Groups, Rings and Galois Theory [2nd Revised ed.] 9812385762, 9789812385765

Table of contents :
Preface
Preface to the Second Edition
Contents
Chapter 1. Group Theory
1.1 The concept of a group
1.2 Exercises
1.3 New groups from old
1.4 Exercises
1.5 Quotient groups and normal subgroups
1.6 Exercises
1.7 Finitely generated abelian groups
1.8 Exercises
1.9 Abelian groups and codes
1.10 Exercises
1.11 Sylow's theorems
1.12 Exercises
1.13 Groups of permutations
1.14 Exercises
Chapter 2. Ring Theory
2.1 Basic definitions
2.2 Exercises
2.3 Factorisation
2.4 Exercises
2.5 Unique factorisation
2.6 Exercises
Chapter 3. Galois Theory
3.1 Fields
3.2 Exercises
3.3 Group characters
3.4 Exercises
3.5 Finite fields
3.6 Exercises
3.7 Further results
3.8 Exercises
3.9 Solution of equations by radicals
3.10 Exercises
3.11 Tensor products
3.12 Exercises
Chapter 4. Rings and Modules
4.1 Basic definitions
4.2 Finitely generated modules
4.3 Tensor products over rings
4.4 Exercises
Chapter 5. Dedekind Domains
5.1 Integrality
5.2 Prime factorisation of ideals
5.3 Finitely generated modules
5.4 Galois extensions
5.5 Exercises
Bibliography
Index

Citation preview

Groups, Rings and Galois Theory Second Edition

Victor P Snaith University of Southampton, UK

\\!) World Scientific ~I

NewJersey •London• Singapore• Hong Kong

Groups, Rings and Galois Theory Second Edition

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224

USA office: Suite 202, 1060 Main Street, River Edge, NJ 07661 UK office: 51 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

The first edition was published in 1998.

GROUPS, RINGS AND GALOIS THEORY Second Edition Copyright© 2003 by World Scientific Publishing Co. Pte. Ltd.

All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 981-238-576-2 ISBN 981-238-600-9 (pbk)

Printed in Singapore.

Preface This book was tailor-made for the third-year algebra course at McMaster University. There is very little that is special about the course - every university must have one like it - except perhaps for the fact that it comes in two halves, permitting the disenchanted student to bail out midway through the material. Those who leave at that point come away with an introductory course on group theory and, it seems to me, are entitled to see groups in action as something other than self-fulfilling exercise fodder. Accordingly - and not very originally - I included a little material on error-correcting codes and on Rubik's cube for them. Students soldiering on into the second term of the course are treated to an introduction to the theory of rings. For them, I thought it appropriate to present enough material so that they could see how the two subjects - rings and groups - eventually joined forces in a major undertaking. In this book that example is Galois theory - the jewel in the crown of algebra. I hope that this succinctly explains the ingredients of the book and that the table of contents, the index and some browsing through the text will do the rest. I am very grateful to my son, Daniel, who typed this manuscript into Latex and to Carolyn, my wife, who proof-read the typescript. They did such a thorough job that any blunders which remain are entirely my responsiblity! It only remains for me to say a few sentences on why I wrote yet another book on this amply covered material. Usually the "book of the course" for this material would be a rather magnificent, expansive tome. I, on the other hand, have tried to give a rather concise treatment, because I have found that demanding proofs and exercises tend to encourage more profitable discussion between the instructor and class. Making the text more terse, providing one does not attempt to cover too much material, made the course more enjoyable - at least for me! Had I made the pace of the book too leisurely I would have run the risk of rendering the instructor superfluous. Had I made the exercises too easy or repetitious I would have V

vi

PREFACE

run the risk of superannuating the students! Learning mathematics at this level seems best accomplished by pondering problems on which one gets stuck rather than repeating finger-exercises with the themes one finds easy. The McMaster students were good-natured enough to indulge me in these whims and I responded by trying to strike a workable compromise vis a vis the exercises. Accordingly, 1the reader - should there be one - will find this book terse enough here 'and there that there will be no alternative but to discuss the subject with others, preferably fellow students and the lecturer. The exercises will occasionally be too hard or too few and there will be no alternative but to ask the instructor for suggestions about where to look elsewhere. This book was used as the basis of my 1995/7 courses at McMaster University during which time I tried to correct as many misprints and errors as I could. I am particularly grateful to Hayssam "Sam" Hulays, who was a student in that class and pointed out a number of errors to me and to Matt Valeriote, who taught the course from this text in 1997 /8 and his student, Kee Ip, who kindly corrected a few more misprints en route. I have also added some extra exercises which suggested themselves to me while I was teaching from the "first edition".

Victor Snaith McMaster University December 1997

Preface to the Second Edition The second edition of this book differs from the first only by the addition of two chapters concerned with the modules over rings. In particular, Chapter Four introduces the notion of a module, imitating the classification of finitely generated abelian groups in Chapter One in order to classify finitely generated modules over a principal ideal domain. In Chapter Five, Dedekind domains are introduced and developed to the point where a classification of their finitely generated modules can be given. Chapter Five concludes with the analysis of how the primes of a Dedekind domain behave under a Galois extension of the field of fractions. Chapters Four and Five take the reader through the first steps beyond undergraduate algebra which are essential in order to study algebraic number theory, for example. Elementary number theory is a very popular undergraduate course at the University of Southampton, which made this material particularly suitable for a graduate course given there in the Autumn of 1998 to a rather heterogeneous audience consisting of fourth year undergraduates, PhD students and staff.

Victor Snaith University of Southampton May 2003

vii

Contents 1

2

3

Group Theory 1.1 The concept of a group . 1.2 Exercises ....... 1.3 New groups from old .. 1.4 Exercises ........ 1.5 Quotient groups and normal subgroups . 1.6 Exercises .............. 1.7 Finitely generated abelian groups . 1.8 Exercises ......... 1.9 Abelian groups and codes 1. 10 Exercises .... 1.11 Sylow's theorems . . . . 1.12 Exercises ........ 1.13 Groups of permutations 1.14 Exercises ........

1

1 11 14 22 25 35 39 51 55 62 62 69 71 80

Ring Theory 2.1 Basic definitions 2.2 Exercises .. 2.3 Factorisation .. 2.4 Exercises .... 2.5 Unique factorisation ..... 2.6 Exercises

83 90 93 98 100 105

Galois Theory 3.1 Fields ... 3.2 Exercises 3.3 Group characters 3.4 Exercises 3.5 Finite fields . . 3.6 Exercises ... 3.7 Further results

107 107 113 113 122 123 126 127

83

ix

CONTENTS

X

3.8 3.9 3.10 3.11 3.12 4

5

Exercises . . . . . . . . . . . . . Solution of equations by radicals Exercises . . . . Tensor products. Exercises . . . .

135 137 143 144 148

Rings and Modules 4.1 Basic definitions . . . . . . 4.2 Finitely generated modules 4.3 Tensor products over rings . 4.4 Exercises . . . . . . . . . .

153

Dedekind Domains 5.1 Integrality . . . . . . . . . . 5.2 Prime factorisation of ideals 5.3 Finitely generated modules 5.4 Galois extensions 5.5 Exercises . . . . . . . . . .

181

153 155 170

177 181 186

191 202 206

Bibliography

209

Index

211

Chapter 1

Group Theory 1.1

The concept of a group

The concept of a group, which we are about to study in considerable detail, is one of the many axiomatic structures which constitute the area of abstract algebra. As the name suggests, this collection of mathematical gadgets has arisen in response to the desire to construct algebraic abstractions of familiar phenomena. Although groups are used nowadays in a number of applications ranging from vibrations of chemical molecules to error-correcting codes, the fundamental origins of the subject arise from the algebraicisation of the notion of symmetry. The following examples will serve to illustrate what this algebraic abstraction is required and expected to do. Example 1.1.1 Suppose that we are given a rigid three-dimensional solid which we will fondly call X, for brevity. Imagine X firmly implanted stably in some position; for example, X might be a regular tetrahedron resting on a table. A symmetry of X is any operation consisting of picking X up, juggling it around in some manner and then replacing it so as to occupy exactly the same space as before. In abstract algebra it is fashionable (and sensible) to denote things by algebraic symbols; in particular, let us denote the symmetries of X by the symbols s1, s2, ....

A symmetry of X is not required to return each point of X to the place from which it started. In fact, in order to keep track of what a symmetry of X does, it is a good idea to decorate X in some manner with markers. For example, if X is an equilateral triangle we might number the vertices. Having done this, the six symmetries of the triangle, X, would look as follows:

1

2

CHAPTER 1. GROUP THEORY

1.1.2 Symmetries of an equilateral triangle

1 1 ~ 6 362 3 2 1 3 ~ 6 362 2 1 1 2 362 163 1 1 ~ 6 362 2 3 1 362 ~ 36 1 _83. .

2

3

~6 1

2

Question: Why is this list complete? 1.1.3 The symmetries of an equilateral triangle, while only a very simple example, can give us some suggestions of what an algebraic abstraction would be needed for and what it should include. Firstly, it should be capable of systematising and simplifying the description of all the possible symmetries of X. Secondly, as part of such a simplification, it should organise what happens when we take several symmetries of X and perform these operations one after another in sequence; the result must, after all, be another in the list of symmetries of X. In the simple example of §1.1.3 this organisation and description may be accomplished by listing all the possibilities but, for a general X, this process will be prohibitively lengthy. Alternatively, we might express the same symmetry information by tabulating all the symmetries and their pairwise compositions algebraically. In the following table the entry in the row labelled si and column labelled s1 is the symmetry obtained by performing first s1 and then Si on the equilateral triangle, X, of §1.1.3.

1.1. THE CONCEPT OF A GROUP

3

1.1.4 Compositions of symmetries of an equilateral triangle St S2 83 84 S5 85 St St S2 83 S4 S5 S5 S2 S2 83 St S5 85 S4 83 83 St S2 S5 84 S5 84 84 85 S5 St 83 S2

S5 S5 84 85 S2 St 83 85 S5 S5 84 83 S2 St

1.1.5 Notice that the table of compositions given in §1.14 is already a

considerable abstraction of the information embodied in the list of §3.1.3. For example, if this table were a little larger we might not be able to guess from it the identity of X, the equilateral triangle, having precisely six symmetries, composing according to §1.1.4. Let us repeat the process of listing all the symmetries and tabulating their compositions for the case in which X is a line-interval and a square. 1.1.6 Symmetries of an interval

There are two symmetries s1

1--2 - 1 - - 2

1--2

s2

-

2--1

whose compositions yield the following simple table: St S2 St St S2 S2 S2 St

1.1. 7 Symmetries of a square

In this case there are eight symmetries.

CHAPTER 1. GROUP THEORY

4

Question: Why is this list complete? Compositions of these eight symmetries are given by the following table. s1 s2 83 S4 85 85 87 Ss S1 S1 s2 83 84 85 S5 87 Ss s2 s2 83 84 S1 85 87 Ss 85 83 83 84 S1 S2 87 ss 85 85 84 84 s1 s2 83 Ss 85 85 87 85 85 Ss 87 S5 S1 84 83 S2 S5 85 85 Ss 87 s2 s1 84 83

87 87 85 85 Ss 83 s2 S1 84

ss Ss 87 85 85 84 83 s2 s1

1.1.8 Observation

Notice that in each of the tables of §§1.1.4, 1.1.6 and 1.1.7 the entries on each row and column are distinct. In fact, each row and column is a rearrangement (or permutation) of the set of all the symmetries of X. Question: Can you explain this observation?

1.1. THE CONCEPT OF A GROUP

5

1.1.9 Consider now the three-dimensional example in which Xis a regular

tetrahedron whose four vertices are numbered. 2

Question: What are the symmetries of this tetrahedron? This example is more complicated than the earlier, two-dimensional examples. However, as a first approximation to an answer, we might at least try to list some symmetries. Looking back at the examples of §§1.1.3 and 1.1.7, one sees that it can be helpful to classify the symmetries into types. For example, in the case of X being a equilateral triangle or a square, half the symmetries flip X onto its back and half of them do not. Questions: What observations can you make about the behaviour of the 'flips' and 'non-flips' in the tables of §§1.1.4 and 1.1.7? Can you explain these observations? In the case of the regular tetrahedron some types of symmetries which come to mind are:

or 4; about an axis from a vertex to the centroid of the opposite side. There are two such symmetries for each vertex, making eight rotations in all.

(a) clockwise rotations through

2;

(b) the symmetry which leaves every point of the tetrahedron where it is. This is sometimes called trivial symmetry.

(c) rotations through

7r about an axis joining midpoints of two opposite sides. The tetrahedron has six sides but each side has precisely one opposite side, making three pairs of opposite sides in all.

The total number of symmetries listed in (a) to (c) is twelve. Question: Are these twelve the only symmetries of the regular tetrahedron? (Hint: When studying symmetries of a polyhedron it is sometimes helpful to classify them by how many vertices, edges, etc. remain fixed under the symmetry.) The following definition of a group is one attempt at an axiomatic algebraic structure that is abstracted from the symmetries of X and their behaviour under composition. The idea is to make an algebraic gadget whose 'multiplication' operation imitates the composing of two symmetries of X to obtain a third.

CHAPTER 1. GROUP THEORY

6

Definition 1.1.10 A group is a set, G, of elements (denoted by lower case letters of the alphabet - a, b, c, ... ) together with a law of composition (often called the multiplication in G) which is a map

GxG--tG which sends an ordered pair (a, b) E G x G to their product, denoted by ab E G. The multiplication in G satisfies the following axioms: Gl: (Associativity) a, b, c E G, Given any three elements belonging to G say- then (ab)c = a(bc)

G2: (Identity or Neutral element) There exists an element, e E G, such that ae =a= ea for all a E G. G3: (Inverses) Given any a E G there exists an element b E G, called the inverse of a, such that ab= e = ba. (Usually we shall write a- 1 for the inverse of a, because it is a notationally suggestive and convienient convention.) Axioms Gl - G3 are slightly redundant and can be replaced by the equivalent, meaner, leaner Axioms Gl, G'2, G'3 in which G'2 guarantees only a "left neutral" element, e, and G'3 guarantees only a "left inverse" for each a E G. Since these axioms are simpler to verify we shall pause to prove this result. Lemma 1.1.11 Let G be a set with a "product", as in §1.1.10, which satisfies Axiom Gl and G'2: (left identity) There exists an element, e E G, such that ea = a for all a E G.

G'3: Given any a E G there exists an element b E G such that ba = e. With these axioms, the product in G satisfies axioms Gl - G3 of §1.1.10 and G is a group. Proof To prove G2 we must show that ae = a for all a E G. We, at least, know that this is true for all a = e since, in this case, ee = e, is the same equation as that of G'2. By G'3, we have ba = e, so that we may substitute for two of the e's in the equation ee = e to obtain (ba)e

= ba.

Now choose c such that cb = e (it does not matter that, with our depleted axiom scheme, we do not yet know that c = a.). Multiplying on

1.1. THE CONCEPT OF A GROUP

7

the left by c yields

ae = (ea)e = ((cb)a)e = (c(ba))e = c((ba)e) = c(ba) = (cb)a = ea

=a

by by by by

G'2 GI Gl Gl

by Gl by G'2.

To verify G3 we must show that, if ba fore in eb = b we obtain

= e,

then ab

= e.

Substituting

b = (ba)b = b(ab). Now choose c such that cb = e and multiply this equation on the left by c to obtain e = cb = c(b(ab)) = (cb)(ab) by Gl = e(ab) = ab by G'2 which completes the proof of the lemma.D Remark 1.1.12 (i) Since the definition of §1.1.10 was intended to imitate the set of symmetries, with "multiplication" given by composition, we should verify that axioms Gl - G3 are true for this example.

Firstly, let us agree that if a and b are symmetries of X then ab is to be the symmetry given by first performing b and then performing a. With this convention (ab)c means the symmetry given by first performing c and then performing the composite symmetry called "first b then a". On the other hand, a(bc) means first perform the composite "first c then b" and then follow the result by performing a. Both these recipes are longwinded ways of describing the composite symmetry "first perform c, then b and then a". Therefore the axiom Gl is true for the symmetries of X. Secondly, let e denote the trivial symmetry, which leaves every point of X where it is. If a is any symmetry of X then the effect of the symmetry ae is first to move each point of X nowhere and then to perform a. This is merely a tortuous description of the symmetry, a. Similarly ea stands for the symmetry "first perform a and then leave every point of X where it is" , which is another roundabout description of a. Therefore the axiom G2 is true for symmetries of X.

8

CHAPTER 1. GROUP THEORY

The verification of G3 for symmetries of X is connected with the observation of §1.1.8 and is easily seen in those examples in terms of the "multiplication" tables of §§1.1.4, 1.1.6 and 1.1.7. In each of those examples the row labelled by Si contains each symmetry just once. In particular the trivial symmetry, s 1 , will appear in some column; the one labelled by Sj, say. In terms of "multiplication" of symmetries this is equivalent to the equation SiSj = s1 = e. Now, consulting the entry in the j-th row and i-th column is seen to yield the equation

In these examples, this process serves to verify axiom G3. A closer look at these examples shows that the inverse of the symmetry, a, is nothing more than the symmetry, b, of X which takes each point of X back to the point from which a moved it. For this choice of b it is clear that ba denotes "first move points of X by a and then put them back", which is an elaborate description of the trivial symmetry under which each point of X stays where it is; that is, ba = e. Similarly, for this choice of b, ab denotes "take a point of X back to where a found it and then return it to its original position by a"; that is, ab = e. (ii) To tabulate the table of compositions of symmetries of X, as in §§1.1.4, 1.1.6 and 1.1.7, is to tabulate the multiplication table of the group of symmetries of X. In general, given a group, G, and a modicum of determination we could depict the group by writing out a similar multiplication table. (iii) In axioms G2 and G3 reference is made to the identity element of G and the inverse of a E G. Before we go any further we should derive from the axioms the fact that those references to unique elements were not merely faux pas. That is, we should convince ourselves that a group does not have several neutral elements and that an element does not have several inverses. Let us pause to record the result of the discussion of §1.1.12 (i). Theorem 1.1.13 The set of all symmetries of X, as defined in §1.1.1, with "multiplication" given by composition in the manner of §1.1.12 (i), satisfies the axioms of a group. Henceforth this group will be referred to as the symmetry group of X. Lemma 1.1.14 (i) In axiom G2 of §1.1.10 the neutral element is unique. (ii) In axiom G3 of §1.1.10 the inverse of a E G is uniquely determined by a.

1.1. THE CONCEPT OF A GROUP

9

Proof For part (i), suppose that e1 and e2 satisfy axiom G2. Hence, setting e = e1 and a = e2 we have the equation

However, setting e = e2 and a= e1 in axiom G2 yields another equation

Combining these equations shows that

as required. For part (ii), we commence in a similar manner. Suppose that b1 and b2 are both candidates to be the inverse of a E G. From axiom G3 this yields two sets of equations,

and

ab2

= e = b2a.

We wish to show that b1 = b2 so we must find a way to solve these equations for b1 and b2 • If these were numerical equations we would multiply by the inverse of a but in our hypothetical group we have two inverses, b1 and b2 . Not knowing which is the better, we multiply by each of these inverses to obtain eight equations of which one is

However eb2

= b2

by axiom G2 and (b1 a)b2

Here we obtain b2

= b1 (ab2) = b1e = b1

= eb2 = (b1a)b 2 = b1 ,

by Gl, by G3, by G2. as required.D

1.1.15 The axiom of associativity was stated in §1.1.10 Gl only for products of three elements. In this case, by the fact that (ab)c = a(bc), we are justified in omitting the brackets and merely writing abc to denote the common value of (ab)c and a(bc). The omission of brackets is a considerable convenience. For this reason we next prove a result which licences us to omit brackets in more complicated products, too.

CHAPTER 1. GROUP THEORY

10

Lemma 1.1.16 Let G be a group and let a1, a2, ... , an be a set of elements of G. In this case all iterated products in G which are denoted by a1 a2 ... an punctuated by the (legitimate) insertion of n - 2 pairs of brackets represent the same element of G. This common value will be denoted by a1 ... an.

Proof When n = 3 this statement is simply axiom Gl of §1.1.10. Suppose, by induction, that the result is true for all iterated products of fewer than n elements of G. Consider a bracket-punctuated expression in which n - 2 brackets are inserted into the "word", a1 ... an. This expression will look, for some 1 ~ r ~ n, like (a1 ... ar)(ar+l ... an) where the word a1 .. , ar is infested with brackets and so is ar+l .. , an, However, by induction, the brackets within a1 ... ar and within ar+l ... an are irrelevant. In fact, all we have to do is show that, for 1 ~ r ~ s ~ n,

By induction, we may rewrite the left-hand side of this equation as (a1 .. ,ar)((ar+l ... as)(as+l .. . an))

= ((a1 .. ,ar)(ar+l ... as))(as+l ... an), by Gl, =

(a1 .. , arar+l ... as)(as+l ... an),

as required. D 1.1.17 Notation and terminology While we are engaged in making our notation more convenient, let us agree to abbreviate the expression

a ... a

(n copies of a E G)

to an, called then-th power of a. As we have seen in the multiplication tables of §§1.1.4, 1.1.7, it is not generally true that ab= ba for a, b E G. A group for which every pair a, b E G satisfies ab = ba is called an abelian group (after N.H. Abel (1802-29)) or a group whose multiplication is commutative. When ab = ba the elements a, b E G are said to commute. Sometimes, in an abelian group, we shall write the "multiplication" additively; that is, as a + b rather than ab. Question: Let e E G be the neutral element. Explain the equation en = e. Is it possible that an = e with a E G different from the neutral element?

1.2. EXERCISES

11

Definition 1.1.18 If G is a group containing only a finite number of elements we define the order of G, denoted by IGI, to be the number of distinct elements of G. By contrast, if a E G the order of a is the least integer, n, such that an = e. If no such integer exists we say that a has infinite order. Question: Why does every element of a finite group have finite order?

1. 2

Exercises

Exercise 1.2.1 Let a, b, c be elements of a group in which ab= ac. Show that b = c. Exercise 1.2.2 Prove that each element of a finite group appears once and only once on each row and column of the multiplication table. Exercise 1.2.3 The following table is part of the multiplication table for a finite group whose elements are called v, w, x, y. Complete the rest of the multiplication table. - vw X y V vw X y w w V - -

x-

V

-

y y- -

-

X

Exercise 1.2.4 The following table is part of the multiplication table for a finite group whose elements are called e, v, w, x, y, z. Complete the rest of the multiplication table.

-

e e e

V V

w X y z w X y z w e y - V

V

w w - - - - X X

z - - -

V

y y

- -

-

z z - - -

-

Exercise 1.2.5 Let G be a group consisting of eight elements denoted by the numbers {O, 1, 2, 3, 4, 5, 6, 7}. Suppose that the product of v and win G is denoted by v#w and that (i) v#w ~ v + w for all v, w E G and (ii) v#v = 0 for all v E G. From this information, reconstruct the multiplication table for G.

12

CHAPTER 1. GROUP THEORY

Exercise 1.2.6 (a) If X is a regular, plane n-gon show that the group of symmetries of X has 2n elements. This group is called the dihedral group of order 2n, denoted by D2n, (b) If a is a rotation clockwise through an angle of 2; and b is a symmetry which rotates X through 7r about a bisecting line (that is, reflects X in a bisecting line) show that an= e

= b2 and bab = an-l.

Exercise 1.2. 7 Describe the group of symmetries of a circular disc. In particular, show that this symmetry group is infinite. Exercise 1.2.8 Let G denote the set of symbols e, a, a 2 , ••• , an-l where n ::::: 1 is an integer. Define a "multiplication" on G by the formula . . { ai+j if i + j < n - 1 a'a1 = ar if i + j = n + r

where a0 = e and a 1 = a. Prove that G, with this multiplication, is a group. G is called the cyclic group of order n, generated by a E G. Exercise 1.2.9 Let Ji, h, h, f4, fs, f5 be the rational functions of a real variable, x E R, given by

fi(x) = x, h(x) = h(x) = l~x f4(x) = fs(x) = 1 - x f5(x) =

x::;;-l

½

x:.

1 .

(a) Show that the set G = {!1, h, ... , f5}, together with the "multiplication" defined by composition of functions of x (for example, (fs) 2 (x) = fsUs(x)) = 1- (1-x) = fi(x) so that ft= Ji and, similarly, h(h(x)) = h((x - 1)/x) = 1/(1 - x) = h(x) so that Ji = h) forms a group. (b) Show that this group has a multiplication table which is identical to that of the symmetries of an equilateral triangle (see §1.1.4). Two groups with identical multiplication tables are called isomorphic ( after attempting (b) see §§1.3.3, 4.2.5). Exercise 1.2.10 Show that the set, GL2(R), of 2 x 2 matrices with nonzero determinant form a group under matrix multiplication. This group is called the general linear group of 2 x 2 matrices. Exercise 1.2.11 (a) Let G be a group containing elements a, b and c. Show that ab = ac if and only if b = c. (b) Why does each element of a finite group, G, appear once and only once in each row and each column of the multiplication table of G.

1.2. EXERCISES

13

Exercise 1.2.12 Let c be a real constant (the speed of light!) and let v be a real variable such that -c < v < c. Write A( v) for the 2 x 2 matrix

-v)

1 A(v) = >.(v) ( ~ 1

where >.( v) = (1 _ ( ~: ))(-1/2). (Note that ( 1 - ( ~:)) is equal to the determinant of (

!v c2

-v) 1

.

(a) Show that the product of these matrices satisfies

where V1

+ V2

v3----

- l+~ C

(b) Use (a) to show that matrix multiplication makes the set

G = {A(v)I - c < v < c} into a group.This group is the Lorentz group; it is the group of symmetries of two-dimensional special relativity. Exercise 1.2.13 (i) Show, for all a, b E G, that b- 1 a- 1 is the inverse of ab (that is, (ab)- 1 = b- 1 a- 1 ). (ii) Show also that (a- 1 )- 1 = a. Exercise 1.2.14 (a) In the group, GL 2 (R) (see §1.2.10), show that

has order three and

b= (_~

~)

has order four. (b) Show that ab has infinite order. Exercise 1.2.15 In the world of real numbers the equation x 2 = e can have at most two solutions for x given e. Give an example of a group Gin which there are at least three elements, x E G, such that x 2 = e, where e is the neutral element. Exercise 1.2.16 Prove that every non-abelian group, G, has order at least six; hence every group of order 2, 3, 4 or 5 is abelian. (Hint: If a, b E G and ab is not equal to ba then the elements { e, a, b, ab, ba} are all distinct. Show that either a 2 is not in this set or a 2 = e. In the latter case verify that aba in not in this set. If you prefer, you may prove directly, case by case, that every group of order 2, 3, 4 or 5 is abelian.)

CHAPTER 1. GROUP THEORY

14

1.3

New groups from old

In this section we shall examine two elementary methods for making more examples of groups from our current store. The first of these procedures is the product. Definition 1.3.1 Let H and G be groups. On the product set, H x G, consisting of all ordered pairs, ( h, 9), with h E H, 9 E G, we may define a "multiplication" by the formula (h, h1 E H; 9, 91 E G)

This coordinate-wise multiplication group. In fact, the following result holds.

makes

H

x G

into

a

Proposition 1.3.2 With the notation of §1.3.1, (i) H x G is a group, (ii) if IHI = n and IGI = m then IH x GI = nm and (iii) if H and G are abelian (see §1.1.17) so is H x G. Proof To prove (i) we have to verify axioms Gl - G3 G, assuming them for H and G separately. This is very easy. If h1,h2,h3 EH and 91,92,93 E G then

( (h1, 91)(h2, 92) )(h3, 93)

,

= = = =

(h1 h2, 9192)(h3, 93) ((h1h2)h3, (9192)93) (h1(h2h3),91(9293)) (h1, 91)(h2h3, 9293) = (h1,91)((h2,92)(h3,h3))

by by by by by

for

H x

§1.3.1, §1.3.1,

Gl, §1.3.1, §1.3.1,

which verifies the associativity axiom. If es, ec are the neutral elements of H and G respectively then, for all h EH, 9 E G, (h,9)(es,ec)

= (hes,9ec) = = =

by (h, 9) by (esh, ec9) by (es,ec)(h,9) by

§1.3.1, G2, G2, §1.3.1,

so that (es, ec) is the identity element of H x G. If h- 1 and 9- 1 are the inverses of hand 9 then (h,9)(h- 1,9- 1)

= = = =

(hh- 1,g9- 1) (es,ec) (h- 1h,9- 19) (h- 1,9- 1)(h,9)

by by by by

§1.3.1, G2, G2, §1.3.1,

1.3. NEW GROUPS FROM OLD

15

so that (h,g)- 1 = (h- 1,9- 1), which verifies axiom G3. Part (ii) is clear, being an elementary property of the cardinality of the product of two finite sets. If H and G are abelian then (h,g)(h1,91)

= (hh1,gg1) = (h1h,g1g) = (h1, g1)(h, g)

for all h, h1 E H and g, 91 E G.D 1.3.3 Relations between groups It is possible for two groups, G1 and G2, which have been constructed or discovered in very different ways, to have identical multiplication tables. That is, it may be possible to define a one-one correspondence,¢ : G1 ---+ G2, between the elements of G1 and G2 so that applying ¢ to each element of the multiplication table of G1 gives that of G2. In such a case it would be foolish to waste energy independently studying the group-theoretic aspects of G1 and G2 separately, since any group-theoretic fact about G1 could readily be transformed into a similar fact about G2, via¢. When such a one-one correspondence (or bijection), ¢ G1 ---+ G2, exists we call ¢ an isomorphism (this terminology is easy to remember from its etymology, iso-morphism = identical-form). One may press this observation a little further. If we have any map of sets, ¢ : G1 ---+ G2, which sends each product ab E G1 to ¢(a)¢(b) E G2 then we may be able to use ¢ to deduce group-theoretic facts about G2 from G1 or vice versa. A map, ¢ : G1 ---+ G2, with this useful property' is called a homomorphism. The formal definitions of an isomorphism and a homomorphism will follow shortly, in §4.2.5; this discussion is intended to explain why such notions are useful. Example 1.3.4 In §§3.1.3 and 1.1.4 we find the group of symmetries of a plane equilateral triangle. It has six elements, G 1 = {s1, s2, S3, S4, ss, 86}. In Exercise 1.2.9 we find a group consisting of six functions, G2 = {!1, h, /3, f4, fs, !6}. I claim that the map given by ¢(si) = Ji for 1 ~ i ~ 6 is a bijection, ¢: G1 ---+ G2, which transforms the multiplication table of G1 into that of G2. In order to verify this we could compute the thirty-six products in G2 and compare the resulting table with that of §1.1.4. Such a method is rather laborious. Instead we will 'use generators and relations.

A set of generators for a group G is any set, S, of elements of G having the property that all elements of G may be obtained by taking iterated products of elements of Sand their inverses. A relation between members of S is simply a formula which equates two iterated products of elements of S or their inverses. For example, in G1 the subset {s2, s4} = S1 is a set of generators because, from §1.1.4, we find that

CHAPTER 1. GROUP THEORY

16

In addition there are relations which include

which enables us to compute all the products between any pair from the set As an example, 8586

= 848~8482 = 8482(8284)82 = 84 ( 8284)8~82 = 84848~8~ = 8~ = 83,

as expected. Since the generators, {8 2, 84}, and the relations betweeen them enable us to calculate all the products in G 1 we will verify the same formulae for {h, /4}. It will then be obvious that ¢ transforms the multiplication table of G1 into that of G2. Here we go, without further ado:

J'd(x) = h(h(x))

by definition,

_

h(x)-1

=

(x-1)-x / x-1

c.f. Exercise 1.2.9,

-h(x) X

X

1 - x-1 1 = 1-x = /3(x) , fi(x) = hU'i(x))

=h(1~x) -_ (-1 1-x _ 1)/-11-x

= (1-1 + x) = fi(x) ,

1.3. NEW GROUPS FROM OLD

17

==

_L

x-1 = f5(x) , f4Ji(x) = f4(1~x) =1-x = f5(x) and for the remaining relations

Jl(x) = f4(¾) = 1/(¾) = x = fi(x),

f2f4(x) = h(¾) = ((¾) -1)/(¾) =1-x = h(x) = f4Ji(x) , as required.

Example 1.3.5 Let G 1 = {e, a, a 2, ... , an-I} denote the cyclic group of order n, generated by a, as defined in Exercise 1.2.8. Let G 2 consist of the set of remainders modulo n. That is,

G2 = {O, 1, 2, ... , n - 1} with the "multiplication" operation denoted by # and defined by

r#s = t where

r

+ s =kn+ t for

some integer, k. In plain terms, for O ~ i,j i# . = J

{i +j t

~

n- 1

if i + j ~ n - 1 'f.i+J=n+. . t

1

Hence the map, ¢ : G1 - - ; G2 , defined by ¢(ai) multiplication table of G 1 into that of G 2 .

= i transforms the

Definition 1.3.6 Let G1 and G2 be groups. Let ¢ : G1 --; G2 be a map of sets, sending a E G 1 to ¢(a) E G 2 . Then¢ is called a homomorphism if it satisfies the following condition: (i) for all a, b E G, ¢(ab)= ¢(a)¢(b) in G2. A homomorphism which is also a bijection of sets is called an isomorphism. Hence an isomorphism is a homomorphism, ¢ : G 1 - - ; G 2 , such

18

CHAPTER 1. GROUP THEORY

that ¢(a) = cp(b) implies a= band, given any c E G 2, there exists a E G 1 G2 to indicate that the homosuch that ¢(a) = c. We write ¢ : G1 morphism, ¢ : G1 ---. G2, is an isomorphism. The following result records the most basic properties of homomorphisms and isomorphisms.

-=-.

Proposition 1.3. 7 With the notation of §4.2.5, let ¢ : G 1 ---. G 2 be a homomorphism. (i) If e1 E G1 is the neutral element then ¢(e 1) is equal to e 2, the neutral element of G 2. (ii) For all a E G 1, ¢(a- 1) is the inverse of ¢(a) in G 2. (iii) Let ¢ : G1 ---. G2 be an isomorphism and define a map of sets 'lj; : G2 ---. G1 by 'lj;(c) = a where a E G1 is the unique element such that ¢(a) = c E G 2. Then 'ljJ is also an isomorphism (called the inverse isomorphism of¢) and 'lj;(¢(a)) = a, ¢('1j;(c)) = c for all a E G 1, c E G 2. Proof For part (i) we have to show that ¢(e 1) = e2. By Lemma l.l.14(i) we know that the neutral element of G2 is unique. In fact, if we can find any elements, x, y E G2 satisfying xy = x then y = e2. This is because xy = x implies that

e2 = x- 1x = x- 1(xy) = (x- 1x)y = e2y = y. But e1e1 = e1 in G1 so that ¢(e1)¢(e1) = ¢(e1) in G2. Therefore, choosing x = ¢(e1) =yin the previous argument shows that ¢(e 1) = e 2. For part (ii) we have to show, for a E G 1, cp(a)cp(a- 1 ) = e 2 = ¢(a- 1)¢(a) where ei E Gi is the neutral element. However,

cp(a)cp(a- 1 ) = cp(aa- 1 ) by §4.2.5 (i), = ¢(e1) by G3, = e2 by §1.3.7 (i). Similarly ¢(a- 1)¢(a) = e2 E G2. For part (iii), we have to show that if 'lj;(c 1) = a 1 and 'lj;(c 2) = a 2 then 'lj;(c1c2) = a1a2. However,by definition of 'lj;, 'lj;(c1c2) = a1a2 if and only if ¢(a1a2) = c1c2. However, this is true, since we have ¢(a1a2) = ¢(a1)¢(a2) = c1c2. The fact that 'lj;(e2) = e 1 follows from §4.2.5 (ii) for¢; namely, ¢(e 1) = e2. The equation 'lj;(¢(a)) = a holds if and only if, by definition of 'lj;, ¢(a) = ¢(a). Finally if ¢(a) = c then a= 'lj;(c) so that c = ¢(a) = ¢('1j;(c)), which concludes the proof.D

1.3. NEW GROUPS FROM OLD

19

Example 1.3.8 Let { ±1} denote the group consisting of the two integers, +1 and -1, with "multiplication" in the group being given by the usual multiplication of integers. Let D2n (see Exercise 1.2.6) denote the symmetries of a plane, regular n-gon. Define a map of sets

¢:

D2n-->

{±1}

in the following manner. Imagine the regular n-gon lying on a horizontal table and having its upper and lower faces coloured in different shades. If g E D2n is a symmetry which changes the colour of the uppermost face, set ¢(g) = -1, and set ¢(g) = 1 otherwise. Question: Is¢: D2n--> {±1} a homomorphism? 1.3.9 Subgroups A second manner in which to manufacture new groups from our current stock is to look for subsets, H, of a given group, G, which are themselves groups, when endowed with the "multiplication" operation of G. Such a subset is called a subgroup of G and the following conditions ensure that H together with the multiplication and the inverse operations of G forms a group. A subset, H, of the group G is called a subgroup if (i) for all a, b E H the product, ab E G, lies in H, (ii) for all a EH the inverse, a- 1 E G, lies in H. Notice that the neutral element of G will lie in H if §1.3.9 (i) / (ii) are true, since e = aa- 1 for any a E H. Example 1.3.10 The subset {Ji, h, h} is a subgroup of the group, G2, in the Example 1.3.4. Question: Why is this subgroup isomorphic to the cyclic group of order three (see Exercise 1.2.8)? Question: If n divides m, how may the group of symmetries of a plane, regular n-gon be considered as a subgroup of the group of symmetries a plane, regular m-gon? Question: If His a subgroup of G, define¢: H--> G by ¢(a) = a for a E H; this is the inclusion mapping. Is ¢ a homomorphism? 1.3.11 Lagrange's theorem In order to make further progress in the study of groups and subgroups and isomorphisms between them we need some more powerful results. For finite groups the first of these will be Lagrange's Theorem (§1.3.13) which states that if His a subgroup of a finite group, G, then the order of H must divide the order of G. The quotient of these two numbers is called the index of Hin G and is written [G: HJ = IGI/IHI, You will be able to appreciate the usefulness of this result by considering the following question with and without the aid of Lagrange's Theorem.

CHAPTER 1. GROUP THEORY

20

Question: What are all the subgroups of D 8 (see §1.1.7 and Exercise 1.2.6)? The next problem is: How to show that IHI divides IGI? We are going to adopt a very simple-minded approach to this, an approach which is forced upon us by the fact that we know very little about G. Namely, we are going to find a process of dividing the set, G, into a number of disjoint subsets called cosets of H; each coset will have precisely IHI elements. As an obvious corollary we shall obtain Lagrange's Theorem. A right coset of Hin G is a subset of the form, for a fixed g E G, Hg= {hglh EH}

:s; G.

That is, the right coset Hg consists of all the elements z E G which may be written z = hg for some h EH. Similarly, given g E G and H :s; G, we may define left cosets, which are subsets of the form

gH = {ghlh E H}

:s; G.

Now we shall prove the technical result which will yield Lagrange's Theorem. Proposition 1.3.12 Let H be a subgroup of a group G. (i) For g E G the map>.: H Hg given by >.(h) = hg is a bijection. (ii) For 91,92 E G either Hg1 = Hg2 or Hg1 is disjoint from Hg2. Proof For part (i) we must show that >. : H Hg is both one-one and onto. Defineµ : Hg H by µ(z) = zg- 1. If z E Hg then z = hg for some h EH and so zg- 1 = (hg)g- 1 = h(gg- 1) =he= h which lies in H. Also >.(µ(z)) = >.(zg- 1) = (zg- 1)g = z and µ(>.(h)) = µ(hg) = (hg)g- 1 = h so that the compositions, >.µ and µ>., are both equal to the identity map. Hence >. is a bijection whose inverse bijection is µ. For part (ii), we must show that if z E Hg1 n Hg2 then Hg1 = Hg2. To this end, write such an element, z, as z = h1g1 = h2g2 with h1, h2 EH. If hg1 E H 91 is an arbitrary element then

hg1

which lies in Hg2 so that Hg1 Hg1 = Hg2.D

= heg1 = hh1 1h1g1 = (hh1 1h2)92 :s; Hg2. Similarly, Hg2 :s; Hg1, so that

Theorem 1.3.13 (Lagrange) Let G be a finite group and let H be a subgroup. Then there exists a positive integer, [G: HJ, the index of Hin G, such that IGI = IHl[G: HJ.

1.3. NEW GROUPS FROM OLD

21

Proof Given any g E G then g = eg E Hg so that, by §1.3.12 (ii), G is the disjoint union of a finite number of right cosets. By §1.3.12 (i) each of these right cosets possesses precisely JHI elements, which yields the result.D

Example 1.3.14 Let G denote the group of symmetries of the regular tetrahedron, as in §1.1.9. Since G contains a subgroup of order three consisting of the neutral element and two rotations which fix a chosen vertex, we see that 3 divides JGI. Also the three rotations through 1r about the lines joining midpoints of opposite sides, together with the neutral element, form a subgroup of order four. Hence 4 divides IGI and therefore 12 divides JGJ. This is as expected, since JGI = 12. Corollary 1.3.15 Let G be a .finite group and let g E G. The order of g (see §1.1.18) divides IGJ, the order of G. Proof If n is the order of g then the subset H

of G with n

= {e, g, g 2 , ••• , gn-l} is a subgroup

= IHl,D

1.3.16 Application: Classification of all groups of order eight We shall conclude this section by applying §1.3.15 to the problem of determining all groups of order eight. Let Cn denote the cyclic group of order n. Using the product we may construct the following abelian groups of order eight: C2 x C2 x C2, C2 x C4, Cs.

In addition, we have the dihedral group of order eight, Ds (see Exercise 1.2.6), which is generated by a and b satisfying a4 =e=b2 , ab=ba 3 .

The elements of Ds are Ds

= {e,a,a 2 ,a3,b,ba,ba2 ,ba3 }.

Also the set of quaternions of the form { ± 1, ±i, ±j, ±k} with quaternion multiplication as the group operation yields a non-abelian group, Qs, called the quaternion group of order eight. If we set a = i and b = j we see that a 2 = b2 = -1 which is not the neutral element and ab= ij = k = -ji = ji3 = ba3 •

Hence Qs is generated by a and b satisfying a2 = b2 , a4 = e, ab

= ba 3 •

CHAPTER 1. GROUP THEORY

22 Again the elements of Qs are Qs

= {e, a, a 2 , a 3 , b, ba, ba 2 , ba3 }.

If G is a non-abelian group of order eight then it cannot contain an element of order eight - for then it would be isomorphic to Cs. Also, being non-abelian, G must have an element of order four, a E G, by Exercise 1.4.6. Now choose b E G- H where H = {e,a,a 2 ,a3 }. It is easy to see that the eight elements of Gare HU bH. Similarly G =HU Hb so that Hb = bH. Therefore ab= bai for some O ~ i ~ 3. However, we have an inner automorphism (see Exercise 1.4.2) ¢: G ~ G given by cp(z) = bzb- 1 and so ¢(a)= ai. Since¢ is an isomorphism, the order of a must equal that of ai. The order a 0 = e is zero and that of a 2 is two soi= 1 or 3. If i = 1 then ab= ba and G is abelian so we must have ab=ba 3 .

Next we consider b2 E G = HUHb. If b2 = ajb then b = b2b- 1 = ajbb- 1 = aj E H, which is a contradiction. Also b4 = e, since the order of b is two or four. If b2 = e then G = Ds and if b2 -/:- e then b2 = a 2 and G = Qs. From this discussion we conclude that, up to isomorphism, there are five groups or order eight; namely

C2 x C2 x C2, C2 x C4, Cs, Ds, and Qs.

1. 4

Exercises

Exercise 1.4.1 Let Hand K be subgroups of the group, G. (a) Prove that the intersection, H n K, consisting of all elements lying both in Hand in K, is a subgroup of G. (b) Prove, by giving an example, that the union, H U K, consisting of all elements lying in either H or in K or in both, need not be a subgroup of G. Exercise 1.4.2 Let G be a group and let g E G. Define ¢ : G--, G by cp(x) = gxg- 1 for all x E G. Show that ¢ is an isomorphism. (Such an isomorphism is called the inner automorphism of G given by conjugation by g.) Exercise 1.4.3 (i) Prove that the group of symmetries of a cube has order divisible by 24. (ii) The regular octahedron is the solid, diamond-shaped polyhedron whose eight vertices are the centroids of the eight faces of a cube. Show that the groups of symmetries of the cube and the octahedron are isomorphic. Exercise 1.4.4 (i) Show that the group of symmetries of the regular dodecahedron has order divisible by 60.

1.4. EXERCISES

23

20 vertices each with 3 edges at 3n:/5 30 edges 12 faces each a regular pentagon

(ii) The regular icosahedron is the solid convex polyhedron whose twelve vertices are the twelve centroids of the twelve faces of the dodecahedron. Prove that the symmetry groups of the dodecahedron and icosahedron are isomorphic. Icosahedron

12 vertices each with 5 edges at 7r/3 30edges 20 faces each an equilaterat triangle

Exercise 1.4.5 (i) Show that the set of 2 x 2 matrices

H = {(

~:) I

a, b E R, a

~ 0}

is a subgroup of GL2R. (ii) Let R * denote the set of non-zero real numbers with multiplication as the group composition operation. Let R denote all the real numbers with addition as the group composition operation. Find an isomorphism of the form ¢:R* xR~H.

CHAPTER 1. GROUP THEORY

24

Exercise 1.4.6 Let G be a group in which every element is equal to its own inverse. (i) Show that g 2 = e for every g E G. (ii) Show that ab= ba for all a, b E G, so that G is abelian. Exercise 1.4. 7 Let H 0 ::;: G be a set of subgroups, indexeq by a E A. Show that the intersection of all Ho.,

H=

nH

0

o.EA

is a subgroup of G. Exercise 1.4.8 Let Z denote the group given by the integers with "multiplication" given by addition. Show that, for each k E Z,

kZ = {ak E Zia= 0,±1,±2, ... } is a subgroup of Z. Exercise 1.4.9 (i) Let G denote the set of non-zero real numbers with a "multiplication" given by

a#b = { ab if a > 0 a/b if a< 0 Prove that G is a group. (ii) Does there exist an isomorphism

.(a, z) = az for all a EA, z E ker(¢). Proof We must first show that >. is a homomorphism, which is where the fact that Bis abelian is used. For a, a1 EA and z, z1 E ker(¢) we have

= >.(a, z)>.(a1, z1), which verifies the conditions of §4.2.5(i). To see that>. is injective we must show that ker(>.) = {(e, e)}. However, for an arbitrary element of ker(>.) we have e

= >.(a, z) = az

so that a= z- 1 E ker(¢). Therefore e =¢(a)= a and z = a- 1 = e which implies (a, z) = (e, e), as required. To see that >. is surjective, take b E B and write b = ¢(b) (¢(b)- 1b) so that ¢(b) EA. Finally, ¢(b)- 1 b E ker(¢) since ¢(¢(b)- 1b) = (¢(¢(b)))- 1¢(b) = ¢(b)- 1¢(b) = e, which shows that b E im(>.), as required.D Definition 1.7.8 Let A be an abelian group. Define Tors(A) to be the subset of A given by

Tors(A) = { all a E AJ am= e for some integer, m}. By Exercise 1.8.5, Tors(A) is a subgroup of A. For example, in 1.7.6, Tors(Z x ... x Z x (Z/ s1Z) x ... x (Z/ StZ)) = (Z/ s1Z) x ... x (Z/ StZ). The following result will permit us to concentrate on the structure of finite abelian groups.

44

CHAPTER 1. GROUP THEORY

Proposition 1. 7.9 Let A be a finitely generated abelian group then there is an isomorphism of the form

Tors(A) x Z x ... x Z

~

A,

in which the number of Z-factors depends only on A and is called the rank of A. Tors(A) is a finite abelian group. Proof Let us assume temporarily that there exists a non-trivial homomorphism, ¢ : A --+ Z. The image of ¢ is a subgroup of Z so that im( ¢) = sZ for some integer, s ~ 1. If we replace¢ by¢' given by¢' (a) = ¢(a)s- 1 E Z we obtain a surjective homomorphism, ¢' : A --+ Z. Choose any a E A s1;1ch that ¢' (a) = 1 then a cannot have finite order because ¢' (an) ¢ (a)+ ... +¢ (a)= n. Hence the subgroup of A given by

C = {... , a- 2 , a- 1 , e, a, a2 , .. . }

--=-.

is isomorphic to Z, via >. : Z composition

C given by >.(n)

q/

an. Hence the

.X

:A--+Z--+C

satisfies (an) = >.(¢ (an)) = >.(n) = an and 1

A~ C x ker()

~

Z x ker(),

by Lemma 1.7.7. As long as non-trivial homomorphisms, = Q., the neutral element,

for all 'M. EC.} From the linear theory of algebra for vector spaces over Z/pZ we know that JC1- J= pn-k. Hence, we can choose vectors which generate c1- and form an (n - k) x n generator matrix for CJ_

H -

c~:'.' ~.:-,., :::~•J .

Since the rows of G represent elements C < (Z/pZ)n and those of H represent elements of c1- then, if Ht denotes the transpose of H, GHt is a k x (n - k) matrix each of whose entries are divisible by p. The matrix, H, may be used to test for errors, by means of the following result.

1.9. ABELIAN GROUPS AND CODES

61

Proposition 1.9.14 The vector±= (pZ + x 1 , ... ,pZ + xn) E (Z/pz)n lies in C if and only if (pZ + Y1, ... ,pZ + Yn-k) = Q, the neutral element of(Z/pz)n-k, where (Y1,.,.,Yn-k) = (x1, ... ,xn)Ht. H is called a parity check matrix for C, by virtue of its role in Proposition 1.9.14. Example 1.9.15 If G = (h, P) where h is the k x k identity matrix and P is a k x (n - k) matrix of integers then

1.9.16 Decoding The decoding step merely consists of taking a receiving word, ;r_ E (Z/pz)n, and finding an element y E C such that d(±, y) is minimal. We shall describe a systematic process for accomplishing-this which uses a parity check matrix, H, of C. Given±= (pZ + x1, ... ,pZ + Xn) E (Z/pZ)n we may calculate (z 1, ... , Zn-k) = (x1, ... , Xn)Ht and examine the element

± = (pZ

+ Z1,,,, ,pZ + Zn-k)

E

(Z/pZ)n-k.

This element, ± E (Z/pZ)n-k is called the syndrome of±· If Q = ± E (Z/pzr-k then ± E C, by Proposition 1.9.14. The syndrome of± determines the coset C + ±:::: (Z/pZ)n. If (pZ + c1, ... ,pZ + cn) EC then (c 1 , ... , cn)Ht is a vector which represents the neutral element in (Z/pz)n. Hence

represents the same element of (Z/pZ)n as does±· A standard array associated to a code C :::: (Z/pZ)n with IOI = pk is an array of pn-k rows in which each row has 1 + pk entries. The first entry of each row is an element of (Z/pzr-k given by the syndrome of a coset C + ± and the remaining entries of that row consist of the pk elements of the coset C + ± with the first of these being ± E C + ± such that d(±, Q) is minimal. That is, d(±, Q) :::: d(± + .Q, Q) for all .Q E C. ~An element is a coset which is of a minimal distance from Q E (Z/pz)n is called a coset leader. In order to decode a word,± E (Z/pZ)n, we inspect the standard array to see to which coset ± belongs. If ± is the chosen coset leader for this coset then ± - ~ E C and d(L ± - ±) = d(Q, ±). However, for any .Q E C then d(Q, ±) :::: d(Q, ± - r) and d(±, ± - ±) :::: d(±, (± - ±) + r) so that ± - ± is an element of C whose distance to± is minimal (although there may be other elements of C which are the same distance from ±).

62

CHAPTER 1. GROUP THEORY

Example 1.9.17 The following code was one of the first linear codes to be discovered, in 1947. Let C:::; (Z/2Z) 7 be given by the generator matrix

G- (

1000011) 0100101 0010110

.

0001111 Hence ICI = 24 . A parity check matrix is given by (the minus sign in _pt is omitted since p = 2)

0111100) H= ( 1011010 . 1101001 The syndrome of (x1, ... , X7) is represented by

(X1, ... , X7) Ht = (X2 +

X3

+x2 +

+

X4

X4

+

X5,

X1 +

X3

+

X4

+ X6, X1

+ X7)

in (Z/2Z) 3 . There are eight cosets.

1.10

Exercises

Exercise 1.10.1 Verify the metric properties of d(2:,_, y) in §1.9.2. Exercise 1.10.2 Prove Proposition 1.9.14. Exercise 1.10.3 Justify the claims of Example 1.9.15. Exercise 1.10.4 Find a coset leader for each syndrome-value in the Hamming code of Example 1.9.17. What is d(C) for this code? Is Can optimal code?

1.11

Sylow's theorems

1.11.1 Lagrange's Theorem (§1.3.13) tells us that if G is a finite group and H is a subgroup then IHI divides IGI, This considerably restricts the number of possibilities for subgroups, H. However, suppose that m is an integer which divides IGI, what can we say about the existence of a subgroup, H:::; G, with IHI= m? To answer these existential questions is considerably more difficult than merely obtaining restrictions on IHI, One of the first results in this direction was the following:

1.11. SYLOW'S THEOREMS

63

1.11.2 Cauchy's Theorem Let G be a finite group and suppose that pis a prime which divides Then there exists an element of G whose order is equal to p.

IGI.

Remark 1.11.3 Cauchy's Theorem is a corollary of Sylow's First Theorem (§1.11.4). Notice that it is the best result possible concerning orders of elements. More precisely take, for example, the group of isometries of a regular tetrahedron (see §1.1.9). It has order twelve. An element of order three is given by rotation through 2n /3 about an axis joining a vertex to the centroid of the opposite face. A rotation through 1r about an axis joining the midpoints of two opposite sides has order two. However, even though four divides twelve, there are no elements of order four in this group. 1.11.4 Sylow's First Theorem [2) Let G be a finite group with IGI = pkm and HCF(p, m) = 1. Then there exists a subgroup, H ::; G, with IHI =pk. In this case, H is called a Sylow p-subgroup of G.

Proof Let a E G and recall, from §1.5.17, that the conjugacy class of a (i.e. the set of all conjugates of a) is given by

Ca= {all z E

GI z = bab- 1 for some b E G}.

Recall also that, if Za is the centraliser of a,

I za = az}

Za = { all z E G

then Za is a subgroup of G and ICal = [G: Za]· Finally, recall (from the proof of Theorem 1.5.19) that G is the disjoint union of the distinct Oa's so that 1.11.5 V

IGI = IZ(G)I + I)G: ZaJ j=l

In 1.11.5 Z(G) is the centre of G (see Definition 1.5.16) so that !Cal= 1 if and only if a E Z(G). Also a1, ... , av in 1.11.5 do not lie in Z(G). Let us

assume that the result is true for all groups of strictly smaller order than G. If there exists 1 ::; j ::; t such that p does not divide [G : ZaJ] then pk must be the highest power of p to divide IZaJ I, by Lagrange's Theorem (§1.3.13). Since aj (/. Z(G), we have a strict inequality IZaJI < IGI and, by induction, ZaJ has a subgroup of order pk and therefore so does G.

CHAPTER 1. GROUP THEORY

64

The only case remaining to consider is that in which each [G : Za 3 ] is divisible by p. Hence p divides IZ(G)I. Since Z(G) is a finite abelian group, Theorem 1.7.19 implies that Z(G) contains at least one element, x, of order p 8 • Hence y = xPs-i E Z(G) has order p (see Exercise 1.12.3). Let N :S: Z(G) be the cyclic group generated by y, N = {e,y,y 2, ... ,yP- 1}. Since aya- 1 = y for all a E G, N 0 for j lb;181bjl = l81I we see that f.j = 0 if and only if

= 2, ... , v. Since

bJ 181bi::; Na81. Being a p-group, b;181bj is in the kernel of 1r: Na81---+ (Na81)/81 (c.f. §1.11.8) but this kernel equals 81. Hence b-; 1 81bj = 81 and therefore bi E Na81 and 81bjNa81 = 81eNa81, which is a contradiction. Hence f.j > 0 for j = 2, ... , v and k = 1 + up, as required.D The remainder of this section will be devoted to miscellaneous applications of the Sylow Theorems. 1.11.17 Proof of Theorem 1.11.2 If IGI = pkm with HCF(m,p) = 1 and k ~ 1 then there exists a Sylow p-subgroup, 8::; G. If e-=/- x E 8 then the order of xis pd for some d ~ 1. d-1 Hence the order of xP is p.D 1.11.18 Corollary to 1.11.6 Let 8::; G be a Sylow p-subgroup of G. Then 8 1) roots in E - F. We assume, by induction, that our result is true for all pairs of fields and separable polynomials with fewer than n roots outside the subfield. Let f(x) = p1(x) .. ·Pr(x) with

120

CHAPTER 3. GALOIS THEORY

irreducible Pi(x) E F[x]. We may suppose deg(p1(x)) = s > 1. Let 0:1 EE be a root of P1(x) so that [F(o:1) : F] = s. Now f(x) E F(o:i)[x] and E/F(o:1) is a splitting field for f(x) so that E/F(o:1) is normal. Hence, for any z E E - F(o:1) there exists O' : E ~ E such that O'(z) =/- z and O' I F(o:1) = 1. Since f (x) is separable the roots of Pl (x), 0:1, ... , o: 8 E E, are distinct elements of E. By Theorem 3.1.10 there exist 0'1, 0'2, ... , O'n such that O'i : F(o:1) ~ F(o:i) and O'i I F = l. By Theorem 3.1.14, E / F (0:1) is a splitting field for f (x) and so is E / F (O:i). Therefore each O'i : F(o:1) ~ F(o:i) extends to O'i : E ~ E. Now let z EE be fixed by all O': E ~ E such that O' J F = l. By induction we know that z E F(o:1) so that z =Co+ c10:1 + ... + C8 -10:i-l with each Cj E F. Now, applying z = O'i(z) = Co +c10:i + ... +cs-10::- 1

so c8 _1ts-l + ... + (co - z) has s roots in E and has degree less than or equal to s - 1. Therefore it is zero and z E F. Now, for the converse, suppose that E / F is normal and let {0'1, ... , O' 8 } = G F and F = Ea F. If o: E E then the minimal polynomial of o: has distinct roots and splits into linear factors in E[x]. This is seen as follows. Firstly, there exists f(x) E F[x] such that f (o:) = 0 since [E : F] < oo and so 1, o:, o: 2, ... are linearly dependent. Let f(x) be the minimal polynomial of o: over F. Let 0:1, 0:2, ... , O:u be the distinct elements in the set {O'i (o:)} then O'j (o:i) equals another O:j and therefore p(x) = (x-0:1 )(x-0:2) ... (x-o:u) E E°F[x] = F[x]. Then the minimal polynomial, f(x), of o: divides p(x) so it too has distinct linear factors in E[x]. In fact, deg(f(x)) 2'. u since all the O:i are roots of f(x) and so p(x) = f(x). Now let w1, w2, ... , Wv E E generate E/ F so that E = F(w1, w2, ... , wv), Then, if fi(x) E F[x] is the minimal polynomial Wi, it is separable and we may set f(x) = TI~=l fi(x). Clearly E is the splitting field for this f(x).D Definition 3.3.16 If f (x) E F[x] is separable and E / F is its splitting field, then Gp = {O' : E ~ E I (O' I F) = 1} is called the group of the equation f (x) = 0. Theorem 3.3.17 (Fundamental Theorem of Galois Theory) Let f(x) E F[x] be a separable polynomial and let E/F be its splitting field. Let G be the group of the equation f(x) = 0. For any subfield, F :'.SB :'.SE, set GB= {O': E ~ E I (O' I B) = 1}. For H :'.S G set EH= {e EE I O'(e) = e for all O' EH}. (i) Then we have a one-one correspondence of the form

{subfields, B}

f-----t

{subgroups, H}

3.3. GROUP CHARACTERS

121

given by

B

~ GB,

EH~H.

(ii) B / F is normal if and only if GB .

= 1/2.)

Exercise 3.10.6 Let f(x) E Q[x] be an irreducible polynomial of degree p, a prime, having precisely p- 2 real roots. Let E/Q denote the splitting field of f(x). Generalise Example 3.9.7 to show that the Galois group of E/Q is isomorphic to the symmetric group, Sp, consisting of all permutations of a set of p elements.

3.11

Tensor products

Definition 3.11.1 If A is a ring then a left A-module is an abelian group, B, with a multiplication AxB---+B such that, for all.>., .>.1, .>.2 EA and b, b1, b2 EB, (i) (.>.1 + .>.2) · b = .>.1 · b + .>.2 · b, (ii) A• (b1 + b2) = A · b1 + A · b2, (iii) .>.1 · (.>.2 · b) = (.>.1 · .>.2) ·band (iv) if A has an identity element then 1 · b = b.

3.11. TENSOR PRODUCTS

145

Similarly we may define a right A-module. We shall be mainly concerned with the case when A is a field and A is a subfield. If A is a right A-module then we define the tensor product of A and B over A, A 0A B, to be an abelian group with the following properties: (i) There is a map of sets

1r:AxB---.A@AB written 1r(a, b) = a 0 b. (ii) The map, 1r, is biadditive in the sense that

1r(a1 + a2, b) = 1r(a1, b) + 1r(a2, b), 1r(a·>.,b)

=1r(a,>.·b),

1r(a, b1 + b2) = 1r(a, bi)+ 1r(a, b2). (iii) Let

f:AxB---.C be any other map of sets in which C is an abelian group and biadditivity analogous to (ii),

f

satisfies

f(a1 + a2, b) = f(a1, b) + f(a2, b), f(a · >., b)

= f(a, >. · b),

f(a, bi+ b2) = f(a, b1) + f(a, b2). Then there is a unique homomorphism of abelian groups

f

:[email protected]

such that J(a 0 b) = f(a, b) for all a EA, b EB. Such a group, A@AB, exists and it may be constructed in the following manner. Form the free abelian group on the set Ax B. This is the abelian group consisting of all formal, finite sums

with ai EA, bi E Band ni E Z. Two such formal sums are added together by the coefficients of each (ai, bi) EA x B,

L ni(ai, bi)+ L mi(ai, bi)= L(ni + mi)(ai, bi). i

Temporarily denote this free abelian group by X(A, B). Define a subgroup, Y(A, B) ~ X(A, B), to be the smallest subgroup containing all formal

146

CHAPTER 3. GALOIS THEORY

sums of any of the following types

(a1

+ a2, b) -

(a1, b) - (a2, b),

(a·,\, b) - (a,,\· b), (a, b1 + b2) - (a, b1) - (a, b2). Set A ®AB= X(A, B)/Y(A, B). Question: Does this A ®AB satisfy the requisite conditions? Why do these conditions determine A ® A B up to isomorphism? Example 3.11.2 Let E/F and L/F be field extensions. We may consider E as a right F-module via multiplication in Eby elements of F. Similarly Lis a left F-module. Hence we may form E@p L. In fact, E ®FL, is a vector space over E and over L (i.e. an E-module and an £-module). If z EE we may define a map

z:ExL-E@pL by the formula

z(a, b) = (z ·a)® b. This map satisfies the conditions off in Definition 3.11.1 with C Hence there is a unique map

= E®pL.

such that z(a ® b) = (z ·a)® b. This will define multiplication by z E Eon E@p L. Similarly we define multiplication by L using multiplication by L on the £-factor of Ex L. Notice that these two multiplications agree on F ~ En L since, if z E F, (z ·a)® b = (a· z) ® b =a® (z · b). Theorem 3. 11 .3 Let E / F be a normal extension then there is a ring isomorphism

,\: E@p E -

IT

E

g;EG(E/F)

given by

,\(a® b)

= (g1(a)b,g2(a)b, ... , 9n(a)b)

where a, b EE and G(E/F) = {g1 , ... ,gn}. The ring structure on E@pE is given by (a® b)(a1 ® b1) = aa1 ® bb1. Proof We begin by verifying that the map, ,\, makes sense. For each i the map, f : E x E E, ·given by f(a, b) = gi(a)b satisfies the conditions of Definition 3.11.1 and so induces a homomorphism of abelian groups which is the i-th component of,\, Hence ,\ is a well-defined homomorphism of

3.11. TENSOR PRODUCTS

147

abelian groups. Also >.((a © b) · (a1 © b1)) has i-th coordinate equal to 9i(a · a1)b · b1 = (gi(a) · b) · (gi(a1) · b1) so that >. is a ring homomorphism. Consider E ©FE as an E-vector space using the right multiplication, z(a © b) =a© bz for a, b, z EE. By the Normal Basis Theorem (§3.7.21) there exists e E E such that {gi(8)} is an F-basis for E. Hence the elements {gi(8) © 1} will certainly generate E©p E as an E-vector space. Now consider, for bi E E,

>.

(tgi(e)

® bi)

:;:: (t919i(8)bi, · · ·, t9n9i(8)bi) . In the proof of Theorem 3.7.21 we saw (at least when F is infinite) that det(gjgi(8)) was non-zero in E. Therefore the only solution of

is b1 = 0 = b2 = ... = bn. Hence>. is one-one and {gi(8)©1} is a basis for E ©FE as in E-vector space. To show that>. is an isomorphism we may either observe that >. is an E-linear injective map between n-dimensional vector spaces or that the non-vanishing determinant guarantees that we may always solve the linear equations

for {bi} given {yi}.D Theorem 3.11.4 Let E/F be a normal extension and let L/F be any finite extension. Then

>.:E©pL--t

II

EL

g;EG(E/ F)/G(E/ LnE)

given by >.(a© b)

= (91 (a)b, g2(a)b, ... , 9n(a)b) is an isomorphism of rings.

Proof This proof will only be sketched. Since [L : F] is finite we may find a Galois extension, W / F, with L / F and E / F as subextensions. Hence A : w ©p

w-

IT hEG(W/F)

w

148

CHAPTER 3. GALOIS THEORY

is an isomorphism, by Theorem 3.11.3. We have a homomorphism of abelian groups, i : E ®FL --; W ®F W, induced by the inclusion map Ex L --; W x W. Also, if a E E, b E L, then ,\(a® b) E TI EL and ,\(i(a ® b)) E TI W have the same coordinates - those of the latter being those of the former, possibly repeated a finite number of times (because G(W/F) has G(E/F) as a quotient). In addition, the map i is injective. Hence, if z E E®FL and ,\(z) = 0 in TI EL then .>.(i(z)) = 0 in TI Wand so i(z) = 0 and hence z = 0. The result now follows by counting dimensions of the domain and range of .>. as Fvector spaces. It is not hard to show that dimF(E ®FL)= [E: F][L: F]. On the other hand dimF(Tig;EG(E/F)/G(E/LnE) EL) n EJ- 1

=[EL: Fl[E: Fl[E: L

=[EL: Ll[L: Fl[E: Fl[E: L

n EJ- 1

= [L : Fl[E : F] by Theorem 3.7.23(ii).D

3 .12

Exercises

Exercise 3.12.1 Let F be a field which contains n distinct n-th roots of unity. If a E F, let E denote the splitting field of xn - a E F[x]. Show that the Galois group of E/F, G(E/F), is cyclic and that I G(E/F) I= n. Exercise 3.12.2 Let F be a field which contains n distinct n-th roots of unity where n is prime to the characteristic of F. Let E / F be a normal extension whose Galois group, G(E / F) , is cyclic of order n. Show that E = F(~) where ~n = a E F.

Exercise 3.12.3 What is the Galois group of the following polynomials? (i) x 3 - x - l over Q , the rationals. (ii) x 3 - 10 over Q. (iii) x 4 - 5 over Q. (iv) x 4 - 5 over Q( v'5). (v) x 4 - 5 over Q(v-5). (vi) x 4 - 5 over Q(i) where i 2 = -1. Exercise 3.12.4 Let CT1, ••• , CTn be distinct field homomorphisms from F to E. Using the independence of the {CTi} , show that there exist a1, ... , an in F such that

149

3.12. EXERCISES

o-1(a1) o-1(a2) · · · o-1(an) o-2(a1) o-2(a2) · · · o-2(an) det

is non-zero. Exercise 3.12.5 (Kronecker) Let f(x),g(x) E F[x] be irreducible. Let E/F be an extension containing roots a, b of f(x), g(x) respectively. Let f(x) = Ji (x)h(x) ... f m(x) be a factorisation of f(x) as a product of irreducible polynomials in F(b)[x] and let g(x) = g1(x)g2(x) .. ,gn(x) be a factorisation of g(x) as a product of irreducible polynomials in F(a)[x]. Show that m = n and that, after suitably permuting the indices, deg(f(x))deg(gi(x)) = deg(g(x))deg(fi(x)) for all i. (Hint: Tensor products may help you here.) Exercise 3.12.6 (Cyclotomic polynomials) In C, the complex numbers, let ~1, ... , ~(n) be the distinct, primitive n-th roots of unity, where ¢(n) is equal to the order of the multiplicative group of units in Z/nZ, the integers modulo n. Let (n)

fn(x) =

IT (x -

~i),

i=l

(i) Show that fn(x) E Q[x]. (ii) Show that xn - 1 = IL1n fd(x). (iii) Show that fp(x) = xP-l + ... + x prime.

+ 1 E Q[x]

is irreducible if pis a

Exercise 3.12.7 Let E/F be a normal extension with Galois group G (E / F). Let F L E be an intermediate extension and let H denote the subgroup of G(E / F) consisting of those o- E G(E / F) such that o-(L) = L. Show that His the normaliser of G(E/L) in G(E/F).

s

s

Exercise 3.12.8 Let F P denote the field with p elements, where p is a prime. Let Fp denote the field which is obtained by adjoining to Fp all the n-th roots of unity for n prime top. (i) Show that Fp is algebraically closed (that is, every irreducible polynomial in Fp[x] has a root in Fp), (ii) Let Ebe the field obtained from F P by adjoining all then-th roots of unity for n a prime different from p. Show that E is algebraically closed.

150

CHAPTER 3. GALOIS THEORY

(Hint: Show that if q is a prime and r 2: 1 is an integer then there exists a prime, l , such that the multiplicative order of p modulo l is equal to qr. Let l be a prime dividing the number

b -_

(

(pqr - 1) _ ( qr-1 )q-1 - 1 pq r-1 -1 ) - P

( qr-1

+q P

- 1

)q-2

+ ... + q.

If l does not divide (pqr- i - 1) the proof is complete. Otherwise l = q, but in that case q2 does not divide b. If ( is an l-th root of unity then [Fv((): Fv] = qr and so E = E'p.) Exercise 3.12.9 Throughout this question let f(x) = x 3 - x - 1 E Q[x] where Q denotes the field of rational numbers. (i) Show that f(x) is separable. (ii) Using calculus, or otherwise, show that f (x) has only one real root (and two complex roots). (iii) Let E/Q denote the splitting field extension for f(x). Hence E/Q is a normal, separable extension with Galois group, G = {all field automorphisms of E fixing Q}. By considering the way in which G permutes the roots of f(x) prove that G is isomorphic to the dihedral group of order 6 - or, if you do not believe this, disprove it! Exercise 3.12.10 (i) Let f(x) E Q[x] and set h(x) = f(l + x). Show that f(x) is irreducible in Q[x] if and only if h(x) is. (ii) Show that, if p is a prime, f (x) = 1 + x + ... + xv-l E Z[x] is irreducible in Q[x]. Exercise 3.12.11 Let G be a finite group and H a subgroup. The normaliser of H in G is the subgroup consisting of all g E G such that gHg- 1 c H. Let E/F be a normal extension with Galois group G(E/F). Let F s; L s; E be an intermediate extension and let H denote the subgroup of G(E/F) consisting of those CT E G(E/F) such that cr(L) = L. Show that His the normaliser of G(E/L) in G(E/F). Exercise 3.12.12 (Witt [4]) Let K be a field and let u, v E K* be nonsquares. Set L = K(a,/3) where a 2 = u, {3 2 = v. Suppose that there is a 3 x 3 matrix, P = (Pij) with det(P) = 1 and such that pt DP= I, where I is the identity matrix, pt is the transpose of P and 0 V

0 0

0

1/uv

) .

3.12. EXERCISES Show that E

151

= L(o), where 15 2 = 1 + pna + P22f3 + (p33/af3),

is a Galois extension of K with

G(E/K) ~ Qs = {x,y I x 2 = y 2 ,xyx = y,x 4 = 1}, the quaternion group of order eight.

Chapter 4

Rings and Modules 4.1

Basic definitions

In this chapter we are going to develop a little further the notions of a ring and a field as treated in Chapters 2 and 3. We are going to examine two cases of the classification of finitely generated modules over well-behaved commutative rings. The first case, the classification of finitely generated modules over principal ideal domains, will be slightly familiar, at least in the special case of the classification of finitely generated abelian groups (see Theorem 1. 7.19). In the second case we shall replace principal ideal domains by the more general case of a Dedekind domain. The study of modules over Dedekind domains is fundamental in algebraic number theory (for example, see [7]). First of all, let us introduce some terminology and definitions. Definition 4.1.1 Let R be a ring. Then a left R-module is a non-empty set, M, together with two operations +:MxM---+M and (- · - ) : R x M

---+

M,

called "addition" and "multiplication", respectively. Hence, if a, b E M and x E R then a + b E M and x · b E M denote, respectively, the appropriate sum and product. These operations are required to satisfy the following axioms: (i) (M, +) is an abelian group. (ii) For all a, b E M and x, y E R x · (y ·a)= (x · y) · a,

153

154

CHAPTER 4. RINGS AND MODULES (x + y) ·a= x ·a+ y · a

and X.

(a+ b)

=

X.

a+

X.

b.

A right R-module is defined similarly. If R is a ring with identity then we shall require that 1 · a = a for all a EM, in the case of a left R-module and a· 1 = a for right R-modules. If Mis a left R-module then NC Mis a submodule if (N, +) ~ (M, +) is a subgroup and x · n EN for all a ER and n EN. Submodules of right R-modules are defined in the obvious manner. If N ~Mis a submodule of a (left) R-module then the quotient group, M/N, becomes a (left) R-module with respect to the multiplication Rx M/N---+M/N

defined by x · (m + N) = xm + N. The R-module, M/N, is called the quotient of M modulo N. Right R-modules behave similarly. Definition 4.1.2 From now on, the term "R-module" will mean a left R-module unless there is some good reason to specify otherwise. Also we shall henceforth almost always assume that all rings have an identity element, 1 E R. As far as the study of R-modules goes we lose nothing by this assumption, as is shown in Exercise 4.4.9. Let {Ma I a E A} be a set of R-modules and let (EBaEA Ma,+) denote the abelian group given by the direct sum of the (Ma,+ )'s. Hence an element of EBaEAMa consists of a family of elements, ma E Ma, of which all but finitely many of the ma's are zero. Addition is defined coordinate-wise, i.e.

Then the coordinate-wise product

x · (ma)

= (x · ma)

makes EBaEca!AMa into an R-module, the direct sum of the Ma's. In the special case in which one takes each Ma to be equal to R-module given by R itself, the resulting direct sum module, EBaEca!A R, is called a free R-module. Any R-module which is isomorphic to a direct sum of this type is also called a free module. If M is a free module which is isomorphic to a finite direct sum of copies of R then the rank of M is defined to be the smallest number of copies of R in such an isomorphism. For completeness, free modules which are not isomorphic to a finite direct sum of copies of R will be said to have infinite rank. An R-module, M, is called projective if there exists a free module, F and another module N together with an isomorphism of R-modules of the form F ~ M EB N.

4.2. FINITELY GENERATED MODULES

155

A finitely generated R-module, M, is one such that there exists a finite set of elements of M, {m 1, ... , mt}, such that every m E M may be written in at least one way in the form

with Xi E R for 1 ::; i ::; t. If M is a finitely generated R-module then there is a surjective R-module homomorphism of the form

f : EB~=l R ___, M given by f (x1, ... , Xt) = E~=l Ximi. If R is a ring with identity these formulations are equivalent, since we may set mi equal to the image under f of the element whose i-th coordinate is 1 and all others are zero. When M is a free R-module of finite rank, t, then there exists a set of elements of M, {m1, ... , mt}, such that the homomorphism, f, is an isomorphism. Any such set, {m 1, ... , mt}, is called a basis for M.

4.2

Finitely generated modules

Just let me remind the reader that, for convenience we shall almost always assume that all rings have an identity element. This is justified by Exercise 4.4.9. Now that we have refreshed our memories as to definitions it is time to prove something. We shall begin by studying modules. A module over the familiar ring given by the rational integers, Z, is just another name for an abelian group. Finitely generated abelian groups have a very satisfying classification (see Theorem 1.7.19). The backbone of that classification is the fact that every subgroup of the integers has the form

nZ

= {z

E

Z

I z = nx for

some x E Z}

for some integer, n. This is just saying that Z is a principal ideal domain, in the terminology of Definition 2.3.3. Therefore, setting our sights on a modest objective to begin with, in this section we shall generalise the classification of finitely generated abelian groups to a classification of finitely generated modules over a principal ideal domain. Theorem 4.2.1 (Chinese Remainder Theorem) Let R be a commutative ring with identity. For 1 ::; j ::; n let Ii 1 consider the homomorphism, h" : M - - R, given by h"(u) = b2 if u = 1 biYi· Then h"(N) = Ra2 and if Ra2 ~ Ra1 we have nothing to prove. If not, we observe that h"(v) = 0, by construction, so that h111 = h + h": M - - R has Ra1 + Ra2 ~ h111 (N). By maximality this implies that h 111 (N) = Ra1 and therefore a1/a 2, which completes the induction.D We shall apply Theorem 4.2.8 to accomplish the classification which was advertised at the beginning of this section.

E:

Theorem 4.2.9 Let R be a principal ideal domain. Suppose that Mis a finitely generated R-module. Then (i) There is an isomorphism of R-modules of the form M ~ ( EBf= 1R) EB R/ Ra1 EB R/ Ra2 EB ... EB R/ Ran

for some integer, k > 0, and some non-zero elements, a2, ... , an of R such that ai/ai+I for 1 ~ i ~ n - 1. (ii) There is an isomorphism of R-modules of the form

a1,

Tors(M) ~ R/ Ra1 EB R/ Ra2 EB ... EB R/ Ran for the same k 2::: 0 and a1, a2, ... , an as in (i). (iii) The R-module, M, is torsion free if and only if it is free. (iv) The R-module, M, is a torsion module (that is, M = Tors(M)) if and only if k = 0. In this case Ann(M) = Ran. Proof As explained in Definition 4.1.2, there is a surjective R-module homomorphism of the form f: EB!=1 R-M

and we may apply Theorem 4.2.8 to the free module of rank t, EB!=l R, and its submodule, given by the kernel off, ker(f). Therefore there exists a basis for EB:=l R, {y1, ... ,Yt}, and a set of non-zero elements of R, {a1, ... , an} with n ~ t, such that ai/ai+I for all 1 ~ i ~ n - 1 and {a1Y1, ... , anYn} is a basis for ker(f). If R · y denotes the cyclic R-module consisting of all multiples of y, this means that we have direct sum decompositions of EB:= 1 R and ker(f) of the form

EB!=1 R

= R · YI EB R · Y2 EB ... EB R · Yt

and ker(f)

= R · a1Y1 EB R · a2Y2 EB ... EB R · anYn·

4.2. FINITELY GENERATED MODULES

163

Hence we have isomorphisms EBj=J R ,::,, ker(f)

-

R-y1EBR·y2ffi ... EBR-y, R-a1 Yl EBR·a2y2EB ... EBR-an Yn

= __B::ui_ R·a1Y1 CV

ffi ffi \.D ' · ' \.D

~ R-anYn

ffi \.D

R 'Yn+l · · · \lJ ffi R 'Yt

~ R/ Ra1 EB ... EB R/ Ran EB (EB~;:r R)

The First Isomorphism Theorem implies that ( EB}= 1 R) /ker(f) ~ M and setting k = t - n completes the proof of part (i). For part (ii) we observe that, in R/ Rai, an(x + Rai) = anx + Rai, which is the zero coset because Ran ~ Rai for all 1 s; i s; n. Therefore R/ Ra1 EB ... EB R/ Ran ~ Tors( R/ Ra1 EB ... EB R/ Ran EB (EB~;:r R)). On the other hand, if x E R is non-zero and z E R/ Ra1 EB ... EB R/ Ran EB (EB~;:r R) has one of the last t - n coordinates equal to a non-zero element, w ER, then z.w is non-zero and it is the corresponding coordinate of xz. Therefore z can only be a torsion element if all its last t - n coordinates are zero, which completes the proof of (ii). Part (iii) is obvious from parts (i) and (ii), since M is torsion free if and only if Tors(M) = {O}. For part (iv) we have only to show that Ann(R/ Ra1 EB R/ Ra2 EB ... EB R/ Ran) = Ran in the situation of part (ii). Furthermore the proof of part (ii) shows that the annihilator contains Ran, Conversely, it is easy to see that Ann(R/ Ra1 EB R/ Ra2 EB ... EB R/ Ran) ~ Ann(R/ Ran)

= Ran.

D

Remark 4.2.10 Theorem 4.2.9 is a nice "'tidying up" result. Now we know what all the finitely generated modules over a principal ideal domain look like, up to isomorphism. However, a classification is rather more than that. To classify some sort of mathematical object is to list all of them once and only once. Therefore, before we are finished, we shall show (Theorem 4.2.15) that the module, M, in Theorem 4.2.9 uniquely determines the integer, k, and the set of elements, a1,a2, .. ,,an, up to replacing any of the ai by its product with a unit of R. Remark 4.2.11 Let R be an integral domain. Recall (see Definition 2.3.9) that a non-zero element, x E R, is called irreducible if x = uv for some u, v E R implies that either u or v is a unit in R. A non-zero element, x E R, is called prime if xluv implies that xlu or xiv, If R is a principal ideal domain then x E R is irreducible if and only if Rx .- 1(IN) = >.- 1(IN+t) for all t 2'.'. 0. Therefore IN = >.(>.- 1(IN)) = >.(>.- 1(IN+t)) = IN+t for all t 2:: 0, as required.D

5.2

Prime factorisation of ideals

Definition 5.2.1 An integral domain, R, is called a Dedekind domain if it satisfies the following conditions: (i) R is a Noetherian ring, (ii) R is integrally closed (c.f. Definition 5.1.1), and (iii) every non-zero prime ideal, I