Fundamentals Of Modern Algebra: A Global Perspective 9814730297, 9789814730297

Table of contents :
Title
Preface
Contents
1. Groups
1.1 Introduction to Groups
1.2 Subgroups
1.3 Homomorphisms of Groups
1.4 Group Structure
1.5 The Sylow Theorems
1.6 Exercises
2. Rings
2.1 Introduction to Rings
2.2 Polynomial Rings
2.3 The Group of Units of a Ring
2.4 Ideals
2.5 Quotient Rings and Ring Homomorphisms
2.6 Localization
2.7 Absolute Values and Completions
2.8 Exercises
3. Modules
3.1 Vector Spaces
3.2 Modules
3.3 Projective Modules
3.4 Tensor Products
3.5 Algebras
3.6 Discriminants
3.7 Exercises
4. Simple Algebraic Extension Fields
4.1 Simple Algebraic Extensions
4.2 Some Galois Theory
4.3 The Ring of Integers
4.4 The Noetherian Property of the Ring of Integers
4.5 Dedekind Domains
4.6 Unique Factorization of Ideals
4.7 Extensions of Q_p
4.8 Exercises
5. Finite Fields
5.1 Invented Roots
5.2 Finite Fields
5.3 Linearly Recursive Sequences
5.4 Exercises
Bibliography
Index

Citation preview

FUNDAMENTALS OF

MBIIERN AlliEBllA

A Global Perspective

Robert G Underwood

p

\‘9 World Scientific

FUNDAMENTALS OF

MIIIIEBN ALGEBBA A Global Perspective

This page intentionally left blank

FUNDAMENTALS OF

IIIIllIIEIIIII AIEEBIIA A Global Perspective

Robert G Underwood Auburn University at Montgomery, USA

We. World Scientific NEW JERSEY - LONDON - SINGAPORE . BEIJING - SHANGHAI - HONG KONG - TAIPEI - CHENNAI - TOKYO

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA ofiice: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK oflice: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging—in—Publication Data Names: Underwood, Robert G. (Robert Gene)

Title: Fundamentals of modern algebra : a global perspective / by Robert G. Underwood (Auburn University at Montgomery, USA). Description: New Jersey : World Scientific, 2016. | Includes bibliographical references and index. Identifiers: LCCN 2015040815 | ISBN 9789814730280 (hardcover : alk. paper) | ISBN 9789814730297 (pbk. : alk. paper) Subjects: LCSH: Algebra, Abstract--Textbooks. | Algebra--Textbooks. Classification: LCC QA162 .U53 2016 | DDC 512--dc23 LC record available at http://lccn.loc.gov/2015040815

British Library Cataloguing—in—Publication Data

A catalogue record for this book is available from the British Library.

Copyright © 2016 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereojf may not be reproduced in anyform or by any means, electronic or mechanical, includingphotocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

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is not required from the publisher.

In-house Editor: Bai Li

Printed in Singapore

to my son, Andre

Preface

The purpose of this book is to provide a concise yet detailed account of fundamental concepts in modern algebra. The target audience for this book is first-year graduate students in mathematics, though the first two chapters are probably accessible to well-prepared undergraduates. The book contains five chapters. In Chapter 1 we cover groups, subgroups, quotient groups, homomorphisms of groups, and group structure, including cyclic groups, the Structure Theorem for finitely generated Abelian groups, Cauchy’s Theorem, and Sylow’s Theorems. In Chapter 2 we consider rings, the group of units of a ring, ideals, quotient rings, and ring homomorphisms. Included also are sections on localizations and com— pletions. In Chapter 3 we turn to modules. We begin with a review of both finite and infinite dimensional vector spaces, and then generalize to modules over PIDs and Noetherian rings. We include sections on projec— tive modules, tensor products of modules, algebras, and the discriminant of modules over an integral domain. In Chapter 4 we define simple algebraic extensions of Q and introduce the Galois group of the splitting field of a monic irreducible polynomial over Q. We state and prove the Fundamental Theorem of Galois Theory. We then follow with an introduction (essentially) to algebraic number theory: we include material on the ring of integers of an algebraic extension, the Noetherian propery of the ring of integers, Dedekind domains and unique factorization of ideals. In the final chapter (Chapter 5) we cover the basic theory of finite fields and linearly recursive sequences. We begin each chapter with an overview of the material to be covered. At the end of each chapter we give an extensive list of exercises which range from basic applications of the theory, to problems designed to challenge the reader. We also include some “Questions for Further Study”, which are

Vii

viii

Fundamentals of Modern Algebra

advanced problems suitable for master’s level research projects. I would like to thank the fellow algebraists who read and commented on earlier drafts of the manuscript. Their suggestions, especially those regard— ing the organization of the sections, have been duly noted and incorporated into the book. My appreciation is also extended to E. H. Chionh and Li Bai, at World Scientific, who have skillfully guided me through the publication process. To my wife, Rebecca Brower, who is also an academic, and who certainly understands the challenge of a writing project of this sort, I thank you for your patience, kindness and companionship. Any finally, to my son Andre, to Whom this book is dedicated, I thank you for understanding that although writing takes a lot of time, in the end it is a worthy endeavor.

Robert G. Underwood

Contents

Preface 1.

Groups 1.1 1.2 1.3 1.4 1.5 1.6

2.

1

Introduction to Groups .................... Subgroups ........................... Homomorphisms of Groups ................. Group Structure ....................... The Sylow Theorems ..................... Exercises ............................

1 11 16 22 31 35 41

Rings 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

3.

Vii

Introduction to Rings ..................... Polynomial Rings ....................... The Group of Units of a Ring ................ Ideals ............................. Quotient Rings and Ring Homomorphisms ......... Localization .......................... Absolute Values and Completions .............. Exercises ............................

42 48 53 60 70 78 82 93

Modules

101

3.1 3.2 3.3 3.4 3.5

102 109 117 122 128

Vector Spaces ......................... Modules ............................ Projective Modules ...................... Tensor Products ....................... Algebras ............................ ix

X

4.

5.

Fundamentals of Modern Algebra

3.6

Discriminants .........................

3.7

Exercises ............................

140

Simple Algebraic Extension Fields

143

4.1 4.2 4.3 4.4 4.5 4.6 4.7

Simple Algebraic Extensions ................. Some Galois Theory ..................... The Ring of Integers ..................... The Noetherian Property of the Ring of Integers ..... Dedekind Domains ...................... Unique Factorization of Ideals ................ Extensions of Qp .......................

4.8

Exercises ............................

144 152 158 169 172 178 181 186

191

Finite Fields 5.1

Invented Roots

........................

5.2

Finite Fields ..........................

5.3

Linearly Recursive Sequences ................

5.4

Exercises ............................

191 193 200 212

Bibliography

215

Index

217

Chapter 1

Groups

In this chapter we introduce semigroups, monoids, and groups, give some basic examples of groups and discuss some of their elementary properties. We then consider subgroups, cosets and Lagrange’s theorem, normal sub— groups and the quotient group. We next turn to the basic maps between groups: homomorphisms and isomorphisms and their kernels. (Throughout this book, map = function.) We give the First, Second and Third Isomorphism theorems and the Universal Mapping Property for Kernels. We close the chapter with the study of group structure, including generating sets for groups and subgroups and the notion of a cyclic group. From the cyclicity of the additive group of integers Z we obtain greatest com— mon divisors, least common multiples, Bezout’s Lemma and the Chinese Remainder Theorem. We state the structure theorem for finitely generated abelian groups. Regarding the structure of groups in general, we introduce G—sets, and give Cauchy’s Theorem and Sylow’s First, Second, and Third Theorems.

1.1

Introduction to Groups

In this section we define semigroups and monoids and give some examples, including the monoid of words on a finite alphabet. From semigroups and monoids, we develop the concept of a group, discuss finite, infinite and abelian groups, and prove some elementary properties of groups. We intro— duce examples of groups that we will appear throughout this book, includ— ing the additive group of integers, Z, the multiplicative group of non—zero real numbers, RX and the group of residue classes modulo n, Zn. For further examples of groups we construct the 3rd and 4th dihedral groups, D3, D4 as the groups of symmetries of the equilateral triangle and the square,

2

Fundamentals of Modern Algebra

as well as the symmetric group on n letters, Sn. *

*

*

Let S be a non—empty set of elements. The cartesian product on S is

defined as S X S 2 {(a,b): a,b E 3}. Definition 1.1. A binary operation on S is a function B : S X S —> S;

we denote the image of (a, b) by ab. A binary operation is commutative if for all a,b 6 S, ab 2 ba. A

binary operation is associative if for all a, b, c E S, a(bc) = (ab)c. Definition 1.2. A semigroup is a set 3 together with an associative binary operation S X S —> S. Let S be a semigroup and let a1,a2,a3 E S. We define the product a1a2a3 to be the common value of the expressions (a1a2)a3 and a1(a2a3). For n 2 4 we define the product of elements a1,a2, . . . ,an E S induc—

tively to be n

n—l

Ha: H

In defining H2121 a; in this way we are asserting that we can insert parenthe— ses into the product in any manner we choose without changing its value. For example, a1a2a3a4 is the common value of the expressions

(a1a2a3)a47 (01(a2a3))a4, ((a1a2)a3)a47 01((a2a3)a4), (a1a2)(a3a4)a a1(a2(a3a4)), a1(a2a3)a4, (aia2)a3a4, ala2(a3a4), a1(a20b3a4)Definition 1.3. A monoid is a semigroup S in which there exists an element 6 E S with ea 2 a = ae,Va E S. Such an element 6 is called an identity element for the monoid. For example, the set of integers Z together with ordinary multiplication is a monoid with identity element 6 = 1 and the set of natural numbers N = {1, 2, 3, . . . } together with ordinary addition is a semigroup. Note that N together with + is not a monoid, however. Here is an example of a monoid that is used in computer science. An alphabet 20 is a non—empty set whose elements are the letters of the

Groups

3

alphabet. A word is a finite sequence of letters in 20. For a given alphabet 20, let 2;; denote the collection of all words formed from the alphabet 20.

For w E 23, the length of w denoted by l(w) is the number of letters in w. The empty word 6 is the (unique) word of length 0 in 23. We endow 23 with a binary operation 23 X 23 —> 23 called concatenation. Concatenation (sometimes denoted as ‘-’) is defined as a: - y = cry, for x, y E 23. As the reader can easily verify, 23 together with concatenation is a monoid; the identity element is the empty word.

For example if 20 = {a, b}, then {a, b}* consists of all finite sequences of a’s and b’s. The word :1: = abbab E {a, b}* has length l(m) = 5. Moreover, if y = bab, then x - y = abbab - bab = abbabbab. Definition 1.4. A group is a set G together with a binary operation G X G —> G for which (i) the binary operation is associative,

(ii) there exists an element 8 E G for which ea 2 a 2 ae, for all a E G, (iii) for each a E G, there exists an element 0 E G for which ca = e = ac.

An element 6 satisfying (ii) is an identity element for G; an element 0 satisfying (iii) is called an inverse element of a and is denoted by a‘l. We note immediately that every group is a monoid. The converse is

false, of course (see §1.6, Exercise 6). There are many familiar examples of groups encountered in mathemat— ics. For example, the set of integers Z, together with ordinary addition + is a group, 0 plays the role of e, and —a is the inverse of a E Z. One eas— ily shows that the set of rational numbers Q under ordinary addition and the set of real numbers R under ordinary addition are groups. The set of non—zero real numbers Rx is a group under ordinary multiplication - with e = 1, and a—1 = 1 / a. A further example is the general linear group GL7, (R) consisting of invertible n X n matrices with entries in R, together with matrix multiplication. Recalling some linear algebra, one has

GLn(]R) = {A E Matn(lR) : det(A) ¢ 0}. In the case that n = 1, GL1(lR) = RX. The order of a group G, denoted by IG I, is the number of elements in

G. If |G| is infinite, then G is an infinite group. All of the examples of groups given above are infinite groups. A group G is finite if |G| is finite. In what follows we give an example of a finite group.

4

Fundamentals of Modern Algebra

Let n,a be integers with n > 0. A residue of a modulo n is an integer 7" for which a = nq—i—r for some q E Z. For instance, if n = 3, a = 8, then 11 is a residue of 8 modulo 3 since 8 = 3(—1) + 11, but so is 2 since 8 = 3(2) + 2. The possible least non—negative residues of a modulo n are 0, 1, 2, . . . , n — 1. The least non-negative residue of a modulo n is denoted as amod n. For example, 8mod3 = 2, but also note that —3m0d4 = 1 and 11 m0d4 = 3mod4 = 3. We say that two integers a, b are congruent modulo n if amodn = bmodn and we write a E bmod n. Let a,n be

integers with n > 0. Then n divides a, denoted by n | a, if there exists an integer k for which a = nk. Proposition 1.1. Let a,b,n 6 Z, n > 0. Then a E bmodn if and only if

n | (a — b). Proof. To prove the “only if” part, assume that a E bmod n. Then amodn = bmod n, so there exist integers l,m for which a = nm + r

and b 2 nl + r with 7' = amodn = bmodn. Thus a — b = n(m — I). For the “if” part, assume that a — b = nk for some k. Then (mm + amod n) — (nl + bmod n) = nk for some m,l E Z, so that n divides amodn — bmod 72. Consequently, a mod n — bmodn = 0, hence a E bmod n. III Proposition

1.1

can help us compute

a mod n.

For instance

—14 mod 17 = 3mod 17 = 3 since 17 | (—14 — 3). Likewise —226 mod 17 = 12 mod 17 = 12 since 17 | (—226 — 12). For n > 0 consider the set J = {0, 1, 2, 3, . . . , n— 1} of least non—negative residues modulo n. Note that a = amod n, Va 6 J. On J we define a binary operation +n as follows: for a, b 6 J,

amodn +1, bmodn = (a+b) modn. Then +n gives J the structure of a group, known as the group of residue classes modulo n. We denote this group by Zn; Z, is a finite group of

order |Zn| = n. For example, Z4 2 {0,1,2,3} and one has 1 +4 2 = 3, 3+4 2 = 1, and so on. One nice feature of a small finite group is that all possible group prod— ucts can be arranged in a finite table in which the elements of the group are listed across the top as labels of the columns and down the left side as labels of the rows. For elements a, b in finite group G, the (a, b)th entry in the table is ab. This table is the group table for finite group G. For instance, the group table for Z4 is

Groups

+4 0 1 2 3

0 0 1 2 3

1 1 2 3 0

5

2 2 3 0 1

3 3 0 1 2

We can construct a new group from a finite set of groups.

Let

31, 32, . . . , 5;, be a finite collection of sets. Then the cartesian product

k

H S,- is the collection of all k—tuples {(a1,a2, . . . , ak) : ai 6 5,}.

i=1 Proposition 1.2. Let Gi, i = 1,...,k, be a finite collection of groups. k: Then the cartesian product HG, is a group under the binary operation i=1

defined as

(01.02,. - '7a‘k) - (51,52, - - 'abk) = (a151,a2b2,- - wakbk): where aibi is the image of (a,, bi) under the binary operation Bi : G,- X G'z- —>

Ci of the group G1, 1 S i S k. Proof. We show that the conditions of Definition 1.4 hold. Clearly the binary operation on the cartesian product is associative; for an identity element we take e = (e1,e2,...,ek) where ei is an identity in

Gi. Lastly, for each k-tuple (a1,a2,...,ak) one has (a1,a2,...,ak)_1 = —1 — —1 (a1,a21,...,ak). III k

The group H G, of Proposition 1.2 is the direct product group. i=1

As an illustration we consider the group Z X Z in which the binary operation is given as (m1, mg) + (n1, n2) 2 (m1 +n1, m2 +n2). For another

example, we take Z2 >< Z3; here for instance, (0,1) + (1,2) = (1,0). Note that |Z2 X Z3l = 6.

In any group the identity and the inverse of an element are unique. Proposition 1.3. Let G be a group. Then there exists a unique element e for which ea 2 a 2 ae, and for each a E G, there exists a unique element a‘1 for which a‘la = e = aa‘l.

6

Fundamentals of Modem Algebra

Proof. Suppose there are two identities 61 and 62. Then with 61 acting on the left, 6182 = 62. Also, with 62 acting on the right, 6162 = 61. Thus 81 = 82.

Now suppose there exist two inverses (11—1 and 012—1 for a given element a E G. Then afla = e = agla. Now multiplying on the right by (11—1 yields —1 _ —1 a1 — a2 .

D

Since (ab) (b—la—l) = e = (ab) (ab)—1, uniqueness of the inverse yields the rule for inverses of products in a group, that is: (ab)_1 = b‘lcfl. In a group the binary operation is by definition associative. It may or may not be commutative.

Definition 1.5. A group for which the binary operation is commutative is an abelian group.

For example, the residue class group Zn is an abelian group, as are Z, Q, and R. The easiest example of a non-abelian group is GL2 (R). In this group, for example, we have

(31) (‘33) 4‘33) (31)For a finite non-abelian group, we consider the 3rd order dihedral group, which is denoted by D3. The elements of D3 are the six “symmetries” of the equilateral triangle AABC (Figure 1.1) and consist of three clockwise rotations of 0°, 120°, and 240° about the center 0 of the triangle, represented by the elements p0, p1, p2, together with three reflections through the perpendicular lines 61, 62, (3, represented by the elements M1,u2,pg, respectively. It is critical to realize that the rotations move the vertices of the triangle, yet the perpendicular lines remain fixed and do not move with the rotation of the triangle.

Groups

7

(—)

Fig. 1.1

20 that is both one—to—one and onto (a bijection). Let Sn denote the set of all permutations on the set of n letters 20. On 5,,

we define a binary operation 0 which is ordinary function composition: for

Groups

9

0,7' 6 Sn, i E 20,

(007')(i) = 0(T(i)). Proposition 1.4. Let 20 denote the set ofn letters and let Sn denote the set of all permutations of 20. Then Sn together with the binary operation 0 is a group.

Proof. We show that conditions of Definition 1.4 hold. For (i): Let a, 7', p 6 Sn. Then

00(Top)=(oor)op, since function composition is associative. For (ii), one takes an identity element e to be the identity permutation L : 20 —> 20, :1: I—> x,V:I3 E 20.

Then L o a = o = o o L, as required. For (iii), since a : 20 —> 20 is a bijection, the inverse map 0— 1 exists and satisfies the property 0— 1 o o = L=0’OO'_1.

E]

The group Sn given in Proposition 1.4 is the symmetric group on n letters. Observe that there are n! possible permutations of the n letter set 20 and so |Sn| = n!. An element a 6 Sn can be written in permutation notation as 0_
|
|
G. If B|H(H X H) g H, then H is closed under the binary operation B. In other words, H is closed under B

if B(a, b) = ab 6 H for all a,b E H. IfH is closed under B, then B|H is a binary operation on H. Closure is fundamental to the next definition. Definition 1.6. Let H be a subset of a group G that satisfies the following conditions.

(i) H is closed under the binary operation of G, (ii) 6 E H,

(iii) for all a E H, a‘1 E H. Then H is a subgroup of G, which we denote by H S G.

For example, 2Z = {2n : n E Z} S Z. The subset {0,3,6,9} is a subgroup of Z12. The subset {p0,,u1} is a subgroup of D3. The set of integers Z is a subgroup of the additive group R. Every group G admits at least two subgroups: the trivial subgroup

{e} g G, and the group G Which is a subgroup of itself. If H g G and H is a proper subset of G, then H is a proper subgroup of G and we write H < G. Observe that the notation H S G implies that H is a group under the restricted binary operation of G. Definition 1.7. Let H be a subgroup of G with a E G. The set of group

products aH = {ah : h E H} is the left coset of H in G represented by a. The collection Ha 2 {ha : h E H} is the right coset of H in G represented by a. Let aH be a left coset. The element as E G is a representative of aH if 33H 2 aH.

12

Fundamentals of Modern Algebra

Since 6H 2 He 2 H, the subgroup H is always a left and right coset of

itself in G represented by e. Observe that 1 + {0, 3, 6, 9} = {1, 4, 7, 10} and 2 + {0, 3, 6, 9} = {2, 5, 7, 11} are left cosets of {0, 3, 6, 9} in Z12 represented by 1 and 2, respectively. Also, {p0,u1} and {p0, H1}M3 = {#3: p2} are right cosets of {p0, M1} in D3 represented by p0 and M3, respectively. Considering Z as a subgroup of R, one has the left coset r + Z, for r E R, as illustrated in Figure 1.4.

—4—3—2—1 Fig. 1.4

0

1

2

3

4

The additive left coset r + Z of Z in IR represented by r 6 1R.

Note that each real number 7' can be decomposed as r = as + n for some

a: E [0, 1) and some n E Z. Thus [0, 1) is a complete set of representatives for the left cosets of Z in R. Proposition 1.6. Let H g G, and let aH, bH be left cosets. Then there

ewists a bijection ¢ : aH —> bH defined as ¢(ah) = bh for h E H. Proof. To show that (t is 1—1 (one—to—one), suppose that gb(ah1) = ¢(ah2), for h1,h2 E H.

Then bhl = bh2, so that h1 = M, and consequently,

ahl = ahg. Next let bh E bH. Then clearly, ¢(ah) = bh, so that gt is onto.

E]

The following corollary is immediate.

Corollary 1.1. Suppose |G| < 00. Then |aH| = |bH| = |H| for all a,b E G. Let G be a group, let H be a subgroup of G and let I be an arbitrary

set. A subset S = {anlnEI of G is a left transversal of H if the family {anH}n€I constitutes the collection of all distinct left cosets of H in G. Proposition 1.7. Let G be a group, let H be a subgroup of G, and let S = {an}nEI be a transversal of H. Then the collection of distinct left cosets of H in G forms a partition of the set G, that is,

G = U anH, 17E]

Groups

13

with anH fl avH = 0 whenever anH yé avH. Proof. Let g E G. Then 9H 2 anH for some 77 E I, thus G Q Una anH. Clearly, Una anH Q G, and so, G = UneI anH. Suppose there exists an element a: E anH fl aVH for 77,7 6 I. Then a: = anhl = awhg for some h1,h2 E H. Consequently, a,7 = a7h2h1—1 E aVH. Now, for any h E H, anh = awhlhglh E aqH, and so, anH Q aVH. By a similar argument

a7H Q anH, and so, anH = aqH. Thus the collection {anH}n€I is a partition of G.

III

To illustrate Proposition 1.7, let G = D3, H 2 {p0, #1}. Then H = {p0, p1}, p1H = {p1,u3} and n = {pm/12} are the distinct left cosets of H which form the partition of D3,

{H,p1H,p2H}. For another example, let G = Z, H = 3Z. Then the collection of distinct

left cosets is {3Z,1 + 3Z, 2 + 32} which forms a partition of Z. In many cases, even if the group G is infinite, there may be only a finite number of left cosets. When this occurs we define the number of left cosets

of H in G to be the index [G : H] of H in G. For instance, [Z : 3Z] = 3. Proposition 1.8. Let K g H g G with [G : H] < 00, [H : K] < 00. Then [G2K]=[G:H][H:K]. Proof. Let [G : H] = m, [H : K] = n. By Proposition 1.7, there are partitions

G=a1HUa2HU---UamH, and

H=b1KUb2KU---Ua. Thus

j=1

j=1

j=1

i,j=1

with aii fl arbsK = (0 if and only ifz' = 7' and j = s. It follows that

[G : K] 2 mn.

I]

14

Fundamentals of Mode'rn Algebra

If the group G is finite, we have the following classical result attributed to Lagrange.

Proposition 1.9 (Lagrange’s Theorem). Suppose H S G with |G| < 00. Then |H| divides |G|. Proof. By Corollary 1.1 any two left cosets have the same number of ele-

ments. Since the left cosets partition G, we have |H|[G 2 H] = |G|.

III

We haven’t said much about right cosets and their relationship to left cosets. The method of Proposition 1.7 applies to show that the collection of distinct right cosets of H in G forms a partition of G with H as one of the cells. We also have the following propositions. Proposition 1.10. Let aH,Hb be left and right cosets ofH in G. There

is a bijection gb : aH —> Hb defined as ¢(ah) = hb for all h E H. Proof. Exercise.

III

Let L denote the collection of all left cosets of H in G, and let R denote the collection of right cosets of H in G.

Proposition 1.11. There exists a bijection gb : L —> R defined as ¢(aH) = Ha‘ 1 . Proof. We first Show that ¢ is well-defined on left cosets. Suppose that a: is a representative of aH, that is, suppose that 50H 2 aH. Then m = ah, for some h E H, so that 33—1 = h‘lcf1 E Ha‘l. Thus ¢(:I:H) 2 Hrs—1 =

Ha‘l, so that (Z) is well—defined (d) is a function on cosets). Now suppose ¢(aH) = ¢(bH). Then Ha‘1 = Hb‘l, hence a‘1 = hb‘1 for some h E H. Thus a = bh‘1 E bH, so that aH = bH. Thus gt is

injective. Let HI) 6 R, then ¢(b_1H) = Hb, so qb is surjective.

E]

If G is abelian, then it is easy to see that aH 2 Ha for all a E G. But if G is non—abelian, one could have aH 75 Ha for some H S G, a E G. For

example, consider {p0,,u1} S D3. Then

p1{p0,u1} ={,01,M3} #{po,u1}p1 ={p1,M2}Definition 1.8. A subgroup H S G is normal if aH 2 Ha for all a E G. In this case, we write H 0. Then (i : Z/nZ —> Zn defined by a + nZ I—> a mod n is an isomorphism of groups. Proof. We first show that ¢ is well—defined, that is, we show that (15 is a

function. If a + nZ = b + nZ, then n I (a — b) and so, by Proposition 1.1, amodn = bmod n. Thus (/5 is well-defined.

Groups

19

Now,

¢((a+nZ) + (b+nZ)) = q5(a+b+nZ) = (a+b)modn = amodn+nbmodn

= ¢(a + M) +.. 45(1) + nZ), so that ¢ is a homomorphism. Moreover, (15(a + nZ) 2 45(1) + nZ) implies that a E bmod n. Thus a = b + nm for some m E Z. It follows that a + nZ = b + nZ, so that qt is an injection. Clearly, gb is surjective. I]

Proposition 1.20. Let n > 0. Then ¢ : Z —> nZ with ¢(a) = no is an isomorphism of additive groups. Proof. Note that

¢ G’ be a group homomorphism with N = ker(¢). Suppose that K S N and K ¢(G) defined by ib(aK) = 45(0). Proof. We only need to check that it is well-defined. Suppose that aK =

bK for a,b E G. Then a = bk for some k E K, and so, ¢(a) = qb(bk) = ¢(b)q§(k) = ¢(b) since k E K S N.

III

Let (b : G —> G’ be a group homomorphism with N = ker(¢). The Uni— versal Mapping Property for Kernels (UMPK) says that given a subgroup K S N, K G’ so that

1P8 = G/K is the canonical surjection; we say that qb “factors through” G/K and the following diagram commutes:

G

>G’

0/}; Here are two applications of the First Isomorphism Theorem.

Groups

21

Proposition 1.24 (Second Isomorphism Theorem). Let H and K be subgroups of G and suppose that K is normal in G. Then

(2-) K1 in which im('y,-) = kerb/1+1) for 2' = 0,1,2 is a short exact sequence of homomorphisms and groups. As an example, we have the short exact sequence of additive abelian groups

22

Fundamentals of Mode'rn Algebra

0—>ZZ—>Z—>Z/2Z—>0, where the maps are the obvious ones. For another example, consider the short exact sequence

0 —> {0,3} —> 26 i> Z6/{0,3} —> 0, again with the obvious maps. This sequence is special, it is “split”, that is, there exists an injective group homomorphism l : 26 / {0, 3} —> 26 for which

”2

take l : Z6/{0,3} —> Z6 defined as l({0,3}) = 0, l({1,4}) = 4, l({2,5}) 2. Consequently, Z6 g {0,3} X Z6/{0,3}. Since Z2 E {0,3} and Z3

II

the composition .3 Cl is the identity map on Z6 / {0, 3}. Indeed, one may

Ze/{0, 3}, this becomes Z6 g Z2 X Z3, as we shall see in the next section.

1.4

Group Structure

In this section we discuss the structure of groups, define generating sets for groups and subgroups, and the concept of a cyclic group. The addi— tive group Z is cyclic and from this we obtain greatest common divisors, least common multiples and Bezout’s Lemma. Using the First Isomor— phism Theorem we proof the Chinese Remainder Theorem, leading to the decomposition theorem for Zn. We state the structure theorem for finitely generated abelian groups.

Let G be a group, and let n > 0 be an integer. We set a” = \_.v_/ aa - - - a, and n

a‘” = a_1a_1 - - - a_1. Since 6 = a— 1a, we put a0 = e. In additive notation, W

one writes na = a+a+~~~+a, —na 2 (—a)+(—a)+---+ (—a), and W

¥

Y

4

0a 2 e. n Let S be a non—empty subset of G and let 5—1 be the set of inverses of

elements of 8'. Let (S U S‘l)* denote the collection of all words of finite length built from the letters in S U S ‘1. For instance, if S = {a, b}, then the elements

a2, ab, ba_1,a_3, a2b, ab2, a_1b_2a2, (flab, b3, ba2

Groups

23

are in (S U S‘1)*. As in §1.1, (S U S‘1)* is a monoid under concatenation Which is now identified With the group product in G. Proposition 1.26. Let G be a group and let S be a non-empty subset of

G. Then (SU S‘1)* is a subgroup of G. Proof. We Show that the conditions of Definition 1.6 hold. Let a: = alagn-ak,

y = b1b2,...,bl

be elements in (S U S‘1)*. Then my 2 men - - -akb1b2, . . .,bl

is in (S U S—1)*, and so (S U S—1)* is closed under the binary operation of G. For a E S, e = (La—1 6 (S U S‘1)*. Lastly, by the rule for inverses of products, :10 1 =(a1a2 - - .ak)_1 —1 —1 _ 1’ _ak—1 ak_1...a

which is an element of (S U S‘1)*.

I]

The subgroup (S US‘1)* of G is called the subgroup of G generated by S and is denoted by (S). For example, if G 2 D3, S = {p2,u1}, then

(S) 2 D3 since

po = p3, p1 = pg, p2 = p2, #1 = M1, M2 = mm, #3 = #192If S = {p2} Q D3, then

(S) = {.03: n E Z} = {pm/11,92}If G is abelian, then (S) has a somewhat simpler definition.

Let S =

{ag}5e I be a non-empty subset of G indexed by the set I. Then (S) consists of all group products of the form

x=Haza fiEI

Where the n5 are integers that are 0 for all but a finite number of subscripts

fl. If G is an additive group (and hence necessarily abelian), (S) has a form familiar to students of linear algebra: it is the collection of all quantities of the form

m = Z ngag, ,BEI

24

Fundamentals of Mode'rn Algebra

where 72,3 2 0 for all but a finite number of ,8, known as the Z-linear combinations of the set S.

Better yet, if G is abelian and S = {a1,a2, . . .,ak} is finite, then (S) consists of all group products of the form k _

n'

11:1

for integers 11,-. In additive notation this is: k: 58:

E

niai.

i=1

Let G be any group and let S = {a} be a singleton subset of G. Then the subgroup (S) is the cyclic subgroup of G generated by a. Note

that (S) =

Fig. 1.5

4—)

Non—square rectangle DABC'D, £3 J_DC, £4 _LAD.

Much number theory results from the next proposition. Proposition 1.31. Every subgroup of a cyclic group is cyclic.

Proof. Let G = (a) be cyclic, and let H S G. If H = {c} then H is cyclic and the proposition is proved. So we assume that H has at least two elements 6 and 1) 3E 6. Since b is non-trivial and H S (a), there exists a positive integer k so that ak E H. By the well-ordering principal for natural numbers, there exists a smallest positive integer m for which am 6 H. Let h E H. Then h = a” for some integer n. Now by the division algorithm for integers, n = mg + r for integers q, r with 0 g r < m. Thus

h = (am)qa", and so, a’" = h(am)_q E H. But this says that r = 0 since m was chosen to be minimal. Consequently, h = (am)q 6 (am), and so H = (am). I] Let {n1,n2,- - - ,nk} be a finite set of integers, at least one of which is non—zero, and consider the subgroup H = (n1,n2,- - - ,nk) of Z. By Proposition 1.31, H is cyclic, that is, H is of the form (d) for some integer d. We may assume d > 0. This integer d is the greatest common divisor

of the set of integers {711, ng, . . . , nk}, and is denoted by gcd(n1, n2, - - - ,nk).

28

Fundamentals of Modern Algebra

The greatest common divisor is an appropriate name: if d

=

gcd(n1,n2,--- ,nk), then n,- E (d),Vi, and so, dkz- = n,- for some inte— ger 16,-.

Consequently, d divides each ni.

Also, if c is a common divi—

sor of the n,, then (d) = (n1,n2, . . . ,nk) Q (0), thus 0 divides d; in this sense, d is the “greatest” common divisor. Conversely, if d is a common divisor of the ni which is divisible by any other common divisor, then d = gcd(n1,n2, . . . ,nk). The greatest common divisor is generalized to commutative rings with unity in §2.1. Readers already know how to compute gcd(n1,n2) for integers n1, n2. For instance, to find gcd(1750, 1176) we perform a series of divisions:

1750 = 1176- 1 + 574, 1176 = 574. 2 + 28, 574 = 28-20 + 14, 28:14-2+0, and the last non—zero residue 14 is the greatest common divisor.

Proposition 1.32 (Bezout’s Lemma). Let n1,n2 be integers with d = gcd(n1,n2). Then there exists integers x, y for which nlx + ngy = d.

Proof. Since gcd(n1,n2) = d, the cyclic subgroup (n1,n2) 3 Z is generated by (1. Thus there exists integers {17,3} so that nlac + ngy = d.

I]

But how do we find a; and 3/? Are they unique? The computation of a: and y follows from the calculation of gcd(n1,n2). For instance, to find :10 and y for which 1750110 + 11763; = 14, we solve for 14, 28, and 574 in the equations above and substitute, thus: 14:574—28-20 = 574 — (1176 — 574 - 2) - 20 = 1176- —20 + 574 - 41 = 1176- —20 + (1750 — 1176- 1) - 41 = 1750 - 41 + 1176- —61, and so, :13 = 41, y = —61.

Integers m, n are coprime if gcd(m, n) = 1. For n 2 1, the number of

integers m, 1 g m S n, for which gcd(m, n) = 1 is Euler’s function G is the binary operation on G.

32

Fundamentals of Modern Algebra

Example 1.2. Let H S G. Then G is an H-set Where the action H X G —> G is defined as by = hgh‘l. Example 1.3. Let H g G and let L denote the collection of left cosets of H

in G. Then L is a G—set With action G X L —> L defined as g(aH) = (ga)H (one should verify that this action is well—defined on left cosets). Example 1.4. Here is an example from linear algebra: Let V be a vector

space over R. Then V is an RX—set, Viewing Rx as a multiplicative group, With action given as scalar multiplication. Let X be a G-set and let 9 E G. Let

X9={m€X: gm=x}. The subset X9 consists of the elements a: E X fixed by 9. For instance, for the G—set X = G of Example 1.1, G6 = G and G9 = 0 if g 75 e. Let

XG={€B€X2 gx=x,Vg€G}, X G is the set of x E X that are fixed by all elements of G. In Example 1.2, for G abelian, GH 2 G, as one can check. Let X be a G-set and let x E X. Let

Gm={g€G: gm=m}. Proposition 1.39. G9, is a subgroup of G. Proof. Let a,b 6 G33. Then

(ab)m = a(b$) 2 am = .73, so that Gm is closed under the binary operation of G. Clearly, e 6 G96. Finally, for a 6 G90, a‘ 1 x = 9: since as: = x, thus a‘1 6 G33. El For :1: E X, the subgroup Gm g G is the isotropy subgroup of ac. Let X be a G-set and let x E X. The orbit of a: in X under G is

Gat={ga:: gEG}. Proposition 1.40. The collection of orbits Ga: of a: as a: ranges over X forms a partition of X.

Proof. (Sketch) Define a relation on X as follows: :3 N y if and only if y 6 G33. Then as one can check, N is an equivalence relation on X, Where the equivalences classes are precisely the orbits in X. Consequently, the orbits form a partition of X, for details see [Rotman (2002), §1.3, Proposition

1.54].

CI

Groups

Proposition 1.41. Let G be a, finite group.

33

Let X be a G-set and let

an E X. Then |Gx| = [G : Gm]. Proof. Let x1 6 G33.

Then x1 = 91:10 for some 91 E G.

Let L be the

collection of left cosets of Gm in G. Let ¢ : Gas —> L be defined by ¢(x1) = gl. Suppose that 9 is some other element of G With x1 = 9:13. Then

(9—191)x = as, hence 91—19 6 G93, and so 9G2c = gl. Consequently, 93 is well—defined, that is, d) is a function.

We show that qb is a 1—1 correspondence. Suppose that ¢(ac1) = $032). Then 91Gx = 92Gm for some 92 E G With x2 = 92:12, Where 92 = 919 for

some 9 6 G96. NOW, $2 = 92$ = (919)33 = 9137 = 961,

thus gb is 1—1. Next, let 9 6 G96 6 L. Put 9 = 951:. Then My) 2 9G,”, so that gb is onto.

El

In the case that X is finite (IX I < 00), there are a finite number of orbits Gw1,Gx2, . . . ,Gc in X represented by a finite subset {331,5102, . . .,CL’S} of X. One has the formula 8

|X| = 2 lil-

(1.3)

i=1

Observe that X G is the collection of one-element orbits in X; let r = |XG|. Then upon renumbering the m if necessary, formula (1.3) is

|X| = IXGI + Z lz-l-

(1-4)

i=r+1

Proposition 1.42. Let X be a finite G-set with G a finite group of order

p”. Then |X| E |XG| modp. Proof. By Proposition 1.41, |Gwi| divides |G| = p” for 1 g i g s, and so, p divides |Gaci| for r + 1 g i g 3. Thus p divides |X| — |XGI. |:I We can now state and prove Cauchy’s Theorem.

Proposition 1.43 (Cauchy’s Theorem). Let p be a prime number and let G be a finite group. If p divides |GI, then G contains a subgroup of order p.

34

Fundamentals of Modern Algebra

Proof. Let X be the set of all p—tuples (91,92, . . . ,gp), 92‘ E G, for which 9192 ' ”9p = 8- Since 9p = (9192 ' ' 3919—1)—1 , an element of X is completely determined by choosing arbitrary elements 91,92, . . . ,gp_1 in G. Hence

|X| = |G|P_1, and so, p divides |X|. Let a = (1,2,. . . ,p) be the cycle in 5,0 of length p. Let H = (0), so that |H| = p. Now X is an H—set With action defined as Ui(glig27 ' ' ' 79p) 2 (gai(l)7gai(2)7 - - - agai(p))7

for 0 S i S p — 1. By Proposition 1.42, |X| E |XH| modp, hence p divides |XH|, and so, X H contains at least two elements. But X H consists of precisely those p—tuples (gl, g2, . . . ,gp) in X for which 91 = 92 = - - - = gp. It follows that there exists a non-trivial element 9 E G for which

(g,g, . . . ,g) E X. Thus 91’ = e and (g) is a subgroup of G of order p. \_\,_/

III

P

Cauchy’s Theorem Will help us prove Sylow’s First Theorem. But first, we need another application of Proposition 1.42. Proposition 1.44. Let G be a finite group and let H be a subgroup of G of

order p”. Let N(H) be the normalizer ofH (see §1.6, Exercise 27). Then [N(H) : H] E [G : H] modp. Proof. Let L be the set of left cosets of H in G, one has |L| = [G : H]. L is an H-set as in Example 1.3, and by Proposition 1.42,

[G : H] E |LH|modp. By definition,

LH = {gH: h(gH) = gH, Vh e H}, that is, LH consists of those left cosets 9H 6 L for which g—n = H. Consequently, LH consists of precisely the left cosets of H in N (H) Thus

|LH| = [N(H) : H]. The result follows.

III

Proposition 1.45 (First Sylow Theorem). Let G be a finite group with |G| =pnm, n 2 1, pfm. Then: (2') G contains a subgroup of order pi, for 1 S i S n. (it) IfH is a subgroup ofG of order pz for 1 S i S n — 1, then H is a normal subgroup of a subgroup of order pH’l.

Groups

35

Proof. (i) We use induction on i: (The trivial case) By Cauchy’s Theorem, G contains a subgroup of order

10-

. (The induction step) Suppose that G contains a subgroup H of order pZ

for l g i 3 11—1. By Proposition 1.44, p | [N(H) : H], and since H N (H ) /H be the canonical surjection. Then ib'1(K) is a subgroup of G of order pi+1.

For (ii), just note that the subgroup H of (i) is normal in ¢_1(K).

III

A subgroup of order p” in G given in Proposition l.45(i) is called a Sylow p—subgroup of G. Sylow stated two other theorems. We defer their proofs, but see [Rot—

man (2002), §5.2]. Proposition 1.46 (Second Sylow Theorem). Let H1, H2 be two Sylow p—subgroups of the group G. Then H1 and H2 are congugate, that is, there exists 9 E G for which ng_1 2 H2.

Proposition 1.47 (Third Sylow Theorem). Let G be a finite group and suppose that p divides |G|. Let s be the number of Sylow p-subgroups

of G. Then 3 E 1 modp and s divides |G|. 1.6

Exercises

Exercises for §1.1

(1) Let S be a finite set with k elements. Compute the number of binary operations on the set S. ( 2 ) How may semigroups with exactly two elements exist? ( 3 ) Give an example of a semigroup that is not a monoid. (4) Let S be a monoid with identity element e. Show that e is unique. ( 5 ) Determine whether the following sets are groups under the indicated binary operations.

(a) Z, together with ordinary subtraction (b) The subset of Z defined as 2Z = {2m : m E Z}, together with ordinary addition

(0) R, together with ordinary multiplication (d) R+, together with ordinary multiplication

(e) RX, together with ordinary multiplication

36

Fundamentals of Mode'rn Algebra

(f) Z X Z, together With the binary operation on Z X Z defined by ((a,b), (c, d)) I—> (a, d), Va, b, c,d E Z. (6) Find an example of a monoid that is not a group. (7) Let D4 denote the 4th dihedral group. (a) Compute the group product p1/J,10'2. (b) Compute (p301p2)_1. (8) Let _

123456

_

123456

”- 254631 ’ 7‘ 321465 be elements in 5'6, the symmetric group on 6 letters.

(a) Compute 0—2. (b) Compute T o a 2 . (c) Decompose a into a product of transpositions. Is a an even or odd permutation?

(9) Prove Proposition 1.5. (10) (Alternate Definition of a Group) Definition 1.4’ A group is a set G together with a binary operation G X G —> G for Which

(i) the binary operation is associative, (ii) there exists an element 6 E G for Which ea 2 a, for all a E G, (iii) for each a E G, there exists an element 0 E G for Which ca 2 6. Show that this definition of a group is equivalent to Definition 1.4. Exercises for §1.2 (11) Find all of the subgroups of Z15.

(12) Find all of the subgroups of Z3 X Z3. (13) Prove that K g H and H g G implies that K g G.

(14) Let H g G, K g G. Prove that (Hfl K) g G. (15) Let the the (16) Let not

G be a group and let H be a finite non—empty subset of G. Prove following: H is a subgroup of G if and only if H is closed under binary operation of G. H be a subgroup of a group G. Let g be an element of G that is an element of H and let h be an element of H. Show that gh ¢ H.

Groups

37

(17) Show that H = {0, 2,4, 6} g Z8. Compute all of the left cosets of H in Z8.

(18) Verify that H 2 {p0, p1, pg} 3 D3. Show that the left cosets of H in D3 form a partition of D3.

(19) Prove Proposition 1.10. (20) Let 5,, denote the symmetric group on n letters. Let H = {a 6 Sn :

0(n) = n}. Prove that H 3 Sn. HOW many elements are in H? (21) Let An denote the collection of all of the even permutations in Sn, n 2 2. Prove that An 457,. (An is the alternating subgroup of Sn.)

(22) Assume that G is finite with H 4 G. Show that |G/H | divides |G|. (23) Let H 4 G and suppose that aH and bH are left cosets of H in G. Define

aHbH = {ahbh’z h,h' E H}. Prove that aHbH = abH.

(24) Compute the group table for the quotient group Z15 /H Where H =

{0,3,6,9,12}. (25) Compute the group table for the quotient group Z/4Z .

(26) Assume that H 4 G. (a) Prove that 6H 2 H is the identity element in the quotient group G/ H .

(b) Prove that for all aH E G/H, (aH)—1 = a‘lH.

(27) Let HS G. Let N(H) ={gE G: gHg‘1 2H}. (a) Prove that N (H) is a subgroup of G. (b) Prove that H 4 N (H) In fact, show that N (H) is the largest subgroup of G which H is normal in (N (H) is the normalizer of

Jr'nlcyy (28) Let R denote the group of real numbers under ordinary addition. Show that Z 4 R and Q 4 R. Describe the elements of the quotient group

IR/Z. Is there any reasonable description of R/Q? (29) Let H g G with G abelian. Prove that G/H is abelian. (30) Suppose H g G with [G : H] = 2. Show that H 4 G. Suppose that [G : H] = 3. Is it necessarily true that H 4 G? (31) Suppose that K 4 H and H 4 G. Prove or disprove: K 4 G. Exercises for §1.3

(32) Prove Proposition 1.16.

38

Fundamentals of Mode'rn Algebra

(33) Show that the map f : R —> R+ defined as f (:13) = e“: is a homomor— phism of groups.

(34) Prove that the map gb : Z —> Zn defined as (Ma) 2 amodn is a group homomorphism.

(35) ( ) (37) (38)

Prove that Z4 is isomorphic to the subgroup {p0, p1, p2, p3} of D4. Find all of the homomorphisms gb : Zp —> Zp. Find all of the homomorphisms (b : D3 —> D3. Let qb : Z6 —> Z3 be the map defined as ¢(a) = amod3 for all a E Z6. (a) Prove that ¢ is a group homomorphism.

(b) Determine ker(¢). (39) Use the Universal Mapping Property for Kernels to prove that there exists a group homomorphism Z/6Z —> Z3.

(40) Let qb : Z6 —> Z5 be the map defined as 45(0.) 2 amod 5. Determine Whether qb is a group homomorphism. (41) Determine the kernel of the homomorphism 45 : Z —> Z/nZ given by

¢(a) = a + nZ.

(42) Determine the kernel of the homomorphism qb : GLn(R) —> RX given by ¢(A) = det(A). (43) Compute the kernel of the homomorphism 45 : R+ —> R defined by

q5(a:) = ln(a7). (44) Suppose ¢ : G —> G’ is an injective group homomorphism. Prove that

kerw) = {e}.

(45) Let G be any group. Prove that G/{e} 2 G and G/G g {e}. (46) Suppose N, N’ are two normal subgroups of G with G/N 2 G/N’. Does it follow that N g N’ ? (47) Prove that Z $ Q as groups. (48) Let H , K, J be subgroups of abelian group G and suppose that K g

H, Jfl H = {0} and JflK = {0}. Prove that (HJ)/(KJ) § H/K. (49) Let g!) : G —> G’ be a surjection of abelian groups, and let H be a subgroup of G. Show that there is a surjection of groups a : G/H —>

G'/¢ On defined as 2' I—> gi is an isomorphism of groups.

Groups

39

(53) Compute all of the group homomorphisms d) : Z —> Z. (54) Let p be prime. Find all of the generators for the cyclic group Zps. ( ) Suppose Zm has exactly m — 1 generators. Prove that m is prime.

(56) Find a set of generators for the Klein 4—group V. (57) Show that V cannot be isomorphic to a subgroup of a cyclic group.

(58) Compute d = gcd(28, 124). Find integers x, y for which 2811: + 124g 2 d. (59) Let 711,712, . . . ,nk be a finite set of integers, not all zero, with least common multiple [n1,n2, . . . ,nk]. Let m be a common multiple of n1, n2, . . . ,nk. Prove that [n1,n2, . . . ,nk] | m. (60) Let n1,n2,...,nk be a finite set of integers, not all zero, and let

d = gcd(n1,n2,...,nk). Suppose that d’ is a common divisor of 721,712, . . . ,nk. Prove that d’ | d. (61) Let 721,722, . . . ,nk be a finite set of integers, not all zero, With least common multiple [111, n2, . . . , nk]. Let 1“ Z 0 be an integer. Prove that

r[n1,n2, . . . ,nk] = [rn1,rn2, . . . ,rnk].

(62) Prove by example that [n1,n2,n3] 75 (n1n2n3)/gcd(n1,n2,n3). (63) Let U_ _

1234 4123

’

T_ _

1234 3214

be elements in S4, the symmetric group on 4 letters.

(a) Compute the order of the cyclic subgroup (a). (b) Compute the order of the subgroup (a, 7'). (64) Prove Proposition 1.28. (65) Let G be any group and let G be the subset of G defined as

C ={aba—1b_1 : a, b 6 G}. (a) Prove that (G) is a normal subgroup of G. ((C’) is the commutator subgroup of G.) (b) Show that G/ (C) is abelian. (66) Prove Corollary 1.2. (67) Let 014 denote the cyclic group of order 14 generated by 9. List all of the generators of 014.

(68) Compute the number of elements in the subgroup ((1, 2)) of Z4 >< Z6.

(69) List the elements in the quotient group (Z2 >< Z8) / ((0, 2)). (70) Compute the number of elements in the subgroup ((1,2), (2,3)) of Z4 X Z6.

40

Fundamentals of Mode'rn Algebra

(71) Find the unique solution in Z4300 of the system of congruences

as E 30 mod 43

:1: E 87 mod 100.

(72) Find three abelian groups of order 8, no two of which are isomorphic. Find four groups of order 8, no two of which are isomorphic. (73) Decompose the group Z40 into a product of cyclic groups. (74) Suppose that there are a: isomorphism classes of abelian groups of order n, n odd. Compute the number of isomorphism classes of abelian groups of order 2n. Exercises for §1.5

(75) Let X be a G—set. The group G is transitive on X if for each 331,1:2 E X, there exists an element 9 6 G with 9301 = x2. Determine which G—sets in Example 1.1—Example 1.4 have groups that are transitive on X.

(76) Let X beaG—set. Let N={g€ G: gm=m,Va¢€X}. (a) Show that N (3"

defined as 7,-(gj) = (if, for 0 S i,j S p— 1. For example, consider G = 03 = (9). Then there are 3 characters of G3 given in the following tables.

The collection of characters of GP, denoted by GP, forms a group of characters under the binary operation defined as

(ViVjXQk) = 71(gk)7j(gk) : (gum

Rings

57

for 0 S i,j,k: S p— 1. Since 71 is so that 7% =7,- for 0 SiSp—1,Gp is cyclic of order p, and is generated by 'y = 71; we have 'yi(gj) = (I? and '70 is the trivial character which we denote by 1. Clearly, Gp g GP, the isomorphism being given as 'y I—> 9. As another example, we compute the character group of the Klein 4group V = {e,a, b, 0}. Here a2 = b2 = c 2=eandab=c,ac=b,bc=a.

Now a homomorphism 'y : V —> (CX must satisfy the relations ('y(a))2 = 1,

('y(b))2 = 1 and (7(0))2 2 1 in (3". Thus the only possibilities for 7(a), 7(b) and 7(0) are :l:1. Moreover, 7(a)*y(b) = 7(0), 7(a)*y(c) = 7(1)) and 7(b)’y(c) = 7(a). Thus the characters are as follows.

The characters of V form the character group V under the product

(Vil/j)(a:) = l/i($)I/j (at). We have V g V. Let G be a finite group and let 7 : G —> (Cx be a character. The group

(Cx can be identified with the elements in GL1 ((C), the invertible 1 X 1 matrices with entries in (C. Thus the character 7 can be written as the group homomorphism

p : G —> GL1((C). This is a 1-dimensional linear representation of G. Generalizing a bit, let G be any group and let K be a field. An n-dimensional linear representation of G is a group homomorphism p : G —> GLn(K). Linear representations of groups can be used to represent group elements as matrices so that the group operation can be represented by matrix multiplication. Representations of groups are important because they allow many group-theoretic problems to be reduced to problems in linear algebra which is well—understood. We give two examples of group representations (readers may need to recall some elementary linear algebra). Example 2.8. Let KCg denote the group ring with (g) = C3. Then K03 is a vector space of dimension 3 over K on the basis {1, g, 92}. It determines a 3—dimensional linear representation of Cg:

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Fundamentals of Mode'rn Algebra

p: 03 —> GL3(K) as follows. For i, j = 0, 1, 2 define p(gi) to be the invertible 3 X 3 matrix over K whose jth column is the coordinate vector of gigj = gi+j with respect

to the basis {1,9,92}. Thus

100 001 010 p(1)= 010 , 10(9): 100 , p(92)= 001 001 010 100 This is the left regular representation of C3. Example 2.9. Let D3 denote the 3rd order dihedral group, the group of symmetries of the equilateral triangle consisting of 3 rotations p0, p1, p2 and 3 reflections ,ul, pg, 113 (§1.1). The triangle 1s embedded in the nay—coordinate

plane with origin 0— — (0,0) and verticesA =(— —, —%), B = (0 1) and C = (i, ——); the line £1 has equation y: ‘rx; 62 has equation y— — —‘/?§a3, see Figure 2.1.

Fig. 2.1

Equilateral AABC in the any—plane.

We can represent D3 as a group of 6 plane isometries, cf. [Martin (1982),

Chapter 9]: p0 = rotation of a point (:17, y) in the plane 0° clockwise about the origin 0 is given by the matrix multiplication

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59

(3 ‘3) (Z) = (:1), p1 = rotation of a point (x, y) in the plane 120° clockwise about the origin 0 is given by the matrix multiplication

fi/2

—1/2

—\/§/2 —1/2

x _ 30'

y — y’ ’

p2 = rotation of a point (:10, y) in the plane 240° clockwise about the origin 0 is given by the matrix multiplication

—1/2

—\/§/2

fi/2

-1/2

a; _ (5’

y _ y’ ’

p1 = reflection of a point (ac, y) in the plane through the line 61 is given by the matrix multiplication

(«152 X33) (3) = (:1), [1,2 = reflection of a point (x, y) in the plane through the line 62 is given by the matrix multiplication

1/2

—\/§/2

—\/§/2

—1/2

a:

_ x’

y _ y’ ’

p3 = reflection of a point (x, y) in the plane through 63 (the y-axis) is given by the matrix multiplication

—1 0

0 l

x y

_ —

a," y’

'

The six matrices define a 2-dimensional representation of D3 p: D3 —> GL2(R),

Where, for example,

_

—1/2

«3/2

p(pl)—(_\/§/2 _1/2),

_

1/2

«3/2

P(M1)—(\/§/2 _1/2)-

It is routine to check that p is a group homomorphism.

60

2.4

Fundamentals of Mode'rn Algebra

Ideals

In this section we introduce ideals in a ring, show that the sum, product and intersection of ideals is an ideal, and apply ideals to the Chinese Remainder Theorem. Assuming that R is a commutative ring, we consider the ideal generated by a subset S Q R, and specialize to finitely generated ideals,

principal ideals, and principal ideal domains (PIDs). We prove that F [:13] is a PID whenever F is a field. We define maximal and prime ideals in a commutative ring With unity and use Zorn’s Lemma to show that every commutative ring contains at least one maximal ideal. Finally, we introduce Noetherian rings and show that these rings are equivalent to rings that satisfy the ascending chain condition (ACC) for ideals. Using the ACC, we prove that every PID is a UFD.

By definition a ring is an abelian group under addition. The additive identity element is denoted by 03, or more simply 0. As an additive abelian group, R has at least two subgroups: {0} and R. Some of the subgroups of R have a special property. Definition 2.10. The additive subgroup N Q R is an ideal of R if aN Q N and NbQ N for all a,b€ R. If the ring R is commutative, then the definition of ideal is somewhat simpler: an additive subgroup N Q R is an ideal if aN Q N for all a E R.

A ring R always has at least two ideals: the trivial ideal {0} and the ring R itself. Here are some more examples. The subgroup nZ of Z is an

ideal since m(nk) = n(mk) 6 NZ for all m, n, k E Z. Let R = Z[:c] and let N be the subset of all polynomials which have constant term 0. Then N is

an additive subgroup of R. Moreover, p(m)q(:1:) has constant term 0 if q(a:) does, hence N is ideal of Z [as]. Here is an example of a subgroup which is not an ideal. Let R = Q with additive subgroup Z. Then Z fails to be an

ideal of Q since %Z Q Z. Proposition 2.16. Let R be a ring with unity. Let I be an ideal of R and let u be a unit in R. Then uI = I. Proof. By the definition of ideal, uI Q I, so it remains to show that

I Q uI. Let a E I. Then a = (uu‘1)a = u(u_1a) with u‘la E I. Hence a E uI.

I]

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61

Proposition 2.17. Let R be a 'ring with unity and suppose that I is an ideal of R which contains a unit. Then I = R. Proof. Since I Q R is immediate, we only need to Show that R Q I. Let u be aunit in I. Then 1 = u‘luE I, and hence r- 1 E I for all 7' E R. |:] By Proposition 2.17, one sees that the only ideals of a field F are {0} and F. There is an arithmetic of ideals which we describe as follows. Let I and

J be ideals of R. Define the sum of ideals I + J to be the collection of all sums a + b, a E I, b E J, and define the product of ideals IJ to be the collection of all finite sums 221:1 aibi where az- 6 I, b,- 6 J.

Proposition 2.18. With the notation as above,

(i)

IJ is an ideal,

(ii)

I + J is an ideal,

(iii)

I D J is an ideal.

Proof. We prove (i), and leave (ii) and (iii) as exercises. We first show that IJ is an additive subgroup if R.

Note that

(2:;1 aibi) + (21:1 cidi) is again a finite sum of the form 2 ab with a E I, b E J. Also, 0 is a finite sum and — 221:1 aibi = zy=1(—ai)bi is a finite sum. Thus IJ S R.

Now, r(IJ) Q IJ, for all 7" E R, since TELI aibi = 221:1 raibi with rai 6 TI Q I, b,- E J. Likewise, (IJ)? Q IJ. Thus the product of ideals IJ

is an ideal of R.

III

Let I be an ideal of R and let a, b be elements of R. We write a E b mod I if and only if a — b E I. We have the following generalization of the Chinese Remainder Theorem. Proposition 2.19 (Chinese Remainder Theorem for Rings). Let R be a commutative ring with unity. Let N1, N2 be ideals of R with N1 +N2 = R and let a1,a2 be elements of R. Then the system of congruences m E a1 modN1 m E a2 modN2 has a unique solution modulo N1N2.

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Fundamentals of Mode'rn Algebra

Proof. There exist elements n1 6 N1, n2 6 N2 for which n1 + n2 2 1. Hence

(a1 — a2)(n1 + n2) = a1 — a2, and so, a1 + 711((12 — a1) = 0,2 + n2(a1 — (1.2).

Thus a: = a1 + n1(a2 — a1) is a solution to the system of congruences. We

show that this solution is unique modulo N1 N2. To this end, suppose that x’ is another solution. Then a: — x’ E 0 mod N1 and x — x’ E 0 mod N2, and so there exists an element m 6 N1 0 N2 for which :10 — ac’ = m. Now,

a: — 30’ 2 (ac — x')(n1 + 77.2) = (ac — x')n1 + (a: — m’)n2 = mnl + mng,

with mnl + min 6 N1N2. It follows that x’ E mmod (N1N2).

E]

Corollary 2.2. Let R be a commutative ring with unity and let I , J be ideals with I+ J = R. Then [J = ID J. Proof. Let 2:;1 aibi E IJ. Then aibi 6 I0 J for each i, thus IJ Q 10 J.

For the reverse containment, suppose that a 6 I H J. Then :L' = a is a solution to the system of congruences a: EOmodI a: EOmodJ

but so is x’ = 0. Thus a = a — 0 E IJ by the Chinese Remainder Theorem for Rings. E] In a commutative ring, each subset of the ring gives rise to an ideal in a natural way.

Definition 2.11. Let S = {a5} be a subset of elements of the commutative ring R. Let N be the collection of all sums of the form 2 mag, where K3

m E R, and where 773 = 0 for all but a finite number of indices fl. Then N is an ideal of R which we call the ideal of R generated by S. If the generating set S is finite, then the ideal is finitely generated. If

the ideal I is generated by S = {a1,a2, . . . ,ak}, then I consists of all the linear combinations 226:1 riai, 7'2- 6 R, and is denoted by (a1, a2, . . . , ak).

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63

Every ideal of a commutative ring is generated by some subset of the ideal. If necessary, one can take the ideal itself as the generating set. A challenging problem in ring theory is to find the smallest generating set for a given ideal. (This is analogous to extracting a basis from a spanning set

for a vector space.) For example, the ideal (2,33,;102) of Z [:11] is the ideal (2,33); the ideal (4, 6) in Z is the ideal (2). Notice that (4,6) = (gcd(4, 6)). When there is exactly one element in a generating set for an ideal, the ideal is given a special name. Definition 2.12. Let R be a commutative ring and let a E R. Then the ideal (a) is the principal ideal generated by a. Not every ideal in a ring is principal, however. For example, the ideal

(2, ac) Q Z[a:] cannot be written in the form (p(x)) for some p(a¢) E Z[a7]. To see this, suppose (2,33) = (p(:1:)) for some p(:c) E Z[:I:]. Then 2 = p(a:)q(x) for some q(:t:) E Z [x], and hence p(:17) has degree 0, and is a constant which is necessarily :l:2. This says that x E (2) Q Z [as], which is impossible. Definition 2.13. A commutative ring R is a principal ideal ring if every ideal in R is principal. Definition 2.14. An integral domain which is a principal ideal ring is a

principal ideal domain (PID). Proposition 1.31 implies that Z is a PID by showing that every subgroup of a cyclic group is cyclic. By essentially the same argument as in Proposition 1.31, we can prove the following proposition.

Proposition 2.20. Let F be a field. Then F[m] is a PID. Proof. Clearly, F [:10] is an integral domain. Let I be a non—zero ideal of

F [It] (clearly, the zero ideal is principal). Let p(ac) be a non-zero polynomial of minimal degree in I. Then every element of I is a multiple of p(.’r), for if f(:1:) is in I, then by the Division Theorem for polynomials over a field,

there exist polynomials q(:1:) and r(m) in F [x] for which

f(w) = P(w)q(w) + TOE), where deg(r(:L')) < deg(p(:c)). Thus r(x) = f(;I:) —p(x)q(x) E I, and so r(a:) = 0.

I]

In a PID we have the following generalization of Bezout’s Lemma. Proposition 2.21. Let R be a PID. Let a,b E R, and suppose that d is

64

Fundamentals of Mode'rn Algebra

a greatest common divisor of a, b. Then there exist elements 56,3; 6 R for which d = a0: + by.

Proof. Let C be the set defined as

C: {ar+bs: 7336 R}. Then C is an ideal of R, and hence C = (d’) for some d' E R. Now d’ | a, d’ | b and c I d’ whenever c | a and c | b, and so d’ is a greatest common divisor of a, b.

Since d divides d’, d’ 2 id for some 7" E R, and since d’ divides d, d = sd’ for s E R. Thus d’ = rsd’. If d’ = 0, then d = 0 and the proposition is proved. So we assume that d’ 75 0. Consequently, rs = 1, and so, 7" is a unit.

By Proposition 2.16, (d’) = (d), and so d = act: + by for some x, y E R.

El

Definition 2.15. Let R be a commutative ring with unity. A maximal ideal is a proper ideal M of R for which there is no proper ideal N of R

with M C N C R. For example, the ideal (2) is a maximal ideal of Z, as is (3). However, since

(4) C (2) C Z, (4) is not a maximal ideal. A commutative ring with unity with a unique maximal ideal is a local ring. Definition 2.16. Let R be a commutative ring with unity. A proper ideal N is prime if ab 6 N implies that either a 6 N or b E N.

For example, in Z, the ideal (5) is prime, but (6) is not prime. How are maximal and prime ideals related? We know that there are non—maximal prime ideals, for example {0} in Z. But we have the following proposition. Proposition 2.22. Every maximal ideal is prime. Proof. Suppose M is a maximal idea] with ab 6 M. We show that either a E M or b E M. By way of contradiction, we assume that a,b 91 M.

Consider the ideals (a) + M and (b) + M, which both contain M. Now if (a) + M is not proper, then (a) + M = R, and (b) +M = R((b) +M) = ((a) +M)((b) +M) g (ab) +M = M, which says that b E M.

If (a)+M is proper, then (a)+M = M, thus a E M. So M is prime.

I]

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65

It is somewhat surprising that every commutative ring with unity contains at least one prime ideal. This is proved using Zorn’s Lemma Which we briefly review. Let S be a non—empty set. A relation on S is a subset -_< of the cartesian product 3 X S. If (at,y) E f, we write a: -_< y. The relation j on S is reflexive if a: -_< x, Vac E S, j is antisymmetric if x j y and y j :1: implies as = y, and -_< is transitive if x j y and y -_< 2 implies that m j 2. We say that S is partially ordered under j if -_< is reflexive, antisymmetric and transitive. An element m E S is a maximal element if x E S and m j :0 implies that m = as. A subset T of a partially ordered set S is a chain if for all 93, y E T, either x j y or y j :13. An upper bound of a chain T is an element u E S for Which a: j u for all a: 6 T.

Zorn’s Lemma. Let S be a non-empty partially ordered set in which each chain has an upper bound. Then S has a maximal element. Proposition 2.23. Let R be a commutative ring with unity. Then every proper ideal of R is contained in a prime ideal. Proof. Let J be a proper ideal of R and let 73 denote the collection of all proper ideals of R Which contain J. Since J 6 73, 73 is a non—empty set Which is partially ordered under set inclusion. Let C be any chain in 73. Then the ideal U I is an upper bound for C. Thus by Zorn’s Lemma, 73 I EC

contains a maximal element M. By construction, M is a maximal ideal containing J. By Proposition 2.22, M is prime. El Since {0} is a proper ideal of R, Proposition 2.23 shows that R has at least one prime ideal. Corollary 2.3. Let R be a commutative ring with unity. Suppose I is an ideal of R which is not contained in any prime ideal. Then I = R. Proof. This is just the contrapositive of Proposition 2.23.

III

In a UFD the prime ideals which are principal can be characterized.

Proposition 2.24. Let R be a UFD, and let a E R, a 73 0. Then (a) is a prime ideal of R if and only if a is an irreducible element of R. Proof. Suppose (a) is prime with factorization a = cb. Then either c E (a) or b E (a). Suppose c = ra, r E R. Then a = rab = arb, and so,

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Fundamentals of Modern Algebra

a(1 — rb) = 0. Since a 7E 0 and R is an integral domain, 1 = rb. Thus b E U (R) which says that a is irreducible. Conversely, suppose that a is irreducible with bc 6 (a). Then bc 2 ra, for some r E R. Let b = b1b2---bk,c = 0102"'Cl, and r = r1r2---rm be

the unique factorizations of the elements b, c, r. Then,

(5152“'bk)(0102"‘cz) = (7‘17“2"'7"m)a, with k+l = m+ 1. Now the m factors of r on the right—hand side correspond to exactly I: + l — 1 factors on the left—hand side, Which includes every factor

of either b or 0. Thus either 0 E (a) or b E (a), and so (a) is prime.

E]

Proposition 2.24 also holds in a PID.

Proposition 2.25. Let R be a PID, and let a E R, a 34$ 0. Then (a) is a prime ideal of R if and only if a is an irreducible element of R.

Proof. Exercise.

D

Definition 2.17. Let R be a commutative ring With unity. Then R satisfies

the ascending chain condition for ideals (ACC) if every ascending chain of ideals in R Io§11§12§’” eventually stops—that is, there exists an integer m 2 0 for which Im = Im-l—l =

m+2 =

The following was proved by E. Noether in 1921.

Proposition 2.26 (Noether). Let R be a commutative ring with unity. The following are equivalent. (i) Every ideal I of R is finitely generated.

(ii) R satisfies the ACC. Proof. (i) => (ii). Let 10911§12§‘” be an increasing sequence of ideals of R and let I = [Js In. Then I is an

ideal of R, and so, I is finitely generated over R. Let S 2 {b1, b2, . . . , bl} be a generating set for I. For 1 S i S l, bi 6 Im, for some integer mi. Let m = max{m,~}. Then S Q Im and so,

121...: m+1= m+2=----

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67

(ii) => (i). Let I be an ideal of R and let .7 denote the collection of all finitely generated ideals of R that are contained in I. Certainly, .7 is non— empty since {0} E .7. We claim that there exists a maximal element in .7, that is, we claim that there is an element J’ E .7 for which there is no J E .7 with J' C J. To this end, assume that no such maximal element exists.

Specifically, I0 = {0} is not maximal, thus there is an element I1 6 .7 with Io C I1. But I1 is not maximal, and so there exists an element I2 with

I0 C I1 C [2. Continuing in this manner, we construct an ascending chain of ideals that does not stop, Violating the ACC. Consequently, the family .7 has a maximal element J. Note that J is finitely generated over R. By construction, J Q I. If J C I, then there is an element at E I\J.

But then J + (m) is an element of .7 with J C J + (3:), contradicting the maximality of J. Consequently, J = I and we conclude that I is finitely generated. E] In honor of Noether, a commutative ring with unity that satisfies either of the two equivalent conditions of Proposition 2.26 is a Noetherian ring. For instance, any PID is Noetherian. Proposition 2.27. Let R be a Noetherian ring and let HR denote the product of a finite number of copies of R. Then H R is Noetherian. Proof. Exercise.

E]

Proposition 2.28 (Hilbert Basis Theorem). Let R be a Noetherian ring. Then the polynomial ring R[a:] is Noetherian. Proof. We show that each ideal N of R[a:] is finitely generated as an Rmodule. For n 2 0, let Jn = {anz anxn+an_1mn_1 +~~+a2x2 +a1m+a0 E N}. Then

J0§J1§J2§'“ is an increasing sequence of ideals of R. By Proposition 2.26, (i) => (ii), there exists an integer m 2 0 for which

Jm= m+1=Jm+2=”' For n 2 0, let {bn,1, bn,2, . . . , a-n} be a generating set for Jn and let {fn,ly fn,2y - - - 7 fn,jn}

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Fundamentals of Modern Algebra

be a set of polynomials of degree n in N so that the leading coefficient of frm'n is bnfln for 1 g in g jn. We claim that the set

B = U {fn,1:fn,27 ' ' ~7fn,jn} 77.20

is a generating set for N as an R[x]—module. To prove the claim we proceed by induction on the degree of the polynomial in N. Clearly, a polynomial of degree 3 0 in N can be written as an R[a:]—linear combination of elements in B. For the induction hypothesis, we assume that all polynomials of degree 3 n — 1 can be written as R[a:]-linear combinations of elements in B. Let

f(:I:) = anx" + an—lm”_1 + . ~ + a2a72 + aw: + (10 be a polynomial of degree n in N. There exists elements r1(.’r), r2 (x), . . . ,

rk(a:) E R[:c] and elements f1(x), f2(x), . . . , fk(ac) E B so that

r1(w)f1(w) + r2(w)f2(w) + - - - + rk(w)fk(w) = anxn + terms of degree 3 n — 1.

Let g(a:) = 2:621 rt(a:)ft(a:).

Then h(a:) = f(x) — 9(33) is an element

of N of degree S n — l, which by the induction hypothesis is an R[£L‘]linear combination of elements in B. It follows that f (:10) is such a linear combination. III Corollary 2.4. Let K be a field and let 371,332, . . . ,mn be indeterminates.

Then K[ac1, x2, . . . , Inn] is Noetherian. Proof. Since any field is Noetherian, K [1101] is Noetherian by Proposition

2.28. Another application of Proposition 2.28 yields K[m1][x2] = K[:L'1, x2] Noetherian, and so on.

E]

Since every PID is Noetherian, every PID satisfies the ACC. This is the key to proving that every PID is a UFD. Proposition 2.29. Let R be a PID, and let a be a non-zero, non—unit of R. Then a is a product of irreducible elements of R. Proof. Suppose there exists a non—zero, non—unit element a E R which is not a product of irreducibles. Then a itself is not irreducible, and thus, a = blcl where both b1 and cl are non—units. Now either b1 or c1 cannot

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69

be written as a product of irreducibles, say d1 2 b1 is one which can’t be. Then all is reducible and d1 = b202 where neither b2 nor 02 is a unit. Now

either ()2 or 02 cannot be written as a product of irreducibles, let us say

that d2 2 b2 is this element. Note that (d1) C ((12). Now the element d2 is reducible: d2 = b303 where neither b3 nor 03 is a unit, and either b3 or 03 (assume b3) cannot be written as a product of

irreducibles. Set d3 2 b3. Now (d1) C (d2) C (d3). Continuing in this manner, one can construct a strictly ascending chain

of ideals {(dz) 29:1. This is impossible by Proposition 2.26.

III

Proposition 2.30. IfR is a PID, then R is a UFD. Proof. By Proposition 2.29, a non-zero, non-unit element of a PID can be factored into irreducibles, so we only need to show that this factorization is unique. Suppose that p1p2...pl

:q1q2...qm,

for irreducible elements 10;, qj. By Proposition 2.25, (p1) is prime and so,

qj 6 (191) for some j, 1 S j S m. Consequently, qj = ulpl for U1 6 U(R). Upon renumbering the factors qj if necessary, one has P1102 ° ° 'Pt = u1101(12 ' ”qm,

thus

P2193 ' ‘ 'P1 = u1(12CI3 ‘ ‘ ‘qm-

By similar reasoning 11,2192 2 qk for some unit 11,2 6 U (R) and some integer k, 2 g k S m, and upon renumbering one obtains

P2P3"'Pl = U1U2P2CI3CI2 ' "film: thus P3"'Pl = U1U2Q3"'qm-

Continuing in this manner yields l = m. It follows that the factorizations are unique in the sense of Definition 2.9. E]

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2.5

Fundamentals of Modern Algebra

Quotient Rings and Ring Homomorphisms

In this section we define the quotient ring R/N of a ring R by an ideal N and show that R/N is a field if and only if N is a maximal ideal and that R/N is an integral domain if and only if N is prime. Next, we introduce the basic maps between two rings — ring homomorphisms, and give some examples of ring homomorphisms. We relate quotient rings and ring homo— morphisms in the First Isomorphism Theorem (for rings) and the Universal Mapping Property for Kernels. As an application we develop the notion of the characteristic of a ring.

Let R be a ring and let N be an ideal of R. Since N is a normal subgroup of the additive group R, the quotient group R/N is defined with group operation

(a+N)+(b+N)=(a+b)+N.

(2.1)

One can endow R/N with the structure of a ring by defining a multi— plication on the left cosets. For this we define a relation

B : R/Nx R/N—>R/N by the rule B(a+N,b+N) =ab+Nfor left cosetsa+N,b+NE R/N. Proposition 2.31. Let N be an ideal of R. Then the relation B defined above is a binary operation on R/N. Proof. We check that B is well—defined on left cosets. Suppose that :1: E a+N, y E b+N. Nowcc =a+n1, y: b+n2 for some n1,n2 E N. Thus

my: (a+n1)(b+n2) =ab+an2+n1b+n1n2 E ab+N, andso,

B(x+N,y+N)=xy+N=ab+N=B(a+N,b-I—N). E]

Proposition 2.32. Let N be an ideal of R. Then R/N is a ring with coset addition defined by (2.1) and coset multiplication defined as in Proposition 2.31.

Proof. It is straightforward to show that R/N satisfies Definition 2.1.

I]

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The ring R/N is the quotient ring of R by N. For an example, let R = Z, N = 4Z. Then the quotient ring Z/4Z consists of the cosets

{42, 1+4Z, 2+4Z, 3+4Z}, together with coset addition and multiplication. The binary operation tables for Z/4Z are as follows

+ 4Z 1+4Z 2+4Z 3+4Z 4Z 1+4Z 2+4Z 3+4Z

|

4Z 4Z 1+4Z 2+4Z 3+4Z 4Z 4Z 4Z 4Z 4Z

1+4Z 1+4z 2+4Z 3+4Z 4Z

1+4Z 4Z 1+4Z 2+4Z 3+4Z

2+4Z 2+4Z 3+4Z 42 1+4Z

2+4Z 4Z 2+4Z 4Z 2+4Z

3+4Z 3+4Z 42 1+4Z 2+4Z

3+4Z 4Z 3+4Z 2+4Z 1+4Z

We point out that the tables for Z/4Z look just like the binary operation tables for the ring Z4 given in §2.1.

For another example, let R = Q[:B] and let N = (x2 — 2), the principal ideal of Q[x] generated by x2 — 2. The elements of the quotient ring

Q[m]/ (x2 — 2) consists of left cosets computed as follows. Let f (as) E Q[a:]. By the Division Theorem there exists polynomials q(a:) and r(a:) for which

f(«'r) = q(«%‘)(962 - 2) + WU), with deg(r(a:)) < deg(ac2 — 2) = 2. Thus 1°(x) = a-l—bx for some a, b E Q. It follows that the elements of (QM/(x2 — 2) are {a+ b:1:+ (x2 — 2) : a, b E Q}. Note that addition in QM /N is given as (a+bx+N)+(c+dx+N) =a+c+(b+d)x+N, while multiplication is given as

(a+ba:+N)(c+da:+N)=(a+b:1:)(c+d:1:)+N = ac+ (ad+bc)ac+bdx2 +N

= ac+ (ad+bc)x+2bd— 2bd+bdx2 +N = ac+2bd+ (ab+bc)m+bd(m2 — 2) +N 2 ac+2bd+ (ab+bc)m+N.

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Let R be a ring with unity, 1. Then the quotient ring R/N is a ring with unity 1 + N. Proposition 2.33. Let R be a commutative ring with unity, let N be a proper ideal ofR and let a E R. Then a + N is a unit of R/N if and only

if (a) + N = R. Proof. Suppose (a) + N = R. Since 1 E R, there exist elements r E R and n6 N so that ra+n= 1, hence ra= 1—n. Now

(r+N)(a+N)=ra+N =(1—n)+N =(1+N)+(—n+N)

=(1+N)+N =1+N, thus r+N = (a+N)_1. Conversely, suppose a + N is a unit of R/N. Then 1 + N 2 ar + N for

some 7' E R, and 80,1 6 ar+N. Thus R Q (a) +N. Since (a) +N Q R, one has (a) + N = R. III Proposition 2.34. Let R be a commutative ring with unity. Then M is a maximal ideal of R if and only if R/M is a field. Proof. Suppose that R/M is a field and let N be a proper ideal of R with M Q N C R. If M 75 N, then there exists an element a E N\M, and hence a + M is a non-zero element of the field R/M. Consequently, a + M

is a unit in R/M, and so, by Proposition 2.33, R = (a) + M Q N. Thus N = R, which is a contradiction. For the converse, we suppose that M is maximal. Since R is a commutative ring with unity, so is R/M. So it remains to show that every

non—zero element of R/M is a unit. Let a + M E R/M, a ¢ M. Then (a) + M is an ideal of R with M g (a) + M g R. But M is maximal, so either (a) + M = M, or (a) + M = R. In the former case, a E M, which is a contradiction. Thus (a) + M = R which says that a + M is a unit in R/M. El Proposition 2.35. Let R be a commutative ring with unity. Then N is a prime ideal of R if and only if R/N is an integral domain.

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Proof. Suppose that R/N is an integral domain, and let ab 6 N. Then

N=ab+N= (a+N)(b+N), and so, either a+N= N, or b+N= N. Thus either a E N or b E N, Which says that N is prime. For the converse, we suppose that N is prime. Since R is a commutative ring with unity, so is R/N. So it remains to show that R/N has no zero

divisors. To this end, let (a + N)(b + N) = N. Then ab + N = N so that ab 6 N. Since N is prime, either a E N or b E N, thus we must have either a+N=N,orb+N=N. III

When we considered functions from one group to another (§1.3), we were particularly interested in functions, namely group homomorphisms, that relate the group operations on the domain and codomain. Now we introduce functions from one ring to another that relate the two operations on each of the rings. Definition 2.18. Let R, R’ be rings. homomorphism if for all a, b E R

A map ¢ : R —> R’ is a ring

(i) ¢(a + b) = (Na) + (Nb), (ii) (Nab) = ¢(a)¢(b)Definition 2.19. Let R, R’ be rings With unity, With unity elements 1 R,

1R/. Then a map ct : R —> R’ is a homomorphism (of rings with unity) if qt is a ring homomorphism and q5(1 R) = 13/. For example, the map 925 : Z —> Z/nZ defined as ¢(a) = a + nZ is a homomorphism of rings With unity. Indeed, we have already seen that qb is a

homomorphism of groups, so (i) holds. For (ii), let a, b E Z. Then ¢>(ab) =

ab+nZ = (a+nZ)(b+nZ) = ¢(a)q5(b). Moreover, 45(1) 2 1+nZ = 1Z/nZHere is another example of a ring homomorphism. Proposition 2.36. Let E/F be a field extension and let oz 6 E. Then the

map da : F[:I:] —> E defined by (Z50: (p(w)) = p(a) is a ring homomorphism.

Proof. Let p(x) = 2:10 aixi, q(x) 2 2?:0 bjwj be polynomials in F [:13] Then

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Fundamentals of Mode'rn Algebra

qba (19(96) + q(w)) = 9M: aiivi + Z 51'i ) i=0 m

2

E

j=0 n

aiozz +

i=0

E

59‘o

j=0

= «>42 m) + m; bjwj) j=0

i=0

and

¢a(P($)CI($)) = ¢a (n,m) is a ring isomorphism. Also, o3 : Z/nZ —> Zn given by a + nZ I—> a modn is a ring isomorphism. A map can be a group isomorphism without being a ring isomorphism. For example, the map gb : Z —> 2Z given by ¢(n) = 2n is an isomorphism if we consider Z and 2Z as groups, but it is not a ring homomorphism since

¢ = 2ab aé = ¢>(a)¢>(b)A map, of course, can be a bijection of sets without being a ring isomorphism. There are many bijective maps between R and C, but these rings are not isomorphic. To see this, we assume that there is an isomor-

phism (of commutative rings with unity) o5 : R —> (C. Note that ¢(1) = ¢(1R) 2 1c 2 1. Since q!) is a group homomorphism, ¢(—1) = —¢(1) = —1. Since (25 is surjective, there exists an element r 6 R with ¢(r) = i, thus

—1 = (q(r))2 = ¢(r2), and so, ¢(—1) = ¢(r2). Now since ct is an injection, r2 = —1 for some real number r, which is impossible. Thus 1R ¥ C. There is an elegant relationship between ring homomorphisms and quo— tient rings expressed in the following propositions (which are the analogs for rings of Proposition 1.21, Proposition 1.22 and Proposition 1.23). Proposition 2.38. Let N be an ideal of R. Then the map 7 : R —> R/N given by 7(a) = a + N is a surjective ring homomorphism with kernel N. Proof. By Proposition 1.21, *y is a surjective homomorphism of additive

groups with ker(7) = N. Now, 7(ab) = ab+N = (a+N)(b+N) = 7(a)7(b), which shows that '7 is also a ring homomorphism.

E]

Proposition 2.39. Let qt : R —> R’ be a ring homomorphism with

ker(¢) = N. Then 7 : R/N —> ¢(R) defined by 7(a + N) = (Ma) is a ring isomorphism.

Proof. One first checks that ¢(R) is a ring. By Proposition 1.22, '7 is an

additive group isomorphism with ker('y) = {N}. Now, 7((a+N)(b+N)) = 7(ab-l—N) : ¢(ab)

= ¢(a)¢(b) =7®+NW@+NL so '7 is also a ring homomorphism.

E]

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Proposition 2.40 (Universal Mapping Property for Kernels). Let

qb : R —> R’ be a ring homomorphism with N = ker(¢). Suppose that K is an ideal of R contained in N. Then there exists a surjeetive homomorphism

of rings 1b : R/K —> (MR) defined by 7(a + K) = ¢(a). Proof. By the UMPK (Proposition 1.23) there exists a surjective homo—

morphism of groups it : R/K —> ¢(R) defined by 7(a + K) = q3(a). Now,

7((a + K)(b + K» = 7(ab + K) = ¢>(ab) = ¢(a)¢(b) = “M + 107(1) + K), so that 7 is a ring homomorphism.

Let qb : R —> R/ be a ring homomorphism With N = ker(gb).

E]

The

Universal Mapping Property for Kernels (UMPK) says that given an ideal

K of R With K Q N, there exists a ring homomorphism 1b : R/K —> R’ so that

¢S=¢ Where s : R —> R/K is the canonical surjection; we say that qb “factors through” R/K and the following diagram commutes:

Rn} Here is a ring isomorphism based on the Chinese Remainder Theorem. Proposition 2.41. Let R be a commutative ring with unity and let N1, N2 be ideals of R with N1 + N2 = R. Then there is a ring isomorphism 2b 2 R/N1N2 —) R/Nl X R/N2

defined by 1b(a + N1N2) = (a + N1,a + N2), for all a E R.

Proof. Let 9b : R —> R/N1 >< R/N2 be the map defined by qb(a) = (a + N1,a+ N2), for all a E R. For a,b E R,

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q3(a+b) = (a+b+N1,a+b+N2) = ((a+N1) + (b+N1),(a+N2) +(b+N2))

= (a+N1,a+N2)+(b+N1,b+N2)

=¢(a)+¢(b), ¢(ab) = (ab + N1, ab + N2)

=< (a + N1)(b + N1), (a + N2)(b + N2)) =< a+N1,a+N2) . (b+N1,b+N2) ¢(a)¢(b), and

¢(1R) = (1R + N1,1R + N2) = 1R/N1xR/N2Thus d is a homomorphism of rings With unity. Note that N1N2 Q ker(q5), and so by the UMPK, there exists a homomorphism of rings With unity, 77b 2 R/N1N2 —> R/Nl X R/N2

defined by 1p(a + N1N2) = 45(a) = (a + N1,a + N2). Let (a1 + N1,a2 + N2) 6 R/N1 X R/Ng. By the Chinese Remainder Theorem for Rings, there exists a unique element a + N1N2 E R/N1N2 for which ¢(a + N1N2) = (a1 + N1, a2 + N2). It follows that it is a ring isomorphism. I]

Corollary 2.5. Let m,n be integers with gcd(m, n) = 1.

Then the map

it) : Z/n —> Z/mZ X Z/nZ defined as 1p(a + n) = (a + mZ,a+ nZ) is an isomorphism of rings (cf. Proposition 1.36). Let p,q be distinct primes. From Corollary 2.5 one obtains the ring isomorphism 1p : q —> Zp >< Zq, defined as amodpq +—> (a modp, amod q). Consequently,

|U(q)| = |U(Zp X Zq)| = |U(Zp)| ' |U(Zq)|, thus yielding the value of Euler’s function at pq: 90(pq) = (p — 1)(q — 1). Proposition 2.42. Let R be a ring with unity, 1R. Then the map 9 : Z —> R defined as 9(n) = a, where a is defined as in §1.4 is a homomorphism of rings with unity.

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Proof. For m,n E Z, g(m+n) = (m+n)1R = 777,113+a : 9(m) + 9(71),

and

g(mn) = ma

= m(9(n)) = m(1RQ(n)) = (mlR)g(n)

by Def. 2.1(iii)

= 9(m)0(n)Also, 9(12) = ll = 13, and so, 9 is a homomorphism of rings with unity.

III

The kernel of g : Z —> R is an ideal of Z of the form rZ for some integer r 2 0. The integer r is the characteristic of the ring R and is denoted as

char(R). Corollary 2.6. Let R be a ring with unity with r = char(R). If r = 0, then R contains a subring isomorphic to Z. If r > 0, then R contains a sabring isomorphic to Z,» (a subring is a ring contained in a larger ring).

Proof. Note that 9(Z) is a subring of R. If r = char(R) = 0, then Z g

Z/ {0} is isomorphic to 9(Z) by Proposition 2.39. If r = char(R) > 0, then Zr g Z/rZ is isomorphic to g(Z) (again by Proposition 2.39). 2.6

D

Localization

In this section we continue the main theme of the chapter — the construction of new rings from existing rings. Given a commutative ring with unity R and a multiplicative set S, we construct a ring of fractions with denomi— nators from S that we call the localization of R at S and denote by S ‘1R. The contruction is a broad generalization of the construction of the field of rational numbers Q from the ring of the integers Z. *

*

>k

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Let R be a commutative ring with unity. Definition 2.22. A subset S Q R is multiplicatively closed (multi-

plicative) if 1 E S and ab E S Whenever a, b E S. Let S Q R be a multiplicative subset of R. On the cartesian product

RX S={(a,b): aER,bE S}

we define an equivalence relation N by the rule (a, b) N (c, d) if and only if there exists an element 8 E S for which 3(ad— be) 2 0. Then N determines a partition of R X S into equivalence classes. The collection of all equivalence classes of N is denoted by S‘1R; we let % denote the equivalence class

containing (a, b). Proposition 2.43. Let R be a commutative ring with unity and let S Q R be a multiplicative subset of R. Then S‘lR is a commutative ring with unity with addition defined as ad + be

0

a

E+E_

bd

and multiplication defined as a

c _ ac

a ' a — a, for %, g E S‘lR. Proof. We first need to show that these relations,

4aS4Rx34R+s*R,

n54RxSARa54R, are actually binary operations on S ‘1R. This amounts to showing that

+, - are well—defined on equivalence classes, that is, if (a’, b') N (a, b) and (c’,d’) N (c,d), then (a’d’ +b’c’,b’d’) N (ad+ bc,bd) and (a’c’,b’d’) N (ac, bd). We leave these straightforward (yet tedious) computations as an exercise.

Now with 03—1 R = I: 0 (S ‘1R, +) is easily shown to be an abelian group. Likewise, the other ring axioms are quickly shown to hold. Finally, S ‘1R is a commutative ring with unity 15—11; = %. El Definition 2.23. The ring S ‘1R given in Proposition 2.43 is the localization of R at S.

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We easily see that if 0 E S, then S_1R = {0}. There is a ring homomorphism A : R —> S_1R defined as A(a) = a/ 1 for a E R. This homomorphism need not be an injection, as we shall see. Proposition 2.44. Let 3—1R be the localization of R at S and let A : R —> 5' ‘1R, a I—> a/l be the associated ring homomorphism. (i) A is injective if and only if S contains no zero divisors,

(ii) A is bijective if and only if 5' consists of units of R. Proof. For (i) it suffices to show that

ker(A)={a€R: a/1=0}={a€R: sa=0,s€S}. But this follows since a/l = 0/1 = 0 if and only if s(a - 1 — 1 ' 0) = sa 2 0 for some 8 E S. For (ii): suppose A is a bijection. Since A is an injection, S contains no zero divisors, by (i). Let b E 5'. Since A is a surjection, there exists an

element a E R with A(a) = a/1= 1/b, thus s(ab — 1) = 0 for some 3 E 5'. Hence, b is a unit of R. We leave the converse to the reader.

I]

The ring homomorphism A : R —> S‘1R has the following universal mapping property. Proposition 2.45 (Universal Mapping Property for Localizations). (UMPL) Let R be a commutative ring with unity, let S be a multiplicative subset of R, let S_1R be the localization ofR at S', and let A : R —> S‘lR be the associated ring homomorphism. Let ¢ : R —> R’ be a homomorphism

of commutative rings with unity for which ¢(S) Q U (R’ )

Then there ea:-

ists a unique ring homomorphism w : S‘lR —> R’ for which the following diagram commutes

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81

Proof. Define w : s—lR —> R’ by the rule ib(a/b) = ¢(a)¢(b)-1. Then as one can check, 1b is well defined on equivalence classes in 8' ‘1R, and hence gives the required map. I] So to construct a localization, we need a multiplicative set. Here is one way to obtain a multiplicative set. An element a E R is nilpotent if a” = 0 for some n 2 1. For instance, 0 is nilpotent in the ring R and 2 is a nilpotent element of Zg. Let f be an element of R which is not nilpotent. Then S = {1, f, f2, . . .} is a multiplicative set. The resulting localization S‘lR is denoted as Rf. Note that 3 is a non-nilpotent element of Z6. Let us construct the

localization of Z6 at S = {3” : n 2 0} = {1,3} (in our notation: this is (Z6)3). One has

Z6 x {1,3} = {(0,1), (1, 1), (2,1), (3, 1), (4, 1), (5, 1) (0, 3), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3)}, and considering the equivalence relation N one easily obtains the classes in

(Ze)3:

{2%} (for instance, (1,1) N (1,3) N (3,1) N (3,3) N (5,1) N (5,3)). Thus (Z6)3 g Z2. Observe that A : Z6 —> (Z6)3 is not injective since 3 is a zero

divisor in 5 (see Proposition 2.44(i)). Prime ideals of R also provide multiplicative sets. For any sets A, B,

let A\B denote the subset of A consisting of all elements of A that are not in B. Proposition 2.46. Let N be a prime ideal of R. multiplicative set.

Then S = R\N is a

Proof. Certainly 1 ¢ N since N is proper. Since N is prime, ab 6 N

implies that either a E N or b E N. Thus a ¢ N and b ¢ N implies that ab g N.

|:I

By Proposition 2.46, for N prime we obtain the localization S‘lR,

S = R\N, which we denote as RN. If R is an integral domain, then (0) is a prime ideal, and we can form the localization R(0)' It is not difficult to show that Rm) is a field, called the field of fractions of R and denoted

as Frac(R). Observe that Z(0) = Frac(Z) = Q. More generally, if F is

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Fundamentals of Mode'rn Algebra

any field then the ring of polynomials in n variables, F [$1, . . . ,xn], is an

integral domain and Frac(F[m1, . . . ,mnD is the field of rational functions, denoted as F(m1, . . . , 1137,). For another example, let (p) be a non—zero prime ideal of Z. Then Z\ (p) is a multiplicative set. The resulting localization 3—12 2 2(1)) consists of

fractions of the form 2b with a,b E Z, 1) ¢ (19); A = 20,) is an integral domain with 14(0) 2 Frac(A) = Q. Proposition 2.47. Let N be a prime ideal of R. Then the localization RN

is a local ring with maximal ideal NRN = {r/s: 1“ E N, S g N}. Proof. Let I be an ideal of RN, and suppose that I contains an element

:1: = 7/3 with 7“ ¢ N. Then 3/? 6 RN, so that a: is a unit of RN. It follows that I 2 RN.

Thus every proper ideal I Q RN is contained in NRN.

Consequently, NRN is a maximal ideal of RN. If Q is some other maximal ideal of RN, then Q Q NRN, and so Q = NRN. III By Proposition 2.47, Z00) is a local ring with unique maximal ideal

(I?) = PZ(p)Localization applied to an integral domain R constructs not only the field of fractions K but also provides a large set of local integral domains RN between R and K. Often when trying to understand phenomena (such

as modules) over R which are understood over K (because modules over a field are vector spaces), it is very helpful to first try to see what’s going on over RN for a non-zero prime ideal of R. (See, for example, the discussion after the proof of Proposition 3.20.) Moreover, understanding the situation over RN is often facilitated by passing to the completion of RN, a subject that we take up in the final section of the chapter. 2.7

Absolute Values and Completions

In this final section of the chapter we introduce the notion of distance in Q; we consider the familiar absolute value | | on Q as well as the p—adic absolute value | |p for a prime p. For each absolute value we construct an extension field of Q called the completion of Q. The completion of Q with

respect to | | is the field of real numbers R, the completion with respect to | |p is the field of p—adic rationals Qp. The valuation ring of Qp is the ring of p—adic integers 2,, which is a ring extension of Z. *

*

>k

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Definition 2.24. An absolute value on Q is a function | | : Q —> R which satisfies for all 33,1; 6 Q,

(i) lxl Z 0, — 0 if and only if x— — 0, ( ii) |a:|—

(iii) lwyl = lmllyl,

(iV) |w+y| < lwl + lyl Consider the rationals Q with the ordinary absolute value | |. A sequence

{an} in Q is a | |-Cauchy sequence if for each L > 0 there exists an integer N for which |am — anl < L Whenever m, n 2 N. For instance, the sequence

11

1 1 1 i: g, Z) ' ' '

is a | l—Cauchy sequence in Q. This sequence converges to a limiting value of 0, which is an element of Q. For another example, consider the sequence

1,1.4,1.41,1.414,1.4142,1.41421,1.414213,1.4142135,1.41421356,. . . where an, n 2 1, is the first 72 digits of the decimal expansion of x/i As one can verify, this is a | |—Cauchy sequence in Q. In this case, the limit of the sequence is x/E which is not an element of Q. We say that Q is not

complete With respect to the absolute value | | since there exists a | |—Cauchy sequence in Q that does not converge to an element in Q. There exists a field extension K of Q with the following properties:

(i) The absolute value | | extends uniquely to an absolute value on K, also denoted as | |, (ii) With respect to | |, Q is dense in K, that is, the closure Q = K, (iii) K is complete with respect to | |, that is, every | |-Cauchy sequence in K converges to an element in K. This field extension K is (of course!) the field of real numbers R, and

since 1R satisfies conditions (i), (ii), (iii), we say that R is the completion of Q with respect to | |. We can define another kind of absolute value on Q. Let x E Q and let p

be a prime number. The p—adic absolute value | |p is defined as |a:|p = 0 if a: = 0 and 1 lmlp : _

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if m = pT(s/t) E Q with gcd(s,p) = 1, gcd(t,p) = 1, r E Z. For instance, |15|3 = %, and |g|3 = 9. One should indeed check that | |p defines an absolute value on Q (see §2.8, Exercise 90). Note that | |p satisfies a stronger condition than (iv) of Definition 2.24. Proposition 2.48. Let p be a prime number and let | IP denote the p-adz’c absolute value on Q. Then for all x, y 6 Q,

(i) la: + ylp S maw{lw|p, lylp}, (ii) If lmlp 75 lylp, then I93 + ylp = ma:17{|93|p,|y|p}Proof. Exercise.

III

A sequence {an} in Q is a | |p-Cauchy sequence if for each L > 0 there exists an integer N for which |am — anlp < L whenever m, n 2 N. For example, the sequence

1,p,p2,p3,p4,-~

(2-2)

is a | |p—Cauchy sequence in Q; the sequence

however, is not a | lp-Cauchy sequence! The | Ip-Cauchy sequence (2.2) converges to the element 0 E Q. As in the case Where Q is endowed With the ordinary absolute value, not every | Ip-Cauchy sequence in Q converges to an element of Q. In other words, the field Q is not complete With respect to the p—adic absolute value. Proposition 2.49. For each prime p there exists a | |p-Cauchy sequence in Q that does not converge to an element of Q. Proof. The prime p = 2 is special, and so we will handle it first. We construct an explicit example of a | lg—Cauchy sequence in Q that does not converge to an element of Q. Consider $3 — 3, which is irreducible over Q. The conguence x3 E 3 mod 2 has solution x1 = 1; (131 = 1 is the first term in

our Cauchy sequence. To compute the second term .732, we find an integer k1 Z 0 so that (1 + 2k1)3 E 3mod 4. Thus:

(1+2k1)3 E 3mod4 1 + 6k1 + 121% + 8k3 E 3mod4

2+l1 E 0mod4 1 + k1 E 0m0d2,

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sothat k1 =1. Nowlet 11:2 =1+2k1 =1+2-1=3.

To obtain :63 we find an integer kg 2 0 so that (3 + 4k2)3 E 3mod 8:

(3 + 4k2)3 E 3mod8 27 + 108k2 + 144193 + 64k2 E 3mod8 24 + 108/92 E 0mod8 6 + 27162 E 0mod2 k2 E 0mod 2,

so that 192 =0. Now, let x3 =3+4k2 =3+4~0=3. To obtain :14, we find kg 2 0 so that (3 + 8/93)3 E 3mod 16:

(3 + 8k3)3 E 3mod 16

27 + 216k3 + 576k; + 512k3 E 3mod 16 24 + 216k3 E 0mod 16 3+ 27kg E 0mod2 k3 E 1 mod 2, so that k3 =1. Now, let x4 =3+8k3 =3+8~1 = 11.

Continuing in this manner, we obtain the | |2—Cauchy sequence

1,3,3,11,... whose limiting value is a root of m3 — 3, and hence cannot be an element of Q. Next we consider the case p > 2. Let a be a non-square integer so that the polynomial x2—a E Z [:13] is irreducible over Q. Assume that a E 0 mod p and that the conguence m2 E a mod p has a solution, which we denote as 361. Now, there exists an integer [cl 2 0 so that the conguence x2 E amodp2 has solution 332 = x1 +pk1. Moreover, there exists an integer kg 2 0 so that

the congruence x2 E amod p3 has solution :33 = x2 + [92/62. Continuing in

this manner, we obtain a I lp—Cauchy sequence {gun} whose limiting value is a root of x2 — a and hence cannot be an element of Q. Try it! (See §2.8,

Exercise 94).

E]

Analogous to the completion R of Q, we seek to construct a field exten— sion K of Q with the following properties:

(i) The p—adic absolute value I I p extends uniquely to an absolute value on K, also denoted as | |p, (ii) with respect to | lp, Q is dense in K, that is, the closure Q = K,

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(iii) K is complete with respect to | lpv that is, every | |p—Cauchy sequence in K converges to an element in K. A field extension K satisfying these conditions is the completion of Q

With respect to | lp. As we shall see, the completion of Q with respect to | |p is the field of p—adic rationals, denoted by Qp. We begin by constructing a certain field extension of Q. Let C denote

the set of all | |p—Cauchy sequences in Q. Let {xn}, {yn} E C. The sum and product of the sequences are defined termWise as

{Con} + {yn} = {wn + yn}, {mm} ' {yn} = {xnyn}

Proposition 2.50. Let {xn}, {yn} E C. Then {xn}+{yn} and {xn} - {yn} a're Cauchy sequences.

Proof. Let L > 0. Since {gun} and {yn} are Cauchy sequences, there exists integers N1, N2 so that m, n 2 N1 implies that |:I:m—a:n|p < L and m, n 2 N2

implies |ym — ynlp < L. Let N = max{N1,N2}. Then

|(wm + ym) - (wn + yn)|p = lwm - wn + ym - ynlp S mMUM - wnlp, lym — ynlp} lxm _ xnlp = max{|xm|p, lxnlpla

a contradiction. Consequently, |xm|p = |xn|p for all m, n 2 N. It follows that limnnoo |xn|p exists. We define

IAIP = 71,1530 lxnlzr In this manner, Qp is endowed with an absolute value which is the unique

extension of | |p to Qp, we denote this extension by | |p.

I]

We next show that Q is dense in Qp.

Proposition 2.53. The field Q is a dense subset of Qp, that is, given

A E Qp and L > 0, there exists an y E Q for which ly — Alp < t. Proof. Recall that /\ is really a left coset, so let {an} represent A. Let

L > 0.

Now, since {an} is Cauchy, there exists N so that m,n Z N

implies Imm — mnlp < L/2. Let {y} be the constant sequence with y = mN. Then for n 2 N, |:I:N — acnlp < L/2. Thus limnnoo lc — xn|p S L/2 < L. Now considering the sequence {a}N — an} as a representative of the element y — A E Qp, one has |y — Alp < L. Thus Q is dense in Qp. I]

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89

Finally, to prove that Qp is the completion of Q, we show that Qp is

complete with respect to the absolute value | |p. Proposition 2.54. The field ofp-adic rationals Qp is complete with respect

to the extended p—ad'ic absolute value | |p.

Proof. Let {An} be a | lp-Cauchy sequence in Qp. (Of course, {An} is really a sequence of cosets in C /N represented by a sequence of Cauchy sequences in Q, A1, A2, . . ., which satisfies the condition, for L > 0, there

exists an integer N fo Which |Am — Anlp < L Whenever m, n 2 N. Here | |p is the extended absolute value.) By Proposition 2.53, for each n 2 1, there exists a constant sequence

y(") E Q for which

and so,

lim |y(”) — An|p = 0.

71—)00

Now, for L > 0, there exists an integer N1 so that

lAn — y(n)|p < L/3, whenever n 2 N1. Since {An} is Cauchy, there exists an integer N2 so that

lAm — Anlp < L/3

whenever m, n 2 N2. Take N = max{N1, N2}. Thus,

|Am — y|p < 2L/3, Whenever m, n 2 N. Observe that

|y — Am|p < L/3 since m 2 N 2 N1, hence

ly(m) — ywlp < LIt follows that

lx(m) _ m(n)|p < L whenever m,n Z N, where mm) 6 Q is the constant term of y(”). Thus

{3:01)} is a Cauchy sequence in Q which converges to the coset A = {m(n)}+ N E Qp. Thus

lim 3,00 = A. 77,—)00

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Fundamentals of Modern Algebra

NOW there exists an integer N3 so that

|y(n) — Alp < 2L/3 Whenever n 2 N3. Let M = max{N1,N3}. Then lAn — Alp < L, Whenever n 2 M , so that lim An = A, n—>oo

thus Qp is complete.

C]

There is a subring of Qp defined as

{x E Qp: |ac|p S 1}. This is the ring of p-adic integers, and is denoted by Zp. Proposition 2.55. Every (coset) :10 E Zp has a representative of the form {tn}, with t1 = a0,

752 = 00 + 0110,

t3 = do + @110 + @2192,

tn 2 a0 + alp + agp2 + a3p2 + - - - + an—1Pn_1,

with 0 S at- g p — 1. We write the coset {tn} +N as the infinite sum a0+a1p+a2p2+a3p3 4"”,

thus

$=a0+alp+a2p2+a3p3+'”-

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91

Proof. Step 1. Let n 2 1. By Proposition 2.53 there exists an element 1% E Q, gcd(a,p) = gcd(b,p) = 1, 7' E Z, so that ’"a

1

p —a: g— 3 be prime and let C6 denote a primitive 6th root of unity. Show that (62 = C6-

(34) Let p be prime. For 2 S a S (p — 1)/2, gcd(a,p) = 1, ShOW that

((1—a (fl) p

1 — Cp

is a real number.

(35) Find all of the characters of D3. (36) Compute the character group of Cg Where the characters are homo-

morphisms ’7 : Cg —> Rx . (37) Compute the character group of D3 Where the characters are homo-

morphisms 'y : D3 —> RX.

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Fundamentals of Mode'rn Algebra

(38) In the manner of Example 2.9, represent the elements of D4 as 8 plane isometries of the square. Exercises for §2.4

(39) Let I, J be ideals of the ring R.

(a) Prove that I D J is an ideal. (b) Prove that I+ J = {a+b: a E I,b E J} is an ideal. (c) Give a counterexample to show that I U J is not an ideal.

(40) Find an example of a ring R and ideals I, J for which IJ C I n J.

(41) Prove that the ideal (2, ac) E Z [3:] is maximal. (42) Prove that the ideal (as) Q Z [:10] is prime, but the ideal (x) Q Z5[x] is not prime. (43) Find a counterexample which shows that the converse of Proposition

2.22 is false.

(44) Let a + be + 002 E Z03, (a) = 03. (a) Verify the formula (a+ba+ca2) ((a2 — bc)+ (c2 —ab)a+(b2—ac)a2) = a3+b3+03— 3abc. (b) Use (a) to show that (7, 2 — a) is a prime ideal of Z03. (c) Prove that (7,2 — a) 7é (7, a + 3). (45) Prove Proposition 2.25. (46) Let R be a commutative ring with unity. By Proposition 2.23, every proper ideal of R is contained in a maximal ideal. This is proved using Zorn’s lemma. Show that the same result holds without using Zorn’s lemma if R is a Noetherian commutative ring with unity.

(47) Prove Proposition 2.27. (48) Let R be a ring, let G be a finite group and let RG’ be the group ring.

Let S = {T296091 7' E R}. Prove that Sis an ideal of RC. (49) Let R be a ring and suppose that

IlgI2g"°nn+lg...

is an ascending sequence of ideals. Show that Ujil Ij is an ideal of R. (50) Let M1 and M2 be maximal ideals of R. Show that either M1+M2 = R 01' M1 = M2.

(51) Prove that the ideal (ac, m2 + 2) in Z [3:] is not principal.

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97

(52) Let K be a field extension of Q. Let p(;L') be an irreducible polynomial in K [as]. Prove that p(:r) has distinct roots. (53) Let p(:c) = a0+a1x+a2x2 be a quadratic polynomial in Q[a:]. Suppose that gcd(p(:r), m3 — 1) = 1. Show that 19(0) is a unit in the group ring Q03, 03 = (0').

Exercises for §2.5 (54) Find a group that is isomorphic to the additive group of the quotient ring Z10/{0, 5}.

(55) Let N be the principal ideal of Z X Z generated by (2,4). List the

elements in the quotient ring (Z X Z) /N . (56) In the ring Q[:r], let N = (a:2 — 3m+2) be the principal ideal generated by x2 — 333+ 2. Prove that ((m — 1) +N)2 2 (:1:— 1) + N in Q[;I:]/N. (57) Prove that Z5[:I:]/(x3 + 2x + 1) is a field.

(58) Prove that Q[:I:] /(x3 — x2 + 2:1: — 1) is an integral domain. (59) Let 0'3 denote the cyclic group of order 3 generated by a. Prove that

ZC3/(7, 2 — a) is a field. (60) Let R be a ring and let I be an ideal of R which is not maximal. Show that R/ I has at least three ideals.

(61) Let I be an ideal of R with quotient ring R/I. Let a be a unit of R. Show that a + I is a unit of R/I. (62) Suppose a + I is a unit in the quotient ring R/I. Give an example Which shows that a need not be a unit of R.

(63) Let K be a field and let K [1.0, x, y, 2] be the ring of polynomi— als in the indeterminates w, x, y, z.

Prove that the quotient ring

K [10, cc, 3/, z] / (wx — yz) is not a UFD. Is it an integral domain? (64) True or False: R g Q as rings. (65) Consider the rings Q and Q X Q.

(a) Show that there exists a bijective map f : Q —> Q X Q. (b) True or False: Q g Q X Q as rings.

(66) Let (b : R —> R’ be a surjective homomorphism of commutative rings. If both rings R, R’ have unity elements 13,131, prove that ¢(IR) = 1 R’ .

(67) Let R and R’ be commutative rings with unity. Find an example of a

ring homomorphism ¢ : R —> R’ for which q5(1R) 7E 1 R’(68) Let qb : Q —> C be an injective homomorphism of commutative rings

with unity. Show that (Mm) = ac,Va: E Q.

q5|Z:Z—>(Cmapsnton,forn€Z.)

(Hint: First show that

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Fundamentals of Mode'rn Algebra

(69) Let qt : F —> R be a homomorphism of rings with F an infinite field and R finite. Show that ¢>(x) = 0 for all a: E F. (70) Let d) : R —> (C be the function defined by ¢('r) 2 Ti, i = x/jl, for all r E R. Determine whether gb is a ring homomorphism. (71) Compute the kernel of the evaluation homomorphism ¢2 : 229 [:13] —>

Z29 (72) Compute the kernel of the evaluation homomorphism (151+ fl : Q[a:] —> R. (73) Let R be a commutative ring with unity and let G' be a finite group.

Let 6 : RG —> R be the map defined as 6(2960 agg) = da ag. (a) Show that e is a surjective ring homomorphism.

(b) Find ker(e).

(74) Let H be a normal subgroup of G. Let N = G/H be the quotient group, and let qfi : RG —> RN be the map of group rings defined by g I—> gH. Show that ¢ is a surjective ring homomorphism.

(75) Compute the characteristic of the ring Z3[x]. (76) Suppose that R is an integral domain with r = char(R) > 0. Prove that 7' is a prime number. (77) Let p be a prime number, let n 2 1 be an integer and let F be a

field with char(F) = p. Prove that the map gb : F —> F defined as q5(a:) = mpn,‘v’ar E F, is a homomorphism of fields. (78) Give another proof of the Eisenstein Criterion (Proposition 2.8) by completing the steps below. As in the proof already given, we assume

that the degree n polynomial f(a:) factors as f(x) = g(ac)h(ac) in Z[a:], with deg(g(a:) = k, deg(h(a:)) = l, k,l 2 1.

(a) Let c3 : Z[ac] —> Zp[:c] be the map defined as ¢(p(a:)) 2 13(3)), where the coefficients of flat) are those of p(x) taken modulo p. Show that 95 is a homomorphism of rings.

(b) Apply (a) to show that §(w) 2 wk and ECU) 2 ml. (c) From (b) conclude that f (0) E 0modp2, a contradiction. Exercises for §2.6 (79) Let R be a commutative ring with unity. Show that the set of nilpo— tents elements of R is an ideal of R. (80) Find all of the nilpotent elements in the group ring Z303.

(81) List all of the elements in the localization (Z6)2.

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99

(82) Show that as is a non—nilpotent element of Z [:13] List four elements in the localization Z [1193. (83) Prove that (as) is a prime ideal of Z [an] Prove that the localization Z[ac](m) is not the same as Z[x]m. (84) Let R be a Noetherian commutative ring With unity and let S be a multiplicative subset of R. Show that the localization S ‘1R is Noethe— rian.

(85) Let R be a local ring with maximal ideal m. Let b E R with b ¢ m. Prove that b is a unit of R. (86) Let S = {1,3, 9, 7} Q Z10. Compute the kernel of A : Z10 —> 5-1210,

a I—> a/ l. (87) Prove that every non-zero integer is a unit in Z(p) for an infinite number of primes p.

(88) Verify the formulas: (a)

Z:

n

Z(p),

p prime

(b) Zf = fl Z(p) for f a non—nilpotent integer. pf f Exercises for §2.7

(89) Show that x/2 is the limit of a | |—Cauchy sequence in Q. (90) Prove that | |p is an absolute value on Q. (91) Prove that the sequence an — 51 is a | |-Cauchy sequence in Q. Is {an} | |3-Cauchy? (92) Prove Proposition 2.48. (93) Suppose Q is endowed with the p—adic absolute value. Let :1),y,z be

distinct non—zero elements in Q. Let a = |cc — z|p, b = |cL' — y|p, c: |y—zlp.

(a) Show that there exists a triangle Whose sides have lengths a, b, c.

(b) Show that any triangle constructed in (a) is isosceles. (94) (95) (96) (97)

Construct a root of x2 — 2 in Q7. Referring to Proposition 2.51, show that N is an ideal of C. Prove that Z(p) = Q 0 Zp. Let a: be an element of Zp with w 6 Z(p). Prove that there exist a,b E Z such that the equation :1: = 2:0 ari holds in Zp where 7" 2 bp.

(98) Prove that ZP is a PID.

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Fundamentals of Mode'rn Algebra.

(99) Verify that

0=1+1+2+4+8+16+--in Z2.

(100) Let {pn} denote the sequence of prime numbers 2,3, 5, 7, 11, 13, . . .. Let m be a positive integer.

(a) Compute ”1:120 lmlpn. (b) Discuss the value of lim lim lmlpn. m—)oo 71—)00

(101) Show that for each a: E Q, l-

H |m|p = 1. p prime

Questions for Further Study (1) Let R be a commutative ring with unity and let G, G’ be finite groups of order n.

(a) If G g 0’, show that RG g RG’ as rings. (b) Prove that the converse of (a) is false, that is, find two groups G, G’ for which RG g RG’, yet G ¥ G’.

Chapter 3

Modules

In this chapter we introduce the concept of a module over a commutative ring with unity. We begin With the familiar notion of a vector space, which is a module over a field. We include both finite and infinite dimensional vec— tor spaces, subspaces and quotient spaces, linear transformations of vector spaces and the linear dual of a vector space. We generalize the definition of a vector space V over a field K to that of a module M over a commutative ring with unity R. An underlying theme is to determine which theorems and properties of vector spaces carry over to the more general setting of modules. For instance, the concept of a finite dimensional vector space over K generalizes to a free R—module M of finite rank. The notion of a free R-module is then generalized to that of a projective module; we show that a finitely generated and projective module over a local ring is free of finite rank. Next, we define tensor products of two modules, specialize to tensor products of vector spaces, and relate these to the linear dual. We next consider algebras A over a ring R, an object that is a ring A as well as an R—module, where the scalar multiplication is given through a ring homomorphism A : R —> A. We specialize to commutative R—algebras, and show that every finitely generated R—algebra is the quotient of a polynomial ring. We compute all of the R—algebra homomorphisms A —> B Where A is finitely generated. Finally, we discuss bilinear forms on a finite dimensional vector space V over the field K, where K is the field of fractions of an integral domain R. For an R—submodule M of V, which is free over R and which satisfies

KM = V, we define the dual module of MD and the discriminants disc(M)

and disc(MD ). As an example we compute disc(RG') and disc(RG'D ), where G is the Klein 4—group. We shall use discriminants in Chapter 4.

101

102

Fundamentals of Modem Algebra.

3. 1

Vector Spaces

In this section we begin with the definition of a vector space V over a field K. Vector spaces (at least finite dimensional vector spaces) should be familiar to readers of this book. We review linear independence of vectors, spanning sets, bases, and prove that every vector space V admits a basis. We define finite and infinite dimensional vector spaces, and consider subspaces and quotient spaces. Next, we turn to linear transformations of vector spaces, specializing t0 the linear dual.

Definition 3.1. Let K be a field. A vector space over K is an additive abelian group V together with a scalar multiplication K x V —> V, denoted

by (731)) I—> rv, that satisfies, for all 7“, s E K, v, w E V,

(i )

r(v+w)=rv+rw,

(ii) (7' + s)v— — 7'?) + 312, (iii) (r3)v— — r(sv),

( iv)1 An element 1) E V is a vector. Note that the conditions of the definition imply that 02) = 0, V1). The most familiar example of a vector space is Euclidean n—space R” which consists of n—tuples (a1, a2, . . . ,an), az- E R, equipped with the usual vector addition and scalar multiplication. Other examples include the K—

Vector space of polynomials K[a:] with scalar multiplication K x K [x] —> K[a:] defined as (r,f(:c)) I—> rf(a:), for ’r 6 K, f(a:) E K[:L‘], and the group ring KG, with scalar multiplication K X KG —> KG defined by (7", g) I—> rg for r E K, g E G. In the field extension L/ K , L is a vector space over K with scalar multiplication K X L —> L defined as (ac, y) I—> my for x E K, g E L. Let V be a vector space over a field K and let S = {va}a€I be a subset

of V indexed by the set I. Then S is linearly independent if Zrava = 0,

Ta 6 K,

(161

where Ta = 0 for all but a finite number of a, implies that Ta = 0 for all

04 E I. If '0 7E 0, then the singleton subset {v} is a linearly independent

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103

subset of V.

Definition 3.2. Let S = {va}a€1 be a subset of a K—vector space V.

Then S is a generating set for V (or S spans V) if each 1) E V has a representation 1) = 20,61 rava for ra E K With ra = 0 for all but a finite number of indices a.

A vector space is finitely generated if there exists a generating set S that is finite. For example, R" is a finitely generated vector space over R. Definition 3.3. Let S = {’00,}ael be a subset of the K-vector space V. Then S is a basis for V if each 1) E V has a unique representation 1) = 20,61 rava for ra E K with ra = 0 for all but a finite number of indices oz.

Proposition 3.1. Let V be a K-vector space. The subset S = {va}a61 is a basis for V if and only if S is a linearly independent spanning set for V.

Proof. Assume that S = {va}a€I is a basis. Then certainly S spans V. Now suppose that 20,61 rava = 0, for ra 6 K. Then 20,6,- rava is a representation of the zero vector, but since ZaEI 0va also represents zero, we conclude that ra = 0,Va. Thus S is linearly independent. For the converse suppose that v = 20,61 rava = 20,61 sava for some

1) E V. Then Ea€I(ra — sa)va = 0, hence ra — sa 2 0,Va E I, and so, S is a basis for V. E] Proposition 3.2. Let V be a vector space over a field K. Then there exists a basis for V. Proof. We employ a Zorn’s Lemma argument (Chapter 2, §2.4). Let S be a generating set for V over K (such S always exists, one could take S = V).

Assume that S 75 {0}, so that S contains a linearly independent subset {v} of V. NOW let T denote the collection of linearly independent subsets of S

that contain {11}. Then T is a non—empty set partially ordered under set inclusion. Moreover, every chain {T2} in T has an upper bound T 2 U1. Ti. Thus by Zorn’s Lemma, T has a maximal element B. By its construction, B is a linearly independent subset of V. We claim that B is a basis for V. To this end let W be the subset of V consisting of all vectors of the form Zbe B rbb, Where rb = 0 for all but a finite number of b E B (W is the subspace of V generated by B). If V = W, then B is a basis for V. Else, there exists an element a: E V\W. Suppose that 21,63 rbb + rm 2 0 for some rb,r E K. If r 75 0,

then x = — ZbEB(rb/r)b E W, which is a contradiction. Consequently,

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Fundamentals of Modern Algebra

EbeB rbb + rs: = 0 implies that rb = r = 0,Vb E B, and so, B U {ac} is a linearly independent subset of V, containing {11} and containing B as a proper subset. This contradicts the maximality of B. I] Moreover, if the vector space V is finitely generated by a set with n elements, then there is a basis for V that contains no more than n elements. Proposition 3.3. Let V be a vector space over K that is finitely generated by a set S with n elements. Then there exists a basis for V that contains m g n elements.

Proof. Let S = {121, v2, . . . ,vn} be a generating set for V over K, and let B be the smallest subset of S for which B generates V over K. Renumbering the elements of S if necessary, write B = {111,112, . . . ,vm} with m g n. We claim that B is a basis for V. By way of contradiction, suppose there exists an element 1) E V for which v=a1v1+a2v2+~~+amvm,

aiEK,

v=b1v1+b2v2+---+bmvm,

biEK,

are two distinct representations of 2). Then there exists an integer i’ for which ai/ yé biz. Thus

(ai’ _ bi’ )vi’_ — ;(bi_

i)'Ui,

ig—fi;

and so, b1 _ ai

vil_ — 7; (a—z/ —b,i/ )Uz.

iyéi;

Hence B\{vi/} generates V, which contradicts the minimality of B.

E]

Thus every finitely generated vector space admits a finite basis. The next proposition shows that the number of elements in any finite basis is the same. Proposition 3.4. Let V be a finitely generated vector space. Then any two bases for V have the same number of elements.

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105

Proof. Suppose that {b1, b2, . . . , bm} and {01, 02, . . . , on} are two different bases for V with m 7E n. Without loss of generality, we may assume that m < n. For 1 g j g n, write cj = 2:1 aid-bi for unique elements am- 6 K.

Let A 2 (am), and let A’ = (a; 3.) denote the reduced row echelon form of A. Since m < n, there exists an integer j*, m < j * S n so that _

/

I

of — “10* c1 + a2’j*02 +

I

+ ammcm.

Hence {01, 02, . . . , on} is not linearly independent, a contradiction.

III

In view of Proposition 3.4, if V finitely generated, then we define the dimension of V to be the number of elements in any basis for V over K. We denote the dimension of V by dim(V); we say that the vector space V

is finite dimensional. If L/ K is a field extension in which L is finitely generated over K, hence

dim(L) = n for some n, then L/K is a finite extension of fields. If L/K is a finite extension of fields, then the degree of L over K, denoted as

[L : K] is the dimension of L as a vector space over K.

Note that the zero vector space {0} is finite dimensional with dim({0}) = 0. If the vector space V is not finitely generated and hence does not admit a finite basis, then V is infinite dimensional. Let V be a vector space over K. A subspace W of V is an additive subgroup of V for which mu 6 W for all r 6 K, w E W. The subspace W is a K—vector space Whose scalar multiplication is the scalar multiplication of V restricted to W. One can always construct a subspace of V as follows. Let S = {Ua}ael be a subset of V and let W consist of all elements of V the form Zael rava where Ta 6 K and Ta = 0 for all but a finite number

of oz. Then W is a subspace of V called the subspace of V generated

(or spanned) by S. Proposition 3.5. Suppose V is a finite dimensional vector space and W

is a subspace of V. Then dim(W) S dim(V). Proof. As a vector space, W is finitely generated by a set containing 3 dim(V) elements. By Proposition 3.3, there exists a basis for W with

g dim(V) elements.

I]

Let W be a subspace of V. Since W is a normal subgroup of V, we may form the quotient group V/ W. Now since 7(1) + W) Q 7‘1) + W for all r E K, v E V, the quotient group V/ W is a vector space with scalar

multiplication K x V/W —> V/ W defined as r(v + W) = 7‘?) + W. The vector space V/ W is the quotient space of V by W. (One should check

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Fundamentals of Modern Algebra

that the map K x V/ W —> V/ W is well—defined on left cosets.) If V/ W is

finite dimensional then the subspace W is cofinite and dim(V/ W) < 00 is the codimension of W. Definition 3.4. Let V, V’ be vector spaces. A linear transformation is a map qt : V —> V’ that satisfies, for all x, y E V, r E K,

(i) ¢(w + y) = 0X93) + My), (ii) ¢(rx) = r¢(x). The linear transformation (b : V —> V’ is an isomorphism of vector spaces if (b is a bijection. Let V be a vector space over a field K. We consider K as a vector space over itself. A linear functional on V is a linear transformation f : V —> K. The collection of all linear functionals on V is the linear dual of V and is denoted as V*. The linear dual V* is a vector space over K With vector addition defined as

(f + 9)(v) = f(’U) + 9(7)), for all f, g E V*, v E V, and scalar multiplication given as

(7', f) '—> Tf, With (rf)(v) = rf('u) for r E K, f E V*, v E V. Let (b : V —> W be a linear transformation of K-Vector spaces. Then qfi induces a map

45* : W* —> V*, defined as ¢*(f)(v) = f(¢(v)) for f E W*, ’U E V. The map ¢* is the transpose of Q5. Proposition 3.6. 45* : W* —> V* is a linear transformation of vector spaces. Proof. Exercise.

E]

Proposition 3.7. A finite dimensional vector space and its dual have the same dimension.

Proof. By hypothesis, V has a finite basis {(91, b2, . . . , bn}. We shall con— struct a K—basis for V* With n elements. For each i, 1 g i g n, define a map fi : V —> K by

fi("°1b1 + T252 + "'+ rnbn) = 7‘1"

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107

for n- E K, 1 g i S n. The map fZ is the ith coordinate map. The collection of coordinate maps {f1, f2, . . . , fn} is a subset of V*. We Show

that {fi} is a basis for V*. Let f E V*. Then

f = f(b1)f1 + f(b2)f2 + ' ' ' + f(bn)bna and so {fi} generates V*. Now suppose that 7'1f1 +7'2f2 +---+7“nfn = 0,

for m- E K, Where 0 is the zero functional: 0(1)) 2 0 for all ’1} E V. Then for all i,

0 = 0(b,-) = (7'1f1 + 7'2f2 + - - - + Tnfn)(bi) : Tia

and so {f1} is a linearly independent subset of V*. It follows that {f,} is a K—basis for V*.

E]

The collection {fi y=1 is the basis for V* dual to the basis {bi}?=1 for V; {f,-} is the dual basis for V* with respect to {bi}. We next consider a more general construction than the linear dual. Let X be a non-empty set of elements and let Map(X, K) denote the set of all functions f : X —> K (here, X is not necessarily a vector space and the

maps need not be linear transformations). Then Map(X, K) is a vector space With vector addition defined pointWise: for f, g E Map(X, K), x E X,

(f +g)(a:) = f(a:) +g(a:) and scalar multiplication given as (rf)(:1:) = rf(a:) for "r E K. Proposition 3.8. Let W be a subspace of Map(X, K) of finite dimension n. Then there exist elements £131,532, . . . , an 6 X and a basis {f1, f2, . . . , fn}

of W for which f1-(xj) = 6M, 1 S i,j S n, where 5131' is Kronecker’s delta. Proof. We proceed by induction on n. For the trivial case n = 1, let {hl}

be a basis for W, dim(W) = 1. Since hl 75 0, there exists an element 1:1 6 X so that h1(m1) 75 0. Define a function f1 : X —> K by the rule f1 = h1(x1)_1h1. Then {fl} is a K—basis for W With f1(ac1) = 1 as required. For the induction hypothesis, we assume that Whenever W Q Map(X, K) is a subspace of dimension n — 1, there exist elements $1,:B2, . . .,xn_1 E X and a basis {f1,f2, . . .,fn_1} for W so that f,(a:j) =

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Fundamentals of Mode’rn Algebra.

NOW suppose that dim(W) = n and let {h1, h2, . . . , hn} be a basis for W. Let W’ be the subspace of Map(X, K) spanned by {h1, h2, . . . , hn_1}. Clearly, dim(W’) = n — 1 and so by the induction hypothesis, there exist elements $1,332, . . . ,xn_1 E X and a basis {f{, f5, . . . ,f7’,_1} for W/ so that fflxj) = (Sm-,1 g i,j g n — 1. Let 9 : hn _ hn(x1)fi _ hn(x2)fé _ ' ' ' _ hn(wn—l)f7,z—l'

Suppose that g = 0. Then hn(x) = hn(m1)f{(:1;) + hn(m2)fé(m) + ' ' ’ + hn($n—1)f7/1—1($)a

for all a: E X, and so, hn 6 W’.

It follows that dim(W) = n — 1, a

contradiction. Hence, there exists an element xn 6 X With g(:cn) 75 0. Let

fn = g(a:n)_1g and f2 = f;—f{(mn)fn for 1 S 2' S n—l. Then fz(acj) = (5M, 1 g i, j S n. We claim that {fZ 2‘21 is a basis for W. To this end, suppose there exists 71,72, . . . , T'n E K for Which

7'1f1 +T2f2+’”+rnfn = 0Then evaluation at xi yields 13 = 0, and hence, {f1, f2, . . . , fn} is linearly independent. Note that hi 6 span(f1,f2, . . .,fn) for 1 S i S n — 1 since

f2, = fz + fi’(xn)fn for 1 $1“ 3 n — 1. Moreover, hn E span(f1,f2, . . .,fn) since

hn = g + hn(w1)f{ + hn(w2)fé + - - - + hn(xn—1)f.’._1 = 9(xn)fn + hn($1)fi + hn(x2)fé + ’ ’ ’ + hn(-’En—1)f£—1

= 9(xn)fn + hn(x1)(f1 + f{(xn)fn) + hn(-’v2)(f2 + fé(wn)fn) + - ~ - + hn(wn—1)(fn—1 + f.’._1(wn)fn)Consequently, W = span(f1, f2, . . . , fn), Which says that {f1, f2, . . . , fn} is a basis for W, as required.

I]

Of course, in the case that X = V, W = V* is a subspace of Map(X, K). Furthermore, if dim(V) = n, then dim(V*) = n and Proposition 3.8 applies to yield elements $1, $2, . . . ,1?” 6 V and a basis {f1, f2, . . . , fn} for V* With

fi(:cj) = 61,3” Proposition 3.7, however, asserts that {mj} is a basis for V; {f1} is the dual basis.

Modules

3.2

109

Modules

In this section we generalize the definition of a vector space V over a field K to that of a module M over a commutative ring with unity R. An underlying theme here is to determine which properties of vector spaces carry over to the more general setting of modules. We consider generating sets for R—modules, bases for free modules, as well as submodules and quotient modules. As with vector spaces (Proposition 3.4) the rank of a free R— module is well—defined. Next, we study homomorphisms of modules and give R—module analogs for the First Isomorphism Theorem (for groups) and the Universal Mapping Property for Kernels. We show that the subspace property of vector spaces (Proposition 3.5) generalizes to PIDs and a still weaker version of Proposition 3.5 is valid for Noetherian rings. *

*

*

Throughout this section R is a commutative ring with unity. Definition 3.5. Let R be a commutative ring with unity. Then a left module over R is an additive abelian group M together with a scalar

multiplication R X M —> M, denoted by (r, m) I—> rm, that satisfies, for all 73.3 E R, m,n€ M,

(i) r(m + n) = ’r'm + Tn,

— rm + sm, (ii) (7' + s)m—

(111) (rs)m—— 1(sm) ( iv)1

=m.

Note that the conditions of the definition imply that 0m 2 0, Vm. Of course, a vector space V over a field K is a K-module. Any ideal I of a ring R is an R—module, the scalar multiplication being the multiplication in R — specifically R is a module over itself. Any additive abelian group G is a Z—module with scalar multiplication defined as 09 = 0,

mg 2 g+g+~~+g, for n > 0, and my =(—g)+(—g)+---+(—g), for v

"

lnl

n < 0. The ring of polynomials R[a:] is an R—module with scalar multiplica-

tion R X R[$] —> R[:c] defined as (r,f(a:)) I—> rf(r1:), for 7‘ E R, f($) E R[a:]. The group ring RC is an R—module; the scalar multiplication R X RG —> RC

is given by (739) = r9.

110

Fundamentals of Modem Algebra.

Let M and N be R—modules. The direct sum M 69 N is the R—module

consisting of all pairs (at, b) E M X N Where addition is defined as (a, b) + (c, d) = (a + c, b + d) and scalar multiplication is given as r(a, b) = (ra, Tb) for 7" E R. More generally, let M1, M2, . . . , Mk be a finite collection of R— modules. The direct sum M1 63 M2 69 - - - 69 Mk is the R—module consisting

of all k-tuples in H1721 M.- Where addition is given as (a1,a2,...,ak)+(b1,b2,...,bk) = (a1+b1,a2+b2,...,ak+bk),

and scalar multiplication is given as r(a1, a2, . . . , ak) = (ral, rag, . . . ,rak). Definition 3.6. Let S = {ma}a€1 be a subset of the R—module M. Then S is a generating set for M if each m E M has a representation m = Zael rama for Ta 6 R With Ta = 0 for all but a finite number of indices oz.

Definition 3.7. The R-module M is free if there exists a subset B = {ma}a€I of M for Which each m E M has a unique representation m = Zael rama for Ta 6 R with Ta = 0 for all but a finite number of indices

oz. The subset B is a basis for M over R.

For example, R[x] is free over R on the basis B = {1,x,ac2,ac3, . . . }. Here is a general construction of a free module. Let S be a non-empty set. The

free R module on S, F (S) is the collection of all formal sums E

rss

365'

Where rs = 0 for all but a finite number of elements 3 6 S. The addition

in F (S) is defined as es + Ztss = 203 + ts)s, 365'

363

368

and scalar mulitplication is given as t - Z 7'33 2 :(trs)s, 865

368

for t E R. Clearly, S is a basis for F M’ of R—modules that satisfies, for all m, y E M, r E R,

(i) ¢(w + y) = 09GB) + My), (ii) Mm) = Mix)The R—module homomorphism (15 : M —> M’ is an isomorphism of R-modules if (b is a bijection. In this case we write M g M’ as Rmodules. If M is a free R-module of rank k, then M g R619 R EB - - - EB R. \—v—/

To see this, let x1,x2, . . . ,xk be a basis for M and note that kthe map

1:,- I—> (0,. ..,0,1,0,. . . ,0), 1 in the ith place, 0’s elsewhere, is an R-module isomorphism. Condition (1) of Definition 3.10 says that an R—module homomorphism is a homomorphism of additive abelian groups. The kernel of the R-module homomorphism qt : M —> M’ is its kernel as a homomorphism of additive

abelian groups, that is, ker(¢) = {11: E M : (M53) 2 0}. Let N = ker(¢). Since ¢(ra:) = r¢(m) = r0 = 0 for r E R, a: E N, the kernel is an Rsubmodule of M. We have easy analogs for Proposition 1.21 and Proposition 1.22. Proposition 3.10. Let N be a submodule of M. Then the map ¢ : M —> M/N defined by a I—> a + N is a surjective homomorphism of R-modules with kernel N. Proof. By Proposition 1.21, 9b is a homomorphism of additive groups, so we only need to check that qt respects scalar multiplication, but this is easy:

forrER,:c€M,¢>(r:z:)=ras+N=r(x+N)=rqb(:c).

D

Proposition 3.11. Let ¢ : M —> M’ be a homomorphism of R-modules

with kernel N. Then the map 7 : R/N —> ¢(M) defined by ’y(a+N) = ¢(a) is an isomorphism of R—modules. Proof. By Proposition 1.22, (13 is an isomorphism of additive groups, so we only need to check that qS respects scalar multiplication, but this is easy: for

r e 1w e M, 7(r($+N)) = 7(m‘+N) = ¢ = W) = mew). D Here is an application of Proposition 3.11. Proposition 3.12. Every finitely generated R-module M is the quotient of a free module of finite rank.

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1 13

Proof. Let {77%}l denote a set of generators for M, let $1,332, . . .,xk be indeterminates and let F denote the collection of all sums of the form

2:l rim, with r, E R. Then F is a free R—module with basis {30,}. There exists a surjective homomorphism of R—modules (b : F —> M defined as 7'2m2 + . . . + Tic-TIC ¢(T1$1 + T2$2 + ' ' ' + Tkmk.) = Tlml +

for r1,r2, . . . ,rk E R. Let N = ker(¢). Now F/N is a quotient module isomorphic to ¢(F) = M, since ¢ is a surjection.

E]

To illustrate Proposition 3.12, let R = Z [(1 + \/—23) /2]. (As we shall see in Chapter 4, R is the ring of integers of the finite field extension

Q(\/——23)/Q-) Let M = 3R + (2 + \/——23)R be the ideal of R generated by 3 and 2 + \/——23; M is a finitely generated module over R. Let F denote the free R-module on the basis {$1,332}, F is a free rank 2 R-module. Now, there exists a surjective homomorphism of R-modules ob : F —> M defined as .231 I—> 3, x2 I—> 2 + \/——23. Since ¢

is surjective, F/N E’ M, as required. Note that N = ker(q5) is non—zero: it contains the element —9ac1 + (2 — \/——23)x2. Moreover, R is not a PID and M is not a principal ideal. Consequently, M is not a free rank one R-module. We have a “universal mapping property for kernels” as in group and ring theory. Proposition 3.13 (Universal Mapping Property for Kernels).

Let (t : M —> M’ be a homomorphism of R-modules with N = ker(¢). Suppose that K is a submodule of M with K Q N. Then there exists a surjective homomorphism of R-modules ¢ : M/K —> ¢(M) defined by

Ma + K) = (15(0)Proof. Since we already know that w is a surjective homomorphism of additive abelian groups, we only need to check that 1p respects scalar mul-

tiplication. Let a + K E M/K, r E R. Then 1/1(r(a + K)) = ib(ra + K) = qb(ra) = r¢(a) = r1,b(a + K).

B Let ((5 : M —> M’ be a homomorphism of R—modules with N = ker(q§). The Universal Mapping Property for Kernels (UMPK) says that given a

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Fundamentals of Modem Algebra

submodule K of M with K Q N, there exists an R—module homomorphism

1b : M/K —> M’ so that

1P8 = (#5 where s : M —> M/K is the canonical surjection; we say that gb “factors through” M/K and the following diagram commutes:

M/k As we have seen, if V is a finite dimensional vector space over a field K and W is a subspace of V, then W is finite dimensional with dim(W) S

dim(V) (Proposition 3.5). If we generalize to arbitrary rings, we cannot hope for the proposition to remain true (can you think of an example where it fails?) However, in the case that R is a PID we have the following. Proposition 3.14. Let R be a PID, let F be a free R-mod'ale of rank m, and let M be a submodule of F. Then M is a free R-mod'ale 0f ran/cl g m. Proof. Our proof is by induction on m. For the trivial case, assume that

m = 1, and let {em} be an R—basis for F. Let M be an R—submodule of F. Let I be defined as

I={r€R:rm1€M}. Then I is an ideal of R, and since R is a PID, I = Ra for some a E R. If a = 0, then M = 0, and consequently, M has rank 0. If a 75 0, then

M = R(aw1), and so, M has rank 1. For the induction hypothesis, we assume that every submodule of a free R-module of rank m — 1 is free of rank l S m — 1. Let F be a free R—module

of rank m on the basis {$1, 172, . . . ,azm}. Let M be a submodule of F. Let

I={TERI{17:71:31+T2$2+---+7‘m_1$m_1+7'$mEM, for some r1,'r'2, . . . ,rm_1 E R}.

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1 15

Then I is an ideal of R so that I = Ra for some a E R. If a = 0, then M is a submodule of the free rank m — 1 R—module R331 69 - - - 69 Rxm_1 and

so by the induction hypothesis, M is free of rank 3 m — 1. If a 75 0, put N=Mfl(Rw1$Rw2$-~€BRmm_1).

Then N is a submodule of the free module Rm 6} - - - 69 Rxm_1, and by the

induction hypothesis, N is free of rank 3 m — 1. Let w be an element of M of the form w 2 mm + T2372 + - -- + rm_1xm_1 + awm,

for some 7‘1,'r2, . . .,7‘m_1 E R, and let 06 = 31$1 + 82392 + - - ’ + sm—lmm—l + mm,

for 81, .32, . . .,sm_1,7' E R, be an element of M. Then there exists 0 E R for which x — cw E N, thus M=N+Rw.

Nowif'wEN,thenM=NandsoMisfreeofranklSm—l M2 With 723 = idMa, Where id M3 : M3 —> M3 is the identity map.

Example 3.1. Let R = Z and let M be a Z—module in Which there exist elements m1,m2,m3 that satisfy the relations 3m1 — m2 = 0 and

—m1 — m3 = 0, only. Let T denote the Z—submodule of M generated by m1, m2, m3. Then by Proposition 3.12, T is a quotient of a free Z-module F of finite rank; F is the free Z-module on the basis {$1,332, .103}. There exists a surjective Z-module homomorphism '72 : F —> T, given as as, I—> mi,

for 2' = 1, 2,3, and F/N E T with N = ker(72). By Proposition 3.14, N is free over Z of rank 3 3, in fact, N is a free rank 2 Z—module on the basis {3x1 — 11:2, —x1 — 5133}. Let 71 : N —> F be the inclusion. Then there exists a short exact sequence of Z—modules

0

’70

>N

’71

2

>F7>T

73

>0

With 70, 73 the obvious maps. This short exact sequence is split since the map 3 : T —> F defined by 3(m1) = $1, 3(m2) 2 3x1, 3(m3) 2 —m1 induces a Z—module homomorphism With 725 = idT.

Example 3.2. Let Q5 : Z6 —> Z3 be the ring homomorphism defined as (Mm) = mmod 3. Then Z3 is a Z6—module With scalar multiplication de-

fined as m - n = q5(m)n. As a Z6-module, Z3 is generated by {1}. By

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Fundamentals of Modem Algebra

Proposition 3.12, Z3 is the quotient of a free rank one Z6 module, indeed

Z6/{0, 3} g Z3. One has the short exact sequence of Z6—modules

0&{0,3}l>z6i>23£>0 With 70, 73 the obvious maps. Now, 3 : Z3 —> Z6 defined as 3(0) 2 0, 3(1) 2 4, 3(2) 2 2 is a Z6—module homomorphism that satisfies (b3 2 idZ3Thus the short exact sequence is split. Definition 3.11. An R—module P is projective if every short exact se-

quence

is split. Proposition 3.16. An R-module M is projective if and only if there exist

elements {an}n€J of M, and R-module homomorphisms {fin}, B7, : M —> R, 17 E J that satisfy

(i) for each :1: E M, fin(a:) = 0 for all but a finite number of 77,

(ii) for each :1; E M, a: = 2,76] ,Bn(a:)a,7. Proof. We prove the “if” part and leave the converse for an exercise.

Observe that (ii) implies that {an}n€J is a generating set for M. Let F be the free R-module on the set of indeterminates {wn}n€J. Then the R-module homomorphism (15 : F —> M defined by 9/)(xn) = a7, is surjective.

Define a map 3 : M —> F by the rule 3(m) = Z776, Bn(m)xn. Then 3 is a homomorphism of R—modules With tbs = idM, and thus the short exact sequence

0—>ker(¢)—>Fi>M—>0 is split. NOW let 0

>A

be any short exact sequence.

>B

¢>M

>0

Let g : F —> B be the map defined as

g(a:,,) 2 an where a,7 is so that ib(an) = $01)”) for all 17 E J. Then 9 is an R-module map for which Q5 2 mtg, cf. [Rotman (2002), Theorem 7.52]. It follows that gs : M —> B is an R—module map for which 7,2193 2 id M, and thus M is projective. I]

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1 19

Proposition 3.16 implies that any free R—module is projective. Proof:

Let {an},,€J be an R—basis for M and set 57(0‘77) = 6%,]. Consequently, the Z—module T of Example 3.1 is projective since T is free rank one on the

basis {m1}. Regarding the Z6—module Z3 of Example 3.2, one sees that the element

{1} of Z3, together with Z5—module map .3 : Z3 —> Z6 satisfy conditions (i) and (ii) of Proposition 3.16. Consequently, Z3 is a projective Zg—module. Z3 is not free over Z6, however, and so here we have an example of a projective module that is not free. Proposition 3.17. Suppose N is a submodule of M. If the quotient module M/N is projective then there exists a submodule A of M with A g M/N

and M = N 63 A. Proof. Let ¢ : M —> M/N denote the canonical surjection. There is a short exact sequence of R-modules

0—>N—>Mi>M/N—>0 and so, since M/N is projective, there exists an R—module map .3 : M/N —>

M With $3 = IdM/N. We show that M g N63 M/N. Let m E M. Then

¢(m — S(¢(m))) = MM) — ¢8(¢(m)) = ¢(m) — ¢(m) = 0, thus m = n + s(gb(m)) for some n E N. This says that M Q N + s(M/N).

Now since N + s(M/N) Q M, we conclude that M = N + s(M/N). Now suppose that m = n+r for n E N, r E s(M/N) with r = s(t+N), t+N E M/N. Then ¢(m) = ¢(r) = t+N, and so t+N is uniquely determined by m, and consequently, r and n are uniquely determined by

m. Thus M 2 N69 s(M/N). Now since A = s(M/N) § M/N, the result follows.

E]

Corollary 3.2. Every finitely generated and projective R-module is a direct summand of a free module of finite rank. Proof. Exercise.

El

If R is a local ring, then a finitely generated and projective R-module is free over B. To prove this result we shall use the following special case of Nakayama’s Lemma. Proposition 3.18. Let R be a local ring with maximal ideal m. Let M be a finitely generated R-module.

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Fundamentals of Modern Algebra

(i)

ImzM, thenM=0,

(ii) Let M —> M/mM denote the canonical surjection of R-modules. Let {$1,332, . . . ,xk} be a subset of M, and suppose that the set of images

{51} generates the quotient M/mM as an R-module. Then the set {513,} generates M.

Proof. To prove (i), assume that M 34E 0, and let S = {$1,562, . . . ,xk} be a minimal generating set for M. Since mM 2 M, there exists elements . k: r1,r2, . . . , rk E m for Whlch :131 2 22:1 rixi. Thus, k (1 — r1):z:1 = x1 — rlxl = E riwi. i=2

Since r1 6 m, 1 — r1 ¢ m, and so, 1 — r1 is a unit of R. (Proof: if 1 — r1

is not a unit, then 1 — r1 is contained in some maximal ideal J of R by Proposition 2.23. But since R is local, J = m.) Now, It

11:1 = :(1 — 7'1)—17"i$ia i=2

and so, S\{x1} is a generating set for M Which contradicts the minimality of S.

For (ii), suppose S = {x1,x2,...,mk} is a subset of M for Which {51,52, . . . ,Ek} generates M/mM . Let B be the submodule of M generated by S, that is, B consists of all sums of the form 2L1 rim,- for ri E R.

Then (M/B)/m(M/B) = M/(mM + B) = 0, since mM+ B = M. Thus, (M/B) = m(M/B). NOW by (i), M/B = 0 Which yields M = B, and so, {1:1} generates M. E] Proposition 3.19. Let R be a local ring with maximal ideal m. Let M be a finitely generated and projective R—module. Then M is a free R-module.

Proof. Assume M 75 0, and let S 2 {$1,332,303, . . . ,mk} be a minimal set of generators for M. The quotient module M/mM is a module over R and also over R/m, which is a field. Let E, denote the images of the ac,- under

the canonical surjection M —> M/mM. Then the elements {5,} generate the R—module M/mM . We claim that E,- ¢ mM for all i. Otherwise, if some 51-1 6 mM, then the set {51,52, . . . ,Ek}\{fil}

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121

generates M/mM . Now by Proposition 3.18(ii), the set S\{.’L'Z'I} generates M, contradicting the minimality of S.

We next claim that {E} is a basis for the R/m—module M/mM . To this end, suppose that $151 +72E2 + ' “ +Pk5k 6 mM,

for elements E- E R/m. Suppose that m ¢ m for some 2". Then k

EE, 6 (2 4m) + mM. i=1,

iyéi’ Thus, aczl E (2— (m)‘ 1mm))+ mM, i=1,

iyéi’

and so, {£1,52, . . . ,Ek}\{§iz} generates M/mM , again contradicting the minimality of S (by Proposition 3.18(ii)). Thus 13 E m for all 13 and so, {51‘} is a basis for M/mM. Now let Rk denote the free R-module of rank k with basis {bi}, and let

M be the surjective map of R—modules defined by ¢(bi) 2 53,. Then there is a short exact sequence of R—modules

0—>ker(¢)—>Rk—>M—>0, Which is split since M is projective. By Proposition 3.17, Rk = M’ GB ker(¢) Where M’ E M. NOW M/mM , and so, ker(¢) Q mRk. Moreover, mRk g mM 69 m ker(¢), and so ker(¢) injects into mMEBm ker(gb). It follows that ker(¢) Q mker(¢), and thus ker(¢) = mker(¢). Therefore, by Proposition 3.18(i), ker(¢) = 0. Thus 45 is an isomorphism of R-modules, and consequently, M is free over R. III Let R be an integral domain, let M be an R—module and let P be a prime ideal of R. We can often say something globally about M if we localize at P. This is possible in View of the following proposition. Let Rp denote the localization of R at P and set

Mp={m/s: mEM,s€S=R\P}.

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Fundamentals of Mode’rn Algebra

Then Mp is an Rp—module, with Rp a local ring (Proposition 2.47). Proposition 3.20. With notation as above, and with the identification

fl

M = M/l, one has M =

Mp.

P prime

Proof. One easily obtains M Q

n

Mp.

P prime

Let it 6

fl

Mp and consider the ideal of R defined as

P prime

I={a€R: amEM}. Now :1: 2 m/s with m E M, and s E R with 3 ¢ P for all primes P in R. Since 3 E I, I is not contained in any prime ideal of R. Thus by Corollary 2.3,I=R,andsolm=m€M. III This formula underscores that a module can be locally free yet not free: the Rp—basis B that could exist for each Mp may not be in the intersection

of all of the Mp, and hence may not yield an R—basis for M. As an example, we recall the illustration of Proposition 3.12; we let

R = Z [(1 + \/——23)/2] (this is the ring of integers of the field extension Q(\/——23)). Let M 2 3R + (2 + \/——23)R. By [Rot02, Proposition 11.98], M is a finitely generated and projective R-module and so by Proposition 3.19, Mp is a free Rp-module for each prime ideal P of R. However, (as we will show in Chapter 4) M is not a principal ideal of R, and hence M is not a free R-module. Of course, if the R—module M is R—free With basis B, then B Q fl Mp and so the image of B in each Mp Will be an Rp—basis for Mp. 3.4

Tensor Products

In this section we introduce bilinear maps and show how they can be used to construct the tensor product M ®RN of R—modules M, N. We specialize to the tensor product V ®K W of two vector spaces over K and relate the

linear dual (V ®K W)* to the dual spaces V* and W*. *

*

*

Let R be a commutative ring with unity and let M, N, A be R—modules. Definition 3.12. A function f : M X N —> A is R—bilinear if for all

a,a’€ M, b,b’€N, r6 R,

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123

(i) f(a+a’,b) = f(a,b)+f(a’,b), (ii) f(a,b+b’) = f(a,b)+f(a,b’), (iii) fond: b) : rf(a, b) : f(aa Tb)‘

Definition 3.13. A tensor product of M, N over R is an R—module M (8 R N together with an R—bilinear map f:M>M®RN so that for every R—module A and every R-bilinear map h : M X N —> A there exists a unique module homomorphism h : M ®R N —> A for which hf = h, that is, the following diagram commutes.

IL

M> K defined as h(a®b) = (fXg) (a, b) = f(a)g(b). Set f (X) g = h. This shows the containment V* (X) W* g (V (X) W)*.

I]

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127

When do we have equality in Proposition 3.24? Proposition 3.25. Suppose that V, W are finite dimensional vector spaces over K. Then

V* ®W* = (V®W)*. Proof. Let dim(V) = m and dim(W) = n.

Let {b1,b2,...,bm} be a

basis for V, and let {f1, f2, . . . , fm} denote the corresponding dual basis

for V* with fi(bj) = 6,7,. Let {c1,c2,...,cn} be a basis for W, and let {91, 92, . . . , gn} denote the corresponding dual basis for W* with gi(cj) = 5M. By Proposition 3.23, the set {bi®cj}, 1 S i S m, 1 S j S n, is a basis for V (X) W. By Proposition 3.7, the set {¢Q,I3}, 1 S 04 S m, 1 S fl S n, with wa,,8(bi ® Cj) = 6a,i6fl,ja

is a basis for (V (X) W)*. By Proposition 3.24 for each pair 04, fl, 1 S oz S m, 1 S fl S n, there is a K-linear map fa®932V®W—>K,

defined by

(fa ® gflXbi 8’ Cj) = fa(bi)9fi(cj)Clearly, fa ® 9;; = 1%,5, for all a,,3, and so V* (X) W* = (V (X) W)*.

III

If V and W are infinite dimensional, however, then V* (X) W* is a proper

subset of (V (X) W)*. Proposition 3.26. Suppose that V, W are infinite dimensional vector

spaces over K. Then V* (X) W* C (V (X) W)*. Proof. Any element 2 f (X) g E V*®W* determines an element of (V®W)* defined as

(E f K defined by

95(2 3,100,) = sfl,

sa 6 K,

aEJ

Where 3a 2 0 for all but a finite number of 04. Then {fa} Q V*, {93} Q W*. There is a countably infinite subcollection {123,921 Q {fa} and a countably infinite subcollection {gfij Bil Q {gfl}. Now 2:1 fa, ® 95, is an infinite sum that becomes a finite sum upon evaluation at an element of V (X) W. Thus

Zfai 69m» e (V69 W)*\(V* 63> W*)i=1

El As an example of this phenomenon we take V = W = K [.73], the vector space of polynomials over a field K. Then {1, $17,132,133, . . .} is a basis for

K [51:] Let {60, 61, 62, 63, . . .} be a collection of linear functionals in K [:13]* defined as

64%) = 5m» for all 2', j. Note that {cg-L921 is a linearly independent subset of K [x]*, yet it is not a basis for K [as]* over K. As one can easily check, 2:1 81 ® 6, is

an element of (K [:13] (X) K [x])* that is not in K [a3]* (8) K [ac]* Every element f E K [x]* has a unique representation as an infinite sum 00

f= aen =30€0+S1€1+3262+---, n=0

for 3,, E K, n 2 0. Thus elements of K [x]* can be identified With sequences

{3n}:°=0 of elements in K. 3.5

Algebras

In this section we consider algebras over a commutative ring with unity B. An R—algebra A is an object that is both a ring A and an R-module — the scalar multiplication is given through a ring homomorphism A : R —> A. We specialize to commutative R—algebras, and ShOW that every finitely generated R—algebra is the quotient of a polynomial ring. We compute all of the R—algebra homomorphisms A —> B, Where A is finitely generated.

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129

Definition 3.14. Let R be a commutative ring with unity, 1R and let A be a ring with unity 1A. Then A is an algebra over R if there exists a homomorphism of rings A : R —> A with unity for which A(R) is contained in

the center of A, that is, A(r)a = aA(7") for all 7" E R, a E A. Then A is an R— module with scalar multiplication R X A —> A given as ra = A(r)a 2 GAO"), for all r E R, a E A. The map A is the R—algebra structure map of A, which we sometimes denote as AA- Specializing to a = 1 A yields A(’r) 2 7'1 A, and so A(R) is a subring of A consisting of all R—scalar multiples of 1A- If A is 1-1 (for instance, if R is a field), then A sends R to an isomorphic copy

of R in A; we can then identify R with A(R), and so A0") = 7", for 'r E R. An R—algebra A is commutative if A is a commutative ring with unity. For example, the polynomial ring R[x] is a commutative R—algebra

where the structure map A : R —> R[:I:] is given by A(r) = 7' (A is 1—1; 7“ 7S 0 is a degree 0 polynomial; 'r' = 0 has degree —oo). The monoid ring RS is an R-algebra with R—algebra structure map A : R —> R5 defined as

A(r) = r1 = 7", for all r E R (again A is 1-1). Likewise, the group ring RG for G a finite group, is an R algebra with structure map A : R —> RG, 7' r—> 7". For another example, let A be an R—algebra, and let J be an ideal of A

(as a ring). Then the quotient ring A/ J is an R—algebra. Indeed, if AA is the R-algebra structure map of A, then the R—module structure of A/ J is given

by the ring homomorphism A : R —> A/J defined as r- (a+J) = AA(r)a+ J. Let A and B be algebras over R and let A®RB be the tensor product of A and B as R—modules. Then the R—module A (8) R B can be endowed with

the structure of an R-algebra. First note that A (83 B is a ring: addition is the module addition k l (Zai®bi)+ (ZCj®dj)=Zai®bi+ZCj®dj

j: 1

3'21

i=1

and multiplication 1s given as k

l

k

l

(:01 ® 52%: c,- ‘3’ dj) = 220:9 ® bidj,

i=1 3’21 121 j:1 for ahcj E A, bi,dj E B. The unity in A®B is 1A® 113. Next, let A : R —> A®R B be the ring homomorphism defined as A(7') 2 AA (7') (X) 13, where AA is the R—algebra structure map of A. Then

()Za®b)A(r)=A(r (,)()Za®b

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Fundamentals of Modern Algebra

for all r E R, Z a (X) b E A ®R B. Now A (83 B, together with scalar mul— tiplication defined through A is an R—algebra, called the tensor product R-algebra. What are the maps between R—algebras? Definition 3.15. Let A, B be R—algebras with R—algebra structure maps AA : R —> A, A3 : R —> B. Then d): A —> B is an R—algebra homomorphism if gt is a ring homomorphism and

¢()\A(r)a) = AB(7°)¢(a),

(3-3)

for all r E R, a E A ((15 preserves scalar multiplication).

Condition (3.3) is equivalent to the condition

(MA/1W) = /\B(7")-

(34}

(Proof: in the case a = 1A in (3.3), one has

¢(/\A(7“)) = ¢()\A (701.41) = AB(7")¢(1A) = AB(7')1B = A30"), for r E R. Conversely, multiplying both sides of (3.4) by ¢(a) yields (3.3).) An R-algebra isomorphism is an R-algebra homomorphism which is a bijection. The collection of all R—algebra homomorphisms from A to B is denoted by HomR_a1g(A, B). The following proposition extends Proposition 3.22 to R—algebras. Proposition 3.27. Let A be an R-algebra and let J be any ideal of A. Then there is an isomorphism of R-algebras

A/J®R A/Jg (A®R A)/(J®RA+A®R J). Proof. By Proposition 3.22 the map

z”: A/J®R A/J —> (A M A)/(J ®RA+A®R J), ~

defined as l((a+N)®(b+N)) = a® b+ (J®RA+A®R J) is an isomorphism of R—modules. One then shows that l is a homomorphism of R-algebras.

E]

For the remainder of this section, all algebras are assumed to be commut ative.

Proposition 3.28. Let A be an R-algebra, and let A (83 A be the tensor product R-algebra. Then the map m : A ®R A —> A defined as k:

k

m(Z ai ® bi) = Z aibi; i=1

is a homomorphism of R-algebras.

i=1

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131

Proof. Clearly, addition is preserved under m. We show that multiplication is preserved. Notice the need for A to be commutative:

k

m

l

(2: ai (X) bi)(Z Cj (3) dj) i=1

k:

l

2 m(:: 2 aicj (X) bidj)

j=1

i=1 j=1

j

l

2 772(2 ai ® bi)m(Z Cj ® dj). i=1

j=1

Finally, we Show that m respects the R-algebra structures on A ®R A and A:

m(r(Z a. (3) 11.)) = m; Lama. ® bi) i=1

i=1

= Z AA(T’)aibi

i=1

= A140") 2 alibi

i=1

= Mam; a. ® b.) i=1 k

= rm(Z az- ® bi). El

Definition 3.16. Let R[{ma}] be the R—algebra of polynomials in the in— determinates {ma}aEJ. Let B be an R—algebra. Let {ba}a€J be a family of

elements in B indexed by J and let f ({wa}) be a polynomial in R[{xa}].

132

Fundamentals of Mode’rn Algebra.

Then the evaluation of f ({xa}) at {ba} is the element f ({ba}) E B. The map ¢{ba} :R[{xa}] —> B

defined as f({:I;a}) 1—) f ({ba}) is a homomorphism of R—algebras called the evaluation homomorphism.

Definition 3.17. Let A be an R—algebra, let S = {ba}a€J be a subset of elements of A and let {ma}a6 J be the corresponding set of indeterminates.

Then S is a generating set for A as an R—algebra if the evaluation homomorphism ¢{ba} : R[{.’L‘a}] —> A

is surjective. Equivalently, S = {ba} is a generating set for A if the set of monomials I _

e1

62

ek

S _{ba1ba2 ' ' ' bak}

Where a1,a2, . . .,ak is a finite set of indices in J and el,e2, . . . ,ek 2 0 are

integers, is a generating set for A as an R—module.

If the generating set S is finite, say, 3 = {b1,b2, . . . ,bn}, then A is finitely generated as an R—algebra and we write A = R[b1, b2, . . . , bn]. For example, {9} is a generating set for the Z—algebra Z03, (g) = Cg, since the evaluation homomorphism ¢{g} : Z [x] —> ZC3 is surjective. Every R-algebra A admits a generating set, if necessary one could choose 3 = A. We aim to characterize every finitely generated R—algebra as a quotient r1ng. Proposition 3.29. Every finitely generated R-algebm A is a quotient of a polynomial ring over R.

Proof. Let {b1,b2,...,bn} be a generating set for A as an R—algebra, let {x1,m2,...,wn} be the corresponding set of indeterminates and let R[ac1, x2, . . . ,xn] denote the ring of polynomials in the indeterminates 33,-. Then the evaluation homomorphism

is surjective.

Let N = ker(¢{b1,___,bn}).

R[m1,...,xn]/N.

By Proposition 2.39, A

El “2

¢{b1,...,bn} 1R[$1,...,mn] —> A

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133

Let B be an R—algebra and let N be a subset of polynomials in

R[x1,m2,...,mn]. Then the indexed family {b1,b2,...,bn}, b,- E B, is a common zero in B of the polynomials in N if f({b1, . . . , bn}) = 0 for all f(a:1, . . . , inn) 6 N. For example, {2,1/2} is a common zero in Q of the ideal of polynomials N = (my — 1) in Q[x, y]. But {3} is not a common zero in Z of {w+3,x— 3} Q Z[m]. Proposition

3.30. Let

A

be

an

R-algebra

of the form

A

=

R[:B1,332,...,:1:n]/N, where N is an ideal of polynomials and let B be an R—algebra. Then HomR_alg(A, B) is in a 1-1 correspondence with the set of common zeros in B of the polynomials in N.

Proof. Let (b : R[:I:1, . . . ,xn]/N —> B be an R—algebra homomorphism. Let E,- = w, modN for 1 g i g n. Then ¢ is determined by sending each 5,- to

an element b,- E B. Now for f(x1,...,xn) 6 N,

0 = ¢(f($1,~-a«’13n))= f({bl,---,bn}), and so the family {b1, . . . ,bn} is a common zero in B of the polynomials in N. Conversely, let {b1, . . . , bn} be a common zero of polynomials in N and consider the evaluation homomorphism

¢{b1,...,bn} : R[ac1, . . .,.’L‘n] —> B. Now, ker(q§{b17_u7bn}) is the ideal of all polynomials for Which {b1, . . . , bn} is a zero. Thus ker(¢{b1,_._,bn}) contains N. By the UMPK (Proposition 2.40), there is an R—algebra homomorphism gb : A —> B. E] Let A, B be R-algebras and assume that A is finitely generated. To compute HomR_.c,,1g (A, B), we first use Proposition 3.29 to write

A g

R[x1, 11:2, . . . , xn]/N for some integer n and some ideal N of

R[ac1, x2, . . . ,xn]. We then apply Proposition 3.30. For example, to com-

pute Homz_a1g(ZC'3, Z), 03 = (a), we note that Z03 E“ Z[;c]/(a:3— 1). Now, any Z—algebra homomorphism is of the form a: I—> b E Z with b3 — 1 = 0. Thus HomZ_a1g(ZCg, Z) contains exactly one element: ¢ : Z03 —> Z defined

by U I—> 1. 3.6

Discriminants

If V is a vector space over the field K, then a K—bilinear map B : V X V —> K of §3.4 is called a bilinear form. This final section of the chapter concerns bilinear forms on a finite dimensional K—Vector space V, Where K is the

134

Fundamentals of Modem Algebra.

field of fractions of an integral domain R. For an R—submodule M of V, which is free over R, and Which satisfies KM = V, we define the dual module of MD with respect to a bilinear form B on V. Next we define the discriminant disc(M) and give some properties of the discriminant. As an example, we let V be the Klein 4—group, define a bilinear form on QV

using the characters of V, and compute disc(ZV), ZVD, and disc(ZVD). We shall use discriminants in Chapter 4. *

>|
K which satisfies, for x,y, z E V, a E K,

(i) B(w+y,2) = B(x,2) +B(y,z), (ii) B(m,y + z) = B(x, y) + B(x,z), (iii) B(a:c, y) = aB(x, y) = B(x,ay). If B(a:,y) = 0 for all y E V implies that ac = 0, and B(a:, y) = 0 for all no G V implies that y = 0, then B is non-degenerate. A bilinear form B

is symmetric if B(a:, y) = B(y,a:) for all :13,y 6 V. Lemma 3.1. Let B be a non—degenerate bilinear form on V.

Let

{b1,b2, . . .,bk} be aK—basz'sfor V. Then {B(b1, —),B(b2, —), . . .,B(bk, —)} is a basis for V*. Proof. Clearly, B (b,, —) : V —> K is a linear transformation, so we only

need to check that {B(b,-, —)}f=1 is a linearly independent subset of V*. Suppose that TlB(b1, —) -|- T2B(b2, —) -|- - - - + T'kB(bk, —) = 0

for r1,r2,...,rk E K. Then for all y E V,

713(5179) + T2B(b2,y) + ' ' ' + T'kB(bk73/)

= 30151.3!) + 30252.11) + ' ' ' + Emblem) = B(’I"1b1 + T2122 + ' ' ' + ’I‘kbk,y)

: 0, whencer1=r2=---=rk=0.

I]

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135

Let B be a symmetric, non—degenerate bilinear form on V and let M be a free R—submodule of V (Viewed as an R—module) Which satisfies KM = V. Then M has rank k over R. The dual module of M with respect to

B (ac, y) is the R—module defined as

MD ={vE V: B(v,M) QR}. The dual module can be identified With the set of R—linear maps M —> R, de—

noted as HomR(M, R), through the isomorphism from MD to HomR(M, R) given by oz I—> B(a, —), for a E MD. The dual module MD is free of rank k over R. Proposition 3.31. Suppose {b1,b2,...,bk} is an R-basis for M. There exists a basis {fl1,,82,...,flk} for MD “the dual basis” that satisfies

B(/8iabj) = 5133'Proof. Since KM = V, {bl-”=1 is also a basis for V. Let {f1, f2 . . . , fk}

be the basis for the dual space V* defined by fi(bj) = 5231” Let {BM}: 1 S i, j S k, be the collection of bilinear forms defined as BM (x, y) = fz(a:)fj (y) for ac, y E V. The bilinear form B then can be written uniquely as k

B = Z B(bi,bj)Bi,j. i,j=1 By Lemma 3.1, {B(bi, —)}f:1 is a basis for V*. For 1 S i g k, one has

and so, the k X k matrix A = (B (bi, bj)) is the matrix of the change of basis

from {fj} to B (bi, —). Consequently, A is invertible. Let (0“) = A—1 and set ,8q 2 Qq’lbl + Oq,2b2 + ' ' ' + 0q,kbk7

for 1 S q 3 k. Then

136

Fundamentals of Modem Algebra.

I:

303ml”) = Z B(b,-,bj)B,-,j(flq,bl) m=1 = ZBUM, bl)fi(5q) k

= 2 3(1),, mewi=1

Thus {fi1,flg,...,flk} isabasis for MD.

El

From the proof of Proposition 3.31, one has

51

b1

5 A

.2

b =

5k

.2

7

bk

with A = (B(b,-, (23)). The discriminant of M over R with respect to B and the basis {b1, b2, . . . , bk} is the R-submodule of K defined as disc(M) = Rdet(A). Now suppose M, N are free R—submodules of V which satisfy KM = V,

KN = V. Then necessarily, both M and N have rank k: over R. Let {1),}l be an R-basis for M and let {cl-M“:1 be an R-basis for N. Then {bi} and {Ci} are K-bases for V. Therefore, there exists an invertible k X k matrix

T in Matk(K) for Which T(M) = N. Let

[M : N] = Rdet(T). Then [M : N] is an R-submodule of K called the module index. Proposition 3.32. The module index [M : N] does not depend on the choice of bases for M and N.

Proof. Let {1),} be an R—basis for M, let {c,} be an R—basis for N and let T E Matk(K) be so that T(bi) 2 01, for all i. Let {1);} be some other basis for M and let {0]} be some other basis for N. Let S E Matk (K) be so that 5(1):) = c]. There exist invertible matrices X, Y E Matk(R) for Which

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137

X(b7;) = b;, Y(c,-) = c; for all 2'. Now, Y-lsX(b.-) 2 Ci and so T = Y-lsX. One has

det(T) = det(Y—13X) = det(Y‘1X)det(S), With det(Y_1X) a unit in R. It follows that Rdet(T) = Rdet(S).

El

Proposition 3.33. [M : N] 2 [ND : MD]. Proof. Let {1).} be an R—basis for M and let {,Bi} be the dual basis for MD . Let {c2} be an R—basis for N and let {71‘} be the dual basis for ND . Let T E Matk(K), be so that T(bi) 2 Ci, for all i. Then Tt(%-) = fl, for

all i, Where Tt is the transpose of T. Thus [ND : MD] = Rdet(Tt) =

Rdet(T) = [M : N].

III

The module index and the discriminant are related by the following result. Proposition 3.34. Suppose M, N are free R-submodules of V with KN 2 KM = V. Then

disc(N) = [M : N]2disc(M). Proof. Let {bi} be a basis for M and let {CZ} be a basis for N. Suppose T is the matrix for Which T(bi) = Ci for 1 S i S [9- Then

disc(N) = Rdet((B(CiaCj))) = Rdet((B(T(b-,) T(b ))))

= Rdet(T (B(biaT(bj )))) = Rdet(T)det(T (B(b.~,bj ))) = Rdet(T) det(T) det(B(b- b;-)) = Rdet(T)2 disc(M)

= [M : N]2disc(M). D

Corollary 3.3. The discriminant disc(M) does not depend on the choice of basis for M.

Proof. Indeed, let {bl} and {cz} be two bases for the R—module M. Let T E

Matk(K) be so that T(bz-) 2 Ci for all i. Then T E Matk(R) With det(T) E U (R) Consequently, [M : M] = R. Now by Proposition 3.34, disc(M) computed using {bi} is the same as disc(M) computed using {Ci}. E]

138

Fundamentals of Modem Algebra.

The next proposition provides an efficient way to show that two modules are equal. Proposition 3.35. Let M,N be free R-modules with KM = KN = V.

Suppose N Q M with disc(M) = disc(N). Then M = N.

Proof. First note that N E M implies that [M : N] is an ideal of R. By Proposition 3.34, [M : N[2 = R. Hence the module index [M : N] = R and the matrix T, T(M) = N, has inverse T‘1 in Matk(R). Thus M = T'1(N) Q N and hence, N = M. I] We close this section With some computations. Let R = Z, K = Q,

and let V = {6, a, b, 0} denote the Klein 4—group. Let V = QV. To simplify notation, we rename the elements of V as 91 = e, 92 = a, 93 = b, 94 = c.

From §2.3, the characters of V are V1, V2, V3, V4.

The trace map tr : QV —> Q is defined as tr(a:) = 2:1 14-(33) for ac E QV. Note that tr(z?:1 aigi) = 4a1 for (Li 6 Q. Let

B (at, y) = tlf(93y)Then B is a symmetric, non-degenerate bilinear form on QV (see §3.7,

Exercise 23). We compute disc(ZV), ZVD , and disc(ZVD) with respect to B.

Since {91, 92, 93, 94} is a Z-basis for ZV, one easily computes disc(ZV) = 256Z. For the other calculations, we first consider the group ring QV. The set {91, 92, 93, 94} is a Q-basis for QV, and with respect to this basis the vectors are multiplied according to the group product on V. But is there a basis E for QV for Which the coordinate vectors 125; are multiplied

component-Wise? In other words, we seek a basis E so that 4 UE’LUE:

E

4

4 via,

E

wjaj

=

E

’iiai.

j=1 Such a basis does exist and has the form E = {011, (12, a3, 054} Where

1

4

061': Z Z Vi(gm)gma m=1

for i = 1, 2, 3,4. These elements are the minimal idempotents of QV, see

[Serre (1977), §2.3]. The existence of this basis implies that as rings, 4

QV=€BQanXQXQXQ i=1

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139

We now proceed with the computation of ZVD . By definition,

s = {u e ov: tr(uZV) g Z}. We have

for m,n = 1,2,3,4, and so

5 = {9; 9_2 a 9i} 4 ’ 4 ’ 4 ’ 4

is the dual basis; ZVD is the free rank 4 Z—module on the basis S. In order to compute disc(ZVD ) we need to View ZVD as an Z—subalgebra of QV. On ZVD we define addition to be the addition on ZVD as an R—

submodule of QV, and we define multiplication * on ZVD as follows. For gnEV,1§l,mS4,

g * 97m) (9..) = %gf = 6l,n6m,n = 6l,m6m,n

Thus, 3 is the collection of minimal idempotents of the Z-algebra ZVD endowed with the binary operations + and *. 4 Proposition 3.36. ZVD g 69 Z041, as Z—algebms. i=1

Proof. The map 4

cs : ZVD —> 69 Z05,-, i=1

defined by % I—> 04, is an Z—algebra isomorphism.

C]

We identify ZVD with @2121 Zai through the isomorphism of Proposition 3.36. Thus we consider {a1,a2,a3,a4} as a Z—basis for ZVD Q QV.

One now easily computes tr(a,~aj) = 6i,j and thus, disc(ZVD) = Z.

140

3.7

Fundamentals of Modem Algebra.

Exercises

Exercises for §3.1 (1) Let V be a finite dimensional vector space over a field K With

dim(V) = n. Let {v1,v2,...,vm} be a subset of V With m > n. Show that S is linearly dependent.

(2) Prove Proposition 3.6. (3) Let V be a finite dimensional vector space over the field K With linear dual V*. The double dual is the dual space V** = (V*)*. Show that

there is a “natural” isomorphism ¢ : V —> V** defined as ¢(a) (f) = f(a), for a E V, f E V*. Exercises for §3.2

(4) Let

M’ be an isomorphism of free R—modules.

Suppose that {m1, m2, . . . ,mk} is a basis for M.

Show that

{¢(m1), ¢(m2), . . . ,¢(mk)} is a basis for M’. (5) Let R be a ring and let M be an R-module. Prove that scalar multiplication R X M —> M, (1', m) I—> rm, defines an R—module homomorphism¢:R—>M,rI—>rm.

(6) Let R be an integral domain and let I be an ideal of R Which is free of finite rank as an R—module. Show that I is a principal ideal of R.

(7) Show that an R-submodule of R" is isomorphic to Rm for some m S n. (8) Prove that Zn is finitely generated as a Z—module but not free over Z. (9) Let R be an integral domain and let M be a free R—module. Suppose that rm 2 0 for some 7' E R, m E M. Prove that either 7' = 03 or m = 0. (10) A group is simple if it is non-trivial and has no proper non-trivial normal subgroups. A non-zero R-module M is simple if it has no proper non—zero submodules. Let G be an abelian group which is simple as a Z-module. Show that G is a simple group. Exercises for §3.3 (11) Give an example of a short exact sequence of R—modules that is not split. (12) Show that Zn is not projective as a Z—module. (13) Prove the converse of Proposition 3.16.

(14) Prove Corollary 3.2.

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141

Exercises for §3.4

(15) Let M, N be R—modules and let T : M®RN —> N®RM be the twist map defined as r(a®b) = b®a for a E M, b E N. Prove that 7' is an isomorphism of R—modules.

(16) Let M1,M2,N be R—modules. Prove that (M1 69 M2) (8) N g (M1 69 N) (X) (M2 69 N).

(17) Prove that Zm ®Z Zn 2 {0} if and only if gcd(m, n) = 1. (18) Find an element of Z [x] (X)Z Z [x] that can not be written in the form

me) Q be the trace map defined as tr(x) = 2:1 V‘(x) for x E QV. Let B(x, y) = tr(xy). Show that B is a symmetric, non—degenerate bilinear form on QV.

(24) Let R be an integral domain With field of fractions K. Let V be a vector space over K and let M be a free R—module With KM = V.

Prove that (MD)D = M. (25) Let R be an integral domain with field of fractions K. Let V be a vector space over K, and let M, N be free R—modules with KM = V and KN = V. Prove that N Q M if and only if MD Q ND.

(26) Let R be an integral domain with field of fractions K. Let V be a vector space over K, and let M, N be free R-modules with KM = V

and KN = V and N Q M. Show that disc(N) Q disc(M). (27) Let R be an integral domain with field of fractions K. Let V be a vector space over K and let M be a free R—module with KM = V.

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Fundamentals of Modern Algebra.

Suppose that V 2 V1 69 V2 and M = M1 69 M2 with KM1 2 V1 and

KM2 2 V2. Prove that disc(M) = disc(M1)disc(M2).

Questions for Further Study (1) Let p be a prime number, let (p denote a primitive pth root of unity, and let K = {2(9) denote the simple algebraic field extension, cf. §4.1. Then the ring of integers of K is the Z—algebra R = Z [Cp], see Propo— sition 4.22; R is an integral domain with field of fractions K. (a) Compute the minimal idempotents for the group ring KCp, where 01) denotes the cyclic group of order p.

(b) Compute disc(RCp) and disc(RCIl,) ).

Chapter 4

Simple Algebraic Extension Fields

This chapter concerns simple algebraic extensions of a field F. We special— ize to the case Where F = K is an intermediate field Q Q K Q C. We

construct the Galois group G of the splitting field L/K of an irreducible

monic (hence, separable) polynomial p(x) E K [as] We give some examples of Galois groups and state the Fundamental Theorem of Galois Theory which says that there is a lattice-inverting bijection between the set of intermediate fields between K and L and subgroups of G. A consequence of the Fundamental Theorem is that since G is finite, there is only a finite number of intermediate fields L’, K Q L’ Q L.

This is remarkable

considering that K is an infinite field. The next part of the chapter (§4.3—§4.6) constitutes an introduction to algebraic number theory. We define the ring of integers R of a simple algebraic extension K of Q, and show that every non—zero ideal of R contains

a Q—basis for K, specifically, R is a free Z—module of rank n = [K : Q]. We show how to construct the ring of integers of m) and compute the ring of integers of Q(Cp), where (p is a primitive pth root of unity. We employ Dirichlet’s Unit Theorem to construct the group of units of the ring of integers R. Specifically, we compute U (R) where R is the ring of integers of a quadratic field extension and where R is the ring of integers of Q(CP). We then show that the ring of integers R is a Noetherian ring by showing that R has the ACC. Moreover, R has additional properties: it is a Dedekind domain. Arguably, the most important property of a Dedekind domain is that its localization at any prime ideal is a PID. We use this fact to construct the class group of a Dedekind domain R, which measures the extent to Which R fails to be a PID. We show that every non—zero proper ideal of R factors uniquely into a product of prime ideals, thus generalizing the familiar notion that each integer m E Z factors uniquely into a product

143

144

Fundamentals of Modem Algebra

primes m = pilp? - - - pik. The final section of the chapter concerns field extensions of Qp, the field of p—adic rationals that was constructed in §2.7. We construct finite field extensions of QP by extending the notion of p—adic absolute value and using the method of §2.7. Of course, another way to construct an extension field of Q}, (or any field for that matter) is to take an irreducible polynomial

p(ac) E Qp[a:] of degree n, and “invent” a root 04 = :1: + (p(:c)) of p(m) in the

field Qp[x]/(p(cc)). Then Qp[w]/(p(x)) E Qp(oz) will be a simple algebraic extension of QP, a finite dimensional vector space over QP on the basis

{1, oz, a2, . . . ,Oln_1}. This idea of inventing roots Will be used in Chapter 5

(Finite Fields). 4.1

Simple Algebraic Extensions

In this section we define the notion of a simple algebraic extension of a field F. We specialize to the case where F = K is an intermediate field Q Q K Q C. Then the Fundamental Theorem of Algebra applies to Show that all simple algebraic extensions of K are subfields of (C. Moreover, L is a simple algebraic extension of K if and only if L is a finite dimensional vector space over K. We Show that a simple algebraic extension L of K of degree 77. determines a set of n distinct embeddings 7',- : L —> (C, 1 g i g n.

Furthermore, if E is a simple algebraic extension of L of degree m, then each embedding 7',- : L —> (C extends to m distinct embeddings TM : E —> (C,

1 g j S m. From this extension theorem, we define the norm and trace maps for extensions L/ K .

Let E/F be a field extension with oz 6 E. Assume that 04 is a zero of a

non-constant polynomial f (x) E F [7;] Let $0, : F [at] —> E be the evaluation homomorphism.

Proposition 4.1. Let M = ker(qba). Then F[a"]/M is a field and M is a non—zero maximal ideal of F [as]

Proof. By Proposition 2.39, F[w]/M E’ (be, (F[x]) g E, and so, F[$]/M is isomorphic to a subring of a field, and hence must be an integral domain. Thus M is prime by Proposition 2.35, and non—trivial since f (ac) E M. Since

F [m] is a PID (Proposition 2.20), M = (p(a:)) for some monic polynomial p(a:) = a0 + alx + - - - + an_1$"_1 + a3”

Simple Algebraic Extension Fields

145

of degree n. It follows by Proposition 2.25 that p(:r) is irreducible, and thus

M = (p(a:)) is maximal. Consequently, F [:13] /M is a field.

El

We define the polynomial p(a:) of Proposition 4.1 to be the irreducible polynomial for 04 over F, which we denote as irr(a, F).

But what is the structure of the field F [:L'] / (10010))? Proposition 4.2. The field F [:13]/ (p(x)) g ¢>a (F [33]) is a finite dimensional vector space over F of dimension n = deg(p(a:)) on the “power basis” {1, oz, a2, . . . ,Otn_1}.

Proof. First note that F [x] is a vector space over F, (p(a:)) is a subspace and the quotient space F [:13] / (p(:1:)) is a vector space over F. Moreover, the map (ta is a linear transformation of F—vector spaces, and consequently,

F[;1:]/ (p(£I7)) g Q50, (F [30]) is an isomorphism of F-Vect0r spaces. Thus, it is natural to characterize the field F [x] / (p(x)) as the collection of all polynomials with coefficients in F evaluated at 04, where p(a) = 0, or equivalently, Where the powers of a satisfy the relation oz” 2 —an_1oz"_1 — . - - — (120(2 — aloz — a0.

Thus any power am with m 2 n can be written in terms of lower pow-

ers {1,a,a 2 ,.. .,a”_1}. Consequently, every element of qfia(F[m]) has a representation as a linear combination ro+r1a+r2a 2 +---+’r‘n_1a 11—1 ,

riGF,

that is, {1, a, a2, . . . , a"_1} is a generating set for ¢a(F[x]) over F. More— over, {1,a,a2...,an—1} is a basis for q5a(F[x]) over F since p(x) is ir— reducible. We conclude that F[a:]/(p(:r:)) g ¢a(F [13]) is an n—dimensional vector space over the field F with the power basis {1, a, a2, . . . ,an-l}.

E]

To illustrate Proposition 4.2, let F = Q, E = R With a = 3/? (the

real zero of m3 — 2 E Q[:r].) Then there is an evaluation homomorphism

d) 3/5 : on] —> R with ker(¢ 3/5) = ($3 — 2). We have

Qlflvl/(III3 — 2) E ¢e/§(Ql$l) and q5 %(Q[w]) consists of all quantities of the form a0 + a1(\:3/§)+ 02(fl), a0,a1,a2 E Q.

As a vector space over Q, ¢\3/§(Q[a:]) has dimension 3 with basis

{Le/Z 3/1}.

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Fundamentals of Modern Algebra

As a field, every non—zero element in ¢%(Q[m]) has a multiplicative inverse. How does one compute (2 + \3/2)_1? Consider the polynomial 2 + m. Now a greatest common divisor of 2 + :1: and m3 — 2 is 1, and so, by

Proposition 2.21, there exists polynomials r(a:) and s(:1:) in Q[x] for which (2 + m)r(m) + (x3 — 2)s(m) = 1.

Thus (2 + \3/2)r(\3/2) = 1, so that “(9/2) is the inverse of 2 + \3/2. (In fact,

7W?) = (4 — 25/5 + (C has

kernel ker(qbcp) = (p(a:)) With p(;v) = wp—l + azp—2 + - -- + 3:2 + a‘ + 1. Note that p(a:) is irreducible by Eisenstein’s criterion (Proposition 2.8). We have

Q[a:] / (p(a:)) E 45(1) (QM), and qbcp (QM) consists of all vectors of the form a0 + (1l + 61293 + ”'+ tip—2C5—2 for ai E Q. As a vector space over Q, c341, (QM) has dimension p — 1 With

basis {LC}, ,3, . . .,C£—2}.

Definition 4.1. The field qfia (F [33]) of Proposition 4.1 is the simple algebraic extension of F by 04 and is denoted as F(a). The degree of

F (04) over K, denoted by [F (oz) : F], is the dimension of the vector space F (04) over F. Thus [F (oz) : F] is the degree of the irreducible polynomial irr(a, F) of oz over F. Proposition 4.3. Let F (a) be a simple algebraic extension of F with

[F(a) : F] = n.

Then every element fl in F(a) is a root of a monic

polynomial of degree S n with coeflicients in F. Proof. Since [F (04) : F] = n, there is a smallest integer m g n for Which

the subset {1, 5, B2, . . . , ,Bm} is linearly dependent (cf. Proposition 3.5). Thus, there exist elements a0, a1, . . . ,am of F for Which 5m = 00 + 015 + - - - + am_1,8m_1.

And so, fl is a zero of the polynomial flat) 2 —a0—a1m—- . ‘—an_1a:m—1+a:m With m g n.

E]

In fact, 5 E F (04) is a root of an irreducible monic polynomial over F.

Proposition 4.4. Let F (04) be a simple algebraic extension of F. Let ,8 E

F (oz). Then there exists an irreducible monic polynomial p(ac) E F [3:] for which p(fl) = 0.

Simple Algebraic Extension Fields

147

Proof. Indeed, let €153 : F [x] —> F (04) be the evaluation homomorphism, and take p(x) to be the generator of the principal ideal ker(¢3). I] Proposition 4.5. Let K = F (a) be a simple algebraic extension of F of degree [K : F], and let L = K (5) be a simple algebraic extension of K of

degree [L : K]. Then L = K(fl) = F(a)(,8) is a finite extension of F of degree [L : F] = [L : K][K : F]. We write L = F(a,fl). Proof. Let m = [K : F], n = [L : K]. Then {1,a,a2,...,am_1} is a F—basis for K and that {1,fl,,82, . . .,,8"_1} is an K—basis for L. Clearly, {aiflj}, 0 g i S m — 1, 0 S j S n — 1, is a generating set for L over F, so it remains to show that {aiflj} is linearly independent. Suppose that

Ell—01 23:01 ri,jai,83 = 0 for TM 6 F. Then 23:01 (21:61 rid-a") 59' = 0, and since {1, fl, fl2, . . . ”fin—1} is an F(oz)-basis for L, 2:111 rig-of = 0, for 0 g j g n — 1. It follows that rm 2 0 for all i, j. The result follows.

III

For the remainder this section we assume that K is an intermediate field

Q Q K Q (C, thus any polynomial f (x) E K [x] has coefficients in (C.

Proposition 4.6 (Fundamental Theorem of Algebra). Let f (x) be a non-constant polynomial in C[x]. Then there is a zero of f (x) in (C.

Proof. We choose to omit a proof here, but see [Rotman (2002), Theorem

4.49].

III

Proposition 4.7. Let f (x) be a non-constant monic polynomial of degree

n in C[x]. Then

f(x) = (w — 011W“ — (12) '“(x— an), where a1, a2, . . . , . . . ,an are complex numbers.

Proof. By Proposition 4.6, there exists a root (11 of f (x) in (C. By the Factor Theorem

f(x) = (93 - a1)q1($), for q1(x) E C[x]. Since f (x) is monic, either ql (x) = 1 or q1(x) is nonconstant. Assume that (11 (x) is non—constant. Then by Proposition 4.6 and the Factor Theorem, there exists 042 E C with

f(x) = (w — 041W — a2)q2(x) where either (12 (x) = 1 or (12 (x) is non—constant. Continuing in this manner one obtains

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Fundamentals of Modern Algebra

f(-’B) = (m— al)($—a2)"' (13— an), as required.

El

Proposition 4.8. Let p(a:) be an irreducible monic polynomial of degree

n 2 1 in K [:13] Then the zeros of p(ac) are distinct. Proof. By Proposition 4.7, all of the zeros of p(x) are complex numbers.

Let oz be any one of these zeros and let qba : K [m] —> C be the evaluation

homomorphism defined as f (x) 1—) f (d). Then ker(¢a) = (p(:1:)), and so, p(ac) is the monic polynomial of smallest degree for which a is a zero. By way of contradiction, assume that 04 has multiplicity Z 2. Since K has characteristic 0, the formal derivative p’ (m) is a non—constant polynomial of degree n — 1 < n for which 04 is a root. This contradicts that p(x) is

a polynomial of smallest degree for which p(a) = 0, thus the zeros of p(x) are distinct.

III

NOW, let a,fl E (C, let L = K (a) be the simple algebraic extension of K and let E = L(B) = K(a)(fl) = K(a,fl) be the simple algebraic extension of L. By Proposition 4.5, E is a finite extension of K of degree

[E : L] [L : K]. Remarkably, E is a simple algebraic extension of K. Proposition 4.9. Let E = K (oz, ,8) as above. Then there exists an element

77 E E for which E 2 K07).

Proof. Let q(m) = irr(a, K) be the irreducible polynomial of a.

The

roots of q(x) are distinct (Proposition 4.8), and we may list them as a = a1,a2,...,al. Let r(ac) = irr(,3,K) be the irreducible polynomial of ,8. Again the roots of r(a') are distinct, and we write fl = B1, B2, . . . ,Bk. Since K has characteristic 0, there exists an element t E K for which

taé (oz,—oz)/(fi—Bj) for alli,j, 1 S i S l, 2 Sj S k. Set 77: Oz+tfl. Clearly, K (77) Q K (a, B), so it remains to show the reverse inclusion. To this end, observe that 77 _ tIBj 5A ai:

for all 1 S i S l, 2 S j S k. Note that q(n — tw) E K(n)[a:] and put f(a:) = q(77 — tx). Then f(fl) = q(7] — tfl) = q(a) = 0, and for 2 S j S k, fwj) = q(n—tflj) 75 0 since n—tflj 75 a, for all 1 S i S l, 2 S j S k, and the

ca are the only zeros of q(ar:). It follows that f (ac) and r(ac) share a common factor in K (17) [as] which must be a: — B. Thus ,8 E K (77) Consequently, azn—tflEKM),andsoK(a,,8)§K(n). I]

Simple Algebraic Extension Fields

149

For a field K with Q Q K Q (C, the notions of simple algebraic extension

and finite extension of fields are equivalent.

Proposition 4.10. L /K is a simple algebraic extension if and only if L/K is a finite extension of fields. Proof. It is immediate that the simple algebraic extension L = K (a) is a finite extension of fields. For the converse suppose that L/ K is a finite

extension of fields. Let {b1,b2, . . .,bn} be a K—basis for L. Then L = K (b1)(b2) - - - (bn) and by repeated uses of Proposition 4.9 one arrives at L=K(a) for someaEL. E] Let K be a field with Q Q K Q (C, and let L = K(0z), 04 E (C, be a simple

algebraic extension ofK with q(x) = irr(a, K) and n = deg(q(ac)) = [L : K]. Then by Proposition 4.7 and Proposition 4.8, the roots of q(x) are distinct complex numbers which can be listed as oz 2 041,042, . . .,ozn. For each i, 1 S i g n, there exists an injective homomorphism of rings gizL—HC

defined as gi(oz) = 05,. Note that g, fixes K (of. §2.8, Exercise 68). These homomorphisms are the embeddings of L into the complex numbers. Assume K Q R. If ozz- E R, then g,- is a real embedding, if ozz- E C\R,

then g is a complex embedding. Moreover, if K Q R and 04¢ E (CUR for some 1 g i S n, then there exists an integer j 7E i so that the complex conjugate 07¢ = aj. In other words, zeros in (C\R occur in conjugate pairs. Thus for K Q R, if 7' denotes the number of real embeddings and 3 denotes the number of pairs of conjugate complex embeddings, then n = 7" + 23.

Now, let K be a field with Q Q K Q (C, and let B E L = K(oz), oz 6 C.

By Proposition 4.4 there exists a monic irreducible polynomial

19(1):) 6 K[a:] for which p(,8) = 0; Km) is a finite extension of K of degree 3 = deg(p(a:)) = [K(fl) : K]; K(a) = K(fl)(oz) is a finite extension of K(B) of degree t = [K(a) : K(,8)]. Let 5 = 51, ,32, . . . , BS denote the distinct roots of p(x) corresponding to embeddings Ti'K(/B)_>C7

[Bl—>181}

1 g i g 3.

Proposition 4.11. Let 7',- : Kw) —> (C, 1 g i g 3, denote the set of s embeddings as above. Then each 7',- emtends to a set of t embeddings Tiyj2K(Oz)—>(C, 133525.

150

Fundamentals of Mode’rn Algebra.

Proof. Let (1(30) 2 a0 + on + was2 + - - - + at—1$t_1 + “it,

“2' 6 KW),

be the irreducible monic polynomial of a over K (fl) of degree t = [K(a) : K(B)]. For each i, 1 S i S 3, let q“(:r) = Ti(a0)+Tz-(a1)m+n(a2)x2+~ - ~+Ti(at_1)wt_1+xt,

7,-(aj) E Kwi),

and let ’71-,1,'yi,2, . . . ,7” denote the distinct roots of q”(m) in (C.

Now every element of K (04) /K (B) can be written in the form f (a) where f(;v) =b0+b1x+---+brw’", 7' 2 0, is a polynomial in K(B)[ac]. So for each pair i,j, 1 S i S 3,1S j S t, there is a map 73,]- :K(a) —> (C, defined by

be + bla + - - - + bra" l—> n(bo) + Ti(bl)%,j + ' ' ' + 7450754As one can check these are embeddings which extend Ti.

E]

To illustrate Proposition 4.11, we consider the finite extension of fields Q(\/§, x/g)/Q of degree 4 over Q. An application of Proposition 4.9 shows

that Q(\/§, x/g) = Q(\/§ + x/g). Let fl = \/§ and consider the tower of fields Q Q Q(\/§) Q (Xx/i + x/g). There are two embeddings of Q(\/§) into (C:

71:Q(\/§)—>(C, 7'2 :Q(\/§) —> (C,

fiHfi, fiH —\/§.

Note that the irreducible polynomial of \/§ + x/g over Q(\/§) is

q(ar:) = 3:2 — Zfix — 1. One has q"1(a:) = q(a:) and the roots of q(m) are 7171 2 \/§ + \/§ and 71,2 = \/— — «3. Thus 7'1 extends to two embeddings Q(\/§ + fl) —> (C defined by

T1,1(¢§+¢§) = \mfi, 7'1’2(\/§+ «5) = «5— fl.

Simple Algebraic Extension Fields

151

On the other hand, qT2 (as) = m2 + Zfix — 1, Whose roots are 7271 2 —\/§ + x/g and 72,2 2 —\[ — x/g.

Thus 7'2 extends to two embeddings

QM? + V3) —> (C defined by

News) = —¢§+¢§, T2,2(\/§+ V3) = —\/§— fl.

Proposition 4.12. Let L = K(a), n = [L : K], and let gl,gg,...,gn denote the set of embeddings gz- : L —> (C. Let ,8 E K(a) and let p(:1c) denote the irreducible polynomial of ,8 over K. Then there exists an integer t for which n

p(w)t = H (C for 1 S j g 3. By Proposition 4.11, for each j, 1 g j S 3, there exists a set of

t = [K(a) : K(fl)] embeddings 73,19, 1 g k: g t, that extend 73. For each i, 1 g i S n, there exists a unique pair j, k With 92' = Tc.

Now,

g z

A

m

|

9‘]

q]

”a?

’2? |

:0;

:1 ~

:H-

||

H

Her: — 91.03)): H 111w: — m(5»)

152

Fundamentals of Modem Algebra.

As a consequence of Proposition 4.12 we have [1:121 91(5) 6 K and

22:1 91(5) 6 K for 5 e L. Definition 4.2. The map NormL/K : L —> K defined as

NormL/K(,3) = HQKB), i=1

fl6 L

is the norm map of L over K. The map trL/K : L —> K defined as

trL/K(fl) = 291(3),

[3 E L

i=1

is the trace map of L over K. 4.2

Some Galois Theory

In this section we continue our assumption on K: Q Q K g C, and we consider the splitting field L of an irreducible polynomial p(x) over K. The splitting field L of p(x) is the smallest field extension of K that contains all of the zeros of 19(5):). Necessarily, L is a simple algebraic extension of K. The collection of all automorphisms of L that fix K is a group, called the Galois group of L/K , and to construct this group of automorphisms we will use the extension theorem of §4.1. We give some examples of Galois groups and state the Fundamental Theorem of Galois Theory. *

*

*

Let K be a field Q E K E (C and let p(a:) be a monic irreducible polyno-

mial of degree n over K. Then p(a:) has n distinct roots in (C (Proposition 4.8) which we list as (11,042, . . . , an. Now the finite extension L=K(a1,a2,...,an),

is the smallest field extension of K that contains all of the zeros of p(x). This is the splitting field L/K of p(:1:). A finite extension of fields L/K is a Galois extension if L is the splitting field of some monic irreducible polynomial p(a:) over K. If L/K is a Galois extension, then the set of all automorphisms of L that fix K is a group under function composition called the Galois group of L / K ,

denoted as Gal(L/K) An abelian extension is a Galois extension with an abelian Galois group. Our definition of Galois extension is equivalent to that of other authors, who define a Galois extension L/ K to be a normal,

Simple Algebraic Extension Fields

153

separable extension of K, that is, a Galois extension of K is the splitting field L/ K of some family of polynomials over K which have distinct roots in some algebraic closure of K. Proposition 4.13. Let L/K be a Galois extension of fields with Galois

group G = Gal(L/K). Then [L : K] = |G|. Proof. By definition, L/ K is the splitting field of some monic irreducible

polynomial p(x) E K[x]. Let n1 2 deg(p(x)) and let ,8; be a root of p1 (x) = p(x). Let 7",, : K(fli) —> L, 1 3 i1 3 n1 be the set of n1 embeddings of K(fli) into L that fix K. Let fig be some other root of p(x) not in K(fli) (indeed, if such a root exists) and let ,8; be so that K (,Bé) = K (Biflflg) Let pg (3:) be the irreducible polynomial of [35 over K (,81) Let n2 = deg(p2(x)).

By Proposition 4.11 each Til extends to n2 embeddings 731M : K (/35) —> L, 1 3 i2 S n2 that fix K.

Next, let ,83 be another root of p(x) not in K (e5)

K (35,) = K (flé)(fl3)

Let ,Bé be so that

Let p3(x) be the irreducible polynomial of fig over

K (35) With n3 2 deg(p3(x)). By Proposition 4.11 each Tim-2 extends to n3 embeddings 7317,2713 : K(,B§) —> L, 1 3 i3 3 n3 that fix K. Continuing

in this manner we must have K (,83) = L for some q 2 1. Thus there are n1n2n3 - - -nq = [L : K] automorphisms of L that fix K.

E]

Let L be the splitting field of the monic irreducible polynomial p(x) E

K [x] By Proposition 4.9 there exists a E L With L = K(a). Let q(x) = irr(a,K) of degree m = [L : K]. Let a = 041,012, . . . ,am denote the roots of q(x). As a consequence of Proposition 4.13 the m embeddings g, : L —> L, a +—> a1, 1 S i S m, constitute the elements of Gal(L/K). These embeddings are permutations of the set of roots {a1,a2, . . .,am}, hence

Gal(L / K) is a subgroup of Sm, the symmetric group on in letters. In what follows we compute some Galois groups.

Example 4.1. Let p(x) = x4 — 10x2 + 1 6 Q[x]. We claim that p(x) is an irreducible polynomial over Q. To this end, suppose that p(x) is reducible over Q. By Gauss’ Lemma this implies p(x) is reducible over Z. If p(x) has a linear factor over Z, then necessarily, :tl is a root of p(x), which is not true. Thus p(x) factors into two polynomials each of degree 2:

p(x) = x4 — 10x2 + 1 = (ax2 + bx + c)(dx2 + ex + f), for some a,b,c,d,e,f E Z. But then

154

Fundamentals of Modem Algebra.

ad=1 ae+bd=0 af+be+cd=—10 bf+ce=0 cf=1, whence either 62 = 8, or e2 = 12, both impossibilities.

Now, the zeros of p(x) are: Ot1=\/§+\/§, 042=—\/§—\/§, 013=—\/§+\/§, Oz4=\/§—\/§,

and the splitting field K of p(m) over Q is Q(\/2 + x/3). Thus [K : Q] = 4 = |Gal(K/Q)|. And so, Gal(K/Q) is either isomorphic to Z4 or Z2 X Z2,

cf. §1.4. We can be more precise: The embeddings 71,1, 71,2, 7271, T22 (§4.1) con-

stitute the elements of Gal(K/Q) and so, Gal(K/Q) g Z2 >< Z2. As a subgroup of S4 (With the roots given by their subscripts) Gal(K/ Q) con— sists of

1234 1234

’

1234 4231

’

1234 3214

’

1234 2134

'

Example 4.2. Let K = Q(\/2). The polynomial p(x) = x2 — 2x/2ac — 1 is irreducible over K. (Why?) The roots of p(:c) are a1 = \/2 + V3, a2 = x/2 — x/3, and so the splitting field of p(x) is L = K(\/2 + x/3). One has Gal(L/K) = {711,712} g Z2.

Example 4.3. Let p be a prime number. The polynomial

p(x) =xp_1+acp_2+---+x3+x2+x+1 is irreducible over Q (Proposition 2.8). The roots of p(:L') are 2

3

—1

041 =Cpa 052:9), a3=Cp7"'7ap—1 =91),

Thus the splitting field is K = Q(Cp) and so

(§2.8, Exercise 31).

|Ga1(K/Q)| = p — 1. In fact, the embeddings are given as T; : K —> K, (p +—> (Ii), 1 g 2' g p — 1.

If 7' is a primitive root modulo p, then

Ga1(K/L) = (7,.) g U(zp) g zp_1. Example 4.4. Let p(x) 2 5133—2. By the Eisenstein Criterion, p(x) = 51:3—2 is irreducible over Q.

The roots of p(;r:) are {72, (33/2, ('3? \3/2, and so,

the splitting field of p(;13) over Q is K = Q(C3, \3/2). By Proposition 4.9, K = Q(C3 + (0/2).

Observe that Q g Q(§3) g K.

We shall compute

Simple Algebraic Extension Fields

155

Gal(K/Q) using the Extension Theorem (Proposition 4.11). There are two automorphisms of Q((3) that fix Q defined by

7'1(C3) = C3,

T2(C3) = C3?Now the irreducible polynomial of (3 + \a/i over Q(C3) is

q(x) = m3 — 3C3x2 + 3C§x — 3. We have

(17'1 (x) = m3 — 3C3x2 + 3C§x — 3. The roots of qT1(m) are

C3 + \3/5,

C3 +C3\3/§,

C3 +C§\3/§.

Thus, T1 extends to three automorphisms of K that fix Q defined by

7'1,1(C3 + \3/5) = C3 + {Via 7'1,2(C3 + \3/§) = C3 + C3 {Via 71,3(C3 + \3/§) = (3 + (3 {75. Moreover,

q”(w) = $3 — 343w? + 33.7: — 3, With roots

LH.

Proof. For (i): Suppose H g G. Clearly, K Q LH. Let x, y E LH, and let

h E H. Then h(x + y) = h(x) + h(y) = x + y, and h(xy) = h(x)h(y) 2 xy, and so, LH is a subring of L. Consequently, LH is a subfield of L. We claim that L/ LH is a Galois extension. To this end, write L = K (04) where 04 is a root of an irreducible monic polynomial p(x) over K. By Proposition 4.12,

10013) = H (a? — 9(a))gEG

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157

Put q(:c) = Hh€H(m—h(oz)). Since each h E H permutes the set {h(a)}h€H, the coefficients of q(x) are fixed by H. Thus q($) E LH [:13]. Since one of the factors of q(:c) is ac— 6(a) = x— a, q(a) = 0. Suppose there is a polynomial Mac) over LH of degree smaller than |H| = deg(q(a:)) With r(a) = 0. Then the set {h(oz)}h€ H constitutes distinct zeros of fins), contradicting Propo— sition 2.6. Thus q(x) is irreducible over LH. Since L is the splitting field

of q(x), L/LH is Galois. Moreover, Gal(L/LH) = H. Now assume that H L fixing E which does not fix Q. So oz ¢ LHE. Thus

LHE Q E, and so E = LHE. Hence, by (i) L/E is Galois With group HE. NOW suppose that E/K is a Galois extension With group G’. Define a

map g5 : G —> G’ by the rule ¢(g)(:z:) 2 9(17) for g E G, a: E E. Every element 9 E G is the extension to L of some element 9’ E G’. Thus for

each 9 E G, there exists an element 9' E G’ so that g(x) = g’(a;) for x E E. Thus gb is surjective. Now, for x E E,

¢(9192)(93) = (9192)($)

= 91(g2($)) = 91(9§($))

= gi(gé($)) = ¢(91)(¢(92)($)) = (¢(91)¢(92))(-’B) and so, Q5 is a group homomorphism.

If follows that ker(¢) 8(0) by the rule E I—> HE. We show that LHE = E and HLH = H, thus ‘119 is the identity map on

f(L/K) and @\Il is the identity on 8(G), thus \I1 is a bijection. From (ii) we already know that LHE = E. We show that H = HLH. Since HLH is

the subgroup of G leaving LH fixed, (ii) implies that L /LH is Galois with group HLH. By (i), L/LH is Galois with group H, hence H = HLH. III In the following example we illustrate the fundamental theorem. Example 4.6. Let (7 denote a primitive 7th root of unity and consider the Galois extension K = Q(C7) over Q with Galois group G = Z7“. Since 3 is a primitive root modulo 7, G is generated by 7'3 : K —> K, (7 +—> (59’. Now,

75? : K —> K, ('7 r—> C? generates a normal subgroup H 2 (7'3) of order 2. The fixed field of H is

KH ={x6 K: 73(33) 2x} =Q(C7+C$). Now, K Galois over KH ; K is the splitting field of the monic irreducible polynomial

p = $2 — are» + c?) + 1 over KH and Gal(K/KH) = H. Moreover, KH is Galois over Q; KH is the splitting field of the monic irreducible polynomial

q(X)=:c3+:132 —2x— 1 over Q and Gal(KH/Q) = (7'3)/H g Z3. 4.3

The Ring of Integers

In this section we define the ring of integers R of a simple algebraic extension K = Q(a). We show that every non—zero ideal of R contains a Q—basis for K, specifically, R is a free Z—module of rank n = [K : Q]. We show how to construct the ring of integers of Qfi/fi) and compute the ring of integers of Q(CP), where Q, is a primitive pth root of unity. We employ Dirichlet’s Unit Theorem to construct the group of units of the ring of integers R. Specifically, we compute U (R) where R is the ring of integers of a quadratic field extension and where R is the ring of integers of Q(Cp). *

*

>k

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159

Let K = Q(a) be a simple algebraic extension of Q of degree n. By Proposition 4.3 every element in K is the zero of a monic polynomial of degree 3 n with coefficients in Q. If we restrict coefficients to Z, we deter— mine a special subset of K: An element 5 E K Which is a zero of a monic polynomial with coefficients in Z is integral over Z. The collection of all elements of K which are integral over Z is the integral closure of Z in K. We want to prove that the integral closure of Z in K is a ring and to do this we need a general fact. Proposition 4.15. Let S be a commutative ring that is also a finitely generated module over a commutative ring with unity R. Let S g T be an inclusion of rings with T a commutative ring with unity. Assume that R Q T and that the multiplication RS is the scalar multiplication on S. Assume that for :0 E T, 335' = 0 implies that x = 0. Leta be an element of T for which 045' Q S. Then a is a root of a manic polynomial over R. Proof. Let {b1, b2, . . . , bm} be a generating set for S over R. There exist elements rm- 6 R for which

051 = 71,151 + 71,252 + ° ° ° + T1,mbm, 052 = 72,151 + T2252 + ° ° ° + Tambm,

abm = rm,1b1 + Tm,2b2 + ’ ’ ’ + rm 1 mbm-

Thus,

(Ot — 7'1,1)bl — 71,252 — ”'— 7'1,mbm = 0, —7"2,1bl + (06 — 7'2,2)b2 — ”'— 7‘2,mbm = 0,

—'r'm,1b1 — Tm’gbg — - - ° + (01 — rmjm)bm = 0.

Put C 2 (TM) 6 Matm(R).

Then (OJm — C)B = 0, where B =

(b1,b2,...,bm)T. NOW,

0 = (adj(aIm — C))(aIm — C)B = det(aIm — C)ImB. It follows that det(oJm — C)S' = 0, and so, det(aIm — C) = 0. Hence 0: is a zero of the monic polynomial f (:13) = det (:L'Im — C) E R[:1:] of degree m. D

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Proposition 4.16. Let K be a simple algebraic extension of Q. Then the integral closure of Z in K is closed under the addition and multiplication ofK. Proof. Let 'y, B be integral over Z. We show that 76 is integral over Z. Since 7 is integral over Z, there exists a monic polynomial f (:13) E Z [x] of degree It for which f ('7) = 0. Likewise, there exists a monic polynomial

9(3)) 6 Z[a:] of degree n for which g(,8) = 0. Now let A be the Z—module generated by {1,7,72, . . .,’yk_1} and let

B be the Z—module generated by {1, e, 32, . . ., [an-1}. Let AB be the Zmodule generated by {WW}, 0 g i S k: — 1,0 S j g n — 1. Then ’yBAB Q AB. N0W, AB Q K is an inclusion of rings With the following properties: for a: E K, acAB = 0 implies x = 0, and AB is a finitely generated Z—module. Thus Proposition 4.15 applies to ShOW that 76 is integral over Z. A similar argument shows that 7+6 is integral over Z (see §4.8, Exercise

15).

II]

It follows from Proposition 4.16 that the set of elements of K that are integral over Z is a ring. We define this ring to be the ring of integers of K. We shall usually denote the ring of integers of K as R. The simplest example of a ring of integers is Z: the integral closure of Z in Q is Z (§4.8,

Exercise 16). Let B E K be integral over Z, that is, suppose

a0 + atfl + on? + - - - + am_1flm‘1 + 3’" = 0, for some monic polynomial f(a:) = a0 + alx + - - - am_1mm_1 + mm over Z.

Now for each embedding g,- : K —> (C, 0 = gi(a0 + alfl + a2fl2 + - - - + ant—157,14 + 5m)

= a0 + amt”) + a29t(5)2 + ’ ’ ’ + am—19i(z3)m_l + 9i(fl)mv so that g,(fl) is integral over Z for all i. Proposition 4.17. The norm and trace maps NormK/Q and TrK/Q restrict to the maps NormK/Q : R —> Z and TrK/Q : R —> Z.

Proof. Let 6 E R. Since 91(6) is integral over Z for all i, NormL/K(B) 2 H2121 g,(fl) and TrL/K(fl) = 2:1 g,(,8) are integral over Z by Proposition 4.16. Since these quantities are also in Q, we conclude that NormK/@ (,8)

and TrK/@ (6) are integers.

I]

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161

If ,8 E K is integral over Z, then irreducible polynomial of ,8 has coefficients in Z.

Proposition 4.18. Let K = Q(a), n = [K : Q] and let R denote the ring of integers of K. Let ,8 E R and let p(m) be the irreducible polynomial of ,8

over Q. Then p(:r) E Z[:r]. Proof. Since ,8 is integral over Z, there exists a monic polynomial q(a:) E

Z[a:] With ,8 as a root. NOW, 19(1):) divides q(a:) in Q[:L'], and so, q(a:) =

p(w)s(w) for some 3(510) in Q[x]. By Gauss’ Lemma, there exist polynomials flat), 3(33) over Z, associates of p(a¢) and 3(a) respectively, with q(a3) = p(m)§(x). Comparing leading coefficients, 15(317) and .§(x) must both be monic or the negatives of monic polynomials. Thus p(ac) = 3213010), and hence is in

III

Z [931-

Lemma 4.1. Let K = Q(a), n = [K : Q] and let R denote the ring of integers of K. Every non-zero ideal J of R contains a basis for K over Q.

Proof. Let ,8 E K = Q(a). Then the set {1,8,82,...,8n} is linearly dependent over Q, and so,

a0+a15+a2fl2+“'+an5n =07 for integers a,- E Z, not all 0. Let j be the largest index for Which aj 75 0. Then

a§_1(ao + alfl + a2fl2 + ~-+ ajflj) = 0, and so,

ag—lao + a;_2a1(aj8) + a§_3a2(aj8)2 + ---+ (aj8)j = 0, thus, aj8 is integral over Z. Hence aj8 6 R.

Next, let {b1, b2, . . . , bn} be a basis for K over Q. NOW by the preceding paragraph there exist integers c, for which {any}; Q R. Let a be any non—zero element of J. Then {cibia} Q J is a basis for K/Q.

E]

Proposition 4.19. Let K = Q(a), n = [K : Q] and let R denote the ring of integers of K. Then R is an integral domain with Frac(R) = K. Proof. Clearly, R is an integral domain. Let (15 : R —> K be the inclusion map and let A : R —> Frac(R) be the localization map. By the UMPL

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Fundamentals of Modern Algebra

(Propositon 2.45) there exists a unique inclusion 1b : Frac(R) —> K. Thus

Frac(R) Q K. Since Z Q R, there exists an inclusion (t’ : Z —> Frac(R). Let X : Z —> Q denote the localization map.

By the UMPL, there is a unique

inclusion w' : Q —> Frac(R). Since R contains a Q—basis for K (Lemma 4.1), K Q Frac(R). El Proposition 4.20. Let K be a simple algebraic extension of Q with ring of integers R. Then K = Q(a) for some 04 E R.

Proof. Write K = QM) for some fl 6 K. Now by Proposition 4.19, there exists a non—zero element r 6 R with rfl 6 R. Now K = Q(a) with oz 2 r,8 E R.

E]

Proposition 4.20 says that a simple algebraic extension of Q can be

written in the form K = Q(a) Where oz 6 R, the ring of integers of K. Suppose that [K : Q] = n. By Proposition 4.18, oz is a zero of a monic polynomial of degree n with coefficients in Z. It follows that

Z[a] =ZEBZa$Za2®-~®Zan_l is contained in R. It may be, however, that R is larger than Z [oz], and in certain cases, we can compute the ring of integers precisely. Let m be a square—free integer and let K = Q(\/m). We want to determine which

elements of Qfi/fi) are integral over Z. Note that any a E (QR/R) can be written in the form a + bW c where a, b, c are integers With gcd(a, b, c) = 1. a:

Lemma 4.2. If a, as above, is integral over Z and a 75 0, then either c = 1 or c = 2. Proof. Note that the irreducible polynomial of oz is 2 2a a2 — b2m

p(a:) — ac — ?x + —c2—

Thus by Proposition 4.18, p(m) E Z[:r], that is, 2?“ and “2—232—"1 are in Z. Let p > 2 be a prime number. Then p ’f c. To see this suppose that p | 0.

Then since c | 2a, p | a. Now since 02 | (a2—b2m) and m is square—free, p | b. But this says that gcd(a, b, c) 75 1, a contradiction. So, p f c, and hence

c = 2j for some integer j Z 0. Suppose that j Z 2. Then 23"1 | a, 23"1 | b (again because m is square-free), and thus gcd(a, b, 0) 7E 1. Conclusion: eitherc=10rc=2. I]

Simple Algebraic Extension Fields

163

Proposition 4.21. Let K = (QB/m), where m is a square—free integer. Then:

(i) R = mm] ifm E 2, 3 mod4; (ii) R = mpg/m] ifm E 1 mod4. Proof. For (i), we already know that Z [m Q R, so the problem is to prove the reverse containment. Let 04 E R. Then by Lemma 4.2, oz 2 w with gcd(a, b, c) = 1 and either 0 = 1 or c = 2. If e = 1, then

certainly, 04 E Z [VEL and thus R = Z [m

So we assume that c = 2.

From the proof of Lemma 4.2, we see that 4 | (a2 — b2m), hence a2 E bgm mod 4. Consequently, either both a2 and b2 are congruent to 0 modulo 4, or both a2 and b2 are congruent to 1 modulo 4. In either case, a E bmod 2.

If 0L2,b2 E 0mod 4, then gcd(a,b,c) 75 1, so we conclude that a2,b2 E 1mod 4, whence 1 E mmod 4. Since we’ve assumed that m E 2,3m0d 4, we cannot have c = 2.

For (ii), the condition m E 1 mod4 implies that w E R, and hence Z[1+§/m] Q R. Now suppose that oz 6 R. By Lemma 4.2, a = w with

gcd(a, b, c) = 1 and either 0 = 1 or c = 2. Clearly Oz 6 Z[1+5/fi] if 0 =1. If c = 2 then from (i) we see that a 2 “EM with a E bmod 2. Now, a+§¢fi=a Z, which as one can check is given as

NormK/Q(a + b(1 + \/—23)/2) = (a + b(1 + \/—23)/2)(a + b(1 — \/—23)/2)

= a2 + ab + 6b2,

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Fundamentals of Modem Algebra.

for a, b E Z. By way of contradiction, we assume that T = (r) for some 7' E R. Since 3 E T, 3 = 723 for some .3 E R. Applying the norm map yields

NormK/Q(rs) = NormK/Q(T)NormK/Q(s) = 9.

Thus NormK/Q(r) | 9. Moreover, 1 — \/—23 E T, and so there exists t E R With 1 — V—Z = rt. An application of the norm map yields

NormK/Q(r)NormK/Q(t) = (1 — \/—23)(1 + \/ —23) = 24, and so, NormK/Q(r) = 3. in R other than :l:1 is at PID. Let p be a prime and integers R of K. We first

This is impossible since the norm of any element least 6. Thus T is not principal and R is not a let K = Q(Cp)- We want to compute the ring of prove two lemmas.

Lemma 4.3. R(1 — (p) D Z = pZ. Proof. From the formula 1) = H§;11(1 — (12,) one obtains p = 7°(1 — (p) With

7' = Hfi:21(1—C;,) E R. Thus pZ Q R(1—(MHZ. Note that R(1—CP)I’1Z is an ideal of Z containing the maximal ideal pZ. Thus either pZ = R(1 — (p) D Z or R(1 — (p) D Z = Z. In the latter case, there exists an element a E R With a(1 — (p) = 1. Applying the norm map yields 1 = NormK/Q(a(1— (1’)) = NormK/Q(a)NormK/Q(1 — ('10) 19—1

= NormK/Qm) (1 — cg) 1 7;:

=p-NormK/Q(a),

Where NormK/Q(a) is an integer. This is impossible. Hence pZ = R(1 —

(p) 0 Z.

El

Lemma 4.4. TrK/Q(R(1 — 9)) Q pZ. Proof. Let oz 6 R. Then TrK/Q(O‘(1_Cp)) = 91(0‘(1_Cp))+92(a(1_Cp))+ ' ' ' +9p-1(0‘(1_Cp))

= 91(a)(1—Cp)+92(a)(1—C§)+ ' ' ' +9p—1(a)(1—C£_1) E R(1—(p),

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165

since 9,-(04) are integral elements. Since TrK/Q(a(1 — Cpl) is an integer, one

has TrK/Q(a(1 — Cpl) E R(1 — (p) D Z. Hence TrK/Q(a(1 — Cp)) 6 pZ by Lemma 4.3.

I]

Now we can compute the ring of integers of K = «2(9). Proposition 4.22. Let p be a prime number and let K = QKP).

Then

R = Z [Cp]. Proof. Since (,0 E R, we easily have Z [(1,] g R. The problem is to show the reverse containment. Let a 6 R and write 04 = a0 + al + (124"; + ' ° ° + ap_2q,’_2,

with a0,a1,...,ap_2 E Q. For each i, 0 S i g p— 2, a fill — (p) = (WC—ill — (1)) ‘l‘ MCI—ill — (p) + ' ' ' + ai(1 — Cpl

- - + awe-Ha — C.» + is an element of R(1 — (p). Applying the trace map yields TrK/Q(a—i(l — (20)) = paiBy Lemma 4.4, pai E pZ, and so ai 6 Z. For a different proof, see [Underwood (2011), Proposition 10.2.15].

E]

In fact, we can use the methods of Proposition 4.22 to extend the result. Proposition 4.23. Let p be a prime number, let n 2 1 be an integer and

let K = (@(Cpn). Then R = Z[Cpn]. Proof. See §4.8, Exercise 18.

III

Let K = Q(a) be a simple algebraic extension of Q of degree n. Let R denote the ring of integers in K. We are interested in computing units of

R. In the case K = QKP), R = Z [(1,], p Z 3, we can construct a special set

of units in Z [(10]. Proposition 4.24. Let p be a prime number, p 2 3. For each integer j,

1 Sj SID—2, the element 1+Cp+Cg+---+CIJ, is a unit in Z[Cp]. Proof. Forlfijfip—Z,1+Cp+§§+---+Cghasinverse 1—61)

1— g,“

_1_

k' 1 110+)

1— g,“ = 1 + (5+1 + (304—1) + . . . + (n—l)(j+1),

with k chosen so that k(j + 1) E 1 modp.

I]

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Fundamentals of Modern Algebra

The units constructed in Proposition 4.24 are the circular units in

Z [4p]In general, the structure of the group of units of a ring of integers is given by a classical result of Dirichlet. Proposition 4.25 (Dirichlet’s Unit Theorem). Let K be a finite ea:tension of Q of degree n. Then

U(R) 2 W x Z”+3_1, where r is the number of embeddings of K into IR, and s is the number of pairs of conjugate embeddings of K into (C and W is the subgroup of R>< generated by the roots of unity in R.

Proof. For a proof see [Cassels and Fréhlich (1967), Chapter II, §18], or

[Samuel (2008), §4.4].

III

The factor Z“Ls—1 is a product of r + s — 1 infinite cyclic groups (C. Let B 6 K, let 3 = [Q(fl) : Q]. Let p(a3) be the irreducible polynomial of fl over Q with distinct roots fl = fll, fig, . . . B5 and

corresponding embeddings Tj : Q(,B) —> (C, 1 g j S 3. Let t = [Q(oz) : Q(fl)] and let q(a:) be the irreducible polynomial for oz over Q(fl) with distinct roots 04 = a1,a2, . . .,at.

By Lemma 4.6, the trace map does not depend on the choice of basis,

so choose the basis {fljod}, 0 g j S s — 1, 0 g i S t — 1, for K over Q. Using this basis

tl‘K/QW) = t ‘ 273(5) = 29M), i=1

i=1

which is precisely the trace map given in Definition 4.2.

E]

Proposition 4.29. The trace map defines a symmetric non-degenerate bilinear form on K:

B : K x K —> Q, 300,31) = tr(96:11), for ac,y€K. Proof. This is straightforward if one takes trK/Q : K —> Q defined as

trK/Q = Zi=1 92*

D

Proposition 4.30. Let K = Q(a) be a finite extension of Q with ring of integers R. Then the bilinear form given above restricts to a bilinear form B:R>Z.

Proof. Take TrK/Q = 2221 g,. Let mfl E R. Then 75 E R. Now, by

Proposition 4.29, TrK/Qhfl) = B(’y,fl) E Z.

I]

Now let J be a non—zero ideal of R and let S be the collection of all

bases 8 for K/Q which are contained in J. By Lemma 4.1, 8 is nonempty. For each basis 8 = {b1,b2, . . .,bl} in 8, let

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171

N3=Zb1®Zb2®---69s§=] be the free Z-module with basis 8. We compute disc(N3) with respect to the bilinear form of Proposition 4.29 using the basis {61,b2, . . . , bl}. By Proposition 4.30, disc(N3) is generated by a non—zero integer which we

identify with disc(N3). The collection

{IdiSC(NB)|}zses, is a non—empty set of positive integers, and as such, has a smallest element. Let M = {7721, m2, . . . ,ml} denote a basis in S which corresponds to the

smallest integer |disc(NM)|. Lemma 4.7. J 2 NM.

Proof. We only need to show that J Q NM- Let a E J. Since {m1} is a basis for K/Q, 0: (mm +q2m2+'“+(Ilml, for elements qi E Q. We claim that each qi is an integer. By way of contradiction, let’s assume that qj ¢ Z for some j. Without loss of generality, we can assume that j = 1. Note that q1 = 77 + L for some 77 E Z and L with

0 < L< 1. Set m’l = a—nml andmg =m¢ for2§ i S l. ThenM’ = {mg} is a basis for K/Q which is contained in J.

Let NM; 2 Zm’l 69 2177/2 63 - -- 69 m. Then NM/ Q NM. The matrix

Which multiplies the basis {mi} to give the basis {m2} is Lq2q3...ql 010 0

001

0,

00---01 thus the module index is

[NM : NMI] = LZ. Now by Proposition 3.34

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Fundamentals of Mode’rn Algebra.

disc(NM/) = (L2Z)disc(N,M), which contradicts the minimality of |disc(NM)| since L2 < 1. Thus each q. is an integer, and so J = NMI] The following proposition will imply that R is a Noetherian ring. Proposition 4.31. For any non—zero ideal J of R, R/ J is a finite ring.

Proof. Let a E J 0 2+.

(Why is J D Z+ non—empty?) Since the ring

homomorphism R/(a) —> R/ J is surjective, we only need to show that R/(a) is finite. By Lemma 4.7, J=Zm1®Zm2$~-®Zml,

for some elements mi 6 J. Now, S={a1m1+a2m2+---+alml: 03a.- 3a}

is a set of coset representatives for R/ (a). Note that IS I 2 at < 00, so that

lR/(a)l = al-

'3

Proposition 4.32. Let R be the ring of integers ofK = Q(a). Then R is Noetherian. Proof. We show that R has the ACC. Let [1 Q 12 Q 13 Q

be an

ascending chain of ideals of R. Suppose there is no integer m 2 0 for which Im = Im+1 = Im+2 =

Then In is non—trivial for some n and the

quotient ring R/In has an infinite number of ideals Ij + In for j Z n. This is impossible by Proposition 4.31. D

4.5

Dedekind Domains

In §4.4 we showed that the ring of integers R is Noetherian. In this section we show that R has additional properties: it is a Dedekind domain. We show that if R is any Dedekind domain, then the localization of Rp, P a prime ideal of R, is a PID. From this result we construct the class group of R which is the group of fractional ideals over R modulo the principal fractional ideals over R. We include some examples of class groups of rings of integers.

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173

Let R be an integral domain With K = Frac(R). An element Oz 6 K is integral over R if it is a zero of a monic polynomial with coefficients in

R (cf. §4.3). The set of all elements of K that are integral over R is the integral closure of R in K. The integral domain R is integrally closed if the integral closure of R in K is R. Proposition 4.33. Let R be the ring of integers of a simple algebraic extension K of Q. Then R is an integrally closed Noetherian ring in which each non-zero prime ideal is maximal.

Proof. We already know that R is Noetherian (Proposition 4.32).

By

Proposition 4.19, R is an integral domain with Frac(R) = K. We need to show that R is integrally closed. To this end, let 04 E K be a root of the monic polynomial of degree t over R: f(x) = a0 + alx + agx2 + - - - + at_1xt_1 + xt.

Now, there is an inclusion of rings R[a] Q K with R[a] finitely gener— ated as an R—module (a generating set is {1, 04,03, . . . , cit—1}). Moreover,

aR[a] Q R[oz]. Let {b1,b2, . ..,bn} be a Z-basis for R (which exists by Lemma 4.7).

Now, 04 is an element of the finitely generated Z-module

M = Z[b1,b2,--- ,bn][oz].

One has an inclusion of rings M Q K with

aM Q M and so by Proposition 4.15, a is integral over Z. It follows that a E R. Finally, let J be a non—zero prime ideal. Then by Proposition 4.31, R/ J is a finite integral domain, hence a field by Proposition 2.3. It follows that J is maximal. E] An integral domain that is an integrally closed Noetherian ring in which each non—zero prime ideal is maximal is a Dedekind domain. The ring of integers R is a Dedekind domain but so is the localization Rp. Proposition 4.34. Let R be the ring of integers of a finite extension K of Q. Let P be a prime ideal of R and let Rp be the localization of R at P. Then RP is a Dedekind domain. Proof. We first show that Rp is Noetherian. Let I be an ideal of Rp, then

I = (I D R)Rp, and hence I = JRP for some ideal J of R. Consequently, I is finitely generated over RP since R is Noetherian. Next, let 01 E K be integral over RP. Then oz is a root of the polynomial 3—1$t—1 + wt, + 7-28—1m2 + . _ _ + rt_1 f($) = 7'03—1 + 713—153

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Fundamentals of Modern Algebra

with r, E R, s E R\P. Now g(sa) = 0 Where g(:L') = stf(:1:) E R[:c]. Thus sa is integral over R. Since R is integrally closed, 3a 6 R, whence 06 E Rp and so Rp is integrally closed. Finally, let I be a non—zero prime ideal of Rp. Then I Q PRp since PRp is the unique maximal ideal in Rp. Let J be the non—zero ideal of R

With I = JRP. Then J Q P. But since J is a non—zero prime ideal of R, J is maximal and so J = P. It follows that I 2 PRP and so I is a maximal ideal of Rp.

III

Quite generally, let R be an integral domain With field of fractions K. Then K is an R-module. A fractional ideal over R is a non-zero Rsubmodule J of K of the form J = cI Where c E K X and I is an ideal of R. A principal fractional ideal is a fractional ideal over R of the form

cR for some 0 E KX.

For example, %Z = {n/2 : n E Z} is a fractional ideal over Z, in fact it is principal. Certainly, any ideal I of R is a fractional ideal over R. If I, J are fractional ideals, then I + J and IJ are fractional ideals.

Lemma 4.8. Let R be a Dedekind domain with K = Frac(R). A non-zero submodule J of K is a fractional ideal if and only if it is finitely generated as an R-module. Proof. Let J be a non-zero submodule of K of the form CI, 0 E K X. Then J is finitely generated since I is finitely generated (R is Noetherian). Conversely, suppose J is a non—zero submodule of K Which is finitely generated as an R—module. Write J = q + Rq2 + --+ q for q1,q2,...,ql E K.

There is a generating set for J of the form {a1 /q, a2 /q, . . . , a; /q} for some ai E R, q E R\{0}.

Consequently, J = CI Where c = q‘1 and

I=(a1,a2,...,al).

B

Let R be a Dedekind domain With K = Frac(R) and let J be a fractional ideal over R. Define

J‘1={mEK: nR}, J={a:EK:mJQJ}. Lemma 4.9. J‘1 and J are fractional ideals over R. Proof. Let y be a non—zero element of J. Then J‘1y Q R, hence J‘1 Q y‘lR. Now, J‘1 is a submodule of the finitely generated R—module y‘lR.

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175

Thus J‘1 is finitely generated by Corollary 3.1, and is a fractional ideal by Lemma 4.8. A similar argument shows that J is fractional. E] Here is one reason why Dedekind domains are so important. Proposition 4.35. Let R be a Dedekind domain. Let P be a non-zero prime ideal of R and let RP denote the localization of R at P. Then the unique maximal ideal m 2 PRP of RP is a principal ideal. Proof. Step 1. First note that m is a fractional ideal over RP and so, m = {:I: E K : 93m Q m} is a fractional ideal over RP by Lemma 4.9. We claim that m 2 RP. Certainly RP Q m, so it is a matter of checking the

reverse inclusion. Let a: E m. Then Rp[x] is an Rp—submodule of m. Since Rp is a Noetherian ring (Proposition 4.34) RP [:13] is finitely generated as an Rp—module by Corollary 3.1. Now xRp [:17] Q RP [:17], and so an application of Proposition 4.15 shows that x is integral over RP. Since RP is integrally closed, a: 6 RP. Step 2. We show that m‘1 aé Rp, where m‘1 = {as E K: 5cm Q RP}. To this end, let j be the collection of all non-zero ideals J of RP that

satisfy

.14 7A RP. Then J is non-empty set since (7") E J for any non-zero ’I“ E m. Suppose that .7 has no maximal element. Then (7') is not maximal, and so there is an element J1 in .7 with (7“) C J1. But J1 is not maximal, and so there is an element J2 E J With (1“) C J1 C J2. Continuing in this manner, we create

an ascending sequence of ideals of Rp that does not stop, a contradiction of the ACC of the Noetherian ring RP. So, .7 has a maximal element Which

we denote as M. We claim that M = m. To this end, we show that M is a prime ideal of RP. Let 30,3; 6 RP with my 6 M. Assume that x ¢ M. Since M E .7, there exists an element z E K with z E M‘l, z ¢ RP. We claim that yz 6 RP.

Now, yz(a:Rp + M) Q R and so, if zy 75 RP, then acRp + M E j with M = m + M. Consequently, a: E M, a contradiction. Now 3/2 6 RP, and so, 2(p + M) = yzRP +zM Q RP with z ¢ Rp. Thus the fractional ideal yRp + M satisfies the requirements for membership in .7. Thus M = yRP + M, and s0 y E M. So M is prime and hence maximal by Proposition 4.34. Thus m = M E J and so,m_1 aé RP.

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Step 3. We show that m is principal. Note that m Q m_1m Q RP,

and since m is maximal, either m = m‘ 1 m or m‘ 1 m 2 RP. In the former case, m—1 Q m and thus m‘ 1 = in since m Q m‘l. But m‘1 2 RP by

Step 1, contradicting Step 2. And so, we conclude that m—lm 2 RP. Now consider the ideal 00

N = n m”. n=1

If N 7E 0, then N is a fractional ideal over RP and hence, N = {:13 E K : xN Q N}, is a fractional ideal over RP by Lemma 4.9. Now the method of Step 1 shows that N 2 RP. Observe that m—1 Q N. Thus, R]: Q m‘1 Q RP, and so, m‘1 = RP, a contradiction. Thus N = {0} Which says that there exists an element 7r 6 m for which 77R]: Q m and

77R]: Q m2. Thus 7rm‘1 Q RP; nm‘1 is an ideal of RP with 7rm‘1 Q m. Thus nm‘1 2 RP and so, m = 7e.

El

Proposition 4.36. Let R be a Dedekind domain and let P be a prime ideal of R. Then RP is a PID. Proof. Let PRP 2 7e and let I be a non-zero ideal of RP. For n 2 0,

I7r‘” is a finitely generated Rp—module With I7F” Q In_"_1. Suppose

that I7r‘" = I7r_"_1 for some n. Then 7r_1(I7r_") = In‘", and so 77—1 is integral over RP by Proposition 4.15. Hence 7r‘1 6 RP since RP is inte— grally closed. Now 1 = 7r7r‘1 6 7e, which is impossible. Consequently, there is a strictly increasing sequence ICI7T_1CI7T—2C"

Since Rp is Noetherian, there exists an integer n 2 0 for which I7F” is an ideal of RP and In_"_1 Q RP. Hence 177—" Q a, thus, 177—" 2 RP by Corollary 2.3. It follows that I = 71'a; I is principal. E] Proposition 4.37. Let R be a Dedekind domain and let J be a fractional ideal over R. Then J‘lJ = R.

Proof. Let Jp = JRp, J,:1 = J—lRp. Then Jp and J;1 are fractional ideals over Rp. Since RP is a PID, there exists 0 E K X for which Jp 2 CRp and J;1 = c‘lRp. Thus

.113n = cRpc_1Rp = RP.

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177

Consequently,

JJ—1 2 fl .113n = flRp = R, P

P

by Proposition 3.20.

I]

Proposition 4.38. Let R be a Dedekind domain and let .7(R) denote the collection of fractional ideals over R. Then .7(R) is an abelian group under the binary operation * defined as I * J = IJ, for I, J E .7(R). Proof. Clearly * is associative and commutative. For an identity element, take e = R. Lastly, by Proposition 4.37 we can take J‘1 to be the inverse of J under *. I]

Proposition 4.39. Let R be a Dedekind domain. Let ’P]:(R) denote the collection of principal fractional ideals over R. Then Pf(R) S .7:(P) Proof. Exercise.

D

The quotient group .7:(R) /”PF (R) is the class group of R, denoted as C (R) Given an fractional ideal J over R in K, the left coset J’PF (R) E C (R) is the ideal Class of J. The class group C (R) is a finite abelian group [Samuel (2008), §4.3]; the order of C (R), denoted as hR, is the class number of R. The significance of the class group C (R) is immediate: a Dedekind domain R is a PID if and only if hR = 1. In this sense, the value of hR measures the degree to which R fails to be a PID.

Example 4.10. Let m be a square-free integer, m < 0 and let K = Q(\/m). Then hR = 1 if and only if

m = —1,—2,—3,—7,—11,—19,—43,—67,—163. This result is due to [Stark (1967)]. In the case that m = —5, R = Z[\/—5] is not a PID and we have hR = 2.

In the case that m = —23, R =

Z[(1 + \/—23)/2] is not a PID, as we have seen. In fact, in; = 3. Example 4.11. Let m be a square—free integer 2 S m g 50. Then hR = 1

for the following values of m [Sloane (2015), A003172]:

m = 2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33, 37, 38, 41, 43, 46, 47. In the case that m = 10, one has hR = 2 [Sloane (2015), A094619].

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Example 4.12. Let p Z 2 be a prime number and let K = Q(CP). Then as we have shown, R = Z [(1,]. The complete list of primes for which hR = 1 is:

p = 2, 3,5,7, 11,13,17, 19 [Washington (1997), Theorem 11.1]. 4.6

Unique Factorization of Ideals

In this section we show that each non—zero proper ideal of a Dedekind domain can be factored into a product of prime ideals in a unique way; this factorization generalizes the familiar notion that each positive integer

m E Z factors uniquely into a product primes m = pflpgz - - 102’”. We introduce ramified and unramified primes and the Hilbert class field. Proposition 4.40. Let R be a Dedekind domain, and let J be a non-trivial proper ideal of R. Then J can be factored into a product of prime ideals of R

J=Pf1P;2---P,fk, where e,- are positive integers. Moreover, this factorization is unique up to . 67; a re-ordering of the factors Pi . Proof. In the first part of the proof we find a prime ideal P1 and an ideal J1 for which J = J1P1 with J C J1.

Since R is Noetherian, §2.8, Exercise 46 applies to show that J is contained in a prime ideal P1 of R.

P1 is a fractional ideal of R With

PflPl = R, thus there exists elements qi 6 P1— 1 and ri 6 P1 for which 1 = ELI qiri. Now J g P1 implies P1_1J g R, thus P1_1J is an ideal of

R, with J = JR = J(P1_1P1) = (Pl—1.1)P1. Set J1 = PflJ. Let a E J, then a = 2:21 aqiri, and so, J Q J1. Now, if J 2 J1, then P1J = P1J1 = P1P1_1J = J. Multiplying on the right by J‘1 yields P1 = R. Thus J C J1.

If J1 = R, then we have proved the proposition. If J1 is proper, then since R is Noetherian, J1 is contained in a prime ideal P2, and as above, J1 = Jn, for some J2 with J1 C J2. Now, J = J2P1P2. Again, we are

done if J2 = R. Else, we repeat the process obtaining a collection of ideals

J,J1,J2...

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179

We claim there exists an integer k for which Jk = R and consequently, the process stops with J = P1P2 - - - Pk. Suppose no such k exists. Then there is an ascending chain of ideals JCJ1CJ2CJ3C'”.

Each containment is proper and the chain does not stop; the ACC fails. This is impossible since R is Noetherian. We leave the statement regarding uniqueness of the factorization as an exercise. E] Example 4.13. Let K = Q(\/—5). By Proposition 4.21, the ring of in—

tegers of K is R = Z [V—S]. As one can check, the ideal (2) in R factors uniquely into prime ideals as

(2) = (2,1 + \/——5)2, and the ideal (3) factors uniquely into prime ideals as

(3) = (3,1 + \/—_5)(3, 1 — \/—_5). Moreover, the ideals (1 + \/——5) and (1 — \/——5) factor uniquely as

(1 + ¢——5) = (2,1 + \/—_5)(3, 1 + V75), (1— \/——5) = (2,1 — \/——5)(3,1— \/——5). Observe that the non-zero non-unit element 6 E R factors into irreducible

elements of R in two different ways:

6:2-3, 6: (1+\/——5)(1—\/—_5), and so R is not a UFD (thus R is not a PID). The principal ideal (6) however, does factor uniquely into prime ideals,

(6): (2,1 + \/——5)(2, 1 — \/——5)(3, 1 + \/——5)(3, 1 — \/—_5). Example 4.14. Let K = Q(Cpn). By Proposition 4.23, the ring of integers

of K is R = Z [Cp'n]. From the factorization mp” — 1 = (mil—1)? — 1

=(xP"‘ —1>((wP >P—1+(xp”‘1>P—2+~-+xp +1) 1

n—1

n—1

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Fundamentals of Modern Algebra

we obtain the formula

19:

H

0—93»)-

19912—1, gcd(i,P)=1

In fact, for all i, 1 g i g p” — 1, gcd(i,p) = 1, the principal ideal (1 — C13,.) equals the prinicipal ideal (1 — (p12) which is prime. Thus the ideal (p) in R has unique factorization

(p) = P"‘1. Let K be a finite extension of Q with ring of integers R; let L be a finite extension of K With ring of integers 3. Let Q be a prime ideal of R. The ideal SQ of S factors uniquely into prime ideals of S,

SQ=Pf1P§2~-Pem. m

If e. > 1 for some i, 1 g i g m, then Q is a ramified prime of R. If e, = 1

for all 1 g i S m, then Q is an unramified prime of R. If every prime Q of R is unramified, then L is an unramified extension of K. Proposition 4.41. Let K be a finite extension of Q, [K : Q] > 1, with ring of integers R. Then there exists a prime number p E Z for which 192 is a ramified prime of Z. Proof. The proof is beyond the scope of this book. The interested reader

should consult [Neukirch (1999), Theorem III.2.17].

I]

Consequently, Q is the only unramified extension of Q. Let K be a finite extension of Q. The Hilbert Class Field L of K is the maximal abelian unramified extension of K. The Hilbert Class Field L of K is a Galois extension with Galois group isomorphic to the class group

C (R) of K, see [Ireland and Rosen (1990), Notes, p. 184]. Example 4.15. By Proposition 4.41, the Hilbert Class Field of Q is Q.

Example 4.16. Let K = (ow—5). Then R = Z[\/——5] with Class group C(R) g C2. The Hilbert Class Field of K is L = K (1) With ring of integers S = R[(i + \/—5) / 2]; L is a Galois extension of K With Galois group 02. The notions of UFD and PID are equivalent in Dedekind domains. Proposition 4.42. Let R be a Dedekind domain. Then R is a PID if and only ifR is a UFD.

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181

Proof. Suppose that R is a PID. Then by Proposition 2.30, R is a UFD. To prove the converse, assume that R is UFD. Let J be a non—trivial proper ideal of R. We claim that J is principal. By Proposition 4.40 there exists a unique factorization

J=Pf1P§2-~P,fk, where P1,P2, . ..,Pk are prime ideals of R. So we can show that J is principal by proving that every prime ideal of R is principal. To this end, let P be a non-zero prime ideal of R. Let 7" be a non—zero element of P. Then if (7") = P we are done. Else, assume that (7') C P. From the proof of Proposition 4.40, we conclude that there exists a non—trivial proper ideal

I for which (7') = IP. Let

1=Q§1 32... ft be the factorization of I into prime ideals Q1. Now

(7") = 1362? S2

16‘-

But since R is a UFD, 1" = q1q2---qm for irreducible elements qi E R. By

Proposition 2.24 each ideal (qz) is prime. Thus (7°) 2 (q1)(q2) - - - (qm) is a prime factorization. By uniqueness of prime factorizations, P = (q1) for some 73. Thus P is principal which proves the proposition. E] 4.7

Extensions of Qp

In this final section of the chapter we construct field extensions of the p— adic rationals Qp. Beginning with a simple algebraic extension K/Q with ring of integers R, we use the unique factorization of (p) in R to extend the

p—adic absolute value | |p to an absolute value | | p on K , where P is a prime in the factorization of (p). Analogous to §2.7, we complete K with respect

to the extension | | p, resulting in a finite extension of fields Kp /Qp. *

*

*

Let p be a prime number, let K be a simple algebraic extension of Q of degree n = [K : Q]. Let R be the ring of integers of K. By Proposition 4.40 (p) factors uniquely into a product of prime ideals of R,

(p) = 1.1..s - - - P59.

(4.1)

For each i, 1 S i S 9, we define an absolute value | | pi on K as follows.

Forx=0,let|x|pi=0. Formzr/SEK,T,SER,T7E0,s%0,lettrbe

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the integer tr 2 0 for which (r) Q Pit’", (r) Q Pin—H; let ts be the integer ts Z 0 for which (3) Q Pits, (s) Q PitSl'I, and define |x

1

p, = W

Then as one can verify | lp, is an absolute value on K which extends | [1,. Indeed, |p p, = 1% as required.

As we did with Q endowed with | |p, we ask: does every | [pi—Cauchy sequence in K converge to a limit in K? In other words is K complete with respect to the absolute value | |p,? Not surprisingly, the answer is “no”. Proposition 4.43. Let K/Q be a finite extension of degree n and suppose that | | p, is an extension of the p-adic absolute value. The K is not complete

with respect to | | p,. Proof. We construct a | | pi—Cauchy sequence in K that does not converge to a limit in K. There exists a prime q yé p for Which q > n and integers x1 and a so that x‘i’ E amodp, a sé OmOdp, a is not a qth power in Q. Then p(x) = xq — a is irreducible over Q. Using the method of Proposition 2.49

we construct a | | pi-Cauchy sequence {3:1, x2, x3, . . .} in K that converges

to a limiting value 04 that is a root of p(x). Now if oz 6 K, then [K : Q] 2 q > n, a contradiction. We conclude that K is not complete with respect to l l“

[I

We form the completion of K With respect to | |p, using the recipe of §2.7: First, we construct a field extension of K defined as

K131. = CPi/NP“ where Cp, is the set of all | | pi-Cauchy sequences {xn} in K and N13,. = {{xn} 6 C12,: lim xn = 0}. 77,—)00

We then show that the absolute value | | pt. extends uniquely to an absolute value on Kp, Which we also denote as | | p,. Next, we show that K is dense in Kp,, and finally, we show that Kp, is complete with respect to | | p,. The field Kp, is the completion of K With respect to | |p,.. Proposition 4.44. Let K be a simple algebraic extension of Q of degree

n = [K : Q]. Letp be a prime number and let (p) = P161 262 - - ~Pgeg be the prime factorization of (p). For each i, 1 S i S g, the completion Kp, is a finite extension of Q1) of degree ni. We have the formula .9

i=1

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183

Proof. Write K = Q(a) for some 04 E K and let p(a:) be the monic irre— ducible polynomial for oz. Let p(x) 2 H221 pj (:L') be the factorization of p(:r) into monic irreducible polynomials over Qp. Let nj 2 deg(pj (33)). We have

K 6% Qp % Qlwl/(p(w)) 6% Qp % Qplxl/(ptvD l g H Q10 [11:] / (pj (33))

by Proposition 2.19

j=1 l =

j=1

Li:

(4'2)

Where Lj = Q10 [an] / (pj (33)). Each Lj is a finite extension of Qp of degree nj since the set of left cosets

{1 + WW)?!” + (pj(w)),w2 + (MW), - - - ’30?” + (Pj($))} is a QP—basis for Lj. Note that the absolute value I pi restricted to Q is

the p—adic absolute value | lp. Thus the completion of Q With respect to

| | 121. is the completion of Q With respect to | |p, and this is precisely Qp. Thus passing to the completion With respect to | | 132. on the left hand side of (4.2) yields the field Kpi. And so, when we pass to the completion on the right hand side of (4.2), we must also obtain a field. Thus all but one of the factors on the right hand side, say, Lj must vanish upon passing to the completion With respect to | lpi. This says that for each i, 1 g i S 9, Kpi = Lj for some j, 1 g j g I. It follows that the collections {Li} and {Kpi} coincide; g = l. Thus Kp1. is a finite field extension of Qp of degree m = [K1:1. : Qp]. We have n = l m, as required. E]

The degree 7n = [Kpi : Qp] is the local degree of K at Pi. Since oz 6 K is a root of p(ac) E Q[a:], 0: is a root of pj (:10) E Qp[x] for exactly one j, 1 g j g l, say j*. Let K133* be the completion corresponding to j*. Now, oz 6 Kpjfig7 and so, Kp], is a simple algebraic extension of Qp through the

evaluation homomorphism gba : Qp[x] —> K13,“ The subset RPi : {m E K131:

lmlpi S 1}

is a subring of Kpi called the valuation ring of Kpi. The group of units in Rm is U(Rpi) = {:66 Kpi: |m

P11 : 1};

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R131. is a local ring with maximal ideal mis-Rpi ={xEKpiz la: P1. < 1}. Proposition 4.45. RE. is a PID. Proof. It is enough to show that the maximal ideal mi is principal. Observe that Rpi = K D Rpi and so, R121. is a subring of Rpi. Now, PiRpi is

principal in Rpi by Proposition 4.35, thus mi 2 PiRpi is a principal ideal in Rpi.

El

An element 71';- 6 R131. with mi = (m) is called a uniformizing param-

eter for Kpi. The “local form” of the factorization (4.1) is

(p) = m? = (we: Note that Im|pi = 1%. Proposition 4.46. Every element of a: E Kp1. can be written as

a: 2 any, for some unit it E U(Rpi) and some n E Z U 00; an element at E Rpi can be written as

with n 2 0.

Proof. Let at E Kpi. If x = 0, then a: = 1 - W30. So assume that :1: 75 0. Then |x| p1. = 1% for some integer n. NOW the element «[71:10 is so that

lflfnxPi=|7TfnPimP.-=P

n/ei; = 1

1311/61

Thus 77—”:10 = u for some unit in RP” whence a: = turf”. Note that x 6 R131. if and only if n 2 0. E] Let a: E Kpi. The element n E Z Uoo for which at 2 air? is the m-order

of x, denoted as ordm(:c). Proposition 4.47. Rpi is an integral domain with Frac(Rpi) = Kp. 1

Proof. Clear.

I]

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185

Example 4.17. Let K = Q(i); one has irr(i, Q) = x2 + 1. By Proposition 4.21, R = Z[i], a PID. Let p be a prime number with p E 1mod 4. Then the unique factorization of (p) is (p) I P1P27

where P1 = (a— bi), P2 = (a+bi) for some a, b E Z, cf. [Underwood (2011), §1.2]. In this case 9 = 2, e,- = 1 for 2' = 1, 2, and there are two prime ideals

lying above (p). Thus there are exactly two extensions of | |p to K: | |p, and | | p2. By Proposition 4.44, the polynomial .702 + 1 factors over QP as

$2+1 = (x—i)(x+i). Thus

K ®Q Qp E mad/($2 + 1) ®Q Qp

E @p[x1/ Kp. Example 4.18. Let K = Q(Cpn), where (pn denotes a primitive pnth root of unity. The irreducible polynomial of (pr: over Q is

10(1L')=(a:10n_1)p_1 -}-(:17pn_1)10_2 + - - - + mph—1 + 1. The ring of integers of K is Z [Cpn]. From Example 4.14, we have the unique factorization

(p) = P"‘1. In this case, 9 = 1, el = p""1(p — 1) = [K : Q]. There is exactly one prime

ideal P = (1 — Cpn) lying above (p) and so there is exactly one extension ||pof| |ptoK. The polynomial p(x) remains irreducible over Qp; one has

K ®Q Qp g QM/ (1900)) ®Q Qp % QpM/ (10(30):

(4-3)

186

Fundamentals of Mode’rn Algebra.

thus Kp g Qp(Cpn); Kp is a simple algebraic extension of Qp since Cpn E Kp; the evaluation homomorphism is $9.7» : Qp[.’L'] —> Kp. The extension of Qp to Kp has local degree [Kp : Q10] 2 [K : Q] = Ian—1(1) — 1). We have RP 2 Zpn]. The unique maximal ideal (1 — Cpn)Zp[Cpn] is principal, generated by 7r; the local factorization is

(p) = P”‘1. 4.8

Exercises

Exercises for §4.1 (1) Let 04 = 1 + V 1 + x/i E R and let qba : Q[a:] —> R be the evauation homomorphism.

(a) Find the monic polynomial that generates ker(¢a).

(b) Compute [Q(a) : Q] and find a basis for Q(a) over Q. (2) Show that R(i) = (C. (3) Let K = Q(C3,C5).

(a) Find an element 77 6 (C for Which K = Q(7)).

(b) Show that 4 3 [K : Q] g 8. (4) Let L be a simple algebraic extension of the field K. Prove the following formulas.

(a) NormL/K(xy) = NormL/K(m)NormL/K(y), for {13,3/ 6 L. (b) NormL/K(r) = rlL‘K], for 7" E K. (c) TrL/K(r;1; + y) = rTrL/K(m) + TrL/K(y), for 7“ E K, m,y E L. (5) Let K = Q(\/§+ x/g). Compute trK/Q(\/§+ x/g) and trK/Q(\/§). (6) Let K = Q(C6). Compute irr(C6,Q) and calculate NormK/Q((6),

TrK/Q(C6)-

(7) Let K = com) and let 5 = (7 + c? e K.

(a) Compute irr (C as an extension of an embed—

ding ow) —> C. (8) Let p be a prime number, let K be a finite extension of Q and suppose that a E K is not a pth power in K, that is, suppose there is no b E K

for which bp 2 a. Let p(x) 2 mp — a.

Simple Algebraic Extension Fields

187

(a) Show that the roots of p(:r) are

«a, w, (3%,..., 1. By Proposition 5.1 there exists an

invented root 04 of qj (x) in some field extension L/F . Over L, f (x) factors

as

f (x) = (11(x)q2(w) ' ‘ ‘qj—1(-’v)(-’E — a)r(w)qj+1 ' ' ' (Mm), for r(x) E L[x]. Put 9(3)) = Q1($)Q2($) ' ' 'qj—1(90)7'(90)qj+1 ' ' @1430)-

Finite Fields

193

Then deg(g(:r)) = d — 1. By the induction hypothesis there exists a field extension E/ L so that that g(:I;) factors into a product of linear factors in E [x] Since f (ac) = g(a:) (x — a), f (:13) factors into a product of linear factors in E [as] I] Proposition 5.2 says that E/F contains d roots of f (as) which may or may not be distinct. These are the only possible zeros of f (x) since a degree d polynomial over a field can have at most d roots in the field (Proposition

2.6). 5.2

Finite Fields

Let p be a prime number and let n 2 1 be an integer. In this section we prove the existence of a field with exactly p" elements. We show that there is essentially only one field with p” elements, which we call the Galois field

GF(pn) of order p”. If f (x) is an irreducible polynomial of degree k over GF(pn) and 04 is an invented root of f (m), then GF(pn)(oz) is the Galois

field GF(pnk). We define the order of a polynomial f (:10) E GF(p")[ac]; a primitive polynomial over GF (pn) is an irreducible polynomial whose order

is maximal. We show that if f (x) is primitive over GF(pn) and 04 is a root of f (ac), then a generates the cyclic group GF(p”k)x. *

*

*

A finite field is a field with a finite number of elements. By §2.8, Exercise 3, we know that Z1) is a finite field with exactly p elements for each prime number p. It turns out that the number of elements in any finite field is always a power of a prime number. Proposition 5.3. Let F be a field with a finite number of elements. Then F is isomorphic to a simple algebraic extension of Zp for some prime number p. Consequently, |F| = p” where n is the degree of the simple algebraic extension of Zp.

Proof. Since F is finite, r = char(F) > 0, hence by Corollary 2.6, F contains a subring B isomorphic to Zn. Henceforth, we identify B with Zr. Since F is a field, r must be a prime number p, hence F contains the field Zp. As F is finite, it is certainly a finite dimensional vector space over Zp, with scalar multiplication Zp X F —> F given by multiplication in F. Thus

F = ZP G3 ZP EB - - - EB Zp, where n = dim(F), whence |F| = p". \—\,—/ n

194

Fundamentals of Mode’rn Algebra.

By Proposition 2.15, the group of units of F, F X, is cyclic of order p” — 1; let oz generate F X. Thus apn_1 = 1, that is, 01 is a root of the polynomial

Mac) 2 xpn_1 — 1 E Zp[x]. We have the evaluation homomorphism ¢a : Zp[x] —> F with ker(¢a) = (q(:13)), Where q(:1:) is the polynomial of smallest degree m in Zp[a:] for which 04 is a root. Then q(x) = irr(a, Zp), so that Zp[:r] /(q(;c)) g Zp (a) is a simple algebraic extension of Zp.

Let it : Zp[x]/(q(x)) —> F be defined by f(:c) + (q(ac)) +—> f(a). Since 9(a)) — f(x) 6 (q(:r)) implies that 9(a) = f(a), ¢ is well—defined on cosets. Clearly, it is onto, and since Zp[a:]/(q(m)) is a field, it is 1—1. Hence ¢ is an isomorphism of fields. The set {1,352, . . . ,Em_1} is a Zp—basis for

Zp[cc]/((q(w)) and so, |F| 2 pm. But we already know that |F| = p”, hence m = n.

El

Let p be a prime number and let n 2 1 be an integer. In the case n = 1, there exists a field With exactly p1 = p elements, namely, Zp. We now show that this is true for n > 1: we show that there exists a field With exactly

1)" elements for all n 2 1. Consider f (ac) = asp” — a: E Zp[x]. Then by Proposition 5.2 there exists a field extension E/Zp which contains p” zeros

of f (ac) (counting multiplicities). These constitute all of the zeros of f (as) Proposition 5.4. The zeros of f(a:) = mp” — x in E are distinct. Proof. Let F = {on}, 1 g i g p" be the set of roots of f (cc), and suppose that some root a,- has multiplicity 2 2. Then f’(ai) = 0. But this is

impossible since f’(:z:) = —1 in Zp[ac].

E]

Proposition 5.5. Let F = {on}, 1 g i S p", be the set of roots of f(:r) 2 mp” — 51:. Then F is a field, with operations induced from E. Proof. By Corollary 2.1 the elements of Zp are roots of f (.73) Thus Zp Q F and char(F) = p. Let ahaj E F. Then (on + cry)?” 2 a?1 + a? = oz, +aj. (The first equality follows from §2.8, Exercise 77.) Thus F is closed

under addition. Moreover, (—04,)1’" = (—1)Pnoz§n 2 —oz,-, since (—1)?

11.

=

—1 modp by Corollary 2.1. Hence F is an additive subgroup of E. Also, (aiozj )P" = a?” a?” = aiaj, so that F is closed under multiplication. Thus F is a commutative ring with unity. For 04,- E F non—zero, (xi—1 6 E. But

also, (04171)?" = 04,71. Thus F is a field.

C]

By Proposition 5.5, there is a field F of order p", consisting of the p” distinct roots of mp” — a). The following proposition shows that there is

Finite Fields

195

essentially only one finite field of order p”. Proposition 5.6. Let F1, F2 be finite fields of order p”. Then F1 E F2. Proof. We Show that F1 g F, Where F is the field constructed in Propo— sition 5.5 consisting of the roots of xpn — :1: E Zp [:13] By Proposition 5.3, F1

is isomorphic to a simple algebraic extension of Zp, Zp[x] /(q(m)) g Zp (a), Where deg(q(a:)) = n, and a is a root of both q(:v) and 33(513pn—1 — 1) 2

at?" — ac. Hence oz 6 F. Define (b : Zp[x]/(q(m)) —> F by the rule f (w)+(q(x)) +—> f (04). Then is well—defined on cosets (Q5 is a function). Moreover, qfi is 1—1 since Zp[x]/(q(m)) is a field, and onto since both Zp[x]/(q(ac)) and F have the same number of elements. Thus (15 is an isomorphism. It follows that F1 g F. Similarily, F2 2 F, which proves the result. III The unique (up to isomorphism) finite field of order p" is called the

Galois field of order p” and is denoted by GF(p"). We next discuss polynomials over the Galois field GF(p").

Proposition 5.7. Let f(x) 6 GF(p")[x] with deg(f(m)) = k 2 1. Assume that f (0) 79 0. Then there exists an integer t, 1 S t g pnk — 1, for which

f(w) I act — 1.

Proof. First note that the quotient ring GF(p”) [a3] / (f (x)) contains pnk — 1 left cosets other than (f (33)) The collection {at2 + (f (x)) : i = 0, 1, 2, . . . ,pnk — 1} is a set of pnk left cosets not containing (f(m)). Thus

90% + (f(~’0)) = $9 + (flail), for some i,j, 0 S i < j S pnk — 1. Since f(0) 3E 0, gcd(x,f(m) = 1. Thus

by Proposition 2.21 there exists polynomials r(a:), 8(x) E GF(p")[a:] so that mr(a:) + f(a:)s(a:) = 1.

Consequently, xi + (f(x)) is a unit in GF(p")[x]/(f(x)). Set t = j — i. It follows that act 6 1+(f(ac)) and hence f(:r) | act—1 With 1 S t g pnk—l. D The smallest positive integer e for which f (:13) | we — 1 is the order of

f(a:), denoted as order(f(ac)). For example, over GF(2), x4 + a: + 1 has order 15 and x4 + x3 + x2 + a: + 1 has order 5; over GF(9) = GF(3) (04), a2 + 1 = 0, ac — oz has order 4 since

5104 — 1 = (:1:— 1)(ac+ 1)(w— oz)(ac+oz) over GF(9). Proposition 5.8. Let f (:13) be an irreducible polynomial in GF(p")[m] of degree k. Then f(:c) divides mpnm — x if and only ifk | m.

196

Fundamentals of Modern Algebra

Proof. Suppose that f(:1:) divides mpnm — 3:. By Proposition 5.1, there exists a field extension E/ GF (1)") that contains a zero or of f (m) Thus

04”

— oz 2 0, and so, 04 E GF(p”m). Moreover, the elements of GF(pn)

are precisely the zeros of mp” — :r and any zero of asp” — at is also a zero of

at?

nm

— 51:. Thus, F = GF(p")(oz) is a subfield of GF(p”m). Note that F has

pm“ elements.

Now let B be a generator of the cyclic group GF(10"m)>< and let q(:1:) be

the irreducible polynomial of ,8 over F. Let t = deg(q(x)). Then F (B) = GF(p"m). NOW F(B) has pm“t elements and so pnl‘“t 2 pm”. Hence kt = m, that is, k | m. For the converse, suppose that k | m. Let a be a zero of f (51:) in some extension field E/GF(p"). Then GF(p")(a) is a field With 10"]6 elements, and oz satisfies the relation apnk — 04 = 0. Let s be so that ks = m. Then apnks — 04 = 0, hence, ozpmn — a = 0. Consequently, oz is a root of

a)? m — :13. It follows that xpnm — a: is in (f (a7)), the kernel of the evaluation

homomorphism qba : GF(p")[:r] —> E, and so f(x) | xpnm — 3:.

III

Proposition 5.9. Let f (x) be an irreducible polynomial in GF(pn)[:1:] of degree k. Let a be a zero of f (x) in an extension field E/G’F(p”). Then the zeros of f(m) are of the form a,apn,ap2n,...,ap(k_1)n. Moreover, the zeros are distinct.

Proof. We already know that a E E is one zero of f (:0)

Since the char—

acteristic of GF(p") is p,

flap”) = new” = o, for 1 g j S k — 1. NOW suppose that apan = apbn for integers 0 S a < b g k — 1. Then

oz

pn(k+a—b)

= (0,1)“ )1) n(k—b) =(ozp nb )p n(k—b) =ozp nk =04.

Thus, f (m) divides mp

n(k+a—b)

— :13, and so by Proposition 5.8, k divides k: + a — b, Which is impossible. It follows that the zeros of f (ac) are distinct. I]

Proposition 5.10. Let f (x) be an irreducible polynomial in GF(pn)[a:] of degree 1:. Let 04 be a zero of f (ac) in an extension field E/ GF(p”). The smallest field extension containing all of the zeros of f (x) is GF(p”)(oz), which is isomorphic to the Galois field GF(pnk). Proof. By Proposition 5.9, GF(p")(a) contains all of the zeros of f (3:)

Clearly, |GF(pn)(a)| 2 pm“.

GF(p"k).

Now, by Proposition 5.6, GF(p”)(oz) g

El

Finite Fields

197

Let f (ac) be an irreducible polynomial in GF(p”)[.r] of degree k 2 1, and let oz be a zero of f (x) As we have seen in Proposition 5.10, GF(p")(oz) contains all of the roots of f (x)

The following proposition computes

order(f(ac)). Proposition 5.11. With f(:r) as above, order(f(ac)) equals the order of any root of f(:s) in the group of units of GF(p")(oz). Proof. Observe that GF (19")(0z)>< is cyclic of order pnk — 1, generated by some element fl. Put oz 2 ,8l for some integer l. Now a typical zero of f (x) can be written filpmn for 0 S m S k — 1. By Proposition 1.34,

|< has prime order for k = 3, 5, 7. We give some examples of Galois fields of order p" and provide examples of primitive and non-primitive polynomials. Example 5.1. Let p = 5, n = 1. The Galois field GF(5) is the finite field Z5; Z5 contains all of the distinct roots of x5 — :1: E Z5[.r] by Corollary 2.1. Indeed, over Z5,

3:5 — x 2 37(37 — 1)(a: — 2)(a‘ — 3)(:r — 4). Observe that 2 has order 4 in Z; and 4 has order 2 in 25X. Hence order(m —

2) = 4 and order(:r — 4) = 2; a; — 2 is a primitive polynomial over Z5. Example 5.2. Let p = 3, n = 2. Then GF(9) consists of all of the roots of $9 — ac. In the UFD Z3[.r], x9 — a: factors into irreducible elements as:

51:9 — :1: = ac(ac — 1)(:L' + 1)(ac2 + 1)(ac2 — x — 1)(a:2 +50 — 1).

198

Fundamentals of Mode’rn Algebra.

We take oz to be a root of $2 + 1 (the other root is 043 = 204). Thus GF(9) g Z3[w]/(x2 + 1) g Z3(a). A Z3-basis for Z3(a) is {1,oz}. The 9 elements of GF(9) are thus: 0=0-1+0-a, a=0-1+1-a, 2a=0-1+2-a, 1=1~1+0-a, 1+a=1~1+1-a, 1+2a=1~1+2~a, 2=2~1+0-a, 2+a=2~1+1-a, 2+20z=2~1+2-oz.

Both 04 and a3 have order 4 in GF(9)X, and so order(ac2 + 1) = 4. Thus :32 + 1 is a non-primitive polynomial over GF(3). On the other hand, the

root fl of m2 — :17 — 1 E GF(3)[$] has order 8 in GF(9)X, and so (172 — :1: — 1 is a primitive polynomial over GF(3).

Example 5.3. In this example, we take GF(9) = GF(3)(a), a2 + 1 = 0, as our base field. Let f (m) = x2 + a: + B 6 GF(9)[£B] With B as in Example 5.2. Then one checks directly that f (x) is irreducible over GF(9). By Proposition 5.9, the (distinct) roots of f(a:) are 7,79; GF(9)(*y) = GF(81). We have

(w-7)(w—79)=w2+w+fi, so that 710 = 5. Since 5 has order 8 in GF(9), 7 has order 80 in GF(81)X. Thus f (as) is primitive over GF(9). The finite fields GF(2"), n 2 1, can be applied to computer science since their base field GF(Z) = {0,1} represents the collection of binary

digits (bits).

Example 5.4. Consider GF(16). The polynomial x16 — a: E Z2[w] factors into irreducibles as

ac(a: + 1)(;1r:2 + :12 + 1)(ac4 + a: + 1)($4 +w3 + 1)(ac4 +x3 +132 + a: + 1) (5.1)

Finite Fields

199

So GF(16) can be constructed as (S‘rF(2)[:I:]/(ar;4 + 36+ 1) = GF(2)(a), Where 04 is a root of $4 + :1: + 1. A GF(2)—basis for GF(16) is {1, 04,042, a3} and so the 16 elements are

0=0-1+0-a+0-a2+0-a3, a3=0-1+0-a+0-a2+1-a3, a2=0-1+0.a+1-a2+0-a3, a2+a3=0.1+0.a+1-a2+1-a3,

a=0.1+1.a+0-a2+0.a3, a+a3=0-1+1-a+0-a2+1-a3, a+a2=0-1+1-a+1-a2+0-a3, a+a2+a3=0-1+1-a+1-a2+1-a3, 1=1~1+0~a+0-a2+0-a3, 1+a3=1.1+0.a+0.a2+1.a3, 1+a2=1.1+0.a+1.a2+0.a3, 1+a2+a3=1.1+0.a+1-a2+1.a3, 1+a=1.1+1.a+0-a2+0.a3, 1+a+a3=1.1+1.a+0.a2+1-a3, 1+a+a2=1-1+1-a+1-a2+0-a3, 1+a+a2+a3=1-1+1-a+1-a2+1-a3.

In fact, f(x) = x4+x+ 1 is a primitive polynomial over GF(2): From the factorization (5.1), f (as) is irreducible. By Proposition 5.11, order( f (33)) = 3

or 5 or 15. But clearly, f(x) 1’ m3— 1 and f(x) (.735— 1, thus order(f(a:) = 15 Which says that f (as) is primitive. GF(16) is the field consisting of all possible half—bytes (strings of 0’s and 1’s of length 4). The addition is given by bit-Wise addition modulo 2 and the multiplication is induced by the relation (14 = 1 + 04. For example, 0110 + 1100 = 1010, and

0110- 1001 = 1100,

since (a + a2)(1 + a3) = 1 + oz. The analogous bit—string multiplication in GF(28) = GF(256) is used in the construction of the symmetric key cryptosystem AES (the Advanced

Encryption Standard), see [Mao (2004), §7.7].

200

5.3

Fundamentals of Modem Algebra.

Linearly Recursive Sequences

In this final section we define kth—order linearly recursive sequences over an arbitrary field K and give some examples, including the arithmetic se— quence, the geometric sequence and the Fibonacci sequence. We specialize to homogeneous sequences and construct the matrix A of a homogeneous sequence and the characteristic polynomial of the sequence. From the Cayley—Hamilton Theorem, we deduce that the minimal polynomial of A is the characteristic polynomial of A. We next consider kth—order linearly recursive sequences {Sn} over the

Galois field GF(pm). We prove that if the characteristic polynomial f (:13)

of {Sn} is primitive over GF(pm), then {3n} has maximal period pmk — 1. *

*

*

Definition 5.1. Let K be a field and let k > 0 be a positive integer. A kth—order linearly recursive sequence in K is a sequence {3n} for which

3n+k = ak—18n+k—1 + ak—2Sn+k—2 + ' ' ' + aosn + a

(5-2)

for some elements a,a0,a1,a2, . . .,ak_1 E K and all n 2 0. The relation

(5.2) is the recurrence relation of the sequence. The linearly recursive sequence {3n} is homogeneous if a = 0. The

sequence {3n} is eventually periodic if there exist integers N 2 0, t > 0 for which sn+t = 5n, for all n 2 N. The sequence {3”} is periodic if {Sn} is eventually periodic with N = 0, that is, {Sn} is periodic if there exists an integer t > 0 so that sn+t = 3,, for all n 2 0. Suppose {Sn} is eventually

periodic. Then the smallest positive integer 7' for which 3",.” = 37, for all

n 2 N is the period of {Sn}. For n 2 0, the vector Sn 2 (3n, sn+1, sn+2, . . . , sn+k_1) is the nth state

vector of {37,}; so = (30,31,32,...,sk_1) is the initial state vector. A linearly recursive sequence is completely determined by specifying the recurrence relation (5.2) and initial state vector. Here are two basic examples of lst—order linearly recursive sequences. Let so = (so) for some 30 E K and let sn+1 = 3n +a

for a E K. Then 81 = 30+a, 32 = 31+a = 30+2a, and so on. The

resulting sequence {37,} is the arithmetic sequence with initial term so and common difference a.

Note that a formula for the nth term of the

Finite Fields

201

sequence is 3,, = so + na. For another example, let s0 = (so) be the initial state vector and let 3n+1 = 0/0311,

for some a0 6 K. Now the sequence is 30, $1 = a030, .32 = agsl = a330,

and so on. This is the geometric sequence With initial term so and ratio a0. The formula for the nth term of the geometric sequence is Sn 2 soag.

Perhaps the most well-known 2nd—0rder linearly recursive sequence is the sequence attributed to Fibonacci. Let so = (so, 31) be the initial state

vector and put 3n+2 = 3n+1 + 3n

for n 2 0. Then the sequence {37,} is the Fibonacci sequence. If K = Q and the initial state vector is so = (0,1), then the Fibonacci sequence is

0,1,1,2,3,5,8,13,21,... (The Fibonacci sequence appeared in Leonardo of Pisa’s Liber Abaci (1202),

but was known centuries earlier in India, see [Singh (1985)].) Homogeneous linearly recursive sequences can be described in terms of matrices. Let {37,} be a homogeneous kth—order linearly recursive sequence

With recurrence relation (5.2). Put 0 1 0--0 01---

A:

0 0

0 0

1 0

0 1

: 0 0 00 0 0-

a0 a1 a2

ak—2 ale—1

Let MT denote the transpose of a matrix M. Proposition 5.12. With A defined as above,

55 = Ansg for all n 2 0. Proof. Use induction on n. The trivial case is n = 0: $31 = Iksg. For

the induction hypothesis, assume that s£_1 = fin—1831. Then As£_1 = AA”_lsg, hence SE = Ansg. I]

202

Fundamentals of Modern Algebra

The matrix A is the matrix of the homogeneous linearly recursive sequence. Let {3”} be a homogeneous kth—order linearly recursive sequence

with matrix A. The characteristic polynomial of {3”} is the characteris— tic polynomial of A in the usual sense, that is, the characteristic polynomial

of {Sn} is

f(x) = det(:I:Ik — A) Where Ik denotes the k X k identity matrix. Proposition 5.13. Let {3n} be a homogeneous lath-order linearly recursive

sequence defined by (5.2). Let f (at) be the characteristic polynomial of {Sn}. Then f(w)=a:k—ak_1w k1 _ —ak_2;v k2 _ —---—ao€K[x]. Proof. We proceed by induction on the order k of the linearly recursive sequence. The trivial case is k = 1. The lst—order linearly recursive se— quence {Sn} with sn+1 = aosn, n 2 0, has associated matrix A = (a0).

Clearly, the characteristic polynomial f (m) = det(x11 — A) = a; — a0. Thus the trivial case holds. For the induction hypothesis, we assume that a (k — 1)th-order linearly

recursive sequence {3”} With matrix 010--001---

B:

0 0

0 0

:::

:

000--000---

1 0

0 1

big—3 bk_2

b0 b1 b2 has characteristic polynomial

g = dew/31,64 — B) = xk—l _ W—z _ W-s _ . .. _ b0. Now for a given kth—order linearly recursive sequence {Sn}, one has

k_A:

x

—1

0

0

0

0

a:

—1

0

0

.

.

.

.

.

0

0

0

—1

0

0

0

0

at

—1

—ao —al —a2

-

_ak—2 90 — ails—1

Finite Fields

203

Computing det(acIk —A) by cofactor expansion about the first column yields

det(a:Ik — A) = :1;-(—1)1+1-det(M1,1)+(—ao)-(—1)k+1-(—1)k_1 = a: ' det(M1,1) — a0,

Where M1,1 is the 1,1—minor matrix of $116 — A.

Of course, M14 is the

(k — 1) X (k — 1) matrix of the form k_1 — B With 0

B:

1

0-~

0

0

0 0

1

0

0

:

:

.

.

.

0 0 0-~

1

0

0 0 0---

0

1

:

a1 a2 a3

,

ale—2 ale—1

and so by the induction hypothesis,

f(a:) = det(mIk — A) = a: - det(a:Ik_1 — B) — a0 = x( mic—1 — ak_1x k _ 2 — ak_2x k _ 3 2 wk _ ak_1xk—1 _ ak_2xk—2 _

The matrix of the Fibonacci sequence {3”} is

“(31) and the characteristic polynomial is f (x) = x 2 — a: — 1. From Proposition 5.12, one has 0 1 1 1

n

so

_

3n

31

—

Sn+1

.

Proposition 5.14. Let K be a field with char(K) yé 5.

Then f(ac) =

1:2 — :1: — 1 E K [:17] has distinct roots in some extension field L/ K . Proof. By Proposition 5.2, there exists a field extension L/K so that f(x) factors into linear factors

f(:v) = (a: — a1)(m — a2), Where 041,012 are the zeros of f (m) Since gcd(f (x), f’ (517)) = 1, these zeros are distinct.

I]

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Fundamentals of Mode’rn Algebra

In fact, if a is one zero of f (m), then 1 — 01 is the other zero since

(l—oz)2—(1—a)—1=0. Using some elementary linear algebra, we can obtain an explicit formula for the Fibonacci sequence.

Proposition 5.15. Let K be a field with char(K) 75 5. Let {37,} be the Fibonacci sequence in K with initial state vector so = (so, 31). Let 04 be a

zero of f (ac) = x2 — :1: — 1 in some extension field L/K . Then

s. 2 3 own-loan — 1) + (1 — eon-1a — 2a) 0< fornZ 0.

+31