A First Course in Rings and Ideals 0201007312, 9780201007312

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A First Course in RINGS AND IDEALS

DAVID M. BURTON University of New Hampshire

...

~~

ADDISON-WESLEY PUBIJSmNG COMPANY

Reading, Massachusetts · Menlo Park, California · London · Don Mills, Ontario

This book is in the ADDISON-WESLEY SERIES IN MATHEMATICS Consulting Editor: Lynn H. Loomis

Standard Book Number 201.00731.2 AMS 1968 Subject Classifications 1610, 1620. Copyright ©1970 by Add1son-Wesley Publishing Company, Inc Philippines copyright 1970 by Addison-Wesley Publishing Company, Inc. All rights reserved. No part of this publication may be reproduced. stored in a retrieval system, or transmitted, in any form or by any means. electromc. mechanical, photocopymg, recording, or otherwise, without the prior wntten permission of the publisher. Prmted in the United States of America. Published simultaneously in Canada. Library of Congress Catalog Card No. 73-100855.

To my Father Frank Howard Burton

To my Father Frank Howard Burton

vi

PREFACE

of topics not treated in the body of the text, as well as impart additional information about material covered earlier; some, especially in the later chapters, provide substantial extensions of the theory. We have, on the whole, resisted the temptation to use the exercises to develop results that will subsequently be needed (although this is not hard and fast). Those problems whose solutions do not appear straightforward are often accompanied by hints. The text is not intended to be encyclopedic in nature; many fascinating aspects of this subject vie for inclusion and some choice is imperative. To this end, we merely followed our own tastes, condensing or omitting entirely a number of topics that might have been encompassed by a book of the same title. Despite some notable omissions, the coverage should provide a firm foundation on which to build. A great deal of valuable criticism was received in the preparation of this work and our moments of complacence have admitted many improvements. Of those students who helped, consciously or otherwise, we should like particularly to mention Francis Chevarley, Delman Grapes, Cynthia Kennett, Kenneth Lidman, Roy Morell, Brenda Phalen, David Smith, and John Sundstrom; we valued their critical reading of sections of the manuscript and incorporated a number of their suggestions into the text. It is a pleasure, likewise, to record our indebtedness to Professor James Clay of the University of Arizona, who reviewed the final draft and offered helpful comments leading to its correction and improvement. We also profited from many conversations with our colleagues at the University of New Hampshire, especially Professors Edward Batho, Homer Bechtell, Robb Jacoby, and Richard Johnson. In this regard, special thanks are due Professor William Witthoft, who was kind enough to read portions of the galleys; his eagle-eyed attention saved us from embarrassment more than once. We enjoyed the luxury of unusually good secretarial help and take this occasion to express our appreciation to Nancy Buchanan and Solange Larochelle for their joint labors on the typescript. To my wife must go the largest debt of gratitude, not only for generous assistance with the text at all stages of development, but for her patience and understanding on those occasions when nothing would go as we wished. Finally, we should like to acknowledge the fine cooperation of the staff of Addison-Wesley and the usual high quality of their work. The author, needless to say, must accept the full responsibility for any shortcomings and errors which remain.

Durham, New Hampshire January 1970

D.M.B.

CONTENTS Chapter I

Introductory Concepts .

Chapter 2 Ideals and Their Operations

16

Chapter 3 The Classical Isomorphism Theorems .

39

Chapter 4 Integral Domains and Fields

52

Chapter 5 Maximal, Prime, and Primary Ideals

71

Chapter 6 Divisibility Theory in Integral Domains

90

Chapter 7 Polynomial Rings

112

Chapter B Certain Radicals of a Ring .

157

Chapter 9 Two Classic Theorems

180

Chapter 10 Direct Sums of Rings

204

Chapter 11

217

Rings with Chain Conditions

Chapter 12 Further Results on Noetherian Rings .

234

Chapter 13 Some Noncommutative Theory

262

Appendix A. Relatio•.

287

Appendix B. Zorn's Lemma

296

Bibliography

300

Index of Special Symbols

303

Index .

305

vii

CONVENTIONS Here we shall set forth certain conventions in notation and terminology used throughout the text: the standard symbols of set theory will be employed, namely, E, u, n, -, and 0 for the empty set. In particular, A - B = {xlx E A and x rf= B}. As regards inclusion, the symbols ~ and 2 mean ordinary inclusion between sets (they do not exclude the possibility of equality), whereas c and => indicate proper inclusion. When we deal with an indexed collection of sets, say {A; liE I}, the cumbersome notations u {A; liE I} and n {A; liE I} will generally be abbreviated to u A; and n A;; it being understood that the operations are always over the full domain on which the index is defined. Fallowing custom, {a} denotes the set whose only member is a. Provided that there is no risk of confusion, a one-element set will be identified with the element itself. A function f (synonymous with mapping) is indicated by a straight arrow going from domain to range, as in the case f: X -.. Y, and the notation always signifies that f has domain X. Under these circumstances,/ is said to be a function on X, or from X, into Y. In representing functional values, we adopt the convention of writing the function on the left, so that j(x), or occasionally fx, denotes the image of an element x E X. The restriction of f to a subset A of X is the function f!A from A into Y defined by U!A)(x) = f(x) for all x in A. For the composition of two functions f: X -.. Y and g: Y-.. Z, we will write go f; that is, g "f: X -.. Z satisfies (go f)(x) = g(f(x)) for each x EX. (It is important to bear in mind that our policy is to apply the functions from right to left.) Some knowledge of elementary number theory is assumed. We simply remark that the term "prime number" is taken to mean a positive prime; in other words, an integer n > 1 whose only divisors are ± 1 and ±n. Finally, let us reserve the symbol Z for the set of all integers, Z + for the set of positive integers, Q for the set of rational numbers, and R# for the set of real numbers. viii

ONE

INTRODUCTORY CONCEPTS The present chapter sets the stage for much that follows, by reviewing some of the basic elements of ring theory. It also serves as an appropriate vehicle for codifying certain notation and technical vocabulary used throughout the text. With an eye to the beginning student (as well as to minimize a sense of vagueness), we have also included a number of pertinent examples of rings. The mathematically mature reader who finds the pace somewhat tedious may prefer to bypass this section, referring to it for terminology when necessary. As a starting point, it would seem appropriate formally to define the principal object of interest in this book, the notion of a ring. A ring is an ordered triple (R, +, ·) consisting of a nonempty set R and two binary operations + and · defined on R such that

Definition 1-1.

1) (R, +) is a commutative group, 2) (R, ·) is a semigroup, and 3) the operation · is distributive (on both sides) over the operation +.

The reader should understand clearly that + and · represent abstract, unspecified, operations and not ordinary addition and multiplication. For convenience, however, one invariably refers to the operation + as addition and to the operation · as multiplication. In the light of this terminology, it is natural then to speak of the commutative group (R, +) as the additive group of the ring and of (R, ·) as the multiplicative semigroup of the ring. By analogy with the integers, the unique identity element for addition is called the zero element of the ring and is denoted by the usual symbol 0. The unique additive inverse of an element a E R will hereafter be written as -a. (See Problem 1 for justification of the adjective .. unique".) In order to minimize the use of parentheses in expressions involving both operations, we shall stipulate thar multiplication is to be performed before addition. Accordingly, the expression a·b + c stands for (a·b) + c and not for a ·(b + c). Because of the general associative law, parentheses

2

FIRST COURSE IN RINGS AND IDEALS

can also be omitted when writing out sums and products of more than two elements. With these remarks in mind, we can now giVe a more elaborate definition of a ring. A ring (R, +, ·) consists of a nonempty set R together with two binary operations + and · of addition and multiplication on R for which the following conditions are satisfied : 1) 2) 3) 4) 5) 6)

a + b = b +a, (a + b) + c = a + (b + c), there exists an element 0 in R such that a + 0 = a for every a E R, for each a E R, there exists an element -a E R such that a + (-a) 0, (a·b)·c = a·(b·c), and a·(b + c) = a·b + a·c and (b + c)·a = b·a + c·a,

where it is understood that a, b, c represent arbitrary elements of R. A ring (R, +,·)is said to be afinite ring if, naturally enough, the set R of its elements is a finite set. By placing restrictions on the multiplication operation, several other specialized types of rings are obtained. Definition 1-2. 1) A commutative ring is a ring (R, +. ·) in which multiplication is a commutative operation, a·b = b·a for all a. bE R. (In case a·b = b·a for a particular pair a, b, we express this fact by saying that a and b commute.) 2) A ring with identity is a ring (R, +, ·) in which there exists an identity element for the operation of multiplication, normally represented by the symboll, so that a·l = l·a = a for all a E R.

Given a ring (R, +, ·) with identity 1, an element a E R is said to be invertible, or to be a unit, whenever a possesses a (two-sided) inverse with

respect to multiplication. The multiplicative inverse of a is unique, whenever it exists, and will be denoted by a- 1 , so that a· a- 1 = a- 1 ·a = 1. In the future, the set of all invertible elements of the ring will be designated by the symbol R*. It follows easily that the system (R*, ·) forms a group, known as the group of invertible elements. In this connection, notice that R* is certainly nonempty, for, if nothing else, 1 and -1 belong toR*. (One must not assume, however, that 1 and -1 are necessarily distinct.) A consideration of several concrete examples will serve to bring these ideas into focus.

Example 1-1. If Z, Q, R# denote the sets of integers, rational, and real numbers, respectively, then the systems (Z,

+, ·),

(Q,

+, ·),

(R#,

+, ·)

are all examples of rings (here, + and · are taken to be ordinary addition and multiplication). In each of these cases, the ring is commutative and has the integer 1 for an identity element.

INTRODUCTORY CONCEPTS

3

Example 1-2 Let X be a given set and P(X) be the collection of all subsets of X. The symmetric difference of two subsets A, B ~ X is the set A L\ B, where A A B = (A - B) u (B - A). If we define addition and multiplication in P(X) by

A

+

B = A A B,

A· B

=

A n B,

then the system (P(X), +, ·) forms a commutative ring with identity. The empty set 0 serves as the zero element, whereas the multiplicative identity is X. Furthermore, each set in P(X) is its own additive inverse. It is interesting to note that if X is nonempty, then neither (P(X), u, n) nor (P(X), n, u) constitutes a ring. Example 1-3. Given a ring (R, +, · ), we may consider the set M,.(R) of n x n matrices over R. If I, = {1, 2, ... , n}, a typical member of M,.(R) is a function f: I, x I, -J R. In practice, one identifies such a function with its values a;i = f(i,j), which are displayed as the n x n rectangular array (

~. 11

.. •

~. 1n )

a, 1

•••

a,,

(aii E R).

For the sake of simplicity, let us hereafter abbreviate the n x n matrix whose (i,j) entry is aii to (aiJ). The operations required to make (M,(R), +.·)a ring are provided by the familiar formulas and where ciJ

= L" a1t·bti· /c=1

(We shall often indulge in this harmless duplication of symbols whereby + and · are used with two different meanings.) The zero element of the resulting ring is the n x n matrix all of whose entries are 0; and -(aii) = ( -aii). The ring (M,(R), +.·)fails to be commutative for n > 1. It is equally easy to show that if (R, +, ·) has an identity element 1, then the matrix with 1's down the main diagonal (that is, au = 1) and O's elsewhere will act as identity for matrix multiplication. In terms of the Kronecker delta symbol t5ii• which is defined by t)ii =

lOfl ifif ii ==!= jj

(i, j = 1• 2, · •• ' n),

the identity matrix can be written concisely as (J;i).

4

FIRST COURSE IN RINGS AND IDEALS

Example 1-4. To develop our next example, let X be an arbitrary (nonempty) set and (R, +,·)be a ring. We adopt the notation map(X, R) for the set consisting of all mappings from X into R; in symbols,

{!If: X

map(X, R) =

--. R}.

(For ease of notation, let us also agree to write map R in place ofmap(R, R).) Now, the elements of map(X, R) can be combined by performing algebraic operations on their functional values. More specifically, the pointwise sum and product off and g, denoted by f + g and f" g, respectively, are the functions which satisfy (f

+

g)(x) = f(x)

+

g(x),

(x eX).

(f·g)(x) = f(x)·g(x),

It is readily verified that the above definitions provide map(X, R) with the structure of a ring. We simply point out that the zero element of this ring is the constant function whose sole value is 0, and the additive inverse - f off is characterized by the rule (-f)(x) = - f(x). Notice that the algebraic properties of map(X, R) are determined by what happens in the ring (R, +, ·) (the set X furnishes only the points for the pointwise operations). For instance, if (R, +, ·) has a multiplicative identity 1, then the ring (map(X, R), +, ·) likewise possesses an identity element; namely, the constant function defined by 1(x) = 1 for all x e X.

Example 1-5. Our final example is that of the ring of integers modulo n, where n is a fixed positive integer. In order to describe this system, we first introduce the notion of congruence: two integers a and b are said to be congruent modulo n, written a = b (mod n), if and only if the difference a - b is divisible by n; in other words, a b (mod n) if and only if a - b = kn for some k e z. We leave the reader to convince himself that the relation "congruent modulo n" defines an equivalence relation on the set Z of integers. As such, it partitions Z into disjoint classes of congruent elements, called congruence classes. For each integer a, let the congruence class to which a belongs be denoted by [a] :

=

[a] = {x e Zlx

=a (mod n)}

={a+ knlkeZ}. Of course, the same congruence class may very well arise from another integer; any integer a' for which [a'] = [a] is said to be a representative of [a]. One final, purely notational, remark: the collection of all congruence classes of integers modulo n will be designated by z,. It can be shown that the congruence classes [0], [1], ... , [n - 1] exhaust the elements of Z,.. Given an arbitrary integer a, the division algorithm asserts that there exist unique q, r e Z, with 0 s; r < n, such that a = qn + r. By the definition of congruence, a r (mod n), or

=

CONVENTIONS Here we shall set forth certain conventions in notation and terminology used throughout the text: the standard symbols of set theory will be employed, namely, E, u, n, -, and 0 for the empty set. In particular, A - B = {xlx E A and x rf= B}. As regards inclusion, the symbols ~ and 2 mean ordinary inclusion between sets (they do not exclude the possibility of equality), whereas c and => indicate proper inclusion. When we deal with an indexed collection of sets, say {A; liE I}, the cumbersome notations u {A; liE I} and n {A; liE I} will generally be abbreviated to u A; and n A;; it being understood that the operations are always over the full domain on which the index is defined. Fallowing custom, {a} denotes the set whose only member is a. Provided that there is no risk of confusion, a one-element set will be identified with the element itself. A function f (synonymous with mapping) is indicated by a straight arrow going from domain to range, as in the case f: X -.. Y, and the notation always signifies that f has domain X. Under these circumstances,/ is said to be a function on X, or from X, into Y. In representing functional values, we adopt the convention of writing the function on the left, so that j(x), or occasionally fx, denotes the image of an element x E X. The restriction of f to a subset A of X is the function f!A from A into Y defined by U!A)(x) = f(x) for all x in A. For the composition of two functions f: X -.. Y and g: Y-.. Z, we will write go f; that is, g "f: X -.. Z satisfies (go f)(x) = g(f(x)) for each x EX. (It is important to bear in mind that our policy is to apply the functions from right to left.) Some knowledge of elementary number theory is assumed. We simply remark that the term "prime number" is taken to mean a positive prime; in other words, an integer n > 1 whose only divisors are ± 1 and ±n. Finally, let us reserve the symbol Z for the set of all integers, Z + for the set of positive integers, Q for the set of rational numbers, and R# for the set of real numbers. viii

6

FIRST COURSE IN RINGS AND IDEALS

between a congruence class and its smallest nonnegative representative; under this convention, Zn = {0, 1, ... , n - 1}. It is perhaps worth commenting that, since Z 1 = Z, a number of texts specifically exclude the value 1 for n. Although it is logically correct {and often convenient) to speak of a ring as an ordered triple, the notation becomes unwieldy as one progresses further into the theory. We shall therefore adopt the usual convention of designating a ring, say {R, +, · ), simply by the set symbol R and assume that + and · are known. The reader should realize, however, that a given set may perfectly well be the underlying set of several different rings. Let us also agree to abbreviate a + {-b) as a - b and subsequently refer to this expression as the difference between a and b. As a final concession to brevity, juxtaposition without a median dot will be used to denote the product of two ring elements. With these conventions on record, let us begin our formal development of ring theory. The material covered in the next several pages will probably be familiar to most readers and is included more to assure completeness than to present new ideas. Theorem 1-1. If R is a ring, then for any a, b, c E R 1) Oa = aO = 0, 2)a{-b)={-a)b= -(ab), 3) {-a){ -b) = ab, and 4) a(b - c) = ab - ac, (b - c)a = ba - ca. Proof These turn out, in the main, to be simple consequences of the distributive laws. For instance, from 0 + 0 = 0, it follows that Oa = (0

+ O)a =

Oa

+ Oa.

Thus, by the cancellation law for the additive group (R, +),we have Oa = 0. In a like manner, one obtains aO = 0. The proof-of (2) requires the fact that each element of R has a unique additive inverse (Problem 1). Since b + (-b) = 0, ab + a(-b) = a(b +(-b))= aO = 0, which then implies that - (ab) = a(- b). The argument that (- a)b is also the additive inverse of ab proceeds similarly. This leads immediately to (3): (-a)( -b) = -( -a)b = -·( -(ab)) = ab. The last assertion is all but obvious. There is one very simple ring that consists only of the additive identity 0, with addition and multiplication given by 0 + 0 = 0, 00 = 0; this ring is usually called the trivial ring.

INTRODUCTORY CONCEPTS

7

Corollary. Let R be a ring with identity 1. If R is not the trivial ring, then the elements 0 and 1 are distinct. Proof Since R =I= {0}, there exists some nonzero element a E R. If 0 and 1 were equal, it would follow that a = a1 = aO = 0, an obvious contradic-

tion. CONVENTION: Let us assume, once and for all, that any ring with identity contains more than one element. This will rule out the possibility that 0 and 1 coincide. We now make several remarks about the concept of zero divisors (the term "divisors of zero" is also in common use): Definition 1-3. If R is a ring and 0 =I= a E R, then a is called a left (right) zero divisor in R if there exists some b =I= 0 in R such that ab = 0 (ba = 0). A zero divisor is any element of R that is either a left or right zero divisor. According to this definition, 0 is not a zero divisor, and if R contains an identity 1, then 1 is not a zero divisor nor is any element of R which happens to possess a multiplicative inverse. An obvious example of a ring with zero divisors is Z", where the integer n > 1 is composite; if n = n 1 n2 in z (0 < nl, n2 < n), then the product nl ·,.n2 = 0 in z". For the most part, we shall be studying rings without zero divisors. In such rings it is possible to conclude from the equation ab = 0 that either a = 0 or b = 0. One can express the property of being with or without zero divisors in the following useful way. Theorem 1-2. A ring R is without zero divisors if and only if it satisfies the cancellation laws for multiplication; that is, for all a, b, c E R, ab = ac and ba = ca, where a f 0, implies b = c. Proof Suppose that R is without zero divisors and let ab = ac, a =I= 0. Then, the product a(b - c) = 0, which means that b - c = 0 and b = c. The argument is the same for the equation ba = ca. Conversely, let R satisfy the cancellation laws and assume that ab = 0, with a =I= 0. We then have ab = aO, whence by cancellation b = 0. Similarly, b =I= 0 implies a = 0, proving that there are no zero divisors in R.

By an integral domain is meant a commutative ring with identity which has no zero divisors Perhaps the best-known example of an integral domain is the ring of integers; hence the choice ofterminology. Theorem 1-2 shows that the cancellation laws for multiplication hold in any integral domain. The reader should be warned that many authors do not insist on the presence of a multiplicative identity when defining integral domains; and

ONE

INTRODUCTORY CONCEPTS The present chapter sets the stage for much that follows, by reviewing some of the basic elements of ring theory. It also serves as an appropriate vehicle for codifying certain notation and technical vocabulary used throughout the text. With an eye to the beginning student (as well as to minimize a sense of vagueness), we have also included a number of pertinent examples of rings. The mathematically mature reader who finds the pace somewhat tedious may prefer to bypass this section, referring to it for terminology when necessary. As a starting point, it would seem appropriate formally to define the principal object of interest in this book, the notion of a ring. A ring is an ordered triple (R, +, ·) consisting of a nonempty set R and two binary operations + and · defined on R such that

Definition 1-1.

1) (R, +) is a commutative group, 2) (R, ·) is a semigroup, and 3) the operation · is distributive (on both sides) over the operation +.

The reader should understand clearly that + and · represent abstract, unspecified, operations and not ordinary addition and multiplication. For convenience, however, one invariably refers to the operation + as addition and to the operation · as multiplication. In the light of this terminology, it is natural then to speak of the commutative group (R, +) as the additive group of the ring and of (R, ·) as the multiplicative semigroup of the ring. By analogy with the integers, the unique identity element for addition is called the zero element of the ring and is denoted by the usual symbol 0. The unique additive inverse of an element a E R will hereafter be written as -a. (See Problem 1 for justification of the adjective .. unique".) In order to minimize the use of parentheses in expressions involving both operations, we shall stipulate thar multiplication is to be performed before addition. Accordingly, the expression a·b + c stands for (a·b) + c and not for a ·(b + c). Because of the general associative law, parentheses

INTRODUCTORY CONCEPTS

9

Example 1-7. The set Ze of even integers forms a subring of the ring Z of integers, for 2n - 2m = 2(n - m) E Ze, (2n)(2m) = 2(2nm) E Ze.

This example also illustrates a fact worth bearing in mind: in a ring with identity, a subring need not contain the identity element. Prior to stating our next theorem, let us define the center of a ring R, denoted by cent R, to be the set cent R

=

{a

E

Rlar

=

ra for all r E R}.

Phrased otherwise, cent R consists of those elements which commute with every member of R. It should be apparent that a ring R is commutative if and only if cent R = R. Theorem 1-4. For any ring R, cent R is a subring of R. Proof. To be conscientious about details, first observe that cent R is nonempty; for, at the very least, the zero element 0 E R. Now pick any two elements a, b in cent R. By the definition of center, we know that ar = ra and br = rb for every choice of r E R. Thus, for arbitrary r E R., (a - b)r

=

ar - br

=

ra - rb

=

r(a - b),

which implies that a - bE cent R. A similar argument affirms that the product ab also lies in cent R. In the light of Theorem 1-3, these are sufficient conditions for the center to be a subring of R. It has already been remarked that, when a ring has an identity, this need not be true of its subrings. Other interesting situations may arise.

1) Some subring has a multiplicative identity, but the entire ring does not. 2) Both the ring and one of its subrings possess identity elements, but they are distinct. In each of the cited cases the identity for the subring is necessarily a divisor of zero in the larger ring. To justify this claim, let 1' =I= 0 denote the identity element of the subring S; we assume further that 1' does not act as an identity for the whole ring R. Accordingly, there exists some element a E R for which a1' =/:: a. It is clear that (a1')1'

= a(l'l') = a1',

or (a1' - a)1' = 0. Since neither a1' - a nor 1' is zero, the ring R has zero divisors, and in particular 1' is a zero divisor. Example 1-8. To present a simple illustration of a ring in which the second of the aforementioned possibilities occurs, consider the set R = Z x Z,

10

FIRST COURSE IN RINGS AND IDEALS

consisting of ordered pairs of integers. One converts R into a ring by defining addition and multiplication componentwise: (a, b)

+ (c, d) =

(a

+

c, b

+

d),

(a, b)(c, d) = (ac, bd).

A routine calculation will show that Z x (0} = ((a, O)la E Z) forms a subring with identity element (1, 0). This obviously differs from the identity of the entire ring R, which turns out to be the ordered pair (1, 1). By our previousremarks,(l, O)mustbeazerodivisorinR;infact,(l, 0)(0, 1) = (0, 0), where (0, 0) serves as the zero element of R. If R is an arbitrary nng and n a positive integer, then the nth power a" of an element a E R is defined by the inductive conditions a 1 = a and a" = a"- 1 a. From this the usual laws of exponents follow at once:

a"am = a"+m, (a")m = a"m

(n, mE Z+>·

To establish these rules, fix m and proceed by induction on n. Observe also that if two elements a, bE R happen to commute, so do all powers of a and b, whence (ab)" = a"b" for each positive integer n. In the event that R possesses an identity element 1 and a- 1 exists, negative powers of a can be introduced by interpreting a-" as (a- 1 )", where n > 0. With the definition a0 = 1, the symbol a" now has a well-defined meaning for every integer n (at least when attention is restricted to invertible elements). Paralleling the exponent notation for powers, there is the notation of integral multiples of an element a E R. For each positive integer n, we define the nth natural multiple na recursively as follows: l'a = a

and

l)a

na = (n -

+ a,

when

n

> 1.

If it is also agreed to let Oa = 0 and ( -n)a = - (na). then the definition of na can be extended to all integers. Integral multiples satisfy several identities which are easy to establish: (n

+

m)a

=

na

+

ma,

(nm)a = n(ma), n(a

+

b)

= na + nb,

for a, b E R and arbitrary integers n and m. In addition to these rules, there are two further properties resulting from the distributive law, namely, n(ab) = (na)b = a(nb),

and

(na)(mb) = (nm)(ab).

Experience impels us to emphasize that the expression na should not be regarded as a ring product (indeed, the integer n may not even be a member of R); the entire symbol na is just a convenient way of indicating

INTRODUCTORY CONCEPTS

11

a certain sum of elements of R. However, when there is an identity for multiplication, it is possible to represent na as a product of two ring elements, namely, na = (nl)a. To proceed further with our limited program, we must first frame a definition. Definition 1-5. Let R be an arbitrary ring. If there exists a positive integer n such that na = 0 for all a E R, then the smallest positive integer with this property is called the characteristic of the ring. If no such positive integer exists (that is, n = 0 is the only integer for which na = 0 for all a in R), then R is said to be of characteristic zero. We shall write char R for the characteristic of R. The rings of integers, rational numbers, and real numbers are all standard examples of systems having characteristic zero (some writers prefer the expression "characteristic infinity"). On the other hand, the ring P(X) of subsets of a fixed set X is of characteristic 2, since 2A

= A .::\ A

= (A -

A) u (A - A)

= ¢

for every subset A £: X. Although the definition of characteristic makes an assertion about every element of the ring, in rings with identity the characteristic is completely determined by the identity element. We reach this conclusion below. Theorem 1-5. If R is any ring with identity 1, then R has characteristic n > 0 if and only if n is the least positive integer for which nl = 0. Proof If char R = n > 0, then na = 0 for every a E Rand so, in particular, n1 = 0. Were ml = 0, where 0 < m < n, it would necessarily follow that ma

= m(la) = (ml)a

=

Oa

= 0

for every element a E R. The implication is that char R < n, which is impossible. One establishes the converse in much the same way. As we have seen, multiplication exerts a strong influence on the additive structure of a ring through the distributive law. The following corollary to Theorem 1-5 shows that by sufficiently restricting the multiplication in a ring R it is possible to reach some interesting conclusions regarding the characteristic of R. Corollary 1. In an integral domain R all the nonzero elements have the same additive order; this order is the characteristic of the domain when char R > 0 and infinite when char R = 0. Proof To verify this assertion, suppose first that char R = n > 0. According to the definition of characteristic, each element 0 =I= a E R will then possess a finite additive order m, with m ~ n. (Recall that for an element

12

FIRST COURSE IN RINGS AND IDEALS

a =1= 0 of the group (R, +) to have order m means that ma = 0 and ka =I= 0 ifO < k < m.) But the relation 0 = ma = (ml)a implies that ml = 0, for R is assumed to be free of zero divisors. We therefore conclude from the theorem that n ~ m, whence m and n are equal. In consequence, every nonzero element of R has additive order n. A somewhat similar argument can be employed when char R = 0. The equation ma = 0 would lead, as before, to ml = 0 or m = 0. In this case every nonzero element a E R must be of infinite order.

The last result serves to bring out another useful point, which we place on record as Corollary 2. An integral domain R has positive characteristic if and only if na = 0 for some 0 =1= a E Rand some integer n E Z+. Continuing in this vein, let us next show that not any commutative group can serve as the additive group of an integral domain. Theorem 1-6. The characteristic of an integral domain is either zero or a prime number. Proof Let R be of positive characteristic nand assume that n is not a prime. Then, n has a nontrivial factorization n = n 1 n 2 , with 1 < n 1, n 2 < n. It follows that 0 = nl = (n 1 n 2 )1 = (n 1 n2 )P = (n 1 l)(n 2 1).

By supposition, R is without zero divisors, so that either n 1 1 = 0 or n2 1 = 0. Since both n 1 and n2 are less than n, this contradicts the choice of n as the least positive integer for which nl = 0. We therefore conclude that char R must be prime. Corollary. If R is a finite integral domain, then char R

=

p, a prime.

Turning again to the general theory, let R be any ring with identity and consider the set Zl of integral multiples of the iden.tity; stated symbolically, Zl = {nlin E Z}. From the relations nl - ml = (n - m)l,

(nl)(ml) = (nm)1

one can easily infer that Zl itself forms a (commutative) ring with identity. The order of the additive cyclic group (Zl, +) is simply the characteristic of the given ring R. When R happens to be an integral domain, then Zl is a subdomain of R (that is, Zl is also an integral domain with respect to the operations in R). In fact, Zl is the smallest subdomain of R, in the sense that it is contained in every other subdomain of R. If R is a domain of characteristic p,

PROBLEMS

13

where p is a prime, then we are able to deduce considerably more: each nonzero element of Zl is invertible. Before establishing this, first observe that the set Zl, regarded as an additive cyclic group of order p, consists of p distinct elements, namely, the p sums nl, where n = 0, 1, ... , p - 1. Now let nl be any nonzero element of Zl (0 < n < p). Since nand pare relatively prime, there exist integers r and s for which rp + sn = 1. But then 1 = (rp + sn)1 = r(p1) + (sl)(n1). As pi = 0, we obtain the equation 1 = (sl)(n1), so that sl serves as the multiplicative inverse of nl in Zl. The value of these remarks will have to await further developments (in particular, see Chapter 4). PROBLEMS I. Verify that the zero element of a ring R is unique, as is the additive inverse of each element a e R.

l. Let R be an additive commutative group. If the product of every pair of elements is defined to be zero, show that the resulting system forms a commutative ring (this is sometimes called the zero ring). 3. Prove that any ring R in which the two operations are equal (that is, a for all a, be R) must be the trivial ring R = {0}. 4. In a) b) c) d)

+ b = ab

a ring R with identity, establish each of the following: the identity element for multiplication is unique, if a e R has a multiplicative inverse, then a-t is unique, if the element a is invertible, then so also is -a, no divisor of zero can possess a multiplicative inverse in R.

5. a) If the set X contains more than one element, prove that every nonempty proper subset of X is a zero divisor in the ring P(X). b) Show that, if n > 1, the matrix ring M.(R) has zero divisors even though the ring R may not. 6. Suppose that R is a ring with identity 1 and having no divisors of zero. For a, b e R, verify that a) ab = 1 if and only if ba = 1, b) if az = 1, then either a = 1 or a = -1. 7. Let a, b be two elements of the ring R. If n e Z + and a and b commute, derive the binomial expansion

(a

+ br =

fi'

+ (i)a"- 1b + ··· +

(:)a"-~ 0}, determine whether s. is a subring of R. 18. Establish the following assertions concerning the characteristic of a ring R :

15

PROBLEMS

a) if there exists an integer k such that ka = 0 for all a E R, then k is divisible by char R; b) if char R > 0, then charS ::;; char R for any subring S of R; c) if R is an integral domain and S is a subdomam of R, then char S = char R. 19. Let R be a ring with a finite number of elements, say a 1 , a2 , ••• , a., and let n; be the order of a; regarded as a member of the additive group of R. Prove that the characteristic of R is the least common multiple of the integers n; (i = 1, 2, ... , n). 20. Suppose that R is a ring with identity such that char R show that R has divisors of zero.

=

n > 0. If n is not prime,

21. If R is a ring which has no nonzero nilpotent elements, deduce that all the idempotent elements of R belong to cent R. [Hint: If a 2 = a, then (ara - ar) 2 = (ara - ra) 2 = 0 for all r E R.] 22. Assume that R is a ring with the property that a 2 + a E cent R for every element a in R. Prove that R is necessarily a commutative ring. [Hint: Utilize the expression (a + b) 2 + (a + b) to show first that ab + ba lies in the center for all a, bE R.] 23. Let (G, +)be a commutative group and R be the set of all (group) homomorphisms of G into itself. The pointwise sum f + g and composition f o g of two functions f, g E R are defined by the usual rules (f

+ g)(x) = f(x) + g(x),

(/ o g)(x)

= f(g(x))

(x E G).

Show that the resulting system (R, +,a) forms a ring. At the same time determine the invertible elements of R. 24. Let (G, ·)be a finite group (written multiplicatively), say with elements x 1, x 2 , and let R be an arbitrary ring. Consider the set R(G) of all formal sums

•.. ,

x.,

.

Lrx i=l

1 1

Two such expressions are regarded as equal if they have the same coefficients. Addition and multiplication can be defined in R(G) by taking

L• r x + L• s x = L• (r1 + s1)x1 1 1

1 1

i=l

and

1=1

1=1

(.± r1x;) (.± S;X;) .± =

r= 1

r=l

where t1

=

t 1x 1,

t-=1

L

rist.

XJXfc=X·

(The meaning of the last-written sum is that the summation is to be extended over all subscripts j and k for which xixt = x 1.) Prove that, with respect to these operations, R(G) constitutes a ring, the so-called group ring of G over R.

TWO

IDEALS AND THEIR OPERATIONS

Although it is possible to obtain some interesting conclusions concerning subrings, this concept, if unrestricted, is too general for most purposes. To derive certain highly desirable results (for instance, the fundamental isomorphism theorems), additional assumptions that go beyond Definition 1-4 must be imposed. Thus, in the present chapter we narrow the field and focus attention on a class ofsubrings with a stronger type of multiplicative closure, namely, closure under multiplication by an arbitrary ring element. Definition 2-1. A subring I of the ring R is said to be a two-sided ideal of R if and only if r e R and a e I imply both rae I and arE I. Viewed otherwise, Definition 2-1 asserts that whenever one of the factors in a product belongs to I, then the product itself must be in I. (This may be roughly summarized by saying that the set I "captures" products.) Taking stock of Theorem 1-3, which gives a minimal set of conditions to be a subring, our current definition of a two-sided ideal may be reformulated as follows. Definition 2-2. Let I be a nonempty subset of a ring R. Then I is a two-sided ideal of R if and only if 1) a, be I imply a - be I, and 2) r e Rand a e I imply both products ra, are I. If condition (2) of the above definition is weakened so as to require only that the product ra belongs to I for every choice of r e R and a e I, we are led to the notion of a left ideal; right ideals are defined in a symmetric way. Needless to say, if the ring R happens to be commutative (the most important case so far as we shall be concerned), then there is no distinction between left, right, and two-sided ideals.

In what follows, let us agree that the term "ideal", unmodified, will always mean two-sided ideal.

CoNVENTION

Before proceeding further, we pause to examine this concept by means of several specific examples. 16

4

FIRST COURSE IN RINGS AND IDEALS

Example 1-4. To develop our next example, let X be an arbitrary (nonempty) set and (R, +,·)be a ring. We adopt the notation map(X, R) for the set consisting of all mappings from X into R; in symbols,

{!If: X

map(X, R) =

--. R}.

(For ease of notation, let us also agree to write map R in place ofmap(R, R).) Now, the elements of map(X, R) can be combined by performing algebraic operations on their functional values. More specifically, the pointwise sum and product off and g, denoted by f + g and f" g, respectively, are the functions which satisfy (f

+

g)(x) = f(x)

+

g(x),

(x eX).

(f·g)(x) = f(x)·g(x),

It is readily verified that the above definitions provide map(X, R) with the structure of a ring. We simply point out that the zero element of this ring is the constant function whose sole value is 0, and the additive inverse - f off is characterized by the rule (-f)(x) = - f(x). Notice that the algebraic properties of map(X, R) are determined by what happens in the ring (R, +, ·) (the set X furnishes only the points for the pointwise operations). For instance, if (R, +, ·) has a multiplicative identity 1, then the ring (map(X, R), +, ·) likewise possesses an identity element; namely, the constant function defined by 1(x) = 1 for all x e X.

Example 1-5. Our final example is that of the ring of integers modulo n, where n is a fixed positive integer. In order to describe this system, we first introduce the notion of congruence: two integers a and b are said to be congruent modulo n, written a = b (mod n), if and only if the difference a - b is divisible by n; in other words, a b (mod n) if and only if a - b = kn for some k e z. We leave the reader to convince himself that the relation "congruent modulo n" defines an equivalence relation on the set Z of integers. As such, it partitions Z into disjoint classes of congruent elements, called congruence classes. For each integer a, let the congruence class to which a belongs be denoted by [a] :

=

[a] = {x e Zlx

=a (mod n)}

={a+ knlkeZ}. Of course, the same congruence class may very well arise from another integer; any integer a' for which [a'] = [a] is said to be a representative of [a]. One final, purely notational, remark: the collection of all congruence classes of integers modulo n will be designated by z,. It can be shown that the congruence classes [0], [1], ... , [n - 1] exhaust the elements of Z,.. Given an arbitrary integer a, the division algorithm asserts that there exist unique q, r e Z, with 0 s; r < n, such that a = qn + r. By the definition of congruence, a r (mod n), or

=

18

FIRST COURSE IN RINGS AND IDEALS

that is, R s I, whence the equality I that I is a proper subset of R.

= R.

This contradicts the hypothesis

Notice that, en route, we have also established Corollary. In a ring with identity, no proper (right, left, two-sided) ideal

contains the identity element. Example 2-3. This example is given to show that the ring M .(R #) of n x n matrices over the real numbers has no nontrivial ideals. As a notational device, let us define Eii to be the n x n matrix having 1 as its ijth entry and zeroes elsewhere. Now, suppose that I =F {0} is any ideal of the ring M.(R#). Then I must contain some nonzero matrix (a;;), with, say, rsth entry a,. =F 0. Since I is a two-sided Ideal, the product Err(bij)(a;j)E••

is a member of/, where the matrix (b;) is chosen to have the element a;; 1 down its main diagonal and zeroes everywhere else. As a result of all the zero entries in the various factors, it is easy to verify that this product is equal to E,.. Knowing this, the relation (i,j = 1, 2•... , n)

implies that all n 2 of the matrices Eu are contained in /. The clinching point is that the identity matrix (..5;i) can be written as (o;i)

=

E 11

+ E 22 + ··· + E,,

which leads to the conclusion that (..5ii) E I and, appealing to the above corollary, that I = M.(R#). In other words, M,.(R #)possesses no nonzero proper ideals. as asserted. As a matter of definition, let us call a ring R =F {0} simple if R has no two-sided ideals other than {0} and R. In the light of Example 2-4. the matrix ring M,.(R#) is a simple ring. We now take up some of the standard methods for constructing new ideals from given ones. To begin with simpler things: Theorem 2-2. Let {/;} be an arbitrary collection of (right, left, twosided) ideals of the ring R. where i ranges over some index set. Then n / 1 is also a (right, left, two-sided) ideal of R. Proof We give the proof for the case in which the ideals are two-sided. First, observe that the intersection n I 1 is nonempty, for each of the ideals I 1 must contain the zero element of the ring. Suppose that the elements a, bE n 11 andrE R. Then a and bare members of / 1, where i varies over

the indexing set. Inasmuch as / 1 is assumed to be an Ideal of R, it follows that a - b, ar and ra all lie in the set I 1• But this is true for every value of

IDEALS AND THEIR OPERATIONS

19

i, whence the elements a - b, ar and ra belong to n I;, making n I; an ideal of R.

Consider, for the moment, an arbitrary ring R and a nonempty subset S of R. By the symbol (S) we shall mean the set

(S) = n

{II S

~I; I is an ideal of R}.

The collection of all ideals which contain S is not empty, since the entire ring itself is an ideal containing any subset of R; thus, the set (S) exists and satisfies the inclusionS ~ (S). By virtue of Theorem 2-2, (S) forms an ideal of R, known as the ideal generated by the set S. It is noteworthy that whenever I is any ideal of R with S ~ I, then necessarily (S) s I. For this reason, one often speaks of (S) as being the smallest ideal of R to contain the set S. It should be apparent that corresponding remarks apply to the one-sided ideals generated by S. If S consists of a finite number of elements, say a 1, a2 , ••. , am then the ideal which they generate is customarily denoted by (a 1 , a 2 , .•• , an). Such an ideal is said to be finitely generated with the given elements a; as its generators. An ideal (a) generated by just one ring element is termed a principal ideal.

A natural undertaking is to determine the precise form of the members of the various ideals (right, left, two-sided) generated by a single element, say a, of an arbitrary ring R. The right ideal generated by a is called a principal right ideal and is denoted by (a),. Being closed with respect to multiplication on the right, (a), necessarily contains all products ar (r E R), as well as the elements na (n an integer), and, hence, includes their sum ar + na. (As usual, the notation na represents the n-fold sum of a.) It is a fairly simple matter to check that the set of elements of the form ar + na constitutes a right ideal of R. Observe, too, that the element a is a member of the ideal, .since a = aO + la. These remarks make it clear that (a)r

= {a~+

Z}.

nair E R; n e

When there is an identity element present, the term na becomes superfluous, for, in this setting, we may write the expression ar + na more simply as ar

+

na = ar

+ a(nl)

= a(r

+

nl) = ar',

where r' = r + nl is some ring element. Thus, the set (a), consists of all right multiples of a by elements of R. If R is a ring with identity, we shall frequently employ the more suggestive notation aR in place of (a),; that is, (a), = aR = {arir E R}.

Similar remarks apply, of course, to the principal left ideal (a) 1 generated by a.

20

FIRST COURSE IN RINGS AND IDEALS

As a general comment, observe that the products ar (r E R) comprise the set of elements of a right ideal of R even when the ring does not possess an identity. The difficulty, however, is that this ideal need not contain a itself. With regard to the two-sided ideal (a) generated by a, the situation is more complicated. Certainly the elements ras, ra, as and na must all belong to the ideal (a) for every choice of r, s E R and n E Z. In general, the sum of two elements ras and r' as' is no longer of the same form, so that, in order to have closure under addition, any finite sum L r;as;, where r., s; E R, is also required to be in (a). The reader will experience no difficulty in showing that the principal ideal generated by a is given by (a)

=

{na

+ ra + as + L

r1as;Jr, s,

r1, s1 E R; n E Z}.

fimte

In case R happens to have an identity, this description of (a) reduces to the set of all finite sums L r;as;. A particularly important type of ring is a principal ideal ring, which we now define. Definition 2-3. A ring R is said to be a principal ideal ring if every ideal I of R is of the form I = (a) for some a E R.

The following theorem furnishes an example of such rings. Theorem 2-3. The ring Z of integers is a principal ideal ring; in fact, if I is an ideal of Z, then I = (n) for some nonnegative integer n. Proof If I = {0}, the theorem is trivially true, since the zero ideal {0} is the principal ideal generated by 0. Suppose then that I does not consist of the zero element alone. Now, if mE I, - m also lies in I, so that the set I contains positive integers. Let n designate the least positive integer in I. As I forms an ideal of Z, each integral multiple of n must belong to I, whence (n) s;;; I. To establish the inclusion I s;;; (n), let k be an arbitrary element of I. By the division algorithm there exist integers q and r for which k = qn + r, with 0 :::; r < n. Since k and qn are both members of I, it follows that r = k - qn E I. lfr > 0, we would have a contradiction to the assumption that n is the smallest positive integer in I. Accordingly, r = 0 and k = qn E (n). Thus, only multiples of n belong to I, implying that I s;;; (n). The two inclusions show that I = (n) and the argument is complete.

Let us now describe certain binary operations on the set of all ideals of R. (Similar considerations apply to the sets of right and left ideals, but for economy of effort we concentrate on two-sided ideals.) Given a finite number of ideals I 1 , I 2 , •.• , In of the ring R, one defines their sum in the natural way:

I1

+ I 2 + ... + In = {al + a2 + ... + anJa1E I 1}.

IDEALS AND THEIR OPERATIONS

21

Then I 1 + I 2 + · · · + In is likewise an ideal of R and is the smallest ideal of R which contains every I;; phrased in another way, I 1 + I 2 + ··· +In is the ideal generated by the union I 1 u I 2 u · · · u In. In the special case oftwo ideals I and J, our definition reduces to I+ J ={a+ blaei; beJ}. More generally, let {I;} be an arbitrary indexed collection of ideals of R. The sum of this collection may be denoted by L I 1 and is the ideal of R whose members are all possible finite sums of elements from the various ideals I 1 : l:I1 = {l:a1la1 ei1}. finite

The reader will take care to remember that, although {I;} may be an infinite family of ideals, only finite sums of elements of R are involved in the definition above. An alternative description of L I 1 could be given by L I 1 = {La; la1 e I 1 ; all but a finite number of the a1 are 0}, where it is understood that L represents an arbitrary sum with one or more terms. Just as n I 1 can be interpreted as the largest ideal of R contained in every I 1, the sum L I 1 supplies the dual notion of the smallest ideal containing every I 1• If R = I 1 + I 2 + · · · + In, then each element x e R can be expressed in the form x = a 1 + a 2 + ··· + an, where a 1 lies in I 1• There is no guarantee, however, that this representation of xis unique. To ensure that every member of R is uniquely expressible as a sum of elements from the ideals I 1, an auxiliary definition is required. Definition 2-4. Let I~> I 2 , ••• , In be ideals of the ring R. We call R the internal direct sum of I 1 , I 2 , ... , I", and write R = I 1 (f) I 2 (f) · .. (f) In, provided that a) R = I 1 + I 2 + ... + I,, and b) I 1 n (I 1 + ... + I 1_ 1 + I 1+ 1 + ··· +In)= {0} for each i. As was heralded by our remarks, we are now in a position to prove Theorem 2-4. Let I 1 , I 2 , ... , In be ideals of the ring R. following statements are equivalent: 1) R is the internal direct sum of I 1, I 2 , ... , I,.. 2) Each element x of R is uniquely expressible in the form x = a 1 + a2 + ... +a,., where a 1 ei1•

Then the

Proof. There is no loss in confining ourselves to the case n = 2; the general

argument proceeds along similar lines. We begin by assuming that R = I 1 (f) I 2 • Suppose further that an element x e R has two representations x = a1

+ b 1 = a2 + b2

(a1 e I 1 , b1 e I 2 ).

6

FIRST COURSE IN RINGS AND IDEALS

between a congruence class and its smallest nonnegative representative; under this convention, Zn = {0, 1, ... , n - 1}. It is perhaps worth commenting that, since Z 1 = Z, a number of texts specifically exclude the value 1 for n. Although it is logically correct {and often convenient) to speak of a ring as an ordered triple, the notation becomes unwieldy as one progresses further into the theory. We shall therefore adopt the usual convention of designating a ring, say {R, +, · ), simply by the set symbol R and assume that + and · are known. The reader should realize, however, that a given set may perfectly well be the underlying set of several different rings. Let us also agree to abbreviate a + {-b) as a - b and subsequently refer to this expression as the difference between a and b. As a final concession to brevity, juxtaposition without a median dot will be used to denote the product of two ring elements. With these conventions on record, let us begin our formal development of ring theory. The material covered in the next several pages will probably be familiar to most readers and is included more to assure completeness than to present new ideas. Theorem 1-1. If R is a ring, then for any a, b, c E R 1) Oa = aO = 0, 2)a{-b)={-a)b= -(ab), 3) {-a){ -b) = ab, and 4) a(b - c) = ab - ac, (b - c)a = ba - ca. Proof These turn out, in the main, to be simple consequences of the distributive laws. For instance, from 0 + 0 = 0, it follows that Oa = (0

+ O)a =

Oa

+ Oa.

Thus, by the cancellation law for the additive group (R, +),we have Oa = 0. In a like manner, one obtains aO = 0. The proof-of (2) requires the fact that each element of R has a unique additive inverse (Problem 1). Since b + (-b) = 0, ab + a(-b) = a(b +(-b))= aO = 0, which then implies that - (ab) = a(- b). The argument that (- a)b is also the additive inverse of ab proceeds similarly. This leads immediately to (3): (-a)( -b) = -( -a)b = -·( -(ab)) = ab. The last assertion is all but obvious. There is one very simple ring that consists only of the additive identity 0, with addition and multiplication given by 0 + 0 = 0, 00 = 0; this ring is usually called the trivial ring.

IDEALS AND THEIR OPERATIONS

23

In this connection, it is important to observe that

forms a decreasing chain of ideals. Remark. If I is a right ideal and S a nonempty subset of the ring R, then Sl =

0:: a,rda; e S; r; e J} finite

forms a right ideal of R. In particular, if S prefer to {a} J) is given by

=

{a}, then

a]

(a notation· we

al = {arlr e I}.

Analogous statements can be made when I is a left ideal of R, but not, of course, a two-sided ideal. The last ideal-theoretic operation which we wish to consider is that of the quotient (or residual), defined below. Definition 2-5. Let I and J be two ideals of the ring R. The right (left) quotient of I by J, denoted by the symbol I :. J (I :1 J), consists of all elements a E R such that aJ s I (Ja s I). In the event R is a commutative ring, we simply write l:J. It is by no means obvious that the set

I:,J ={aeRials I}

actually forms an ideal of R, whenever 1 and J are ideals. To verify this, suppose that the elements a, be I :. J and r E R. For any x e J, we clearly have (a - b)x = ax - bx E I, since ax and bx both belong to I by definition. This establishes the inclusion (a - b)J s I, which in turn signifies that a - b E I :. J. Likewise, the relations raJ s rl s I and arJ s aJ s I imply that ra, arE I:. J. In consequence, I:. J comprises an ideal of R in its own right, and that I :1 J is also an ideal follows similarly. The purpose of the coming theorem is to point out the connection between the quotient ideal and the operations defined previously. This result, although it might seem to be quite special, will serve us in good stead when we develop the theory of Noetherian rings. Theorem 2-5. The following relations hold for ideals in a ring R (capital

letters indicate ideals of R):

1) (n I;) :,J = n (I, :,J), 2) I :, ""£ 1; = n (I :, 1;), 3) I :,(JK) =(I :,K) :,J.

24

FIRST COURSE IN RINGS AND IDEALS

Proof. Concerning (1), we have

(n / 1) :, J = {a e RjaJ £ n I;} = {a e RlaJ £ /;for all i} = n {a e RJai £ Ii} = n (1 1 :,J). With an eye to proving (2), notice that the inclusion 1; £ I: I; implies a(l: J;) £ I if and only if aJ1 £ I for all i; thus,

I:, l: J 1

=

{a e RJa(I: J 1)

=

{a e RjaJ1

£

I} I for all i} £

=

n (I:, 1 1).

Confirmation of the final assertion follows from I :,(JK) = {a e Rja(JK) £ I} = {a e RJ(aJ)K £ I}

= {aeRJal £ I:,K} = (I:,K):,J.

Remark. Similar results hold for left quotients; the sole difference being that, instead of(3), one now has I :1 (JK) =(I :1 J) :1 K.

This may be a good place to observe that if I is an ideal of the ring R and J is an ideal of I, then J need not be an ideal of the entire ring R. For an illustration, we turn to the ring map R # and let R be the subring consisting of all continuous functions from R # into itself. Consider the sets I = {fijfe R;f(O) = 0}, J = {fi 2

+

ni 2 Jfe R;f(O) = 0; n e Z},

where i denotes the identity function on R # (that is, i(x) = x for all x e R # ). A routine calculation verifies that J is an ideal of I, which, in turn, forms an ideal of R. However, J fails to be an ideal of R, since i 2 E J, while ti 2 ¢ J. (The symbol ! is used in this setting to represent the constant function whose value at each real number is!.) We assume that !i2 E J and derive a contradiction. Then, !i2 = fi2

+

ni2

for a suitable choice off e R and n e Z, with f(O) = 0. In consequence, fi 2 = (! - n)i 2 , implying that f(x) = ! - n =I= 0 for every 0 =I= x e R#; in other words, f is a nonzero constant function on R # - {0}. But this obviously violates the continuity off at 0. A condition which will ensure that J is also an ideal of R is to take R to be a regular ring, a notion introduced by Von Neumann [52]. Definition 2-6. A ring R is said to be regular if for each element a e R there exists some a' e R such that aa'a = a. If the element a happens to have a multiplicative inverse, then the regularity condition is satisfied by setting a' = a- 1 ; in view of this, a' is

IDEALS AND THEIR OPERATIONS

25

often referred to as the pseudo-inverse of a. In the commutative case, the equation aa'a = a may, of course, be written as a 2 a' = a. The result which we have in mind now follows. Theorem 2-6. Let I be an ideal of the regular ring R. Then any ideal J of I is likewise an ideal of R. Proof. To start, notice that I itself may be regarded as a regular ring. Indeed, if a E I, then aa'a = a for some a' in R. Setting b = a'aa', the

element b belongs to I and has the property that

=

=

(aa'a)a'a = aa'a = a. Our aim is to show that whenever a E J s I and r E R, then both ar and ra lie in J. We already know that arE I; hence, by the above, there exists an element x in I for which arxar = ar. Since rxar is a member of I and J is assumed to be an ideal of I, it follows that the product a(rxar) must belong to J, or, equivalently, arE J. A symmetric argument confirms that raE J. aba

a(a'aa')a

Although Definition 2-6 appears to have a somewhat artificial air, we might remark that the set of all linear transformations on a finite dimensional vector space over a field forms a regular ring (Problem 20, Chapter 9). This in itself would amply justify the study of such rings. We now turn our attention to functions between rings and, more specifically, to functions which preserve both the ring operations. Definition 2-7. Let Rand R' be two rings. By a (ring) homomorphism, or homomorphic mapping, from R into R' is meant a function f: R -+ R' such that f(a + b) = f(a) + f(b), f(ab) = f(a)f(b) for every pair of elements a, b E R. A homomorphism which is also one-to-one as a map on the underlying sets is called an isomorphism. We emphasize that the + and ·occurring on the left-hand sides of the equations in Definition 2-7 are those of R, whereas the + and· occurring on the right-hand sides are those of R'. This use of the same symbols for the operations of addition and multiplication in two different rings should cause no ambiguity if the reader attends closely to the context in which the notation is employed. Ifjis a homomorphism of R into R', then the imagef(R) of R underf will be called the homomorphic image of R. When R = R', so that the two rings are the same, we say that f is a homomorphism of R into itself. In this connection, a homomorphism of R into itself is frequently referred to as an endomorphism of the ring R or, if an isomorphism onto R, an automorphism of R.

26

FIRST COURSE IN RINGS AND IDEALS

For future use, we shall label the set of all homomorphisms from the ring R into the ring R' by the symbol hom(R, R'). In the event that R = R', the simpler notation hom R will be used in place of hom(R, R). (Some authors prefer to write end R, for endomorphism, in place of hom R; both notations have a certain suggestive power and it reduces to a matter of personal preference.) A knowledge of a few simple-minded examples will help to fix ideas. Example 2-4. Let Rand R' be arbitrary rings andf: R ---+ R' be the function which sends each element of R to the zero element of R'. Then.

= 0 = 0 + 0 = f(a) + f(b), f(ab) = 0 = 0 0 = f(a) f(b) f(a

+

b)

(a, bE R),

so thatfis a homomorphic mapping. This particular mapping, the so-called trivial homomorphism, is the only constant function which satisfies Definition 2-7. Example 2-5. Consider the ring Z of integers and the ring Z,. of integers modulo n. Definef: Z ---+ Z,. by takingf(a) = [a]; that is, map each integer into the congruence class containing it. Thatfis a homomorphism follows directly from the definition of the operations in Z,.:

= [a] +,. [b] = f(a) +,.f(b), f(ab) = [ab] = [a]",.[b] = f(a)· .. f(b). f(a +b)= [a+ b]

Example 2-6. In the ring map(X, R). define ra to be the function which assigns to each f E map(X, R) its value at a fixed element a E X; in other words, ra(f) = f(a). Then ra is a homomorphism from map(X, R) into R, known as the evaluation homomorphism at a. We need only observe that

r,.(f + g) •a(fg)

= (f + g)(a) = f(a) + g(a) =

= •,.(f)

+

•,.(g),

(fg)(a) = f(a)g(a) = 7:,.(f)ra(g).

We now list some of the structural features preserved under homomorphisms. Theorem 2-7. Letfbe a homomorphism from the ring R into the ring R'. Then the following hold: 1) f(O) = 0, 2) f(- a) = -f(a) for all a E R. If, in addition, R and R' are both rings with identity and f(R) = R', then

3) f(l) = 1, 4) f(a- 1) = f(a)- 1 for each invertible element a E R.

IDEALS AND THEIR OPERATIONS

27

Proof From f(O) = f(O + 0) = f(O) + f(O), we obtain f(O) = 0. The fact that f(a) + f(- a) = f(a + (-a)) = f(O) = 0 yields f(- a) = -f(a). As regards (3), let the element a E R satisfy f(a) = 1; then,f(l) = f(a)f(l) = f(al) = f(a) = 1. Finally, the equationf(a)f(a- 1) = f(aa- 1 ) = f(l) = 1 shows thatf(a)- 1 = f(a- 1 ), whenever a E R has a multiplicative inverse.

Two comments regarding part (3) of the above theorem are in order. First, it is evident that f(a)l

=

f(a) = f(al) = f(a)f(l)

for any a in R. Knowing this, one might be tempted to appeal (incorrectly) to the cancellation law to conclude thatf(1) = 1; what is actually required is the fact that multiplicative identities are unique. Second, if the hypothesis that f map onto the set R' is omitted, then it can only be inferred that f(l) is the identity for the homomorphic image f(R). The element f(l) need not serve as an identity for the entire ring R' and, indeed, it may very well happen thatf(l) =/= 1. We also obs~rve, in passing, that, by virtue of statement (2), f(a - b) = f(a)

+ f(- b)

= f(a) - f(b).

In short, any ring homomorphism preserves differences as well as sums and products. The next theorem indicates the algebraic nature of direct and inverse images of subrings under homomorphisms. Among other things, we shall see that iff is a homomorphism from the ring R into the ring R', then f(R) forms a subring of R'. The complete story is told below. Theorem 2-8. Let f be a homomorphism from the ring R into the ring R'. Then, 1) for each subring S of R,J(S) is a subring of R'; and

2) for each subring S' of R', f

- 1(S') is a subring of R.

Proof To obtain the first part of the theorem, recall that, by definition, the imagef(S) = { f(a)ia E S}. Now, suppose thatf(a) andf(b) are arbitrary elements of .fl.S). Then both a and b belong to the set S, as do a - b and ab (S being a subring of R). Hence, f(a) - f(b) = f(a - b) ef(S)

and f(a)f(b) = f(ab) ef(S).

According to Theorem l-3, these are sufficient conditions for f(S) to be a subring of R'. The proof of the second assertion proceeds similarly. First, remember thatf- 1(S') = {a E R!f(a) E S'}. Thus, if a, b E/- 1(S'), the imagesf(a) and

28

FIRST COURSE IN RINGS AND IDEALS

f(b) must be members of S'. Since S' is a subring of R', it follows at once

that f(a - b) = f(a) - f(b) e S'

and f(ab)

=

f(a)f(b) e S'.

This means that a - b and ab lie in f- 1 (S'), from which we conclude that f- 1 (S') forms a subring of R. Left unresolved is the matter of replacing the term "subring" in Theorem 2-8 by "ideal". It is not difficult to show that part (2) of the theorem remains true under such a substitution. More precisely: if I' is an ideal of R', then the subring f- 1(1') is an ideal of R. For instance, suppose that a e f- 1(1'), so that f(a) e I', and let r be an arbitrary element of R. Then, f(ra) = f(r)f(a) e I'; in other words, the product ra is inf- 1(1'). Likewise, are f- 1(1'), which helps to make f - 1 (1') an ideal of R. Without further restriction, it cannot be inferred that the image f(I) will be an ideal of R', whenever I is an ideal of R. One would need to know that r'f(a) ef(I) for all r' e R' and a e I. In general, there is no way of replacing r' by some f(r) in order to exploit the fact that I is an ideal. The answer is obvious: just take f to be an onto mapping. Summarizing these remarks, we may now state: Corollary. 1) For each ideal I' of R', the subring f- 1(1') is an ideal of R. 2) If f(R) of R'.

=

R', then for each ideal I of R, the subring f(l) is an ideal

To go still further, we need to introduce a new idea. Definition 2-8. Let f be a homomorphism from the ring R into the ring R'. The kernel off, denoted by kerf, consists of those elements in R which are mapped by f onto the zero element of the ring R': kerf= {a e Rlf(a) = 0}. Theorem 2-7 indicates that kerf is a nonempty subset of R, since, if nothing else, 0 e kerf. Except for the case of the trivial homomorphism, the kernel will always turn out to be a proper subset of R. As one might suspect, the kernel of a ring homomorphism forms an ideal. Theorem 2-9. The kernel kerf of a homomorphism f from a ring R into a ring R' is an ideal of R. Proof. We already know that the trivial subring {0} forms an ideal of R'. Since kerf = f- 1 (0), the conclusion follows from the last corollary.

INTRODUCTORY CONCEPTS

9

Example 1-7. The set Ze of even integers forms a subring of the ring Z of integers, for 2n - 2m = 2(n - m) E Ze, (2n)(2m) = 2(2nm) E Ze.

This example also illustrates a fact worth bearing in mind: in a ring with identity, a subring need not contain the identity element. Prior to stating our next theorem, let us define the center of a ring R, denoted by cent R, to be the set cent R

=

{a

E

Rlar

=

ra for all r E R}.

Phrased otherwise, cent R consists of those elements which commute with every member of R. It should be apparent that a ring R is commutative if and only if cent R = R. Theorem 1-4. For any ring R, cent R is a subring of R. Proof. To be conscientious about details, first observe that cent R is nonempty; for, at the very least, the zero element 0 E R. Now pick any two elements a, b in cent R. By the definition of center, we know that ar = ra and br = rb for every choice of r E R. Thus, for arbitrary r E R., (a - b)r

=

ar - br

=

ra - rb

=

r(a - b),

which implies that a - bE cent R. A similar argument affirms that the product ab also lies in cent R. In the light of Theorem 1-3, these are sufficient conditions for the center to be a subring of R. It has already been remarked that, when a ring has an identity, this need not be true of its subrings. Other interesting situations may arise.

1) Some subring has a multiplicative identity, but the entire ring does not. 2) Both the ring and one of its subrings possess identity elements, but they are distinct. In each of the cited cases the identity for the subring is necessarily a divisor of zero in the larger ring. To justify this claim, let 1' =I= 0 denote the identity element of the subring S; we assume further that 1' does not act as an identity for the whole ring R. Accordingly, there exists some element a E R for which a1' =/:: a. It is clear that (a1')1'

= a(l'l') = a1',

or (a1' - a)1' = 0. Since neither a1' - a nor 1' is zero, the ring R has zero divisors, and in particular 1' is a zero divisor. Example 1-8. To present a simple illustration of a ring in which the second of the aforementioned possibilities occurs, consider the set R = Z x Z,

30

FIRST COURSE IN RINGS AND IDEALS

any ring R with identity which is of characteristic zero will contain a subring isomorphic to the integers; more specifically, Z ~ Zl, where 1 is the identity of R. Suppose that f is a homomorphism from the ring R onto the ring R'. We have already observed that each ideal I of the ring R determines an ideal f(I) of the ring R'. It goes without saying that ring theory would be considerably simplified ifthe ideals of R were in a one-to-one correspondence with those of R' in this manner. Unfortunately, this need not be the case. The difficulty is reflected in the fact that if I and J are two ideals of R with I s J s I + kerf, then f(/) = f(J). The quickest way to see this is to notice that f(I) S f(J) S f(I

+

kerf) = f(I)

+ f(ker f)

=

f(I),

from which we conclude that all the inclusions are actually equalities. In brief, distinct ideals of R may have the same image in R'. This disconcerting situation could be remedied by either demanding that kerf= {0} or else narrowing our view to consider only ideals I with kerfs I. In either event, it follows that I S J S I + kerf= I and, in consequence, I = J. The first of the restrictions just cited has the effect of making the function f one-to-one, in which case R and R' are isomorphic rings (and it then comes as no surprise to find their ideals in one-to-one correspondence). The second possibility is the subject of our next theorem. We turn aside briefly to establish a preliminary lemma which will provide the key to later success. Lemma. Let f be a homomorphism from the ring R onto the ring R'. If I is any ideal of R such that kerfs I, then I = f- 1 (!(1)).

r

1(/{I)). so that j{a) E./{/). Then Proof Suppose that the element a E f(a) = f(r) for some choice ofr in I. As a result, we will havef(a - r) = 0, or, what amounts to the same thing, a - r E ker j s I. This implies that aEI, yielding the inclusionf- 1 (/(I)) s I. Since the reverse inclusion always holds, the desired equality follows.

Here now is one of the main results of this section. Theorem 2-11. (Correspondence Theorem). Let f be a homomorphism from the ring R onto the ring R'. Then there is a one-to-one correspondence between those ideals I of R such that kerf s I and the set of all ideals I' of R'; specifically, I' is given by I' = f(I). Proof Our first concern is to show that the indicated correspondence actually maps onto the set of all ideals of R'. In other words, starting with an ideal I' of R', we must produce some ideal I of the ring R, with kerfs I,

IDEALS AND THEIR OPERATIONS

31

such that f(J) = 1'. To accomplish this, it is sufficient to take I = f - 1(1'). By the corollary to Theorem 2-8,f- 1 (1') certainly forms an ideal of R and, since 0 E I', kerf= f- 1 (0)

5;

f- 1 (I').

Inasmuch as the function f is assumed to be an onto map, it also follows that f(l) = f(f - 1 (1')) = I'. Next, we argue that this correspondence is one-to-one. To make things more specific, let ideals I and J of R be given, where kerf 5; I, kerf 5; J, and satisfying f(l) = f(J). From the elementary lemma just established, we see that I=

f-t(f(J)) = f-t(f(J)) =

J.

One finds in this way that the correspondence I - f(l), where kerf 5; I, is indeed one-to-one, completing the proo( Before announcing our next result, another definition is necessary. Definition 2-9. A ring R is said to be imbedded in a ring R' if there exists some subring S' of R' such that R ~ S'. In general, if a ring R is imbedded in a ring R', then R' is referred to as an extension of R and we say that R can be extended to R'. The most important cases are those in which one passes from a given ring R to an extension possessing some property not present in R. As a simple application, let us prove that an arbitrary ring can be imbedded in an extension ring with identity. Theorem 2-12. (Dorroh Extension Theorem). Any ring R can be imbedded in a ring with identity.

Proof. Consider the Cartesian product R x Z, where RxZ= {(r,n)lreR;neZ}.

If addition and multiplication are defined by (a, n) + (b, m) (a, n)(b, m)

=

(a

+ b, n +

m),

= (ab + ma + nb, nm),

then it is a simple matter to verify that R x Z forms a ring; we leave the actual details as an exercise. Notice that this system has a multiplicative identity, namely, the pair (0, 1); for (a, n)(O, 1)

=

(aO

+

1a

+

nO, n1)

and, similarly, (0, 1)(a, n) = (a, n).

=

(a, n),

32

FIRST COURSE IN RINGS AND IDEALS

Next, consider the subset R x {0} of R x Z consisting of all pairs of the form (a, 0). Since (a, 0) - (b 0) = (a - b, 0), (a, O)(b, 0) = (ab, 0),

it is evident that R x {0} constitutes a subring of R x z. A straightforward calculation, which we omit, shows that R x {0} is isomorphic to the given ring R under the mapping f: R --+ R x {0} defined by f(a) = (a, 0). This process of extension therefore imbeds R in R x Z, a ring with identity. A point to be made in connection with the preceding theorem is that the imbedding process may be carried out even if the given ring has an identity to start with. Of course, in this case the construction has no particular merit; indeed, the original identity element only serves to introduce divisors of zero into the extended ring. Although Theorem 2-12 shows that we could confine our study to rings with identity, it is nonetheless desirable to develop as much of the theory as possible without the assumption of such an element. Thus, unless an explicit statement is made to the contrary, the subsequent discussions will not presuppose the existence of a multiplicative identity. We now take a brief look at a different problem, namely, the problem of extending a function from a subring to the entire ring. In practice, one is usually concerned with extensions which retain the characteristic features of the given function. The theorem below, for instance, presents a situation in which it is possible to extend a homomorphism in such a way that the extended function also preserves both ring operations. Theorem 2-13. Let I be an ideal of the ring R and fa homomorphism from I onto R', a ring with identity. If I s cent R, then there is a unique homomorphic extension off to all of R. Proof As a start, we choose the element u E I so that f(u) = 1. Since I constitutes an ideal of R, the product au will lie in the set I for each choice of a E R. It is therefore possible to define a new function g: R --+ R' by setting g(a) = f(au) for all a in R. If the element a happens to belong to I, then g(a)

= f(au) = f(a)f(u)

= f(a)l = f(a),

showing that g actually extends the original function! The next thing to confirm is that both ring operations are preserved by g. The case of addition is fairly obvious: if a, b E R, then g(a

+

b)

= f((a + b)u) =

f(au

= f(au)

= g(a)

+ f(bu)

+ bu) + g(b).

33

IDEALS AND THEIR OPERATIONS

As a preliminary step to demonstrating that g also preserves multiplication, notice that f((ab)u 2 ) = f(abu)f(u) = f(abu). From this we are able to conclude that

g(ab)

= f(abu) = f(abu 2 ) = f((au)(bu)) = f(au)f(bu) = g(a)g(b).

The crucial third equality is justified by the fact that u e cent R, hence, commutes with b. As regards the uniqueness assertion, let us assume that there is another homomorphic extension off to the set R; call it h. Since f and h must agree on I and, more specifically, at the element u, h(u) = f(u) = 1. With this in mind, it follows that

h(a) = h(a)h(u) = h(au)

= f(au) = g(a)

for all a E R and so h and g are the same function. Hence, there is one and only one way of extendingfhomomorphically from the ideal I to the whole ring R. Before closing the present chapter, there is another type of direct sum which deserves mention. To this purpose, let R 1, R 2 , •.. , R,. be a finite number of rings (not necessarily subrings of a common ring) and consider their Cartesian product R = X R; consisting of all ordered n-tuples (a 1, a 2 , ... , a,.), with a; E R;. One can easily convert R into a ring by performing the ring operations componentwise; in other words, if (a 1, a 2 , ••• , a,.) and (b 1 , b2 , ••• , b,) are two elements of R, simply define

(a 1, a2 ,

••• ,

a.,)

+

(b 1, b2 ,

••• ,

b,) = (a 1

+ b1, a2 + b2 , ••• , a, +

b,)

and

(a 1 , a 2 ,

••• ,

a,)(b 1, b2 ,

••• ,

b,) = (a 1 b 1 , a 2 b 2 ,

••• ,

a,b,.).

The ring so obtained is called the external direct sum of R 1 , R 2 , ••• , R, and is conveniently written R = R 1 R2 R,.. (Let us caution that the notation is not standard in this matter.) In brief, the situation is this : An external direct sum is a new ring constructed from a given set of rings, and an internal direct sum is a representation of a given ring as a sum of certain of its ideals. The connection between these two types of direct sums will be made clear in the next paragraph. If R is the external direct sum of the rings R; (i = 1, 2, ... , n), then the individual R; need not be subrings, or even subsets, of R. However, there is an ideal of R which is the isomorphic image of R;. A straightforward calculation will convince the reader that the set

+

+ ··· +

I 1 = {(0, ... , 0, a;, 0, ... , O)la; e R;}

34

FIRST COURSE IN RINGS AND IDEALS

(that is, the set consisting of all n-tuples with zeroes in all places but the ith) forms an ideal of R naturally isomorphic to R; under the mapping which sends (0, ... , 0, a;, 0, ... , 0) to the element a;. Since (a 1, a 2 ,

••• ,

a,.) = (a 1 , 0, 0, ... , 0)

+

(0, a 2 , 0, ... , 0)

+ ·· · +

(0, 0, ... , 0, a,.),

it should also be clear that every member of R is uniquely representable as a sum of elements from the ideals I;. Taking note of Theorem 2-4, this means that R is the internal direct sum of the ideals I; and so

In summary, the external direct sum R of the rings R 1 , R 2 , •.• , R,. is also the internal direct sum of the ideals I 1 , I 2 , ..• , I,. and, for each i, R; and I; are isomorphic. ln view of the isomorphism just explained, we shall henceforth refer to the ring R as being a direct sum, not qualifying it with the adjective "internal" or "external", and rely exclusively on the EB-notation. The term "internal" merely reflects the fact that the individual summands, and not isomorphic copies of them, lie in R. We take this opportunity to introduce the simple, but nonetheless useful, notion of a direct summand of a ring. In formal terms, an ideal I of the ring R is said to be a direct summand of R if there exists another ideal J of R such that R = I EB J. For future use, let us note that should the ideal I happen to have an identity element, say the element e E I, then it will automatically be a direct summand of R. The argument proceeds as follows. For any choice of r E R, the product re E I. The assumption that e serves as an identity for I then ensures that e(re) = re. At the same time (and for the same reasons), (er)e = er. Combining these pieces, we get re = ere = er, which makes it plain that the element e lies in the center of R. This is the key point in showing that the set J = {r- reire R} forms an ideal of R; the details are left to the reader. We contend that the ring R is actually the direct sum of I and J. Certainly, each element r.of R may be written as r = re + (r - re), where re E I and r - re E J. Since I n J = {0}, this is the only way r can be expressed as a sum of elements of I and J. (A moment's thought shows that if a E I n J, say a = r - re, then a = ae = (r - re)e = r(e - e 2 ) = 0.) It is also true that the ideal/ = eR = Re, but we did not need this fact here. As a further application of the idea of a direct summand, let us record Theorem 2-14. If the ring R is a direct summand in every extension ring containing it as an ideal, then R has an identity.

Proof To set this result in evidence, we first imbed R in the extension ring R' = R x Z in the standard way (see Theorem 2-12). Then, R ~ R x {0}, where, as is easily verified, R x {0} constitutes an ideal of R'. We may

PROBLEMS

35

therefore regard R as being an ideal of the ring R'. Our hypothesis now comes into play and asserts that R' = R EB J for a suitable ideal J of R'. It is thus possible to choose an element (e, n) in J so that (0, -1) = (r, 0) + (e, n), for some r E R. The last-written equation tells us that e = - r and n = -1; what is important is the resulting conclusion that (e, -1) E J. For arbitrary r E R, the product (r, O)(e, -1) = (re - r, 0) will consequently be in both Rand J (each being an ideal of R'). The fact that R n J = {0} forces (re - r, 0) = (0, 0); hence, re = r. In a like fashion, we obtain er = r, proving that R admits the element e as an identity.

PROBLEMS 1. If I is a right ideal and J a left ideal of the ring R such that I n J that ab = 0 for all a E I, b E J.

=

{0}, prove

2. Given an ideal I of the ring R, define the set C(I) by C(I) = {r E Rfra - arE I for all a E R}. Verify that C(I) forms a subring of R. 3. a) Show by example that if I and J are both ideals of the ring R, then I u J need not be an ideal of R. b) If {I;} (i = 1, 2, ... ) is a collection of ideals ofthe ring R such that/ 1 s I 2 s · ·· s I. £ ···,prove that v I 1 is also an ideal of R. 4. Consider the ring M.(R) of n x n matrices over R, a ring with identity. A square matrix (a;) is said to be upper triangular if aii = 0 for i > j and strictly upper triangular if aii = 0 for i ~ j. Let T,(R) and T!(R) denote the sets of all upper triangular and strictly upper triangular matrices in M.(R), respectively. Prove each of the following: a) T,(R) and T!(R) are both subrings of M.(R). b) T!(R) is an ideal of the ring T,(R). c) A matrix (a;) E T,(R) is invertible in T,(R) if and only if aii is invertible in R for i = 1, 2, ... , n. [Hint: Induct on the order n.] d) Any matrix (aii) E T!(R) is nilpotent; in particular, (aut = 0. 5. Let I be an ideal of R, a commutative ring with identity. For an element a E R, the ideal generated by the set I u {a} is denoted by (I, a). Assuming that a¢ I, show that

(I, a)= {i

+ raliel,reR}.

6. In the ring Z of integers consider the principal ideals (n) and (m) generated by the integers n and m. Using the notation of the previous problem, verify that

((n), m) = ((m), n) = (n)

+ (m) = (n, m) =

where d is the greatest common divisor of n and m.

(d),

36

FIRST COURSE IN RINGS AND IDEALS

7. Suppose that I is a left ideal and J a right ideal of the ring R. Consider the set

1J

= {I: a1b1 ia1 E I; b1 E J},

where I: represents a finite sum of one or more terms. Establish that IJ is a twosided ideal of R and, whenever I and J are themselves two-sided, that IJ S I (") J.

8. If S is any given nonempty subset of the ring R, then ann,S

= {r E Riar =

0 for all a e S}

is called the right annihilator of S (in R); similarly, ann1S

= {r E Rira

= 0 for all a

E

S}

is the left annihilator of S. When R is a commutative ring, we simply speak of the annihilator of S and use the notation ann S. Prove the assertions below: a) ann,S (ann1 S) is a right (left) ideal of R. b) If S is a right (left) ideal of R, then ann,S (ann S) is an ideal of R. c) If Sis an ideal of R, then ann,S and ann 1S are both ideals of R. d) When R has an identity element, ann,R = ann 1R = {0}. 9. Let I 1, I 2 , ••• , I. be ideals of the ring R with R = I 1 + I 2 + ··· + I.. Show that this sum is direct if and only if a 1 + a 2 + ··· + a. = 0, with a 1 E I 1, implies that each a1 = 0. 10. If P(X) is the ring of all subsets of a given set X, prove that a) the collection of all finite subsets of X forms an ideal of P(X); b) for each subset Y s X, P(Y) and P(X - Y) are both principal ideals of P(X), with P(X) = P(Y) ffi P(X - Y). 11. Suppose that R is a commutative ring with identity and that the element a E R is an idempotent different from 0 or l. Prove that R is the direct sum of the pnnctpal ideals (a) and (1 - a).

12. Let I, J and K be ideals of the ring R. Prove that a) I(J + K) = lJ + IK, (I + J)K = IK + JK; b) if I 2 J, then I (") (J + K) = J + (I n K). . 13. Establish that in the ring Z, if I

= (12) and J = (21), then

= (84), IJ = (252), I:J [Hint: In general, (a):(b) = (c), where c = afgcd (a, b).] I

+J=

(3),

In J

= (4),

J:I=(7).

14. Given ideals I and J of the ring R, verify that a) 0:,/ = ann 1 /, and 0: 1 / = ann,/ (notation as in Problem 8); b) I:,J (I: 1 J) is the largest ideal of R with the property that (/:,J)J (J(I :1 J) s I).

s I

15. Let I, J and K be ideals of R, a commutative ring with identity. Prove the following assertions:

PROBLEMS

37

a) If I ~ J, then I:K ~ J:K and K:I 2 K:J. b) I:J"+l = (I:J"):J = (I:J):J" for any n E z+. c) J:J = R if and only if J ~I. d) I:J = I:(I + J).

16. If I is a right ideal of R, a ring with identity, show that I :1 R is the largest two-sided ideal of R contained in I.

=

{a e RJRa ~ I}

17. Given that f is a homomorphism from the ring R onto the ring R', prove that a) f(cent R) ~ cent R'. b) If R is a principal ideal ring, then the same is true of R'. [Hint: For any a e R,

f((a))

=

(!(a)).]

c) If the element a e R is nilpotent, then its image f(a) is nilpotent in R'. 18. Let R be a ring with identity. For each invertible element a e R, show that the function f.: R-+ R defined by .f.(x) = axa- 1 is an automorphism of R.

19. Letfbe a homomorphism from the ring R into itself and S be the set of elements that are left fixed by f; in symbols, S

= {a e RJJ(a) = a}.

Establish that S forms a subring of R.

20. Iff is a homomorphism from the ring R into the ring R', where R has positive characteristic, verify that char f(R) ::;;; char R. 21. Letfbe a homomorphism from the commutative ring R onto the ring R'. If I and J are ideals of R, verify each of the following: a) f(I + J) = f(l) + f(J); b) f(IJ) = f(l)f(J); c) f(I (') J) ~ f(l) (') f(J), with equality if either I ~ kerf or J ~ kerf; d) f(I:J) £;; f(I):f(J), with equality if I ~kerf 22. Show that the relation R

~ R'

is an equivalence relation on any set of rings.

23. Let R be an arbitrary ring. For each fixed element a E R, define the left-multiplication function T,: R -+ R by taking T,(x) = ax. If TR denotes the set of all such functions, prove the following: a) T, is a (group) homomorphism of the additive group of R into itself; b) TR forms a ring, where multiplication is taken to be functional composition; c) the mappingf(a) = T, determines a homomorphism of R onto the ring TR; d) the kernel ofjis the ideal ann 1 R; e) if for each 0 =/= a E R, there exists some b E R such that ab =/= 0, then R ~ TR. (In particular, part(e) holds whenever R has an identity element.)

24.. Let R be an arbitrary ring and R x Z be the extension ring constructed in Theorem 2-12. Establish that a) R x {0} is an ideal of R x Z; b) Z ~ {O}xZ; c) if a is an idempotent element of R, then the pair (-a, 1) is idempotent in R x Z, while (a, 0) is a zero divisor.

38

FIRST COURSE IN RINGS AND IDEALS

25. Suppose that R is a ring of characteristic n. If addition and multiplication are defined in R X = {(x, a)Jx E R; a E Z.} by

z.

+ (y, b) = (x + y, a +.b), (x, a)(y, b) = (xy + ay + bx, a·.b),

(x, a)

prove that R x z. is an extension ring of R of characteristic n. Also show that Rx has an identity element.

z.

26. Let R = R 1 Ef) R 2 Ef) ... Ef) R. be the (external) direct sum of a finite number of rings R; with identity (i = 1, 2, ... , n). a) For fixed i, define the mapping n;: R---+ R; as follows: If a= (a 1, a 2 , ... , a.), where ai e Ri, then n;(a) = a;. Prove that n; is a homomorphism from the ring R onto R;. b) Show that every ideal I of R is of the form I= 1 1 Ee 12 Ee ... Ee 1., with I; an ideal of R;. [Hint: Take I; = n;(/). If b; e I;. then there exists some (b 1, ... , b1, ... , b.) e I. It follows that (b 1, ... , b1, ... , b.)(O, ... , I, ... , 0) = (0, ... , b 1, ... , 0) E

I.]

27. A nonempty subset A of a ring R is termed an adeal of R if (i) a, b E A imply a + b E A, (ii) r e R and a e A imply both ar e A and ra e A. Prove that a) An adeal A of R is an ideal of R if for each a e A there is an integer n =t= depending upon a, such that na e aR + Ra. (This condition is satisfied, particular, if R has a multiplicative identity.) b) Whenever R is a commutative ring, the condition in part (a) is a necessary well as sufficient condition for an adeal to be an ideal. [Hint: For any a e the set A = {naJn e Z+} + aR is an adeal of R; hence, an ideal of R.]

0, m as R,

28. Let R be a ring with identity and M.(R) be the ring of n x n matrices over R. Prove the following: a) If I is an ideal of the ring R, then M.(l) is an ideal of the matrix ring M.(R). b) Every ideal of M.(R) is of the form M.(I), where I is an ideal of R. [Hint: Let F;i(a) denote the matrix in M.(R) having a as its ijth entry and zeroes elsewhere. For any ideal ..It in M.(R), let I be the set of elements in R whtch appear as entries for the matrices in .If. Given any a e I, say a is the rsth entry of a matrix A e .It, it follows that F;i(a) = F~r(l)AFsJ(l) e .If.] c) If R is a simple ring, then so is the matrix ring M.(R). 29. Let R be a ring with the property that every subring of R is necessarily an ideal of R. (The ring Z, for instance, enjoys this property.) If R contains no divisors of zero, a e R, consider the prove that multiplication is commutative. [Hint: Given 0 subring S generated by a. For arbitrary be R, ab = reS, so that ar = ra.]

+

PROBLEMS

13

where p is a prime, then we are able to deduce considerably more: each nonzero element of Zl is invertible. Before establishing this, first observe that the set Zl, regarded as an additive cyclic group of order p, consists of p distinct elements, namely, the p sums nl, where n = 0, 1, ... , p - 1. Now let nl be any nonzero element of Zl (0 < n < p). Since nand pare relatively prime, there exist integers r and s for which rp + sn = 1. But then 1 = (rp + sn)1 = r(p1) + (sl)(n1). As pi = 0, we obtain the equation 1 = (sl)(n1), so that sl serves as the multiplicative inverse of nl in Zl. The value of these remarks will have to await further developments (in particular, see Chapter 4). PROBLEMS I. Verify that the zero element of a ring R is unique, as is the additive inverse of each element a e R.

l. Let R be an additive commutative group. If the product of every pair of elements is defined to be zero, show that the resulting system forms a commutative ring (this is sometimes called the zero ring). 3. Prove that any ring R in which the two operations are equal (that is, a for all a, be R) must be the trivial ring R = {0}. 4. In a) b) c) d)

+ b = ab

a ring R with identity, establish each of the following: the identity element for multiplication is unique, if a e R has a multiplicative inverse, then a-t is unique, if the element a is invertible, then so also is -a, no divisor of zero can possess a multiplicative inverse in R.

5. a) If the set X contains more than one element, prove that every nonempty proper subset of X is a zero divisor in the ring P(X). b) Show that, if n > 1, the matrix ring M.(R) has zero divisors even though the ring R may not. 6. Suppose that R is a ring with identity 1 and having no divisors of zero. For a, b e R, verify that a) ab = 1 if and only if ba = 1, b) if az = 1, then either a = 1 or a = -1. 7. Let a, b be two elements of the ring R. If n e Z + and a and b commute, derive the binomial expansion

(a

+ br =

fi'

+ (i)a"- 1b + ··· +

(:)a"-~ I. By virtue of the remarks on page 42, ZP" has exactly one ideal for each positive divisor of p" and no other ideals; these are simply the principal ideals (pk) = pkZP" (0 :s:;; k :s:;; n). For 0 < k ~ n, we have

{p")"

= {p"/c) = (0) = {0},

so that each proper ideal of ZP" is nilpotent. Before leaving this chapter, we should present an example to show that, in general, nil and nilpotent are different concepts. Example 3-3. For a fixed prime p, let S be the collection of sequences = {a,} with the property that the nth term a, E ZP" (n 2 1). S can be made into a ring by performing the operations of addition and multiplication term by term:

a

{a,.}

+ {b,}

=

{a,.

+ b,.},

The reader will find that the zero element of this ring is just the sequence formed by the zero elements of the various ZP" and the negative of {a,} is {-a,}. Now, consider the set R of all sequences inS which become zero

49

PROBLEMS

after a certain, but not fixed, point. One may easily check that R constitutes a subring of the ring S (in fact, R is not only a subring, but actually an ideal of S). It is in the ring R that we propose to construct our example of a non-nilpotent nil ideal. Let us denote by I the set of sequences in R whose nth term belongs to the principal ideal in ZP" generated by p; in other words, the sequence a e I if and only if it is of the form a = (pr 1 , pr2 ,

... ,

pr,., 0, 0, ... )

(r,. e z,.).

A routine calculation confirms that I is an ideal of R. Since each term of a is nilpotent in the appropriate ring, it follows that ~

=

0

=

(0, 0, 0, ... )

for n large enough, making I a nil ideal. (This also depends on the fact that a has only a finite number of nonzero terms.) At the present stage, it is still conceivable that I might be a nilpotent ideal of R. However, we can show that for each positive integer n there exist elements (sequences) a e I for which a" =I= 0. For instance, define a = {a,.} by taking a,. = p if k = 1, 2, ... , n + 1 and a" = 0 if k > n + 1; that is, with a = (p, ... , p, p, 0, ... ) n + 1 p's. One then obtains

a" = (0, ... , 0, p", 0, ... ), where all the terms are zero except the (n + l)st, which is p". Since p" is a nonzero element of the ring Z pn. '• the sequence a" =I= 0, implying that I" =1= {0}. As this argument holds for any n e Z+, the ideal I cannot be nilpotent. We shall return to these ideas at the appropriate place in the sequel, at which time their importance will become clear.

PROBLEMS

L Let

=

=

be an equivalence relation on the ring R. We say that is compatible (with the ring operations) if and only if a b implies a + c = b + c, ac be, ca = cb for all a, b, c E R. Prove that there i~ a one-to-one correspondence between the ideals of R and the set of compatible equivalence relations on R.

=

2. If R is an arbitrary ring and n E Z+, prove that a) the sets I. = {naia E R} and J. = {a E Rlna = 0} are both ideals of R; b) char {R/1,.) divides n; c) if char R ::/= 0, then char R divides n char {R/J,.).

=

50

FIRST COURSE IN RINGS AND IDEALS

3. Let I be an ideal of the ring R. Establish each of the following: a) R/I has no divisors of zero tf and only if abE I implies that either a orb belongs to I. b) R/I is commutative if and only if ab - ba e I for all a, b in R. c) R/I has an identity element if and only if there is some e e R such that ae - a e I and ea - a e I for all a in R. d) Whenever R is a commutative ring with identity, then so is the quotient ring R/I.

4. Let R be a commutative ring with identity and let N denote the set of all mlpotent elements in R. Verify that a) The set N forms an Ideal of R. [Hint: If a"= bm = 0 for mtegers nand m, consider (a - br+"'.] b) The quotient ring R/N has no nonzero nilpotent elements. 5. Prove the following generalization of the Factorization Theorem: Let ! 1 and ! 2 be homomorphisms from the ring R onto the rmgs R 1 and R 2 , respectively. If kerf1 ~ kerf2 , then there exists a unique homomorphism!: R 1 -+ R 2 satisfymg f 2 = Joj~. [Hint: Mimic the argument of Theorem 3 6; that is, for any element f 1(a) E R 1, definel(II(a)) = f 2 (a).] 6. Let I be an ideal of the ring R. Assume further that J and K are two subnngs of R with I ~ J, I ~ K. Show that a) J ~Kif and only ifnat1 J ~ nat1 K b) nat1 (J f"\ K) = nat 1J n nat1K. 7. If I is an ideal of the ring R, prove that a) R/I is a simple ring if and only if there is no ideal J of R satisfying I c J c R; b) if R is a principal ideal ring, then so is the quotient rmg R1l; m particular, z. is a principal Ideal ring for each n e Z+. [Hint: Problem 17, Chapter 2.] 8. a) Given a homomorphism f from the ring R onto the rmg R', show that 1(bJib E R'} constitutes a partition of R into the cosets of the Ideal kerf [Hint: If b = f(a), then the coset a+ kerf= f- 1(b).] b) Verify that (up to isomorphism) the only homomorphic images of the ring Z of i9tegers are the rings z., n ~ 0, and {0}. . .:;'

u-

9. Suppose that S ts a subring and I an ideal of the ring R. If S n I = {0}, prove that S is isomorphic to a subring of the quotient ring Rjl. [Hint: Utilize the mappingf(a) = a + I, where a E S.J 10. A commutator in a ring R is defined to be any element of the form [a, b] = ab - ba. The commutator ideal of R. denoted by [R, R], is the ideal of R generated by the set of all commutators. Prove that a) R is a commutative ring if and only if [ R, R] = { 0} (in a sense, the size of [ R, R] provides a measure of the noncommutativity of R); b) for an ideal I of R, the quotient ring R/1 is commutative if and only if

[R, R]

~I.

11, Assuming that f is a homomorphism from the ring R onto the commutative ring R', establish the assertions below:

PROBLEMS

51

a) [R, R] ~ kerf; b) f = Jo nat1R RJ• whereJis the induced mapping; c) if kerf~ [R, R], then R/[R, R] ~ R'/[R', R']. 12. a) Suppose that I 1 and I 2 are ideals of the ring R for which R = I 1 ffi I 2 • Prove that R/I 1 ~ Iz, and R/I2 ~ I 1 • b) Let R be the direct sum of the rings R; (i = 1, 2, ... , n). If I; is an ideal of R; and I= I 1 ffi I 2 ffi ··· ffi I., show that R/I ~ (R 1/I 1 ) ffi (R 2 /I2 ) ffi ... ffi (R./I.). [Hint: Find the kernel of the homomorphism[: R----> l: ffi (R;/1;) that sends = (a 1, a2, ... , a.) to f(a) = (a 1 + II> a2 + I 2 , ... , a. + I.).]

a

13. For a proof of Theorem 3-9 that does not depend on the Fundamental Homomorphism Theorem, define the function h: R/I----> R'/f(I) by taking h(a + I) = f(a) + f(I~ a) Show that his a well-defined isomorphism onto R'/f(I); hence, R/I =::: R'/f(l). b) Establish that h is the unique mapping that makes the diagram below commutative: R

nat1

LR'=f(R)

l

1 R/I

h

nat f(I)

R'/f(I)

14. Given integers m, n e Z +• establish that a) if m divides n, then z.j(m)/(n) ~ Z,.; b) if m and n are relatively prime, then z_

~

Z,. ®

z•.

15. If I is an ideal of the ring R, prove that the matrix ring M.(R/1) is isomorphic to M.(R)/M.(I). [Hint: Consider the mapping f: M.(R)----> M.(R/I) defined by f({aiJ)) = (au + I).] 16. Let R be a ring without divisors of zero. Imbed R in the ring R' = R x Z, as illustrates that R' may contain descnbed in Theorem 2-12. (The case R = zero divisors even though R does not.) Assuming that I denotes the left annihilator of R in R',

z.

I

= {a e R'lar = 0 for all r e R},

verify that . a) I forms an ideal of R'. [Hint: R is an ideal of R'.] b) R'/I is a ring with identity which has no divisors of zero. c) R'/I contains a subring isomorphic toR. [Hint: Utilize Problem 9.]

FOUR

INTEGRAL DOMAINS AND FIELDS

In the preceding chapters a hierarchy of special rings has been established by imposing more and more restrictions on the multiplicative semigroup of a ring. At first glance, one might be tempted to require that the multiplicative semigroup actually be a group; such an assumption would be far too demanding in that this situation can only take place in the trivial ring consisting of zero alone. A less stringent condition would be the following: the nonzero elements comprise a group under multiplication. This leads to the notion of a field. Definition 4-1. A ring F is said to be a field provided that the set F - {0} is a commutative group under the multiplication of F (the identity of this group will be written as 1). Definition 4-1 implicitly assumes that any field F contains at least one element different from zero, for F - {0} must be nonempty, serving as the set of elements of a group. It is also to be remarked that, since aO = 0 = Oa for any a E F, all the members ofF commute under multiplication and not merely the nonzero elements. Similarly, the relation lO = 0 = 01 implies that 1 is the identity for the entire ring F. Viewed otherwise: a field is a commutative ring with identity in which each nonzero element possesses an inverse under multiplication. Occasionally, we shall find it convenient to drop the requirement of commutativity in the consideration of a field, in which case the resulting system is called a division ring or skew field. That is to say, a ring is a division ring if its nonzero elements form a group (not necessarily commutative) with respect to multiplication. After this preamble, let us look at several examples. Example 4-1. Here are some of the more standard illustrations of fields: the set Q of all rational numbers, the set F = {a + bJ2Ia, bE Q}, and the set R # of all real numbers. In each case the operations are ordinary addition and multiplication. 52

INTEGRAL DOMAINS AND FIELDS

53

Example 4-2. Consider the set C = R # x R # of ordered pairs of real numbers. To turn C into a field, we define addition and multiplication by

+ (c, d) = (a + c, b + d), (a, b)(c, d) = (ac - bd, ad + be).

(a, b)

The reader may verify without difficulty that C, together with these operations, is a commutative ring with identity. In this setting, the pair (1, 0) serves as the multiplicative identity, and (0, 0) is the zero element of the ring. Given any nonzero element (a, b) of C, either a =!= 0 or b =1= 0, so that a2 + b2 > 0; thus,

exists inC and has the property that

This shows that each nonzero member of C has an inverse under multiplication, thereby proving the system C to be a field. It is worth pointing out that the field C contains a subring isomorphic to the field of real numbers. For, if R# x {0} = {(a, O)la e R#},

it follows that R# ~ R# x {0} via the mapping f defined by f(a) = (a, 0). Inasmuch as the distinction between these systems is only one of notation, we customarily identify the real number a with the corresponding ordered pair (a, 0); in this sense, R # may be regarded as a subring of C. Now, the definition of the operations in C enables us to express an arbitrary element (a, b) e C as (a, b) = (a, 0)

+ (b, 0)(0, 1),

where the pair (0, 1) is such that (0, 1) 2 = (0, 1)(0, 1) = ( -1, 0). Introducing the symbol i as an abbreviation for (0, 1), we have (a, b) = (a, 0)

+ (b, O)i.

Finally, if it is agreed to replace pairs of the form (a, 0) by the first component a (this is justified by the preceding paragraph), the displayed representation becomes (a, b) = a

+

bi,

with

i2

=

-1.

In other words, the field C as defined initially is nothing more than a disguised version of the familiar complex number system.

54

FIRST COURSE IN RINGS AND IDEALS

Example 4-3. For an illustration of a division ring which is not a field, we turn to the ring of (Hamilton's) real quat ern ions. To introduce this ring, let the setH consist of all ordered 4-tuples of real numbers:

H

=

{(a, b, c, d)ja, b, c, dE R#}.

Addition and multiplication of the elements of H are defined by the rules

(a, b, c, d) + (a', b', e', d') = (a + a', b + b', c + e', d + d'), (a, b, c, d)(a', b', e', d') = (aa' - bb' - ce' - dd', ab' + ba' + ed' - de', ac' - bd' + ea' + db', ad' + be' - eb' + da'). A certain amount of tedious, but nonetheless straightforward, calculation shows that the resulting system is a ring (known as the ring of real quaternions) in which (0, 0, 0, 0) and (1, 0, 0, 0) act as the zero and identity elements, respectively. Let us next introduce some special symbols by putting 1

=

(1, 0, 0, 0),

i

=

(0, 1, 0, 0),

j = (0, 0, 0, 1, 0),

k = (0, 0, 0, 1).

The elements 1, i, j, k have a number of distinctive properties; specifically, 1 is the multiplicative identity of H and j2 = /

ij = k,

jk = i,

= k2 = -1,

ki = j, ji = - k,

kj = - i,

ik = - j.

These relations demonstrate that the commutative law for multiplication fails to hold in H, so that H definitely falls short of being a field. As in Example 4-2, the definition of the algebraic operations in H permits us to write each quaternion in the form

(a, b, c, d)

=

(a, 0, 0, 0)1

+

(b, 0, 0, O)i

+

(c, 0, 0, O}j

+ (d, 0, 0, O)k

Since the subring {(r, 0, 0, O)jr E R#} is isomorphic to R #,the notation can be further simplified on replacing (r, 0, 0, 0) by the.element r itself; adopting these conventions, the real quaternions may henceforth be regarded as the set

H ={a+ bi + ej + dkla,b,e,deR"'}, with addition and multiplication performed as for polynomials (subject to the rules of the last paragraph). The reader versed in linear algebra should recognize that H comprises a four-dimensional vector space over R # having

{1, i,j, k} as a basis. The main point in our investigation is that any nonzero quaternion q = a + bi + cj + dk (in other words, one of a, b, e, d must be different from zero) is a multiplicatively invertible element. By analogy with the complex numbers, each quaternion has a conjugate, defined as follows:

q = a - bi - cj - dk.

IDEALS AND THEIR OPERATIONS

19

i, whence the elements a - b, ar and ra belong to n I;, making n I; an ideal of R.

Consider, for the moment, an arbitrary ring R and a nonempty subset S of R. By the symbol (S) we shall mean the set

(S) = n

{II S

~I; I is an ideal of R}.

The collection of all ideals which contain S is not empty, since the entire ring itself is an ideal containing any subset of R; thus, the set (S) exists and satisfies the inclusionS ~ (S). By virtue of Theorem 2-2, (S) forms an ideal of R, known as the ideal generated by the set S. It is noteworthy that whenever I is any ideal of R with S ~ I, then necessarily (S) s I. For this reason, one often speaks of (S) as being the smallest ideal of R to contain the set S. It should be apparent that corresponding remarks apply to the one-sided ideals generated by S. If S consists of a finite number of elements, say a 1, a2 , ••. , am then the ideal which they generate is customarily denoted by (a 1 , a 2 , .•• , an). Such an ideal is said to be finitely generated with the given elements a; as its generators. An ideal (a) generated by just one ring element is termed a principal ideal.

A natural undertaking is to determine the precise form of the members of the various ideals (right, left, two-sided) generated by a single element, say a, of an arbitrary ring R. The right ideal generated by a is called a principal right ideal and is denoted by (a),. Being closed with respect to multiplication on the right, (a), necessarily contains all products ar (r E R), as well as the elements na (n an integer), and, hence, includes their sum ar + na. (As usual, the notation na represents the n-fold sum of a.) It is a fairly simple matter to check that the set of elements of the form ar + na constitutes a right ideal of R. Observe, too, that the element a is a member of the ideal, .since a = aO + la. These remarks make it clear that (a)r

= {a~+

Z}.

nair E R; n e

When there is an identity element present, the term na becomes superfluous, for, in this setting, we may write the expression ar + na more simply as ar

+

na = ar

+ a(nl)

= a(r

+

nl) = ar',

where r' = r + nl is some ring element. Thus, the set (a), consists of all right multiples of a by elements of R. If R is a ring with identity, we shall frequently employ the more suggestive notation aR in place of (a),; that is, (a), = aR = {arir E R}.

Similar remarks apply, of course, to the principal left ideal (a) 1 generated by a.

56

FIRST COURSE IN RINGS AND IDEALS

Corollary. Any finite integral domain is a field. Because the aforementioned corollary is so basic a result, we offer a second proof. The "counting argument" involved in this latter proof adapts to a variety of situations in which the underlying ring is finite. The reasoning proceeds as follows. Suppose that a 1 , a 2 , .•• , an are the members of the integral domain R. For a fixed nonzero element a E R, we consider the n products aa 1 , aa 2 , ••• , aan. These products are all distinct, for if aa; = aai, the cancellation law (valid in any integral domain) would yield a; = ai. It follows that each element of R must be of the form aa; for some choice ofi. In particular, there exists some ii; E R such that aii; = 1. From the commutativity of multiplication, we infer that a- 1 = ii;, whence every nonzero element of R possesses a multiplicative inverse. There are no finite division rings which are not fields. To put it another way, in a finite system in which all the field properties except the commutativity of multiplication are assumed, the multiplication must also be commutative. Proving this renowned result is far from being as elementary as the case of a finite integral domain and is deferred until Chapter 9. For the moment, let us take a closer look at the multiplication structure of zn. It has been previously shown that, for each positive integer n, zn comprises a commutative ring with identity. A reasonable question is: For precisely what values of n, if any at all, will this ring turn out to be a field? For a quick answer: n must be a prime number. (What could be simpler or more natural?) This fact is brought out by the coming theorem. Theorem 4-3. A nonzero element [a] E Zn is invertible in the ring Zn if and only if a and n are relatively prime integers (in the sense that gcd(a, n) = 1). Proof If a and n are relatively prime, then there exist integers r and s such that ar + ns = 1. This implies that

[1] = [ar + ns]

= [ar] +n [ns] =

[ar] +n [OJ

= [a] "n [r],

showing the congruence class [r] to be the multiplicative inverse of [a]. Now to the "only if" part. Assume [a] to be a multiplicatively invertible elementofZn;say,withinverse[b]. Wethushave[ab] =[a] ·n [b] = [1], so that there exists an integer k for which ab - 1 = kn. But then ab + n(- k) = 1 ; hence, a and n are relatively prime integers. Corollary. The zero divisors of Zn are precisely the nonzero elements of zn which are not invertible. Proof Naturally, no zero divisor of Zn can possess a multiplicative inverse. On the other hand, suppose that [a] =1= [OJ is not invertible in Zn, so that

INTEGRAL DOMAINS AND FIELDS

57

gcd(a, n) = d, where 1 < d < n. Then, a = rd and n = sd for suitable nonzero integers r and s. This leads to

[a]·,. [s]

=

[as]

=

[rds]

=

[rn]

=

[OJ.

Since the defining properties of s rule out the possibility that [s] = [0], it follows that [a] is a zero divisor of Z,.. These results may be conveniently summarized in the following statement. Theorem 4-4. The ring Z,. of integers modulo n is a field if and only if n is a prime number. If n is composite, then z. is not an integral domain and the zero divisors of Z,. are those nonzero elements [a J for which gcd(a, n) =I= 1. Every field necessarily has at least two elements (1 being different from 0); Theorem 4-4 indicates that there is a field having this minimum number as its number of elements, viz. Z 2 . As an interesting application of these ideas, consider the following assertion: If there exists a homomorphism f: Z --+ F of the ring Z of integers onto a field F, then F is necessarily a finite field with a prime number of elements. For, by the Fundamental Homomorphism Theorem, Z/ker f ~ F. But kerf = (n) for some positive integer n, since Z is a principal ideal domain. (In this connection, observe that n =I= 0, for otherwise Z would be isomorphic to a field, an impossibility.) Taking stock of the fact that Z/(n) = z., we are thus able to conclude that z. ~ F, in consequence of which F has n elements. At this point Theorem 4-4 comes to our aid; since F, and in turn its isomorphic image Z,., forms a field, n must be a prime number. A useful counting function is the so-called Euler phi-function (totient), defined as follows: ¢(1) = 1 and, for each integer n > 1, ¢(n) is the number of invertible elements in the ring Z,.. By virtue of Theorem 4-~~-may also be_characterized_as....the number of positive_ integers < n_which arerel~tiv~ly prime to n. For instance, -¢(6) = f. ¢(9) = 6, and ¢(12) = 4; it should beequallyclear that whenever pis a prime number, then ¢(p) = p - 1. Lemma. If G,. is the subset of Z,. defined by G,. = {[a] e Z,.la is relatively prime to n}, then (G,., ·,.)forms a finite group of order ¢(n). Proof In the light of the preceding remarks, (G,., ·,.) is simply the group of invertible elements of Z,..

This leads at once to a classical result of Euler concerning the phifunction; the simplicity of the argument illustrates the advantage of the algebraic approach to number theory.

58

FIRST COURSE IN RINGS AND IDEALS

Theorem 4-5. (Euler-Fermat). Ifn is a positive integer and a is relatively prime to n, then atf>(n), it follows that [a]4> on R and so determined uniquely, if it exists at all. These remarks suggest that, in attempting to extend , we should consider the assignment: for all

INTEGRAL DOMAINS AND FIELDS

65

For a verification that is a well-defined function, let ab- 1 = cd- 1 in Qci(R); that is to say, ad = be in R. Then. the equation c/J(a)(d) = c/J(b)c/J(c) holds in R' £ Qc 1(R') or, viewed otherwise, (a)(b)- 1 = c/J(c)c/J(dr 1 . But this means (ab - 1 ) = (cr 1 ), so that does not depend on how an element in Qc1(R) is expressed as a quotient. One verifies routinely that , as defined above, is a homomorphism of Qci(R) into Qc1(R'). This homomorphism certainly extends cfJ; indeed, if a is an arbitrary element of R and b is a non-zero-divisor of R, maps a= (tib)b-Litito (a) = c/J(ab)(b)- 1

= c/J(a)c/J(b)c/J(b)- 1

=

c/J(a).

To see that is a one-to-one function, we examine its kernel. Now, if (ab- 1 ) = 0, then c/J(a) = 0. But, cjJ being an isomorphism, this implies that a = 0, whence ab- 1 = 0. Accordingly, ker = {0}, which forces to be one-to-one. Without going into the details, we also point out that carries Qc~(R) onto Qc 1(R') (this stems from the fact that cjJ maps onto R'). Therefore, is the desired extension of cp. A special case of particular importance occurs when R and R' are the same ring and cjJ is taken to be the identity isomorphism on R. Corollary. Any two quotient rings of a commutative ring R with at least one non-zero-divisor are isomorphic by a unique mapping fixing all the elements of R.

At this point we leave the theory of quotients and turn to prime fields. Clearly, any field F has at least one subfield, namely, F itself; a field which does not possess any proper subfields is called a prime field. Example 4-4. The field Q of rational numbers is the simplest example of a prime field. To see this, suppose that F is any subfield of Q and let a E F be any nonzero element. Since F is a subfield of Q, it must contain the product aa- 1 = l. In turn, n = n1 E F for any n in Z; in other words, F contains all the integers. It then follows that every rational number nfm = nm - 1 (m =I= 0) also lies in F, so that F = Q. Example 4-5. For each prime p, the field ZP of integers modulo pis a prime field. The reasoning here depends on the fact that the additive group (Zp, +p) is a finite group of prime order and therefore by Lagrange's theorem has no non-trivial subgroups.

An observation which will not detain us long is that each field F contains a unique prime subfield. To make things more specific, let {F;} be the collection of all subfields of F. Then the intersection n F; is also a subfield of F. Now, ifF' is any subfield of the field n F;, then F' E {F;}. whence n F; s; F'; the implication is that F' = n F;, forcing n F; to be a prime

66

FIRST COURSE IN RINGS AND IDEALS

field. As regards the uniqueness assertion, suppose that K 1 and K 2 are both prime subfields of F. Then K 1 n K 2 is a subfield ofF as well as K 1 , with K 1 ~ K 1 n K 2 . But K 1 can possess no proper subfields, which signifiesthatK 1 = K 1 n K 2 . Likewise,K 2 = K 1 n K 2 ,whenceK 1 = K 2 • We conclude this chapter by showing that, to within isomorphism, the rational number field and the fields Z P are the only prime fields. Theorem 4-12. Any prime field F is isomorphic either to Q, the field of rational numbers, or to one of the fields ZP of integers modulo a prime p. Proof To begin, let 1 be the multiplicative identity of F and define the mapping f: Z--+ F by f(n) = n1 for any integer n. Then f is a homomorphism from Z onto the subring Zl of integral multiples of 1. In compliance with Theorem 3-7, we therefore have Z/ker f ~ Zl. But kerf is an ideal of Z, a principal ideal domain, whence kerf= (n) for some nonnegative integer n. The possibility that n = 1 can be ruled out, for otherwise 1 = f(l) = 0 or, what amounts to the same thing, F = {0}. Notice further that if n =F 0, then n must in fact be a prime number. Suppose to the contrary that n = n 1n2 , where 1 < n; < n. Since n E kerf, it follows that (n 1 l)(n 2 1) = (n 1 n2 )1 = nl = 0, yielding the contradiction that the field F has divisors of zero. (This result is not entirely unexpected, because the integer n is the characteristic of F and as such must be a prime, whenever n =f 0.) The preceding discussion indicates that two possibilities arise: either 1) Zl 2) Zl

~ ~

Z/(p) = ZP for some prime p, or Z/(0) = Z.

Turning to a closer analysis of these cases, assume first that Zl ~ ZP, with p prime. Inasmuch as the ring of integers modulo a prime forms a field, the subring Zl must itself be a field. But· F, being a prime field, contains no proper subfields. Accordingly, Zl = F, which leads to the isomorphism F ~ ZP. For the final stage of the proof, consider the situation where Zl ~ z. Under these circumstances, the subring Zl is an integral domain, but not a field. Taking stock of Theorem 4-8, as well as the hypothesis that F is a prime field, we conclude that F

= {ab- 1 la,beZl;b =f 0} = {(nl)(ml)- 1 ln, me Z; m

+ 0}.

It is now a purely routine matter to verify that the fields F and Q are isomorphic under the mapping g(n/m) = (nl)(ml)- 1 ; we leave the details as

an exercise.

PROBLEMS

67

Since every field contains a unique prime subfield, the following subsidiary result is of interest. Corollary l. Every field contains a subfield which is isomorphic either to the field Q or to one of the fields ZP. Theorem 4-12 also provides some information regarding field automorphisms. Corollary 2. Iff is an automorphism of the field F, then f(a) = a for each element a in the prime subfield of F (hence, a prime field has no automorphism except the identity). Proof. The prime subfield ofF is either

F 1 = {(n1)(m1)- 1 ln,meZ; m =I= 0}

or F 2 = {n1ln = 0,1, ... ,p- 1}, according as the characteristic ofF is 0 or a prime p. Since any automorphism of a field carries the identity 1 onto itself, the result should be clear.

PROBLEMS I. a) Assuming that R is a division ring, show that cent R forms a field. b) Prove that every subring, with identity, of a field is an integral domain. 2. Let R be an integral domain and consider the set Zl of all integral multiples of the identity element:

Zl

= {nljn E Z}.

Establish that Zl is a field if and only if R has positive characteristic. 3. In the field C, define a mapping f: C --+ C by sending each complex number to its conjugate; that is,f(a + bi) = a - bi. Verify that f is an automorphism of C. 4. Find the center of the quatemion ring H.

5. Let R be the subring of M 2 (C) consisting of all matrices of the form

(_j

~}=

(-:: ::

~ ~ :) (~b,c,deR#).

Prove that R is a division ring isomorphic to the division ring of real quaternions.

6. By the quaternions over a field F is meant the set of all q = a + bi + cj + dk, where a, b, c, d e F and where addition and multiplication are carried out as with real quaternions. Given that F is a field in which a 2 + b2 + c 2 + d2 = 0 if and only if a = b = c = d = 0, establish that the quatemions over F form a division ring.

68

FIRST COURSE IN RINGS AND IDEALS

7. Establish the following facts concernmg the Euler phi-function: a) If n and mare relatively prime integers, then rj>(nm) = rp(n)rp(m). b) For any prime p and n > 0, rj>(p") = p"(l- 1/p) = p"- p"- 1 • [Hint: The integers k such that 0 < k < p" and gcd (k, pn) =I= 1 are p, 2p, ... , p"- 1 p.] c) If p 1, p 2 , ... , Pk are the distinct pnme divisors of an mteger n > 1, then rf>(n) = n(l - l/ptl(1 - l/p2 ) .. • (1 - 1/pk). d) n = Lain rf>(a). 8. Let r(n) denote the number of (distinct) positive divisors of an integer n > 1. Prove that a) If n has the prime factorization n = p~'p~1 . . . p~". where the P; are distinct primes and n1 e Z+, then -r(n) = (n 1 + l)(n 2 + 1) ... (nk + 1). b) The number of ideals of z. is -r(n). c) -r(n)r/>(n) ~ n. [Hint: D(n1 + l)TI(l - 1/pJ ~ 2kJJ(l/2n 9. Given that the set H. = {[a] E z.j[a] is not a zero divisor of z.]. prove that (H., ·.) forms a finite group of order rf>(n).

10. a) Derive Fermat's Little Theorem: If pis a prime number and a=!= 0 (mod p), then ap-l = 1 (mod p). b) If gcd (a, n) = 1, show that the equation ax = b (mod n) has a umque solution modulo n. [Hint: All solutions are given by x = bali>l•-~l+ kn.]

ll. a) Prove that every field is a principal ideal domain. ', jr· b) Show that the ring R = {a + bJ2ja, bE Z} is not a field by exhibiting a nontrivial ideal of R. 12. Let f be a homomorphism from the ring R into the ring R' and suppose that R has a subring F which is a field. Establish that either F ~ kerf or else R' contains a subring isomorphic to F. 13. Derive the following results: a) The identity element of a subfield is the same as that of the field. b) If {F;} is an index collection of subfields of the peld F, then n F; is also a subfield of F. c) A subring F' of a field F is a subfield ofF if and only ifF' contains at least one nonzero element and a- 1 E F' for every nonzero a e F'. d) A subset F' of a finite field F is a subfield of F if and only if F' contains more than one element and is closed under addition and multiplication.

14. a) Consider the subsetS of R# defined by

S = {a + b.J.Pia, be Q; p a fixed prime}. Show that S is a subfield of R #. b) Prove that any subfield of the field R# must contain the rational numbers.

15. Prove that if the field F is of characteristic p > 0, then every subfield of F has characteristic p.

PROBLEMS

16. Let F be a field of characteristic p > 0. Show that for fixed n E

69

z +}

F' = {aeFiaP" =a} is a subfield of F. 17. Let F be a field, F' a subfield ofF, and fan automorphism of F. We say that[ fixes an element a E F in case j (a) = a. Prove the following assertions: a) The set of all automorphisms ofF form a group (in which the binary operation is composition of functions). b) The set of all automorphisms ofF which fix each element ofF' comprise a group. c) If G is a group of automorphisms ofF, then the set of all elements of F that are fixed by G (that is, the set F(G) = {a e Fif(a) = a for allfe G}) is a subfield of F, known as the fixed field of G. viS. Let R be a commutative ring containing at least one non-zero-divisor. Prove that a) An element ab - I is a non-zero-divisor of Q., 1(R) if and only if a is a non-zerodivisor of R. b) If R has an identity and every non-zero-divisor of R is invertible in R, then R = Q.1(R); in particular, F = Q.1(F) for any field F. c) Q. 1(Q.1(R)) = Qc~(R). d) If R IS fimte, then R = Qt(R). [Hint: For any non-zero-divisor a E R, there is some b e R such that a 2 b = a; ab is idempotent; thus, R has an identity element 1 and ab = 1 by Problem 12, Chapter 1.] .,19. Utilize part (d) of the preceding problem to give another proof that any finite integral domain is a field. 20. Show that any field containing the integral domain R as a subring contains the field of quotients Q.1(R); in this sense, Q.1(R) is the smallest field containing R. 21. a) If R = {a + b.J2ia, bE Z}, then R forms an integral domain under ordinary addition and multiplication, but not a field. Obtain the field of quotients of R. Do the same for the domain z •. b) If K is a field of quotients of an integral domain R, prove that K is also a field of quotients of every subdomain of K containing R. 22. Let R be an arbitrary ring (not necessarily commutative) with at least one non-zerodivisor. Prove that R possesses a classical ring of quotients if and only If it satisfies the so-called Ore condition: for all a, b e R, b being a non-zero-divisor, there exist elements c, d e R, with d a non-zero-divisor such that ad = be. 23. Prove that any automorphism of an integral domain R admits a unique extension to the field of quotients Q.1(R~ 24. Let F be a field and Zl the set of integral multiples of the identity. Verify that the prime subfield ofF coincides with Q. 1(Zl). [Hint: Problem 20.] 25. Establish the following assertion, thereby completing the proof of Theorem 4--12: If F is a field of characteristic zero and K = {(n1)(ml)- 1 ln, me Z; m 0}

+

is the prime subfield ofF, then K ~ Q via the mappingf(n/m)

=

(nl)(m1)- 1 •

70

FIRST COURSE IN RINGS AND IDEALS

In Problems 26-29, R is assumed to be a commutative ring. 26. Let S be any multiplicatively closed subset of the ring R (that is, the product of any two elements of S again lies in S) which contains no zero divisor of R and 0 ¢ S. If the set R8 is defined by R8

=

{ab- 1 E Qc1(R)Ia e R, be S},

prove that R 8 is a subring of the ring of quotients quotients of R relative to S.

Qc~(R),

known as the ring of

27. a) Show that the set S = {n e ZIP .f n; p a fixed prime} is multiplicatively closed and determine Z 8 , the ring of quotients of Z relative to S. b) If R is any ring satisfying Z 5:; R 5:; Q, prove that R = Z 8 for a suitable multiplicatively closed subsetS 5:; Z. [Hint: Constder the setS = {me Zlfor some n e Z, n/m e R; gcd (n, m) = 1}.] 28. Let S be a multiplicatively closed subset of the ring R with 0 ¢ S. Prove the statements below: a) The set I = {a e Rias = 0 for somes e S} is an ideal of R. b) S/I = nat1 S is a multiplicatively closed subset of the quotient nng R/l. c) No element of S/1 is a zero divisor of R/I. (Thus, one can form the ring of quotients of RJI relative to S/I; the result is called the generalized ring of quotients of R relative to S.)

d) If S contains no zero divisor of R, then (R/1)811

=

R8 •

29. Let S be a multiplicatively closed subset of the ring R which contains no zero divisor of R nor zero. a) If I is an ideal of R, verify that the set JS- 1 = {ab- 1 e Qc1(R)Ia e I, be S} is anidealofQ z. is a maximal ideal of the external direct sum Z EB Z. b) Show that the ring R is a field if and only if {0} is a maximal ideal of R.

2. Prove that a proper ideal M of the ring R is maximal if and only if, for every element r ¢ M, there exists some a e R such that 1 + ra e M. 3. Letfbe a homomorphism from the ring R onto the ring R'. Prove that a) if M is a maximal ideal of R with M 2 kerf, thenf(M) is a maximal ideal of R', b) if M' is a maximal ideal of R', thenf- 1(M') is a maximal ideal of R, c) the mapping M -> f(M) defines a one-to-one correspondence between the set of maximal ideals of R which contain kerf and the set of all maximal ideals of R'.

4. If M 1 and M 2 are distinct maximal ideals of the ring R, establish the equality M,M 2 = M 1 n M2 • 5. Let M be a proper ideal of the ring R. Prove that M is a maximal ideal if and only if, for each ideal I of R, either I ~ M or else I -f. M = R. 6. An ideal I of the ring R is said to be minimal if I 1= {0} and there exists no ideal J of R such that {0} c J c I. a) Prove that a nonzero ideal I of R is a minimal ideal if and only if (a) = I for each nonzero element a E I. b) Verify that the ring Z of integers has no minimal ideals. 7. Let I be a proper ideal of the ring R. Show that I is a prime ideal if and only if the complement of I is a multiplicatively closed subset of R. 8. In the ring R

= map R #, define the set I by I= {feRJJ(l) =f(-1)

= 0}.

Establish that I is an ideal of R, but not a prime ideal.

PROBLEMS

87

9. a) With the aid of Theorem 5-5 and Example 5-1, obtain another proof of the fact that ZP is a field if and only if p is a prime number. b) Prove that in z. the maximal ideals are the principal ideals (p) = pZ., where p is a prime dividing n. 10. Given that f is a homomorphism from the ring R onto the ring R', verify that a) R' is a field if and only if kerf is a maximal ideal of R, b) R' is an integral domain if and only ifkerfis a prime ideal of R.

11. a) Show that if P 1 and P 2 are two ideals of the ring R such that P 1 'J P 2 and P 2 $ P 1 , then the ideal P 1 n P 2 is not prime. b) Let {P;} be a chain of prime ideals of the ring R. Prove that u P; and n P; are both prime ideals of R.

12. Prove that if I is an ideal of the ring R and P is a prime ideal of I, then P is an ideal of the whole ring R. 13. Let R denote the set of all infinite sequences {a.} of rational numbers (that is, a. e Q for every n). R becomes a commutative ring with identity if the ring operations are defined termwise:

{a,} + {b,}

=

{a, + b.},

Verify each of the following statements: a) the set B of bounded sequences is a subring (with identity) of R, b) the set C of convergent sequences is a subring (with identity) of B, c) the set C 0 of sequences which converge to zero is a subring of C, d) C0 is an ideal of B, but not a prime idea~ e) C0 is a maximal ideal of C, f) the set D of Cauchy sequences is a subring (with identity) of B, g) C0 is a maximal ideal of D. Remark. Since the field D/C 0 is isomorphic to R#, this provides an alternative procedure for constructing the real numbers [16].

14. Assume that P is a proper prime ideal of the ring R with the property that the quotient ring R/P is finite. Show that P must be a maximal ideal of R. 15. Let R = R 1 Efl R 2 Efl ... ffi R. be the direct sum of a finite number of rings R;. Establish that a proper ideal I of R is a maximal ideal if and only if, for some i (1 ~ i ~ n), I is of the form

I

= R1

E9 ••• ffi R1_ 1 ffi M 1 E9 R1+ 1 E9 ··• E9 R.,

where M; is a maximal ideal of R;. [Hint: Problem 26, Chapter 2.]

16. Let P and I be ideals of the ring R, with P prime. If I $ P, prove that the quotient ideal P:I = P. 17. Letfbe a homomorphism from the ring R onto the ring R'. Prove that a) if Pis a prime {primary) ideal of R with P ~ kerf, thenf(P) is a prime {primary) ideal of R';

IDEALS AND THEIR OPERATIONS

31

such that f(J) = 1'. To accomplish this, it is sufficient to take I = f - 1(1'). By the corollary to Theorem 2-8,f- 1 (1') certainly forms an ideal of R and, since 0 E I', kerf= f- 1 (0)

5;

f- 1 (I').

Inasmuch as the function f is assumed to be an onto map, it also follows that f(l) = f(f - 1 (1')) = I'. Next, we argue that this correspondence is one-to-one. To make things more specific, let ideals I and J of R be given, where kerf 5; I, kerf 5; J, and satisfying f(l) = f(J). From the elementary lemma just established, we see that I=

f-t(f(J)) = f-t(f(J)) =

J.

One finds in this way that the correspondence I - f(l), where kerf 5; I, is indeed one-to-one, completing the proo( Before announcing our next result, another definition is necessary. Definition 2-9. A ring R is said to be imbedded in a ring R' if there exists some subring S' of R' such that R ~ S'. In general, if a ring R is imbedded in a ring R', then R' is referred to as an extension of R and we say that R can be extended to R'. The most important cases are those in which one passes from a given ring R to an extension possessing some property not present in R. As a simple application, let us prove that an arbitrary ring can be imbedded in an extension ring with identity. Theorem 2-12. (Dorroh Extension Theorem). Any ring R can be imbedded in a ring with identity.

Proof. Consider the Cartesian product R x Z, where RxZ= {(r,n)lreR;neZ}.

If addition and multiplication are defined by (a, n) + (b, m) (a, n)(b, m)

=

(a

+ b, n +

m),

= (ab + ma + nb, nm),

then it is a simple matter to verify that R x Z forms a ring; we leave the actual details as an exercise. Notice that this system has a multiplicative identity, namely, the pair (0, 1); for (a, n)(O, 1)

=

(aO

+

1a

+

nO, n1)

and, similarly, (0, 1)(a, n) = (a, n).

=

(a, n),

PROBLEMS

89

d) If I is a proper ideal of R and b e R - I, then I s; bR. 29. For a fixed pnme p, consider the subset of rat10nal numbers defined by

vp =

{alb e Qjp ~ b}.

Show that a) ~ is a valuation ring of Q; b) the unique maximal ideal of~ is MP = {alb e Qjp {b, but pja}; c) the field ~IMP~ ZP. [Hint: Let the homomorphism/: VP--> ZP be defined by f(alb) = [a][br 1 .] 30. a) Let I 1 , I 2 , •.• , I. be arbitrary ideals of the ring R and P be a prime ideal of R. If I 1 I 2 ···I. s; P, establish that I; s; P for at least one value of i. [Hint: If I; $ P for all i, choose a; e I; - P and consider the element a = a 1a 2 ···a•. ] b) Assume that M is a maximal Ideal of R. Prove that, for each integer n e Z +> the only prime ideal containing M" is M. 31. Let R be an integral domain w1th the property that every proper ideal is the product of maximal ideals. Prove that a) If M is a maximal ideal of R, then there exists an element a E R and ideal K {0} such that MK = (a). [Hint: If M i= {0}, pick 0 a eM. Then M 1M 2 ••• M. =(a) s; M for suitable max1mal ideals M;; hence, M = M; for some i.] b) If I, J, Mare ideals of R, with M maximal, then IM = JM implies I = J.

+

+

32. a) If I is an ideal of the ring R such that I s; (a), show that there exists an ideal J of R for which aJ = I. [Hint: Take J = (I: (a)).] b) Prove that 1f a principal ideal (a) of the ring R properly contains a prime ideal P, then P s; {j (a"). n=l

33. Let I be a primary tdeal of the ring R. Prove that I has exactly one minimal prime ideal, namely, Jl. [Hint: Problem 19.]

SIX

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

As the title suggests, this chapter is concerned with the problem of factoring elements of an integral domain as products of irreducible elements. The particular impetus is furnished by the ring of integers, where the Fundamental Theorem of Arithmetic states that every integer n > 1 can be written, in an essentially unique way, as a product of prime numbers; for example, the integer 360 = 2·2·2·3·3·5. We are interested here in the possibility of extending the factorization theory of the ring Z and, in particular, the aforementioned Fundamental Theorem of Arithmetic to a more general setting. Needless to say, any reasonable abstraction of these numbertheoretic ideas depends on a suitable interpretation of prime elements (the building blocks for the study of divisibility questions in Z) in integral domains. Except for certain definitions, which we prefer to have available for arbitrary rings, our hypothesis will, for the most part, restrict us to integral domains. The plan is to proceed from the most general results about divisibility, prime elements, and uniqueness offactorization to stronger results concerning specific classes of integral domains. Throughout this chapter, the rings considered are assumed to be commutative; and it is supposed that each possesses an identity element. Definition 6-l. If a =fo 0 and b are elements·of the ring R, then a is said to divide b, in symbols alb, provided that there exists some c E R such that b = ac. In case a does not divide b, we shall write a J b. Other language for the divisibility property alb is that a is a factor of b, that b is divisible by a, and that b is a multiple of a. Whenever the notation alb is employed, it is to be understood (even if not explicitly mentioned) that the element a =fo 0; on the other hand, not only may b = 0, but in such instances we always have divisibility. Some immediate consequences of this definition are listed below; the reader should convince himself of each of them. Theorem 6-1. Let the elements a, b, c e R. Then, 1) a!O, t!a, ala; 90

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

2) 3) 4) 5)

91

ajl if and only if a is invertible; if ajb, then acjbc; if alb and bjc, then ajc; if cia and cjb, then cj(ax + by) for every x, y E R.

Division of elements in a ring R is closely related to ideal inclusion:

ajb if and only if (b) s; (a). Indeed, ajb means that b = acfor some c E R; thus, bE (a), so that( b) s; (a). Conversely, if (b) s; (a), then there exists an element c in R for which b = ac, implying that ai b. Questions concerning divisibility are complicated somewhat by the presence of invertible elements. For, if u has a multiplicative inverse, any element of a E R can be expressed in the form a = a(uu- 1), so that both uja and u- 1 1a. An extreme situation occurs in the case of fields, where every nonzero element divides every other element. On the other hand, in the ring ze of even integers, the element 2 has no divisors at all. In order to overcome the difficulty that is produced by invertible elements, we introduce the following definition. Definition 6-2. Two elements a, bE Rare said to be associated elements or simply associates if a = bu, where u is an invertible element of R. A simple argument shows that the relation -, defined on R by taking a - b if and only if a is an associate of b, is an equivalence relation with

equivalence classes which are sets of associated elements. The associates of the identity are just the invertible elements of R. Example 6-1.

In the case of the ring Z, the only associates of an integer n E Z are ± n, since ± 1 are the only invertible elements.

Example 6-2. Consider the domain Z(i) of Gaussian integers, a subdomain of the complex number field, whose elements form the set

Z(i) ={a+ bija,bEZ; i2 = -1}. Here, the only invertible elements are ± I and ± i. For, suppose a hasamultiplicativeinversec + di. Then,wemusthave(a + bi)(c so that (a - bi)(c - di) = 1. Therefore,

+ bi)(c + di)(a = (a2 + bz)(cz + dz).

1 = (a

+ bi E Z(i) + di) = 1,

bi)(c - di)

From the fact that a, b, c, d are all integers, it follows that a2 + b2 = 1. The only solutions of this last equation are a = ± 1, b = 0 or a = 0, b = ± 1. This leads to the four invertible elements ± 1, ± i. In con-

92

FIRST COURSE IN RINGS AND IDEALS

sequence, the class of associates determined by any Gaussian integer a consists of exactly four members: a

+

bi, -a - bi, - b

+

+

bi

ai, b - ai.

Since associated elements are rather closely related, it is not surprising that they have similar properties; for instance: Theorem 6-2. Let a, b be nonzero elements of an integral domain R. Then the following statements are equivalent: 1) a and b are associates, 2) both alb and bla, 3) (a) = (b). Proof To prove the equivalence of(l) and (2), suppose that a = bu. where u is an invertible element; then, also, b = au- 1 , so that both alb and bla. Going in the opposite direction, if alb, we can write b = ax for some x E R; while, from bla, it follows that a = by withy E R. Therefore, b = (by)x = b(yx). Since b =fo 0, the cancellation law implies that 1 = yx. Hence, J' is an invertible element of R. with a = by, proving that a and b must be

associates. The equivalence of (2) and (3) stems from our earlier remarks relating division of ring elements to ideal inclusion. We next examine the notion of a greatest common divisor. Definition 6-3. Let a 1, a 2 , .•• , an be nonzero elements of the ring R. An element d E R is a greatest common dil'isor of a 1, a 2 , ... , a. if it possesses the properties 1) d!a; fur i = 1, 2, ... , n (dis a common divisor), 2) cia; for i

= 1, 2, ... , n implies that cid.

The use of the superlative adjective "greatest" in this definition does not imply that d has greater magnitude than any other common divisor c, but only that d is a multiple of any such c. A natural question to ask is whether the elements a 1, a 2 , ... , an E R can possess two different greatest common divisors. For an answer, suppose that there are two elements d and d' in R satisfying the conditions of Definition 6-3. Then, by (2), we must have did' as well as d'ld; according to Theorem 6-2, this implies that d and d' are associates. Thus, the greatest common divisor of a 1, a 2 , ••• , a,. is unique, whenever it exists, up to arbitrary invertible factors. We shall find it convenient to denote any greatest common divisor of a 1, a 2 , ... , a,. by gcd (a 1 , a 2 , ... , a,.). The next theorem will prove, at least for principal ideal rings, that any finite set of nonzero elements actually does have a greatest common divisor.

33

IDEALS AND THEIR OPERATIONS

As a preliminary step to demonstrating that g also preserves multiplication, notice that f((ab)u 2 ) = f(abu)f(u) = f(abu). From this we are able to conclude that

g(ab)

= f(abu) = f(abu 2 ) = f((au)(bu)) = f(au)f(bu) = g(a)g(b).

The crucial third equality is justified by the fact that u e cent R, hence, commutes with b. As regards the uniqueness assertion, let us assume that there is another homomorphic extension off to the set R; call it h. Since f and h must agree on I and, more specifically, at the element u, h(u) = f(u) = 1. With this in mind, it follows that

h(a) = h(a)h(u) = h(au)

= f(au) = g(a)

for all a E R and so h and g are the same function. Hence, there is one and only one way of extendingfhomomorphically from the ideal I to the whole ring R. Before closing the present chapter, there is another type of direct sum which deserves mention. To this purpose, let R 1, R 2 , •.. , R,. be a finite number of rings (not necessarily subrings of a common ring) and consider their Cartesian product R = X R; consisting of all ordered n-tuples (a 1, a 2 , ... , a,.), with a; E R;. One can easily convert R into a ring by performing the ring operations componentwise; in other words, if (a 1, a 2 , ••• , a,.) and (b 1 , b2 , ••• , b,) are two elements of R, simply define

(a 1, a2 ,

••• ,

a.,)

+

(b 1, b2 ,

••• ,

b,) = (a 1

+ b1, a2 + b2 , ••• , a, +

b,)

and

(a 1 , a 2 ,

••• ,

a,)(b 1, b2 ,

••• ,

b,) = (a 1 b 1 , a 2 b 2 ,

••• ,

a,b,.).

The ring so obtained is called the external direct sum of R 1 , R 2 , ••• , R, and is conveniently written R = R 1 R2 R,.. (Let us caution that the notation is not standard in this matter.) In brief, the situation is this : An external direct sum is a new ring constructed from a given set of rings, and an internal direct sum is a representation of a given ring as a sum of certain of its ideals. The connection between these two types of direct sums will be made clear in the next paragraph. If R is the external direct sum of the rings R; (i = 1, 2, ... , n), then the individual R; need not be subrings, or even subsets, of R. However, there is an ideal of R which is the isomorphic image of R;. A straightforward calculation will convince the reader that the set

+

+ ··· +

I 1 = {(0, ... , 0, a;, 0, ... , O)la; e R;}

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FIRST COURSE IN RINGS AND IDEALS

Theorem 6-4. Let a, b, c be elements of the principal ideal ring R. If c!ab, with a and c relatively prime, then c!b. Proof. Since a and care relatively prime, so that gcd (a, c) = 1, there exist elements r, s e R satisfying 1 = ra + sc; hence, b

= 1b = rab + scb.

As c!ab and cic, Theorem 6-1(5) guarantees that ci(rab + scb), or rather, cib.

Dual to the notion of greatest common divisor there is the idea of a least common multiple, defined below. Definition 6-4. Let a 1, a2 , ... , a. be nonzero elements of a ring R. An element de R is a least common multiple of a 1 , a 2 , ... , a. if 1) a1id fori

= 1, 2, ... , n (dis a common multiple),

2) adc fori

=

1, 2, ... , n implies d!c.

In brief, an element d e R is a least common multiple of a 1 , a2 , ••• , a. if it is a common multiple of a 1, a 2 , ••• , a. which divides any other common multiple. The reader should note that a least common multiple, in case it exists, is unique apart from the distinction between associates; indeed, if d and d' are both least common multiples of a 1 , a 2 , .•. , a•• then did' and d'!d; hence, d and d' are associates. We hereafter adopt the standard notation Icm (a 1, a 2 , ... , a.) to represent any least common multiple of a 1 , a 2 , ••• , a•. The next result is a useful companion to Theorem 6-3. Theorem 6-5. Let a 1, a 2 , ••• , a. be nonzero elements of the ring R. Then a 1, a2 , ... , a. have a least common multiple if and only if the ideal n (ad is principal. We begin by assuming that d = lcm (a 1,- a 2 , . •• , a.) exists. Then the element d lies in each of the principal ideals (a;). for i = l, 2, ... , n, whence in the intersection n (a;). This means that (d) s; n (a;). On the other hand, any element r e n (a;) is a common multiple of each of the ai' But dis a least common multiple, so that dlr, or, equivalently, r e (d). This leads to the inclusion n (a;) s; (d) and the subsequent equality. Going in the opposite direction, suppose that the intersection n (a;) is a principal ideal of R, say n (a;) = (a). Since (a) s; (a 1), it follows that a 1 !a for every i, making a a common multiple of a 1 , a 2 , ... , a •. Given any other common multiple b of a 1 , a2 , ... , a., the condition a;lb implies that (b) s; (a;) for each value of i. As a result, (b) £ n (a;) = (a) and so alb. Our argument establishes that a = Icm (a 1, a2 , ... , a.), completing the proot ·

Proof.

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

95

At this point, we introduce two additional definitions. These will help to describe, in a fairly concise manner, certain situations which will occur in the sequel. Definition 6-5. A ring R is said to have the gcd-property (/em-property) provided that any finite number of nonzero elements of R admit a greatest common divisor (least common multiple). The content of Theorem 6-3 is that a ring R has the gcd-property if and only if every finitely generated ideal of R is principal. Likewise, Theorem 6--5 tells us that R possesses the !em-property if and only if the intersection of any finite number of principal ideals of R is again principal. Suffice it to say, every principal ideal ring satisfies both these properties. The immediate task is to prove that any integral domain has the gcdproperty if and only if it has the )em-property. In the process, we shall acquire certain other facts which have significance for our subsequent investigation. So as to avoid becoming submerged in minor details at a critical stage of the discussion, let us first establish a lemma. Lemma. Let a 1 , a2 , domain R.

..• ,

1) Iflcm (al> a 2 ,

a,.) exists, then lcm (ra 1, ra 2 ,

••• ,

a,. and r be nonzero elements of an integral

lcm (ra 1 , ra2 , 2) If gcd (ra 1 , ra 2 ,

••• ,

••• ,

ra.. )

••• ,

= r lcm (a 1 , a2 ,

ra,.) exists, then gcd (a 1 , a2 ,

ra,.) also exists and

••• ,

••• ,

a,.).

a,.) also exists and

Proof First, assume that d = lcm (a 1 , a2 , ••• , a,.) exists. Then a;ld for each value of i, whence ra;!rd. Now, let d' be any common multiple of ra 2 , ra 2 , ••• , ra,.. Then rjd', say d' = rs, where s E R. It follows that a;ls for every i and so dis. As a result, rdirs or rdid'. But this means that lcm (ra 1 , ra 2 , ••• , ra,.) exists and equals rd = r lcm (a 1 , a2 , ••• , a,.). As regards the second assertion, suppose that c = gcd (ra 1, ra 2 • ••• , ra,.) exists. Then ric; hence, c = rt for suitable t E R. Since cjra;. we have tja; for every i, signifying that t is a common divisor of the a;. Now, consider an arbitrary common divisor t' of a 1, a 2 , ••• , a,.. Then rt'ira; fori = 1, 2, ... , n and therefore rt'jc. But c = rt, so that rt'lrt or t'lt. The implication is that gcd (a 1 , a 2 , ••• , a,.) exists and equals t. This proves what we wanted: gcd (ra 1 , ra 2 ,

••• ,

ra,.)

=c

= rt = r gcd (a 1 , a2 ,

... ,

a.,).

Remark. It is entirely possible for gcd (a 1 , a 2 , ... , a,.) to exist without the existence of gcd (ra 1 , ra 2 , ... , ra,.); this accounts for the lack of symmetry in the statement of the above lemma. (See Example 6-4.)

96

FIRST COURSE IN RINGS AND IDEALS

Although the coming theorem is somewhat specialized in character, the information it contains is frequently useful. Theorem 6-6. Let a 1 , a 2 , •.. , a. and b 1, b 2 , ... , b. be nonzero elements of an integral domain R such that a 1 b 1 = a 2 b 2 = ··· = a. b. = x. 1) If lcm (a 1, a 2 , ••• , a.) exists, then gcd (b 1, b 2 , ... , b.) also exists and satisfies lcm (a 1, a 2 , ... , a.) gcd (b 1 , b 2 , ... , b.) = x. 2) If gcd (ra 1, ra 2 , ... , ra.) exists for all 0 =I= r E R, then lcm (b 1, b2 , b,.) also exists and satisfies gcd (a 1, a2 ,

••• ,

a.) lcm (b 1, b2 ,

••• ,

....

b.) = x.

Proof. For a proof of statement (1), set a= lcm (a 1 , a2 , ... , a.). Then a;ia for i = 1, 2, ... , n, say a = r;a;. From the relation x = a;b;. we see that ab; = (r;a;)b; = r;x and so xlab;. On the other hand, consider any divisor y of the ab;. Then ya;i(ab;)a;. or ya;ixa, making xa a common multiple of ya 1, ya 2 , ... , ya•. According to the lemma, Icm (ya 1, ya 2 , ... , ya.) exists and equals ya. Thus, by the definition of least common multiple, we conclude that yaixa, whence ylx. To recapitulate, we have shown that xlab; for each i and whenever yiab;, then ylx. This simply asserts that x = gcd (ab 1 , ab 2 , ••• , ab,.) = a gcd (b 1, b 2 , ••• , b.) = lcm (a 1 , a 2 ,

... ,

a.) gcd (b 1, b 2 ,

... ,

b.).

where, once again, the lemma has been invoked. We omit the proof of the other half of the theorem, which follows by much the same reasoning. In order to apply the lemma, it is now necessary to assume n only that gcd (a 1 , a2 , ... , a.) exists but, more generally, the existence of g~c (ra 1, ra 2 , ••• , ra.) for all r =I= 0. 4

Our next result is rather striking in that it tells us that, at least for integral domains, the gcd-property implies the Icm-property, and conversely. Theorem 6-7. An integral domain R has the gcd-property if and only if R has the lcm-property.

Proof Let b 1, b 2 , ... , b. be nonzero elements of R and suppose that R possesses the lcm-property. Taking x = b 1 b 2 ... b. and ak = b 1 ... bk_ 1 bk+ 1 .. · b. for k = 1, 2, ... , n, we may appeal to the first part of Theorem 6--6 to conclude that gcd (b 1 , b 2, ... , b,.} exists; hence, the gcd-property holds in R. Conversely, if it is hypothesized that any finite number of nonzero elements of R admit a greatest common divisor, then the existence of Icm (b 1 , b2 , ... , b.) can be inferred in the same way. We now have quite a bit of information about divisibility in integral domains, but the basic question remains unanswered: when does a ring

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

97

possess a factorization theory in which the analog of the Fundamental Theorem of Arithmetic holds? To this end, let us introduce two new classes of elements, prime and irreducible elements; when the ring is specialized to the ring of integers, these concepts are equivalent and yield the usual notion of a prime number. Definition 6-6. 1) A nonzero element p E R is called a prime if and only if p is not invertible and plab implies that either pia or else plb. 2) A nonzero element q E R is said to be irreducible (or norifactorizable) if and only if q is not invertible and in every factorization q = be with b, c E R, either b or c is invertible. Briefly, an irreducible element q is an element which cannot be factored in R in a nontrivial way; the only factors of q are its associates and the invertible elements of R. In such rings as division rings and fields, where each nonzero element possesses a multiplicative inverse, the concept of an irreducible element is of no significance. Observe also that every element which is an associate of an irreducible (prime) element is itself irreducible {prime). It follows by an easy induction argument that if a product a 1a 2 ... an is divisible by a prime p, then p must divide at least one of the factors a; (i = 1, 2, ... , n). Lemma. In an integral domain R, any prime element p is irreducible. Proof Suppose that p = ab for some a, bE R. Since pis prime, either pia or pI b; say p divides b, so that there exists some element c in R for which b = pc. We then have abc = pc = b. It follows from the cancellation law that ac = 1; hence, a is invertible. This allows us to conclude that p must be an irreducible element of R. Although prime elements are irreducible in integral domains, the converse is not always true, as we shall see later on. In the context of principal ideal domains (our primary interest in this chapter), the notions of an irreducible element and a prime element coincide. This is brought out in the theorem below.

Theorem 6-8. Let R be a principal ideal domain. A nonzero element p E R is irreducible if and only if it is prime. Proof By what we have just proved, p prime always implies p irreducible. So, assume that p is an irreducible element and that p divides the product ab, say pc = ab, with c E R. As R is a principal ideal ring, the ideal generated by p and a. (p, a) = (d) for some choice of d in R; hence, p = rd, for suitable r E R. But p is irreducible by hypothesis, so that either r or d must be an invertible element.

98

FIRST COURSE IN RINGS AND IDEALS

If d happened to possess an inverse, we would have (p, a) = R. Thus, there would exist elements s, t e R for which 1 = sp + ta. Then,

b

= bl = bsp + bta = bsp + pet = p(bs + ct),

which implies that Plb. On the other hand, if r is invertible in R, then d = r- 1p E (p), whence (d) s:: (p). It follows that the element a E (p) and, in consequence, pia. At any rate, if plab, then p must divide one of the factors, making p a prime element of R. We next take up two theorems having to do with the ideal structure of a principal ideal domain; the first result has considerable theoretical importance and will, in particular, serve as our basic tool for this section. Theorem 6-9. Let R be a principal ideal domain. If {I.}. n any infinite sequence of ideals of R satisfying

E

Z +, is

then there exists an integer m such that I. = Im for all n > m.

= u I" is an ideal of R (see the argument of Theorem 5-2). Being an ideal of a principal ideal ring, I = (a) for suitable choice of a E R. Now, the element a must lie in one of the ideals of the union, say the ideal I m· For n > m, it then follows that Proof It is an easy matter to verify that I

I= (a)

hence, I.

s:: Im s:: I. s:: I;

= Im, as asserted.

In asserting the equivalence of maximal and prime ideals in principal ideal domains, Theorem 5-9 failed to identify these ideals; this situation is taken care of by our next theorem. First, let us define a principal ideal of the ring R to be a maximal principal ideal if it is maximal (with respect to inclusion) in the set of proper principal ideals of R. · Lemma. Let R be an integral domain. For a nonzero element p e R, the following hold : a) p is an irreducible element of R if and only if (p) is a maximal principal ideal ; b) p is a prime element of R if and only if the principal ideal (p) =!= R is prime. Proof To begin, we suppose that pis an irreducible element of Rand that (a) is any principal ideal for which (p) c (a) £ R. As p e (a), we must have p = ra for some r e R. The fact that p is an irreducible element implies that either r or a is invertible. Were r allowed to possess a multiplicative

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

99

inverse, then a = r- 1p E (p), from which it follows that (a) £;;;; (p), an obvious contradiction. Accordingly, the element a is invertible, whence (a) = R. This argument shows that no principal ideal lies between (p) and the whole ring R, so that (p) is a maximal principal ideal. On the other hand, let (p) be a maximal principal ideal of R. For a proof by contradiction, assume that p is not an irreducible element. Then p admits a factorization p = ab where a, b E R and neither a nor b is invertible (the alternative possibility that p has an inverse implies (p) = R, so may be ruled out). Now, if the element a were in (p), then a = rp for some choice of r E R; hence, p = ab = (rp)b. Using the cancellation law, we could deduce that 1 = rb; but this results in the contradiction that b is invertible. Therefore, a¢ (p), yielding the proper inclusion (p) c (a). Next, observe that if (a) = R, then a will possess an inverse, contrary to assumption. We thus conclude that (p) c (a) c R, which denies that (p) is a maximal principal ideal. Our original supposition is false and a must be an irreducible element of R. With regard to the second assertion of the lemma, suppose that p is any prime element of R. To see that the principal ideal (p) is in fact a prime ideal, we let the product abE (p). Then there exists an element r E R for which ab = rp; hence, plab. By hypothesis, pis a prime element, so that either pia or plb. Translating this into ideals, either a E (p) orb E (p); in consequence, (p) is a prime ideal of R. The converse is proved in much the same way. Let (p) be a prime ideal and plab. Then abE (p). Using the fact that (p) is a prime ideal, it follows that one of a orb lies in (p). This means that either pia or else plb, and makes p a prime element of R. For principal ideal domains, all of this may be summarized by the following theorem. Theorem 6-10. Let R be a principal ideal domain. The nontrivial ideal (p) is a maximal (prime) ideal of R if and only if pis an irreducible (prime)

element of R. An immediate consequence of this theorem is that every nonzero noninvertible element of R is divisible by some prime. Corollary. Let a =1= 0 be a noninvertible element of the principal ideal domain R. Then there exists a prime pER such that pia. Proof. Since a is not invertible, the principal ideal (a) =I= R. Thus, by Theorem S-2, there exists a maximal ideal M of R such that (a) £;;;; M. But the preceding result tells us that every maximal ideal is of the form M = (p), where p is a prime element of R (in this setting, there is no distinction between prime and irreducible elements). Thus, (a) £;;;; (p), which is to say that pja.

100

FIRST COURSE IN RINGS AND IDEALS

Many authors do not insist that an integral domain possess an identity element; for this reason, let us sketch a second proof of the foregoing corollary which avoids the use of Theorem 5-2. First, put (a) = I 1 . If I 1 is not already a maximal ideal, then there exists an ideal I 2 of R such that I 1 c I 2 • By the same reasoning, if I 2 is not maximal, then I 1 c I 2 c I 3 for some ideal I 3 . Appealing to Theorem 6--9, this process must terminate after a finite number of steps; in other words, we can eventually find a maximal ideal M of R containing I 1 = (a). As before, M = (p), with p a prime in R; the remainder of the proof is like that above. If R is an integral domain with the property that every noninvertible element of R can be expressed uniquely (up to invertible elements as factors and the order of factors) as a product of irreducible elements, then we say that R is a unique factorization domain; for a more formal definition:

Definition 6-7. An integral domain R is a unique factori=ation domain in case the following two conditions hold : 1) every element a E R, which is neither zero nor invertible, can be factored into a finite product of irreducible elements; 2) if a = p 1 p 2 ... Pn = q 1 q 2 ... qm are two factorizations of a into irreducible elements, then n = m and there is a permutation n of the indices such that p1 and q"lil are associates (i = 1, 2, ... , n). In short, an integral domain is a unique factorization domain if it possesses a factorization theory in which the analog of the Fundamental Theorem of Arithmetic holds. We intend to show that any principal ideal domain is a unique factorization domain; towards this goal, let us first prove: Theorem 6-11. If R is a principal ideal domain, then every element of R which is neither zero nor invertible has a factorization into a finite product of primes. Proof Consider any nonzero noninvertible element a E R. By the last corollary, there exists a prime p 1 in R with p 1 la. Then a = p 1 a 1 for some (nonzero) a 1 E R, whence (a) £ (a 1 ). Were (a) = (a 1 ), we would have a 1 = raforsuitablerER;itwouldfollowthata = p 1 a 1 = p 1 ra,orl = p 1 r, resulting in the contradiction that p 1 is invertible. Consequently, we have the proper inclusion (a) c (a 1 ). Repeat the procedure, now starting with a 1 , to obtain an increasing chain of principal ideals

(a) c (a 1 ) c (a 2 ) c ... c (an) c: ... , with an-t = p"a" for some prime Pn E R. This process goes on as long as a" is not an invertible element of R. But Theorem 6--9 asserts that the

36

FIRST COURSE IN RINGS AND IDEALS

7. Suppose that I is a left ideal and J a right ideal of the ring R. Consider the set

1J

= {I: a1b1 ia1 E I; b1 E J},

where I: represents a finite sum of one or more terms. Establish that IJ is a twosided ideal of R and, whenever I and J are themselves two-sided, that IJ S I (") J.

8. If S is any given nonempty subset of the ring R, then ann,S

= {r E Riar =

0 for all a e S}

is called the right annihilator of S (in R); similarly, ann1S

= {r E Rira

= 0 for all a

E

S}

is the left annihilator of S. When R is a commutative ring, we simply speak of the annihilator of S and use the notation ann S. Prove the assertions below: a) ann,S (ann1 S) is a right (left) ideal of R. b) If S is a right (left) ideal of R, then ann,S (ann S) is an ideal of R. c) If Sis an ideal of R, then ann,S and ann 1S are both ideals of R. d) When R has an identity element, ann,R = ann 1R = {0}. 9. Let I 1, I 2 , ••• , I. be ideals of the ring R with R = I 1 + I 2 + ··· + I.. Show that this sum is direct if and only if a 1 + a 2 + ··· + a. = 0, with a 1 E I 1, implies that each a1 = 0. 10. If P(X) is the ring of all subsets of a given set X, prove that a) the collection of all finite subsets of X forms an ideal of P(X); b) for each subset Y s X, P(Y) and P(X - Y) are both principal ideals of P(X), with P(X) = P(Y) ffi P(X - Y). 11. Suppose that R is a commutative ring with identity and that the element a E R is an idempotent different from 0 or l. Prove that R is the direct sum of the pnnctpal ideals (a) and (1 - a).

12. Let I, J and K be ideals of the ring R. Prove that a) I(J + K) = lJ + IK, (I + J)K = IK + JK; b) if I 2 J, then I (") (J + K) = J + (I n K). . 13. Establish that in the ring Z, if I

= (12) and J = (21), then

= (84), IJ = (252), I:J [Hint: In general, (a):(b) = (c), where c = afgcd (a, b).] I

+J=

(3),

In J

= (4),

J:I=(7).

14. Given ideals I and J of the ring R, verify that a) 0:,/ = ann 1 /, and 0: 1 / = ann,/ (notation as in Problem 8); b) I:,J (I: 1 J) is the largest ideal of R with the property that (/:,J)J (J(I :1 J) s I).

s I

15. Let I, J and K be ideals of R, a commutative ring with identity. Prove the following assertions:

102

FIRST COURSE IN RINGS AND IDEALS

where the Pi and qi are all primes. Since p 1 i(q 1 q 2 •·• qm), it follows that p 1 divides some qi (1 ~ i ~ m); renumbering, if necessary, we may suppose that p 1 lq 1 • Now, p 1 and q 1 are both prime elements of R, with p 1 lq 1 , so they must be associates: q 1 = p 1 u 1 for some invertible element u 1 E R. Cancelling the common factor p 1 , we are left with P2 ••· Pn = ulq2 ··· qm.

Continuing this argument, we arrive (after n steps) at 1

=

ulu2 •.. unqn+l •..

qm.

Since the qi are not invertible, this forces m = n. It has also been shown that every Pi has some qi as an associate and conversely. Thus, the two prime factorizations are identical, apart from the order in which the factors appear and from replacement of factors by associates. Attention is called to the fact that the converse of Theorem 6--13 is not true; in the next chapter, we shall give an example of a unique factorizatiOn domain which is not a principal ideal domain. A useful fact to bear in mind is that in a unique factorization domain R, any irreducible element pER is necessarily prime. For, suppose that p divides the product ab, say pc = ab. Let

a = p1

...

Pn•

b

= q1 ... qm,

and

c

= t 1 ... t.

be the unique factorizations of a, b, and c into irreducible factors. We then have Pt ... Pnq 1 ... qm = ab = pt 1 ... t•. Since the factorization of ab into irreducibles is unique, the element p must be an associate of one of the Pi or qi, and, consequently, p divides either a or b. Another interesting class of integral domains, w)lich we propose to look at now, is provided by the so-called Euclidean domains; these arose out of attempts to generalize the familiar Division Algorithm for ordinary integers to arbitrary rings. The precise definition follows. Definition ~.

An integral domain R is said to be Euclidean if there exists a function !5 (the Euclidean valuation) such that the following conditions are satisfied : 1) o(a) is a nonnegative integer for every 0 =I= a E R; 2) for any a, b E R, both nonzero, o(ab) ~ o(a); 3) for any a, bE R, with b =I= 0, there exist elements q, r E R (the quotient and remainder) such that a = qb + r, where either r = 0 or else o(r)


-· < 1 + J2. Assuming that u(1 + J'i.)-• = a + bJ'i, show that a = 1, b = 0.] Factor each of the following into primes: 11 + 1i in Z(i); 4 + 7J2 in Z(J'i.); 1

17. a)

4- .,r-::'JinZ(..,r-::3).

=:;;;

u(l

PROBLEMS

b) Show that in the quadratic domain Z(j6), the relation 6 = (j6)2 not violate unique factorization.

=

111

3·2 does

18. Prove that the domain Z(FfJ) is not a unique factorization domain by discovering two distmct factonzations of the element 10. Do the same for element 9 m the domain Z(.J- 7). 19. Show that the quadratic domain Z(.J- 5) is not a principal ideal domain. [Hint: Consider the ideal (3, 2 + .J- 5).] 20. Describe the field of quotients of the quadratic domain Z(.jn) where n is a squarefree integer.

SEVEN

POLYNOMIAL RINGS

The next step in our program is to apply some of the previously developed theory to a particular class of rings, the so-called polynomial rings. For the moment, we shall merely remark that these are rings whose elements consist of "polynomials" with coefficients from a fixed, but otherwise arbitrary, ring. (The most interesting results occur when the coefficients are specialized to a field.) As a first order of business, we seek to formalize the intuitive idea of what is meant by a polynomial. This involves an excursion around the fringes of the more general question of rings of formal power series. Out of the veritable multitude of results concerning polynomials, we have attempted to assemble those facets of the theory whose discussion reinforces the concepts and theorems expounded earlier; it is hoped thereby to convey a rough idea of how the classical arithmetic of polynomials fits into ideal theory. Our investigation concludes with a brief survey of some of the rudimentary facts relating roots of polynomials to field extensions. To begin with simpler things, given an arbitrary ring R, let seq R denote the totality of all infinite sequences

f =

(ao, at, a2, ··· • ak, · · .)

of elements ak E R. Such sequences are called formal power series, or merely power series, over R. (Our choice of terminology will be justified shortly.) We intend to introduce suitable operations in the set seq R so that the resulting system forms a ring containing R as a subring. At the outset, it should be made perfectly clear that two power series f = (a 0, at, a 2, ... ) and g = (b 0, b 1, b2, ... ) are considered to be equal if and only if they are equal term by term:

f

= g

if and only if ak = bk for all k

~

0.

Now, power series may themselves be added and multiplied as follows:

f +g

= (a 0

+ b 0 , at + b1, ...),

fg = (co, ct, c2, ... ), 112

POLYNOMIAL RINGS

113

where, for each k ;;::: 0, ck is given by ck

=

L

i+j=k

aibJ

=

a 0 bk

+

a 1 bk-t

+ ··· +

a,._

1 b1

+

akb 0 •

(It is understood that the above summation runs over all integers i, j ;;::: 0 subject to the restriction that i + j = k.) A routine check establishes that with these two definitions seq R becomes a ring. To verify a distributive law, for instance, take

f

= (a 0 , a 1,

... ),

g = (b 0 , b1 ,

... ),

h

=

(c 0 , c 1 ,

•.. ).

One finds quickly that f(g

+

h)

= (a 0 , a 1 ,

... )(b 0

+

c0 , b 1

+ c 1,

... )

= (d 0 , d1 ,

... ),

where

A similar calculation of fg + fh leads to the same general term, so that + h) = fg + fh. The rest of the details are left to the reader's care. We simply point out that the sequence (0, 0, 0, ... ) serves as the zero element of this ring, while the additive inverse of an arbitrary member (a 0 , a 1 , a 2 , ••• ) of seq R is, of course, ( -a 0 , -a 1 , -a 2 , ••• ). To summarize what we know so far: f(g

Theorem 7-I. The system seq R forms a ring, known as the ring of (formal) power series over R. Furthermore, the ring seq R is commutative with identity if and only if the given ring R has these properties. If S represents the subset of all sequences having 0 for every term beyond the first, that is, the set S = {(a, 0, 0, ... )Ia E R},

then it is not particularly difficult to show that S constitutes a subring of seq R which is isomorphic to R ; one need only consider the mapping that sends the sequence (a, 0, 0, ... )to the element a. In this sense, seq R contains the original ring R as a subring. Having reached this stage, we shall no longer distinguish between an element a E R and the special sequence (a, 0, 0, ... ) of seq R. The elements of R, regarded as power series, are hereafter called constant series, or just constants.

With the aid of some additional notation, it is possible to represent power series the way we would like them to look. As a first step in this direction, we let ax designate the sequence (0, a, 0, 0, ... ).

114

FIRST COURSE IN RINGS AND IDEALS

That is, ax is the specific member of seq R which has the element a for its second term and 0 for all other terms. More generally, the symbol ax", n ~ 1, will denote the sequence (0, ... , 0, a, 0, ... ),

where the element a appears as the (n example, we have

+

l)st term in this sequence; for

ax 2 = (0, 0, a, 0, ... ) and

ax 3 = (0, 0, 0, a, 0, ... ). By use of these definitions, each power series

f

=

(ao, al, a2, ... 'an, ... )

may be uniquely expressed in the form

f = (ao, 0, 0, ... ) + (0, al, 0, ... ) + ... + (0, = a0 + a1 x + a2x 2 + .. · + a"~ + ···

... '0, an,

o.... ) + ...

with the obvious identification of a 0 with the sequence (a 0 , 0, 0, ... ). Thus, there is no loss in regarding the power series ring seq R as consisting of all formal expressions

f = a0 + a 1x + a 2 x 2 + ... +

an~

+ .. ·,

where the elements a 0 , a 1 , ... , a", ... (the coefficients of f) lie in R. As a notational device, we shall often write this as f = L akxk (the summation symbol is not an actual sum and convergence is not at issue here). Using sigma notation, the definitions of addition and multiplication of power series assume the form

L a"x"

+L

b~cx" = L (a"

+

b")X',

(L akxk)(L bkxk) = L ckxk, where

We should emphasize that, according to our definition, x is simply a new symbol, or indeterminant, totally unrelated to the ring R and in no sense represents an element of R. To indicate the indeterminant x, it is common practice to write R[[x]] for the set seq R, andf(x) for any member of the same. From now on, we shall make exclusive use of this notation. Remark. If the ring R happens to have a multiplicative identity 1, many authors will identify the power series 0 + lx + Ox 2 + Ox 3 + ··· with x, thereby treating x itself as a special member of R[[ x ]] ; namely, the sequence

POLYNOMIAL RINGS

115

x = (0, I, 0, 0, ... ). From this view, ax becomes an actual product of members of R[[ x ]] : ax = (a, 0, 0, ... )(0, I, 0, 0, ... ).

Concerning the notation of power series, it is customary to omit terms with zero coefficients and to replace ( -ak)xk by -akxk. Although xis not to be considered as an element of R[[x]J, we shall nonetheless take the liberty of writing the term lxk as xk (k ~ 1). With these conventions, one should view, for example, the power series

1 + x 2 + x 4 + ··· + x 2 " + ··· e Z[[ x ]] as representing the sequence (1, 0, 1, 0, ... ). An important definition in connection with power series is that of order, given below. Definition 7-1. If f(x) = L akxk is a nonzero power series (that is, if not all the ak = 0) in R[[ x ]], then the smallest integer n such that a., =I= 0 is called the order ofj(x) and denoted by ordf(x). Suppose f(x), g(x) e R[[ x ]], with ord f(x) = n and ord g(x) = m, so that f(x) = a,. X' + a,.+ 1 X'+l + ··· g(x) = bmx"' + bm+lx"'+l + ···

(a,. =I= 0), (bm =/= 0).

From the definition of multiplication in R[[ x ]], the reader may easily check that all coefficients of f(x)g(x) up to the (n + m)th are zero, whence f(x)g(x)

=

a.,bm:x!'+m

+ (an+lbm + a,.bm+l):x!'+m+l + ···.

If we assume that one of a., or bm is not a divisor of zero in R, then a.,bm and

ord (f(x)g(x))

=n +m =

ord f(x)

=I=

0

+ ord g(x).

This certainly holds if R is taken to be an integral domain, or again if R has an identity and one of a. or bm is the identity element. The foregoing argument serves to establish the first part of the next theorem ; the proof of the second assertion is left as an exercise. Theorem 7-2. If f(x) and g(x) are nonzero power series in R[[x ]], then

= 0 or ord (f(x)g(x)) ~ ord f(x) with equality if R is an integral domain ; 2) either f(x) + g(x) = 0 or I) either f(x)g(x)

ord (f(x)

+ g(x))

+ ord g(x),

~ min {ord f(x), ord g(x)}.

The notation of order can be used to prove the following corollary.

116

FIRST COURSE IN RINGS AND IDEALS

Corollary. If the ring R is an integral domain, then so also is its power series ring R[[ x ]] .

Proof We observed earlier that whenever R is a commutative ring with identity, these properties carry over to R[[ x]]. To see that R[[x]] has no zero divisors, select f(x) =I= 0, g(x) =I= 0 in R[[ x ]]. Then, ord (f(x)g(x)) = ordf(x)

+

ord g(x) > 0;

hence, the product f(x)g(x) cannot be the zero series. Although arbitrary power series rings are of some interest, the most __ important consequences arise on specializing the discussion to power series whose coefficients are taken from a field. These will be seen to form principal ideal domains and, in consequence, unique factorization domains. The following intermediate result is directed towards establishing this fact. Lemma. Let R be a commutative ring with identity. A formal power series f(x) = L Gkxk is invertible in R[[x]J if and only if the constant term G 0 has an inverse in R.

Proof If f(x)g(x) = 1, where g(x) = L bkx\ then the definition of multiplication in R[[ x ]] shows that G 0 b0 = 1 ; hence, G 0 is invertible as an element of R. For the converse, suppose that the element G 0 has an inverse in R. We proceed inductively to define the coefficients of a power series I bk xk in R[[x ]] which is the inverse ofj(x). To do this, simply take b0 = a0 1 and. assuming b 1, b2 , ••• , bk-t have already been defined, let bk = -a()t(atbk-1

Then a0 b0 =,1, while, fork ck

=

L

i+j=k

~

+

a2bk-2

+ ... +

akbo).

1,

aibi = a 0 bk

+

a 1 bk-l

+ ··· +

akbo = 0.

By our choice of the bk's, we evidently must have c~:>kxk)(I bkxk) and so L akxk possesses an inverse in R[[ x ]].

=

1,

Corollary. A power seriesf(x) = Iakxk E F[[x]J, where F is a field, has an inverse in F[[ x ]] if and only if its constant term a 0 =I= 0. Having dealt with these preliminaries, we are now ready to proceed to describe the ideal structure ofF[[ x]]. Theorem 7-3. For any field F, the power series ring F[[x]J is a principal ideal domain; in fact, the nontrivial ideals of F[[ x ]] are of the form (xk), Where k E Z+.

Proof Let I be any proper ideal of F[[x]J. Either I = {0}, in which case I is just the principal ideal (0), or else I contains nonzero elements. In the

POLYNOMIAL RINGS

latter event, choose a nonzero power series f(x) Suppose thatf(x) is of order k, so that f(x) = akx"

+

ak+ 1 xk+ 1

+ ···

= xk(ak

+

E

ll7

I of minimal order.

ak+ 1x

+ ···).

Since the coefficient ak f. 0, the previous lemma insures that the power series ak + ak+ Ix + ... is an invertible element ofF[[ x]]; in other words, f(x) = xkg(x), where g(x) has an inverse in F[[x ]]. But, then, xk

= f(x)g(x) -I

E

I,

which leads to the inclusion (xk) s;;; I. On the other hand, take h(x) to be any nonzero power series in I, say of order n. Since f(x) is assumed to have least order among all members of I, it is clear that k :::;; n; thus, h(x) can be written in the form h(x)

=

xk(b.x•-k

+

b.+lx•-k+ 1

+ ... ) E (xk).

This implies that I s;;; (xk), and the equality I = (xk) follows. Corollary 1. The ring F[[ x ]] is a local ring with (x) as its maximal ideal. Proof Inasmuch as the ideals of F[[x]] form a chain F[[x]] => (x) => (x 2 ) => ••• =>

{0},

the conclusion is obvious. Corollary 2. Any nonzero element f(x) E F[[ x ]] can be written in the formf(x) = g(x)x", where g(x) is invertible and k ~ 0. To this we add, for future reference, the following assertion regarding the maximal ideals of a power series ring over a commutative ring with identity. Theorem 7-4 Let R be a commutative ring with identity. There is a one-to-one correspondence between the maximal ideals M of the ring Rand the maximal ideals M' of R[[ x ]] in such a way that M' corresponds to M if and only if M' is generated by M and x; that is, M' = (M, x). Proof Assume that M is a maximal ideal of R. To see that M' = (M, x) forms a maximal ideal of the ring R[[x]J, we need only show that for any power series f(x) = L akxk ¢ M', the element 1 + g(x)f(x) EM' for some g(x) in R[[ x ]] (Problem 2, Chapter 5). Since the series f(x) does not lie in M', its constant term a 0 ¢ M; hence, there exists an element r E R such that 1 + ra0 E M. This implies that

1

+

rf(x) = (1

+

ra 0 )

+ r(a 1 +

a2 x

+ ··· +

and so M' is a maximal ideal, as required.

a.x"- 1

+ ···)x E (M, x),

118

FIRST COURSE IN RINGS AND IDEALS

Next, take M' to be any maximal ideal of R[[x]J and define the set M to consist of the constant terms of power series in M':

= {a 0 E Rl Lakx" EM'}.

M

The reader can painlessly supply a proof that M forms a maximal ideal of the ring R. Notice incidentally that M must be a proper ideal. Were M = R, then there would exist a power series b" x" in M' with constant term b0 = 1. By the last lemma, L bnx" would then be an invertible element, so that M' = R[[ x ]], which is impossible. Owing to the inclusion M' s (M, x) and the fact that M' is maximal in R[[ x ]], it now follows that M' = (M, x). To verify that the correspondence in question is indeed one-to-one, suppose that (M, x) = (M, x), where M, M are both maximal ideals of the ring R; what we want to prove is that M = M. Let rEM be arbitrary. Given f(x) E R[[x]J, the sum r + f(x)x E (M, x) = (M, x), so that

L

r

+ f(x)x

=

r + g(x)x

for appropriate rEM and g(x) E R[[x]J. Hence, r If g(x) - f(x) =I= 0, then, upon taking orders, 0

=

ord(r - r) = ord (g(x) - f(x))

r

+ ord x

=

(g(x) - f(x) )x.

~ l,

an absurdity. In consequence, we must have g(x) - f(x) = 0 which, in its turn, forces r = r E M. The implication is that M s M and, since M is maximal, we end up with M = M. This completes the proof of the theorem. Power series have so far received all the attention, but our primary concern is with polynomials. Definition 7-2. Let R[ x] denote the set of all power series in R[[ x]] whose coefficients are zero from some index onward (the particular index varies from series to series):

R[x] = {a 0

+

a 1x

+ ··· + a.x"la,.~ R; n;;:::: 0}.

An element of R[x] is called a polynomial (in x) over the ring R. In essence, we are defining a polynomial to be a finitely nonzero sequence of elements of R. Thus, the sequence (1, 1, 1, 0, 0, ... ) would be a polynomial over Z 2 , but (1, 0, 1, 0, ... , 1, 0, ... )would not. It is easily verified that R[ x] constitutes a subring of R[[ x]], the socalled ring of polynomials over R (in an indeterminant x); indeed, if f(x) = L a,.x", g(x) = L b"x" are in R[x], with a" = 0 for all k ~ nand bk = 0 for all k ;;:::: m, then

a" + bk

= 0 for k ~ max {m, n},

L

0 for k ;;:::: m

i+j=k

ai b1 =

+ ...n,

POLYNOMIAL RINGS

119

so that both the sum f(x) + g(x) and product f(x)g(x) belong to R[x]. Running parallel to the idea of the order of a power series is that of the degree of a polynomial, which we introduce at this time. Definition 7-3. Given a nonzero polynomial

in R[ x ], we call an the leading coefficient of f(x); and the integer n, the degree of the polynomial. The degree of any nonzero polynomial is therefore a nonnegative integer; no degree is assigned to the zero polynomial. Notice that the polynomials of degree 0 are precisely the nonzero constant polynomials. If R is a ring with identity, a polynomial whose leading coefficient is 1 is said to be a monic polynomial. As a matter of notation, we shall hereafter write degf(x) for the degree of any nonzero polynomial f(x) E R[ x]. The result below is similar to that given for power series and its proof is left for the reader to provide; the only change of consequence is that we now use the notion of degree rather than order. Theorem 7-5.

If f(x) and g(x) are nonzero polynomials in R[x], then

1) either f(x)g(x) = 0 or deg (f(x)g(x)} :-::;; degf(x)

+ deg g(x),

with

equality whenever R is an integral domain; 2) either f(x) + g(x) = 0 or deg (f(x)

+ g(x))

:-::;; max {degf{x), deg g(x)}.

Knowing this, one could proceed along the lines of the corollary to Theorem 7-2 to establish CoroUary. If the ring R is an integral domain, then so is its polynomial ring R[x]. Example 7-1. As an illustration of what might happen if R has zero divisors, consider Z 8 , the ring of integers modulo 8. Taking

g(x) = 4 + x + + 2x, 4 + x + 6x2 , so that

f(x) = 1 we obtainf(x)g(x) =

deg (f(x)g(x)) = 2 < 1

+

2 = degf(x)

4x 2 ,

+

deg g(x).

Although many properties of the ring R carry over to the associated polynomial ring R[ x ], it should be pointed out that for no ring R does R[ x] form a field. In fact, when R is a field (or, for that matter, an integral domain), no element of R[ x] which has positive degree can possess a

120

FIRST COURSE IN RINGS AND IDEALS

multiplicative inverse. For, suppose that f(x) E R[x], with degf(x) > 0; if f(x)g(x) = 1 for some g(x) in R[ x ], we could obtain the contradiction 0

= deg 1 = deg (f(x)g(x)) = degf(x) + deg g(x)

=/= 0.

The degree of a polynomial is used in the factorization theory of R[ x] in much the same way as the absolute value is employed in Z. For. it is through the degree concept that induction can be utilized in R[ x] to develop a polynomial counterpart of the familiar division algorithm. One can subsequently establish that the ring F[x] with coefficients in a field forms a Euclidean domain in which the degree function is taken to be the Euclidean valuation. Before embarking on this program, we wish to introduce several new ideas. To this purpose, let R be a ring with identity; assume further that R' is any ring containing R as a subring (that is, R' is an extensiOn of R) and let r be an arbitrary element of R'. For each polynomial

f(x) = a 0

+

+ ··· +

a 1x

a,x!'

in R[x], we may definef(r) E R' by taking

f(r) = a 0

+

a 1r

+ .. · +

a,r".

The element f(r) is said to be the result of substituting r for x inf(x). Suffice it to say, the addition and multiplication used m defining f(r) are those of the ring R', not those of R[x]. Now, suppose that f(x), g(x) are polynomials in R[ x] and recent R'. We leave the reader to prove that if

h(x)

= f(x) + g(x),

k(x) = f(x)g(x),

= f(r) + g(r),

k(r) = f(r)g(r).

then h(r)

This being so, it may be concluded that the mapping ¢,: R[x] -+ R' which sendsf(x) tof(r)is a homomorphism of R[x] into R'. Such a homomorphism will be called the substitution homomorphism determined by r and its range denoted by the symbol R[r] :

R[r] = {!(r)jf(x) E R[x]} = {a0

+ a 1r + ··· +

a,r"Ja~cE R; n ;;::::: 0}.

It is a simple matter to show that R[r] constitutes a subring of R'; in fact, R[r] is the subring of R' generated by the set R u {r}. (Since R has an identity element 1, lx = x E R[ x ], and so r e R[r ].) Notice also that R[r] = R if and only if r e R. The foregoing remarks justify part of the next theorem.

POLYNOMIAL RINGS

121

Theorem 7-6. Let R be a ring with identity, R' an extension ring of R, and the element rEcent R'. Then there is a umque homomorphism l/>,: R[ x] -+ R' such that l/>,(x) = r, lj>,(a) = a for all a E R. Proof We need only verify that ¢,is unique. Suppose, then, that there is another homomorphism •: R[ x] -+ R satisfy in the indicated conditions and consider any polynomial f(x) = a 0 + a 1 x + ·.. + an x" E R[ x]. By assumption, 0. Then f(x) is a primitive polynomial in R[x] if and only if, for each prime element p E R, the reduction of f(x) modulo the principal ideal (p) is nonzero. Proof By definition, the reduction of.f(x) modulo (p) is

v(f(x)) = (a 0

+ (p}) +

(a 1

+

(p))x

+ ... + (a,. +

(p))x".

Thus, to say that v(f(x)) = 0 for some prime pER is equivalent to asserting that ak E (p), or rather, plak for all k. But the latter condition signifies that contf(x) =(=., 1; hence,f(x) is not primitive. One of the most crucial facts concerning primitive polynomials is Gauss's Lemma, which we prove next. Theorem 7-16. (Gauss's Lemma). Let R be a unique factorization domain. Iff(x), g(x) are both primitive polynomials in R[ x ], then their productf(x)g(x) is also primitive in R[x]. Proof Given a prime element pER, (p) is a prime ideal of R, whence the quotient ring R' = R/(p) forms an integral domain. We next consider the

reduction homomorphism v modulo the principal ideal (p). Since R'[x] is an integral domain, it follows that the reduction of f(x)g(x) cannot be the zero polynomial: v(f(x)g(x)) = v(f(x) )v(g(x)) =I= 0. The assertion of the theorem is now a direct consequence of our last result.

THE CLASSICAL ISOMORPHISM THEOREMS

47

From this factorization, it is easy to see that f is a homomorphism with f(I) = (I + J)jJ. To confirm that the kernel off is precisely the set I n J, notice that the coset J serves as the zero element of (I + J)/J, and so kerf= {a e Iif(a) = J} = {a e Iia + J = J}

= {a e Iia e J}

= I n J.

The asserted isomorphism should now be evident from the Fundamental Homomorphism Theorem. We conclude this chapter with a brief excursion into the theory of nil and nilpotent ideals: a (right, left, two-sided) ideal I of the ring R is said to be a nil ideal if each element x in I is nilpotent; that is to say, if there exists a positive integer n for which x" = 0, where n depends upon the particular element x. As one might expect, the ideal I will be termed nilpotent provided fR = {0} for some positive integer n. By definition, I" denotes the set of all finite sums of products of n elements taken from I, so that I" = {0} is equivalent to requiring that for every choice of n elements a 1 , a 2 , ••• , a. e I (distmct or not), the product a 1 a 2 ···a.= 0; in particular, a" = 0 for all a in I, whence every nilpotent ideal is automatically a nil ideal. We speak of the ring R as being nil (nilpotent) if it is nil (nilpotent) when regarded as an ideal. Notice, too, that any ideal containing a nonzero idempotent element cannot be nilpotent. With these definitions at our disposal, we can now prove two lemmas. Lemma. 1) If R is a nil (nilpotent) ring, then every subring and every homomorphic image of R is nil (nilpotent). 2) If R contains an ideal I such that I and R/1 are both nil (nilpotent), then R is a nil (nilpotent) ring. Proof The proof of assertion (1) follows immediately from the definitions and Problem 2-17. To verify (2), assume that I and R/1 are nil rings and that a e R. Then there exists some positive integer n for which the coset (a

+ /)" =

a"

+

I

=

I,

signifying that the element a" e I. Inasmuch as I is a nil ideal, (a")m = anm = 0 for some me Z+. This implies that a is nilpotent as a member of Rand, in consequence, R is a nil ring. The remainder of the proof is left to the reader's care.

Lemma. If N 1 and N 2 are two nil (nilpotent) ideals of the ring R, then their sum N 1 + N 2 is likewise a nil (nilpotent) ideal. Proof With reference to Theorem 3-10, we have (N 1 + N 2 )/N 1 ~ N 2 /(N 1 n N 2 ). The right-hand side (hence, the left-hand side) of this equation is a nil ring, being the homomorphic image of the nil ideal N 2 •

132

FIRST COURSE IN RINGS AND IDEALS

1'heorem 7-17. Let R be a unique factorization domain, with field of quotients K. lf f(x) E R[ x J is an irreducible primitive polynomial, then it is also irreducible as an element of K[x].

Proof Assume to the contrary that f(x) is reducible over K. Then, f(x) = g(x)h(x), where the polynomials g(x), h(x) are in K[x] and are of positive degree. By virtue of the lemma just proven,

with a, b, c, dE Rand g 1 (x), h 1(x) primitive in R[x]. Thus,

bdf(x)

=

acg 1 (x)h 1(x).

Now, Gauss's Lemma asserts that the product g 1(x)h 1(x) is a primitive polynomial in R[x], whencef(x)and g 1 (x)h 1(x) differ by an invertible element of R: f(x) = ug 1 (x)h 1(x). Since deg g 1(x) = deg g(x) > 0, deg h 1(x) = deg h(x) > 0, the outcome is a nontrivial factorization of f(x) in R[ x ], contrary to hypothesis. There is an obvious converse to Theorem 7-17, viz.: if the primitive polynomial f(x) E R[ x J is irreducible as an element of K[x ], it is also irreducible in R[ x]. This is justified by the fact that R[ xJ(or an isomorphic copy thereof) appears naturally as a subring of K[x]; thus, if f(x) were reducible in R[x], it would obviously be reducible in the larger ring K[x]. Our remarks lead to the following conclusion: Given a primitive polynomial f(x) E R[ x ], R a unique factorization domain, f(x) is irreducible in R[ x J if and only if f(x) is irreducible in K[ x]. Our next concern is a generalization of a famous theorem of Eisenstein dealing with the problem of irreducibility (this result is of fundamental importance in the classical theory of polynomials with integral coefficients). The generalization which we have in mind is formulated below. Theorem 7-18. Let R be an integral domain and the nonconstant polynomialf{x) = a0 + a 1 x + ... + anx" E R[x]. Suppose that there exists a prime ideal P of R such that 1) an¢ P,

2) ale E P for 0

~

k < n,

Thenf(x) is irreducible in R[x].

Proof Assume that, contrary to assertion, f(x) is reducible in R[x]; say, f(x) = g(x)h(x) for polynomials g(x), h(x) E R[x], where g(x)

= b0 + b 1x + ··· + b,x',

h(x) = c0

+ c1x + ··· +

c,:x:'

(r

+ s = n; r, s >

0).

POLYNOMIAL RINGS

133

Now consider the reduction off(x) modulo the ideal P. Invoking hypothesis (2), it can be inferred that v(g(x))v(h(x)) = v(f(x)) = (a,.

+ P)x".

Since the polynomial ring (R/P)[x] comprises an integral domain, the only possible factorizations of (a,. + P)x" are into linear factors. This being so, a moment's reflection shows that v(g(x)) = (b,

v(h(x)) = (c8

+ P)x', + P)r.

This means that the constant terms of these reductions are zero; that is, b0

+P=

c0

+P=

P.

Altogether we have proved that both b0 , c0 E P, revealing at the same time that a0 = b0 c0 E P 2 , which is untenable by (3). Accordingly, no such factorization of f(x) can occur, and f(x) is indeed irreducible in R[ x]. Theorem 7-18 leads almost immediately to the Eisenstein test for irreducibility. Corollary. (The Eisenstein Criterion). Let R be a unique factorization domain and K be its field of quotients. Letf(x) = a 0 + a 1 x + ... +

a. x" be a nonconstant polynomial in R[x]. Suppose further that for some prime pER, p fa,., pjak for 0 ~ k < n, and p 2 f a 0 • Then,f(x) is irreducible in K[x]. Proof We already know that (p) is a prime ideal of R. Taking stock of the theorem, f(x) is an irreducible polynomial of R[ x]; hence, of K[ x] (at this point a direct appeal is made to Theorem 7-17).

This is probably a good time at which to examine some examples. Example 7-5. Consider the monic polynomial

f(x) = x"

+

a e Z[x]

(n > 1),

where a =f. ± 1 is a nonzero square-free integer. For any prime p dividing a, p is certainly a factor of all the coefficients except the leading one, and our hypothesis ensures that p'- {a. Thus, f(x) fulfils Eisenstein's criterion, and so is irreducible over Q. Incidentally, this example shows that there are irreducible polynomials in Q[ x] of every degree. Ontheotherhand,noticethatx4 + 4 = (x 2 + 2x + 2)(x 2 - 2x + 2); one should not expect Theorem 7-18 to lead to a decision in this case, since, of course, 4 fails to be a square-free integer. Example 7-6. polynomial

Eisenstein's test is not directly applicable to the cyclotonic

48

FIRST COURSE IN RINGS AND IDEALS

Since {N 1 + N 2 )/N 1 and N 1 are both nil, it follows from the previous lemma that N 1 + N 2 is necessarily a nil ideal. Similar reasoning applies to the nilpotent case. Corollary. The sum of any finite number of nil (nilpotent) ideals of the ring R is again nil {nilpotent). Having completed the necessary preliminaries, let us now establish Theorem 3-11. The sum nil ideal.

L N; of all the nil ideals N; of the ring R is a

L

Proof If the element a E N;, then, by definition, a lies in some finite sum of nil ideals of R; say, a E N 1 + N 2 + ... + N,, where each Nk is nil. By virtue of the last corollary, the sum N 1 + N 2 + .. · + N, must be a nil ideal; hence, the element a is nilpotent. This argument shows that L: N; is a nil ideal. It is possible to deduce somewhat more, namely,

Corollary. The sum of all the nilpotent ideals of the ring R is a nil ideal. Proof Since each nilpotent ideal is a nil ideal, the sum N of all nilpotent

ideals of R is contained in a nil ideal, making N nil.

L: N;, the sum of all nil ideals.

But

L N; is itself

Example 3-2. For examples of nilpotent ideals, let us turn to the rings ZP"' where pis a fixed prime and n > I. By virtue of the remarks on page 42, ZP" has exactly one ideal for each positive divisor of p" and no other ideals; these are simply the principal ideals (pk) = pkZP" (0 :s:;; k :s:;; n). For 0 < k ~ n, we have

{p")"

= {p"/c) = (0) = {0},

so that each proper ideal of ZP" is nilpotent. Before leaving this chapter, we should present an example to show that, in general, nil and nilpotent are different concepts. Example 3-3. For a fixed prime p, let S be the collection of sequences = {a,} with the property that the nth term a, E ZP" (n 2 1). S can be made into a ring by performing the operations of addition and multiplication term by term:

a

{a,.}

+ {b,}

=

{a,.

+ b,.},

The reader will find that the zero element of this ring is just the sequence formed by the zero elements of the various ZP" and the negative of {a,} is {-a,}. Now, consider the set R of all sequences inS which become zero

POLYNOMIAL RINGS

135

With these operations, the set R[ x, y] becomes a ring containing R (or rather an isomorphic copy of R) as a subring. The (total) degree of a nonzero polynomial m

f(x, y)

=

n

L L a;ixiyi i=O j=O

is the largest of the integers i + j for which the coefficient aii =1= 0 and is denoted, as before, by degf(x, y). Without going into details here, let us simply state that it is possible to obtain inequalities involving degrees analogous to those of Theorem 7-5; in particular, if R is an integral domain, we still have deg (f(x, y)g(x, y))

= degf(x, y) + deg g(x, y).

From this rule, one can subsequently establish that whenever R forms an integral domain, then so does the polynomial ring R[ x, y]. Rather than get involved in an elaborate discussion of these matters, we content ourselves with looking at two examples. Example 7-7. To illustrate that the ideal structure of the ring F[ x, y] (Fa field) is more complicated than that of F[x], let us show that F[x, y] is not a principal ideal domain. This is accomplished by establishing that the ideal I = (x, y) is not principal, where I = {f(x, y)x

+ g(x, y)yjf(x, y), g(x, y) e F[x, y]}.

Notice that the elements of this ideal are just the polynomials in F[ x, y] having zero constant term. Suppose that I was actually principal, say I = (h(x, y) ), where deg h(x, y) ;;::: 1. Since both x, y e /, there would exist polynomials f(x, y), g(x, y) ip. F[x, y] satisfying x

= f(x, y)h(x, y),

y = g(x, y)h(x, y).

Now,degx = degy = l,whichimpliesthatdegh(x,y) = l,anddegf(x,y) = deg g(x, y) = 0; what amounts to the same thing, x

=

ah(x, y),

y = bh(x, y)

(a, bE F).

Moreover, h(x, y) must be a linear polynomial; for instance,

But if the coefficient c2 =/= 0, then x cannot be a multiple of h(x, y), and if c 1 =f. 0, y is not a multiple of h(x, y). This being the case, we conclude that c 1 = c 2 = 0, a contradiction to the linearity of h(x, y), and so I does not form a principal ideal.

136

FIRST COURSE IN RINGS AND IDEALS

Another point worth mentioning is that since F[ x J constitutes a unique factorization domain, so does (F[ x ])[y J = F[ x, y] (Theorem 7-11 ). The present situation thus furnishes us with an illustration of a unique factorization domain which is not a principal ideal domain. Example 7-8. This example is given to substantiate a claim made earlier that a primary ideal need not be a power of a prime ideal. Once again, consider the ideal I = (x, y) of the ring F[ x, y ], where F is a field. If the polynomial f(x, y) ¢I, f(x, y) necessarily has a nonzero constant term a 0 • But a 0 lies in the ideal (I,f(x, y)) and so 1 = a 0 1 a 0 E (I,f(x, y)), forcing this ideal to be the entire ring. In consequence, I = (x, y) is a maximal (hence, prime) ideal of the ring F[ x, y]. (The maximality of I could otherwise be deduced from the fact that it is the kernel of the substitution homomorphism f(x, y) -+ f(O, 0).) Next, let us look at the ideal (x 2 , y) ofF[ x, y]. As the reader rna y verify I2

=

(x2, xy, y2)

!;;

(x2, y)

!;;

I.

Inasmuch as .j(x2 , y) = I, Problem 25, Chapter 5, guarantees that (x 2 , y) is primary. A straightforward argument shows that (x 2 , y) is not the power of any prime ideal of F[x, y]. For, in the contrary case, (x 2 , y) = P", where P is a prime ideal and n 2: 1. Since P" ~ I, with I prime, we may appeal to Problem 30, Chapter 5, to conclude that P ~ I. By the same token, the inclusions I 2 !;; P" ~ P, coupled with the fact that Pis a prime ideal, yield I s P. Hence, I = P, so that r = (x 2 , y). Now, the element x E I, while x ¢ (x 2 , y), implying that n 2: 1. On the other hand, y E (x 2 , y), but y ¢ I 2 = (x 2 , xy, y 2 ), which means I 2 c: (x 2 , y) c I. These inclusion relations show that it is impossible to have I" = (x 2 , y) for any n 2: 1. Let us close this phase of our investigation by saying that there is no difficulty in extending the above remarks to polynomials in a finite set of indeterminants. For any ring R, just define recursively R[x 1 , x 2 ,

••• ,

x,]

=

(R[x 1, ~2 , ... , x,_ 1])[x,].

It would not be out of place to devote the remainder of this chapter to the matter of field extensions (most notably, algebraic extensions) and splitting fields. The concepts are presented here partly for their own interest and partly to lay a foundation for a proof of the celebrated Wedderburn theorem on finite division rings (Theorem 9-11). We shall have neither occasion nor space for more than a passing study, and certain topics are touched upon lightly. By an extension F' of a field F, we simply mean any field which contains F as a subfield. For instance, the field of real numbers is an extension of Q, the rational number field. In view of Theorem 4-12, it may be remarked

POLYNOMIAL RINGS

137

that every field F is an extension of a field isomorphic to Q or to ZP, according as the characteristic ofF is 0 or a prime p. Assume that F' is an extension field of a field F and let the elements r 1 , r 2 , ... , rn all lie in F'. The subfield of F' generated by the subset F u {r1 , r 2 , ... , r.. } is customarily denoted by F(r 1 , r 2 , ... , rn):

F(r 1 , r 2 ,

... ,

r.,) = n {KIK is a subfield ofF'; F s;; K; r 1 E K}.

Thus, F(r 1 , r2 , ... , rn) is an extension field ofF containing the elements r 1 (clearly it is the smallest such extension) and one speaks of F(r 1, r 2 , ... , rn) as being obtained by adjoining the r1 to F, or by adjunction of the elements r 1 to F. The purpose of the coming theorem is to determine, up to isomorphism, the structure of all simple extension fields, that is, extension fields F(r) arising from a field F by the adjunction of a single element r. Now, for each element rEF', we have at our disposal the substitution homomorphism F is just the degree of its minimum polynomial (this degree is 1 if and only if rEF). In the course of proving Theorem 7-19, we established the surprising fact that, whenever r is algebraic over F, the integral domain F[r] becomes a field, so that F(r) = F[r]. This amounts to saying that every element of F(r) is of the form f(r), where f(x) is a polynomial in F[x]. Example 7-9. If n =I= 1 is any square-free integer, the element .jii e R # 2 Q is a root of the quadratic polynomial x 2 - n e Q[ x] and, hence, is algebraic over Q. From the preceding paragraph we know that Q[y'!z] is a field and so Q[.Jn] = Q(y'n); in other words, every member of Q(Jn) is of the form f(Jn), where f(x) is a polynomial in Q[x]:

Q(/n)

=

{a 0

+ aly'n + a2 (Jn) 2 + ... + ak(.jii)kja; E Q;

But (Jn) 2 = n, (/n) 3 simply as

=

k ~ 0}.

ny'n, ... , so that Q(y'n) can be described more

Q(y'n)

=

{a

+ by'nja, be Q}.

That is to say, the simple field extension Q(Jn) is what we referred to as a _ quadratic field in Chapter 6. Notice also that an arbitrary element a + E Q(Jn) satisfies the polynomial

bJn

50

FIRST COURSE IN RINGS AND IDEALS

3. Let I be an ideal of the ring R. Establish each of the following: a) R/I has no divisors of zero tf and only if abE I implies that either a orb belongs to I. b) R/I is commutative if and only if ab - ba e I for all a, b in R. c) R/I has an identity element if and only if there is some e e R such that ae - a e I and ea - a e I for all a in R. d) Whenever R is a commutative ring with identity, then so is the quotient ring R/I.

4. Let R be a commutative ring with identity and let N denote the set of all mlpotent elements in R. Verify that a) The set N forms an Ideal of R. [Hint: If a"= bm = 0 for mtegers nand m, consider (a - br+"'.] b) The quotient ring R/N has no nonzero nilpotent elements. 5. Prove the following generalization of the Factorization Theorem: Let ! 1 and ! 2 be homomorphisms from the ring R onto the rmgs R 1 and R 2 , respectively. If kerf1 ~ kerf2 , then there exists a unique homomorphism!: R 1 -+ R 2 satisfymg f 2 = Joj~. [Hint: Mimic the argument of Theorem 3 6; that is, for any element f 1(a) E R 1, definel(II(a)) = f 2 (a).] 6. Let I be an ideal of the ring R. Assume further that J and K are two subnngs of R with I ~ J, I ~ K. Show that a) J ~Kif and only ifnat1 J ~ nat1 K b) nat1 (J f"\ K) = nat 1J n nat1K. 7. If I is an ideal of the ring R, prove that a) R/I is a simple ring if and only if there is no ideal J of R satisfying I c J c R; b) if R is a principal ideal ring, then so is the quotient rmg R1l; m particular, z. is a principal Ideal ring for each n e Z+. [Hint: Problem 17, Chapter 2.] 8. a) Given a homomorphism f from the ring R onto the rmg R', show that 1(bJib E R'} constitutes a partition of R into the cosets of the Ideal kerf [Hint: If b = f(a), then the coset a+ kerf= f- 1(b).] b) Verify that (up to isomorphism) the only homomorphic images of the ring Z of i9tegers are the rings z., n ~ 0, and {0}. . .:;'

u-

9. Suppose that S ts a subring and I an ideal of the ring R. If S n I = {0}, prove that S is isomorphic to a subring of the quotient ring Rjl. [Hint: Utilize the mappingf(a) = a + I, where a E S.J 10. A commutator in a ring R is defined to be any element of the form [a, b] = ab - ba. The commutator ideal of R. denoted by [R, R], is the ideal of R generated by the set of all commutators. Prove that a) R is a commutative ring if and only if [ R, R] = { 0} (in a sense, the size of [ R, R] provides a measure of the noncommutativity of R); b) for an ideal I of R, the quotient ring R/1 is commutative if and only if

[R, R]

~I.

11, Assuming that f is a homomorphism from the ring R onto the commutative ring R', establish the assertions below:

POLYNOMIAL RINGS

141

Let us establish a simple, but nonetheless effective, result about successive extensions. Theorem 7-24. IfF' is a finite extension ofF and F" is a finite extension ofF'; then F" is a finite extension of F. Furthermore, [F":F] = [F":F'][F':F]. Proof An abbreviated proof runs as follows. Suppose that [ F': F] = n and [ F": F'] = m. If { a 1 , a 2 , ... , a,.} is a basis for F' as a vector space over F and {b 1, b2 , ••• , bm} is a basis for F" over F', then the set of mn elements of the form a;bi constitutes a basis for F" over F. This implies that

[F":F] = mn

= [F":F'][F':F].

We still have a few loose ends to tie together, including a more precise description of F(r), when r is algebraic over F. Theorem 7-25. Let F' be an extension field ofF and r E F' be algebraic over F of degree n. Then the elements 1, r, ... , r"- 1 form a basis for F(r) (considered as a vector space over F). Proof Let a be any element of F(r) = F[r]. Then there exists a polynomial g(x) E F[x] such that a = g(r). Applying the division algorithm to g(x) and

the minimum polynomial f(x) of r, we can find q(x) and s(x) in F[ x] satisfying g(x) = q(x)f(x) + s(x), where either s(x) = 0 or deg s(x) < n. Since f(r) = 0, it follows that = g(r) = a. If s(x) = 0, necessarily a = 0, while if s(x) = b0 + b 1 x + .. · + bm~ is a nonzero polynomial of degree m < n, then a = b0 + b 1 r + ... + bm,m· Therefore, the elements 1, r, ... , r"- 1 generate F(r) as a vector space over F. It remains to show that the set {1, r, ... , r"- 1 } is linearly independent. Pursuing this aim, let us assume that c0 1 + c 1 r + ... + c,_ 1r"- 1 = 0, where the c,. E F. Then the polynomial

s(r)

h(x) =

Co

+

c 1X

+ ... +

C,_ 1x"-l E

F(x]

and clearly h(r) = 0, so that h(x) E ker (j>, = (f(x) ). This being the case, = f(x)k(x) for some polynomial k(x) in F[ x ]. But if h(x) =!= 0, we obtain n > deg h(x) = degf(x) + deg k(x) ~ degf(x) = n, h(x)

a manifestly false conclusion. Thus, the polynomial h(x) = 0, which forces the coefficients c0 = c 1 = ... = c,_ 1 = 0. The proof that the n elements 1, r, ... , r"- 1 constitute a basis for F(r) over F is now complete. The statement of Theorem 7-25 can be rephrased in several ways.

142

FIRST COURSE IN RINGS AND IDEALS

Corollary 1. If rEF' 2 F is algebraic of degree n, then every element of the simple extension F(r) is of the form f(r), where f(x) E F[ x] is a polynomial of degree less than n : F(r)

= {a0 + a 1r + ··· + a"_ 1r"- 1iak E F}.

Corollary 2. If r E F' 2 F is algebraic of degree n, then F(r) is a finite (hence, algebraic) extension with [ F(r): F] = n. We include the next theorem for completeness; it is an immediate consequence of Theorem 7-22 and Corollary 2 above. Theorem 7-26. Let F' be an extension of the field F. An element r E F' is algebraic over F if and only if F(r) is a finite extension of F. From this, it is a short step to Corollary. Let r E F' 2 F be algebraic and [ F': F] finite. F' = F(r) if and only if [F(r): F] = [F' :F].

Then

Proof By the last-written theorem, F(r) has a finite degree [ F(r): F]. Now, F(r) is a subspace (over F) of the vector space F'. This corollary is equivalent to asserting that a subspace is the entire space if and only if the dimensions of the two are equal. Example 7-10. Consider the element r = ..)2 + i E C 2 Q, C as usual being the complex number field. Then r 2 = 1 + 2..}2i, so that (r 2 - 1)2 = -8 or r 4 - 2r 2 + 9 = 0. Thus, r is a root of the polynomial f(x)

= x4

-

2x 2

+

9 E Q[ x]

and, hence, is an algebraic element over Q. Now, f(x) has the irreducible factorization over C, f(x) = (x -

J2 + i)(x -

.j2 -

i)(x

+ .j2 +

i)(x

+ .j2 -

i),

which indicates thatf(x) has no linear or quadratic factors in Q[x]. Therefore, f(x) is irreducible as a member of Q[x J and serves as the minimum polynomial of r over Q; in particular, the element r has degree 4. By Theorem 7-25, the simple extension Q(r) is a four-dimensional vector space over Q, with basis

1,

r =

J2

+

i,

r2

= 1 + 2..}2i,

r3

=

-J2

+ 5i. + 3 E R # [ x ],

At the sam~ time r is a root of the polynomial x 2 - 2J2x with x 2 - 2J2x + 3 irreducible over R # ; thus, r is of degree 2 over R #.

Example 7-11. For a second illustration, we turn to the extension field Q(.j2, ..)3). The elements ..)2 and J3 are clearly algebraic over Q, being roots of the polynomials x 2 - 2, x 2 - 3 E Q[ x ], respectively. Our contention is

POLYNOMIAL RINGS

143

that the field Q(-)2, J3) is actually a simple algebraic extension of Q; in fact, Q(J'l, -j3) = Q(J'l _+ -j3),_ with J2 + J3 algebraic over Q. Since t~e element J2 + J3 belongs to Q(J2, -)3), we certainly have Q(J'l + J3) ~ Q(J'l, -j3). As regards the reverse inclusion, a simple computation shows that 2-J2 = (J'l is a member of Q(-)2

+

-)3) 3

-

9(-)2

+

J3)

+ J3), and therefore so is J2. But then,

J3

= (J'l +

JJ> - J2

also _lies i!! Q(-)2 + -)3). This leads to the inclusion Q(J'l., J3) ~ Q(-)2 + -)3) and the asserted equality. To see that r = -)2 + J3 is an algebraic element over Q, notice that r 2 = 5 + 2J6, (r 2 - 5f = 24, and, hence, the polynomial f(x) = x4 -

10x2 + 1 has rasa root. One may verify thatf(x) is irreducible in Q[x], making it the minimum polynomial of r. Perhaps the quickest way to see this is as f~llows. Let F' = Q(J2); th~n [F':Q] = 2, with basis {1, -)2}, and [F'(-,j3):F'] = 2, with basis {1, -)3}. From Theorem 7-24, it follows that [ F'(.J3): Q] = 4and a basis for F'(.J3)over Q is given by {1, J2, .J3,-J6}. But F'(-j3) = Q(J'l)(-j3) = Q(J'l, -j3) = Q(J'l + -j3), and we know that the dimension of Q(-)2 + J3) is equal to the degree of the minimum polynomial of r = J2 + JJ. Incjdentalry, there are five fields_ between Q and Q(J'l., J3), namdy, Q, Q(J2), Q(.J3), Q(j6), and Q(J'l, J3). Taking stock of Steinitz's theorem (page 140), it should come as no surprise that Q(..jL, -j3) can be generated by a single element. • Until now, we have always begun by assuming the existence of an extension field F' of F and then studied the structure of simple extensions F(r) within F'. The subject can be approached from a somewhat different standpoint. Given a field F and an irreducible polynomialf(x) E F[x], one may ask whether it is possible to construct a simple extension F' of F in which f(x), thought of as a member ofF'[ x ], has a root. (If degf(x) = I, then, in a trivial sense, F is itself the required extension). To answer this question, we take our cue from Theorem 7-19. For if such an extension ofF can be found at all, it must be of the form F(r), with r algebraic over F. As pointed out in our earlier discussion, r will possess a minimum polynomial g(x) which is irreducible in F[x] and such that F(r) ~ F[x]j(g(x)). This suggests that, when starting with a prescribed irreducible polynomial f(x) E F[ x ], the natural object of interest should be the associated quotient ring F[x]/(f(x)). After this preamble, let us proceed to some pertinent details.

144

FIRST COURSE IN RINGS AND IDEALS

Theorem 7-27. (Kronecker). If f(x) is an irreducible polynomial in F[ x ], then there is an extension field ofF in which f(x) has a root.

Proof For brevity, we shall write I in place of the principal ideal of F[x] generated by polynomial f(x); that is to say, I = (f(x)). Since f(x) is assumed to be irreducible, the associated quotient ring F' = F[x]/I is a field. To see that F' constitutes an extension ofF, consider the natural mapping nat 1 : F[ x] --+ F'. According to Theorem 4-7, either the restriction nat 1 IF is the trivial homomorphism or else nat 1 (F) forms a field isomorphic to F, where as usual nat 1(F) = {a + II a E F}. The first possibility is immediately excluded by the fact that nat1 (1) = 1

+

I =I= I,

which is the zero element ofF'. Therefore, F is imbeddable in the (quotient) field F' and, in this sense, F' becomes an extension of F. It remains to be established that the polynomialf(x) actually has a root in F'. Assuming that f(x) = a0 + a 1 x + ... + anx", then, from the definitions of coset addition and multiplication,

+ I)(x + I) + .. · + (a .. + I)(x + I)" = a0 + a 1 x + ··· + a.. x" + I = f(x) + I = 0 + I. If we now identify an element ak E F with the coset ak + I which it determines (a 0

+

I)

+

(a 1

in F' (the fact that F is isomorphic to nat 1(F) permits this). we obtain

a0

+

a 1(x

+

I)

+ ... +

a,.(x

+

I)" = 0,

which is equivalent to asserting that f(x + I) = 0. In other words. the coset x + I ,., lx + I is the root off(x) sought in F'. Since each polynomial of positive degree has an irreducible factor (Theorem 7-13), we may drop the restriction thatf(x) be irreducible. Corollary. If the polynomialf(x) E F[x] is of positive degree, then there exists an extension field ofF containing a root of f(x). To go back to Theorem 7-27 for a moment, let us take a closer look at the nature of the co sets of I = (f(x)) in F[ x ], with the aim of expressing the extension field F' = F[x]/1 in a more convenient way. As usual, these cosets are of the form g(x) + I, with g(x) E F[ x]. Invoking the division algorithm, for each such g(x) there is a unique polynomial r(x) in F[x] satisfying g(x) = q(x)f(x) + r(x), where r(x) = 0 or deg r(x) < degf(x). Now g(x) - r(x) = q(x)f(x) E I, so that g(x) and r(x) determine the same coset; g(x) + I = r(x) + I. From this, it is possible to draw the following conclusion: each coset of I in F[x] contains exactly one polynomial which

POLYNOMIAL RINGS

145

has degree less than that of f(x) or else is the zero polynomial. In fact, the cosets of I are uniquely determined by remainders on division by f(x) in the sense that g(x) + I = h(x) + I if and only if g(x) and h(x) leave the same remainder when divided by f(x). Thus, if degf(x) = n > 1 (for instance,f(x) = a 0 + a 1x + ··· + a.x"), then the extension field F' may be described by F'

= {b 0 +

b 1x

+ ··· +

b._ 1x"- 1

+

lib" E F}.

Identifying b" + I with the element b", we see as before that a typical coset can be uniquely represented in the form

+

+ I) + ··· + b._ 1(x + I)"- 1 . As a final simplification, let us replace x + I by some new symbol A, so that b0

b 1(x

the elements ofF' become polynomials in A: F' = {bo + b1 A. + ··· + b._ 1 A."- 1 jbk E F}. Observe that since A. = x + I is a root off(x) in F', calculations are carried out with the aid of the relation a0 + a 1 A. + .. · + a.A." = 0. The last paragraph serves to bring out the point that F' is a finite extension ofF with basis {1, ).,. ).2 , ••• , )."- 1 }; in particular, we infer that

[F':F]

=

n

=

degf(x).

To recapitulate: if f(x) E F[ x] is an irreducible polynomial over F, then there exists a finite extension F' ofF, such that [ F': F] = degf(x), in which f(x) has a root. Moreover, F' is a simple algebraic extension generated by a root of f(x). (Admittedly, some work could be saved by an appeal to Theorems 7-21 and 7-25, but our object here is to present an alternative approach to the subject.) We pause now to examine two concrete examples of the ideas just presented. Example 7-12. Consider Z 2 , the field of integers modulo 2, and the polynomial f(x) = x 3 + x + 1 E Z 2 [x]. Since neither of the elements 0 and 1 is a root of x 3 + x + l,f(x) must be irreducible in Z 2 [x]. Theorem 7-27 thus guarantees the existence of an extension of Z 2 , specifically, the field Z 2 [x]/(/(x) ), in which the given polynomial has a root. Denoting this root by )., the discussion above tells us that

+ b). + d 2 la, b, c E Z 2 } = {0, 1, A., 1 + A., l 2 , 1 + A. 2 , A. + A.2 , 1 + )., + ). 2 }, where, of course, A. 3 + )., + 1 = 0. Z 2 [x]/(f(x)) = {a

As an example of operating in this field, let us calculate the inverse of 1 + A. + ). 2 • Before starting, observe that by using the relations ;_3 =

-(A.

+

1) = ;.,

+

1,

).,4 =

;.z + ;.,

146

FIRST COURSE IN RINGS AND IDEALS

(our coefficients come from Z 2 , where -I = l), the degree of any product can be kept less than 3. Now, the problem is to determine elements a, b, C E Z 2 for which

Carrying out the multiplication and substituting for A. 3 , A. 4 in terms of I, A., and .A?, we obtain

+

(a

+

b

c)

+

a.A.

+

(a

+

b)A. 2 = 1.

This yields the system of linear equations

+

a+b=O, a= 0, = 1, with solution a = b = 0, c = 1; therefore, (1 + A. + ..1. 2 )- 1 = A. 2 • It is worth noting that x 3 + x + 1 factors completely mto factors in Z 2 [ x ]/(f(x)) and has the three roots A., )_ 2 , and ). + ;_ 2 : a+ b

x3

+

+

x

c

linear

+ -1.2)).

1 = (x - A.)(x - ). 2 )(x - (..l.

Example 7-13. The quadratic polynomial x 2 + 1 is irreducible in R # [x]. For, if x 2 + 1 were reducible, it would be of the form

+

x2

1 = (ax

+ b)(cx + d) + (ad + bc)x +

acx 2

=

bd,

where a, b, c, dE R #. It follows at once that ac = bd = 1 and ad Therefore, be = -(ad), and

+ be =

0.

1 = (ac)(bd) = (ad)(bc) = -(ad) 2 , or rather, (ad) 2 = -1, which is impossible. In this instance, the extension field R # [x]/(x 2 R#[x]/(x 2

+

1)

=

{a

+

+

b..l.Ja, bE R# ;_Xz

1) is described by

+

1 = 0}.

Performing the usual operations for polynomials, we see that (a

+

b.A.)

+

(c

+

d..l.) = (a

+

+ +

(ad

c)

+

(b

+

d)..l.

bc)A.

+

bd(A. 2

and (a

+

b..l.)(c

+

d..l.) = (ac - bd)

=

(ac - bd)

(ad

+ +

+

1)

bd)..l..

The similarity of these formulas to the usual rules for addition and multiplication of complex numbers should be apparent. As a matter of fact, R # [x ]/(x2 + 1) is isomorphic to the field C of complex numbers under the mapping : R # [ x ]/(x 2 + 1) --. C given by ( a + b..l.) = a + bi. This provides an elegant way of constructing C from R #.

POLYNOMIAL RINGS

147

Before proceeding further, two comments are in order. First, Example 7-12 shows that there exist finite fields other than the fields Z of integers modulo a prime p. The fact that the field of this example hasP2 3 = 8 elements is typical of the general situation: ifF is a fimte field, then F contains p" elements, where the prime pis the characteristic ofF (Theorem 9-7). In the second place, the construction of Theorem 7-27 yields an extension of the field F in which a given (nonconstant) polynomial f(x) E F[ x] splits off one linear factor. By repeated application of this procedure, we can build up an extension F' ofF in whichf(x), thought of as a member of F'[ x], factors into a product of linear factors; that is, the field F' is large enough to contain all the roots ofj(x) (technically speaking, the polynomial splits completely in F'[x]). We present this result in the form ofan existence theorem. Theorem 7-28. If f(x) E F[ x] is a polynomial of positive degree, then there exists an extension field F' of F in which f(x) factors completely into linear polynomials. Proof The proof is by induction on n = degf(x). If n = 1, then f(x) is already linear and F itself is the required extension. Therefore, assume that n > 1 and that the theorem is true for all fields and for all polynomials of degree less than n. Now, the polynomial f(x) must have some irreducible factor g(x). By Theorem 7-27, there is an extension field K ofF in which g(x) and, hence, f(x), has a root r 1 ; specifically, the field K = F[x]j(g(x)). Thus,f(x) can be written in K[x] asf(x) = (x - r 1 )h(x), where deg h(x) = n - l. By our induction assumption, there is an extension F' of Kin which h(x) splits completely; say h(x) = a(x - r 2 )(x - r 3 ) · • · (x - r "),with ri E F', a =f- 0. From this, we see that f(x) can be factored into linear factors in

F'[x].

Corollary. Let f(x) E F[x], degf(x) = n > 0. Then there exists an extension ofF in whichf(x) has n (not necessarily distinct) roots. Example 7-14. To illustrate this situation, let us look at the polynomial f(x) = (x 2 - 2)(x 2 - 3) over the field Q of rational numbers. From Example 7-4, x 2 - 2 (and by similar reasoning, x 2 - 3) is already known to be irreducible in Q[x]. So we begin by extending Q to the field F 1 , where F 1 = Q[x]j(x 2

-

2) = {a

+ bJ.Ia, bE Q; J. 2

and obtain the factorization f(x)

= (x - J.)(x + J.)(x 2 = (x - ..j2)(x

-

3)

+ ..j2)(x2

(As ).2 = 2, one customarily identifies ). with

J2.)

-

3).

-

2 = 0};

148

FIRST COURSE IN RINGS AND IDEALS

However, f(x) does not split completely, since the polynomial x 2 - 3 remains irreducible in F 1 [ x]. For, suppose to the contrary that x 2 - 3 has a root in F 1 ; say c + d..)2, with c, dE Q. Substituting, we find that

+

(c 2

2d2

-

+ 2cdJ2

3)

= 0,

or, what amounts to the same thing,

cd = 0. The latter equation implies that either c = 0 or d = 0. But neither c nord can be zero, since this would mean that d2 = 3/2 or c 2 = 3, which is clearly impossible. Accordingly, x 2 - 3 does not split in F 1 [ x]. In order to factor f(x) into linear factors, it becomes necessary to extend the coefficient field further. We therefore construct a second extension F 2 , where F 2 = F 1 [x]/(x 2

-

3) ={a+ PJLia,peF 1 ;Jt 2

-

3 = 0}.

The elements ofF 2 can be expressed alternatively in the form

(a + bJ'i) + (c

+ dJ2).j3

= a

+

bJ2

+ c.j3 +

dJ6,

where, of course, the coefficients a, b, c, d all lie in Q. It follows without difficulty that the original polynomial now factors in F 2 [ x] as

f(x) = (x - .A.)(x = (x -

+

J'i)(x

.A.)(x - p)(x

+

+ p)

J2)(x - J3)(x

+ J3).

Let a field F be given and consider a nonconstant polynomialf(x) E F[ x]. An extension field F' ofF is said to be a splitting field for f(x) m·er F provided thatf(x) can be factored completely into linear factors in F'[x], but not so factored over any proper subfield ofF' containing F (this minimum nature of the splitting field is not required by all authors). Loosely speaking, a splitting field is the smallest extension field F' in which the prescribed polynomial factors linearly: /(x) = a(x - r 1 )(x - r 2 )

•••

(x - rn)

(r 1 e F').

To obtain a splitting field for f(x), we need only consider the family {F;} of all extension fields F; in which f(x) can be decomposed as a product of linear factors (Theorem 7-27 guarantees the existence of such extensions); then n F 1 serves as a splitting field for f(x) over F. Having thus indicated the existence of a splitting field for an arbitrary polynomial in F[ x ], it is natural to follow this up with a query as to uniqueness. For a final topic, we shall prove that any two splitting fields of the same (nonconstant) polynomial are isomorphic; this being so, one is justified in using the definite article and speaking of the splitting field of a given polynomial.

POLYNOMIAL RINGS

149

Before presenting the main theorem, two preparatory results of a somewhat technical nature are needed. Lemma. Letf(x) be an irreducible polynomial in F[x] and r be a root of f(x) in some extension field K of F. Then F(r) ~ F[ x]/(f(x)) under an isomorphism whereby the element r corresponds to the coset x + (f(x)). Proof Since the element r is algebraic over F, it follows directly from Theorem 7-19 that F(r) ~ F[ x ]/(f(x)) via an isomorphism () with the property that ¢, = ()a nat: F(r) -+ F'(r'); this situation

IS

portrayed m the diagram below:

$ F(r) - - - - + F'(r')

a

j

~-• j

F[x ]j(f(x)) ---> F'[y]/(f'()')) Certainly, Cl> is an isomorphism of F(r) onto F'(r'), for the individual mappings a, r, p- 1 are themselves isomorphisms. If a is an arbitrary element of F. then Cl>(a)

= (p- 1 o r)(a:(a)) = (p- 1 o r)(a + = p- 1 (u(a) + (f'(y))) = u(a),

(f(x)))

whence Cl> is actually an extension of u to all of F(r). Finally, we point out that Cl>(r) = (p- 1 o r) (a(r)) = (p- 1 a r) (x + (f(x))

= p-t(y + (f'(y))) = r', as required, and the theorem is proved in its entirety. For a simple, but nonetheless satisfying, illustration of this last result, take both F and F' to be the real number field R t~>; let f(x) e R # [x] be the irreducible polynomial l(x) = x 2 + I, so that f'(.v) = y 2 + I (recall that the identity map is the only isomorphism of R# onto itself). Finally, choose r = i and r' = - i. Theorem 7-29 then asserts that R #(i) ~ R #(- i) under an isomorphism which carries i onto - i. Inasmuch as R # (i) = R # ( - i) = C, the isomorphism in question is just the correspondence between a complex number and its conjugate. We now have the mathematical machinery to show the uniqueness (to within isomorphism) of splitting fields. Actually, we shall prove a somewhat more general result. Theorem 7-30. Let u be an isomorphism of the field F onto the field F'. Let /(x) = a0 + a 1x + ··· + a"x" E F[x] and f'(y) = u(a 0 ) +

PROBLEMS

151

+ ··· + u(a. )y" be the corresponding polynomial in F'[y]. If K is a splitting field of f(x) and K' a splitting field of f'(y), then u can be extended to an isomorphism of K onto K'.

u(a 1 )y

Proof Our argument will be by induction on the number n of roots ofJ(x) that lie outside F, but (needless to say) in K. When n = 0, all the roots of f(x) belong to F and F is itself the splitting field of f(x); that is, K = F. This in turn induces a splitting of the polynomial f'(y) into a product of linear factors in F'[y ], so that K' = F'. Thus, when it happens that n = 0, the isomorphism u is, in a trivial sense, the desired extension to the splitting fields. Let us next assume, inductively, that the theorem holds true for any pair of corresponding polynomials f(x) and f'(y) over isomorphic fields E and E', provided that the number of roots of roots of f(x) outside of E is less than n (n ~ 1). IfJ(x) E F[x] is a polynomial having n roots outside ofF, then not all of the irreducible factors of f(x) can be linear in F[x]; for, otherwise,f(x) would split completely in F, contrary to assumption. Accordingly, f(x) must have some factor g(x) of degree m > 1 which is irreducible in F[x]. Let g'(y) denote the corresponding irreducible factor of f'(y). Since K is a splitting field of f(x) over F, g(x) in particular must have a root in K; call it r. Similarly, one of the roots of the polynomialf'(y), say r', is a root of g'(y) inK'. By Theorem 7-29, u can be extended to an isomorphism u' between the fields F(r) and F'(r'). Now, K is a splitting field of f(x), viewed as a polynomial with coefficients from F(r); in a like manner, K' can be regarded as a splitting field off'(y) over the field F'(r'). Because the number of roots of f(x) lying outside of F(r) is less than n, the induction hypothesis permits us to extend u' (itself an extension of u) to an isomorphism of K' onto K. This completes the induction step and the proof of the theorem as well, for u has been suitably extended.

With the corollary below, we achieve our objective. Corollary. Any two splitting fields of a nonconstant polynomial f(x) E F[x] are isomorphic via an isomorphism such that the restriction j F is the identity mapping.

Proof. This is an immediate consequence of the theorem on taking F = F' and u to be the identity isomorphism iF. PROBLEMS

1. If R is a commutative ring with identity, prove that a) The set I = {f(x) E R[[xJ]Iordf(x) > 0} u {0} forms an ideal of the ring R[[x]]; in fact, I= (x).

152

FIRST COURSE IN RINGS AND IDEALS

b) The ideal!" consists of all power series having order 2 n, together with 0. c) z + 1" = {0}.

n...

2. For any field F, consider the set F(x) consisting of all expressions of the form

~

akx!' = a_.x-•

+ a-n+tx-•+l + ... + a_,x- 1 + a0 + a 1x + a 2x 2 + ··

k=-n

where all the ak E F and n

~

0 varies.

If addition and multiplicatiOn are defined in the obvious way, F(x) becomes a ring, known as the ring of extended (formal) power series otw F. Show that F(x) is in fact the field of quotients of the domain F[[ x ]]. [Hint: Given n E Z +, Qc~(F([x]]) must contain x-•.J 3. Let R be a commutative ring with identity. If R is a local ring, prove that the power series ring R[[x]] is also local. 4. Given that R is a commutative ring with identity, deduce that a) No monic polynomial is R[x] is a zero divisor. b) If the polynomial f(x) = a 0 + a 1 x + ... + a.x" is a zero divisor in R[x], then there exists an element 0 1- r E R such that rf(x) = 0. [Hint: Assume that f(x)g(x) = 0. Use the polynomials akg(x) to obtain 0 1- h(x) E R[x], with deg h(x) < deg f(x), satisfying h(x)f(x) = 0.] 5. If R is a commutative ring with ident1ty, verify that the polynomial 1 + ax is invertible in R[x] if and only ifthe element a is nilpotent in R. [Hint: Problem 10, Chapter 1.] 6. For an arbitrary ring R, prove that a) If I is an ideal of R, then I[ x] forms an ideal of the polynomial ring R[ x]. b) If R and R' are isomorphic rings, then R[ x] is ISomorphic to R'[ x]. c) char R = char R[x] = char R[[x]]. d) If I is a nil ideal of R, then I[ x] is a nil ideal of R[ x]. [Hint: Induct on the degree of polynomials in I[ x ].] 7. Establish the following assertions concerning the polynomial ring Z[x]: a) The ideal (x)

= {a 1x + a 2xz + ... + a.x"!at E Z; n ~ 1}

is a prime ideal of Z[x], but not a maximal ideal. Incidentally, (x) is maximal in F[ x ], where F is a field. b) Z[x] is not a principal ideal domain. [Hint: Consider (x, 2), the (maximal) ideal of polynomials with even constant terms.] c) The primary ideal (x, 4) is not the power of any prime ideal of Z[x]. [Hint: (x, 2) is the only prime ideal containing (x, 4).] 8. Let P be a prime ideal of R, a commutative ring with identity. Prove that P[ x] is a prime ideal of the polynomial ring R[x]. If M is a maximal ideal of R, is M[x] a maximal ideal of R[x]?

153

PROBLEMS

9. Consider the polynomial domain F[ x ], where F is a field, and a fixed element reF. Show that the set of all polynomials having rasa root, M,

= {J(x) e F[x]jf(r)

=

0},

forms a maximal ideal of F[ x ], with F[ x ]IM, ::>: F. [Hint: M, -+ F is the substitution homomorphism induced by r.]

=

ker 1/J., where

(nm) = rp(n)rp(m). b) For any prime p and n > 0, rj>(p") = p"(l- 1/p) = p"- p"- 1 • [Hint: The integers k such that 0 < k < p" and gcd (k, pn) =I= 1 are p, 2p, ... , p"- 1 p.] c) If p 1, p 2 , ... , Pk are the distinct pnme divisors of an mteger n > 1, then rf>(n) = n(l - l/ptl(1 - l/p2 ) .. • (1 - 1/pk). d) n = Lain rf>(a). 8. Let r(n) denote the number of (distinct) positive divisors of an integer n > 1. Prove that a) If n has the prime factorization n = p~'p~1 . . . p~". where the P; are distinct primes and n1 e Z+, then -r(n) = (n 1 + l)(n 2 + 1) ... (nk + 1). b) The number of ideals of z. is -r(n). c) -r(n)r/>(n) ~ n. [Hint: D(n1 + l)TI(l - 1/pJ ~ 2kJJ(l/2n 9. Given that the set H. = {[a] E z.j[a] is not a zero divisor of z.]. prove that (H., ·.) forms a finite group of order rf>(n).

10. a) Derive Fermat's Little Theorem: If pis a prime number and a=!= 0 (mod p), then ap-l = 1 (mod p). b) If gcd (a, n) = 1, show that the equation ax = b (mod n) has a umque solution modulo n. [Hint: All solutions are given by x = bali>l•-~l+ kn.]

ll. a) Prove that every field is a principal ideal domain. ', jr· b) Show that the ring R = {a + bJ2ja, bE Z} is not a field by exhibiting a nontrivial ideal of R. 12. Let f be a homomorphism from the ring R into the ring R' and suppose that R has a subring F which is a field. Establish that either F ~ kerf or else R' contains a subring isomorphic to F. 13. Derive the following results: a) The identity element of a subfield is the same as that of the field. b) If {F;} is an index collection of subfields of the peld F, then n F; is also a subfield of F. c) A subring F' of a field F is a subfield ofF if and only ifF' contains at least one nonzero element and a- 1 E F' for every nonzero a e F'. d) A subset F' of a finite field F is a subfield of F if and only if F' contains more than one element and is closed under addition and multiplication.

14. a) Consider the subsetS of R# defined by

S = {a + b.J.Pia, be Q; p a fixed prime}. Show that S is a subfield of R #. b) Prove that any subfield of the field R# must contain the rational numbers.

15. Prove that if the field F is of characteristic p > 0, then every subfield of F has characteristic p.

190

FIRST COURSE IN RINGS AND IDEALS

Proof. Let F be a finite field with p" elements and F* be its multiplicative group of nonzero elements; this group has order p" - 1. The argument about to be presented hinges on finding an element in F* of order h = p" - 1. To this end, we first consider the prime factorization of h:

where the q; are distinct primes and r; E Z +· For i = 1, 2, ... , m, set h; = h/q;. Now, there exists a nonzero element a; E F which is not a root of the polynomial xh' - 1 E F[ x J; for this polynomial has at most h; distinct roots in F and h; < h, the number of nonzero elements of F. Next, take b; =

and define b = b 1 b2

a71•', where s; = qj' •••

(i = 1, 2, · · ·

m)

bm. We certainly have b~'

= a7

= 1,

so that the order b; must divide s; = qi'. On the other hand, if b~·-l

=

1,

then contrary to our original choice of the element a;. The implication is that b; has order f{;'. To settle the whole affair, we will show that the element b is of order h. In the contrary case, the order of b must be a proper divisor of h (since bh = 1, the order of b certainly divides h) and therefore divides at least one of the integers h; (i = 1, 2, ... , m), say hp We then have

1 = bht =

b~·b~·

... b~··

If2::;; i ::;; m, then qi'lh 1 , which implies that b7• := 1 and sob~· = 1. This means that q;' (the order of b 1) divides h1 , which is impossible. Thus, the element b has order h and, in consequence, the cyclic subgroup of F* generated by b will also be of order h ; since F* contains only p" - 1 = h elements, this cyclic group must be all ofF*. It is not surprising and is quite easy to prove: Corollary. Any finite field F with p" elements is a simple algebraic extension of the field Z,. · Proof. We already know that F is an algebraic extension of degree n of its prime subfield ZP. The theorem above indicates that the p" elements ofF can be written as 0, 1, b, b2 , ... ,bP"- 2 for some bE F*; in other words, the field F = Z,(b).

TWO CLASSIC THEOREMS

191

As an application of these ideas, let us prove a statement made earlier to the effect that, for any finite field F, the polynomial domain F[x] contains irreducible polynomials of arbitrary order. Let F be a finite field. For each positive integer n, there exists an irreducible polynomialf(x) E F[x] with degf(x) = n.

Theorem 9-10.

Proof Suppose that F' is an extension ofF with [ F': F] = n. As was just seen, there exists an element b in F' such that F' = F(b). If f(x) is the

minimum polynomial of b over F, then (invoking Corollary 2 of Theorem 7-25) degf(x) = [F':F] = n. Therefore, f(x) E F[x] is the required irreducible polynomial of degree n and the theorem follows. Finite fields are called Galois fields after the French mathematician Evariste Galois, who first discovered the existence of finite fields aside from those of the form Zr The (essentially unique) field with p" elements is commonly denoted by the symbol GF(p"). To construct GF(p"), we need only determine an irreducible polynomial f(x) of degree n in ZP[x]; then ZP[x]j(f(x)) is the required Galois field with p" elements. It is now time to redeem a promise made earlier to provide a proof that every finite division ring is a field (Wedderburn's Theorem). Our approach is founded on a treatment by Herstein [ 43]. Although this is perhaps the most elementary, other proofs of Wedderburn's Theorem are common; an entirely different one requiring the concept of cyclotonic polynomials appears in [5]. The argument which we are about to give is lengthy and will be prefaced by two simplifying lemmas (the student who is pressed for time may wish to omit all this on a first reading). Much of our success, both with Wedderburn's Theorem and its applications, inevitably flows from the result below. Lemma 1. Let R be a division ring of characteristic p > 0, p a prime. Suppose that the element a E R, a ¢ cent R, is such that aPm = a for some m > 0. Then there exists an x E R for which

1) xax- 1 f a, 2) xax- 1 E ZP(a), the extension field obtained by adjoining a to ZP. Proof Let Z Pbe the prime subfield of R. Since aPm - a = 0, a is algebraic

over ZP. By Theorem 7-25, we know that the extension ZP(a) is a finite field and therefore must have p" elements for some n E Z +· Furthermore, each r e ZP(a) satisfies rP" = r.

192

x in

FIRST COURSE IN RINGS AND IDEALS

Now, define the function f: R ----.. R by setting f(x) = xa - ax for all R. Using induction, it is not difficult to show that the composite Jk(x)

=

k

L (-1) 1(~)a 1 xak-l

(k ~ 1).

i=O

When k = p, the foregoing equation reduces simply to fP(x) = xaP - aPx, because pi (f) for 0 < i < p {recall also that char R = p). Another routine induction argument extends this to fP"(x)

=

xaP" - aP"x.

But aP" = a, whence fP"(x) = xa - ax = f(x) for all x E R, which is equivalent to asserting thatfP" = f. For each element r E ZP(a), consider the function T,. on R defined by T,.(x) = rx. Our contention is that f commutes with all such T,.. The reasoning proceeds as follows: Being a field, ZP(a) is commutative, so that, ifxeR, {f o T,.){x) = f(rx) = (rx)a - a(rx) = rxa - rax

= r(xa - ax) = (T,. o f)(x). This, in short, means thatf o T,. = T,. o /for every r in ZP{a). From the corollary on page 188, the polynomial yP" - y E Z P[y] factors completely in ZP{a); in other words, we have

rr ()' - r),

yP" - y =

rEZp(a)

or, what amounts to the same thing, yP" _ y

= y

rr

(y - r).

0 'frEZp(a)

This formal identity requires only that y commute with all elements r E ZP(a). Taking stock of the fact that f o T,. = T,. of, as well as the relation fP" = f, we thereby obtain

o=

r - 1 = 1° rr

O'freZp(a)

(J·- T..>·

(In essence, one applies the substitution homomorphism cjJ 1 to the ring of polynomials whose coefficients are homomorphisms on (R, + ).) If, for every r =I= 0 in ZP(a), it happens that (f- T,.)(x) = 0 implies x = 0, then the last-written equation would necessarily lead to f = 0. This would mean that xa - ax = 0 for all x E R, forcing a to lie in the center of R, contrary to hypothesis. Consequently, there must exist some 0 =I= reZP(a) and some element x =I= 0 in R for which(/- T,.)(x) = 0; that is to say, xa - ax = rx and so xax- 1 = r

+ a e ZP(a).

Since r =I= 0, certainly the product xax- 1

=1= a.

TWO CLASSIC THEOREMS

193

Corollary. In the lemma, xax- 1 = ak =I= a for some integer k E Z +· Proof Since aP"- 1 = 1, the element a has finite order as a member of the multiplicative group R*. Lets be the order of a. Then, in the field ZP(a), each of the s elements 1, a, a 2 , ••• , a•-t is a root of the polynomial y• y E ZP[ x]. This polynomial can possess at most s roots in ZP(a) and 1, a, ... , as- 1 are all distinct. But xax- 1 E Zp(a) and clearly (xax- 1)11

In consequence, xax- 1

=

= xa•x- 1 = xx- 1 = 1.

a" for some k., with 2 ;5;; k ::;; s - 1.

To cope with the problem at hand, we shall also need the following: Lemma 2. IfF is a finite field and 0 =I= a E F, then there exist elements a, bE F such that a: = a2 + b2 • Proof We first dispense with the case where char F = 2. In this special situation, F has 2n elements and any element of F satisfies the equation x 2 " = x. Thus, every nonzero member a ofF is a square and, in particular, a:

=

a:l"

=

(a:l"- ')2.

The lemma is thereby established on taking a = a 2 "_, and b = 0. Now, if the characteristic ofF is an odd prime p, then F will contain p" elements. Let f be the mapping of F* into itself defined by f(x) = x 2 (as usual, F* denotes the multiplicative group of F). Then f is a group homomorphism, with kerf= {xEF*Jx2 = 1}

= {1, -1}.

Since char F =I= 2, 1 and - 1 are necessarily distinct. This implies that, for each p Ej(F*), there exist exactly twoelementsa 1 , a 2 inF* with ai = a~ = P; in fact, a 2 = -a 1 • To put it another way, for each pair of elements a 1 and - a 1 in F*, we get one element which is a square. Hence, half the elements of F* will be squares, call these P1, P2 , ••• , pk, where the integer k = (pn - 1)/2. Given 0 =I= a E F, assume that a is not a square and consider the set s = {a: - P1li = 1, 2, ... , k}. If it turns out that a - pi is not a square for any value of i, the set S (which contains k distinct elements) must coincide with the k nonsquares ofF*. But then a will lie in S, yielding a = a - P; for some choice of i; whence ft; = 0, an obvious contradiction. This being the case, we conclude that a - ft; = pi for suitable integers i and j, or a = P; + pi. Thus, a is the sum of two squares in R and the requisite equation holds.

Corollary. IfF is a finite field and 0 =I= a E F, then there exist elements a, b in F such that 1 + a2 - ab 2 = 0.

194

FIRST COURSE IN RINGS AND IDEALS

After this preparation, we now undertake the task of proving the theorem which serves as the focal point of the present chapter. Theorem 9-11. (Wedderburn). Every finite division ring is a field. Proof Suppose, for purposes of contradiction, that the theorem is not true for all finite division rings. Let R have minimal order among the set of noncommutative division rings, so that any division ring with fewer elements than R will be commutative. Before becoming involved in the technical argument of the proof, let us note that if the elements a, b E R satisfy abk = bka, but ab =I= ba, then b" E cent R. For, consider the centralizer of b" in R: C(b")

=

{x

E

Rlxb"

=

b"x}.

It follows without difficulty that C(b") comprises a division ring (a division subring of R). If CW) =I= R, then by our hypothesis CW) would necessarily be commutative. But a, b both lie in C(b") and these elements clearly do not commute. This entails that C(bk) = R, which is scarcely more than a restatement that b" E cent R. Now to the proof proper. Since the multiplicative group R* is finite, every nonzero element of R must have finite order and, as a result, some power of it belongs to the center of R. By virtue of this circumstance, the set S = {mE Z+lfor some c ¢cent R, em E cent R} is not empty. Pick the integer n to be minimal in S. Then there exists an element a¢ cent R such that a" E cent R. We assert that n is a prime number. Indeed,weren = n 1 n 2 ,with1 < n 1,n 2 < n,itwouldfollowthata"'¢centR, yet (a"'t 2 = a" E cent R. In other words, the integer n2 is a member of S, a contradiction to the minimal nature of n. Next apply Lemma 1 to obtain an element x E Rand an integer k such that xax- 1 = a" ::/= a. At the outset, observe that x 2 ax- 2

=

x(xax- 1 )x- 1 = xa"x- 1

= .(xax- 1 )" =

a" 2 ,

so, by induction, x"- 1 ax- 2, then (n - 1)/2 is an integer and so tn(n-1)/2 = (t")(n-1)/2 = 1, which implies that (c- 1 d)" = 1. Being a solution of the equation y" = 1, it follows from earlier reasoning that c- 1 d = r; E cent R for some choice of i. But then d- 1c = (c- 1d)- 1 E cent Rand so (using (2) above), t = c- 1 tc = (dc- 1 d- 1 )c

= d(r 1 c)c- 1

an obvious contradiction. Thus, the theorem is proved, at least when n is an odd prime. Turning to the more troublesome possibility, we now suppose that n = 2. In this event, t 2 = 1 and, of course, t =1= l, whence t = -1. Then, cd = -de =I= de; consequently, the characteristic of R is different from 2. Applying Lemma 2 to the field cent R, we can find elements X; (i = 1, 2) in cent R satisfying 1

+ xf

-ax~ = 0

Armed with this, a direct computation shows that (c

+ dx 1 +

cdx 2 ) 2 = c2 (1

+ xf

- ax~) = 0,

which, because R is a division ring, leads to c + dx 1 + cdx 2 = 0. clinch matters, since char R =/= 2, 0 =/= 2c 2 = c(c

+

dx 1

+ cdx 2 ) + (c +

dx 1

+

To

cdx 2 )c = 0,

an absurdity. This contradiction finally completes the proof of Wedderburn's Theorem. We next proceed to take up a class of rings introduced by Jacobson. Definition 9-l. A ring R with identity is called a J-ring if, for each x E R, there exists an integer n(x) > 1 (depending on x) such that x"(x) =X.

Our immediate goal is to prove that every J-ring is commutative. (In a very natural way, this can be regarded as a generalization of Wedderburn's Theorem). Before establishing the quoted result in full generality, we first settle the question for the special case of division rings; the argument relies heavily on the Wedderburn Theorem.

TWO CLASSIC THEOREMS

197

Theorem 9-12. Let R be a J-ring. If R forms a division ring, then R is commutative (hence, a field).

Proof As a first step, let us show that R is of characteristic p > 0, p a prime. If char R = 2, then there is nothing to prove; thus, it may be assumed that char R =I= 2. Consider any element a =1= 0 in R. By hypothesis, there exist integers h, k > 1 for which ah = a, (2a)k = 2a. Setting q = (h - 1)(k - 1) + 1 > I, it follows that both aq = a and (2a)q = 2a. From this, we obtain (2q - 2)a = 0, with 2q - 2 =1= 0. Therefore, there exists a least positive integer p such that pa = 0, which implies that char R = p, p a prime (Theorem 1-6). Let Z P be the prime subfield of R. Since ah = a, the element a is algebraic over ZP and, hence, the extension ZP(a) constitutes a finite field; say with p" elements. In particular, a itself lies in ZP(a), so that aP" = a. If we now assume that a ¢ cent R, then all the hypothesis of Lemma I will be satisfied; thus, there exists an element bE Rand integer k > 1 satisfying bab- 1 = ak =I= a. Similar reasoning applied to the extension field ZP(b) indicates that bPm = b for some integer m > 1. At this point we turn our attention to the set of finite sums W =

t~l

phi

rljai!Jijrtj

Ezp}·

It should be apparent that W is a finite set which is closed under addition. Since the relation akb = ba allows us to bring the a's and b's together in a product, W is also closed under multiplication. Whatever further it may be, W has at least been shown to be a ring. As a finite subring of a division ring, W is more than just a ring; it is, in fact, a finite division ring (Problem 32). Hence, by Wedderburn's Theorem, we know that W is necessarily commutative. In particular, a and b are both members of W, so that ab = ba, contradicting the relation bab- 1 = ak =I= a. Having arrived at a suitable contradiction, we infer that the choice of a ¢ cent R is impossible and R must be commutative.

The transition of Theorem 9-12 from the division rings case arbitrary rings is accomplished by two lemmas. Lemma 1. Let R be a J-ring. sided ideal of R.

Then every right ideal I of R is a two-

Proof To begin with, we assert that R can possess no nonzero nilpotent elements. Indeed, if x =I= 0, the condition x"1-"l = x necessarily implies that :x!" =1= 0 for all m ::2:: 1. Now, suppose that e is any idempotent element of R; then, for any x E R, (xe - exe) 2 = (ex - exe) 2 = 0,

198

FIRST COURSE IN RINGS AND IDEALS

so that xe - exe = 0 = ex - exe. Therefore, ex = exe = xe, in consequence of which e E cent R. It follows that every idempotent of R must be in the center. Given that a E I, with a" = a (n > 1), it is easy to show that e = a"- 1 is an idempotent element of R: (a"-1)2

= a2n-2

= a"d'-2 = aan-2

= d'-1.

Hence, d'- 1 E cent Rand so, for any r in R,

where r' = a"- 2 ra. Since ar' E I, this shows that raE I also, making I a two-sided ideal of R. Lemma 2. in rad R.

Let R be a J-ring. For all a, bE R, the element ab - ba lies

Proof A standard argument, using Zorn's Lemma, shows that R is endowed with maximal right ideals M, which are two-sided from Lemma 1 (the presence of an identity element in R enters here). By virtue of the fact that R/ M has no nontrivial ideals, the quotient ring R/ M becomes a division ring. Being a homomorphic image of R, R/ M inherits the property that x"(xl = x. Thus, we are thrown back to a situation where Theorem 9-12 can act, and the quotient ring R/M is thereby rendered commutative. In other words, (a

+

M)(b

+

M) = (b

+

M)(a

+

M)

for all a, bin R, or, equivalently, ab - ba EM. As this last relation holds for every maximal ideal of R, it follows that ab - ba E rad R. With these preliminaries established, we now have the constituent pieces to prove Theorem 9-13. {Jacobson). If R is a J-ring, then R is commutative. Proof Suppose that the element x E rad R. As in the proof of Lemma 1,

some power of x is an idempotent; to be quite explicit, if x" = x, then e = x"- 1 turns out to be idempotent. Since rad R forms an ideal of R, the element e will lie in rad R. But, according to the corollary of Theorem 8-2, Ois the only idempotent belonging to rad R; hence, the element e = x"- 1 = 0 and so x = x" = xx"- 1 = 0. The implication of this is that R comprises a semisimple ring. Lemma 2 tells us that ab - ba E rad R = {0} for all a, b in R. The net result is that any two elements of R commute, thereby completing the proof. As an interesting application of Jacobson's Theorem, we cite

PROBLEMS

199

Corollary. Let R be a ring with the property that every nonzero subring of R forms a division ring. Then R is a field. Proof Observe first that the ring R has prime characteristic. Indeed, if R were of characteristic zero, it would contain a proper subfi.eld isomorphic to Q and, hence, a proper subring isomorphic to z. Since the ring Z of integers is not a division ring, we obtain a contradiction. Now, letS be the subring of R generated by any nonzero element a E R. Then S consists of all polynomials in a over the prime subfield of R; that is to say, S = ZP[a], for some prime p. Since the element a- 1 E S, a- 1 must be a polynomial in a, which implies that a is a root of some polynomial with coefficients from ZP. In consequence, S forms a simple algebraic extension (field) of ZP. By Theorem 7-26, we also know that S is a finite field. This being the case, an = a, where n(a) is the number of elements inS. From Jacobson's result, it follows that R is necessarily commutative; hence, a field.

There are a number of other fairly general assumptions which at a glance seem quite far removed from commutativity, but when imposed on a given ring render it commutative. In this connection, we might mention without proof Theorem 9-14. (Herstein). Let R be a ring with the property that, for each x E R, there exists an integer n(x) > 1 dep.ending on x such that _x" 1 to conclude that x""- xlxP"- x.] 23. Establish the following assertions: a) given that an irreducible polynomial f(x) E ZP[x], then f(x)lxP" - x if and only if degf(x)ln; b) if an irreducible polynomialf(x) E ZP[x] has a root in GF(p"), thenf(x) splits completely in GF(p"); c) xP" - x is the product of all the irreducible monic polynomials f(x) E Z P[ x] such that degf(x)ln. 24. If p is an odd prime, prove that the Galois field GF(p") contains an element which is not a square. 25. Let P be a prime ideal of R, a commutative ring with identity. If the quotient ring R/P has only a finite number of elements, verify that R/P is a Galois field. 26. Prove that if F is a finite field and K is a subfield of F, then F forms a simple extension field of K. [Hint: Any generator ofF* will generate F as a vector space over K.] 27. Let F be a finite field with p" elements. Prove that the mapping a P: F -+ F defined by taking up(a) = aP is an automorphism, the so-called Frobenius auromorphism of F; furthermore, = iF"

a;

28. a) Suppose that R is a ring with identity (not necessarily commutative). If R has no nontrivial ideals, establish that R is a division ring. b) Show that iff is a homomorphism from a ring R onto a division nng, then kerf forms a maximal ideal of R. 29. Prove that any finite subring of a division ring is again a division ring. 30. For any element a E R, a division ring, define C(a) by

C(a) = {r E Rlra

= ar}.

a) Show that C(a) is a division subring of R containing cent R. b) If R is finite and there are q elements in cent R, prove that there are q• elements in C(a) for some n e Z+. [Hint: C(a) may be regarded as a vector space over the finite field cent R.] 31. If R is a division ring, show that its dimension as a vector space over cent R cannot equal2. 32. If an integral domain R is finite dimensional as a vector space over its center, prove that R forms a division ring. [Hint: For fixed a =I= 0, the linear mapping T..,x = ax is one-to-one; hence, onto R.] 33. a) Prove that every finite field is a J-ring. b) More generally, establish that a field F is a J-ring if and only ifF is of prime characteristic and is an algebraic extension of its prime subfield.

PROBLEMS

203

34. Show that the assumption of an identity element is unnecessary in proving that J-rings are commutative; in other words, if R is a ring with the property that for every a E R there is an integer n(a) > l for which a• a2, ..• 'a,)(b1, b2, ... 'b,) = (albl, a2b2, ... 'anbn).

Now, let us define I; to be the set of all n-tuples (a 1, a 2 , ••• , a,) E R with the property that ak = 0 for k =I= i. It is easily checked that I; constitutes an ideal of R, which is isomorphic to the ring R; under the assignment a 1 --+ (0, ... , 0, a1, 0, ... , 0)

Furthermore, every element of R has a unique representation in the form (a 1, a 2 ,

••• ,

a,.) = (at> 0, ... , 0)

+

(0, a 2 ,

••• ,

0)

+ ·· · + (0, ... , 0, a,.).

This feature throws us back into the situation described in Chapter 2 (see page 21). If we invoke Theorem 2-4, it follows that the ring R is the direct sum (in the sense of Definition 2-4) of the ideals I;. The point which we wish to make is that the concept of complete direct sum extends our previously defined direct sum; in the finite case, the two notions coincide up to isomorphism of components. The particular ring so obtained is customarily denoted by either L?= 1 EB R; or R 1 EB R 2 EB ··· EB R,. We might also mention in passing that if J is the positive integers, then L EB R; may be viewed as the set of all infinite sequences (a 1 , a 2 , ••• , a,, ... ) such that a1 e R 1 for each i e J. Since the generality of the complete direct sum confronts the imagination with such a hurdle, we shall seldom have occasion to use it. Certain subrings of the complete direct sum are more manageable and more interesting. For instance, the discrete direct sum of the rings R 1 is the subring of L EB R 1 consisting of those functions which are zero for almost all i; here the phrase "for almost all i" is short for "for all i with at most a finite number of

206

FIRST COURSE IN RINGS AND IDEALS

exceptions." It would not be too far removed from traditional connotations to represent the discrete direct sum of the rings R; by Ld EB R;: Ld EB R; = {a E

L EB Rda(i)

=

0 for all but a finite number of i}.

Again, if the index set J is taken to be finite, say J = {1, 2, ... , n}, then the stipulation "for almost all i" is redundant and may be dropped from the Ef) R1 ; in this latter setting, description of

Id

Id Ef) R; =

Rl Ef) R2 Ef) ••• Ef) R,.

Another special subring of the complete direct sum L EB R; which is worthy of consideration is the so-called subdirect sum. Let us proceed to examine this particular concept in some detail. First, observe that for a fixed index i, we may define a function n:;: L Ef) R; -+ R; by the equation n:;(a) = a(i).

One can verify that n:; is a homomorphism of I EB R; onto the ring R;. called the ith component projection. If S is any subring of L EB R;, the restriction n;IS defines a homomorphism of S into R; and, hence, onto a subring n:;(S) of R;. The case of principal interest is that in which 1t;(S) = R; for each index i; in this event, we call S a subdirect sum of the rings R;. Let us record these remarks as a formal definition. Definition 10-2. A subring S of the complete direct sum I EB R; is said to be a subdirect sum of the rings R;. written S = EB R;, if the induced projection n;IS: S -+ R; is an onto mapping for each i. The subdirect sum is nontrivial if none of the mappings n;IS is one-to-one (hence, S is not isomorphic to any R;).

I'

In effect, a subring S ~ I Ef) R; is a subdirect sum of the rings R; if and only if, for each index i, every element of R; appears as the functional value at i of some function in S. Definition 10-2 raises a rather significant question: What necessary and sufficient conditions upon a ring R will enable us to write it (up to isomorphism) as the subdirect sum of more tractable rings R;? Up to this point, everything has been a matter of definition and observation; with the needed preliminaries finally compiled, let us make a start at providing an answer to the above problem. Lemma. A ring R is isomorphic to a subdirect sum of the rings R; if and only if there exists an isomorphism f: R -+ I Ef) R; such that, for each i, n:1 of is a homomorphism of R onto R;.

I•

Proof. Given an isomorphism f of R onto a subdirect sum Ef) R; of the rings R;. the composition n:; a f: R -+ R; defines a homomorphism of R into

207

DIRECT SUMS OF RINGS

R; (n; itself being a homomorphism). Since L" ffi R; is a subdirect sum, of actually carries R onto R;. On the other hand, if there happens to exist an isomorphism f satisfying the indicated conditions, then we certainly have lr;

R ~ f(R) =

L' ffi R;.

It is helpful to translate the foregoing lemma into a condition on the ideals of a given ring; in what follows we describe just such a condition. Theorem 10-1. A ring R is isomorphic to a subdirect sum of rings R;

if and only if R contains a collection of ideals {I;} such that R/I; and n I 1 = {0}.

~

R;

Proof To start, we assume that R ~ :L• ffi R;. Then there exists an isoffi R; such that the "natural" homomorphisms morphism f: R-+ n; of: R-+ R; are all onto mappings. Using the Fundamental Homomorphism Theorem, this implies that R/I; ~ R;, where I; = ker(n; of). Note further that

I•

kerf= {r e Rjf(r) = 0} = {r e Rj(n; o f)(r) = 0 for all

i}

= n I;.

Since f is a one-to-one function, kerf= {0}, from which it follows that n / 1 = {0}.

Going in the other direction, suppose that we are given a set of ideals {I;} of R with R/I; ~ R; and n I; = {0}. Define a functionf: R -+ ffi R; by requiringf(a) to be such that its ith projection n;( (f(a)) = a + I;. (The essential point here is that any element of L ffi R; is completely determined by its projections.) Then R is isomorphic by means offto a subring of the direct sum L ffi R;. To see thatfis one-to-one, for instance, simply observe that

L

kerf= {a E Rj(n1 o f)(a) = 11 for all i} = {a e Rja =

+

I1

=

I 1 for all i}

n I 1 = {0}.

We leave the checking of the remaining details as an exercise. Most applications depend more directly on the following version of Theorem 10-1. Corollary. A ring R is isomorphic to a subdirect sum of the quotient

rings R/I; if and only if R contains a collection of ideals {I;} such that n I; = {0}. Furthermore, the subdirect sum is nontrivial if and only if I 1 =!- {0} for all i.

208

FIRST COURSE IN RINGS AND IDEALS

If a ring R is isomorphic to a subdirect sum L' EB R; of rings R;. it is convenient to speak of EB R; as being a representation of R (as a subdirect sum of the rings R;). The last corollary, although satisfying in the sense that it reduces the problem of finding such representations to that of establishing the existence of certain ideals, is actually a stepping stone to the more fruitful results below. These theorems tell us under what conditions a ring R is isomorphic to a subdirect sum of rings whose structure is well known.

z:•

Theorem 10-2. A ring R is isomorphic to a s4bdirect sum of fields if and only if R is semisimple. Proof A ring R is semisimple if and only if the intersection of all its maximal ideals M; is the zero ideal. By the previous corollary, this latter condition is a necessary and sufficient condition that R be isomorphic to a subdirect sum of the quotient rings R/M;, each of which is a field.

Corollary. For any ring R, R/rad R is isomorphic to a subdirect sum of fields. Going one more step in this direction, we also have Theorem 10-3. A ring R is isomorphic to a subdirect sum of integral domains if and only if R is without prime radical. Corollary. For any ring R, R/Rad R is isomorphic to a subdirect sum of integral domains. Since any integral domain can be imbedded in a field, Theorem 10-3 implies the following: a (commutative) ring R with no nonzero nilpotent elements is isomorphic to a subdirect sum of fields. Example 10-1. The ring Z of integers furnishes a simple illustration of the lack of any kind of uniqueness in the representation of a ring as a subdirect sum. Since Z is semisimple, Theorem 10-1 ensnres that it is isomorphic to a subdirect sum of the rings Z/(p) = Z P' where p is a prime number:

z

~

~· EB

p pnme

zp,

it being understood that the summation runs over all primes. At the same time, Z can be represented as a subdirect sum of the rings Z P,, since the intersection of the ideals (p 2 ) is also the zero ideal:

z ~ I:"

p pnme

EB

zpl·

All the component rings in the first representation are fields, while none is a field in the second. This shows that a given ring may be representable as a subdirect sum of rings having quite different properties.

DIRECT SUMS OF RINGS

209

Example 10-2. For another application of Theorem 10-1, consider the ring map R # of real-valued functions on R #. As we know, each of the ideals Mx

= {femapR#If(x) =

0},

xeR#

is maximal in mapR#. Since nxeRMx = {0}, it follows that mapR# is the subdirect sum of uncountably many copies of the real field-one for each point of R #. (This should come as no surprise, being essentially the definition of map R # .) Simply as an application of the foregoing ideas (for we shall make no subsequent use of the result), let us establish Theorem 10-4. A ring R is isomorphic to a subdirect sum of fields if and only if for each nonzero ideal I of R, there exists an ideal J =1= R such that I + J = R. Proof Let I =I= {0} be an ideal of R, where R is isomorphic to a subdirect sum of fields. Then R contains a collection {MJ of maximal ideals with n M; = {0}. Since I is nonzero, this entails that I $ M; for some value of i; for any such i, we necessarily have I + M, = R.

Conversely, assume that the indicated condition holds. We shall argue that each nonzero element is excluded by some maximal ideal of R, whence rad R = {0}. Pursuing this end, let 0 =I= a e R, so that the principal ideal (a) =I= {0} (there is no loss in supposing also that (a) =I= R). By our hypothesis, (a) + J = R for some proper ideal J of R. Now, Zorn's Lemma implies the existence of an ideal M which is chosen maximal in the set of ideals satisfying (i) J £:; M and (ii) a¢ M. To see that M is actually a maximal ideal of R, consider any ideal K with M c K £:; R. Then, by the maximal nature of M, the element a e K; hence, R = (a) + J £:; (a) + K £:; K, or R = K. The outcome is that the intersection of all the maximal ideals of R is zero. This being so, Theorem 10-2 allows us to conclude that R is isomorphic to a subdirect sum of fields. One direction of Theorem 10-3 can be sharpened considerably, as the next result shows. Theorem 10-5. Let R be a ring containing no nonzero nil ideals. Then R is isomorphic to a subdirect sum of integral domains. Proof For each nonnilpotent element a e R, the set

sll

= {a,a 2 , ••• 'a", ... }

is closed under multiplication and does not contain 0. Thus, there exists a prime ideal Pa of R. with Pan Sa = 0 (corollary on page 164). We assert that R ~ Ef> (R/P11 ), where the summation ranges over all the nonnilpotent elements of R.

I•

210

FIRST COURSE IN RINGS AND IDEALS

Clearly, I = n Pa comprises an ideal of R and is not nil by hypothesis. If I =I= {0}, we can select some nonnilpotent element bE I. But then I ~ Pb,

while b ¢ Pb, an obvious contradiction. This being the case, we must have I = n Pa = {0}. It follows from Theorem 10-1 that R is isomorphic to a subdirect sum of the quotient rings (actually integral domains) R/Pa. Before pressing forward with the main line of investigation, let us look at a special case which will prove useful when, at a later stage, we study Artinian rings. Theorem 10-6. Let I 1 , I 2 , •.. , I" be a finite set of (nontrivial) ideals of the ring R. If I; + Ii = R whenever i =I= j, then R/n I; :::,: L Ee (R!I;). Proof To start, we define a mapping/: R f(x) = (x

+

11, x

+

->

12 ,

I

••• ,

Ee (R/I;) by x

+ /").

The reader can painlessly supply a proof that f is a homomorphism with kerf= n I;. Our problem is to show that, under this homomorphism, any element (x 1 + I 1 , x 2 + I 2 , ••• , x" + I.) of the complete direct sum I Ee (R/I;) appears as the image of some element in R; the stated result then hinges upon an application of the Fundamental Homomorphism Theorem. Fix the indexj for the moment. Using the fact that I; + Ii = R whenever i =I= j, there exist elements a; E 1;, b; Eli with a; + b; = 1. This ensures that the product r 1 = a 1a 2 ... a1 _ 1aj+l ... a,.

En /1. i'fj

Furthermore, since 1 - a 1 Eli, the coset a; + Ii = l + Ii for all i =1= j, whence r1 + 11 = 1 + 11. Now, pick arbitrary elements x; E R (i = 1, 2, ... , n); our contention is that

where x =

I x

r;X;- To see this, observe that we may write x

+

11 =

I

i'fj

(r1

+ I1)(x1 + 11) +

(r1

+

I 1 )(x1

+

+ Ii as 11).

But r; e I1 for i =I= j, while ri + Ii = 1 + Ii, so the displayed equation reduces to x + Ii = xi + Ii U = 1, 2, ... , n). This substantiates the claim that f is actually an onto mapping, leading to the isomorphism Rfn 11 ~

I

ffi (R/11).

Careful scrutiny of the above argument shows that we have proved a subresult of independent interest; namely,

DIRECT SUMS OF RINGS

211

Corollary. Let I 1, I 2 , ... , I. be a finite set of ideals of the ring R with the property that I; + Ii = R whenever i =I= j. Given any n elements X1, X2o ••• , x. E R, there exists some x E R such that x - X; E I; for i = 1, 2, ... , n.

This corollary may be applied to the ring Z of integers and to the principal ideals (m 1), (m 2 ), ••• , (m.), where the integers m1 are relatively prime in pairs. One then obtains an old and famous theorem about congruences which goes by the name of the Chinese Remainder Theorem (the result being known to Chinese mathematicians as early as A.D. 250): Theorem 10-7. (Chinese Remainder Theorem). Let m1 , m2 , • .. , m. be positive integers such that gcd (m;, mi) = 1 for i =I= j. If a 1, a 2 , ... , a. are any n integers, then the system of congruences

admits a simultaneous solution. Furthermore, this solution is unique modulo m = m1 m2 ••• m•. The hypothesis in Theorem 10-6 is conveniently expressed in terms of the following: a finite set of ideals I 1 , I 2 , ••• , I. of a ring R is said to be pairwise comaximal (or pairwise relatively prime, in the older terminology) if I 1 =I= Rand I; + Ii = R fori =I= j; when n = 2, we simply term I 1 and I 2 comaximal. Thus, the condition on the ideals in Theorem 10-6 is that they be pairwise comaximal. Evidently, the definition of pairwise comaximal implies that I 1 =I= Ii for i =I= j, as well as I 1 =I= {0} for all i. If, in the representation of a ring R as a subdirect sum of the rings R 1, the "natural" homomorphism of R onto R 1 happens to be an isomorphism for some i, then the representation is termed trivial; in the contrary case it is nontrivial. (A nontrivial representation does not rule out the possibility that R : : : : R1 by way of some mapping other than the "natural" homomorphism of R onto R;.) A ring R is called subdirectly irreducible if there is no nontrivial representation of R as a subdirect sum. Let us summarize these remarks in a definition. Definition 10-3. A ring R is said to be subdirectly irreducible if, in any representation of Rasa subdirect sum ofthe rings R1, at least one of the associated homomorphisms of R onto R 1 is actually an isomorphism; otherwise, R is subdirectly reducible.

The corollary to Theorem 10-1 may be taken as asserting that R is subdirectly reducible if and only if there exists in R a set of nonzero ideals with zero intersection. An equivalent and often handier formulation is the following: a ring R is subdirectly irreducible if and only if the intersection of all the nonzero ideals of R is different from the zero ideal.

212

FIRST COURSE IN RINGS AND IDEALS

The importance of subdirectly irreducible rings is demonstrated by the following representation theorem due to Birkhoff. (Birkhoff). Every ring R is isomorphic to a subdirect sum of subdirectly irreducible rings.

Theorem 10-8.

Proof For each element a =I= 0 of R, Zorn's Lemma can be used to select an ideal I a which is maximal in the family of all ideals of R contained in R - {a}; this family is evidently nonempty, since the zero ideal belongs to it. Our definitwn of Ia imphes that tf I is any ideal of R with the property that Ia c I, then a E /. We should also point out that the intersection of the ideals Ia (where a runs over all nonzero elements of R) is zero. Indeed, if bE natO Ia with b =I= 0, then b must, in particular, lie in the ideal Ib; this contradicts the fact that Ib was originally chosen so as to exclude the element b; hence, natO Ia = {0}. It now follows from the corollary to Theorem 10-1 that R is isomorphic to a subdirect sum of the quotient rings R/Ia. The proof is completed upon showing that each ring R/Ia is itself subdirectly irreducible or, more to the point, that the intersection of all the nonzero ideals of R/Ia is nonzero. By the Correspondence Theorem, it suffices to establish that the intersection of all the ideals of R properly containing I a again contains I a as a proper subset. In light of the maximality of I a• the element a must belong to all such ideals; therefore, their intersection contains a and, hence, contains I a properly. The implication is that the coset a + Ia is nonzero and lies in every nonzero ideal of Rfla. Thus, our goal is achieved.

Before announcing the next result, let us introduce some convenient notation. Definition 10-4. For any ring R, the heart of R is the ideal

R" = n {IJI is a nonzero ideal of R}. We observe that R v is a minimal ideal of R which is contained in each nonzero ideal of R; for this reason, R v is frequently called the minimal ideal of R. When R v =I= {0}, it is not hard to see that R v constitutes a principal ideal with any of its nonzero elements serving as a generator. The relation of this notion to the concept of a subdirect sum should be fairly obvious: a ring R is subdirectly irreducible if and only if R" =1= {0}. A definition deserves a theorem, so we oblige with the following: Theorem 10-9. (McCoy). If R is a ring for which R" 1) ann R v is a maximal ideal of R;

=1=

2) ann R v consists of all zero divisors of R, plus zero; 3) whenever R is without prime radical, R forms a field.

{0}, then

DIRECT SUMS OF RINGS

213

Proof First, suppose that the element r ¢ ann R v. Then ra =!= 0 for some choice of a in R v. Since ra lies in R v. it will serve as a generator for R v ; that is, R v = (ra). Thus, we can find an element s E R satisfying a = ras, whence the product (1 - rs}a = 0. It follows that 1 - rs E ann R v, in consequence ofwhich ann Rv is a maximal ideal of R (Problem 2, Chapter 5).

Regarding the second assertion of the theorem, choose r to be any zero divisor of R. Then ann (r) =!= {0} and, since R v is contained in every nonzero ideal of R, ann (r) ~ R v. This last inclusion simply asserts that r E ann (ann (r)) s ann R v, so that ann R v consists of all zero divisors, together with 0. We now pass to a proof of (3). According to the hypothesis, the ideal (R v) 2 =!= {0} (by Problem 14, Chapter 8, {0} is the only nilpotent ideal of R). Thus, there exists some element r E R v for which rR v =1=- {0}. The implication of this fact is that R v $ ann R v. Inasmuch as R v is minimal in the set of nonzero ideals of R, we conclude at once that ann R v = {0}. The rest follows from (1): {0} is a maximal ideal of Rand so R forms a field. As a special case of part (3) above, we might point out that any subdirectly irreducible Boolean ring must be a field, which is clearly isomorphic to Z 2 (Theorem 9-2). There is a corollary to Theorem 10-9 that will be useful later on. Corollary. If R v

=1=-

{0}, then the annihilator of the set of zero divisors

of R is precisely R v. Proof With reference to the theorem, it is enough to prove that ann (ann R v) = R v. Since one always has R v s ann (ann R v ), let us concentrate on the reverse inclusion. If a is any nonzero element of ann (ann R "), then R v s (a) and, hence, 0 =1=- ar E R v for some choice of r ¢= ann R v (in other words, r is not a zero divisor of R). As in the proof of Theorem 10-9, we can find an element s E R for which l - rs E ann R v. This means that a(l - rs) = 0 and so a = (ar)s E R v. It follows that ann (ann R v) s R v, which completes the argument.

There are a number of situations where the hypothesis of Theorem 10-9 occurs quite naturally. By way of example, the hypothesis is certainly

fulfilled in any field. A more interesting illustration is provided by the ring R = Zp" of integers modulo a power of a prime; in this setting, one has = (pn-l) and annRv = (p). Although no further attempt is made to discuss the subject of subdirect sums systematically, we shall continue to throw sidelong glances in this direction (for a more thorough treatment, the reader is invited to consult [ 49]). Some of these ideas will be put to work in the next section when rings with chain conditions are discussed.

Rv

MAXIMAL, PRIME, AND PRIMARY IDEALS

77

ring with identity, so is the quotient ring R/1. It remains therefore only to verify that Rjl is free ofzero divisors. For this, assume that (a

+

I)(b

+

I) = I.

In other words, the product of these two cosets is the zero element of the ring Rjl. The foregoing equation is plainly equivalent to requiring that ab + I = I, or what amounts to the same thing, abE I. Since I is assumed to be a prime ideal, one of the factors a orb must be in I. But this means that either the coset a + I = I or else b + I = I; hence, R/I is without zero divisors. To prove the converse, we simply reverse the argument. Accordingly, suppose that R/1 is an integral domain and the product abE I. In terms of cosets, this means that (a

+

I)(b

+ /)

= ab

+

I = I.

By hypothesis R/I contains no divisors of zero, so that a + I = I or b + I = I. In any event, one of a orb belongs to I, forcmg I to be a prime ideal of R. There is an important class of ideals which are always prime, namely, the maximal ideals. From the several ways of proving this result, we choose the argument given below; another approach involves the use of Theorems 5-5 and 5-6. Theorem 5-7. In a commutative ring with identity, every maximal ideal is a prime ideal.

Proof Assume that I is a maximal ideal of the ring R, a commutative ring with identity, and the product abE I with a¢ I. We propose to show that b E I. The maximality of I implies that the ideal generated by I and a must be the whole ring: R = (I, a). Hence, there exist elements i E I, r E R such that I = i + ra. Since both ab and i belong to I, we conclude that b

=

lb

= (i + ra)b = ib + r(ab) E I,

from which it follows that I is a prime ideal of R. We should point out that without the assumption of an identity element this last result does not remain valid; a specific illustration is the ring Ze of even integers, where (4) forms a maximal ideal which is not prime. More generally, one can prove the following: if R is a commutative ring without identity, but having a single generator, then R contains a nonprime maximal ideal. To establish this, suppose that R = (a). First observe that the principal ideal (a 2 ) is a proper ideal of R, since the generator a ¢ (a 2 ). Indeed, were a in (a 2 ), we could write a = ra 2 + na 2 for some r E R and n E z; it is a simple matter to check that the element e = ra + na would

215

PROBLEMS

8. Prove that an irredundant subdirect sum of a finite number of simple rings is their direct sum.

9. If R is a (commutative) regular ring, verify that R is isomorphic to a subdirect sum of fields. 10. a) Prove that a ring R is isomorphic to the complete direct sum of a finite number of fields if and only if (i) R contains only a finite number of ideals and (1i) rad R = {0}. b) Prove that a finite ring R is a direct sum of fields if and only if it has no nonzero nilpotent elements. 11. Demonstrate that the conclusion of Theorem 10-6 is false if an infinite number of 1deals I 1 arc allowed. [Hint: Consider the ring Z and the ideals I 1 = (pJ, where p1 is the ith prime.]

12. a) Let a 1, a2 , .•• , an be a finite set of nonzero elements of the principal ideal domain R such that a 1 and ai arc relatively prime fori =fo j. If a = lcm (a 1, a 2 , •.. , an), show that R/(a) ~ I Ef) (R/(a;)). b) Prove that if the integer n > l has the prime factorization n = pt'f~ 2 then ~I EB ZP.''

•••

z.

13. Let I 1 , I 2 , ••• , I. be a finite set of ideals of the ring R. Prove that a) the ideals I, are pairwise comaximal if and only if their nil radicals pairwise comaximal ; b) if the ideals I 1 are pairwise comaxin1al, then their product

[Hint: Use induction on n. Notice that In is comaximal with I 1 since

R = Rn = II(/.

+

I1)

JT;

p:',

are

11 ··· 11 I.-~>

s In + (n I 1) s R

for 1 : n, the leading coefficient be I, = I,. and one may write

for suitable choice of c1 e R. Then the polynomial

belongs to I and has degree less than r; indeed, the coefficient of x' in this polynomial is mn

b -

L c;a,. = i=l 1

0.

(Notice particularly thatf1 (x) differs fromf(x) by an element of J.) At this point, the inductive assumption can be applied to f 1 (x) to conclude that / 1 (x) and, in turn, f(x) lie in the ideal J. If r s n, a similar line of reasoning can be employed. Indeed, since be I., we can always find elements d 1 , d2 , ••. , dm, in R such that the polynomial f 2 (x)

=

f(x) - (dtf. 1 (x)

+ d2 /. 2 (x) + ··· + d,,.J,.,.,.(x))

is an element of I with degree r - 1 or less. In either case, our argument leads to the inclusion I s; J and the subsequent equality I = J.

By induction, Hilbert's Theorem can be extended to polynomials in several indeterminants.

222

FIRST COURSE IN RINGS AND IDEALS

Corollary. If R is a Noetherian ring, then so is the polynomial ring xnJ in a finite number ofindeterminants X1, X2, xn. R[x1, X2, 000'

000

'

We recall that an ideal I is nilpotent provided that there exists an integer n for which In = {0}, whereas I is said to be a nil ideal if every element of I is nilpotent. It is not hard to see that any nilpotent ideal is a nil ideal. Levitsky proved that for Noetherian rings the converse also holds: nil ideals are nilpotent. This fact is brought out as a corollary to our next theorem. Theorem 11-4. (Levitsky). In a Noetherian ring R, the prime rad1cal Rad R is the largest nilpotent ideal of R. Proof At the outset, observe that since R is Noetherian, we can use the maximum condition to select an ideal N of R which is maximal with respect to being nilpotent. Our contention is that N is the largest nilpotent ideal of R (in the sense of containing all other nilpotent ideals). To set this in evidence, let N 1 be an arbitrary nilpotent ideal of R, say N~ = {0}; assume further that Ni = {0}. Then (N + N di+k = {0}, so that the ideal N + N 1 is nilpotent. From the inclusion N s; N + N 1 and the maximal property of N, it follows that N = N + N 1. One is then left with N 1 s; N, which settles the point. Now every nilpotent ideal must also be nil and thus N s; Rad R by the corollary to Theorem 8-8. To derive the reverse inclusion, assume that a + N is any nilpotent element of the quotient ring R/N. Then an + N = (a + N)" = N for some n E Z +• implying that a" EN. Because N is a nil ideal, there exists a positive integer m for which (a"t = 0, and so a is nilpotent as an element of R. This being the case, we conclude that the principal ideal (a) is nilpotent; hence, (a) s; N, by the maximality of N. The rest should be clear: since a E (a) s; N, the coset a+ N = N. Our reasoning shows that the quotient ring R/N contains no nonzero nilpotent elements, which is to say that R/N has zero prime radical. But it is already known that Rad R is the smallest ideal !Jf R possessing a quotient ring without prime radical (Theorem 8-12). Therefore, Rad R s; N, which yields the desired equality N = Rad R; the theorem is now established.

As corollaries we have Corollary 1. In a Noetherian ring, any nil ideal is nilpotent. Proof The proof amounts to the observation that any nil ideal is contained in the prime radical of a ring.

Corollary 2. A semisimple Noetherian ring contains no nonzero nilpotent ideals. The breakdown of Levitsky's Theorem is rather dramatic when one replaces the prime radical by the Jacobson radical. The following example

RINGS WITH CHAIN CONDITIONS

223

will serve as a simple illustration: in a power series ring F[[ x ]] over a field

F, rad F[[x]] = (x), but (x)" = (x")

=1=

{0}

for all n E Z + (Problem 1, Chapter 7). Let us now broaden the outlook by considering rings with the descending chain condition. Definition 11-3. A ring R is said to satisfy the descending chain condition

for ideals if, given any descending chain of ideals of R,

there exists an integer n such that/" = Jn+t = Jn+ 2 = As in Theorem 11-1, this definition leads to The following statements concerning the ideals of a ring R are equivalent: 1) R satisfies the descending chain condition for ideals. 2) Every nonempty set of ideals of R, partially ordered by inclusion, contains a minimal element (the minimum condition holds). A ring satisfying either of these conditions is said to be Artinian (after Emil Artin). Theorem 11-5.

It would be repetitious to prove this modified version of Theorem 11-1 and we shall refrain from doing so. However, lest some reader try to obtain the exact analog for Art in ian rings of Theorem 11-1, we hasten to point out that every ideal in the ring Z of integers is finitely generated, but Z is not Artinian. Indeed, if(n) is any nonzero ideal of Z, then (2n) is a nonzero ideal properly contained in (n); thus, the set of all nonzero ideals of Z has no minimal element. In the light of the equivalence of the ascending (descending) chain condition with the maximum (minimum) condition, the two will be used interchangeably. Certain results are more easily proved in terms of one than the other, and convenience will be our guide. Example 11-5. The statement of the Hilbert Basis Theorem is no longer true if Artinian is subtituted for Noetherian. For example, ifF is any field,

then

is a strictly descending chain of principal ideals of F[ x]. descending chain condition fails to be satisfied in F[ x].

Thus, the

224

FIRST COURSE IN RINGS AND IDEALS

Example 11-6. Consider the ring R = mapR# of real-valued functions on R#. Given an arbitrary real number r > 0, we define I,

= {IE Rlf(x)

=

0 for -r ~ x ~ r}.

Then I, is an ideal of Rand it is not difficult to see that ... c I 3 c I 2 c I 1 c

I 112 c I 113 c

···.

The implication is that R contains ascending and descending chains that do not become stationary, whence R is neither Artinian nor Noetherian. It is perhaps appropriate to call attention to the fact that each of the ideals I, is properly contained in the maximal ideal M = {IE Rl/(0) = 0}. Example 11-7. We next give an example of an Artinian ring which is not Noetherian. For this purpose, let p be a fixed prime. Consider the group

Z(p"') of all rational numbers r between 0 and 1 of the form r = m/p", where m is an arbitrary integer and n runs through the nonnegative integers, under the operation of addition modulo 1 : Z(p"') = {m/p"IO ~ m < p"; mE Z; n

=

0, 1, 2, ... }.

We make Z(p 00 ) into a ring (without identity) by defining the product ab to be zero for all a, b E Z(p"'). It is important to observe that the ideals of the resulting ring are simply the subgroups of the additive group of Z(p""). Now, let I be any nontrivial ideal of Z(p 00 ) and choose k to be the smallest positive integer such that for some a, a/pk ¢ I; we implicitly assume that a and p are relatively prime. Then I must contain all the elements 0, 1/pk- 1 , 2/pk- I, ... , (pk- 1 - 1)/pk- 1 • Our contention is that these are the only members of I. To support this, suppose to the contrary that b/pi E I, where i ~ k and, of course, band p are relatively prime. One can then find integers r, s for which rb + sp = 1. Since both the rational numbers (reduced modulo 1) and lie in I, it follows that (rb + sp)fp" = 1/pk also belongs to I, contradicting the minimality of k. Thus, the ideal I is finite and is given by

Representing I by the symbol Ik_ 1 , we conclude that the only ideals of Z(p 00 ) are those which appear in the chain

{0} c I 1 c I 2 c ··· c I" c ... c Z(p«l). Therefore, Z(p 00 ) possesses an infinite (strictly) ascending chain of ideals, but any descending chain is of finite length.

RINGS WITH CHAIN CONDITIONS

225

In the sequel, there occur certain results which hold for both Noetherian and Artinian rings. Where the proofs are virtually the same, our policy will be to establish the theorem in question only in the Noetherian case. Let us first show that the chain conditions are not destroyed by homomorphisms. Theorem 11-6. If R is a Noetherian (Artinian) ring, then any homomorphic image of R is also Noetherian (Artinian). Proof Let f be a homomorphism of the Noetherian ring R onto the ring R' and consider any ascending chain 1'1 ~ I~ ~ · · · ~ I~ ~ · · · of ideals of R'. Put Ik = f- 1 (I~). fork = 1, 2, ... . Then I 1 ~ I 2 ~ •·· ~ I. ~ ··· forms an ascending chain of ideals of R which, according to our hypothesis, must eventually be constant; that is, there is some index n such that I m = I" for all m ~ n. Taking stock of the fact that f is an onto mapping, we have f(I 1..) = I~. Hence, I',. = I~, whenever m ~ n, so that the original chain

also stabilizes at some point. Lettingfbe the natural mapping, we have as a corollary: Corollary. If I is an ideal of the Noetherian (Artinian) ring R, then the quotient ring R/I is Noetherian (Artinian). Further progress will be facilitated by the technical lemma below. Lemma. If I, J, and K are ideals of a ring R such that (1) J s;;; K, (2) J n I = K then J = K

11

and

I,

(3) J/I = K/I,

Proof Evidently, we need only establish the inclusion K s;;; J. To this purpose, select any member k of K. On the basis of (3), there exists an element j e J for which j + I = k + I, which signifies that k - j = i for some choice of i in I. But, since J ~ K, the difference k - j also lies in the

ideal K. Using condition (2), we thus find that i

and, in consequence, k

=k - jEI = i + j e J.

II

K

=

J II I,

This fact is enough to enable us to prove a partial converse of the last corollary. Theorem 11-7. Let I be an ideal of the ring R. If I and R/I are both Noetherian (Artinian) rings, then R is also Noetherian (Artinian). Proof To begin, let J 1 ~ J 2 s;;; ••• ~ J" ~ · · · be any ascending chain of ideals of R. From this, we may construct a chain of ideals of I,

11

II

I

£;

12

II

I s;;; •••

£;

Jn

II

I s;;;

MAXIMAL, PRIME, AND PRIMARY IDEALS

81

Corollary. If P is a prime ideal of the ring R, then Pis semiprime. Proof Because Pis prime, R/P possesses no zero divisors and, in particular, no nonzero nilpotent elements.

This corollary provides another good reason why a semiprime ideal was termed as it was; being a semi prime ideal in a ring is a bit weaker than being prime. There is much more that could be said about semiprime ideals, and more will be said later in the text, but let us now turn our attention to primary ideals. In Chapter 11 we shall show that the ideals of a rather wide class of rings (to be quite explicit, the Noetherian rings) obey factorization laws which are roughly similar to the prime factorization laws for the positive integers. It will turn out that the primary ideals, which we are about to introduce, play a role analogous to the powers of prime numbers in ordinary arithmetic. Definition 5-5. An ideal/ of the ring R is called primary if the conditions ab E I and a ¢= I together imply b" E I for some positive integer n. Clearly, any prime ideal satisfies this definition with n = 1, and thus, the concept of a primary ideal may be viewed as a natural generalization of that of a prime ideal. Lest the reader jump to false conclusions, we hasten to point out that a primary ideal is not necessarily a power of a prime ideal (see Example 8, Chapter 7). Notice too that Definition 5-5 may be stated in another way: an ideal I is primary if ab E I and a ¢= I imply b E Jf; this formulation in terms of the nil radical is frequently useful. In the ring Z, the primary ideals are precisely the ideals (p"), where p is a prime number and n 2 1, together with the two trivial ideals. Our first theorem on primary ideals is simple enough; it shows that to every primary ideal there corresponds a specific prime ideaL Theorem 5-12. If Q is a primary ideal of the ring R, then its nil radical

.JQ is a prime ideal, known as the associated prime ideal of Q.

Proof Suppose that abE .JQ, with a¢= .JQ. Then (ab)n = anb:_E Q for some positive integer n. But d' ¢= Q, for otherwise a would lie in .JQ. Since Q is assumed to be primary, we must therefore have (bnr E Q for suitable choice of mE Z +, and so b E .JQ. This is simply the statement that .JQ is a prime ideal of R. It may very well happen that different primary ideals will have the same associated prime ideal. This is demonstrated rather strikingly in the ring of integers where, for any n E Z +• (p) is the prime ideal associated with each of the primary ideals (p"). _ It might also be of interest to mention that the nil radical .JQ is the smallest prime ideal to contain a given primary ideal Q. For, suppose that

RINGS WITH CHAIN CONDITIONS

227

One can say considerably more about the ideal structure of an Artinian ring R: R has only a finite number of prime (hence, maximal) ideals. For, suppose that there exists an infinite sequence {P;} of distinct proper prime ideals of R. We would then be able to form a descending chain of ideals pl 2 plp2 2 plp2p3 2 .... Since R is Artinian, there exists a positive integer n for which

P1P2 ··· P,. = p1p2 ·•· P,.P,.+t· It follows from this that P 1 P 2 ••• P,. £ P,.+ 1 , whence Pk £ P,.+ 1 for some k ~ n. But Pk is a maximal ideal of R, so that we must have Pk = Pn+l• contrary to the fact that the P; are distinct. These observations are summarized as

Theorem 11-10. Every Artinian ring has only a finite number of proper prime ideals, each of which is maximal. We now come to the interesting part of the theory; namely, the extension of Levitsky's Theorem to Artinian rings. Theorem 11-11. If R is an Artinian ring, then rad R forms a nilpotent ideal.

Proof The descending chain condition applied to the chain rad R 2 (rad R) 2 2 (rad R) 3 2 · · · shows that there exists an integer n for which (rad R)" = (rad R)"+ 1 If we put I = (rad R)", then I £ rad Rand 12 = I. Our contention is that I= {0}. Assume for the moment that I =f. {0} and consider the family of all ideals J of R such that (i) J £ I and (ii) Jl =f. {0}. This collection is not empty since it contains I and, hence, it admits a minimal member K. By (ii), KI =f. {0}, so that al =f. {0} for some nonzero element a e K. Thus, (al)l

= al 2

= al =I=

{0},

with al £ K £ I; hence, al = K by the minimality of K. This being the case, there exists an element b e I such that ab = a. But b e I £ rad R, which implies that l - b must be an invertible element of R (Theorem 8-2); in other words, (1 - b)c = 1 for suitable c e R. We then have a = a(1 - b)c = (a - ab)c = 0,

contradicting the fact that al =f. {0}. I = (rad R)" = {0}, as asserted.

This contradiction signifies that

With little additional effort we can learn a good deal more about nilpotent ideals in rings with the descending chain condition.

228

FIRST COURSE IN RINGS AND IDEALS

Corollary. In any Artinian ring R, the following hold: 1) rad R is the largest nilpotent ideal of R. 2) Every nil ideal of R is nilpotent. 3) R/rad R contains no nonzero nilpotent ideals. Proof By Theorem 8-8, any nilpotent ideal of R is contained in the prime radical and this comcides with rad R. Concerning (2), each nil ideal is contained in Rad R = rad R, which is a nilpotent ideal. The final assertion follows from the fact that R/rad R is a semisimple Artinian ring.

The next theorem is perhaps of secondary interest, but it affords us an opportunity to discuss subdirectly irreducible rings again. The reader will recall that these are rings R possessing a smallest nonzero ideal R v (the heart of R). Clearly, R v is a principal ideal generated by any of its elements, other than zero. We shall, in the proof below, let r v designate a fixed nonzero element of R v, so that Rv = (rv). Observe also that for any nonzero element a E R, (a) is a nonzero ideal of R, and, hence, must contain R v ; thus, there exists an element x in R such that ax = rv. The only other fact which we will require is that the annihilator of the set of zero divisors of R is precisely the ideal R v = (r v ). Theorem ll-12. If R is a subdirectly irreducible ring satisfying either chain condition, then every zero divisor of R is nilpotent (that is, R is a primary ring). Proof In the first place, we take R to be Artinian. Suppose further that a E R is a zero divisor which is not nilpotent and consider the descending chain of principal ideals

By assumption, none of these is the zero ideal and, because of the minimum condition, we must have (a") = (a"+ 1) for some n E z+. This being the case, a" = ra"+ 1 or a"(l - ra) = 0, with r E R. Inasmuch as a" =1= 0, the expression in parentheses is a zero divisor of R and, hence, lies in ann R v by Theorem 10-9. Thus,foranynonzeroelementx E R v, wehavex(l - ra) = 0. But xa = 0, since a also belongs to ann R v, and so x = 0. This contradiction forces the element a to be nilpotent, as desired. We next extend the stated result to rings with the ascending chain condition. As in the previous paragraph, suppose that the element a is a zero divisor of R which is mot nilpotent. Then all the powers a 2 , a 3 , ••• , a", ... are zero divisors and, of course, none is zero. Thus, for every power a", there exists an element x, such that ax 1

= a 2 x 2 = ··· = a"x, = ... =

rv

=f

0.

RINGS WITH CHAIN CONDITIONS

Consider the ideals I k

=

229

ann (ak); clearly, we have

Now, xn is not in In, but a(anxn) = arv = 0, since a is a zero divisor of R. Therefore, xn lies in In+ 1 and the I k form a properly ascending chain. This contradicts the ascending chain condition and no such element a exists. Remark. Over the course of the next several pages, we shall often simply say "the set of zero divisors of R form an ideal" when what is really meant is "the set of zero divisors, together with zero, form an ideal". Corollary. If R is a subdirectly irreducible ring satisfying either chain condition, then the set of zero divisors of R form a nil ideal. Proof. Suppose that a and bare both zero divisors of R. Then ax = 0 = by for some nonzero x, y in R. Inasmuch as the principal ideals (x) and (y) have nonzero intersection, there also exist elements u, v E R such that xu = yv f 0. But then, (a - b)xu

= -bxu = -byv =

0,

in consequence of which a - b is a divisor of zero. Certainly, the product ra will be a zero divisor for any choice of r E R. The implication is that the set of all zero divisors of R constitute an ideal (indeed, this is true in any subdirectly irreducible ring); by the theorem, such an ideal must be nil. In the light of the corollary above, it would appear natural to study rings whose zero divisors form an ideal which is contained in the Jacobson radical (we point out that this condition holds trivially in any integral domain). Our next two results present criteria for these rings to become local rings. Theorem ll-13. Let I be an ideal of the ring R with I R is a local ring if and only if R/I is a local ring.

~

rad R. Then

Proof. One direction is fairly obvious, since the homomorphic image of a local ring is necessarily local. Going the other way, suppose that R/I is local and let a + I be any invertible element of R/I. Then ax + I = 1 + I for some x in R, or, equivalently, ax = 1 + r with rEI. Since the ideal I ~ rad R, Theorem 8-2 tells us that ax is an invertible element of R. But then a itself will possess an inverse in R. Now, let a and b be two non-invertible elements of R. The reasoning of the previous paragraph shows that the cosets a + I and b + I lack inverses in Rji. Since the quotient ring R/I constitutes a local ring, their sum (a + I) + (b + I) = a + b + I is again a non-invertible element (Problem 8, Chapter 8). This means that a + b fails to have an inverse in R, forcing R to be a local ring.

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FIRST COURSE IN RINGS AND IDEALS

Corollary. Let R be a ring in which the set of zero divisors of R forms an ideal D, with D ~ rad R.. Then R is a local ring if and only if R/ D is a local ring. Theorem 11-14. Let R be a ring which satisfies the following conditions: 1) rad R is a nonzero prime ideal of R; 2) all the ideals containing rad R are principal; 3) the set of zero divisors D Then R forms a local ring.

~

rad R.

Proof With reference to Problem 8, Chapter 8, it suffices to show that rad R coincides with the set of all noninvertible elements of R. For this purpose, let us suppose that rad R = (x) and choose an arbitrary a ¢ rad R; the strategy is to show that a has an inverse. Now, our hypothesis signifies that the ideal (rad R, a) must be principal; say (rad R, a) = (b), where b ¢ rad R. Thus, x = by for some choice of y E R. Since rad R is a prime ideal, a further deduction is that y E rad R. Knowing this, we can write y = ex with e E R. But then x = by = bex, or x(l - be) = 0, which implies that 1 - be lies in D ~ rad R. Falling back on Theorem 8-2, the product be and, in turn, the element b are necessarily invertible in R. Accordingly, the ideal (rad R, a) = R. It follows that l - ra E rad R for suitable r E R, making ra an invertible element. From this we may conclude that a itself possesses an inverse in R, as desired. Corollary. Let R be a principal ideal ring with D ~ rad R. is a local ring whenever rad R is a nonzero prime ideal.

Then R

Although the hypotheses of Theorem 11-14 appear somewhat formidable, it is worth remarking that the power series ring F[[x]] (F a field) satisfies the requisite conditions. Our next goal is to describe semisimple Artinian rings; crucial to the discussion is the fact that such rings have only a finite number of maximal ideals with zero intersection. The theorem below is the commutative version of Wedderburn's fundamental result (Theorem 13-3). Theorem 11-15. (Wedderburn). Any semisimple Artinian ring R is the direct sum of a finite number of fields.

Proof Since R has only a finite number of maximal ideals, we may assume that if any one of these ideals is omitted the intersection of the others is different from zero. (If this is not the case, a set with the desired property may be obtained by simply deleting certain ideals.) Accordingly, there exist maximal ideals M 1, M 2 , .•• , M. of R such that n M; = {0}, but the ideals I;= nkriMk =/= {O}foreveryi. lnasmuchasM;ismaximal,R =I;+ M;; moreover, I; n M; = {0}, so that this sum is actually direct. Hence, the

MAXIMAL, PRIME, AND PRIMARY IDEALS

83

which is to say that a + Q is nilpotent. As every zero divisor of the quotient ring R/Q is nilpotent, an appeal to Theorem 5-13 is in order and we may conclude that Q is a primary ideal of R. There is another, frequently useful, criterion for deciding whether a given ideal is actually primary. Theorem 5-14. Let P and Q be ideals of the ring R such that 1) Q £ p £ JQ, 2) if ab e Q with a ¢ P, then b e Q. Under these conditions, Q is a primary ideal of R with P

= JQ..

Proof To see that Q is a primary ideal, suppose that the product ab e Q but b ¢ Q. Using (2), we may conclude that a E P £ JQ, whence a" E Q for some positive integer n; this shows that Q is primary. In order to prove that P = JQ., we need only establish the inclusion JQ £ P, since equality would then follow from (1). For this, let the element b e JQ., so that there exists some n E Z + for which b" e Q; assume that n' is the smallest positive integer with this property. If n' = 1, we would have bE Q £ P, from condition (1). If n' > 1, it follows that b"" = b"'- 1 b E Q, with b"'- 1 ¢ Q; heDfe, b E P by (2). In any event, we have shown that be JQ implies b e P, as required. A relationship between maximal ideals and primary ideals is brought out in the following corollary to Theorem 5-14. Corollary. If M is a maximal ideal of the ring R, then all its powers M" (n ~ 1) are primary ideals. Proof Since M" £ M = J M", we need only verify condition (2) of the foregoing theorem. Suppose, then, that ab E M" with a¢ M. Because M is maximal, the ideal (M, a), generated by Manda must be the whole ring R. Hence, the identity element 1 E (M, a), so, for some mE M and r e R, we must have 1 = m + ra. Now, m" lies in M". Raising the equation 1 = m + ra to the nth power and using the binomial theorem, it follows that 1 = m" + r' a, where r' e R. But then

b = bm"

+ r'(ab)

is an element of M", and M" is primary. Another result which has this same general flavor, but which we leave as an exercise, is the following: If the nil radical JT of an ideal I is a maximal ideal, then I itself is primary. Before closing this chapter, we present two additional theorems regarding prime ideals. The first of these involves the notion of a minimal prime ideal of an ideal.

232

FIRST COURSE IN RINGS AND IDEALS

Proof In conjunction with Theorem 11-15, one needs only the fact that a finite direct sum of Noetherian rings (in this case, fields) is again Noetherian. We shall see later that the imposed semisimplicity condition is unnecessarily stringent; indeed, the foregoing result can be sharpened to read that every commutative Artinian ring with identity is Noetherian. The ring of integers shows that the converse need not hold.

PROBLEMS In all problems, R is a commutative ring with identity. 1. Let I be a nonzero ideal of the principal ideal domain R. Prove that the quotient ring R/I satisfies both chain conditions.

2. Prove that every homomorphism f of a Noetherian ring R onto itself 1s necessanly one-to-one. [Hint: Consider the ascending chain {0} c::; kerf c::; ker j2 c::; • · • of ideals of R.] 3. a) If I is an ideal of the Noethenan ring R, show that .jl/1 forms a nilpotent ideal of R/I. b) Let I and J be two ideals of the Noetherian nng R. Establish that I" c::; J for some integer n e Z+ if and only if c::; .JJ.

Jl

4. Prove that every ideal of a Noetherian ring R contains a product of prime ideals. [Hint: If not, letS be the set of those ideals of R which do not con tam a product of prime ideals and apply the maximum condition.] 5. a) Obtain the converse of the Hilbert Basis Theorem: if R[ x] IS a Noetherian ring, then so also is R. b) Verify that the power series ring R[[x]J is Noetherian if and only if R is a Noetherian ring. [Hint: Min1ic Hilbert's Theorem, now using elements of lowest order when defining the ideals It.] c) Let R' be an extension ring of the Noetherian ring R. For a fixed element r e R', show that the ring

R[r]

=

{f(r)lf(x) e R[x]}

is Noetherian.

6. Prove that if R is a Noetherian ring, then the matrix ring M.(R) is also Noetherian. [Hint.· Problem 28, Chapter 2.]

7. Assuming that R is a Noetherian ring, establish that a) Rad R is the sum of all the nilpotent ideals of R ; b) the quotient ring R/Rad R has no nonzero nilpotent ideals. 8. Let R be a ring with at least one non-zero-divisor. Prove that if R is Noetherian thenitsclassicalringofquouentsQc 1(R)isalsoNoetherian. [Hint:lfJ 1 c::; J 2 c::; •• : is an ascending chain of ideals of Qc~(R), then, by Problem 29, Chapter 4, I 1 c::; I 2 c::; ••• forms an ascending chain of ideals of R, where I k = J t n R.]

PROBLEMS

233

9. Let p be a fixed prime number and put

QP

= {mfp"lm E Z;

n

= 0, 1, 2, ... }.

In QP, define addition to be ordinary addition of rational numbers and multiplication to be the trivial multiplication (i.e. ab = 0 for all a, b E Qv). Establish that a) Z forms an ideal of the resulting ring Qv; b) the quotient ring Qv/Z is isomorphic to Z(p"'); 10. a) Prove that a finite direct sum 2; ffi Ri is Noetherian (Artinian) if and only if each of the component rings Ri is Noetherian (Artinian). [Hint: If n = 2, say R = R 1 ffi R 2 , then R/R 1 ~ R 2 .] b) Let R be a nnghavingafinitenumberofideals I 1> I 2 , ••• , I. such that n Ii = {0}. If each of the quotient rings R/Ii is Noetherian (Artinian), show that R is also Noetherian (Artinian). 11. In an Artinian ring R, prove that the zero ideal is a product of maximal ideals. [Hint: rad R = M 1 n M 2 n ··· n M., where each Mi is maximal; use Theorem 11-11 and Problem 13, Chapter 10, to conclude {OJ = (radR)k = MrM~ ... M! for some integer k.]

12. Establish that an Artinian ring R is tsomorphic to a finite direct sum of Artinian local rings. [Hint: From Problem 11 and the factthat the ideals M:arecomaximal, we have Mf n ··· n M! = Mr ··· M! = {0}. By Theorem 10-1, R ~ 2: ffi RIM~. Now use Problem 5-18.]

13. Prove that any semisimple Artinian rmg possesses only a finite number of ideals. [Hint: Assume that R admits that decomposition R = F1 ffi F2 ffi ··· ffi F., Fi a field; if I is an ideal of R, then I = I 1 ffi ··· ffi I. with I; an ideal ofF,.] 14. a) Let I be a nontrivial minimal ideal of the Artinian ring R. Show that the annihilator ann I forms a prin1e and, hence, maximal, ideal of R. [Hint: If a ri ann I, ai = I.] b) Assume that I is a nonzero ideal of the ring R (no chain conditions). If P is a maxin1al member of the collection {ann (x)IO =/= x E I), deduce that Pis a prime ideal. [Hint: Let abE P = ann (r), with a¢ P; then, P ~ (P, b) £ ann (ar).] 15. A ring R is tem1ed divisible if every non-zero-divisor of R is invertible. Assuming that R is a divisible ring prove the following: a) R is a local ring if and only if the set D of all zero divisors (together with zero) is included in a proper ideal of R; in this case, D itself becomes an ideal. b) If I 1 n 12 =/= {0} for any two nonzero ideals of R, then R is a local ring. [Hint: Show that the sum of two noninvertible elen1ents of R is again noninvertible.] 16. Let R be a principal ideal ring which is not an integral domain. If the set of all zero divisors D = rad R, verify that R is a local ring. 17. a) If R is a finite Boolean ring, prove that R is isomorphic to the direct sum of a finite number of fields Z 2 • [Hint: See the remark following the second proof ofTheorem 11-15.] b) Prove that a finite Boolean ring has 2" elements for some n E Z+.

TWELVE

FURTHER RESULTS ON NOETHERIAN RINGS

In the present chapter, emphasis is laid on certain aspects of ideal theory in which the ascending chain condition plays a dominant role. Although our treatment is rather selective, it may fairly claim to cover most of the high spots, as well as utilize a cross-section of the previously developed material. A special concern will be the proof of a fundamental theorem by Emmy Noether which asserts that every ideal in a Noetherian ring is the intersection of primary ideals; to some extent, this reduces the study of arbitrary ideals in such rings to that of primary ideals. Particular attention will also be paid to a number of results dealing with the intersection of the powers of an ideal in a Noetherian ring. The latter portion of this chapter furnishes the reader with a brief introduction to module theory (roughly speaking, a module is a vector space over a ring rather than a field); the ultimate aim being to prove that every commutative Artinian ring with identity is necessarily Noetherian. Underlying all our arguments is the equivalence of the ascending chain condition for ideals and the maximum condition. Failing any indication to the contrary, all rings considered are assumed to be commutative and have an identity element. Often it is not essential to stipulate both these hypotheses and this will be revealed from a careful examination of the proof in question. . Let us begin our development by showing how primary ideals fit into the theory of Noetherian rings. One of the basic decomposition theorems concerning the ring of integers (itself a Noetherian ring) is that every ideal can be expressed as the intersection of a finite number of primary ideals. Indeed, if n = p~'p~2 • • • p~· is a factorization of the positive integer n into distinct primes Pi• then an integer m is divisible by n if and only if m is divisible by each p~ 1 ; in the notation of principal ideals, this amounts to asserting that (n) = (p~') II (p~2 ) II .. · II (p!•), where each of the (p~ 1 ) is a primary ideal of Z. Our immediate aim is to prove that a representation of the above type (that is, as a finite intersection of primary ideals) is valid for ideals in an 234

FURTHER RESULTS ON NOETHERIAN RINGS

235

arbitrary Noetherian ring. A convenient vehicle for this discussion is the notion of an irreducible ideal. Definition 12-1. Let I be an ideal of the ring R. Then I is said to be irreducible if it is not a finite intersection of ideals of R properly containing I; otherwise, I is termed reducible. As a general comment, it is worth remarking that any prime ideal Pis always irreducible. For, suppose that there exist ideals I and J of R satisfying P = I n J, P c: I, P c: J. We can then select elements a E I - P and bE J - P. Now, the product ab lies in both I and J, whence it is a member of P. From this it follows that P cannot be a prime ideal. On the other hand, we note that there exist (non-prime) primary ideals which are not irreducible. A simple illustration is furnished by the polynomial ring F[ x, y], where F is a field. Here the ideal M = (x, y) is maximal, so that its square M 2 = (x 2 , xy, y 2 ) must be primary (see Example 7-8); however, M 2 has the following representation as an intersection of proper ideals ofF[ x, y]: M 2 = (M 2 , x) n (M 2 , y). Our program is somewhat lengthy and will be completed in Theorem 12-5; we prepare the way by first establishing two auxiliary results. Lemma 1. Every ideal in a Noetherian ring R is a finite intersection of irreducible ideals. Proof Let F be the family of all ideals of R which are not finite intersections of irreducible ideals. If it happens that F =I= 0, then Theorem 11-1 asserts the existence of an ideal I which is maximal in the set F (this is where the Noetherian hypothesis enters). Then any ideal of R properly containing I must be a finite intersection of irreducible ideals. Since IE F, I is not itself irreducible. Thus, we can write I = J n K, where J and K are ideals of R strictly containing I. The maximal nature of I implies that J and K both are finite intersections of irreducible ideals; hence, I is one also. But this clearly contradicts the fact that IE F. Accordingly, the set F is empty, thereby proving the assertion.

To exploit this situation fully, we also require: Lemma 2. In a Noetherian ring R, every irreducible ideal is primary. Proof Our plan is to prove that any ideal I of R which is not primary is necessarily reducible; in other words, we will deduce the contrapositive form of the theorem. Since I is not primary, there exist a pair of elements a, b in R such that ab E I, b ¢ I and no power of a belongs to I. Now,

I: (a) s;; I: (a2 ) s;; ••· s;; I: (a") s;; · •·

236

FIRST COURSE IN RINGS AND IDEALS

forms an ascending chain of ideals of R; indeed, if ran E I, certainly ran+ 1 E I. Because R is taken to be Noetherian, we can therefore find an integer k for which I: (ak) = I: (ak+ 1 ). The bulk of our argument consists of showing that I can be expressed as

I

= (I, if) n (I, b).

Evidently, each ideal on the right-hand side of this equation contains I, so that I ~ (I, ak) n (I, b). To obtain the opposite inclusion, select an arbitrary X E (I, if) n (I, b). Then,

x

= i+

for suitably chosen elements i, i' ra''+ 1

=

E

rak

= i' +

I and r, r'

(i' - i)a

+

E

r'b

R. Consequently, the product

r'(ab)

E

I,

which in turn implies that r E I: (if+ 1) = I: (ak). But this signifies that = i + rak lies in I, as we wished to show. To complete the proof, observe that I c (I, ak), for our hypothesis assures us that ak ¢I; furthermore, the element b ¢I, whence I c (I, b). Inasmuch as both the ideals in equation (1) properly contain I, it follows that I must be reducible. rakE I and so the element x

These results may now be put into the form of a decomposition theorem, the so-called Primary Decomposition Theorem of Noether. Theorem 12-1. {Noether). Every ideal of a Noetherian ring can be represented as a finite intersection of primary ideals.

n;

Let us call a representation of an ideal I in the form I = Q;, where each Q; is a primary ideal, a primary representation of I; the individual Q; are said to be the primary components of the representation, while .JQ; are the prime ideals associated with I. What was just proved is that, in a Noetherian ring, every ideal admits a finite primary representation. Before announcing the next result concerning primary representations, we wish to introduce a new idea. Definition 12-2. A primary representation I

=

ni'=

1

Q; will be termed

irredundant if it satisfies the following two conditions.

1) No Q; contains the intersection of the other primary components; that is to say, ni'FJ Qi =I= n Q; for any j = 1, 2, ... 'n. 2)

JQ1 =I= ..jQi for i

=I= j.

ni'=

If an ideal I admits a finite primary representation, say I = 1 Q;, then enough of the Q;'s can be omitted to yield an irredundant representation. To make this precise, we first let Q; be the intersection of all those primary components which have the same associated prime ideal; that is,

FURTHER RESULTS ON NOETHERIAN RINGS

237

if JQ;, = JQ; 2 = · .. = JQ;,, simply take Qi = Q;, n Q;, n ... n Q;,· BLthe corollary to Theorem 5-13, the ideal Qi is itself primary with JQi = JQ;, and, of course, we have I = nQ;. In this way, the components of a primary representation can be combined so that their nil radicals are all distinct. Next, strike out one at a time those ideals Qi which contain the intersection of the remaining ones. The result of removing these superfluous primary ideals is that condition (1) of Definition 12-2 now holds. In this manner, the given primary representation can be transformed into an irredundant one. Using the language of irredundance, we can now state our basic representation theorem as

Theorem 12-2. Every ideal in a Noetherian ring has a finite irredundant primary representation. We shall have occasion to use the following lemma.

Lemma. Let R be an arbitrary ring and I an ideal of R having a finite irredundant primary representation I = n~=t Q;. Then a prime ideal P of R contains I if and only if P contains some JQ;. The if part is trivial: .JQ;_r;;. P implies that I r;;. Q; r;;. P. Conversely, assume that there is no JQ; which is contained in P. For each index i, we can then choose an element a; E JQ; with a; ¢ P. There also exist suitable integers k; such that a~' E Q;. Setting a = a:'a~2 ••• a!", it follows that a E n~= 1 Q; = I s; P. Now, the product

Proof

with a 1 ¢ P and so, by the definition of prime ideal, (a~2 • • • a~") E P. Repeating this argument, we finally obtain a=" E P, whence an E P, which is impossible. Recall that a prime ideal of R is said to be a minimal prime of the ideal I if it is minimal in the set of prime ideals containing I. Keeping the same notation, the foregoing lemma asserts that the minimal prime ideals of I are the minimal elements of the family {JQ;}, partially ordered by inclusion. With this in mind, we can now formulate

Theorem 12-3. Any ideal of a Noetherian ring has a finite number of minimal prime ideals. One of the tasks which is still ahead of us is the burden of showing uniqueness (in some sense) of the primary representation. Given an irredundant representation I = 1 Q; of an ideal I as a finite intersection of primary ideals Q;, we do not claim that these primary ideals are uniquely determined by I; the illustrative example below shows that this is not to be expected. All that can be established is that the associated prime ideals

ni=

238

FIRST COURSE IN RINGS AND IDEALS

.jQ1 are unique and are the same in all irredundant primary representations of I; thus, it is the number of primary components that will be unique. To verify this, it is enough to show that the associated prime ideals can be characterized in terms of the properties of I alone, independent of any particular primary representation considered. Before proceeding to the proof, let us illustrate the fact that an ideal in a Noetherian ring need not have a unique irredundant primary representation. Example 12-1. In the polynomial ring F[ x, y], where F is any field, consider the ideal (x 2, xy). It is easy to see that (x 2 , xy) consists of those polynomials which have x as a factor and which do not possess linear terms. Now, the nonzero elements of (x 2 , xy, y 2 ) are precisely the polynomials each of whose terms are of degree ~2; hence the intersection (x 2 , xy, y 2 ) n (x) contains the zero polynomial together with all polynomials of degree 2::::2 which have x as a factor. Thus, we have

(x 2, xy) = (x 2 , xy, y 2 ) n (x). Besides this irredundant representation, there is yet another:

To derive the relation above, notice that for a polynomial to lie in (x 2 , xy), it is sufficient to require that the polynomial be divisible by x and that the coefficient of y be zero. As has been seen in Example 7-8, (x 2 , xy, y 2 ) and (x 2 , y) are both primary ideals, while (x) is prime in F[ x, y] and, hence, also primary. We next determine a characteristic of the primary representation which is uniquely determined by the ideal in question. Theorem 12-4. Suppose that an ideal I of the ring R has a finite irredundant primary representation, say I = (ii= 1 Q1, and let P be any prime ideal of R. Then P = 1 for some i if and only if there exists an element a¢ I such that P = .ji: (a).

JQ

Proof. To start with, assume that P = JQ 1 for a given index i. We shall argue that the hypothesis (and, hence, the conclusion) of Theorem 5-14 holds. Now, the irredundancy of the representation I = (lk Qk implies that there exists an element a E nk'fi Qk, but a¢ I. For any such element a, we must have

I: (a) s:;

.jQ1 £; .jI: (a).

The first inclusion is justified by the fact that, since a( I: (a)) s; I s; Q1 with a¢ Q1, necessarily I: (a) s; .jQ1• To see the second inclusion, simply note that aQ 1 s:; I, whence Q1 s:; I: (a).

86

FIRST COURSE IN RINGS AND IDEALS

for some j =1= i, which is impossible by our original choice of ri. On the other hand, if j =I= i, the element ai necessarily lies in P; (r; being a factor of aj). For the final stage of the proof, put a = L aj. We first note that, because each of a 1, a2 , ••• , an is in I, the element a E I. From the relation a; = a - Ln;ai, with LNi aj E P;, it follows that a¢ P;; otherwise, we would obtain a; E P;, an obvious contradiction. Our construction thus ensures the existence of an element a = L ai which belongs to the ideal I and not to any P;, thereby proving the theorem. Corollary. Let I be an arbitrary ideal of the ring R and P 1 , P 2 , be prime ideals of R. If I £ uP;, then I £ P; for some i.

... ,

Pn

PROBLEMS In the following set of problems, all rings are assumed to be commutative with identity. 1. a) Prove that Z Ef> z. is a maximal ideal of the external direct sum Z EB Z. b) Show that the ring R is a field if and only if {0} is a maximal ideal of R.

2. Prove that a proper ideal M of the ring R is maximal if and only if, for every element r ¢ M, there exists some a e R such that 1 + ra e M. 3. Letfbe a homomorphism from the ring R onto the ring R'. Prove that a) if M is a maximal ideal of R with M 2 kerf, thenf(M) is a maximal ideal of R', b) if M' is a maximal ideal of R', thenf- 1(M') is a maximal ideal of R, c) the mapping M -> f(M) defines a one-to-one correspondence between the set of maximal ideals of R which contain kerf and the set of all maximal ideals of R'.

4. If M 1 and M 2 are distinct maximal ideals of the ring R, establish the equality M,M 2 = M 1 n M2 • 5. Let M be a proper ideal of the ring R. Prove that M is a maximal ideal if and only if, for each ideal I of R, either I ~ M or else I -f. M = R. 6. An ideal I of the ring R is said to be minimal if I 1= {0} and there exists no ideal J of R such that {0} c J c I. a) Prove that a nonzero ideal I of R is a minimal ideal if and only if (a) = I for each nonzero element a E I. b) Verify that the ring Z of integers has no minimal ideals. 7. Let I be a proper ideal of the ring R. Show that I is a prime ideal if and only if the complement of I is a multiplicatively closed subset of R. 8. In the ring R

= map R #, define the set I by I= {feRJJ(l) =f(-1)

= 0}.

Establish that I is an ideal of R, but not a prime ideal.

240

FIRST COURSE IN RINGS AND IDEALS

the nil radical ..)1 of an ideal I in a Noetherian ring is the intersection of the minimal primes of I; granting this fact, one finds that .j(x2 , xy) = (x).

Incidentally, our example has the added advantage of showing that be prime without the ideal I being primary.

,j1 can

At this stage, it is reasonable to inquire under what circumstances (if any at all) the ideals in a given primary representation will turn out to be prime ideals. The following theorem supplies an answer. Theorem 12-5. Let I be an ideal of the ring R with a finite irredundant primary representation I = I Q;. Then I is semiprime (that is, I = ..) I) if and only if each Q; is a prime ideal of R.

n?=

Proof We begin by assuming that all the Q; in the given primary representation of I are prime ideals. If the element a E ..)I, then an E I for some positive integer n; hence, an lies in each Q;. As Q; is taken to be prime, this implies that a itself belongs to Q; for every i and so a E I. Our argument shows that £ I, from which the desired equality follows. With regard to the converse, suppose that the ideal I coincides with its nil radical. Then, using Theorem 5-10 again,

JI

I =

.Ji =

-.Jn Q; nJQ;. =

i

i

It is important to point out that this is actually an irredundant representation of I as an intersection of primary (in fact, prime) ideals. Suppose not; there would then exist some positive integer j such that I = r i Q;. But then

n

n; v

n

JQ; 2 Q; 2 I. i'f j i 'f j which means that I = ni'fi Q1 • This, however, contradicts the hypothesis that the given representation of I is irredundant. I=

Nex~ fix the integer j and let a be any elell!ent ~the ideal .J~. Since

n

ni'fi

j[, we can find some bE .JQ 1 with b ¢ .JQi. Then the product abE JQ; = I £ Qi, whence a is a member of Q .. Therefore, .jQi £ Qi, yielding the subsequent equality Qi = .jQi. This implies that Qi is a prime ideal, which was what had to be proved. ni'fi .jQ; =/=

n

With this information at our disposal, we can now state Corollary. In a Noetherian ring, any semiprime ideal is a finite intersection of prime ideals. Let us change direction now. The reader will no doubt recall that a ring R is Noetherian if and only if every ideal of R is finitely generated (Theorem 11-1). Actually, it is enough to consider just the prime ideals of R, the proof being due to I. S. Cohen.

FURTHER RESULTS ON NOETHERIAN RINGS

24)

Theorem 12-6. (Cohen). A ring R is Noetherian if and only if every prime ideal of R is finitely generated. Proof The "only if" part is an immediate consequence of Theorem 11-1. Passing to the less trivial assertion, assume that every prime ideal of R is finitely generated, but that R is not Noetherian. This assures that the collection .fF of ideals of R which are not finitely generated is nonempty. Appealing to Zorn's lemma, .fF must contain a maximal element, call it I. By virtue of our hypothesis, I cannot be a prime ideal of R. Consequently, there exist elements a, b of R which are not in I such that their product abE I. Now, both the ideals(!, b) and I: (b) properly contain I; in particular, a E I: (b). By the maximal nature of I in ?F, these ideals are finitely generated. For definiteness, let us suppose that

and Then

C;

=

a;

+

br;, where a; E I and r; E R (i = 1, 2, ... , n), so that

(I, b)

=

(al, a2, ... ' an, b).

Next, consider the ideal J generated by the elements a; and bdi; in other words, the ideal

Since bdi E I for every j, the inclusion J £; I certainly holds. What is not so obvious is that I £; J. To see this, let x be an arbitrary member of I; because x E (J, b), it may be written in the form X

=

alyl

+ ... +

anYn

+

by

{y;, y E R).

As each a; lies in the ideal I, so does by, which is simply to assert that y E I: (b). Knowing this, we are able to find elements z; E R such that

y = d 1z 1

+

d2 z 2

+ ... +

d,.z,.,

leading directly to

+ ... + anYn + b(d 1z 1 + ... + d,.z,.) a 1Yl + ... + anYn + (bd 1 )z 1 + .... ~ (bd,.)zm E J.

x = a1 y =

The equality I = J now follows and so one concludes that I itself is finitely generated, an impossibility since I E ?F. This contradiction completes the proof. Scrutiny of the preceding argument reveals a fact which is important enough to be stated independently: Let I be an ideal of the ring R and b an element of R; if the ideals (J, b) and I: (b) are both finitely generated,

242

FIRST COURSE IN RINGS AND IDEALS

then I is also finitely generated. As an application of Cohen's Theorem, we present Corollary. If R is a ring in which each maximal ideal is generated by an idempotent, then R is Noetherian. Proof We first prove that every primary ideal of R is maximal. Suppose otherwise; that is, let I be a primary ideal which is not maximal in R. Now, I will be contained in some maximal ideal M. By hypothesis, M has an idempotent generator; say M = (e), where e is an idempotent different from 0 or 1 (if e = 0, R becomes a field and there is nothing to prove). Then e(1 - e) = 0 E I and, since I is a primary ideal, it follows that (1 - e)"e Is;;; M

for some positive integer n. This implies that 1 - e e M = (e), so that 1 e M, an obvious contradiction. Because every primary ideal of R is maximal, the notions of maximal, prime, and primary ideal all agree. In the light of our hypothesis. every maximal ideal (hence, every prime ideal) is finitely generated. That R is necessarily Noetherian now follows from Cohen's result. We next propose to take a look at several results concerning the intersection of the powers of an ideal in a Noetherian ring. Before any deductions can be made, it will be convenient to establish a technical lemma. Lemma. Let I and J be two ideals of the ring R, with I finitely generated. If lJ = I, then there exists an element r E J such that (1 - r)I = {0}. Proof Suppose that I is generated by the elements a 1 , a 2 , ••• , a.. Let I; denote the ideal (a;. a;+ 1 , ••• , a.) and put I.+ 1 = {0}. By induction on i, we shall prove the existence of an element r; E J such that (l - r;)I ,;: I; (i = 1, 2, ... , n + l); in particular, r.+ 1 will be the element mentioned in the statement of the theorem. When i = 1, the ideali 1 = I and one can simply take r 1 = 0. Using the induction hypothesis that (1 - r;)I ,;: I; for some r; e J, together with the fact that I s;;; IJ, we have (l - r;)I £ (1 - r;)IJ s;;; I;J.

Since each generator a; lies in I, it follows that (1 - r;)a; e I;J and so n

(1 - r 1)a1 =

L

b1tat

(bikE

J).

k=i

In consequence, (1 - r 1

-

b;;)a;

=

•

L b;~:ak e Ii+I· i:=l+l

FURTHER RESULTS ON NOETHERIAN RINGS

243

It now suffices to take 1 - r;+ 1 = (1 - r;)(1 - r; - b;;); clearly, r;+ 1 e J and, as a little computation will show, (1 - r;+ 1 )I = (1 - r;)(l - r; - b;;)I £ (1 -

r; - bii)I;

This proves the lemma. In a moment, we shall appeal to this lemma to characterize the elements which belong to the intersection of the powers of an ideal. Let us temporarily turn aside from this pursuit, however, to call attention to a noteworthy result of Nakayama. Theorem 12-7. (Nakayama's Lemma). Let I be a finitely generated ideal of the ring R. If I(rad R) = I, then I = {0}.

Proof The foregoing lemma tells us that there exists an element r E rad R for which (1 - r)I = {0}. If 1 - r were not invertible in R, then it would be contained in some maximal ideal M. But r E rad R £ M, leading to the contradiction that 1 EM. Accordingly, 1 - r is an invertible element of R, which forces I = {0}. Remark. It is possible to prove somewhat more than is asserted above, for one may replace rad R by any ideal which is contained in rad R. What is important in the present situation is that Nakayama's Lemma holds in any Noetherian ring.

We now come to the theorem that was alluded to earlier. Theorem 12-8 Let I be a proper ideal of the Noetherian ring R. Then

n I" = m, we then have (pm) S (a) S (pk) S (pm), whence the equality (~) = (p"}. The lemma now tells us that pm = p"v, where v is an invertible element of R. This means that a = p"t•u, with vu invertible, contradicting the last assertion of the lemma. Corollary I. Let R be a local ring with principal maximal ideal M = (p). Assume further that n:=t M" = {0}. If I is any nontrivial ideal of R, then I = M" for some k E Z + (hence, R is Noetherian). Proof Clearly, I S M, so that each nonzero element of I can be written as p"u, with u invertible. Take k to be the least integer such that p"u E I. It then follows that I s (p"). On the other hand, since p"u e I, so does p" = (p"u)u- 1 ; this implies that (p") s I and equality follows.

247

FURTHER RESULTS ON NOETHERIAN RINGS

Corollary 2. If R is a Noetherian local ring whose maximal ideal is principal, then R is a principal ideal ring. Our development has now reached a point where, in order to make further progress, we need to bring in certain results that belong primarily to the theory of modules. The concept of a module is the natural generalization of that of a vector space; instead of requiring the "scalars" to be elements of a field, we now allow them to lie in an arbitrary ring with identity. The major theorem to be established is a remarkable result of Hopkins that every commutative Artinian ring with identity is Noetherian. This theorem does not extend to rings lacking an identity; indeed, Z(p shows that it is possible for the descending chain condition to be satisfied in a ring without the ascending chain condition also holding. Apart from some standard results about ideals, Hopkin's argument requires only the JordanHolder Theorem for modules (including the fact that a composition series exists if and only if both chain conditions on submodules are satisfied). The proof will not be given immediately, but only after we assemble some of the module-theoretic prerequisites. Our discussion is not entirely selfcontained in this regard and certain facts will be presented without proof. The reader who is unfamiliar with modules would profit from working out the details. It is time for these somewhat vague preliminaries to give way to a more precise definition. 00 )

Definition 12-3. Let R be a ring with identity. By a left module over R (or a left R-module), we mean a commutative group M (written additively) together with an operation of multiplication which associates with each r E R and a E M a unique element ra E M such that the following conditions are satisfied: 1) (r

+

s)a = ra

+

sa,

+

rb,

2) (rs)a = r(sa), 3) r(a

+

b) = ra

4) la =a, for all r, s E R and a, bE M. The parallel notion of a right R-module can be defined symmetrically. Technically speaking, (left) module multiplication is a function a: R x M -;. M, where a(r, a) is denoted by ra. The element ra is often called the module product of rand a. In effect, Definition 12-3 states that a left R-module is an ordered pair (M, a); this approach gets a little cumbersome and so, when there is no possibility of confusion, we shall lapse into saying "the left R-module M". We pause to look at some typical examples of modules.

248

FIRST COURSE IN RINGS AND IDEALS

Example 12-3. If R = F, where F is any field, then a left R-module is simply a vector space over F. Example 12-4. Every commutative group (G, +) can be considered as a left Z-module in a natural way. For, given an integer nand element a E G, na has a well-defined meaning :

na=a+a+···+a

(n summands).

Example 12-5. If I is a left ideal of a ring R with identity, then the underlying additive group (I, +)of I forms a left R-module. Indeed, the definition of left ideal insures that the ring product raE I for every r E R and a E J. As a special case, any ring R with identity can be viewed as a left (or right) R-module over itself. Example 12-6. Consider the set hom G of all homomorphisms of a commutative group (G, +) into itself (that is, the set of endomorphisms of G). It is already known that (hom G, +, o) constitutes a ring with identity, where o indicates the operation of functional composition. To provide G with the structure of a left module over hom G, we define the module product fa by putting {fE hom G, a E G). fa = f(a) Condition (3) of Definition 12-3 is satisfied by virtue of the fact that f is a homomorphism. To avoid a proliferation of symbols, 0 will be used to designate the additive identity element of the group (M, +) as well as the zero element of R. This convention should lead to no ambiguity if the reader attends closely to the context in which the notation is employed. As with vector spaces, we have the laws (i) Oa = rO = 0, (ii) r(- a) = (- r)a = - (ra), for all r E R and a E M. One can introduce the notions of submodule, quotient module, and module homomorphisms, all by natural definitions. These are of fundamental importance for our theory and from them our ultimate goal, Hopkin's Theorem, will follow easily. In the remainder of this discussion, we shall drop the prefix "left", so that the term "R-module" will always mean "left R-module"; it should be apparent that the entire discussion applies equally well to right R-modules. Of course, when R is a commutative ring, any left R-module can be turned into a right R-module simply by putting ar = ra. Modules over commutative rings are essentially two-sided and all distinction between left and right disappears (it is merely a matter of personal preference whether one writes the ring elements on the left or on the right). A natural starting point is, perhaps, to examine the concept of a submodule. Suppose then that M is an arbitrary module over the ring R. By

FURTHER RESULTS ON NOETHERIAN RINGS

249

an R-submodule of M we shall mean a nonempty subset N of M which is itself a module relative to the addition and module multiplication of M. To make this idea more precise: Definition 12-4. A nonempty subset N of the R-module M is an Rsubmodule (or simply a submodule) of M provided that 1) (N, +) is a subgroup of (M, +); 2) for all r E R and a

E

N, the module product ra

E

N.

Needless to say, the first condition in Definition 12-4 is equivalent to requiring that if a, bEN, then the difference a - beN. Every R-module M clearly has two trivial submodules, namely, {0} and M itself; a submodule distinct from M is termed proper. Paralleling our discussion of rings, we shall call an R-module M simple, if M =!= {0} and the trivial submodules are its only submodules. It is well worth noting that if M is a vector space over a field F, then any F-submodule is just a vector subspace of M. A further illustration arises by considering a ring R as a module over itself; when this is done the (left) ideals of R becomes its R-submodules. The concept of a quotient structure carries over to modules as expected. To be more concrete, let N be a submodule of a given R-module M. Since M is a commutative group, N is automatically normal in M and we can form the quotient group M/N. The elements of this group are just the cosets a + N, with a E M; coset addition is given, as usual, by (a

+ N) +

(b

+ N)

= a

+

b

+

N.

To equip M/N with the structure of a module, a notion of multiplication by elements of R is introduced by writing r(a

+

N) = ra

+

N.

We must first satisfy ourselves that module multiplication is unambiguously defined, depending only on the coset a + N and element r E R. Thisamountstoshowingthatwhenevera + N =a'+ N,thenr(a + N) = r(a' + N), or, rather, ra + N = ra' + N. Our aim would obviously be achieved if we knew that ra- ra' = r(a- a')e N. But this follows directly from the fact that a - a' E N and that N is assumed to be a submodule over R. Thus, the module product in M/N is independent of coset representatives. One can easily check that M/N, with the above operations, forms an R-module (the so-called quotient module of M by its submodule N). When forming quotient rings, it became necessary to introduce a special subsystem (namely, ideals) in order to ensure that the operations of the

SIX

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

As the title suggests, this chapter is concerned with the problem of factoring elements of an integral domain as products of irreducible elements. The particular impetus is furnished by the ring of integers, where the Fundamental Theorem of Arithmetic states that every integer n > 1 can be written, in an essentially unique way, as a product of prime numbers; for example, the integer 360 = 2·2·2·3·3·5. We are interested here in the possibility of extending the factorization theory of the ring Z and, in particular, the aforementioned Fundamental Theorem of Arithmetic to a more general setting. Needless to say, any reasonable abstraction of these numbertheoretic ideas depends on a suitable interpretation of prime elements (the building blocks for the study of divisibility questions in Z) in integral domains. Except for certain definitions, which we prefer to have available for arbitrary rings, our hypothesis will, for the most part, restrict us to integral domains. The plan is to proceed from the most general results about divisibility, prime elements, and uniqueness offactorization to stronger results concerning specific classes of integral domains. Throughout this chapter, the rings considered are assumed to be commutative; and it is supposed that each possesses an identity element. Definition 6-l. If a =fo 0 and b are elements·of the ring R, then a is said to divide b, in symbols alb, provided that there exists some c E R such that b = ac. In case a does not divide b, we shall write a J b. Other language for the divisibility property alb is that a is a factor of b, that b is divisible by a, and that b is a multiple of a. Whenever the notation alb is employed, it is to be understood (even if not explicitly mentioned) that the element a =fo 0; on the other hand, not only may b = 0, but in such instances we always have divisibility. Some immediate consequences of this definition are listed below; the reader should convince himself of each of them. Theorem 6-1. Let the elements a, b, c e R. Then, 1) a!O, t!a, ala; 90

FURTHER RESULTS bN NOETHERIAN RINGS

251

condition can also be applied to R-modules, the sole difference being that, in our earlier definitions, the term "ideal" must now be replaced by the word "submodule". Adapting the argument of Theorem 11-2, it is a simple matter to show that an R-module M satisfies the ascending (descending) chain condition on submodules if and only if M satisfies the maximum (minimum) condition on submodules; we leave the verification of this to the reader. The coming theorem indicates how the chain conditions on submodules are affected by certain operations. Theorem 12-14. 1) If the R-module M satisfies the ascending (descending) chain condition, then so does every homomorphic image of M. 2) Let N be a submodule of the R-module M. Then M satisfies the ascending (descending) chain condition if and only if Nand M/N both satisfy it. For the most part, the stated results are merely a translation ofTheorems 11-6 and 11-7 into the language of modules. What is new in the present situation is that any submodule N of M inherits the ascending (descending) chain condition. This follows from the fact that any submodule of N is itself a submodule of M (a marked contrast to the behavior of ideals). Before the reader collapses under a burden of definitions, let us turn our attention to the matter of normal and composition series. By a normal series for an R-module M is meant a (finite) chain of Rsubmodules running from M to {0}:

M = M0

;;2

M1

;;;;! ••• ;;;;!

Mn_ 1

;;;;!

M,. = {0}.

A given normal series can be lengthened or refined by the insertion of new submodules between those already present. In technical terms, a second normal series

M = N0

;;;;!

N1

is said to be a refinement of M = M 0 ;;;;! M 1

;;;;! ••• ;;;;!

Nm-1

;;;;!

Nm = {0}

;;;;! ••• ;;;;!

M,_ 1

;;;;!

M,. = {0}

provided that there exists a one-to-one function f from {0, 1, ... , n} into {0, 1, ... , m} such that M; = Nl: homR(l, J) -+ eRe by means of the rule 4>(/) = f(e). It is immediately apparent that

4>(/) + 4J(g).

If f is caused by a left multiplication by the element ese and g is a left multiplication by ete, then 4> enjoys the further property

b1 E R}

(here L, represents an arbitrary finite sum) constitutes a two-sided ideal of R different from zero, since 0 =I= e = tel E ReR. Hence, we must have ReR = R. In particular, the identity 1 E ReR, so that it is possible to select elements a 1, b1 E R satisfying 1 = L, a 1eb 1• Now, choose any D-endomorphism fe homv(l, J) and any element x = erE J. A straightforward computation gives f(x) = f(erl)

=

f(er L, a 1eb1)

=

f(L, era1eb 1)

= L,f(era1eb 1)

= L, era1ef(eb1) = er L a1ef(eb 1)

(since era1e ED)

= x L, a1ef(eb1).

From this formula, it appears that f(x) = T.x, where the element s (which does not depend on x) is g1ven by s = L, a 1ef(eb;). Therefore, feR.,, confirming that R., = homv(J, J). Putting our remarks together, we obtain the isomorphism

282

FIRST COURSE IN RINGS AND IDEALS

To clinch the argument, let us show that I is a finite dimensional (left) vector space over D. We suppose this to be false. Then I possesses an infinite basis, from which can be extracted a sequence x 1 , x 2 , ••• of linearly independent elements. That is, for each integer n, the set {x 1 , x 2 , . •• , x.} is independent wtth respect to D. Given an integer n, define I.

=

{a

E

Rlx 1a

= ... =

x.a

= 0}.

Then I 1 ~ I 2 ~ · • · ~ I. ~ · · · forms a descending chain of right ideals of R. Since the ideal I = eR can be regarded as a simple (right) R-module, a direct appeal to the last lemma is permissible. Thus, there exists some element a E R for which

leading to the conclusion that I • is properly contained in I • _ 1 . The point which we wish to make is that I 1 :;:, I 2 :;:, .. • :;:, I, :;:, .. · is a strictly descending chain, in violation of the assumption that R is a right Artinian. Accordingly, the dimension dirnDl < oo and this finally ends the proof of Theorem 13-14. If Vis an n-dimensional vector space over the division ring D, then the ring homD( V, V) is well-known to be isomorphic to the ring M. (D) of n x n matrices over D. To spell out some details, let Vhave the basis {x 1 , x 2 , ••• , x.}. Iff E homD(V, V), each of these basis elements will be mapped by f into a uniquely determined linear combination of x 1, x 2 , ••• , x.. In other words, there exist elements aij ED, uniquely defined by f, such that f(xi) can be expressed in the form

f(xi) =

•

L aijxi i= 1

U=

I, 2, ... , n).

(Observe that the summation is over the first index.) Thus, to each endomorphism /E homD(V, V) there corresponds a unique n x n matrix (aii) with entries from D. There is no problem in showing that the map f-+ (aii) yields a one-toone function from homD(V, V) onto the matrix ring M.(D). Indeed, starting withanarbitrary(aii) E M.(D),onedefinesaD-endomorphismjE homD(V, V) by first settingf(x) = L~=t aijxi and then extending linearly to all of V; it is evident that (aii) is precisely the matrix identified with the resulting endomorphism. If (aii) and (bii) are the matrix representations of two elements of homD(V, V), say f and g, then

(/ + g)(x1 ) and

= /(xJ)

+ g(xJ) =

n

n

n

i=l

i=l

1=1

L a11 x 1 + L b11 x 1 = L (ail + biJ)x1

PROBLEMS

(fa g)(x)

=

1(J

1

bkixk)

283

= kt bkJ(xd

ktl bkj itl aikxi

= itl

ctl

aikbkj}

These relations make clear that

f +g

-+

(ali

+ bii)

= (aii)

+ (b;i),

and, by definition of the product of two matrices, that

The conclusion is that the mapping which associates with eachf E homD(V, V) its matrix representation (aii) (relative to the fixed basis) induces a ring isomorphism homD(V, V) ~ Mn(D). On the strength of these remarks, our various results may be collected to give a description of nil-semisimple right Artinian rings in terms of matrix rings. Theorem 13-15. Let R be a nil-semisimple right Artinian ring. Then there exist division rings D; and suitable integers n; (i = 1, 2, ... , r) such that

R

~

Mn 1(Dt) EB Mn 2(D2) EB ••• EB M,..(D,).

Although we must now close this chapter and thereby conclude our presentation of the theory of rings, we are precisely at the point where one could start delving deeply into the subject. (For a more thorough treatment of the noncommutative aspects, see the excellent account by Herstein [I 5] and the references cited there.) Needless to say, we have merely scratched the surface of this fascinating branch of algebra; nonetheless, the reader should now be in a better position to appreciate the details and the difficulties.

PROBLEMS Unless specified otherwise, R always denotes an arbitrary ring with identity. 1. Let e be an idempotent element of the ring R. For any two-sided ideal/ of R, show that the subring ele = If"' (eRe).

2. If R is a right Artinian ring, prove that whenever there exists two elements a, b e R with ab = 1, then ba = 1. [Hint: Consider the descending chain bR 2 b2 R 2 ... 2 b"R 2 ···of right ideals of R.]

284

FIRST COURSE IN RINGS AND IDEALS

3. Given a right Artinian ring R, establish that a) The sum N of all mlpotent nght 1deals of R 1s again a nilpotent right ideal. [Hint: If N 1s not mlpotent, tt contains a nonzero 1dempotent by Theorem 13-1.] b) N forms a two-stded 1deal of R. [Hint. The 1deal RN is such that (RN)" s; RN", whence RN s; N.] c) The ideal N contains any nilpotent left 1deal I of R. [Hint· I s; RN s; N]

4. Let I be a minimal nght ideal of the ring R. Show that either ! 2 = (0] or [ 2 = I: in the second case, deduce that I = eR = el for some idempotent e =r 0 m I. [Hint: See the proofs of Theorems 13-1 and 13 2; first, estabhsh the existence of an element a E I for which al = I.]

5. Assume that I and 1 are two mm1mal nght ideals of the nng R which are morphic as right R-modules. If I 2 = I, prove that a) Any R-isomorphismf: I -. J is of the formf(x) = ax for some a E J. b) The product JI = J. [Hint: By part {a), J = ai, where a E J.]

ISO-

6. For nonzero idempotents e, u of the ring R, prove that a) eR = uR if and only if eu = u and ue =e. b) eR and uR are isomorphic as R-modules If and only If there exist x, y E R such that xy = u and yx = e. [Hint: Since xe = ux and yu = ey. the function f(er) = x(er) = u(xr) defines an R-Isomorphism f: eR -+ uR, with f- 1(us) = y(us) = e{ys).] c) eR ~ uR, as R-modules, if and only if Re ~ Ru. 7. Let eR and uR be two mmimal nght Ideals of the ring R, where e and u are nonzero idempotents of R. Show that eR and uR are R-isomorphic if and on!} if their product uReR =/= {0}. [Hint. lfure =/= 0 for some r E R. define the R-Isomorphism f: eR -. uR by f(es) = (ure)s.] 8. Let R be a nil-semisimple nng Without Identity. If the element a E R and aR = [ 0}, [Hint: The ideal A(R) = {r E RlrR = {0}} satisfies establish that a = 0. A(R) 2 g. A(R)R = {0}.] 9. Establish the statements below: a) A nght ideal I of the ring R is a direct summand of R if and only if I = eR for some idempotent e E R ; b) a minimal right ideal I of the ring R is a direct summand of R If and only if I2 {0}.

+

10. Prove that a right Artinian ring R is ml-semisimple tf and only 1f I 2 =t= {0} for each minimal right ideal I of R. 11. As.mming that R is a nil-semisimple right Artinian ring, verify the followmg assertions: a) R is right Noetherian; that is, R satisfies the ascending chain condition on right ideals. [Hint: By Theorem 13-2, the right ideals of R are finitely generated.] b) The mapping e -+ eR defines a one-to-one correspondence between the set of all idempotent elements e E cent R and the set of two-sided ideals of R. c) For any two-sided ideal I of R, ann,] = ann 11 and so R = I EfJ ann I. d) Every right ideal I of R is a direct summand of any right ideal containing it.

PROBLEMS

285

12. a) Suppose that the ring R is a fimte direct sum of right ideals I; =f- {0}, say R = I1 EB I2 E9 ··· EB In. If 1 = e1 + e2 + ... + e., where e;e I, prove that the elements e; form a set of orthogonal idempotents and that I; = e;R (i = 1, 2, ... , n). b) If the ideals I; of part (a) are all two-sided, show that e; e cent R and so serves as an identity element for I 1•

13. a) Prove that an idempotent e =f- 0 of the ring R is primitive If and only If R contains no idempotent u e such that eu = ue = u. b) Establish that any idempotent element e =f- 0 of a nil-semisimple right Artinian ring R is the sum of a finite number of orthogonal primitive Idempotents. [Hint: There exists a minimal right idealJ 5:; eR. Write eR = I EB J, where the right ideal J 5:; I. Now, either J = {0}, or else e = e 1 + e2 , with e1 E I a pnnutive idempotent. If e 2 e J is not primitive, repeat this process as applied to J = e2 R.]

+

14. a) If M =f- {0} is a nght R-module, verify that M becomes a left homR(M, M)module on defining the module product fx by fx

= f(x)

(fe homR(M, M);

X

EM).

b) Let M and N be right R-modules which are R-isomorphic under the mapping a: M --+ N. Show that homR (M, M) ::: homR (N, N), as rings, by means of the function that carriesfe homR(M, M) to a o fo a- 1 • 15. Let F be a field and M 2(F) denote the ring of 2 x 2 matnces over F. Prove that

a) The matrices e 1

=

{~ ~)and e

2

= {~

~)are orthogonal idempotents;

b) I;= e;M 2 (F)(i = 1,2)isaminimalrightidealofM2 (F),withM 2 (F) = I 1 EB 12 ; c) e1M 2 (F)e1 ::: F fori = 1, 2. 16. Use Theorem 13-14 to deduce that any commutative semisimple (in the usual sense) Artinian ring is a finite direct sum of fields.

17. Prove that a right Artinian ring R is a regular ring if and only if R is nil-semisimple. [Hint: Problems 19 and 20, Chapter 9.] 18. Let M =f- {0} be a simple right R-module and so, by Theorem 13-9, a vector space over the division ring D = homR(M, M). Prove the following version of the Jacobson Density Theorem. Given any x 1, x 2 , ... , x. eM which are linearly independent with respect to D and arbitrary y" y2, ... , y. E M, there exists some element a e R (equivalently, some D-endomorphism T. E R, 1 ) such that xka = Yk fork = I, 2, ... , n. [Hint: From the lemma preceding Theorem 13-14, it is possible to choose elements a1 e R for which

J= 0 forj + i

xJa'l.f

0 for j

Let r1 e R be an element such that x 1a1r1

=i·

= Y;·

Now, consider a

=

Li~ 1 a,r,.J

286

FIRST COURSE IN RINGS AND IDEALS

19. Given a right R-module M, set M* = homR(M, R). Prove the statements below: a) M* can be made into a left R-module (known as the dual module of M) by defining the module product rf as (r E R, x EM).

(rf)(x) = rf(x)

b) When R is a division ring, so that M* forms a vector space over R, then [M* :R] = [M:R]. [Hint: If x 1 , x 2 , ... , x. is a basis forM, then then functions fi* EM* (i = 1, 2, ... , n) prescnbed by

fi

*

_

_

(xi) - 6u -

{1

for i = j 0 for i =!= j

serve as a basis for M*.J c) M and M** are naturally tsomorphic as nght R-modules, where M** = (M*)* = homR(M*, R). [Hint: For each x EM, define x EM** by .x(g) = g(x), gEM*; then M ~ M** under the R-isomorphtsm that sends x to .x.J 20. Let M be a right R-module and I a right ideal of R. Prove that a) The set M I = {L x;r; IX; E M, r; E I}, where I is an arbttrary fimte sum, constitutes a submodule of M. b) If M and I are both simple (as R-modules) with Ml "f {0}, then M ~ I. [Hint: Since xi = M for some x EM, an R-isomorphism j: I-+ M can be given by f(a) = xa, where a E I.] 21. Let R be a nil-semisimple right Artmian ring. Verify that, up toR-isomorphism, there exist only a finite number of simple right R-modules. [Hint: R is a direct sum R = 1 1 E9 12 E9 ·· · E9 I. of finitely many two-sided ideals. If M of {0} is any simple right R-module, use Problem 20(b) to conclude that M :: I; for some i.] 22. Prove that the ring R is simple if and only if every simple right R-module is faithful. [Hint: If R has nontrivial two-sided ideals, it possesses a maximal one I by Zorn's Lemma. Let J be any maximal right ideal of R with J 2 I and obtain a contradiction by considering the simple right R-module A(R/J) 2 I.] 23. Prove each of the following assertions : a) The radical J(R) = (I :1R), where the mtersectipn runs over all maximal right ideals I of R. b) Any nonzero ideal of a primitive ring of endomorphisms of a commutative group G is also a primitive ring of endomorphisms of G. c) A ring R is primitive if and only if it contains a maximal right ideal/ such that the quotient ideal (I :1R) = {0}.

n

24. An ideal I of the ring R is said to be a primitive ideal if R/1 is a primitive ring. Establish that the radtcal J(R) can be represented as J(R) = P, where the intersection is taken over all primitive ideals P of R. [Hint: If P = A(M), where M is a simpleR-module, then M is a faithful simple module over the ring RjP.]

n

APPENDIX A

RELATIONS

We herein append a few definitions and general results concerning certain types of relations that can be imposed on a set. For the most part, our attention is confined to two relations of particular utility, namely, equivalence relations and order relations. Intuitively, a (binary) relation on a set S provides a criterion such that for each ordered pair (a, b) of elements of S, one can determine whether the statement "a is related to b" is meaningful (in the sense of being true or false according to the choice of elements a and b). The relation is completely characterized once we know the set of all those pairs for which the first component stands in that relation to the second. This idea can best be formulated in set-theoretic language as Definition A-1. A (binary) relation R in a nonempty setS is any subset of the Cartesian product S x S.

If R is a relation, we express the fact that the pair (a, b) E R by saying that a is related to b with respect to the relation R, and we write aRb. For instance, the relation < in R can be represented by all points in the plane lying above the diagonal line y = x; it is customary to write 3 < 4, rather than the awkward (3, 4) E 1). When n < 0, we may view Q(Jn) as a subdomain of the complex number system C and represent its elements in the standard form a + bJn i. It is not difficult to show that if n1 , n2 are square-free integers, then Q(Jn~) = Q(.Jn 2 ) if and only if n 1 = n2 . Each ele!!J.ent a ~ a + E Q(.jn) gives rise to another element a = a - bJn of Q(Jn), which we shall call the conjugate of a (for n < 0, a is the usual complex conjugate of a). A simple argument establishes that the mappingf: Q(Jn) -+ Q(J'.) defined by f(a) = a is an isomorphism. To study divisibility properties of Q(Jn), it is convenient to make use of the concept of the norm of an element (an analog of the absolute value notion in Z):

b.Jn

Definition 6-9. For each element a = a + b.Jn in Q(J'.), the norm N(a) of II is simply the product of a and its conjugate a:

N(II)

= aa =

(a

+ b.Jn)(a

- bJ'.)

=

a2

-

b2 n.

Some properties of the norm function which follow easily from the definition are listed below.

Lemma. For all a, PE Q(.Jn), the following hold: 1) N(II) = 0 if and only if II = 0; · 2) N(II/1) = N(II)N(fJ); 3) N(l) = 1; Proof Given a = a + b.Jn in Q(Jn), N(a) = a 2 - b2 n = 0 if and only if both a = b = 0 (that is, a = 0); otherwise, we would contradict the choice of n as a square-free integer. Since the mapping f(a) = a is an isomorphism, N is a multiplicative function in the sense that N(afJ) = IIpa]J = IIPiilJ = aaPlJ = N(a:)N(fJ)

for all a, PE Q(Jn). The proof of assertion (3) follows from the fact that

N(l)

= N(1 2 )

= N(l)N(l)

= N(l)2 ,

whence N(l) = 1. Although Q(J'.) has been labeled as a field, we actually have not proved this to be the case; it is high time to remedy this situation.

RELATIONS

29)

but not under the other; say a - b, but not a -'b. By Theorem A-1, there is an equivalence class in Sj- containing both a and b, while no such class appears in S/"' '. Accordingly, S/"' and S/ ,...,' differ. Theorem A-2. If f!lJ is a partition of the set S, then there is a unique equivalence relation in S whose equivalence classes are precisely the members of f!JJ. Proof Given a, b E S, we write a - b if and only if a and b both belong to the same subset in f!JJ. (The fact that f!lJ partitions S guarantees that each element of S lies in exactly one member of f!J.) The reader may easily check that the relation -, defined in this way, is indeed an equivalence relation inS. Let us prove that the partition f!lJ has the form S/-. If the subset P E f!J, then a E P for some a inS. Now, the element bE P if and only if b - a, or, what amounts to the same thing, if and only if b E [a]. This demonstrates the equality P = [a] E S/-. Since this holds for each P in f!JJ, it follows that f!lJ £ S/-. On the other hand, let [a] be an arbitrary equivalence class and P be the partition set in f!lJ to which the element a belongs. By similar reasoning, we conclude that [a] = P; hence, Sj- £ f!JJ. Thus, the set of equivalence classes for - coincides with the partition f!JJ. The uniqueness assertion is an immediate consequence of the lemma.

To summarize, there is a natural one-to-one correspondence between the equivalence relations in a set and the partitions of that set; every equivalence relation gives rise to a partition and vice versa. We have a single idea, which has been considered from two different points of view. Another type of relation which occurs in various branches of mathematics is the so-called partial order relation. Just as equivalence generalizes equality, this relation (as we define it below) generalizes the idea of "less than or equal to" on the real line. Definition A-4. A relation R in a nonempty set S is called a partial order in S if the following three conditions are satisfied : 1) aRa (reflexive property), 2) if aRb and bRa, then a = b (antisymmetric property), 3) if aRb and bRc, then aRc (transitive property), where a, b, c denote arbitrary elements of S. From now on, we shall follow custom and adopt the symbol ~ to represent a partial order relation, writing a ~ b in place of aRb; the foregoing axioms then read: (1) a ~ a, (2) if a ~ b and b ~ a, then a = b, and (3) if a ~ b and b ~ c, then a ~ c. As a linguistic convention, let us also agree to say (depending on the circumstance) that "a precedes b" or

292

FIRST COURSE IN RINGS AND IDEALS

"a is a predecessor of b", or "b succeeds a", or "b is a successor of a" if a ~ band a =f b. By a partially ordered set is meant a pair (S, ~)consisting

of a set S and a partial order relation ~ in S. In practice, one tends to ignore the second component and simply speak of the partially ordered set S, or, when more precision is required, say that SIS partially ordered by ~. If A is a subset of a partially ordered set S, then the ordering of S restricted to A is a partial ordering of A, called the induced partial order; in this sense, any subset of a partially ordered set becomes a partially ordered set in its own right. When considering subsets of a partially ordered set as partially ordered sets, it is always the induced order that we have in mind. Let S be a set partially ordered by the relation ~. Two elements a and b of S such that either a ~ b or b ~ a are said to be comparable. In view of the reflexivity of a partial order, each element of S is comparable to itself. There is nothing, however, in Definition A-3 that ensures the comparability of every two elements of S. Indeed, the qualifying adverb "partially" in the phrase "partially ordered set" is intended to emphasize that there may exist pairs of elements in the set which are not comparable. Definition A-5. A partial order ~ in a set S is termed total (sometimes simple, or linear) if any two elements of S are comparable: that is, a ~ b orb ~ a for any two elements a and b of S. A partially ordered set (S, ~ ) whose relation ~ constitutes a total order in S is called a totally ordered set or, for short, a chain. Let us pause to illustrate some of the preceding remarks. Example A-6. In the set R# of real numbers, the relation ~ (taken with the usual meaning) is the most natural example of a total ordering. Example A-7. Given the set Z + of positive integers, define a ~ b if and only if a divides b. This affords a partial ordering of Z +• which is not total; for instance, the integers 4 and 6 are not comparable, since neither divides the other. Example A-8. Let S be the collection of all real-valued functions defined on a nonempty set X. Iff~ g is interpreted to mean .f(x) ~ g(x) for all x EX, then ~ partially, but not totally, orders S. Example A-9. For a final illustration, consider the set P(X) of all subsets of a set X. The relation A ~ B if and only if A s; B is a partial ordering of P(X), but not a total ordering provided that X has at least two elements. For example, if X = {1, 2, 3}, and A = {1, 2}, B = {2, 3}, then neither A ~ B nor B ~ A holds. As regards terminology, any family of sets ordered in this manner wtll be said to be ordered by inclusion. Let (A, ~ ) and (B, ~) be two partially ordered sets (when there is no danger of confusion, we write ~ for the partial orders in both A and B).

RELATIONS

293

A mapping/: A ~ B is said to be order-preserving or an order-homomorphism if for all a, bE A, a :s; b implies f(a) ::::;; f(b) in B. A one-to-one orderhomomorphism/ of the set A onto B whose inverse is also an order-homomorphism (from B onto A) is an order-isomorphism. If such a function exists, we say that the two partially ordered sets (A,::::;; ) and (B, :s; ) are order-isomorphic. When the partial order is the primary object of interest and the nature of the elements plays no essential role, order-isomorphic sets can be regarded as identical. The coming theorem emphasizes the fundamental role of our last example on partially ordered sets {Example A-9), for it allows us to represent any partially ordered set by a family of sets. Theorem A-3. Let A be a set partially ordered by the relation ::::;; . Then A is order-isomorphic to a family of subsets of A, partially ordered by inclusion. Proof. For each a E A, let ]a = {x E Alx ::::;; a}. It is not hard to verify that the mapping f: A ~ P(A) defined by f(a) = la is an order-homomorphism of A into P(A). Indeed, if a ::::;; b, then the condition x ::::;; a implies x :s; band therefore la s; 1&, or, equivalently,f{a) s; f(b). To see that f is one-to-one, suppose a, bE A are such that f(a) = f(b). Then the element a E ]a = lb, and, hence, a ::::;; b by definition of lb; likewise, b ::::;; a, from which it follows that a = b. Finally, the inverse f- 1 is also orderpreserving. For, if the inclusion la s; I& holds, then a E J& and so a ::::;; b. These calculations make it clear that A is order-isomorphic to a certain set of subsets of P(A). CoroUary. For no set A is A order-isomorphic to P(A). Proof. We argue that iff: A ~ P(A) is any order-homomorphism from A into P(A)1 then f cannot map onto P(A). For purposes of contradiction, assume thatfdoes carry A onto P(A). Define B = {aEAia¢f(a)} and B* = {c E Ale :S;; a for some a E B}. By supposition, the set B* = f(b) for some element bE A. If b ¢ B*, then, according to the definition of B, b E B s; B*, a contradiction. Hence, b E B* and so b :s; a for some a in B. From the order-preserving character off, B* = f(b) s; f(a). But then, a E B s; B* s; f(a). The implication is that a¢ B, which is again a contradiction. In an ordered set, there are sometimes elements with special properties that are worth mentioning. Definition A-6. Let S be a set partially ordered by the relation ::::;; . An element x E S is said to be a minimal (maximal) element of S if a E S and a ::s; x (x ::s; a) imply a = x.

294

FIRST COURSE IN RINGS AND IDEALS

In other words, x is a minimal (maximal) element of S if no element of S precedes (exceeds) x. It is not always the case that a partially ordered set possesses a minimal (maximal) element and, when such an element exists, there is no guarantee that it will be unique. Example A-10. The simplest illustration of a partially ordered set without minimal or maximal elements is furnished by the set R # with the ordering ~ in the usual sense. Example A-11. In the collection P(X) - { 0} of all nonempty subsets of a nonempty set S (partially ordered by set-theoretic inclusion), the minimal elements are those subsets consisting of a single element. Example A-12. Consider the set S of all integers greater than 1 and the partial order ~ defined by a ~ b if and only if a divides b. In this setting, the prime numbers serve as minimal elements. It is technically convenient to distinguish between the notion of a minimal (maximal) element and that of a first (last) element.

Definition A- 7. Let S be a set partially ordered by the relation ~ . An element x E S is called the first (last) element of S if x ~ a (a ~ x) for all a E S. Let us point out immediately the important distinction between first (last) elements and minimal (maximal) elements. Definition A-7 asserts that the first (last) element of a partially ordered set S must be comparable to every element of S. On the other hand, as Definition A-6 implies, it is not required that a minimal (maximal) element be comparable to every element of S, only that there be no element in S which precedes (exceeds) it. A minimal (maximal) element has no predecessors (successors), whereas a first (last) element precedes (succeeds) every element. Clearly, any first (last) element is a minimal (maximal) element, but not conversely. First (last) elements of partially ordered sets are unique, if they exist at all. Indeed, suppose that the partially ordered set (S, ~) has two first elements, say x and y; then, x ~ y and y ~ x, so that x = y by the antisymmetric property. Thus, x is unique and we are justified in using the definite article when referring to the first element of S. A similar argument holds for last elements. Let us introduce some additional terminology pertaining to partially ordered sets. Definition A-8. Let S be a set partially ordered by the relation ~ and let A be a subset of S. An element xES is said to be a lower (upper) bound for A if x ::s; a (a ~ x) for all a e A.

RELATIONS

295

We emphasize that a lower (upper) bound for a subset A of a partially ordered set is not required to belong to A itself. If A happens to have a first (last) element, then the same element is a lower (upper) bound for A; conversely, if a lower (upper) bound for A is contained in the set A, then it serves as the first (last) element for A. Notice, too, that a lower (upper) bound for A is a lower (upper) bound for any subset of A. A subset of a partially ordered set need not have upper or lower bounds (just consider Z .:;; R # with respect to :::;;; ) or it may have many. For an example of this latter situation, one may turn to the family P(X) of all subsets of a set X, with the order being given by the inclusion relation; an upper bound for a subfamily .xl .:;; P(X) is any set containing u .xl, while a lower bound is any set contained in n d.

APPENDIX B

ZORN'S LEMMA

In this Appendix, we give a brief account of some of the axioms of set theory, with the primary purpose of introducing Zorn's Lemma. Our presentation is descriptive and most of the facts are merely stated. The reader who is not content with this bird's-eye view should consult [12] for the details. As we know, a given partially ordered set need not have a first element and, if it does, some subset could very well fail to possess one. This prompts the following definition: a partially ordered set (S, ::::;; ) is said to be wellordered if every nonempty subset A £:; S has a first element ("with respect to ::::;; " being understood). The set Z + is well-ordered by the usual ::::;; ; each nonempty subset has a first element, namely, the integer of smallest magnitude in the set. Notice that any well-ordered set (S,::::;; ) is in fact totally ordered. For, each subset {a, b} £:; S must have a first element. According as the first element is a or b, we see that a ::::;; b or b ::::;; a, whence the two elements a and b are comparable. Going in the other direction, any total ordering of a finite set is a well-ordering of that set. Let it also be remarked that a subset of a well-ordered set is again well-ordered (by the restriction of the ordering). Example B-1. Consider the Cartesian product S = Z + x Z +· We partially order S as follows: if (a, b) and (a', b') are ordered pairs of positive integers, (a, b) ::::;; (a', b') means that (l) a < a' (in the usual sense) or (2) a = a' and b ::::;; b'. (This is called the lexicographic order of Z + x Z +• because of its resemblance to the way words are arranged in a dictionary.) For instance, (4, 8) ::::;; (5, 2), while (3, 5) ::::;; (3, 9). To confirm that ::::;; is a well-ordering of S, let 0 =I= A £:; S and define B = {a E Z +I (a, b) E A}. Since A is a nonempty subset of Z+, it has a first element, call it a 0 . Now, let C = {bEZ+i(a 0 ,b)EA}. Again, the well-ordering of Z+ under ::::;; guarantees that C has a first element, say b0 • We leave it to the reader to convince himself that the pair (a 0 , b0 ) serves as the first element of A, thereby making S a well-ordered set relative to ::::;; . 296

ZORN'S LEMMA

297

A fundamental axiom of set theory, which has a surprising variety of logically equivalent formulations, is the so-called Well-Ordering Theorem of Zermelo (1904). The designation "theorem" notwithstanding, we take thrs to be an axiom (assumed and unproven) of our system. We state: Zermelo's Theorem. Any set S can be well-ordered; that is, there is a partial ordering ~ for S such that (S, ~)is a well-ordered set.

Accepting the existence of such orderings, we do not pretend at all to be able to specify them. Indeed, nobody has ever "constructed" an explicit function that well-orders an uncountable set. Moreover, the promised well-ordering may bear no relation to any other ordering that the given set may already possess; the well-ordering of R #, for instance, cannot coincide with its customary ordering. Zermelo based the "proof" of his classical Well-Ordering Theorem on a seemingly innocent property whose validity had never been questioned and which has since become known as the axiom of choice. To state this axiom, we first need the definition of a choice function. Definition B-l. Let rc be a (nonempty) collection of nonempty sets. A function f: rc -+ u rc is called a choice function for rc if /(A) E A for every set A in tC.

Informally, a choice function f can be thought of as "selecting" from each set A E rc a certain representative element f(A) of that set. As a simple illustration, there are two distinct choice functions / 1 and / 2 for the family of nonempty supsets of {1, 2}:

ft({1, 2}) = 1,

ft({l}) = 1,

ft({2}) = 2,

/2({1,2}) = 2,

/ 2 ({1}) =

/ 2 ({2}) =

1,

2.

The question arises whether this selection process can actually be carried out when rc has infinitely many members. The possibility of making such choices is handled by the axiom mentioned above: Axiom of Choice. Every collection ~ of nonempty sets has at least one

choice function. Since this general principle of choice has a way of slipping into proofs unnoticed, the reader should become familiar with its disguised forms. For instance, one often encounters the following wording: if {X;} is a family of nonempty sets indexed by the nonempty set J, then the Cartesian product x ieJ X; is nonempty (it should be clear that the elements of X X; are precisely the choice functions for {X;}). For another common phrasing, which again expresses the idea of selection, let rc be a collection of disjoint, nonempty sets. The axiom of choice, as we have stated it, is equivalent to

298

FIRST COURSE IN RINGS AND IDEALS

asserting the existence of a set S with the property that A n S contains exactly one element, for each A in 1{/. Granting Zermelo's Theorem, it is clear that a choice function f can be defined for any collection Cfi of nonempty sets: having well-ordered u Cfi, simply take f to be the function which assigns to each set A in '6' its first element. As indicated earlier, Zermelo's Theorem was originally derived from the axiom of choice, so that these are in reality two equivalent principles (although seemingly quite different). Although the axiom of choice may strike the reader as being intuitively obvious, the sou:tdness of this principle has aroused more philosophical discussion than any other single question in the foundations of mathematics. At the heart of the controversy is the ancient problem of existence. Some mathematicians believe that a set exists only if each of its elements can be designated specifically, or at the very least if there is a rule by which each of its members can be constructed. A more liberal school of thought is that an axiom about existence of sets may be used if it does not lead to a contradiction. In 1938 Godel demonstrated that the axiom of choice is not in contradiction with the other generally accepted axioms of set theory (assuming that the latter are consistent with one another). It was subsequently established by Cohen (1963) that the denial of this axiom is also consistent with the rest of set theory. Thus, the axiom of choice is in fact an independent axiom, whose use or rejection is a matter of personal inclination. The feeling among most mathematicians today is that the axiom of choice is harmless in principle and indispensable in practice (provided that one calls attention to the occasions of its use). It is also valuable as an heuristic tool, since every proof by means of this assumption represents a result for which we can then seek proofs along other lines. A non-constructive criterion for the existence of maximal elements is given by the so-called "maximality principle", which generally is cited in the literature under the name Zorn's Lemma. (From the point of view of priority, this principle goes back to Hausdorff and Kuratowski, but Zorn gave a formulation of it which is particularly suitable to algebra; he was also the first to state, without proof, that a maximality principle implies the axiom of choice.) Let S be a nonempty set partially ordered by ~ . Suppose that every subset A s; S which is totally ordered by ~ has an upper bound (in S). Then S possesses at least one maximal element.

Zorn's Lemma.

Zorn's Lemma is a particularly handy tool when the underlying set is partially ordered and the required object of interest is characterized by maximality. To demonstrate how it is used in practice, let us prove what is sometimes known as Hausdorff's Theorem (recall that by a chain is meant a totally ordered set):

ZORN'S LEMMA

299

Theorem B-1. Every partially ordered set contains a maximal chain; that is, a chain which is not a proper subset of any other chain. Proof Consider the collection t(J of all chains of a partially ordered set (S, : : ;:; ) ; rr5 is nonempty, since it contains the chains consisting of single elements of S. Partially order t(J by inclusion and let d be any chain of t(J (for the ordering .:;;). We maintain that the union u d belongs to t(J. Given elements a, bE u d, we have a E A Ed and bE BEd, for some A, B. As d is a chain, either A .:;; B or B .:;; A; suppose, for convenience, that A .:;; B. Then a, b both lie in B and, since B is itself a chain (in S), it follows that a : : ;:; b or b : : ;:; a. In consequence, any two elements of u d are comparable, making u d a chain inS. Since u dis clearly an upper bound ford in t(J, Zorn's Lemma implies that (t(J, o:;;) has a maximal member.

As another brief application of Zorn's Lemma, consider the following assertion: if (S, :s;) is a partially ordered set every chain of which has an upper bound, then for each a E S there exists a maximal element x E S with the property that a :s; x; in other words, there exists a maximal element larger than the given element. For a proof of this, first observe that the set Ja = {y E Sla :s; y} satisfies the hypotheses of Zorn's Lemma (under the restriction of $); hence, has a maximal element x. But x is maximal in S, not merely in Ja. For, suppose that s E S with x :s; s. Then a :s; s (since both a $ x and x :s; s) and so s E Ja. From the maximality of x in Ja, it then follows that s = x, completing the argument. Needless to say, we could just as well have phrased Zorn's Lemma in terms of lower bounds and minimal elements. The assertion in this case is that there exists at least one minimal element inS. Before concluding, let us state Theorem B-2. Zermelo's Well-Ordering Theorem, the Axiom of Choice and Zorn's Lemma are all equivalent. The deduction of these equivalences is somewhat involved and the argument is not presented here; the interested reader can find the proofs in any number of texts on set theory.

BIBLIOGRAPHY

GENERAL REFERENCES

Our purpose here is to present a list of suggestions for collateral reading and further study. The specialized sources will carry the reader considerably beyond the point attained in the final pages of this work. I. ADAMSON, 1., Introduction to Field Theory. New York: Interscience. 1964. 2. ARTIN, E., Galois Theory, 2nd Ed. Notre Dame. Ind.· Umversity of Notre Dame Press. 1955. 3. ARTIN, E., C. NESBITT, and R. THRALL, Rings with Minimum Condition. Ann Arbor. Mich.: University of Michigan Press, I944. 4. AuSLANDER, M., Rings, Modules and Homology, Chapters I and II. Waltham. Mass.: Department of Mathematics, Brandeis University (lecture notes). 1960. 5. BARNES, W., Introduction to Abstract Algebra. Boston: Heath. 1963. 6. BoURBAKI, N., Algebra, Chapter 8. Paris: Hermann, 1958. 7. BouRBAKI, N., Algebra Commutative, Chapters 2. 4 and 5. Pans: Hermann. 1961. 8. BuRTON, D., Introduction to Modern Abstract Algebra. Readmg. Mass: Addison-Wesley, 1967. 9. CURTIS, C. and I. REINER, Representation Theory of Finite Groups and Associatil•e Algebras. New York: Interscience, 1962. 10. DIVINSKY, N., Rings and Radicals. Toronto: University of Toronto Press. 1965. 1I. GoLDIE, A., Rings with Maximum Condition. New Haven: Department of Mathematics, Yale University (lecture notes), 1961. 12. HALMos, P., Naive Set Theory. Princeton: Van Nostrand, 1960. 13. HERSTEIN, I. N., Topics in Algebra. New York: Blaisdell, 1964. 14. HERSTEIN, I. N., Theory of Rings. Chicago: Department of Mathematics. University of Chicago (lecture notes), 196 I. I5. HERSTEIN, I. N., Noncommutative Rings. (Carus Monographs). Menascha, Wis.: Mathematical Association of America, I 968. 16. HEWITT, E. and K. STROMBERG, Real and Abstract Analysis. New York: SpringerVerlag, 1965. 17. JACOBSON, N., Lectures in Abstract Algebra, Vol. I, Basic Concepts. Princeton: Van Nostrand, 1951.

300

PROBLEMS

109

of an irreducible element and of a nonzero prime do not always coincide in an arbitrary integral domai~Specifically, we have (2 + J- 5)13 · 3~t (2 + J- 5){ 3, so that 2 + J- 5 cannot be a prime element of Z(,j- 5). PROBLEMS

1. Let R be a commutative ring with identity and the elements a, e E R, with e2 = e. Prove that a) If (a) = (e), then a and e are associates. [Hint: a = (1 - e + a)e.] b) If for some n E Z + the elements a• and e are associates, then am and e are associates for all m 2 n. 2. Given that/, J, and K are ideals of a principal ideal domain R, denve the following relations: a) If I = (a) and J = (b), then IJ = (ab); in particular,/" = (a"). b) I(J n K) c) I + (J n d) I n (J + e) lJ = I n

n JK. K) = (I + J) n (I + K). K) = (I n J) + (I n K). J if and only if I + J = R for all nonzero I, J.

= lJ

3. Suppose that R = R 1 Ef) R 2 Ee ··· Ee R., where each R; ts a principal ideal ring. Verify that R is also a principal ideal ring. 4. Let R be an integral domain having the gcd-property. Assuming that equality holds to within associates, prove that, for nonzero a, b, c E R, a) gcd (a, gcd (b, c)) = gcd (gcd (a, b), c). b) gcd (a, 1) = 1. c) gcd (ca, cb) = c gcd (a, b); in particular, gcd (c, cb) = 1. d) if gcd (a, b) = 1 and gcd (a, c) = 1, then gcd (a, be) = 1. e) if gcd (a, b) = 1, ale and blc, then able. t) gcd (a, b) !em (a, b) = ab. [Hint: Theorem 6-6.] 5. If R is an integral domain having the gcd-property, show that a nonzero element of R is prime if and only if it is irreducible. 6. In a principal ideal domain R, establish that the primary ideals are the two trivial ideals and ideals of the form (p"), where p is a prime element of R and n E Z+. [Hint: If I is primary, then .fi = (p) for some prime element p. Choose n E z+ such that I!;;;;; (p"), but I$ (p"+ 1), and show that I= (p").] 7. If R is a principal ideal doman, we define the length ..l(a) for each nonzero a E R as follows: if a is invertible, then A(a) = 0; otherwise, A(a) is the number of primes (not necessarily distinct) in any factorization of a. Prove the following assertions: a) the length of a is well-defined; b) if alb, then A(a) ~ A(b); c) if alb and A{a) = A(b), then bla; d) if a ~ b and b ~ a, then there exist nonzero p, q E R such that A(pa

+ qb)

~

min {A{a), A(b)}.

302

BIBLIOGRAPHY

46. KovAcs, L., "A Note on Regular Rings," Pub!. Math. Debrecen 4, 465--468 (1956). 47. LUH, J., "On the CommutatlVlty of J-Rings," Canad. J. Math. 19, 1289-1292 (1967). 48. McCoY, N., "Subdirectly Irreducible Commutative Rings," Duke Math. J.12, 381-387 (1945). 49. McCoY, N., "Subdirect Sums of Rings," Bull. Am. Math. Soc. 53, 856-877 (1947). 50. McCOY, N., "A Note on Finite Umons of Ideals and Subgroups," Proc. Am. Math. Soc. 8, 633-637 (1957). 51. NAGATA, M., "On the Theory of Radicals in a Ring," J. Math. Soc. Japan 3, 33~344 (1951). 52. VON, NEUMANN, J., "On Regular Rings," Proc. Nat{. A cad Sci. U.S. 22, 707-713 (1936). 53. NoRTHCOTT, D., "A Note on the Intersection Theorem for Ideals," Proc. Cambridge Phil. Soc. 48, 366-367 (1952). 54. PERLIS, S. "A Characterization of the Radical of an Algebra," Bull. Am. Math. Soc. 48, 128-132 (1942). 55. SAMUEL, P., "On Unique Factorization Domains," Illinois J. Math. 5, 1-17 (1961). 56. SATYANARAYANA, M., "Rings with Primary Ideals as Maximal Ideals." Math. Scand. 20, 52-54 (1967). 57. SATYANARAYANA, M., "Characterization ofLocal Rings," Tohoku Math. J. 19,411--416 (1967). 58. SNAPPER, E., "Completely Primary Rings, I," Ann. Math. 52, 666-693 (1950). 59. STONE, M. H., "The Theory of Representations of Boolean Algebras," Trans. Am. Math. Soc. 40,37-111 (1936).

INDEX OF SPECIAL SYMBOLS

The following is by no means a complete list of all the symbols used in the text, but is rather a listing of certain symbols which occur frequently. Numbers refer to the page where the symbol in question is first found.

{a}

[a] a+l (a)

aR annS alb, a[b a= b (mod n) ajb aob AAB AxB A(M)

c C(a)

cent R char R contf(x) degf(x) f(A)

rF(a) 1

F[a] [F':F]

gcd (a,b) GF{p") hom(R,R')

set consisting of the element a, 8 congruence class determined by the element a, 4 coset of the ideal J, 39 smallest (two-sided) ideal containing the element a, 19 smallest right ideal of R containing the element a, 19 annihilator of the set S, 36 a divides (does not divide) the element b, 90 integer a is a congruent to integer b modulo n, 4 formal fraction of elements a and b, 61 circle-product of elements a and b, 171 symmetric difference of sets A and B, 3 Cartesian product of sets A and B, 9 annihilator of the module M, 275 field of complex numbers, 53 centralizer of the element a, 14 center of the ring R, 9 characteristic of the ring R, 11 content of the polynomial.f{x), 129 degree of the polynomial.f{x), 119 direct image of the set A under f, 27 inverse image of the set A under f, 27 field generated by the element a over F, 137 set of polynomials in the element a, 120 degree of the field F' over the subfield F, 140 greatest common divisor of the elements a and b, 92 Galois field with p" elements, 191 set of ring homomorphisms from R into R', 26 303

304

INDEX OF SPECIAL SYMBOLS

homR(M,M') IJ I+J lff:JJ (J:J) I:-1;

Jl

J(R) kerf l(M) lcm (a,b) Mn(R)

map(X,R) nat 1 ordj{x)

0 (n)

, P(X)

Q Q(J'z) Qc~(R)

R#

R* R" R[x]

R[x,y] R[[x]] R/1 rad R Rad R L' fB R;

LfBR; Z,Ze

z+ Zl Z(i)

z,. Z(J'z)

+n, ·,.

set of R-module homomorphisms from M into M', 272 product of the ideals I and J, 22 sum of the ideals I and J, 21 internal direct sum of the ideals I and J, 21 quotient of the ideal I by the ideal J, 23 sum of a set of ideals / 1, 21 nil radical of the ideal/, 79 J-radical of the ring R, 172 kernel of the homomorphism f, 28 length of the module M, 252 least common multiple of the elements a and b, 94 ring of n x n matrices over R, 3 ring of mappings from X into R, 4 natural mapping determined by the ideal/, 40 order of the power series j{x), 115 the empty set, 3 Euler phi-function, 57 substitution homomorphism induced by the element r, 120 power set of the set X, 3 field of rational numbers, 2 quadratic number field, 105 classical ring of quotients of R, 60 field of real numbers, 2 set of invertible elements of R, 2 heart of the ring R, 212 polynomial ring in one indeterminant over R, 118 polynomial ring in two indeterminants over R, 134 power series ring in one indeterminant over R, 114 quotient ring of R by the ideal I, 40 Jacobson radical of R, 157 prime radical of R. 163 subdirect sum of a set of rings R;, 206 complete direct sum of a set of rings R 1, 204 ring of integers (even integers), 2, 9 set of positive integers, 12 ring of multiples of the identity element, 12 domain of Gaussian integers, 91 ring of integers modulo n, 4 a domain of quadratic integers, 106 addition (multiplication) modulo n, 5 is isomorphic to, 29

INDEX

additive group of a ring, adeal, 38 adjunction (of an element to a field), 137 algebraic element, 138 extension field, 140 number field, 155 algebraically closed field, 156 annihilator of a subset, 36 Artinian ring, 223 ascending chain condition, 217 associated elements, 91 prime ideal of a primary ideal, 81 prime ideal in a Noetherian ring, 236 atom in a Boolean ring, 200 automorphism, 25 axiom of choice, 297 Bezout identity, 93 binomial equation, 13 Boolean ring, 14 cancellation law, 7 center of a ring, 9 centralizer, of an element, 194 of a set of endomorphisms, 277 chain (in a partially ordered set), 292 chain conditions, 217, 223 characteristic of a ring, 11 choice function, 297 classical ring of quotients, 60 coefficients of a power series, 114 comaximal ideals, 211

comparable elements, 292 component rings (in a direct sum), 204 component projection, 206 commutative diagram, 43 commutative ring, 2 complete direct sum, 204 composition series, 251 congruence modulo n, 4 congruence class, 4 representation of, 4 conjugate of an element, 105 content of a polynomial, 129 Correspondence Theorem, 30 coset of an ideal, 39 degree, of an extension field, 140 of a polynomial, 119 derivative function, 153 descending chain condition, 223 direct sum, complete, 204 discrete, 205 external, 33 internal, 21 of modules, 259 direct summand, 34 divides (divisor), 90 division ring, 52 finite, 194 divisor of zero, 7 domain, Euclidean, 102 integral, 7 principal ideal, 20 unique factorization, 100

305

INDEX

element(s), algebraic. 138 associate, 91 conjugate, 105 idempotent, 14 identity, 2 invertible, 2 irreducible, 97 nilpotent, 14 prime, 97 quasi-regular, 170 related to an ideal, 258 relatively prime, 93 transcendental, I 38 torsion, 259 zero, 1 Eisenstein irreducibility criterion, I 33 endomorphism of a module, 272 of a ring, 25 evaluation homorphism, 26 equivalence class, 288 relation, 287 Euler phi-function, 57 Euclidean domain, 102 valuation, 102 extension, algebraic, 140 simple, 137 extension ring, 31 faithful module, 275 field, 52 algebraically closed, 156 extension, 136 Galois, 191 obtained by adjoining an element, of algebraic numbers, 155 of complex numbers, 53 of quadratic numbers, 1OS of rational functions, 138 skew, 52 splitting, 148 finite division ring, 194 integral domain, 56 field, 187 ring, 2 finitely generated, 19 first element, 294

137

306

fixed field, 69 formal fraction, 61 formal power series, 112 Frobenius automorphism, 202 Fundamental Homorphtsm Theorem, 44 Fundamental Theorem of Algebra, 128 Galois field, 191 Gaussian integers, 91 gcd-property, 95 generators (of an ideal), 19 greatest common divisor, 92 group of invertible elements (of a ring), group ring, 15 H-ring, 203 heart of a ring, 212 Hilbert ring, 178 homomorphism. 25 evaluation, 26 kernel of, 28 of modules, 250 of partially ordered sets, ofrings, 25 reduction. 130 substitution, 120 trivial, 26 homomorphic image, 25 associated prime, 236 comaximal. 211 commutator, 50 finitely generated, 19 ideal. 16 irreducible, 235, left (right), 16 maximal, 71 minimal, 86 minimal prime, 84 modular, 173 modular maximal, 174 nil, 47 nilpotent, 47 primary, 81 prime, 76 product of, 22 quotient, 23 regular, 179 sum of, 20

293

2

307

INDEX

semiprime, 80 idempotent Boolean ring, 200 element, 14 orthogonal, 268 primitive, 270 imbedding, 31 induced partial order, 292 irreducible element, 97 ideal, 235 polynomial, 126 irredundant primary representaion, subdirect sum, 214 isomorphism of modules, 250 of partially ordered sets, 293 ofrings, 25 J-radical, 172 J-ring, 196 Jacobson radical,

236

157

kernel of a homomorphism,

28

prime ideal of a ring, 84 minimum condition, 223 polynomial, 139 modular ideal, 173 module, 247 annihilator of, 275 centralizer of, 272 direct sum, 259 dual, 286 endomorphism of, 272 faithful, 275 homomorphism of, 250 indecomposable, 260 isomorphism, 250 quotient, 249 simple, 249 submodule, 249 torsion-free, 259 monic polynomial, 119 multiplicatively closed set, 70 multiplicative semigroup of a ring,

last element, 294 !em-property, 95 least common multiple, 94 left annihilator, 36 ideal, 16 lemma, Fitting's, 256 Gauss', 130 Nakayama's, 243 Schur's, 274 length of an element, 109 of a normal series, 252 of a module, 252 lexicographic order, 296 lifting idempotents, 167 local ring, 88 localization, 88 lower bound (for a partially ordered set), 294

order homomorphism, 293 order isomorphic, 293 order of a power series, 115 Ore condition, 69 orthogonal idempotents, 268

maximal element, 293 ideal, 71 maximum condition, 218 minimal element, 293 ideal, 86 prime ideal of an ideal, 84

partial order, 291 partition, 290 polynomial, 118 content, 129 cyclotomic, 133

natural mapping, 41 nil ideal, 47 nilpotent element, 14 ideal, 47 nil radical of an ideal, 79 of a ring, 79 nil-semisimple ring, 264 Noetherian ring, 219 nontrivial subring, 8 subdirect sum, 206 non-zero-divisor, 60 norm, 105 normal series, 251

INDEX

degree of, 119 function, 153 in two indeterminants, 134 irreducible, 126 leading coefficient, 119 minimum, 139 monic, 119 primitive, 129 root of, 121 primary component, 236 ideal, 81 representation, 236 ring, 169 prime element, 97 field, 65 ideal, 76 radical, 163 primitive idempotent, 270 ideal, 286 polynomial, 129 ring, 278 principal ideal, 19 ideal ring, 20 proper subring, 8 pseudo-inverse, 25 quadratic number field, quasi-inverse, 170 quasi-regular, 170 quaternions,. 54 quotient ideal, 23 field of, 60 module, 249 ring, 40

105

radical, J-, 172 Jacobson, 157 nil, 79 prime, 163 rational function, 138 reduction homomorphism, 130 regular ring, 24 relation (binary), 287 antisymmetric, 291 associated with a function, 288 compatible equivalence, 49 congruence modulo n, 4

308

defined by a partition, 291 equivalence, 287 partial order, 291 reflexive. 287 symmetric, 287 transitive, 287 relatively prime elements, 93 Remainder Theorem. 123 refinement of a normal series, 251 ring, I Artinian. 223 Boolean. 14 commutative. 2 divisible, 233 divisJOn, 52 finite, 2 H-.

203

H1lbert, 178 J-, 196 local, 88 nil-semisimple, 264 Noetherian, 219 of endomorphisms of a module, 272 of extended power series over R, 152 of formal power series over R, 113 of functions between a set and ring, 4 of integers modulo n. 5 of polynomials over R, 118 of matrices over R, 3 primary, 169 quotient, 40 regular, 24 right Artinian, 262 semisimple, 157 simple, 18 subdirectly irreducible, 211 with identity, 2 without radical, 157, 163 zero, 13 ring of quotients, classical, 60 generalized, 70 relative to a set, 70 root of a polynomial, 121 multiple, 153 saturated, 178 semisimple ring,

157

309

INDEX

series, composition, 251 equivalent, 252 normal, 251 length of, 252 refinement of, 251 simple extension field, 137 simple module, 249 ring,. 18 skew field, 52 splitting field, 148 square-free integer, 105 subdirectly irreducible, 211 subdirect sum, 206 subfield, 59 submodule, 249 subring, 8 generated by a set, 14 proper, 8 trivial, 8 substitution homomorphism, in a polynomial, 120 symmetric difference, 3 Theorem, Akizuki-Hopkins, 255 Birkhoff's, 212 Brauer's, 262 Chinese Remainder, 211 Cohen's, 241 Dorroh Extension, 31 Euler-Fermat, 58 Euclid's, 101 Fermat's little, 68 Hausdorff's, 299 Herstein's, 199

120

Hilbert Basis, 220 Jordan-Holder, 252 Jacobson's. 198 Jacobson Density, 285 Kronecker's, 144 Krull Intersection, 244 Krull-Zorn, 74 Levitski's, 222 McCoy's, 212 Noether's, 236 Stone Representation, 183 Wilson's, 188 Wedderburn's, 194, 266, 230 Wedderburn-Arlin, 281 Zermelo's, 297 total ordering, 292 torston element, 259 torsion-free module, 259 transcendental element (over a field}, 138 trivial subring, 8 homomorphism, 26 unique factorization domain, I00 upper bound (for a partially ordered set), 294 valuation ring,

88

well-ordered set, 296 without radical, 157, 163 zero divisor, 7 zero element of a ring, zero ring, 13 Zorn's lemma, 298

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