (It is fun to do this
for yourself, and we won’t spoil it by showing you how.) The case n=4 can
be disposed of by elementary means, and we know that Fermat did prove this
case himself. In any event, the general problem boils down to proving that
x P + yP = zP is impossible if p is an odd prime and xy z =1= O.
A proof for p = 3 was published in 1770 by Leonhard Euler, the most
prolific mathematician of all time. Euler’s proof involved using numbers of
the form a+bv=:3, where a,bElL and v=:3 =V3i. At one point in his
argument, he made a claim about these numbers which was apparently based
on the tacit assumption that they obey unique factorization. His claim was
correct, but the tacit assumption behind it was not, and his proof remained
incomplete until the missing justification was supplied by Legendre some time
later.
A proof for p = 5 was given by Legendre, and independently by Dirichlet,
around 1825. The casep=7 was handled by Lame in 1840. The first general
and by far the most significantattack on the problem was made by E.
Kummer in 1843. Kummer’s basic idea was to consider numbers of the form
where a; E lL and ~P is a complex number =1= I such that ~: = 1. (For example,
~P = cos(2’17″/ p) + i sin(2?T / p ).”t) These numbers form a subring of C, which we
denote by lL[~p]. Using them, it is possible to factor x P + yP completely, and
the equation x P + yP = zP becomes
Assuming that lL[~p] is a UFD, Kummer used this form of the equation to
prove that x P + yP = zP is impossible if xyz =1=0.
Kummer presented his proof to Dirichlet (a more established mathemati
cian), who pointed out that Kummer had neglected to verify the assumption
that factorization into irreducibles is unique in lL[~p]’ (Kummer was later to
point out a similar flaw in an attempt by Lame.) In 1847, Cauchy (after
having made the same mistake himself) pointed out that factorization is not
unique in lL[~23]’ Thus Fermat’s Last Theorem remained unproved.
1’That [cos(2’IT /p)+ isin(2’IT /p)Y’ = 1 follows from De Moivre’s Theorem:
[cos8+ isin8t = cosk8 + isink8.
De Moivre’s Theorem can be proved by induction on k, using the fact that
[cos a + i sin aJ[cos {:I + isin{:l] = (cos a cos{:l sin a sin{:l) + i(sina cos {:I + cos a sin,B)
= cos(a + {:I) + isin(a+ {:I).
Section 21. Unique Factorization Domains 215
Undaunted, Kummer set about trying to modify Z[~p] so as to restore
the uniqueness of factorization. He introduced what he called ideal
numbers, and the theory he developed was a precursor of the modern theory
of ideals. Kummer succeeded in proving Fermat’s Last Theorem for certain
primes which he called regular.
The complete proof that Wiles published in 1995 employed modern and
sophisticated ideas far beyond those known to Kummer (and far beyond the
scope of this book). But Kummer’s method had a lasting impact on the
development of algebra. Attempts by number theorists to exploit the
properties of specific systems more inclusive than Z led naturally to a
study of such systems in general, hence to the emergence of an abstract
theory of rings .
After we have developed some more information about UFDs, we will
discuss in detail the application of another extended number system (the ring
of Gaussian integers) to the proof of Fermat’s classic “two squares” theorem.
But first we want to show you an explicit example of the failure of unique
factorization in a fairly simple domain.
In fact, let us use the set of all complex numbers of the form a + b V3 ,
with a, bEl, referred to above. If we denote this set by l[V3], then
l[v=3] is a subring of C containing 1, hence an integral domain. In
l[v=3] we have the following two factorizations of the number 4:
4=2·2=(1 + V3 )(1 V3).
We are going to show that 2, 1 + V3 , and 1 V3 are all irreducible in
l[v=3] and that no two of them are associates. This will establish that
l[v=3] is not a UFD.
For a=a+bV3 , define the norm N(a) of a to be N(a)=a 2 +3b2• For
every a El[v=3 ], N(a) is a nonnegative integer, and N(a)=O iff a=O. An
easy calculation shows that if a, {3 E l[ v=3 ], then
N(<<{3) = N(a)N( {3).
Using norms, we can readily determine all the units in l[V3]. For if a
is a unit, then a{3 = 1 for some {3, whence
N(a)N({3)=N(I)= 1,
which implies that N(a) = 1, since N(a) and N({3) are both nonnegative
integers. Conversely, if N(a) = 1 and, say, a=a+bV3, then a2 +3b2 = 1,
which implies that a = ± 1 and b =0. Thus a = ± 1, and a is a unit. Thus we
see that a is a unit iff N(a) = 1, and the only units in l[V3] are ± 1.
Now we claim that 2 is irreducible in l[ V3 ]. Clearly, 2 is not 0, and
it is not a unit since 2* ± 1. We must show that if 2= a{3, then either a or {3
must be a unit, that is, either N( a) or N( {3) must be 1. Now if 2 = a{3, then
216 Section 21. Unique Factorization Domains
N(2)=N(a)N(f3)
4=N(a)N(f3),
so all we have to show is that neither N(a) nor N(f3) can be 2. But the norm
of any element of Z [v=3 ] has the form a2 + 3b2, and it is clear that this
never gives us 2.
The same argument shows that both 1 + v=3 and 1  v=3 are irre
ducible, because their norms are both 4.
No two of our three irreducible elements are associates, because the only
units are ± 1. Thus Z[ v=3 ] is not a UFO.
Notice a couple of other interesting things. The equation
2·2=(1 + v=3 )(1 v=3)
shows that 2 divides the product (I + v=3 )(1 v=3), but it obviously
does not divide either factor. Thus 2 is irreducible, but not prime. The same
can be said for I + v=3 and I  v=3 .
Observe that if a EZ[v=3 ] and N(a) is a prime integer, then a is
irreducible. For if a=f3y, then N(a)=N(f3)N(y), so one of N(f3), N(y) is 1.
The converse is false; for example, 2 is irreducible, but N(2) = 4 is not a prime
integer.
By considering norms, it can be shown that every element of Z[v=3]
which is neither ° nor a unit can be factored into a product of irreducibles
(Exercise 21.5), and thus it is only the uniqueness of factorization that fails,
not the existence. On the other hand, it is not true that every element *0, ± 1
can be written as a product of prime elements. For instance, 2 is not a prime,
and if
then
N(2) = 4 = N( al)N( a2)' .. N( ak)'
so some lX; has norm 4 and the others are units, not primes.
We have not yet exhibited a prime in Z[v=3]. An example is given in
Exercise 21.22.
Under what conditions might it be impossible to factor an element into
irreducible factors? Suppose D is a domain, a E D is neither ° nor a unit, and
a cannot be written as the product of a finite number of irreducible elements.
In particular, then, a is not itself irreducible, so we can write a = alb l, where
neither al nor bl is a unit. Furthermore, our assumptions on a imply that a l
and b l cannot both be written as products of irreducible elements. Say b l
cannot be so written. We have
a = albl; al,bl are not units; b l is not a product of irreducible elements.
[21.1]
Section 21. Unique Factorization Domains 217
Now bl has the same properties that a had when we started, so we can write
b l = a2b2, where neither a2 nor b2 is a unit, and, say, b2 cannot be written as a
product of irreducible elements. Looking at b2, we write b2 = a3b3, where
neither a3 nor b3 is a unit, and b3 cannot be written as a product of irreducible
elements. Clearly we can continue like this indefinitely by induction; we have
bn=an+lbn+I' with an+l,bn+1 not units and bn+1 not a product of irreducible
elements. We have the following equations:
a = alb l = a 1a2b2 = a la2a3b3 = ala2a3a4b4 = ....
What conditions on D could possibly rule this out? Think what the recurring
situation [21.1] means. If we have a= alb l, then aEblD, the principal ideal of
D generated bl, and this implies that aD ~ biD, since biD is an ideal. Is
b l D ~ aD too? If so, then in particular b l E aD, so b l = ac for some c. Then
bl =ac=alblc.
Since b l =F 0 and D is a domain, this yields I = a I c, and a I is a unit.
Contradiction!
Thus aD f biD. Since b2 was obtained from b l just as b l was obtained
from a, we get biD f b2D by an identical argument. And, in general,
bnD f bn+ID, so we have a strictly increasing chain of ideals:
aD f biD f b2D f b3D f b4D f ....
The union of this chain, aD U biD U b2D U .'., is easily seen to be an ideal. If
this ideal were principal, then the fact that our chain of ideals is strictly
increasing would be contradicted. For if
aDubIDub2DUb3DU'" =dD,
then dE bjD for some i, hence dD ~ biD, and the chain would stop at biD.
Thus one way to ensure that elements factor into irreducibles is to assume
that every ideal in D is principal. An integral domain with the property that
every ideal is principal is called a principal ideal domain (PID). For example,
71.. is a PID, and so is F[X], for any field F. Our discussion proves
TIfEOREM 21.2 Let D be a PID, and let a E D. If a is neither 0 nor a unit, then
a can be written as the product of finitely many irreducible elements.
This theorem constitutes half of a proof that every PID is a UFD. The
other half deals with the uniqueness of the factorization, and for this we take
a hint from 71... The key to the uniqueness of factorization for 71.. (Exercise 4.28)
was the fact that if p were irreducible and p divided a product ab, then p had
to divide either a or b. In our current terminology this says that every
irreducible element was prime.
We thus rest our hope for unique factorization in PID's on
218 Section 21. Unique Factorization Domains
THEOREM 21.3 Every irreducible element of a PID is prime.
We will establish Theorem 21.3 by proving the following sharper resu~t,
which also generalizes Theorem 20.2.
THEOREM 21.4 Let D be a PID, let d ED, and assume that d is neither OD nor
a unit. Then the following are equivalent:
i) d is prime
ii) d is irreducible
iii) dD is a maximal ideal
iv) dD is a prime ideal.
PROOF. The proof is not much longer than the statement of the theorem; we
will show that (i) ~ (ii) ~ (iii) ~ (iv) ~ (i).
(i) ~ (ii): See Theorem 21.1.
(ii) ~ (iii): Since d is not a unit, It£. dD, so dD is a proper ideal. We must
show that if dD ~ bD, then bD = D. Now if dD ~ bD, then in particular
dE bD, so we have d= be for some e. Since d is irreducible, either b or e must
be a unit. If c is a unit, then b = de , and bE dD, so bD c;;" dD, a contradic
tion. Thus b is a unit, and bD = D.
(iii) ~ (iv): See Corollary 17.8.
(iv) ~ (i): Suppose dD is prime, and suppose d divides a product be. Then
bCEdD, so either bEdD or eEiiD, that is, either dlb or die, as desired. 0
THEOREM 21.5 Every PID is a UFD.
PROOF. By Theorem 21.2, it suffices to show that if
P,P2"'Pr=q,qz"'qs [21.2]
for irreducible elements Pi' qj' then r = s and there is a permutation f of
{l,2,oo.,s} such that Pi and qJ(i) are associates for each iE{1,2,oo.,s}.
Now, from the above equation, p,lq,q2' .. qs' so by Theorem 21.3 and
induction,p, divides some qj' By renaming the q's if necessary, we can assume
j = 1. Thus q, = ap, for some a, and since q, and p, are irreducible, this means
that a is a unit, so q, andp, are associates. We have
P,Pz" 'Pr=ap,qz'" qs'
and since we are in a domain, this becomes
(a  ~2)P3' .. Pr= qZq3' .. qs' [21.31
We now proceed by induction on r. If r= 1, then s must be 1, else
Equation [21.3] would show that the irreducible element q2 is a unit, which is
nonsense. Thus if r= 1, then the original equation [21.2] had one irreducible
on each side, and p, = q,.
Section 21. Unique Factorization Domains 219
Assuming the result for r = n, suppose that r = n + 1. Then by [21.3], s> 1,
and by the inductive hypothesis, r – I = s – I and there is a permutation g of
{2, … ,s} such that Pi and qg(i) are associates for 2 < i
the definition of Euclidean domain is satisfied.
Now for Condition (ii). Let a,f3 E l..[i], 13=1=0. We seek to ·find y and p in
1..[i] such that a = yf3 + p, and either p = 0 or v(p) < v( /3). Since 13=1= 0, a / 13 is a
complex number, so there are real numbers x and y such that
a = (x + yi)f3.
222 Section 21. Unique Factorization Domains
Our idea is to use for 'I an element of l[i] that is close to x + yi. Specifically,
choose integers a and b such that Ix  al <;; i and I y  bl '" i. Then
a =(a+ bi){J+ [(xa) +(y  b)i] {J,
so if we take y=a+bi and p=[(xa)+(yb)i){J, we have
a=y{J+p.
Now if p=O, we are done. Otherwise v[(xa)+(yb)i) is defined, and we
have
as desired.
v(p) = v[(x a) +(y  b)i]v( {J)
= [ (x  a)2 + (y  b)2] v( {J)
< [ ( ~ r + ( ~ r] v( {J)
I
= 2" v( {J) < v( {J),
Unique factorization in land lli) provide a very appealing proof of a
classic theorem of Fermat. Fermat observed that some primes can be ex
pressed in the form a2 +b2 (with a,bEl), while others cannot. For instance,
5=22+ 12, 13=32+22, 17=42+ 12,29=52+22,37=62+ 12; but none of 3,7,
11, 19,23, or 31 can be written as the sum of two squares. It is also true that
5, 13, 17,29, and 37 are all congruent to 1 (mod 4), while 3, 7, 11, 19,23, and
31 are all congruent to 3. Fermat's Two Squares Theorem asserts that, in
general, a positive odd prime p is the sum of two squares iff p = 1 (mod 4).
Fermat claimed to have proved this result in a letter he wrote in 1640. In
accordance with the practice of the time, he never published a proof, and the
first published proof was given by Euler in 1754. It is said that Euler worked
on and off for seven years to find a proof.
THEOREM 21.8 (Fermat) Let p > 0 be an odd prime. Then there exist integers a
and b such thatp=a2 +b2 iff p=1 (mod 4). If p=1 (mod4), thenp can be
written as a2 + b2 in only one way (we do not count things like b2 + a2 or
( ai+( – bi as different ways.)
PROOF. Suppose p = a2 + b2• Then p = a2 + b2 (mod 4), and each of a2,b2 is
congruent (mod4) to one of 02, 12, 22, or 32; that is, each of a2,b2 is =0 or 1
(mod 4). Since p is odd, one of a2,b2 is = 0 and the other is = 1 (mod 4).
Hence p = 1 (mod 4).
That was the easy half. To finish the proof, we must show that if p = I
(mod4), then p can be written (uniquely) as a2 + b2• The idea of our proof is
to view a2 + b2 as being (a + bi)( a – bi) in l[ i), and to shQW that p can be
Section 21. Unique Factorization Domains 223
written as (a + bi)( a – bi) for some integers a, b.
Our first step is to show that p is at least not irreducible in 1[ i]. We know
that if p were irreducible, then since l[i] is a PlD, p would be prime in l[i].
We are going to show that this is not the case by showing that there exists an
integer m such that
(m – i)(m+ i) = kp
for some integer k. Thus pl(m – i)(m + i), but clearly p divides neither (m – i)
nor (m + i) in 1[i], so p is not prime.
We find m by observing that having (m – i)(m + i) = kp for some k is the
same as having m2 + 1 = 0 (mod p), that is, m2 = – 1 (mod p). We know
something that is = 1 (mod p), namely (p I)! (see Wilson’s Theorem,
Exercise 10.19). Thus to fmd a suitable m it suffices to show that (p – I)! = m2
for some m. Now we know that p = 4n + 1 for some n> 0, hence
(p I)! = (1)(2)· .. (2n)(2n + 1)(2n +2)· .. (4n 1)(4n)
=(1)(2)·· · (2n)( 2n)( (2nI»··· (2)( l)(mod p)
=( li”122232 ••• (2nf(modp)
=[(1)(2)(3)··· (2n)]2(modp).
Thus if we take m=(l)(2)(3)· ·· (2n), we have the required m, and we see that
p is not prime, hence not irreducible, in 1[ i].
The rest is easy. Write
p =(a+ bi)(c+ di),
where neither a + bi nor c + di is a unit in 1[i]. Then
v(p) = v(a + bi)v(c + di),
that is,
p2 = (a2 + b2)( c2 + d2).
Since neither a + bi nor c + di is a unit, neither a2 + b2 nor c2 + d 2 is
(Exercise 16.24), so by unique factorization in 1 ,
and we have shown that p can be written as the sum of two squares. (Note
that fromp=(a+bi)(c+di) it follows that c=a and d= b.)
All that remains is to establish the essential uniqueness of the representa
tion. We leave this to you as Exercise 21.19.
EXERCISES
21.1 Let D be a domain. Show that it is not possible to express a unit of D as a
product of prime elements.
224 Section 21. Unique Factorization Domains
21.2 Let D be a domain and let a, bED. Under what conditions is it true that
Da=Db?
21.3 Let R be a UFD, r E R. Show that r is irreducible in R iff it is irreducible in
R[X].
21.4 Complete the proof that Z[X] is a UFD by showing that every element which
is neither ° nor a unit can be written as a product of finitely many irreducible
elements.
21.5 Show that every element *0, ± I in Z[v=3] can be expressed as the product
of finitely many irreducible elements.
21.6 Show that the following domains are not UFDs.
a) D= {a+bViO la,bEZ}. (Consider N(a+bViO )=a2 lOb2, and find
two distinct factorizations of 6.)
b) D= {a+ bY=5 la,bEZy. (Consider N(a+ bY=5 )=a2 +5b2, and find
three distinct factorizations of 21.)
21.7 Let D be a Euclidean domain, with v: D – {OD }~Z+ U {O}.
a) Show’ that if d ED, then d is a unit iff v( d) = vel D)’
b) Show that if v is a constant function then D is a field.
c) Show that, in general, if two nonzero elements a and b of a Euclidean
domain are associates, then v(a)=v(b).
21.8 Show that the following domains are Euclidean with the given function v.
a) D= {a+bY2la,bEZ}, with v(a+bY2 )=la2 2b21
b) D= {a+bv=2la,bEZ}, with v(a+bv=2 )=a2 +2b2
21.9 Let D be a domain with the property that every element of D which is Ineither
o nor a unit can be written as the product of a finite number of irreducibles.
Show that D is a UFD iff every irreducible element of D is prime.
21.10 Let F be a field, and let X and Y be variables. Define the polynomial ring
F[X, Y] in two variables over F by
F[X, Y]=F[X][Y].
Thus F[X, Y] is the ring of polynomials in Y, over F[X].
a) Show that every element of F[X, Y] is a finite sum of terms of the form
aXiyj, where a E F and i,j are nonnegative integers.
b) Show that F[X, Y] is not a PID.
21.11 Let D be a domain, and let a, bED. An element dE D is called a greatest
common divisor (g.c.d.) of a and b if
i) dla and dlb; and
ii) if c is any element such that cia and clb, then cld.
Show that if D is a PID, then any two elements of D have a g.c.d. which can
be written in the form xa + yb, for some x,y ED. (Suggestion: Consider the
ideal {xa+yblx,y ED}.)
Section 21. Unique Factorization Domains 225
21.12 In the context of 7L, the definition in Exercise 21.11 assigns two g.c.d.’s to each
pair of nonzero elements a, b. For instance, the g.c.d.’s of 12 and 15 are 3 and
– 3. (Compare Exercise 4.26.) Show that, in general, if D is a domain, and
dE D is a g.c.d. of a and b, then all the g.c.d.’s of a and b are precisely the
associates of d.
21.13 Show that the elements 4 and 2(1 +V3) in 7L[v=3] have no g.c.d.
21.14 Use the Euclidean algorithm (that is, repeated application of the division
algorithm) in O[X] to find a g.c.d. of the elements 2X3 + 9X2 + 12X + 5 and
2X s+5X4 +8X +20.
21.15 Let a E 7L[i]. Show that a is a prime element in 7L[i] iff either v(a) is a prime
integer or a is an associate of some prime integer p such thatp==:3 (mod4).
21.16 Write each of the following elements of 7L[i] as a product of primes (see Exercise
21.15).
a) 1+ 3i
b)7+8i
c) 99+27i
21.17 a) Prove that if a,bE7L andp is a prime integer,p==:3 (mod4), such that
pl(a2 + b2), then p21(a2 + b2).
(Use the fact that p is prime in 7L[i].)
b) Prove that an integer n;;. 2 is the sum of two (integer) squares iff in the prime
factorization of n (in 7L), every prime p ==: 3 (mod4) occurs to an even power.
21.18 a) Use the Euclidean algorithm in 7L[i] to find a g.c.d. for 53 + 9i and 1+ 7 i.
(See the proof that 7L[i] is a Euclidean domain, and use the fact that for
complex numbers a + bi and c + di,
a+bi = (a+bi)(cdi) )
c+di c2 +d2 •
b) The proof that I[i] is Euclidean provides an upper bound on the number
of steps it will take the Euclidean algorithm to produce a g.c.d. What is this
bound?
21.19 Prove the uniqueness statement of Theorem 21.7.
21.20 Give an example of two domains D and D’ such that D is a UFD, D’ is a
homomorphic image of D, and D’ is not a UFD.
21.21 Let R be a PID and I an ideal of R.
a) Show that every ideal of R/ I is principal. Must R/ I be a PID?
b) Show that RII has only finitely many ideals if I is nontrivial.
21.22 Show that v=3 is prime in 7L[ v=3 ].
21.23 Let
1+v=3
~3 = COS(21T /3) + i sin(21T /3) = – ~ + i(V3 ) /2 = 2
226 Section 21. Unique Factorization Domains
Let
Z[t31 = {a + bt3+ ctlla,b,cEZ},
and observe that tl= 1. Thus Z[t31 is the domain considered by Kummer in
his work on Fermat’s Last Theorem, for p = 3. This exercise will result in a proof
that Z[t31 is a Euclidean domain. [A proof of Fermat’s Last Theorem forp=3,
using Z[t31, can be found in Hardy and Wright, An Introduction to the Theory
of Numbers, Chapter XIII (Oxford: Clarendon Press, 1960).]
a) Show that Z[t3] = {a+ bt3la,bEZ}, and that every element of Z[t3] has a
unique representation in the form a + bh
b) Let us write just t instead of h for simplicity. Define
v(a+ bn= (a + bO( a+ bt 2).
Show that v(a + bn = a2 + b2 – ab, and that v is multiplicative, that is,
v(afJ) = v(a)v(fJ).
c) Show that v maps Zm {OJ into Z+, and that Z[n with the function v,
is a Euclidean domain.
21.24 a) Show that Z[v=3 ]C;Z[t3]’ (See the preceding exercise.)
b) Determine the relationship between Z[v=3 ] and Z[t31 by describing the
elements of Z[tJ] – Z[v=3 ] in terms of integers and ‘\f=3 .
1l.lS See Exercise 21.23.
a) Show that if a EZ[t3], then a is a unit iff v(a)= 1.
b) Show that if a EZ[t3] and v(a) is a prime integer, then a is irreducible.
Conclude that 1 t3 is irreducible.
c) Show that 2 is prime in Z[t3]. (Hint: Show that if 21(a2 + b2 ab), then both
a and b are even.) There are many more primes in Z[t3]; see Hardy and
Wright, Chapter XV.
21.l6 Let D be a Euclidean domain. Show that the q and r in Condition (ii) for a
Euclidean domain will be unique for every choice of a and b iff v has the
property that v(a+b)<:max{v(a),v(b)} for all nonzero a and b such that
a+b*O.
SECTION 22
EXTENSIONS OF FIELDS
In this section we begin an indepth study of fields. The work begun here will
culminate in the presentation of Galois theory in Section 25.
The developments in this and the succeeding sections represent a beautiful
interplay between the ideas we have developed so far. Aside from this, we will see
a number of striking applications, some in Sections 25 and 26 and some even
sooner in Section 23.
If E and F are fields we write F ~ E to indicate that E is an extension of F, i.e.
F is a subfield of E. In this context we are of course using the symbol "~" to mean
more than just "is a subset of'; here the symbol means that (F, +) is a subgroup of
(E, +) and (F  {OJ, ) is a subgroup of(E  {OJ, .).
If F ~ E and a E E we will often study a by considering the smallest subfield
of E containing F u {a}.
Notation. If F ~ E and a E E then F(a) denotes the intersection of all subfields of E
containing F u {a}. (The notation "F(a)" is read "F adjoin a".)
It is easy to see that F(a) is a subfield of E containing F u {a} and IS
contained in every subfield of E that contains F u {a}.
THEOREM 22.1 F(a) "" {g(a) / h(a) I g(x), heX) E F[X] and h(a) * OJ.
PROOF. If we denote the set on the right side of the equation by S then clearly
every element g(a) / h(a) of S is in F(a) since F(a) is a sub field of E containing F
u {a}. On the other hand, S is a subfield of E containing F u {a}, so F(a) is a
subset of S. Thus F(a) = S. 0
There is one very important situation in which the description of F(a) in
Theorem 22.1 can be simplified considerably. This is the situation in which the
element a happens to be a root of some nonzero polynomial in F[X].
227
228 Section 22. Extensions of Fields
DEFINITIONS If F ~ E and a E E then a is algebraic over F if there exists a
nonzero polynomial j{X) E F[X] such thatj{a) == O. We say that E is an algebraic
extension of F (or that E is algebraic over F) if every element of E is algebraic
over F.
If a is algebraic over Fthen the set of polynomials
1= (f(X) E F[X] I I(a) = O}
is a nontrivial ideal in F[X]. By Theorem 20.1 there must existfiX) E I such that 1=
(f(X), the prinicipal ideal generated by fiX). It will be convenient to have several
equivalent ways to describe those polynomials that generate I.
THEOREM 22.2 If F ~ E, a E E is algebraic over F, and I = (f(X) E F[X] I j{a) ==
O} then the following statements about an element j{X) E I are all equivalent to
each other:
i) 1= (t(X)
ii)j{X) has minimal degree among all the nonzero polynomials in I
iii)j{X) is irreducible.
PROOF. That (i) implies (ii) is clear, since if 1= (f(X) then fiX) divides every g(X)
E I, so if g(X) is nonzero then deg(f) :s deg(g) by the degree rule. To see that (ii)
implies (iii) note that if fiX) is reducible and fiX) = g(X)h(X) with g(X) and heX)
each of degree less than deg(f), then 0 = fia) = g(a)h(a) in E, so either g(a) = 0 or
h(a) = O. Thus either g(X) E lor heX) E I, and thus (ii) cannot be true. Finally, (iii)
implies (i) because iffiX) E I is irreducible and g(X) is any generator for I then fiX)
= g(X)h(X) for some heX), so since fiX) is irreducible heX) must be a nonzero
constant. ThusfiX) is a generator for I. 0
A polynomial is called monic if its leading coefficient is 1. We can obtain a
monic generator for the ideal I by starting with any generator and multiplying by
an element of F. I has only one monic generator, because any two generators
divide each other and therefore differ by a constant factor.
DEFINITIONS Let F ~ E and let a E E be algebraic over F. The unique monic
irreducible polynomial in F[X] having a as a root is called the irreducible
polynomial of a over F. We denote it by irr(a/F)' The degree of irr(alF) is called
the degree of a over F and denoted by deg(alF).
It follows from Theorem 22.2 that if j(){) is any nonzero polynomial in F[X]
such thatj{a) = 0 then deg(a/F) :s deg(j{X), and we have deg(alF) = deg(j{X) iff
j{X) is irreducible in F[X].
Examples
Section 22. Extensions of Fields 229
1. Let F be any field and let a E F. Then a is algebraic over F, irr(alF) is X 
a, and deg(alF) = I.
Conversely, if F S;;;; E and a E E is such that deg(alF) = I, then a is a root of
some polynomial of degree I in F[X] and thus a E F because F is a field .
2. Consider JR s;;;;
”
I
= XpI + Xp2 + . .. + X + 1 p X I
‘
so (p is algebraic over Q. We claim that irr«(IQ) =
= p – I.
To see this it suffices to show that
by applying Eisenstein’s Criterion to
(X + IY 1
(X +I)1
X p + pX ,,1 + … + pX
X
where for each r E {I, … , p – I} the coefficient of Xfrr in the numerator is the
binomial coefficient I( P� )1 and is thus an integer divisible by p since p is prime. r. p r .
By Eisenstein’s Criterion,
together with the number 1, are p equally spaced points on the circle of radius 1
around the origin. (The word “cyclotomic” means “circle cutting”.) As we will see
later, the fact that deg«(IQ) = p – 1 is the key to determining for which primes pit
is possible to construct a regular pgon using straightedge and compass
4. If F is a field then by Theorem 22.1 the quotient field of the domain F[X] is
F(X). The element X in F(X) is not algebraic over F, because if ao, ai, … , an E F
and anXn + an_IXn1 + . . . + alX + ao is 0 in F[X] then all the coefficients aj are 0 in
F because this is what it means for anXn + an_IXn1 + … + alX + ao to be 0 in F[X].
In general, any element that is not algebraic over F is called transcendental
over F.
230 Section 22. Extensions of Fields
We can now simplify our description of F(a) in the case where a is algebraic
over F.
THEOREM 22.3 Let F <;;;; E and let a E E be algebraic over F. Then every element
of F(a) has a unique representation in the formj(a), wherej(X) E F[X] is either 0
or of degree less than deg(a/F).
PROOF. Let K = {g(a) I g(X) E F[X]}. We first show that F(a) = K. Every element
of K is in F(a) since F(a) is a subfield of E containing Fu{a}. To establish the
reverse inclusion it suffices to show that K is a field (for then K is a subfield of E
containing Fu {a} ).
We have an evaluation homomorphism cpa : F[X] + K such that
cpa(g(X» = g(a)
for all g(X) E F[X] (Exercise 19.12). The kernel of cpa is {g(X) I g(a) = O},and this
is the principal ideal! = (irr(a/F). We thus have an isomorphism
qJa : F[X]/!+ K,
so to show that K is a field it suffices, by Theorem 17.7, to show that! is a
maximal ideal. But this is so by Theorem 20.2, since irr(a/F) is irreducible over F.
We now know that F(a) = K. We also know (Exercise 20.6) that each element
of Flx]/! is the coset j(X) represented by a unique polynomial that is either 0 or
has degree less than deg(a/F). Since 'Pa ([(X) = f(a), we see that each element of
K = F(a) isf(a) for a unique such polynomialf(.X).o
We have also proved
THEOREM 22.4 Suppose F <;;;; E and a E E is algebraic over F. Let! = (irr(a/F)
and denote each element! +j(X) of F[X]/I by j(X). Then there is an isomorphism
qJ" : F[X]/!+ F(a)
such that 'Pa ([(X» = f(a) for allf(X) E F[X]. In particular 'Pa (c) = c for all c E F
and 'Pa(X) = a.
In Section 20 we saw that if we start with an irreducible polynomialf(X) over
F then we can fmd a field extending F and containing a root of f(X) by forming
F[X]/([(X». Theorem 22.4 shows that this is the only way to adjoin a root for f(X)
to F.
COROLLARY 22.5 Suppose fiX) is irreducible over F and al and a2 are roots of
fiX) in extension fields El and E2 of F. Then there is an isomorphism
Section 22. Extensions of Fields 231
cp : F(al) + F(a2)
such that q:>(c) = c for all c E F and q:>(a l ) = az.
PROOF Note that irr(a/F) = irr(a2 /F) since both are obtained by multiplyingJ{X)
by the multiplicative inverse of its leading coefficient. Thus by Theorem 22.4 q:> =
((J,a 0 q:>a I has the desired properties. 0
2 1
Our purpose in considering F(a) is to use it to draw conclusions about the
element a. To do this we will need to use some ideas that will be familiar to you if
you have studied linear algebra.
In general, if F S; E then a subset S of E is said to be linearly independent
over F if whenever SI~ S2, “‘, Sn are finitely many distinct elements of Sand CI, Cz,
.•. , Cn are elements of F such that CIS) + C282 + … + CnSn = a in E, then CI = C2 = …
= c” = 0 in F. On the other hand, a subset S of E is said to span E over F if for
every element bEE there exist finitely many elements s], S2, … , s” in Sand Cj, cz,
… , Cn in F such that CISI + C2S2 + … + CnSn = b, i.e. b is a linear combination of SI,
sz, … , Sn with coefficients c, E F. S is called a basis for E over F if S spans E over
F and is linearly independent over F.
Example Theorem 22.3 shows that if a E E is algebraic over F and deg(a/F) = n
then the set {I, a, a2, •.• , anI} is a basis for F(a) over F.
As a specific example, consider Q <;:;::: Q(lJ5) <;:;::: R Since lJ5 is a root of X 3  5,
which is irreducible over Q by Eisenstein's Criterion, deg(lJ5/Q) = 3. Thus
{I, 51/3, 52/3} is a basis for QOJS) over Q.
THEOREM 22.6 Suppose F t;;;; E. If {SI' Sz, ... , sn} spans E over F then no subset
of E that is I inearly independent over F can have more than n elements.
PROOF. Suppose for a contradiction that {b l, b2, ... , bn+ I} is a set of n + I elements
linearly independent over F. Since {SI' S2, ... , sn} spans E over F there exist CI, Cz,
... , en in F such that
Since {bl, b2, ... , hn+d is linearly independent, b l * 0 and therefore at least one c,*
O. By renumbering if necessary we can assume CI * 0, and therefore we can
express SI as a linear combination of b], Sz, S3, ... , S". It follows that since {SI' 82, ... ,
Sn} spans E over F, so does {bl, S2, S3, ... , sn}. Therefore there are d], d2, ... , dn in F
such that
If d2 = ... = dn = 0 this contradicts the independence of {bl, b2, ... , btl+ l }, so by
renumbering if necessary we can assume that d2 * 0 and therefore we can express
232 Section 22. Extensions of Fields
S2 as a linear combination of bl, b20 83, ... , Sn. Thus since {bl, 82, ... , 8n } spans E over
F, so does {bl, b20 83, ... , sn}.
Continuing in this way, we conclude that {bl, ... , bn } spans E over F. In
particular, bn+1 is a linear combination of b l , ... , bn , and this contradicts the linear
independence of {b], ... , bn+I }. 0
COROLLARY 22.7 If F ~ E and S is a finite basis for E over F then every basis
for E over F has the same number of elements as S.
PROOF. If B is any basis for E over F then, since S spans E over F and B is
linearly independent over F, Theorem 22.6 shows that the number of elements in B
is no larger than the number in S. But then reversing the roles of Sand B shows
that the number of elements in S is no larger than the number in B. 0
DEFINITION If F ~ E, we say that E is a finite extension of F (or that E is finite
over F) if there exists a finite basis for E over F. The number of elements in such a
basis is called the degree (or dimension) of E over F. We denote this degree by
[E: FJ.
Note that [E : FJ is well defined by Corollary 22.7.
Examples
1. If E = F then {I} is a basis for E over F, so [E : F1 = 1. Conversely, if [E : F1
= I and {b} is a basis for E over F then since {b} spans E over F we can in
particular write 1 = cb for some c E F. Thus b = cl E F, so since every element of
E can be written in the form db with d E F, E = F.
2. If F c;;;, E and a E E is algebraic over F then F(a) is a fmite extension of F
and [F(a) : F1 = deg(a/F). For we have observed that {I, a, if, ... , an l } is a basis for
F(a) over F, where n = deg(a/F).
3. Consider F c;;;, F(X), where F(X) is the quotient field of F[X]. Since the subset
{1,X,X2,X3, ... } is linearly independent over F, theorem 22.6 shows that F(X) is
not a finite extension of F.
We noted previously that F(X) is not an algebraic extension of F, and therefore
the fact that F(X) is not finite over F also follows from
THEOREM 22.S Every finite extension is algebraic.
PROOF. Suppose E is a finite extension of F, with [E: F1 = n. We want to show
that every a E E is algebraic over F.
If d" = d' for two distinct positive integers m and k then a is a root of X'"  x",
Section 22. Extensions of Fields 233
so a is algebraic over F. If am1' d' for all m * k then {a, cl, ... , an+1} is a set of more
than n elements of E and cannot be linearly independent over F by Theorem 22.6.
Therefore there must exist c], C2, ... , Cn+l in F, not all zero, such that cIa + ... +
cn+lan+1 = O. So a is a root of cIX + ... + Cn+lXn+1• 0
COROLLARY 22.9. If F ~ E and a E E is algebraic over F then F(a) is an
algebraic extension of F.
PROOF. F(a) is finite over F, hence algebraic over F. 0
This result is not a triviality. For it asserts that every element of F(a)not just
ais algebraic over F.
Example Here's an example of an algebraic extension that is not a fmite extension.
For each integer n 2 I let an be the positive real2nth root of2. Let Qo = Q and
for n 2 I let Qn = Qnl(an). Then
Since Qn = Q(an) (why?) and an is a root of X 2"  2, Qn is algebraic over Q by
Corollary 22.9. Since every element of E is in some Qn this implies that E is
algebraic over Q.
But E is not fmite over Q. For since X 2n  2 is irreducible over Q by
Eisenstein's Criterion, [Q (an) : Q] = 2n. Thus for every n 2 I there is a subset of E
that has 2n elements and is linearly independent over Q. Therefore by Theorem 22.6
E cannot be finite over Q.
Before we can establish the most useful fact about finite extensions, we need
another result about linear independence.
THEOREM 22.10 Suppose F ~ E and there exists a positive integer n such that
every subset of E that is linearly independent over F has at most n elements. Then
any linearly independent melement subset {aJ, ... , am} of E can be extended to a
basis for E over F, i.e. there is a basis for E over F that contains {aI, ... , am}.
PROOF. If {aI, ... , am} spans E over Fthen it is already a basis for E over F. If {aJ,
... , am} does not span E over F there is some b l E E that is not a linear combination
of a" ... , am with coefficients from F. It follows that {aI, ... , am, bd is independent
over F, for if not we can write
234 Section 22. Extensions of Fields
with CI, ... , em and din F, not all O. Since {aI, .. ,' a"J is independent we must have
d *" 0, and thus we can solve for b l and contradict the fact that b l is not a linear
combination of aI, , .. , am.
If {a" ... , am, btl spans E over F then it is a basis for E over F. If {a" ... , am,
b l } does not span E over F then we can repeat the above argument to show that
there is b2 E E such that {aI, ... , am, b" b2} is independent over F.
If we keep repeating this reasoning then, since no subset of E linearly
independent over F can have more than n elements, we must arrive at some bl, b2"
"', bk such that {aI, ... , am" b l , "', bd is a basis for E over F. 0
The most frequently used fact about finite extensions is the fact that "degrees
multiply",
THEOREM 22.11 Suppose F ~ E ~ K. Then K is finite over F iff E is finite over
F and K is finite over E. Furthermore, if [E : F] = m and [K: E] = n then [K: F] =
mn, and if {a" ... , am} is a basis for E over F and {bl, ... , bn } is a basis for Kover E
thenS= {a;bj [I ::::i::::m, I ::::j::::n} isabasisforKoverF.
PROOF Suppose first that K is finite over F and B is a basis for Kover F having,
say, r elements. Then B spans Kover F and therefore B spans Kover E. Applying
Theorem 22.6 twice, we see that no subset of E that is independent over F can have
more than r elements, and that no subset of K that is independent over E can have
more than r elements. Applying Theorem 22.10 twice, we see that the independent
subset {I} of E can be extended to a basis for E over F with at most r elements,
and that the independent subset {I} of K can be extended to a basis for Kover E
with at most r elements. So E is finite over F and K is finite over E.
Now assume that E is finite over F and K is finite over E. Let {aI, ... , am}, {b"
.. " bn } and S be as given in the statement of the theorem. All our claims will be
established if we can show that S is a basis for Kover F.
For any v E Kthere exist d" .. " dn E E such that
Each di can be written as a linear combination of aJ, ... , am with coefficients in F,
and therefore v can be written as a linear combination of elements of S with
coefficients in F. Thus S spans Kover F.
To see that S is linearly independent over F, suppose we have
for some cij's in F, i.e.
Section 22. Extensions of Fields 235
For every bJ the coefficient c),a)+ ... +c""am on b, is in E, so since {b), .'" bn } is
independent over E and bb ... , bn are distinct we conclude that c),a)+ ... + Cn"am =
o for each}, 1 'S} 'S n. By the independence of {a), ... , am} over F we then conclude
that all cij = 0.0
COROLLARY 22.12 If F c;;:; E and [E : Fl = n then, for every a E E, deg(alF)
divides n.
PROOF. We have F ~ F(a) ~ E, so by Theorem 22.11 [F(a) : F] divides n, i.e.
deg(a/F) divides n.D
Notation. If F c;;:; E and a), ... , an E E then F(a), ... , an) denotes the intersection of all
subfields of E containing F u {a), ... , an}.
Note that for elements a) and az, F(a), az) can also be described as F(a)(az),
the extension obtained by adjoining az to F(a). Equally well, F(ab az) = F(az)(a).
Examples
1. Since irr( Ji IrQ) = XZ  2, deg( Ji IrQ) = 2 and every element of Q( Ji) can
be written uniquely in the form a + b Ji , with a, b E Q. Thus .[3 'l Q( Ji),
because if (a + b Ji )2 = 3 then
If ab "# 0 we can solve for J2 and conclude that J2 E Q, contradicting
deg( J2 IQ) = 2. So a = 0 or b = 0, and thus 2b2 = 3 or a2 = 3. But then J3i2 E Q
or .J3 E Q, again a contradiction
Since .[3 'l rQ( Ji) the polynomial XZ 3 in Q( Ji )[X] is irreducible over
Q( Ji) (it has degree 2 and has no root in Q( Ji », so deg( .[3 IQ( Ji» = 2 and
{1,.[3 } is a basis for E = Q( Ji , .[3 ) over Q( J2). By Theorem 22.11
[E: rQ] = [Q( Ji) : Q] . [E : rQ( Ji )]
and this equals
deg( Ji IQ) . deg( .[3 IQ( Ji» = 2 . 2 = 4.
Again by Theorem 22.11, {I' I, I . .[3, Ji . 1, Ji . .[3} = {I, .J3, Ji,.J6} is a
basis for E over Q.
It is interesting to note that E can be obtained by adjoining one element to Q.
For note that since E is a field containing the number a = Ji + .[3 , Q(a) c;;:; E. On
the other hand, E c;;:; Q(a) because lQ!(a) is a field containing both Ji and .J3 . To
see this, note that
236 Section 22. Extensions of Fields
(Ii + fj/=2+3 +216 =5+216,
and thus 16 E
so since E ~ JR, 1>3(X) has no root in E. Since 1>3(X) has degree 2, 1>3(X) is
therefore irreducible over E, so since
[K :1(£] = [E : QJ][K : E],
[K :
22.10 a) Prove the trigonometric identity cos(3B) = 4cos3 B3 cos B.
b) Use part (a) to show that deg(cos(7V9)/Q) = 3.
22.11 Let p E Z+ be prime and let , 2 = cos(2111p2) + i sin(2111p2). Then , 2 is a root of the
p p
cyclotomic polynomial
Prove that 4> 2 (X) is irreducible over Q and that deg(‘ 2 IQ) = p(PI).
p p
22.12 Suppose £ is an extension field of Zs and £ has exactly 78125 elements.
Find deg(alZs) for every a E £ – Zs.
22.13 Suppose £ is an extension field of Z7 and dEE – ~. Find deg(diZ7) if
a) d3 = 2
b) If = 2
22.14 Prove that ‘5 ~ Q( ‘7)’
22.15 Suppose £ is a finite extension of F and [£: F1 is prime. Prove that F(a) = £ for every
aE£F.
Section 22. Extensions of Fields 239
22.16 Suppose E is an extension field ofQ and E has elements a and b such that a8 + 4a7
+2a l – 6a + 2 = 0 and b9 + 3b6 + 12b2 – 6b + 15 = O. Prove that Q(a) n Q(b) = Q.
22.17 Suppose F c;;;; E, a E E is algebraic over F and deg(a/F) is odd. Prove that F(a) =
F(a2).
22.18 Suppose F c;;;; E and let K = {a EEl a is algebraic over F}.
a) Prove that K is a subfield of E. (K is called the algebraic closure of F in E.)
b) Prove that if bEE and b is algebraic over K then b E K.
22.19 Suppose F c;;;; E and c and d are elements of E that are algebraic over F, such that
deg(clF) and deg(d/F) are relatively prime.
a) Prove that [F(c, d) : F] = deg(clF) . deg(d/F).
b) Prove that irr(d/F) is irreducible over F(c).
22.20 Suppose F c;;;; E and a and b are elements of E that are algebraic over F. Prove that
irr(a/F) is irreducible over F(b) iff irr(b/F) is irreducible over F(a).
22.21 a) If Pn is the nth positive prime, show that for every n
b) Prove that the infinite set {2112, 2113, 21.6, 2117, 21111, … } is linearly independent
over Q.
22.22 Suppose E is an algebraic extension of F and D is a subring of E containing F. Prove
that D is a field.
22.23 Prove that if F c;;;; E and a and b are elements of E that are not both algebraic over F
then ab and a + b cannot both be algebraic over F.
22.24 Suppose F c;;;; E and assume there are ai, … , an in E such that E = F(al, … , a,J Prove
that the following three statements are all equivalent to each other:
i ) E is finite over F.
ii) E is algebraic over F.
iii) Each of ai, … , all is algebraic over F.
22.25 Let E be an algebraic extension of F and let cp : E > E be a onetoone ring
homomorphism such that cp(c) = c for all c E F. Prove that cp is onto.
22.26 Suppose F c;;;; E and {Slo .. ” SII} is a finite subset of E that spans E over F. Prove that
there exists a subset of {SI’ … , SII} that is a basis for E over F.
SECTION 23
CONSTRUCTIONS WITH
STRAIGHTEDGE AND
COMPASS
The geometers of ancient Greece considered the problem of performing
geometric constructions using only a straightedge and compass. (A straightedge is
a ruler with no markings on it. Thus the straightedge can be used to draw the line
through two previously constructed points, but not to measure or mark off
distances. The compass can be used to draw the circle whose center is a previously
constructed point and whose radius is the distance between two previously
constructed points.) The Greeks succeeded, for example, in constructing regular 3
gons (equilateral triangles), 4gons (squares), pentagons and hexagons, and they
were familiar with the method for bisecting a given angle.
When it came to certain other problems, however, the Greeks were baffled.
They were unable to solve any of the following. (We use the word “construct” to
mean “construct with straightedge and compass”.)
Problem I. Given the side of a cube, construct the side of a larger cube with twice
the volume.
Problem II. Give a general method for trisecting angles. That is, show how,
starting with a given angle, to construct an angle onethird the size.
Problem Ill. Construct a regular heptagon (7gon).
The Greeks worked on these problems over 2000 years ago, and in the
intervening centuries many people have sought in vain for solutions. Even today
there are people who still try, unaware that the task is utterly hopeless. For it was
proved in the nineteenth century that the constructions required by these problems
cannot be carried out using only straightedge and compass. Our purpose in this
section is to use what we know about field extensions to prove that the
constructions are impossible.
240
Section 23. Constructions with Straightedge and Compass 241
We shall be working with a certain subfield C, of C, called the field of
constructible numbers. To describe this subfield, we identify C with the set of
points in the xyplane. The field IR of real numbers is identified with the set of
points on the xaxis.
We will define the set C, of constructible (or, more precisely, constructible
from 0 and 1) numbers to consist of all elements of C that can be obtained by
starting with 0 and 1 and using the following methods (A), (8), (C) any finite
number of times:
(A) Let LI be the line determined by two points that we already know are in C,
and let L2 be another such line that is not parallel to L I • Take the point of
intersection of LI and L2.
(8) Let L be the line determined by two points in Cn and let C be a circle whose
center is in C, and whose radius is the distance between some two points of C,.
Take the points of intersection of Land C.
(C) Let CI and C2 be two (distinct) circles satisfying the requirements on the circle
C in (8). Take the points of intersection of C I and C2•
DEFINITION Let z E Co Then z E C,. if and only if z can be obtained by starting
with 0 and 1 and applying methods (A), (B), (C) some finite number oftimes.
To get an idea of how this definition generates points, observe that the points
marked in each of the following diagrams are in C,o :
2
The points in C, correspond in a clear sense to the points we want to call
“constructible”, because we get them by starting with 0 and I and using the
straightedge (for lines) and compass (for circles).
THEOREM 23.1 C, is a subfield ofe.
PROOF. We first show that C, n IR is a subfield ofe. To show this it will suffice,
since 0 and I are in C,o , to establish (i) and (ii):
242 Section 23. Constructions with Straightedge and Compass
(i) If a, b E ICc n JR and b * ° then a – b E IC, n lR.
(ii) If a, bE ICc n JR and b * ° then alb E ICc n JR.
To establish (i) we use method (B), with the xaxis as the line L (the xaxis is
detennined by 0, I E ICc , so we can use it) and the circle with. center at a and
radius Ibl as C. (This circle is legitimate since its radius is the distance between 0,
b E ICc .) One of the two points of intersection of Land C is ab.
To establish (ii) it suffices by (i) to handle the case where both a, b > O.
Notice that there are points of ICc on the yaxis, other than the origin. For example,
1, 2 E ICc by (i), so if we draw the circles of radius 2 centered at ±1, their points
of intersection (± J3i ) are both in ICc . Thus we can use the yaxis as the line in
method (B), and in particular if we intersect the yaxis first with the circle of radius
a centered at the origin, and then with the circle of radius b centered at ai, we see
that the points ai and (a + b)i are both in ICc . By a similar argument we see that the
point 1 + ai is in ICc . If we draw the line L through (a + b)i and 1 + ai, then the
point 1 + c where L intersects the xaxis is in ICc:
Hence, by (i), C E ICc. But, using similar triangles,
a+b a
–=
I +c c
and thus c = alb, so alb E ICc nlR.
We now prove that ICc itself is a subfield of C. We will use the fact that if z =
a + bi E IC then z E ICc iff both a and b are in ICc . (See Exercise 23.1.) Using this
fact we see that if z = a + bi arid w = c + di are in ICc then a, b, c, d E ICc , so a – c
and’b – d are in ICc since ICc n JR is a field, and thus z – w = (a – c) + (b – d)i is
in ICc . Finally, if w :f 0 then
Section 23. Constructions with Straightedge and Compass 243
z/w= a+bi = (a+bi)(cdi) =(aC+bd)+(bCad)i.
c+di (c+di)(cdi) C2 +d2 C2 +d2
Since C, n lR is a field, Exercise 23.1 implies that Z/W E CC, . This concludes the
proof. 0
It follows from Theorem 23.1 that Ql, the prime subfield of CC, is contained in
CC, . The key to the applications of CC, to the construction problems is the following
characterization of the elements of CC” which enables us to apply our results on
field extensions.
THEOREM 23.2 Let z E cc. Then z E CC, iff there exist Z], •.. , Zn E CC such that
ZT E Ql, zJ E Ql(z], … , Zj_]) for 2 ~j ~ n, and Z E Ql(z], … , zn).
PROOF. Assume first that there exist such numbers Z], … , Zn. We want to show
that Z E CC,. Ifwe knew the truth of the implication
then we could argue as follows: First, Z] E CC, because ZT E Ql; hence Ql(z]) ~ CCc
since CC, is a field. Then Z2 E CC,. since z~ E Ql(z]) ~ CCc, so Ql(z]’ Z2) ~ CC, since
CCc is a field. Continuing in this way, we could conclude by induction that Ql(zj, … ,
Zn) ~ CCc and hence z E CC, .
To verify the above implication we first deal with the case where w2 is a
positive real number and thus w E lR. We can assume w > 0 since CC, is a field.
The following diagram indicates how to construct w from w2:
The length of the vertical segment is w, as we see by using the fact that the two
small triangles are similar.
For the general case we use the fact that one square root of c + di is given by
244 Section 23. Constructions with Straightedge and Compass
~ +c + ).Jc2 +d2c
I
2 2
[23.1 )
(we take the “+” if d > 0 and the “” if d < 0). It follows from the special case we
have just dealt with and Exercise 23.1 that if c + di E C, then both square roots of
c + di are in C,. This concludes the first halfofthe proof.
For the second half it will be convenient to call a sequence z" ... , Zn of
complex numbers a square root sequence if zI E Q and z} E Q(ZI, ... , Zjl) for 2 :s
j :s n. We let S be the set of complex numbers Z for which there exists a square root
sequence z" "', Zn such that Z E Q(ZI, ... , Zn). We wantto show that C(" ~ S. We
will frequently use the easily verified fact that if z" ... , Zn and w" ... , Wm are square
root sequences then z" ... , Zm W" ... , Wm is a square root sequence.
To prove that C" <;:;; S it will suffice to establish
if Z = a + bi E C, then a E Sand b E S, [23.2)
for then if a E Q (Z], .. " zn) and b E Q (WI, ... , wm ) for square root sequences ZI, ... ,
Zn and WI, ... , Wm we have a + bi E Q (zJ, .. " Z'" WI, .. " W m, i),
Notice that if a line L is determined by two points that have all their
coordinates in Q(z" .. " zn) then L has an equation of the form ax + by + c = 0 with
a, b, cE Q(ZI, ... , zn). Likewise, if a circle C has center WI and radius the distance
between W2 and W3, and all the coordinates of WI, Wz, W3 are in Q(z" .. " zn), then C
has an equation of the form x2 + l + ax + by + c = 0 with a, b, c E Q(z" ... , zm).
We now prove [23.2] by induction on the number of applications of methods
(A), (8), and (C) used in constructing z. Ifno applications are used then Z = 0 or 1,
and 0, I E Q <;:;; S.
Assume now that Z is obtained by applying method (A) to two lines
determined by points which, by inductive hypothesis, have all their coordinates in
Q(z" .. " zn) for some square root sequence ZI, ... , Zn. Then these lines have
equations
ax + by + c = 0 and a x + by + c' = 0
with all the coefficients in Q(ZI, ... , Zn). The coordinates of Z are obtained by
solving these two equations simultaneously and hence are in Q(z" ... , zn).
If Z is obtained by applying method (8) then the coordinates of z are found by
solving simultaneously two equations
ax + by + c = 0 and x2 + l + a x + by + c '= 0
with all coefficients in Q(ZI, ... , Zn) for some square root sequence Zl, ... , Zn' Finding
the coordinates of Z will require us to use a square root Zn+l of an element of Q(ZI,
"', zn). Then Zl, "', Zn, Zn+l is a square root sequence and the coordinates of Z are in
Q(ZI, ... , Zm Zn+l)'
Section 23. Constructions with Straightedge and Compass 245
Finally, if z is obtained by using method (C) then the coordinates of z are
found by solving simultaneously two equations
x 2 + / + ax + by + c = 0 and x2 + / + a x + by + c' = 0
with all coefficients in Q (zl' ... , zn) for some square root sequence ZI' ... , Zw Solving
these equations is equivalent to solving
(a  a')x + (b  b')y + (c  c') = 0 and x2 + y + a'x + b'y + c' = 0,
so we can conclude the argument as we did for method (B). 0
COROLLARY 23.3 If Z E C, then z is algebraic over Q and deg(zIQ) = 2' for
some nonnegative integer r.
PROOF. If z E Ce then by Theorem 23.2 we have z E Q(z\, ... , zn) for some square
root sequence z\, ... , Zn. Then each extension in the chain
is finite, of degree 1 or 2. Hence [Q(ZI, ... , Zn) : Q] = 2m for some m :S 11. By
Corollary 22.12, deg(zlQ) = 2' for some r :S m. 0
It is worth mentioning that the converse of Corollary 23.2 is false. There exist
z's in C such that deg(zIQ) = 4 but z (I' Ce. (See Exercise 24.28.)
Applications
1. (Doubling the Cube) If we could construct from the side of a cube the side
of a larger cube with twice the volume, then from 1 we could construct iff.. Since
1 E C, this would imply iff. E C e. But deg( iff. IQ) = 3 and therefore by Corollary
23.3 iff. ~ Ce. Thus there is no construction for doubling cubes.
The problem of doubling the cube is sometimes called the Delian problem,
because of a legend concerning its origin. As the story goes, there was a terrible
plague in Athens about 2500 years ago, and the Athenians consulted the oracle at
Delos about what steps they could take to end it. The oracle told them to double the
cubical altar to Apollo. They proceeded to double its side, which of course
multiplied the volume by 8. The plague continued, and it was surmised that
perhaps the oracle had meant that they should double the volume. At this point it
became rather a pressing matter to construct iff..
2. (Trisecting Angles) In order to demonstrate the nonexistence of a general
construction for trisecting angles with straightedge and compass, it suffices to
246 Section 23. Constructions with Straightedge and Compass
exhibit one specific angle that cannot be so trisected. We use an angle of n!3
radians.
It is easy to see that the point (6 = cos(n!3) + isin(n/3) is in C e. If it were
possible to trisect an angle of nl3 radians with straightedge and compass then the
point (18 = cos(n/9)+isin(n/9) would be in Cn and so would cos(n/9). But by
Exercise 22.10 deg(cos(n/9)/Q) = 3, so cos(n/9) ~ C e.
3. (Constructing a Regular Heptagon) We will prove the general result that if P
is prime and P ~ 3 then the constructibility of a regular pgon implies that p = 2'+ I
for some r ~ I. From this it is immediate that it is impossible to construct a 7gon.
The constructibility of a regular pgon implies that (p == cos(2n/p) + i sin(2n/p)
is in C r (Exercise 23.5). But irr«(/Q) ==
ngon is constructible. Then for any ri ~ 2, a regular q/ gon is constructible.
(Why?) Thus
is in Ce so deg( Sq,’ /Q) = 2′ for some r ~ O. But by Exercise 22.11, this says that
q;(q; – 1) = 2′, and therefore q; = 2. Thus for all qi ~ 3 we must have rj = 1 and Sq,
E !CC> so qi must be a Fermat prime, as above. We have proved
THEOREM 23.4 If n ~ 3 and a regular ngon can be constructed with straightedge
and compass then n = 2mPI . Pk where m ~ 0, k ~ 0 and PI, … , Pk are distinct
Fermat primes.
The conditions on n given in Theorem 23.4 are also sufficient for the
constructibility of a regular ngon. (We will prove this in Section 25, using Galois
theory.) The sufficiency was first proved by Gauss towards the end of his now
classic book Disquisitiones Arithmeticae (1801). In this work Gauss also claimed
Section 23. Constructions with Straightedge and Compass 247
that he had a proof of the result in Theorem 23.4, but he never published a proof.
The first published proof was given by Pierre Wantzel in a paper that appeared in
1837. In this paper Wantzel also gave the first proofs of the impossibility of
doubling the cube or trisecting angles with straightedge and compass.
The ancient Greeks also considered constructions with tools other than
straightedge and compass. For example, they considered the possibility of allowing
the distance between any two constructed points to be marked off on the
straightedge. With this added ability it is possible to double the cube, trisect angles,
and construct a heptagon (see Exercises 23.1223.14).
EXERCISES
23.1 Let z = a + bi E IC. Show that z E IC,. iff both a and b are in ICc,
23.2 For each of the following values of n determine whether a regular ngon is
constructible with straightedge and compass: n = II; 13; 15; 18; 19; 112; 340.
23.3 Give an example of a nonzero angle that can be trisected using straightedge and
compass.
23.4 Show how to construct a regular hexagon using straightedge and compass.
23.5 Show that if a regular pgon is constructible anywhere in the xyplane then (p E ICc.
23.6 Prove that a regular decagon (IOgon) is constructible, as follows. Consider the triangle
with vertices at 0, 1, and (10 = cos(n!S) + isin(n!S). Find the length of the side
connecting 1 and (10 by bisecting the angle of the triangle at (10 and using similar
triangles.
23.7 The result of Exercise 23.6 implies that a regular pentagon is constructible. Prove this
in another way by starting from the fact that if B = 2n15 then cos(2B) = cos(3B).
23.8 (Squaring the Circle) Lindemann proved in 1882 that l( is transcendental over Q.
Assuming this result, show that there is no construction with straightedge and compass
that will produce from the radius of a given circle the side of a square that has the same
area as that circle.
23.9 Prove that if r E Z+ and p = 2′ + I is prime then r = 211 for some nonnegative integer n.
23.10 Express cos 5B in terms of cos B and use your result to show that there is no general
method for cutting angles into tive equal pieces with straightedge and compass.
23.11 We have seen that an angle of 20° is not constructible with straightedge and compass.
a) Prove that neither an angle of 1 ° nor an angle of 2° is constructible.
b) Prove that if n E Z+ then an angle of nO is constructible iff 31n.
248 Section 23. Constructions with Straightedge and Compass
23.12 The following construction for trisecting angles using compass and marked
straightedge is due to Archimedes.
a) Consider an acute angle AOB, where A is the point 1, 0 is the origin, and B is a
point in the first quadrant one unit from O. Construct the circle of radius 1 with
center at O. Mark two points one unit apart on the straightedge. Then place
the straightedge so that its edge passes through B and one of the two marked
points is on the circle while the other is on the negative xaxis at some point C
outside the circle. Prove that angle ACB is onethird of angle
AOB.
b) Explain why being able to trisect acute angles enables us to trisect all
angles.
23.13 The following construction for doubling the cube using compass and marked
straightedge is due to Isaac Newton.
Construct a line L with negative slope that passes through the origin and makes an
angle of 30° with the positive xaxis. Mark two points one unit apart on the
straightedge. Then place the straightedge so that its edge passes through the point i
and one of the two marked points is on L while the other is at a point A on the
positive xaxis. Prove that the distance from A to i is ifi.
23.14 The following construction of a regular heptagon using compass and marked
straightedge was published by Crockett Johnson in the Mathematical Gazette,
volume 59 (1975), pages 1721.
Construct a square ABCD with sides of length I. Then construct the circle centered at
B and passing through D, and construct the perpendicular bisector of AB at M. Mark
two points E and F one unit apart on the straightedge, and place the straightedge so
that E lies on the perpendicular bisector, F lies on the circle, and the straightedge
passes through A, with F between A and E.
Let x denote the length of BE and let 8 denote angle AEM. Note that 2x sin 8 = I and
apply the law of cosines to triangle BEF to show that sin 38= cos 48. Conclude that 8
= n/14 and thus we can construct the angle 4(n/14) = 2m’7.
SECTION 24
NORMAL AND
SEPARABLE EXTENSIONS
We will now consider certain special kinds of field extensions that are useful
in studying the roots of polynomials.
If F ~ E andfiX) E F[X] can be written as C(Xal)(Xa2) … (Xan) for some
c and some a;s in E (not necessarily distinct) then we say thatfiX) splits over E
and that E splits fiX). Note that then c E F, because c is the leading coefficient of
f{X) E F[X]. It is apparent that if K is any extension of E then f{X) also splits over
K, so there are many different extensions of F over which f{X) splits. We wish to
single out a standard extension that splits f{X), so that we can discuss f{X) in terms
of this one definite extension. The thing to do is to opt for economy and use a
minimal extension that splitsf{X).
DEFINITION Let F <;;; E, and letf{X) E F[X] be nonconstant. We say that E is a
splitting field for f(X) over F if f{X) splits over E but does not split over any
proper subfield of E containing F.
THEOREM 24.1 Supposef{X) E F[X] has degree n ~ 1 and E is an extension of F
such that in E[X] we have
Then E is a splitting field for f{X) over F if and only if E = F(al, ... , an).
PROOF. If E is a splitting field for f{X) over F then it follows from the fact that
f{X) splits over F(aJ, ... , an) that F(a1' ... , an) cannot be a proper subfield of E.
Hence E = F(aJ, ... , an), Conversely, if E = F(a1' .. " an) then to show that E is a
splitting field for f{X) over F we must show that if K is a subfield of E containing F
andj(X) splits over K, say asf{X) = c(X  bl) ... (X  bn), then every a, is in K (for
then K = E because E = F(al' ... , an». But since 0 = fiai) = c(ai  bl) ... (ai  bn),
some ai  bj = 0 and thus ai = bj E K. 0
249
250 Section 24. Normal and Separable Extensions
It follows from Corollary 20.4 and Theorem 24.1 that for any j(X) E F[X]
there exists a splitting field for j(X) over F, and every splitting field for j(X) over F
is a finite extension of F. But we can say a little more.
THEOREM 24.2 Suppose j(X) E F[X] has degree n :::: l. Then there exists a
splitting field for j(X) over F whose degree over F divides n!.
PROOF. We argue by induction on n. If n = I then F itself is a splitting field for
j(X) over F, of degree lover F.
Assuming the result for polynomials of degree less than n, we consider two
cases. First, if j(X) is irreducible over F of degree n then we consider an extension
F(aj) of F, where aj is a root ofj(X). In F(aj)[X] we can writej(X) = (X  aj)g(X)
with deg(g(X» = n  l. Then by the inductive hypothesis there is a splitting field E
for g(X) over F(aj) such that [E: F(aj)] divides (n  I)L By Theorem 24.1, E =
F(aj)(a2' ... , an) where a2, ... , an are roots of g(X). Sincej(X) splits over E and E =
F(aj, ... , an), E is a splitting field for j(X) over F. Since [F(aj) : F] = n becausej(X)
is irreducible over F, [E : F] = n . [E : F(aj)], and this divides n! because [E :
F(aj)] divides (n  I)L
Second, if j(X) factors in F[X] as j(X) = g(X)h(X), where deg(g(X» = m < n
and deg(h(X» = k < n, then by inductive hypothesis there is a splitting field K =
F(aj, ... , am) for g(X) over F such that [K : F] divides mL Again by inductive
hypothesis there is a splitting field E = K(b j, ••• , bk) for heX) over K such that [E :
K] divides kL Then E = F(aj, ... , am, bj, ... , bk) is a splitting field for j(X) over F,
and [E : F] = [K : F][E : K], which divides m!k!, which divides nL (Note that
n!lm!k! = n!lm!(n  m)! is an integer since it is the number of ways of choosing m
elements from an nelement set.) 0
Examples
1. C is a splitting field for X2 + lover IR, since X2 + I = (X i)(X + i) and C =
JR.(i). We have [C : JR.] = 2.
2. Ifj(X) = X 2 then the roots ofj(X) in Care ifi. (3 ifi, and (~ ifi. so E =
'O( ifi. (3 ifi, (~ ifi) = 'O( ifi. (3) is a splitting field for X  2 over 'O. We saw
in the third example following Corollary 22.12 that [E : '0] = 6.
3. Continuing the preceding example, we note that since ifi E R R«(3) is a
splitting field for X3 2 over R. Since (3 is a root of X2 +X +1, which is irreducible
over R we have [JR.«(3) : R] = 2. R«(3) is also a splitting field for X2 + X + lover
R
4. Ifj(X) = ~  5X2 + 6 = (X2  2)(X2 3) then E = '0 (ji. J3) is a splitting
field for j(X) over 'O. We saw in the first example following Corollary 22.12 that
Section 24. Normal and Separable E:l$tensions 251
[E : Q] = 4. We also saw that E = Q( ..fi + J3) and that irr( ( ..fi + J3 )/ Q) = X'
10X2 + 1. The other roots of this polynomial in E are (..fi + J3) and
±( ..fi  J3 ), and E is a splitting field for X'1 OX2+ lover Q.
5. If n E Z+ and (n = cos(2n!n) + isin(2n!n) then the roots of X'  I in C are ('"
s;,s~, ... ,S:l, I so Q «(n) is a splitting field for X'  I over Q. Ifn is a prime p then,
as we have seen, the splitting field Q«(p) has degree p  lover Q.
We have indicated that we want a splitting field for j{X) over F to serve as a
"standard" extension of Fthat splitsj{X). What we want to show, then, is that if E
and K are both splitting fields for j{X) over F then E and K are essentiaIIy the
"same" extension of F, in the sense that there is an isomorphism ep : E > K such
that ep(e) = e for every e E F.
DEFINITION Let E,K be extension fields of F. E and K are isomorphic over F if
there exists an isomorphism ep from E onto K such that ep(e) = e for every e E F.
Such an isomorphism is caIIed an isomorphism of E and Kover F.
THEOREM 24.3 Let F be a field, j{X) E F[X]. Let E and K be splitting fields for
j{X) over F. Then E and K are isomorphic over F.
The idea of the proof is to proceed by induction on deg(f{X). We take an
irreducible factor g(X) of j{X) in F[X] and let a E E, b E K be roots of g(X). We
show that F(a) and F(b) are isomorphic over F via an isomorphism that sends a to
b, and then we want to apply the inductive hypothesis toj{X)/(X a) E F(a)[X] and
the corresponding polynomialj{X)/(X b) E F(b) [X]. Unfortunately we now have
two polynomials, with coefficients in two different fields. But the fields are
isomorphic, and under this isomorphism the two polynomials match up. To prove
Theorem 24.3, therefore, we first state it in a more general form.
THEOREM 24.4 Let F, F* be fields and let ep : F > F* be an isomorphism. Let
j{X) E F[X] be nonconstant and let f”(X) E F* [X] be the polynomial obtained by
replacing each coefficient e; ofj{X) by ep(e,). Let E be a splitting field for j{X) over
F and let E* be a splitting field for f*(X) over F*. Then there exists an
isomorphism ep+ : E > E* such that ep+(e) = ep(e) for every e E F.
We express the result of this theorem by saying that ep can be extended to an
isomorphism ep+ : E > E*. Note that Theorem 24.3 follows immediately from
Theorem 24.4, by starting with the identity isomorphism from F onto F.
We aim to prove Theorem 24.4 by induction on deg(f{X). We need the
foIIowing result to carry out the induction step.
LEMMA 24.5 Let F, F*, ep, f{X),and f*(X) be as in Theorem 24.4 and assume that
fiX) is irreducible in F[X]. Then if a and b are roots off{X) andf*(X) in extensions
252 Section 24. Normal and Separable Extensions
of F and F*, respectively, there is an isomorphism CPl : F(a) ~ F*(b) such that
cpl(a) = band cpl(e) = cp(e) for every e E F.
PROOF. We use a generalized version of the proof of Corollary 22.5.
First, cP extends in an obvious way to an isomorphism from F[X] onto F*[X],
obtained by mapping every g(X) E F[X] to g*(X). If we let I be the principal ideal
(j(X) in F[X] and let 1* = (j*(X) in F*[X] then I is mapped onto 1* so we have an
isomorphism
If! : F[X]II ~ F*[X]II*
such that for every g(X) in F[X] the coset g(X) of g(X) in F[X]II is mapped to the
coset of g*(X) in F*[X]II*, which we will denote by [g*(X)].
Since.f(X) is irreducible over F, r(X) is irreducible over F* . . Note that I =
(irr(a/F) and 1*= (irr(b/F*». If we let CfJa : F[X]/I ~ F(a) and CfJb : F*[X]/I* ~
F*(b) be the isomorphisms provided by Theorem 22.4, and let CfJl = CfJb 0 lI’ 0 CfJa1,
we see that CfJl : F(a) ~ F*(b) is an an isomorphism. We have
and for every e E F
D
Another version of this proof, using Theorem 22.3 instead of Theorem 22.4, is
requested in Exercise 24.14.
PROOF OF THEOREM 24.4. We proceed by induction on the degree offix). If
deg(j(X) = 1 there is nothing to prove, since E = F and E* = F*. So suppose
deg(j(X) = n and the theorem is true for all fields F, F* and all g(X) E F[X] of
degree less than n. Let heX) be an irreducible factor offiX) in F[X] and let h*(X) be
the corresponding irreducible factor of r(X) in F*[X]. Since fiX) splits over E
there is a root a of heX) in E (see Exercise 24.1). Likewise, there is a root b of
h*(X) in E*. By Lemma 24.5 there is an isomorphiam CPI : F(a) ~ F*(b) that
extends cP and maps a to b.
We now conclude the proof by applying the inductive hypothesis to the
isomorphism CPl. In order to do so, we write fiX) = (Xa)g(X) in F(a) [X]. This
yields r(X) = (X – b)g*(X), where g*(X) is obtained by applying CPI to all the
coefficients of g(X). If we can show that E is a splitting field for g(X) over F(a) and
E* is a splitting field for g*(X) over F*(b) then the inductive hypothesis will imply
that CPI can be extended to an isomorphism cP+ : E ~ E*, and cP+ will extend cP
because CPI does.
Section 24. Normal and Separable Extensions 253
Now g(X) splits over E, because f{X) does (Exercise 24.1). Furthermore, if
g(X) were to split over some proper subfield of E containing F(a) then so would
f{X), and this would contradict the fact that E is a splitting field for f{X) over F.
Thus E is a splitting field for g(X) over F(a), and similarly E* is a splitting field for
g*(X) over F*(b). This completes the proof. D
With this theorem behind us, we can speak of the splitting field of a
polynomial over a given field.
It is worthwhile to record the following consequences of Theorem 24.4 while
the ideas are still fresh.
COROLLARY 24.6 Let F be a field, f{X) E F[X], and let E be the splitting field
for f{X) over F. Let K,K* be subfields of E containing F, and let qJ : K + K* be an
isomorphism over F. Then qJ extends to an automorphism of E over F.
PROOF. E is the splitting field for f{X) over K, and E is also the splitting field for
f{X) over K*. Since qJ fixes every element of F,f{X) is the polynomial obtained by
applying qJ to all the coefficients of fiX). Thus, by Theorem 24.4, qJ extends to an
isomorphism from E onto E. D
DE}’INITION If F ~ E then two elements a and b of E are said to be conjugate
oveJ’ F if a and b are algebraic over F and irr(a/F) =irr(b/F).
COROLLARY 24.7 Let F, f{X), and E be as in the previous corollary and let a, b
E E. Then a and b are conjugate over F iff there exists an automorphism qJ of E
over F such that qJ(a) = b.
PROOF. Since E is finite over F, a and b are algebraic over F. If there exists an
automorphism qJ of E over F such that qJ(a) = b, then denoting irr(alF) by g(X) and
applying qJ to both sides of the equation g(a) = 0 yields g(b) = O. Thus g(X) is
irr(b/F).
Conversely, if a and b are conjugate over F then, applying Lemma 24.5 to the
irreducible polynomial of a and b over F, we see that the identity isomorphism
from F onto itself extends to an isomorphism from F(a) onto F(b) that maps a to b.
By Corollary 24.6 this isomorphism extends to an automorphism of E over F. D
Example The roots of X’ – 2 in C are ±ifi and ±i ifi, so the splitting field for X’
– 2 over Ql is E = Ql( ifi, i), and [E : Ql] = 4 . 2 = 8 since irr(iIQ( ifi» = X2 + 1.
If b is any of the four roots of X – 2 we know by Lemma 24.5 that there is an
isomorphism qJl : Ql( ifi) + Ql(b) over Ql mapping ifi to b. Since irr(±i/Ql( ifi»
= X2 + I and applying qJl to the coefficients of X2 + 1 leaves the polynomial
unchanged, Lemma 24.5 tells us that qJl can be extended to an automorphism of E
over Ql mapping i to either of the roots ±i of X2 + 1. So we have obtained eight
automorphisms of E over Ql. Since any automorphism of E over Ql is completely
254 Section 24. Normal and Separable Extensions
determined by its effects on ifi and i. and has the same effects on ifi and i as one
of the eight automorphisms we have indicated, we have found all the
automorphisms of E over IQ.
As another illustration of Lemma 24.5, consi,der
(i)=>(ii): If at • …. ak are the distinct roots ofj(X) in E then E = F(at. …. ak)’ If
E ~ K and
(iii): Suppose g(X) is irreducible in F[X] and a E E is a root of g(X). To
show that g(X) splits over E we will find an extension K of E such that g(X) splits
over K and K is the splitting field over F for some polynomial heX) E F[X]. We can
then argue as follows: For any root b of g(X) in K, a and b are conjugate over F, so
by Corollary 24.7 there is an automorphism
(i): Let {bl, … , br } be a basis for E over F and let g,(X) = irr(b/F) for
each i. Each g,(X) has a root in E, so by (iii) each g,(X) splits over E. Thus j{X) =
gl(X)g2(X) … g,(X) splits over E, and E is in fact the splitting field for j{X) over F
since if F (;: L (;: E andflX) splits over L then in particular h[, … , hr are all in L so L
= E. 0
DEFINITION A finite extension E of F is called a normal extension of F if it
satisfies the conditions of Theorem 24.8.
The proof that (ii)=>(iii) in Theorem 24.8 includes a proof of the following
fact, which we record for future use.
LEMMA 24.9 If F(bJ, … , br ) is a finite extension of Fthen the splitting field over
F(b[, … , br) of irr(b/F) ….. irr(b/F) is a finite normal extension of F.
It turns out that for algebraic extensions that are not tinite, (ii) and (iii) in
Theorem 24.8 are still equivalent, and (i) gets replaced by the statement that E is
the splitting field over F for a set of polynomials.
Normal extensions will be tied in with normal subgroups when we come to the
main theorem of Galois theory.
Examples
1. Extensions of degree 2 are normal. For let [E : F] = 2 and let g(X) be an
irreducible polynomial over F that has a root a E E. Then [F(a) : fl divides 2, i.e.
deg(g(X)) is I or 2. Thus g(X) splits over E, so E is normal over Fusing
characterization (iii) of normality.
2. A normal extension of a normal extension of F need not be a normal
extension of F. For consider
We have [1Qi( ..fi) : IQiJ = 2 = [1Qi( ifi) : 1Qi( ..fi)], so both extensions are normal.
But iQ( ifi) is not normal over iQ, since X’ 2 is irreducible over IQi and has a root
in iQ( ifi) but does not split over iQ( ifi ).
3. We have seen that there are automorphisms of K = iQ( ifi, i) over iQ that
send ifi to i ifi and thus do not map iQ( ifi) onto itself. Thus property (ii) of
Theorem 24.8 fails for the extension iQ( ifi) of iQ, showing again that iQ( ifi) is
not normal over iQ.
256 Section 24. Normal and Separable Extensions
4. If a is an element of an extension of Z7 such that a3 = 2, then the elements a,
2a, and 4a of Z7(a) are roots of X’ – 2. So X’ – 2 splits over Z7(a) and Z7(a) is the
splitting field for X’ 2 over Z7. Thus Z7(a) is normal over Z7.
The polynomial g(X) = X’ + X + 1 is irreducible over Z7 (since it has degree 3
and has no roots in Z7), but the element a2 + a of Z7(a) is a root of g(X). It follows
from property (iii) of Theorem 24.8 that g(.x) must split over Z7(a). (It turns out
that the other two roots of g(X) in Z7(a) are 4a2 + 2a and 2a2 + 4a.)
If E is the splitting field over F for a nonconstant polynomialj(X) E F[X] and
we write j{X) = c(X – at) … (X – an) in E[X], then at, … , an need not all be
distinct. This is a significant issue when we try to use E to study j{X).
DEFINITION Ifj(X) E F[X] and a is a root ofj{X) in an extension K of F, we say
that a is a root of multiplicity m if in K[X] we can write j{X) = (X – a)mg(X) with
g(a)”* O.
The meaning of the definition is unchanged if we replace K[X] by L[X], for
any extension L of K, because ifj{X) = (X – a)mg(X) in L[X] then g(X) E K[X]. In
particular, if L is the splitting field for j{X) over K andj{X) = c(X – at) … (X – an)
in L[X] then a is a root of multiplicity m if and only if there are exactly m factors X
– ai such that ai = a.
DEFINITIONS Roots of multiplicity 1 are called simple roots, and roots of
multiplicity at least 2 are called multiple roots. A polynomial is said to have
distinct roots if all of its roots are simple.
There is an easy way to determine whether a root a of a polynomial fiX) is a
multiple root, by using the formal derivative fiX). If fiX) = Co +clX +C2X2 +. . .
+cnX”, then the formal derivative off(X) is defined to be
f'(X) = CI + 2C2 X + … + nc,xI.
This is called the “formal” derivative because it is not defined by a limit, as
derivatives are defined in calculus. (We have no notion of “limit” available, since
we are working over an arbitrary field F.) Nevertheless, formal derivatives obey
the usual sum, product, and power rules for derivatives in calculus. (See Exercise
19.17.)
THEOREM 24.10 Letj{X) E F[X] be nonconstant.
i) If F ~ K and a is a root of fiX) in K then a is a mUltiple root if and only if
f(a) = O.
Section 24. Normal and Separable Extensions 257
ii) Suppose fiX) is irreducible over F. Then there exists an extension K of F
containing a multiple root of fiX) if and only if f(X) is the zero polynomial, and in
this case every root of jeX) in every exten.5ion K of F is a multiple root.
PROOF. i) Exercise 19.18(b).
ii) First supposef(X) has a multiple root a in some extension K of F. We want
to show that f(X) is the zero polynomial. By part (i), f(a) = 0, so by Theorem
22.2 fiX) divides f(X) in F[X]. But this is impossible if f(X) is not the zero
polynomial, because then f(X) has a degree that is less than that of fiX). So f(X) is
the zero polynomial.
Now suppose f(X) is the zero polynomial. Then f(a) = 0 for every root a of
fiX) in any extension K of F, so by part (i) every root off(X) is a multiple root. D
COROLLARY 24.11 If F is of characteristic 0 andj(X) E F[X] is irreducible over
F, thenj(X) has distinct roots.
PROOF. If the leading coefficient of fiX) is Cn * 0 then the leading coefficient of
f(X) is nCn * 0 (since F is of characteristic 0), so f(X) is not the zero polynomial.
Thusf(X) has distinct roots. D
The assumption thatj(X) is irreducible is indispensable in Corollary 24.11. For
example, the polynomial (X – 1)2 in Q![X] clearly does not have distinct roots.
In characteristic p, even an irreducible polynomial can have multiple roots.
Example Consider Z2(X), the quotient field of Z2[~\1. The polynomial Y 2 – X in
Z2(X) [ Y] is irreducible over Z2(X) because it has degree 2 and has no roots in
Z2(X)’ (If g(X), heX) E Z2 [X] and (g(X)/h(X)2 = X, then, mUltiplying both sides by
h(X)2 we get an equation in which the left side has even degree and the right side
has odd degree.) Since the formal derivative of y2 – X is 2Y – 0 = 0, every root of
Y 2 – X is a mUltiple root. Indeed, if we adjoin a root a then y2 X = (Y – ai since
we are in characteristic 2.
DEFINITIONS Suppose F <;;; E. If a E E then a is said to be separable over F if a
is algebraic over F and irr(a/F) has distinct roots. E is called a separable extension
of F if every element of E is separable over F.
It follows from Corollary 24.11 that if F has characteristic 0 then every
algebraic extension of F is separable. In characteristic p, some algebraic extensions
are separable and some are not. (The extension Z2(X)(a) of Z2(X) in the preceding
example is not separable.)
DEFINITIONS If F <;;; E then E is called a simple extension of F if there exists a
E E such that E = F(a). The element a is then called a primitive element for E
over F.
258 Section 24. Normal and Separable Extensions
One nice thing about finite separable extensions is that they are always simple
extensions. To see this we need a preliminary result that is interesting in its own
right.
THEOREM 24.12 If F is a field then every finite subgroup of the multiplicative
group F  {OJ is cyclic.
PROOF Let H be a finite subgroupof F  {OJ under multiplication. For every
positive integer n, the polynomial X"  1 has at most n roots in F by Corollary
19.4, so hn = 1 has at most n solutions h E H. Thus H is cyclic by Exercise 5.27.
(Exercise 5.27 can be proved directly, or by applying Theorem 14.2.)0
And now for our result about finite separable extensions.
THEOREM 24.13 (Primitive Element Theorem) If F is a field then every finite
separable extension E of F is simple.
PROOF. If F is a finite field, then since E is finite over F, E has only finitely many
elements. Therefore the multiplicative group E {OJ is cyclic by Theorem 24.12.
If a is a generator for E  {OJ then E = F(a), so E is a simple extension of F.
For the remainder of the proof we can assume that F has infinitely many
elements.
Since E is a finite extension of F, there are elements aI, ... , a r in E such that E
= F(at. ... , ar). If we can show that E is a simple extension of F in the case where r
= 2 then the result for all r's will follow by an easy induction. So suppose E = F(a,
b). Let [F(a) : F] = m and [E : F(a)] = n, so that [E : F] = mn.
We will show that there is some c E F such that E = F(a + cb). The idea is that
we can find c E F such that [F(a + cb) : F] ? mn, and therefore F(a + cb) = E.
Let K be the splitting field for irr(a/F) . irr(b/F) over F. We claim that we can
find embeddings Tt. ... , Tmn of E into Kover F such that for some c E F the
elements Ti(a + cb), 1 ~ i ~ mn, are all distinct. It will then follow that irr«a +
cb)/F) has at least mn distinct roots in K, and therefore deg«a + cb)/F) ? mn, as
desired.
To find Tt. ... , T""" note that irr(a/F) splits over K and has distinct roots aI, ... ,
am in K since a is separable over F. For any ai there is by Lemma 24.5 an
isomorphism qJi : F(a)  F(ai) over F such that qJi(a) = ai. IffiX) = irr(b/F(a» then
there is some g(X) E F(a)[X] such that irr(b/F) = fiX)g(X), so if we denote by fi(X)
and glx) the polynomials obtained by applying qJi to the coefficients of fiX) and
g(X) then irr(b/F) = fi(X)glx), with deg(fi(X) = n. Since irr(b/F) splits over K and
has distinct roots, it follows thatfi(X) has n distinct roots bi!, ... , bin in K. For any bij
there is by Lemma 24.5 an extension of qJi to an embedding qJij of E into K such
that qJij(b) = bij . The embeddings qJij, 1 ~ i ~ m, 1 ~ j ~ n are mn distinct
embeddings of E into Kover F. Call these embeddings Tt. ... , Tmn.
Section 24. Normal and Separable Extensions 259
If i '* j then either r;(a) '* ria) or ri(b) '* r)Cb). We want to show that there is
some e E F such that riCa + cb) '* ria + cb) whenever i '* j, i.e.
rtCa)  rtCa) '* eCr/b)  r,Cb» if either riCa) '* rj{a) or riCb) '* rtCb).
Ifri(b) = rib) this is clearly true for every c, for then riCa) '* ria). So all we need is
and since F is infinite it is clear that such a c exists. 0
The Primitive Element Theorem will be useful to us in our discussion of
Galois theory in the next section. The argument that produced rl, ... , rmn has the
following consequence, which will also be significant.
If j(X) E F[X] we say j(X) is separable over F if every irreducible factor of
j(X) in F[X] has distinct roots.
THEOREM 24.14 If j(X) is separable over F and E is the splitting field for j(X)
over F, then there are exactly [E : F] automorphisms of E over F.
PROOF. We have E = F(ah ... , an), where the a;'s are the roots ofj(X) in E. For
each a;, irr(alF) divides.f{X) in F[X] and thus has distinct roots. If [F(aJ) : F] = mJ
and [F(ah ... , aj) : F(aJ, ... , ajD] = mj for 2 s.j S. n then as in the proof of the
Primitive Element Theorem we see that there are mJ embeddings of F(aJ) into E
over F; that each of these has m2 extensions to an embedding of F(aJ, a2) into E
over F; that each of these extensions has m3 extensions to an embedding of F(ah
aZ, a3) into E over F; and so on. This gives us mrm2 ..... mn = [E : F] embeddings
of E into E over F. Since.f{X) splits over the image of each of these embeddings
(because.f{X) splits over E), the image cannot be a proper subfield of the splitting
field E. So each of these embeddings is onto and is therefore an automorphism of E
over F. By considering aJ, ... , an in that order we see that every automorphism of E
over F must have the same effect on aJ, ... , an as one of our [E : F] embeddings,
hence must be one of these embeddings. 0
We conclude this section by taking a closer look at finite fields.
By Theorem 22.13, there exists a finite field with exactly k elements if and
only if k is a power of a prime. If F is any field with pn elements then the
multiplicative group F  {O} has pn  1 elements, so by Theorem 10.4 every
nonzero element of F is a root of X p"J  1. Therefore every element of F,
including 0, is a root of X p"  X, and F is a splitting field for X p" _·X over the
prime subfield. It follows, by Theorem 24.4, that if two fields each have pn
elements then the obvious isomorphism of their prime subfields extends to an
isomorphism between the fields themselves. We have proved
260 Section 24. Normal and Separable Extensions
THEOREM 24.15 If F is a field with exactly pn elements then every element of F
is a root of X I,n_ X , and F is a splitting field for X p" X over the prime subfield.
If two finite fields have the same number of elements then they are isomorphic.
Notice that if F has pn elements then F has an extension of degree r, for any
integer r ~ 1, obtained as a splitting field E for X pnr  X over F. For since pn  1
divides pnr  1, apnr I = 1 for every nonzero a E F, so every element of F is a root
of X pnr  X . Therefore E is also a splitting field for X pnr  X over the prime
subfield of F, so lEI = pnr . Since iFl = pn , a basis for E over F must have exactly r
elements, so [E : F] = r.
In addition, any extension of F of degree r must have exactly pnr elements, and
must therefore be a splitting field for X pnr_ X over the prime subfield, hence over
F. Thus any two such extensions are isomorphic over F, and we have
THEOREM 24.16 Let F be a field with exactly pn elements. Then for any integer
r ~ 1, F has an extension of degree r, which is obtained as a splitting field for
X pnr_ X over F.
Any extension of F of degree r has exactly pnr elements, and any two
extensions of F of degree r are isomorphic over F. Furthermore, any such
extension is normal and separable over F.
The statement about separability follows from the fact that for every element a
of the extension, irr(a/F) divides X pllr_ X in F[X], hence has distinct roots.
EXERCISES
24.1 Suppose K is a field, fiX) E K[X] and g(X) is a nonconstant factor of fiX) in K[X].
Prove that iff(X) splits over K then g(X) splits over K.
24.2 For each polynomial, find the splitting field over Q and its degree over Q.
a)~l b)X'l
c)XS1 d)X'8
e)~+1 .f}X'+1
g)~+2 h)~+4
i) X'  2, where p is prime j) (X2  3)~  2)
k) 24~  26X2 + 9X  I l);t + 5X2 + 6
m);t5X2+36 n)X'+~+l
24.3 Find all multiple roots of XS + XI in its splitting field over 1Z3.
24.4 Find all multiple roots of 3X' + 2X + 3 in its splitting field over 1Z19·
Section 24. Normal and Separable Extensions 261
24.5 How many automorphisms are there of Q( ifi ) over Q? How many are there of
Q ( ifi, (3) over Q?
24.6 Give an example of a finite extension that is not normal but has the property of normal
extensions indicated in Corollary 24.6.
24.7 Show again that Q( ifi ) is not normal over Q by exhibiting two elements a, b of
Q ( ifi) that are conjugate over Q for which there is no automorphism ofQ( ifi) over
Q mapping a to b.
24.8 Prove that Q( ~2 + ji ) is normal over Q.
24.9 Suppose F ~ E ~ K with K finite and normal over F. Must E be normal over F? Must
K be normal over E?
24.10 Let E be a finite normal extension of F. Must condition (iii) of Theorem 24.8 continue
to hold if we delete the word "irreducible"?
24.11 Give an example of a separable extension that is not normal, and give an example of a
normal extension that is not separable.
24.12 Suppose F ~ E ~ K and K is separable over F. Must E be separable over F? Must K
be separable over E?
24.13 Suppose that E is a separable extension of F and there exists a positive integer n such
that deg(a/F) ~ n for all a E E. Prove that E is a finite extension of F and [E : FJ ~ n.
24.14 Give another proof of Lemma 24.S, by using Theorem 22.3 instead of Theorem 22.4.
24.15 A field F is called perfect if every finite extension of F is separable. Show that every
field of characteristic 0 is perfect and that every finite field is perfect.
24.16 Let Fbe a field of prime characteristic p and letj(X) E F[XJ be irreducible over F.
a) Show that there is a polynomial g(X) E F[XJ such thatj(X) = g(X P') for
some integer r ~ 0 and g(X) has distinct roots.
b) Show that there is an integer r ~ 0 such that all the roots ofj(X) have
multiplicity pro
c) Suppose F ~ E and a E E is algebraic over F. Show that there is an integer r
~ 0 such that apr is separable over F.
24.17 Let F be a field of characteristic p, where p is prime. Prove that F is perfect if and
only if every element of F has a pth root in F.
24.18 Let F be the quotient field of Zz[XJ[Y], and consider the extension F(a, b), where aZ =
Xand bZ = Y.
262 Section 24. Normal and Separable Extensions
a) Show that [F(a, b) : Fl = 4 and F(a, b) is not a simple extension of F.
b) Show that F(a, b) is a normal extension of F, but F(a, b) is not the splitting
field over F of any irreducible polynomial.
24.19 Let F be a finite field and letj{X), g(X) be two irreducible polynomials in F[X] of the
same degree. Prove that the splitting field forj{X) over F is also the splitting field for
g(X) over F.
24.20 Let F be a finite field such that IFI = pn and let k E Z+. Prove that F has a subfield
with exactly pk elements if and only if kin, and that if such a subfield exists it is
unique.
24.21 Let p be prime and let g(X) be irreducible over the field Z[J" Prove that if n E Z+ then
g(X) divides Xpn  X in Zp[X] if and only if deg(g(X)) divides n.
24.22 (Automorphisms of a finite field) Let F be the finite field such that IFI = pn and let G
be the group of automorphisms of F over its prime subfield, under composition.
a) The Frobenius automorphism of F is the mapping (J : F + F given by u(a) = d'.
Verify that (J E G.
b) Show that (J has order n in G.
c) Show that G =< (J> •
d) Let E be an extension of F of degree r. Show that the group of automorphisms of E over
F is cyclic of order r, generated by the automorphism that maps every a E E to apn .
24.23 Let F be a finite field. Prove that every element of F can be written as the sum of two
squares, i.e. for every a E F there exist b, c E F such that a = b2 + c2.
24.24 Suppose E is a finite extension of a field F and K\ and K2 are subfields of E that
contain F and are normal over F. Prove that K\ n K2 is normal over F.
24.25 Suppose F s;;; K\ s;;; E, E is finite over F and K\ is normal over F. Prove that if F s;;; K2
s;;; E and no proper subfield of E contains both K\ and K2 then E is normal over K2.
24.26 Prove that if E is a finite extension of F then E is normal over F if and only if for
every polynomialj{X) E F[X] that is irreducible over F, all the irreducible factors of
j{X) in E[X] have the same degree.
24.27 Let E be a finite extension of F. Prove that E is a simple extension of F if and only if
there are only finitely many distinct subfields of E containing F. [Hints: Show that if
E = F(a) then any subfield K of E containing F is obtained by adjoining to F the
coefficients of irr(a/K), and irr(a/K) divides irr(a/F) in E[X]. For the converse, it
suffices to show that for any two elements b, dEE, F(b, d) is a simple extension of F.
Section 24. Normal and Separable Extensions 263
coefficients of irr(alK), and irr(alK) divides irr(a/f) in E[X]. For the converse, it
suffices to show that for any two elements b, dEE, F(b, d) is a simple extension of F.
Consider the subfields F(b + cd), for c E F. If F is infinite there must exist Cj * C2 in F
such that F(b + Cjd) = F(b + czd).]
24.28 Suppose the complex number z is in Ce, so z E Q(aj, … , an) for some square root
sequence aj, … , an of complex numbers. By Lemma 24.9 let K be an extension of
Q(a\, … , an) that is a finite normal extension ofQ.
a) Prove that for every root z’ of irr(zIQ) in K, there is a sequence d], … , dn of
elements of K such that z’ E Q(d\, … , dn), d l2 E Q, and d: E Q(dj, … , dj\)
for all 2 :5;j :5; n.
b) Prove that if E is the splitting field for irr(zIQ) over Q then [E: Q] = 2m for
some integer m.
c) Prove that ifw E C and deg(wIQ) = 4 and the splitting field for irr(w;Q)
over Q has degree 12 or 24 over Q then w e Ce. This indicates how the
converse of Corollary 23.3 can fail. (For an example where this happens see
Exercise 26.13.)
The remaining sequence of exercises develops further aspects of the idea of
separability.
24.29 Let E be a finite extension of F. By Lemma 24.9 let K be an extension of E that is a
finite normal extension of F. Define the separable degree [E: Fls of E over F to be
the number of embeddings of E into Kover F.
a) Show that [E: Fls:5; [E : Fl·
b) Show that [E: Fls does not depend on our choice of K.
24.30 Prove that if F s:; E s:; K and K is finite over F then
[K: Fl, = [E : F)., .. [K : E).”,
24.31 Prove that if E is a finite extension of F then [E : F)., divides [E: F). [Suggestion:
Consider the case E = F(a) first, and use Exercise 24.16 in characteristic p.]
24.32 Prove that if E is a finite extension of F then E is a separable extension of F if and
only if [E: F), = [E : F).
24.33 Prove that if F s:; E s:; K are finite extensions such that E is separable over F and K is
separable over E then K is separable over F.
24.34 Show that F(aj, … , all) is a separable extension of F if and only if each a, is separable
over F.
264 Section 24. Normal and Separable Extensions
24.35 Let F ~ E. Define the separable closure SE(F) of Fin E to consist of all elements of
E that are separable over F.
a) Prove that SE(F) is a subfield of E containing F.
b) Prove that if E is finite over F then [SE(F) : fl = [E: fls’
SECTION 25
GALOIS THEORY
Galois theory brings together ideas from group theory (subgroups, normal
subgroups, quotient groups, indices) and ideas from field theory (subfields, normal
and separable extensions, degrees, automorphisms). It is often referred to as one of
the most beautiful parts of mathematics, both because it reveals a close and precise
interplay between these two sets of ideas and because this interplay has fascinating
consequences. We will see in this section that it enables us to complete the
determination of which regular polygons are constructible with straightedge and
compass by turning a question about a field extension into an answerable question
about a group. In the next section we will see that the same kind of translation
allows us to answer questions about formulas for finding the roots of polynomials.
Our first step is to introduce the main ingredients of the theory. If F s E then
the set of automorphisms of E over F is a group under composition. We call this
group the Galois group of E over F and denote it by r(E/F). On the other hand, if
S is a set of automorphisms of a field E, then by the fixed field of S we mean the
subset {a EEl cp(a) = a for every cp E S}. It is easy to verify that this subset is a
sub field of E. We denote it by (S).
It is obviously true that
F s (f(E/F» [25.1 J
since every element of F is fixed (not moved) by every automorphism of E that
fixes all of F. The finite extensions that provide the setting for Galois theory are
those for which equality holds in [25.1]. These extensions can be characterized in
several equivalent ways.
THEOREM 25.1 Let E be a finite extension of F. The following are equivalent:
i) F = (f(E/F).
ii) E is normal and separable over F.
265
266 Section 25. Galois Theory
iii) E is the splitting field over F for some separable fiX) E F[X].
PROOF. We show that (i):::;> (ii):::;> (iii):::;>(i).
(i):::;>(ii): Suppose that F =
irr(a/F) has distinct roots, since E is separable over F, and each irr(a/F) splits over
E, since E is normal over F. Thus E is the splitting field over F for the separable
polynomial irr(a/F) ….. irr(ajF).
(iii)=>(i): Suppose E is the splitting field over F for fiX), which is separable
over F. Let K =
f(EIIl>(H» = H. Thus the mapping K ft f(EIK) is a onetoone mapping
of the set of intermediate fields onto the set of subgroups of G and the
mapping H ft 1l>(H) is its inverse.
iii) If K and L are intermediate fields then K ~ L iff f(ElK) :2 f(EIL).
iv) For any subgroup H of G, [E : 1l>(H)] = IHI and [1l>(H) : F] = [G : H].
v) If H is a subgroup of G then H
case
268 Section 25. Galois Theory
f(ct>(H)IF) == GIH.
Before you look at the proof, take a few minutes to absorb what the theorem
says, and to appreciate the symmetry and harmony it expresses. The result is
beautiful because it is so perfect, and everything fits together so well.
PROOF OF THEOREM 25.2. i) This is immediate from Theorem 24.14, since E is
the splitting field over F (and therefore over K) of a polynomial J(X) that is
separable over F (and therefore over K).
ii) Since E is the splitting field over K for a separable polynomial we have
ct>(f(EIK» = K by Theorem 25.1.
Next, for any subgroup H of G, every element of H is clearly in f(Elct>(H), so
if we can show that I f(Elct>(H) I :5IHI we will have f(Elct>(H) = H.
Let K = ct>(H). so that our goal is to show that I f(EIK) I :5 IHI. i.e. [E : K] :5 IHI
by part (i). If (using the Primitive Element Theorem) we choose a E E such that E
= K(a) it will suffice to show that a is a root of some polynomial J(X) E K[X] of
degree IHI. If H = {
every a E E that is fixed by every element of f(EIK) is fixed by every element of
r(EIL). i.e. ct>(f(EIK» ~ ct>(f(EIL». By part (ii) this says K ~ L.
iv) By parts (i) and (ii),
[E : ct>(H)] = I f(Elct>(H) I = IHl·
This yields
[ct>(H) : F] = [E : F]/[E: ct>(H)] = IGIlIHI = [G : H].
v) First suppose H
that for every a E ct>(H), irr(alF) splits over ct>(H). We know it splits over E, so we
need to show that if bEE is a root of irr(alF) then b E ct>(H).
Since E is normal over F, Corollary 24.7 tells us that there is some
f(KlF) with kernel H. We can then apply
the Fundamental Theorem on Group Homomorphisms. We define \fI by restriction:
If (J E G then since K is normal over F we have u(K) = K by Theorem 24.8(ii).
Thus the automorphism (J* : K –> K given by (J*(a) = u(a) for all a E K is in
f(KlF). We define \fI by letting \fI«(J) = (J*. Then \fI is a homomorphism since «(JI 0
(Jz)* = (J 1 0 (J ‘2, and \fI is onto because for any rp E f(KlF), cp extends to an
element (J E G since E is normal over F (Corollary 24.6), and then (J* = rp, i.e. \fI«(J)
= cpo Finally, ker(\fI) = H because for any (J E G, (J E ker(\fI) iff (J* fixes all of K,
i.e. all of CfJ(H), and this is so iff (J E f(E/CfJ(H» = H. 0
DEFINITION If j(X) E F[X] then the Galois group f(f{X)/F) of j(X) over F is
defined to be [(E/F), where E is the splitting field for j(X) over F.
Every element of f(f{X)/F) must permute the distinct roots of j(X) in the
splitting field, and is completely determined by its effect on these roots. Thus
distinct automorphisms give rise to distinct permutations, and we have a oneto
one mapping from f(f{X)/F) into the group of permutations of the distinct roots of
j(X). If (J, r E f(f{X)/F) give rise to permutations gm gr, then (J 0 r gives rise to g” 0
gn so our mapping is an isomorphism from f(f{.X)/F) onto a subgroup of the
symmetric group Sm, where m is the number of distinct roots of j(X). Since Sm is
isomorphic to a subgroup of Sm where n = deg(f(X», we have
THEOREM 25.3 If j(X) E F[X] has degree n then f(f{X)/F) is isomorphic to a
subgroup of Sn.
Even if m = nand j(X) is irreducible, however, the Galois group need not be
isomorphic to Sn itself.
As an illustration, let us consider the case of an irreducible polynomial j(X) of
degree 3 over a field F of characteristic * 2, 3. Since the characteristic is not 3,
I(x) is not the zero polynomial, so j(X) is separable over F. f(f{X)/F) is
isomorphic to a subgroup of S3, and has (by Theorem 25.2(i» order [E : F], where
E is the splitting field for j(X) over F. For any root a ofj(X) in E, [F(a) : F] = 3, so
3 divides [E : F]. Thus f(f{X)/F) is isomorphic to either A3 or S3, and there is an
easy way to find out which.
If aJ, az’ a3 are the three distinct roots ofj(X) in E, let
Then for any (J in the Galois group, u(d) = ±d, and (J(d) = d iff (J gives rise to an
even permutation of the indices I, 2, 3. (This is because any transposition of the
indices sends dto d. We have used the fact that, since the characteristic is not 2, d
* d.) Thus d is in CfJ([(E/F)i.e. dE F, since E is Galois over Fiff f(f{X)/F) ==
A3.
270 Section 25. Galois Theory
lt is clear that if we let D = J then D does not depend on how we order the
roots of f(X), and (J(D) = D for every (J, so D E F. D is called the discriminant of
f(X). We have proved
THEOREM 25.4 LetfiX) be an irreducible polynomial of degree 3 over a field F
whose characteristic is neither 2 nor 3. ThenfiX) is separable over F and nJ(X)/F)
is isomorphic to either A3 or S3. f(j(X)/F) == S3 if and only if the discriminant of
fiX) has no square root in F.
This result makes it interesting to find an easy way to calculate the
discriminant. It can be shown that, under the circumstances of Theorem 25.4, if
fiX) = X3 + aX2 + bX + c then the discriminant is
In general, if fiX) is separable over F then its splitting field over F is a finite
Galois extension and thus we can use Theorem 25.2 to study f(fiX)/F). Of course,
if F has characteristic 0 or is a finite field then (by Corollary 24.11 and Theorem
24.16) all polynomials in F[X] are separable over F, so we can always apply
Theorem 25.2 in these cases.
Examples
1. LetfiX) = X3 – 3X + 1 in Q[X]. By Exercise 19.1 the only possible roots of
fiX) in Q are ±l, neither of which works. So, since fiX) has degree 3, fiX) is
irreducible in Q[X]. The discriminant is D = 4(3)3 27(1)2 = 81, which has a
square root in Q. SO f(fiX)/ Q) == A3.
2. The splitting field of r 2 over Q is E = Q( ifi, i), which has degree 8
over Q. Since E is a Galois extension of Q, Theorem 25.2 guarantees us that if G =
f(E/Q) then IGI = 8. In the example following Corollary 24.7 we confirmed this
directly, by showing that we get all the elements of G by choosing to map ifi to
any of ± ifi or ±i ifi and choosing to map i to ±i.
Let (J E G be such that (J( ifi) = i ifi and (J(i) = i. Then
and likewise (J\ ifi) = i ifi and (J\ ifi) = ifi. So (J has order 4 as an element of
G.
Now let rEG be such that r( ifi) = ifi and r(i) = i. Then i(i) = r( i) =
–i = i, so r has order 2 in G.
Since T It: < (j' >,
Section 25. Galois Theory 271
Since < a > has index 2 in G, < a >
4, TmI must be aOf aI. But
wr –
I
( ifi ) = rer( ifi ) = r (i ifi ) = – i ifi * er( ifi ),
and therefore Tcrrl = uI, which leads to TUk = UkT for all k E Z. Thus G;::; D4.
Since D4 has exactly 10 subgroups it follows from Theorem 25.2(ii) that there
are exactly 10 intermediate subfields in E.
For example, if HI is the subgroup < r> of G then by Theorem 25.2(iv) we
have [<1>(HI) : Q] = [G : Hd = 4. Since ifi E <1>(HI) it follows that <1>(HI) =
Q ( if2).
If H2 =< er> then [<1>(H2) : Q] = [G : H2] = 2. Since i E <1>(H2) it follows that
<1>(H2) = Q(i). Note that Q(i) is normal over Q, corresponding to the fact that H2
<1>(HI) = Q( ifi). Thus <1>(H3) = Q(a), where a E Q( ifi) – Q is fixed by if. Since
if( if2) = – ifi, if( .fi) = (if( ifi »
2
= .fi. Thus <1>(H3) = Q ( .fi).
If H4 =< if > then H4 � H3 so <1>(H4) ;;;;2 <1>(H3). Likewise, since H4 � H2,
<1>(H4) ;;;;2 <1>(H2)’ So <1>(H4) ;;;;2 Q(.fi, i). But [<1>(H4) : Q] = [G : H4] = 4 and
therefore <1>(H4) = Q( .fi, i). Note that H4
by observing that f(Q( .fi, WQ) is a group of order 4 in which every element has
order 1 or 2.
3. Let F be a finite field such that IFI = pn, with prime subfield Zp. By
Theorem 24.16, F is Galois over Z” and [F : Zp] = n. If G = f(FIZ,,) then by
Theorem 25.2(i) IGI = n. If er is the Frobenius automorphism defined by cr(a) = a”
then er has order n in G since al’n = a for all elements a E F and n is the smallest
integer with this property. Thus G =< er> .
For every positive integer k that divides n, G has a unique subgroup < d' > of
order nlk. Thus the intermediate subfields of Fare
By Theorem 25.2(iv)
{<1>« d’ » I kE Z+and kin }.
[<1>« d’ » : Zp] = [G:< d' >] = n/(nlk) = k,
so <1>« d’ » is a field with l elements. The fact that d’ fixes every element of
<1>( <: d' » just says that every element of <1>( < d' » is a root of x,l  X By
Theorem 25.2(ii), f(FI<1>« d’ >)) =< d' >.
272 Section 25. Galois Theory
The facts that G =< (J > and that the subfields of F are exactly the fields with
l elements for kin can of course be established without using Galois theory. (See
Exercises 24.22 and 24.20.)
4. Suppose F is a field that contains n distinct roots of X”l. We claim that for
every c E F the Galois group of X” – cover F is isomorphic to a subgroup of
(Z”,EB) and is therefore cyclic. If c = 0 the Galois group is trivial, so we can
suppose c “* O.
The roots of Xn – 1 form a subgroup of order n in the multiplicative group F –
{O}, and this subgroup must be cyclic by Theorem 24.12. Let b be a generator, so
that the roots of X n – 1 in Fare 1, b, b2, … , bn1. If E is the splitting field for Xn – c
over F and a is any root of Xn – c in E then all the roots are a, ba, b2 a, … , bn1 a, so
E =F(a).
If G = [(ElF) then for any (J E G we have (J(a) = bKua for some k(f E {a, 1, . .. ,
n – I}, and (J is completely determined by ktr So we have a onetoone mapping ‘¥
: G � (Zn,EB) given by ‘P«(J) = k”. We assert that ‘P is a group homomorphism, i.e.
for all (J, rEG we have
This is because
So’¥ is an isomorphism from G onto a subgroup of (Zn,EB).
Note that in Example 2 above the Galois group of X – 2 over rQ was not
cyclic, but rQ did not contain 4 distinct roots of X — I. However, the intermediate
field $(H2) = rQ(i) did contain 4 such roots, and the Galois group of X – 2 over
rQ(i) was r(rQ( 12, i)/$(H2)) = H2, a cyclic group of order 4.
5. For each n � I we will give an example of a polynomial of degree n whose
Galois group is the full symmetric group SIl’ The natural thing to do is to seek a
polynomial whose roots are as unrelated to each other as possible, so that we have
a maximum amount of freedom in permuting the roots by automorphisms.
So let F be any field and let F[X) , … ,Xn] = F[Xd[X2l .. [Xn] be the polynomial
ring in n variables over F. Let E be the quotient field of F[X), … ,Xn], so that every
element of E can be expressed as a quotient
of elements of F[XJ, … ,Xnl We want a polynomial whose roots are X), … ,Xn, so let
X be another variable and let
Section 25. Galois Theory 273
fiX) = (X – XJ)(X – Xz) … (X – Xn)
= Xn – SJ(XJ . … ,Xn)Xn 1 + SZ(XI, … ,Xn)Xn Z – …
+ (_l)nISn_I(XJ, … ,Xn)X + (_l)nSn(XI’ … ,Xn),
where s[, … , Sn E F[X[, … ,Xn] and Sj is the sum of all products of j distinct X;’s. For
instance
The ~/s are called the elementary symmetric polynomials in XJ, … ,Xm because
they remain unchanged if we apply any element of SII to the subscripts of the ~’s.
Clearly j(X) is a separable polynomial over K = F(sJ’ ”” sn). E is a splitting field for
j(X) over K, so E is Galois over K. We get an isomorphism of Sn with f(EIK) by
associating to each permutation the element of f(EIK) whose value at g(XJ,
“”Xn)/h(XIo .. “Xn) is obtained by applying the permutation to the subscripts of the
X/’s.
An element of E is called symmetric if it is left fixed by every permutation of
the ~’s, i.e. by every element off(EIK). Since E is Galois over K, an element of E
is symmetric if and only if it is in K, i.e. it can be written in the form g(sJ,
sn)/h(sJ, … , sn).
We will conclude this section by giving a couple of applications of Galois
theory. Section 26 will be devoted to the application that motivated Galois.
In Theorem 23.4 we showed that if a regular ngon is constructible with
straightedge and compass then n = 2mpJ . , . Pk, where m ~ 0, k ~ ° and the p,’s are
distinct Fermat primes. We can now establish the converse.
THEOREM 25.5 Suppose n ~ 3 and n = 2mpJ . , , Pk, where m ~ 0, k ~ ° and Plo .. ”
Pk are distinct Fermat primes. Then a regular ngon can be constructed with
straightedge and compass.
PROOF. Let r = PI … Pk’ If r = 1 then n = 2m for some m ~ 2, so we can construct
a regular ngon by starting with the square with vertices at ±l, ±i, inscribed in the
unit circle around the origin, and repeatedly bisecting the central angles to double
the number of sides.
If r > 1 then to construct a regular ngon it suffices to construct a regular r
gon. For once we have an rgon we can bisect the central angles repeatedly to
obtain an ngon.
We now claim that if (1′; E CCc for every Pi then (r E CC, so a regular rgon is
constructible. Since the integers rlpl, “‘, riPk have greatest common divisor 1 there
exist a[, … , ak E Z such that
274 Section 25 . Galois Theory
and thus
Using the identity
cos( a + fJ) + isin( a + fJ) = (cos a + isin a)( cos fJ + isin fJ)
repeatedly, we thus obtain
i.e. S;: .. . S;: = (,. Thus if every t;p; E C e then (, E C” so our problem is now
reduced to showing that each t; p, E C e .
We want to show that if p is a Fermat prime then (p E Q(al , .. . , a,) for some
square root sequence ai , “” a” We know that Q«(p) is a Galois extension of Q (as
the splitting field of xP lover Q) and [Q «(p) : Q] = p – 1 = 2′ for some t, since p
is a Fermat prime. Thus r(Q «(p)/Q ) has order 2’, so by the First Sylow Theorem
there are subgroups
such that IHII == 2! , Ifwe let KJ == $(HJ) then by Galois theory
and [Q «(p) : Rjl = I~ 1= 2J, so for 1 ::;. j ::;. t we have [Rj’_1 : Rjl = 2. For any b E
Rj_1 – Rj there are thus c, d E Rj such that b2 + cb + d = 0 and thus
If we let a = b + !:. then Kj – 1 = ‘0(a) and a2 E Kj . Thus there is a square root
sequence a” … , at stch that Q «(p) = Q (al, … , a,), and in particular (p E Q (al, … , at)·
o
We have mentioned previously that the determination (before Galois theory
existed) of which regular polygons are constructible with straightedge and compass
is due to Carl Friedrich Gauss (1777 1855), Gauss was an extraordinary
mathematician; many cons ider him to have been the greatest of all time, He did his
work on polygons while he was still in his teens, and it is said that this success was
Section 25. Galois Theory 275
what made him decide to be a mathematician. He remained very proud of this work
throughout his life.
Our final result for this section is another result of Gauss (often referred to as
the Fundamental Theorem of Algebra), which states that every nonconstantj{X) E
C[X] splits over
a) H == {e, if’, UT,~r}
b) H== < if'r>.
25.5 If F is a field such that IFI == 4 and E is an extension field such that lEI == 4096, find
f(EfF).
25.6 Suppose E is an algebraic extension of F and every nonconstantfiX) E F[X] splits over
E. Prove that E is algebraically closed. (This provides an alternative to the last
paragraph of the proof of Theorem 25.6.)
25.7 Is Q(i) algebraically closed?
Section 25. Galois Theory 277
25.8 Prove that no finite field is algebraically closed.
25.9 Let p be a prime and consider a chain of fields
where, for each n, IF.I = p.!. Prove that the field F obtained by taking the union of
the chain is algebraically closed.
25.10 If F) and F are as defined in Exercise 25.9, prove that r(F/F) is an infinite abelian
group.
25.11 How many elements does rcllttlQ) have?
25.12 Let E be a finite extension of R Show that if E.r IR then E is isomorphic to Cover R
25.13 Suppose that z E C is algebraic over Q and E is the splitting field for irr(ztQ) over Q.
Prove that if [E : Q] = 2m for some integer m then z E Ce. (This is the converse of
Exercise 24.28.)
25.14 Let G be a finite group. Show that there exists a finite Galois extension F k E such
that rcElF) == G. [Hint: Use Example 5 in the text.]
25.15 A finite extension F £;; E is said to be abelian (respectively cyclic) over F if E is
Galois over F and [‘(ElF) is abelian (respectively cyclic).
a) Prove that if E is abelian over F then every intermediate field is abelian over
F.
b) Prove that if E is cyclic over F then every intermediate field is cyclic over F
and the intermediate fields are in onetoone correspondence with the positive
divisors of [E : Fj.
25.16 Let E be a finite Galois extension of F and let G = rcElF). Let H) and H2 be
subgroups of G.
a) Show that
O.
(In those days it was not possible to write all cubic equations in the form
X 3 + aX2 + bX + c = 0, because negative numbers were not accepted. Solving
cubic equations was therefore a matter of dealing with a number of different
cases.) Like other professors of his time, del Ferro kept his job py competing in
mathematical contests and being able to solve problems that others couldn’t solve.
So he did not make his method public, although he did later reveal it to his student
Antonio Fiore.
In 1535 Niccolo Tartaglia announced that he had discovered a method for
solving equations of the form X + aX2 “” b. Fiore challenged him to a contest and
posed problems requiring the solution of equations of the form X+aX = b.
Tartaglia figured out how to solve equations of this type too, and won the contest.
He kept his methods to himself for several more years, but in 1539 was persuaded
to reveal them to Geralamo Cardano, in return for a pledge of secrecy and the
promise of an introduction to a potential patron.
Cardano’s life is one of the strangest and most varied in the history of
mathematics. He was trained as a physician and was for a time a prominent one in
Milan and throughout Europe. At other times he wrote on mathematics and was
professor of mathematics at several universities in Italy. At still other times he was
a philosopher and astrologer. He also had a reputation as a gambler, and at one
point was brought before the Inquisition for casting a horoscope of Christ. His
elder son murdered his own wife and was beheaded for it, and Cardano had his
younger son imprisoned several times and disinherited him. (One story says
Cardano went so far as to cut off the boy’s ears in a fit of anger.) He spent the last
part of his life as astrologer to the papal court, and while in Rome wrote his
autobiography (The Book of My Life), which is still in print today.
In any case, Cardano set about demonstrating that Tartaglia’s method for X +
279
280 Section 26. Solvability
aX = b was correct, and eventually developed methods for solving all cases of the
cubic. He also learned, by gaining access to del Ferro’s unpublished papers in
Bologna, that del Ferro had solved X3 + aX = b about twenty years before Tartaglia.
He felt that this released him from his promise to Tartaglia, and so in 1545 he
published the solutions for cubics in his Ars Magna, acknowledging del Ferro’s
and Tartaglia’s accomplishments. Cardano’s student Ludovico Ferrari had by this
time discovered a method for solving quartics (equations of degree 4), and this was
also included in the Ars Magna.
Tartaglia felt that Cardano had, by publishing the solutions, robbed him of the
credit he deserved for his discoveries. After a bitter public dispute, a contest was
held between Tartaglia and Ferrari. Ferrari won, and as a result Tartaglia lost his
teaching position. Today the formulas for solving cubics bear Cardano’s name.
The methods given in the Ars Magna make it possible to express the roots of
cubic and quartic polynomials in terms of the coefficients by using addition,
subtraction, multiplication, division, and the extraction of radicals. The success of
these methods led mathematicians to seek similar “solutions by radicals” for
equations of the fifth and higher degrees, but centuries went by without any
progress being made. In 1770, Lagrange gave a unified account of why equations
of degree at most 4 are solvable by radicals. He observed that the known methods
reduced the problem of finding the roots of a polynomialj(X) to that of finding the
roots of an associated polynomial of smaller degree. When he tried this for j(X) of
degree 5, however, he was unable to produce such a polynomial of smaller degree.
This impasse was actually a signal pointing in the direction of the truth.
In the years between 1799 and 1813 Paolo Ruffini made several attempts to
prove that it was actually impossible to find a formula for solving quintics
(equations of degree 5) by radicals. His arguments were difficult to follow and
incomplete at some points, so his results were not accepted by his contemporaries.
The matter was finally settled in 1826, when Neils Abel published a proof that for
n ~ 5 there is no general formula for finding the roots of nthdegree polynomials in
terms of radicals.
An earlier (1824) version of Abel’s result for quintics was published at his
own expense and had to be kept very brief to keep down the cost. As a result the
arguments were difficult to follow and the work did not receive much attention.
(Gauss apparently didn’t bother to read the copy that was sent to him.) Even after
the 1826 version appeared, recognition was slow in coming, and Abel supported
himself by taking lowlevel academic jobs while he continued to work on
mathematics. His life came to a premature end in spring 1829, when he died of
tuberculosis at the age of 26. Two days after his death he was sent a letter
informing him that he would be offered a position on the faculty of the University
of Berlin.
The work of Ruffini and Abel demonstrated that there is no general formula
for solving all quintics by radicals. This left open the possibility that every quintic
in, say,
LEMMA 26.6 If F has characteristic 0 and E is a radical extension of F then there
exists an extension K of E that is radical and Galois over F.
PROOF. We have E = F(a” … , ak) for some radical sequence ai, . . . , ak with
associated positive integers nl, … , nk. If K is the splitting field over E for
irr(atlF) ….. irr(alF)
then K is a finite normal extension of F by Lemma 24.9, and K is separable over F
since we are in characteristic O. So K is Galois over F.
To see that K is radical over F, let, for each i, ail, .. ” ail; be the roots of irr(a/F)
in K Then K is obtained by adjoining all of the au’s to F, so it will suffice to show
that each aij is a member of some radical sequence of elements of K (If we string
together such radical sequences for all the au’s we get a radical sequence that,
when adjoined to F, gives us K)
Since ai and au are conjugate over F and K is normal over F we have an
automorphism cp of Kover F such that cp(a;) = aij. Then cp(al), cp(a2), … , cp(ak) is a
radical sequence of elements of K (for example, from aI” E F(a” … , aiI) we get
(QJ(ai))”i E F(cp(al), … , cp(ail))), and its ith member is aij’ 0
PROOF OF THEOREM 26.2 If L is the splitting field for f(X) over F then by
assumption there is an extension E of L that is radical over F. By Lemma 26.6 we
286 Section 26. Solvability
have E k:: K, where K is radical and Galois over F, say K = F(al, … , ak), with a;” E
F and a? E F(al, … , ajI) for 2 Sf S k.
In the course of the following argument we will want to apply Example 4 of
Section 25 to each polynomial X”i – a�’i, so we will need nJ distinct roots of Xn,
I for each j. Accordingly, let ri be the least common multiple of nl, … , nk. The
equation X’ – I has n distinct roots in its splitting field over K since we are in
characteristic 0, and these form a cyclic group under multiplication by Theorem
24.12. If b is a generator then K(b) is still radical over F (since bn = I), and is still
Galois over F since if K is the splitting field for, say, g(X) over F then K(b) is the
splitting field for g(X). (X’I). (Separability is again automatic.) For each nj, K(b)
contains n, distinct n}h roots of I.
We now have F � L � K(b), with both Land K(b) Galois over F. We want to
show that f(L/F) is solvable, and we know by Theorem 25.2(v) that
r(L/F) == f(K(b)/F)Ir(K(b)/L),
so it will suffice, by Lemma 26.5, to show that f(K(b)/F) is solvable. The
advantage of shifting the problem from L to K(b) is that in addition to being Galois
over F as L was, K(b) is also radical over F and contains the needed nth roots of I.
To show that r(K(b)/F) is solvable, we think of K(b) as being obtained by
adjoining first b and then ai, … , ak to F:
with Fo = F(b) and FJ = FJI(a) for I:::;j :::; k. This chain of intermediate fields will
give us the series of subgroups in f(K( b)/ F) needed to demonstrate its solvability.
Let Gj = f(K(b) IFJ) for 0 “‘5.j “‘5. k and G = f(K(b)/F). Then
{e} =Gk �Gk_1 �···�GI �Go �G.
We assert first that Gj
if!(r(Ef F) = F. If we can show that there is an i such that ( bi, c) (j F then this ( bi, c)
will be the element a we are after.
If we add the equations in [26.2] for 0 ::: i ::: pJ, then for J :::) ::: pJ the
coefficient on d(c) is
since lJ is a root of
Thus adding the p equations in [26.2] yields
(bO,c)+W,c)+ … +(bP1,c) = pc.
Since c (2: F and F has characteristic 0, pc (2: F. Therefore at least one W, c) is not
in F. 0
PROOF OF THEOREM 26.7 Let E be the splitting field for j(X) over F. By
assumption, r(E/F) is solvable. If [E : F] = n then since F has characteristic 0 the
polynomial X’ – I has n distinct roots in its splitting field over E, and these form a
cyclic group under multiplication. Let b be a generator for this group and consider
F’ = F(b). (We consider F’ in order to prepare ourselves for an application of
Lemma 26.1 I.) By Lemma 26.10, r(j(X)/F� is isomorphic to a subgroup of
r(j(X)/F), so r(j(X)/F� is solvable by Lemma 26.8. We will show that this implies
that the splitting field E’ for j(X) over F’ is a radical extension of F’, hence a radical
extension of F. Since E’ contains a splitting field for j(X) over F this will yield the
desired result thatj(X) is solvable by radicals over F.
To see that E’ is a radical extension of F’ let G = rcE’IF�. Since G is solvable,
Lemma 26.9 tells us that there is a series of subgroups
{e} = Go <1 G I <1 G2 <1 . . • <1 G, I <1 G, = G
such that G
i
1G
i
1 is cyclic of prime order Pi for 1 ::: i::: s. Let Fi = if!(G
i
) for 0::: i::: s.
Then
For 1 ::: i::: s, E' is Galois over Fi and, by Theorem 25.2(ii), f(E'fFi) = Gi. Since
Gi1
order Pi’ If we can show thatFi contains Pi distinct roots ofXPi 1 then Lemma 26.11
290 Section 26. Solvability
will tell us that Fj_1 = Fj(aj) for some aj such that ajPi E Fj_l> and therefore E’ is a
radical extension of F’.
But Pi divides jf(E1F)j which divides jf(EIF)j = n, since f(E1F) is
isomorphic to a subgroup of f(EIF). Thus bnl Pi has order Pi in (F’ {O}, .), so F’
(and therefore F) contains Pi distinct roots of X Pi – 1. 0
Since the Galois group of the general polynomial of degree n is Sn, and since
S3 and S4 are solvable, Theorem 26.7 implies that in characteristic 0 the general
polynomials of degree 3 and 4 are solvable by radicals. To explicitly obtain
Cardano’s solution for the cubic we first use the fact that ifJ(X) = X’ + aJ(2 + bX +
c then
( a ) 3′ . a2 ab 2a3 f X– =X +pX+qwlthp=b–andq=c–+.
3 3 3 27
[26.31
So if we can find the roots of X’ + pX + q (which is called a reduced cubic because
it has no X2 term) then by subtracting a/3 from each of them we obtain the roots of
J(X).
Although variables were not yet in use when Cardano wrote the Ars Magna
(and he therefore had to write everything out in words) the idea of his strategy for
solvingX’ + pX + q = 0 is to compare it with the identity
(A + Bi – 3AB(A + B) – (A3 + B3) = O.
If we can find A,B such that
[26.4]
then A + B is a root of X’ + pX + q.
We need A3B3 = (pl3i and A3 + B3 = q, so A3 and B3 are required to be
roots of
By the quadratic formula, A3 and B3 are
Therefore
Section 26. Solvability 291
with (by [26.4]) the cube roots chosen so that their product is pI3. If rl = A + B
results from one such cho”ice, and w is such that w 7= 1 but w3 = 1, then r2 = wA +
w2 Band r3 = w2 A + wB are also valid choices, and it can be checked that
so that rl, r2, r3 are all the roots of X + pX + q.
The preceding argument is valid over any field of characteristic 7= 2, 3. Using
[26.3], then, we have proved
THEOREM 26.12 (Cardano’s formula) If/(X) =X+ a.X2+ bX+ C E F[X], with
F of characteristic 7= 2, 3 and
with the cube roots chosen so that their product is pI3, then the roots off(}.) are
A + B !.:, wA+ w2B !.:, and w2A + wB !.:,
3 3 3
where w is such that w 7= 1 but w3 = 1.
Example LetfiX) = X 3 – 2X2 – X + 2 E Ql[X]. Then we find
=_2 = 20 A= ~10+9i.J3 B= ~1O9i.J3
p 3′ q 27′ 3′ 3′
with A,B E C chosen to be complex conjugates so that their product is real (and =
7/9). Using w = (3 E C, the three roots ofjeX) are then
[26.5)
On the other hand, it is easy to check that the roots of j(X) are ±l and 2. It is not
immediately apparent how these match up with the roots in [26.5], but if we
expand out the equation (x+yi.J3 )3 = 10 + 9i.J3 we can see that one cube root of
10 + 9i.J3 in C is 2 + i.J3 , so with
A = 2 + i.J3 and B = 2 – i.J3
3 3
we have AB = 7/9, and the roots in [26.5] are then, respectively, 2, 1, and I.
292 Section 26. Solvability
The cubic formula is significantly more complicated than the quadratic
formula, and the quartic formula even more so. Instead of developing the quartic
formula we will just outline Ferrari’s method of solution.
Ferrari was originally hired by Cardano as a servant but became Cardano’s
student and secretary when Cardano realized how bright he was. Ferrari discovered
his method for solving quartics when Cardano was unable to solve a problem that
had been sent to him requiring the solution of a quartic, and passed the problem on
to Ferrari. Ferrari eventually left Cardano’s service to go out on his own as a
lecturer in mathematics, and he became a professor at the University of Bologna in
1565. Less than a year after he assumed this post he was poisoned to death,
apparently by his own sister.
Ferrari’s method for solving a quartic consists of transforming the equation
into the form
with deg(f{X» = 2 and deg(g(X» = I, and then taking square roots. During the
process a root of a certain cubic equation is needed, and this can be found either by
using Cardano’s formula or by other means.
Example To find the solutions of X’ + 6X + 15X2 + 2X 24 = 0 we write the
equation as
X’ + 6X = – 15X2 – 2X + 24
and then complete the square on the left by adding 9X2 to both sides:
(X2 + 3X)2 = 6X2 – 2X + 24.
Ferrari’s insight was that the left side remains a perfect square if we add Y to X2 +
3X:
(X2 + 3X + Y)2 = 6X2 – 2X + 24 + y2 + 2(X2 + 3X)Y,
i.e.
We want to choose Y to produce on the right side a quadratic in X that is a perfect
square. A quadratic AX2 + BX + C is a perfect square if and only if its two roots
coincide, and by the quadratic formula this will be the case if and only if B2 – 4A C
= o. So we need to choose Y so that
(2 + 6Y)2 – 4(6 + 2Y)(24 + y2) = 0,
i.e.
Section 26. Solvability 293
4(2y3 – 15y2 + 54Y – 145) == O.
We could try to find a root of this cubic by using Cardano’s formula, but it is
simpler to apply Exercise 19.1, according to which any rational root has the form
min with ml145 and n12. By trial and error we find the root 5, and using this in
[26.6] we get
(X2 + 3X + 5)2 == 4X2 + 28X + 49 == (2X + 7/.
Thus
X2 + 3X + 5 == ±(2X + 7).
Solving the resulting quadratic equations
X2 + X – 2 == 0 and X2 + 5X + 12 == 0
5 ± i..J23
we get the roots 1, 2, and .
2
We could also have found the roots 1 and 2 at the outset by using Exercise
19.1, and then divided the original quartic by (X 1)(X +2) to be left with X2 + 5X
+ 12.
EXERCISES
26.1 Let G be a group with a nonnal subgroup H such that both Hand G/H are solvable.
Prove that G is solvable.
26.2 Is every j{X) E lR[X] solvable by radicals over lR?
26.3 Determine whether each of the following polynomials is solvable by radicals over Ql.
a) 5XS – 12xt + 9X2 – 7
b)XS6X+3
c) 3XS – 20x’ + 15
26.4 Find all roots in C for the following cubics.
a) x’ + 54X – 54
b) x’ + 3X2 + 66X – 20
c) x’ + 6X2 + 93X + 8
294 Section 26. Solvability
26.5 In the Ars Magna Cardano solved the equation � + 6X= 20.
a) Find the roots of � + 6X 20 in C by using Cardano’s formula.
b) By finding the cube roots involved in the form x + y J3 . show that the roots
are 2 and I ± 3i.
26.6 a) Find the roots of � 15X 4 in C by using Cardano’s formula.
b) By evaluating the cube roots involved, show that the roots in Care 4 and
2 ± .J3 . [This problem appeared in the Algebra of Rafael Bombelli in
1572. Although Bombelli did not see how to make rigorous sense out of
complex numbers, he was the tirst to calculate with them, and he showed
via these calculations how Cardano’s formula could yield real roots of
cubics even when the formula involved square roots of negative numbers.]
26.7 Let fiX) = � + pX + q in JR[X). Show that ali the roots of fiX) in C are real if and on Iy
if27q2 + 4/ :’S 0, and that they are ali real and distinct if and only if27l + 4/ < O.
26.8 (the "casus irreducibilis") The purpose of this exercise is to show that iffiX) is an
irreducible cubic in Q[XJ with 3 (necessarily distinct) real roots then there is no radical
extension ofQ contained in JR and containing the splitting field E for fiX) over Q. SO
although all the roots of fiX) are real they cannot be expressed in terms of radicals
without involving nonreal numbers. This was troublesome to mathematicians when
they did not yet accept the existence of nonreal numbers.
a) Prove that if F is a field, F � F(a) � JR, p is prime, and aP E F but a e F then
[F(a) : F] = p.
b) Show that the discriminant D offiX) is positive.
c) Now assume for a contradiction that there is a radical extension K of Q such that E
� K � lR. Show that there exist a radical sequence .fi5 , ah .. " ak and primes Ph ... ,
Pk such that E s;; Q( .fi5 , ah ... , ak) S;; JR and
a/i E Q{JD) and a/i E Q(.Ji5, a" '''' aj_l ) for 2 <;'} <;, k.
d) Let Ko = Q(.fi5) and let IS = Ko(aj, ... , a) for 1 <;, } <;, k. Choose} as small as
possible so that IS contains a root ofj(X). Show that} � 1 and that [IS : ISIl = Pj = 3.
e) Show that IS is the splitting field for j(X) over ISI' (Use the fact that .Ji5 E K j_I')
f) Show that �  a� splits over K; and conclude that (3 E KJ s;; JR, a contradiction.
Section 26. Solvability 295
26.9 Use the result of Exercise 26.8 to show that ifj(X) is an irreducible cubic in Q[X] with
three real roots then althoughj(X) is solvable by radicals over Q the splitting field for
j(X) over Q is not a radical extension ofQ.
26.10 Fran�ois Viete (15401603) showed that when a cubic in Q[X) has three distinct real
roots (as in the "casus irreducibilis" of Exercise 26.8) these roots can be expressed in
terms of the cosine function. As in the development of Cardano' s formula, Viete
replaced the cubic by a reduced cubic g(X) = X3 + pX + q, and then he matched up
g(X) with the trigonometric identity
3 1 cos) e   cos e·  cos 3e = 0
(see Exercise 22.10).
4 4
a) Show that for every rand e in lR, r cos e is a root of
b) Note that 27l ; 4p3 < 0 by Exercise 26.7. Using this, show that we can choose r E
IR such that .:l!:.:. = p.
4
c) Sh0;V that with r chosen as in part (b) there is some e in IR such that
r cos 39
= q 4 .
d) With rand e so chosen, show that the roots of g(X) are
r cos e, r cos (e + 2;) . and r cos (e + 4;) .
[By the way, Viete was the first person to use letters to represent unknown quantities
in algebra.]
26.11 Find all roots in
5.8 72
5.9 nlm; one of m,n divides the other
5.12 (b) {I, – I}
5.13 {(~ ~)la~O}
Section 6
6.1 (a) 18; (b) 24; (c) 18; (d) 765
6.2 The group in (d) is cyclic.
6.11 You should end up with 16 distinct subgroups.
Section 7
7.1 (a) Function from S to T; neither onetoone nor onto
(b) Not a function from S to T
( c) Not a function
(d) Function from S to T; neither onetoone nor onto
(e) Function from S to T; onetoone and onto
(f) Not a function from S to T
(g) Not a function from S to T
(h) Function from S to T; onetoone but not onto
(i) Function from S to T; onto but not onetoone
(j) Function from S to T; neither onetoone nor onto
7.4 Onto, not onetoone
7.5 A =X
7.6 Yes
7.8 Yes, on all counts
Section 8
8.1
8.2
4 5
3 2
(a) (31 2 3 4 5
6 142
(c) (51 2 3 4 5
634 1
8.3 (a) Odd r’s
Answers to Selected Exercises 303
~)
~) = (1,3)(2,6,5) = (1,3)(2,5)(2,6). Odd.
~)=(I,5)(2,6). Even.
(b) Factor it into cycles. The permutation is even iff the number of
rcycles with even r’s is even.
(c) Even
8.10 (c) 12
8.15 (b)j; (c) {e}
Section 9
8.12 No; yes 8.14 {e} and {e,f2}
8.19 (a) No; (b) Yes
9.1 The relation in (b) is an equivalence relation.
9.3 The equivalence classes are straight lines with slope 1.
9.5 (a)H= {J,I,J, I} andH· K= {L,K,L,K}
9.6 The right cosets are H= {e,f2g}, Hf= If,fg} = Hfg, Hf2 = {f2, g} = Hg, and Hf3
= {f3,f3g}.
9.7 H= {(O, 0), (I, 0), (2, O)} and H + (0, 1) = {CO, 1), (1, 1), (2, I)} = H + (I, 1) =
H+ (2, 1)
Section 10
10.1 4, 2, and 2 10.2 (a) 16; (b) 6; (c) 4
10.3 (a) 6; (b) 4 10.4 4
10.25 The conjugacy classes are {I}, {I}, {J, J}, {K, K}, and {L, L}.
The class equation is 8 = 2 + 2 + 2 + 2.
Section 11
11.2 No
11.11 6
11.14 (a) 4; (b) No
11.15 It is essentially the same as V.
11.20 Just a remark: You are given more information than you need to answer
the que~tion.
304 Answers to Selected Exercises
SectIon 12
12.1 (a) Epimorphism; (b) isomorphism; (c) epimorphism; (d) epimorphism;
(e) not a homomorphism
12.3 No; no
12.4 (a) No; (b) yes; (c) no; (e) no; (g) yes; (h) yes
12.8 No; yes
12.9 No
12.17 6
SectIon 13
13.1 ker(!p) = {0,4}. The quotient group is isomorphic to (~,ffi).
13.3 D4, V, (12, ffi), and the trivial group
13.6 G has 3 elements.
SectIon 14
14.1 (b) I8 X I9
I8 XI3xI3
I4 XI2xI9
I4xZ2XI3X I3
Z2XI2xI2X I9
I2 X I2 X I2 X I3 X I3
(d) I2 X I9 X I2s
I2XI9XIsXIs
I2 x Z3 X I3 x I2s
I2XZ3xI3X IsXI~
14.8 1:9 x 1:3
Section 15
15.8 I2s X I169
I2s X 1\3 X Il3
Zs xIsxI169
Is X Is X ZI3 X 1\3
Section 16
16.2 (a) No; (b) no
16.4 Yes
16.11 (b) Units: (1,1),(1,2),(2,1),(2,2);
zerodivisors: (0,0), (0, 1), (0, 2), (1,0), (2, 0);
nilpotent elements: (0,0)
16.20 (a) No
Answers to Selected Exercises 305
SectIon 17
17.1 The sets in (b) and (d) are subrings.
17.2 (a) Ideal; (b) not a subring; (c) subring, not an ideal
17.3 The subgroups of G
17.6 (a) {0,3}, the principal ideal generated by 3; or {0,2,4},
the principal ideal generated by 2 or 4.
SectIon 18
18.1 The mapping in (b) is a ring homomorphism.
18.2 No; yes
18.7 Up to isomorphism, they are {ldld a positive divisor of n}.
SectIon 19
19.2 (a) Irreducible
(e) Irreducible
19.3 (a) Irreducible
(d) (X + 1)(X3+2X +2)
SectIon 20
20.2 No
20.5 (ac+bd)+(ad+bc+bd)X
20.7 (a) Let F=l2, letj(X)=X3+X+ 1, and let K be the field
described in Exercise 20.6.
SectIon 21
21.142X+5
21.16 (a) (I + ;)(12;)
21.18 (a) – 1 + ;
Section 22
22.4 4
Section 24
22.12 7 22.13 (a) 3; (b) 2; (c) 4
24.2 (b) 0«(6) = O(i J3 ). degree 2; (c) 0 «(8) = 0 ( ..fi. I). degree 4;
(m) 0 ( .Jfi. ;.J7 ). degree 4; (n) Q «(9). degree 6 .
24.5 1 and 6 24.10 No 24.12 Yes and yes
306 Answers to Selected Exercises
Section 25
25.2 (b) Zp–l; (t) D4
25.7 No
Section 26
26.3
26.11
26.13
(a) Yes; (b) No 1 ± J17
(c)l±i.J5 and —
(a) S4 2
abelian extension, 277
abelian group, 17
action of a group on a set, 98
addition
in an abelian group, 25, 33, 34
in a ring, 154
additive group of integers mod n, 20
additive identity element, 154
algebraically closed field, 275
algebraic closure of a subfield, 239
algebraic element, 228
algebraic extension, 228
alternating group, 72
of degree 4, 93
annihilator, 175
associates, 213
associative operation, 11
automorphism of a group,110
inner, 119
nontrivial, 110
automorphism of a ring, 179
axioms
for groups, 16
for rings, 154
INDEX
basis, 231
Bell, E.T., 17
binary operation, 10
associative, 11
commutative, 11
Boolean ring, 162, 176
Burnside’s formula, 98
cancellation laws, 29
canonical homomorphism
for groups, 122
for rings, 180
cardinality of a set, 89
Cauchy, AugustinLouis, 214
Cayley, Arthur, 66, 168
Cayley’s Theorem, 116
generalized, 132
center
of a group, 46
of a ring, 175
centralizer of an element, 93
characteristic
of a domain, 163
of a field, 183
307
308 Index
characteristic subgroup, 119
class equation, 94, 148
closure
under inverses, 44
under an operation, 44
coefficients of a polynomial, 191
commutative operation. 11
commutative ring, 154
commutator subgroup, 107, 119
composite function, 63
congruent modulo n, 21
conjugacy class, 94
conjugate elements
of a field over a subfield, 253
of a group, 41, 93, 94
conjugate subgroups, 54, 143
constant polynomial, 193
constructible numbers, 241
content of a polynomial, 198
coset, 83
double, 97
left, 85
right, 83
cycle, 67
cyclic extension, 277
cyclic groups, 39
isomorphisms between, 113
subgroups of, 50
cyclotomic polynomial, 229
degree of a field extension, 232
degree of an algebraic element
over a subfield, 228
degree of a polynomial, 193
degree rule, 194
Delian problem, 245
de Moivre’s Theorem, 214
determinant, 19
dihedral group, 77
direct product of groups, 55
direct sum of rings, 156
Dirichlet, P.G. Lejeune, 214
discriminant, 270
disjoint cycles, 68
distributive laws, 153
left and right, 154
division algorithm
for F[X], 194
for Z[i], 221
for Z, 20
division ring, 159
domain, integral, 159
domain of a function, 61
double coset, 97
doubling the cube, 245
Eisenstein Criterion, 200
Eisenstein, Ferdinand, 199,200
element
of finite order, 34
of infinite order, 34
of a set, 1
elementary symmetric
polynomials, 273
embedding, 178, 183
empty set, 2
epimorphism
of groups, 110
of rings, 178
equality
of functions, 63
of sets, 2
equivalence class, 82
equivalence relation, 81
Euclidean algorithm, 35,225
Euclidean domain, 220
Euler, Leonhard, 92, 214
Euler’s function, 52
Euler’s Theorem, 92
evaluation homomorphism, 203
even permutation, 70
extension by radicals, 281
extension of a ring, 187
factor of a direct product, 55
factor group, 104
factor ring, 169
Fermat, Pierre de, 92, 213
Fermat’s Last Theorem, 213
Fermat primes, 246, 273
Fermat’s Theorem, 92
Fibonacci sequence, 9
field, 159
algebraically closed, 275
finite, 237, 259, 260
of characterisitc p, 183
of characteristic zero, 183
perfect, 261
field extension, 227
algebraic, 228
finite, 232
Galois, 266
normal,255
radical, 281
separable, 257
simple, 257
field of quotients, 186
finite abelian groups
Index 309
converse of Lagrange’s Theorem for,
137
fundamental theorem on, 135
invariants of, 135
number of order m, 135
finite extension of a field, 232
First Isomorphism Theorem
for groups, 127
for rings, 181
fixed field, 265
formal derivative, 203, 256
Frobenius automorphism, 262
function, 59
composite, 63
domain of, 61
image of, 61
injective, 61
inverse, 62
onetoone, 60
onto, 60
surjective, 61
Fundamental Theorem of
Algebra, 225
Fundamental Theorem of
Arithmetic, 7, 42, 211
Fundamental Theorem on
Finite Abelian groups, 135
Fundamental Theorem on
Group Homomorphisms, 123
Fundamental Theorem on
Ring Homomorphisms, 181
Galois extension, 266
Galois group
of a field extension, 265
of a polynomial, 269
310 Index
Gauss’ Lemma, 199
Gaussian integers, 163
general linear group
over the real numbers, 19
over the complex numbers, 47
general polynomial of degree fl, 282
generator, 39
greatest common divisor
of elements in a domain, 224
of integers, 35
of polynomials, 209
group, 16
abelian, 17
acting on a set, 98
alternating, 72
cyclic, 39
dihedral, 77
general linear , 19
of integers mod fl, 20
of primepower order, 135
simple, 146
special linear, 52
of subsets of X, 20
symmetric, 66
of symmetries of a square, 75
trivial, 22
of unit quaternions, 47
Hamilton, Sir William Rowan,
101, 166, 167
Hamiltonian group, 10 1, 107
heptagon, 240, 246,248
homomorphic image, 122
homomorphism
of groups, 109
of rings, 177
ideal, 168
improper, 170
maximal, 171, 189
prime, 171, 189
principal, 170
proper, 170
semiprime, 176
trivial, 170
idempotent element, 162
identity element, 16
left, 29
right, 29
identity function, 63
image of a function, 61
improper ideal, 170
index of a subgroup, 89
inductive hypothesis, 6
injective function, 61
inner automorphism, 119
integral domain, 159
intermediate field, 267
intersection
of sets, 2
of subgroups, 49
invariants of a finite abelian group, 135
inverse
of an element in a group, 16
of a function, 62
left, 29
right, 29
inverse image, 115
irreducible element of a domain, 212
irreducible polynomial, 196
isomorphism
of groups, 110
of rings, 178
isomorphism over a field, 251
kernel of a homomorphism
for groups, 122
for rings, 180, 181
Klein’s 4group, 40
Kronecker, Leopold, 198,207
Kronecker’s method, 204
Kummer, Ernst Eduard, 214
Lagrange, JosephLouis, 97
Lagrange Interpolation Theorem, 203
Lagrange resolvent, 288
Lagrange’s Theorem, 88
Lame, Gabriel, 214
leading coefficient, 193
least common mUltiple, 42
left cancellation law, 29
left coset, 85
left identity, 29
left inverse, 29
Legendre, AdrienMarie, 214
Leibniz, Gottfried, 97
linear combination, 231
linear independence, 231
marked straightedge, 247, 248
mathematical induction
first form, 4
second form, 6
matrix, 11
invertible, 18
members of a set, 1
metabelian group, 131
Miller, William, 70
monic polynomial, 228
monomorphism
of groups, 110
of rings, 178
Motzkin, T., 221
multiple root, 203
multiplication
in a group, 25
in a ring, 154
table, 31
Index 311
multiplicative identity element, 154
multiplicative inverse, 156
multiplicity of a root, 256
natural homomorphism, 122
nilpotent element, 155
norm, 215
normal extension, 255
normal subgroup, 99
normalizer of a subgroup, 107, 143
null set, 2
octic group, 75
odd permutation, 70
onetoone function, 60
onto function, 60
orbit, 98
order
of an element, 34
of a group, 40
orderable field, 190
ordered field, 190
312 Index
partition of a positive integer, 135
perfect field, 261
permutation, 66
even, 70
odd, 70
pgroup, 134
polynomial, 191
power set, 20
ppower order, group of, 134
prime element of a domain, 212
prime ideal, 171
prime integer, 6, 211
prime subfield, 183
primepower order, group of, 135
primitive element, 257
Primitive Element Theorem, 258
primitive polynomial, 198
principal ideal, 170
principal ideal domain, 217
product of ideals, 176
proper ideal, 170
proper subset, 2
proper subgroup, 44
pSylow subgroup 144
quaternions
group of unit, 47
ring of, 166
quotient field of a domain, 186
quotient group, 104
quotient ring, 169
radical extension, 281
radical of an ideal, 176
radical sequence, 281
reduced cubic, 290
reduced quartic, 296
reflexive relation, 81
regular heptagon, 240, 246, 248
regular ngon, 246
relation on a set, 80
reflexive, 81
symmetric, 81
transitive, 81
relatively prime integers, 36
remainder of a mod n, 21
representative of an equivalence class,
82
right cancellation law, 29
right coset, 82
right identity, 29
right inverse, 29
ring, 154
Boolean, 162
commutative, 154
of Gaussian integers, 163
of quaternions, 166
trivial, 157
with trivial multiplication, 157
with unity, 154
root of a polynomial, 195
multiple, 256
simple, 256
Second Isomorphism Theorem
for groups, 127
for rings, 182
semidirect product, 151
semiprime ideal, 176
separable closure of a subfield, 264
separable degree, 263
separable element, 257
separable extension, 257
separable polynomial, 259
set, 1
empty, 2
simple extension, 257
simple group, 146
simple root, 256
solvable by radicals, 281
solvable group, 281
span, 231
special linear group, 52
splitting field, 249
square root sequence, 244
squaring the circle, 247
stabilizer, 98
straightedge and compass, 240
subgroup, 43
characteristic, 119
lattice, 44
normal, 99
proper, 44
subring, 164
subset, 2
proper, 2
sum of ideals, 176
surjective function, 61
Sylow Theorems, 144
symmetric difference of sets, 12
symmetric group, 66
symmetric relation, 81
symmetry of a polygon, 75
Index 313
Third Isomorphism Theorem
for groups, 128
for rings, 182
transcendental element, 229
transitive relation, 81
transposition, 69
trisecting angles, 240, 245
trivial group, 22
trivial ideal, 170
trivial multiplication, 157
trivial ring, 157
union
of sets, 2
of subgroups, 49
unique factorization domain, 213
unit in a ring, 155
unity of a ring, 154
Viergruppe,40
Wedderburn, I.H.M., 160
wellordering principle, 4
Wilson’s Theorem, 97
zero of a polynomial, 195
zerodivisor, 155
Title Page
Table of Contents
Preface
Section 0: Sets and Induction
Section 1: Binary Operations
Section 2: Groups
Section 3: Fundamental Theorems about Groups
Section 4: Powers of an Element; Cyclic Groups
Section 5: Subgroups
Section 6: Direct Products
Section 7: Functions
Section 8: Symmetric Groups
Section 9 Equivalence Relations; Cosets
Section 10: Counting the Elements of a Finite Group
Section 11: Normal Subgroups
Section 12: Homomorphisms
Section 13: Homomorphisms and Normal Subgroups
Section 14: Direct Products and Finite Abelian Groups
Section 15: Sylow Theorems
Section 16: Rings
Section 17: Subrings, Ideals, and Quotient Rings
Section 18: Ring Homomorphisms
Section 19: Polynomials
Section 20: From Polynomials to Fields
Section 21: Unique Factorization Domains
Section 22: Extensions of Fields
Section 23: Constructions with Straightedge and Compass
Section 24: Normal and Separable Extensions
Section 25: Galois Theory
Section 26: Solvability
Suggestions for Further Reading
Answers to Selected Exercises
Index
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