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English Pages 317 Year 1970
A First Course in RINGS AND IDEALS
DAVID M. BURTON University of New Hampshire
A VV ADDISONWESLEY PUBLISHING COMPANY Reading, Massachusetts  Menlo Park, California ' London  Don Mills, Ontario
This book is in the ADDISONWESLEY SERIES IN MATHEMATICS Consulting Editor: Lynn H. Loomis
Standard Book Number 201.00731.2 AMS 1968 Subject Classiﬁcations 1610, 1620.
Copyright ©1970 by AddisonWesley Publishing Company, Inc. by Addison’Wesley Publishing Company, Inc.
Philippines copyright 1970
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, electronic, mechanical, photocopying, recording,
or otherwise, without the prior written permission of the publisher. Printed in the United States of America. Published simultaneously in Canada. Library of Congress Catalog Card No. 73100855.
To my Father Frank Howard Burton
PREFACE As the title suggests, this volume is designed to serve as an introduction to the basic ideas and techniques of ring theory; it is intended to be an expository textbook, rather than a treatise on the subject. The mathematical background required for a proper understanding of the contents is not extensive. We assume that the average reader has had some prior contact with abstract algebra, but is still relatively inexperienced in this respect. In consequence, nearly everything herein can be read by a person familiar with basic grouptheoretic concepts and having a nodding acquaintance with linear algebra. The level of material should prove suitable for advanced undergraduates and beginning graduate students. Indeed, a builtin ﬂexibility permits the book to be used, either as the basic text or to be read independently by interested students, in a variety of situations. The reader whose main interest is in ideal theory, for instance, could chart a course through Chapters 2, 3, 5, 8, ll, 12, 13. Taken as a whole, the present work is more nearly a begin
ning than an end. Our hope is that it may serve as a natural point of departure for the study of the advanced treatises on ring theory and, in some aspects of the subject, the periodical literature. As regards treatment, our guiding principle is the strong conviction that intelligibility should be given priority over coverage; that a deeper understanding of a few important topics is preferable to a superﬁcial knowledge of many. This calls for a presentation in which the pace is unhurried and which is complete in the details of proof, particularly of basic results. By adhering to the “theoremproof” style of writing, we hope to achieve greater clarity (perhaps at the sacriﬁce of elegance). Apart from the general knowledge presupposed, an attempt has been made to keep the text technically selfcontained, even to the extent of including some material which is undoubtedly familiar. The mathematically sophisticated reader may prefer
to skip the earlier chapters and refer to them only if the need arises. At the end of each chapter, there will be found a collection of problems of varying degrees of difﬁculty. These constitute an integral part of the book and require the reader’s active participation. They introduce a variety V
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, Delmon Grapes, Cynthia Kennett, Kenneth Lidman, Roy Morell, Brenda Phalen, David Smith, and John Sundstrom; we valued their critical reading of sections of the manu
script 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 Bechtel], 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 eagleeyed 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 AddisonWesley 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_.
Chapter Chapter
Chapter Chapter
Chapter Chapter
WGNIQUIﬁUNlI
CONTENTS Introductory Concepts . Ideals and Their Operations
16
The Classical Isomorphism Theorems .
39
Integral Domains and Fields
52
Maximal, Prime, and Primary Ideals .
71
Divisibility Theory in Integral Domains .
90
Polynomial Rings ‘ .
1 12
Certain Radicals of a Ring .
157
Two Classic Theorems . . .
180
Chapter 10
Direct Sums of Rings
204
Chapter 11
Rings with Chain Conditions
217
Chapter 12
Further Results on Noetherian Rings .
234
Chapter 13
Some Noncommutative Theory
262
Appendix A.
Relations .
287
Appendix B.
Zom’s Lemma
296
Chapter
Chapter Chapter
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, 6, u, n, —, and Q for the empty set. In particular, A — B = {xxeA and x¢B}. As regards inclusion, the symbols E 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 {Ali G I}, the cumbersome notations
u {AilieI} and n {AiieI} 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. Following custom, {a} denotes the set whose only member is a. Provided that there is no risk of confusion, a oneelement 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 casef: X —> Y, and the notation always signiﬁes that f has domain X. Under these circumstances, f 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 f(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 fIA from A into Y deﬁned by (fA)(x) = f(x) for all x in A. For the composition of two functions f: X > Yand g: Y—> Z, we will write g of; that is, g of: X —> Z satisﬁes
(g o f)(x) = g(f(x)) for each x e X. (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 in. 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.
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. Deﬁnition 11. A ring is an ordered triple (R, +, ) consisting of a nonempty set R and two binary operations + and  deﬁned on R such that 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 that multiplication is to be performed before addition. Accordingly, the expression ab + c stands for (ab) + c and not for a(b + c). Because of the general associative law, parentheses l
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) a + b = b + a, 2)(a+b)+c=a+(b+c), 3) there exists an element 0 in R such that a + 0 = a for every a e R, 4) for each a e R, there existsan element —a e R such that a + (—a) = 0, 5) (ab)c = a(bc), and 6) a(b + c) = ab + we and (b + c)'a = ba + ca,
where it is understood that a, b, 0 represent arbitrary elements of R. A ring (R, +, ') is said to be a ﬁnite 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 12. 1) A commutative ring is a ring (R, +, ') in which multiplication is a cormnutative operation . ab = b'a for all a, b e R. (In case ab = 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 ofmultiplication, normally represented by the symbol 1, so that a'l = la = 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 (twosided) inverse with
respect to multiplication. The multiplicative inverse of a is unique, whenever it exists, and will be denoted by a‘ 1, so that aa‘ 1 = a‘ 1a = 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 —I belong to R*. (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 11. If Z, Q, R’“ denote the sets of integers, rational, and real numbers, respectively, then the systems (Z: +9 ')s
(Q: +a ')r
(R#, +a )
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
Example 12
3
Let X be a given set and P(X) be the collection of all subsets
of X. The symmetric diﬂ'erence of two subsets A, B S X is the set A A B, where
AAB= (A —B)u(B— A). If we deﬁne addition and multiplication in P(X) by A+B=AAB,
A~B=AnB,
then the system (P(X), +, ) forms a commutative ring with identity. The empty setZ 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 13. Given a ring (R, +, ), we may consider the set Mn(R) of n x n matrices over R. If In = {1, 2, , n}, a typical member of Mn(R) is a function f: I,I x In a R. In practice, one identiﬁes such a function with its values ail. = f(i, j), which are displayed as the n x n rectangular array a11 a1" 5 5 (at) e R)
a,,1...a,,n For the sake of simplicity, let us hereafter abbreviate the n x n matrix
whose (i, j) entry is a“. to (au).
The operations required to make (Mn(R), +, ) a ring are provided by the familiar formulas
and
(aij) + (bij) = (aij + bij):
(aij)'(bij) = (Caj),
where l
Ci] :
z aik'bkj. k=l
(We shall often indulge in this harmless duplication of symbols whereby + and  are used with two diﬂ'erent meanings.) The zero element of the resulting ring is the n x n matrix all of whose entries are 0; and
—(a,j) = (—11”). The ring (Mn(R), +, )fails to be commutative forn > 1. It is equally easy to show that if (R, +, ) has an identity element 1, then the matrix with l’s down the main diagonal (that is, a” = 1) and 0’s elsewhere will act as identity for matrix multiplication. In terms of the Kronecker delta symbol 6ija which is deﬁned by
_ {Oifi lifi=j #1.
6,1 —
.._
(19] _ 1923'3n)a
the identity matrix can be written concisely as (60).
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,
map(X, R) = {ff:X —> R}. (For ease of notation, let us also agree to write map R in place of map (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 of f and g, denoted by f + g and f g, respectively, are the functions which satisfy
(f + 9W) = f(x) + 906),
(f'g)(x) = f(X)'g(x),
(x 6 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 15, 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 a b (mod n), if and only if the difference a — b is divisible by n; in other words, a E 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] = {elx E 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 2,.
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, reZ, with 0 s r < n, such that a = qn + r. By the deﬁnition of congruence, a E r (mod n), Or
INTRODUCTORY CONCEPTS
5
equivalently, [a] = [r]. Thus, there are at most n different congruence classes in Z", namely, [0], [1], , [n — 1]. But these n classes are themselvesdistinct. ForifO S b < a < n,then0 < a — b < nandsoa — b
cannot be divisible by n, whence [a] 76 [b]. Accordingly, Z,l consists of exactly n elements:
2. = {[0]. [1]. , [n — 11}. The reader should keep in mind that each congruence class listed above is determined by any one of its members; all we have done is to represent the class by its smallest nonnegative representative. Our next step is to deﬁne the manner in which the members of Z" are to be added and multiplied, so that the resulting system will form a ring.
The deﬁnitions are as follows: for each [a], [b] e Z",
[a] +.. [b] = [a + b],
[(03.0)] = [ab]
In other words, the sum and product of two congruence classes [a] and [b] are the unique members of Z,I which contain the ordinary sum (1 + b and ordinary product ab, respectively. Before considering the algebraic properties of these operations, it is necessary to make certain that they are welldeﬁned and do not depend upon which representatives of the congruence classes are chosen. In regard to multiplication, for instance, we want to satisfy
ourselves that if [a'] = [a] and [b’] = [b], then [a’],, [b’] = [a] , [b], or, rather, that [a’b’] = [ab]. Now, a’e [a’] = [a] and b’e [b’] = [b], which signiﬁes that a’ = a + kn and b’ = b + jn for some k, j eZ. But then
a’b’ = (a + kn)(b + jn) = ab + (aj + bk + kjn)n.
Hence, a’b’ E ab (mod n) and so [a’b’] = [ab], as desired. The proof that addition is unambiguously deﬁned proceeds similarly. We omit the detailed veriﬁcation of the fact that (2,, +,,, ,,) is a commutative ring with identity (traditionally known as the ring of integers modulo n), remarking only that the various ring axioms hold in Z,' simply because they hold in Z. The distributive law, for instance, follows in Z,l
from its validity in Z:
[a]'..([b] +.. [6]) = [a]'..[b + c] = [0(1) + c)] = [ab + ac] = [ab] +..[¢w] = [a]°..[b] +. [a]'..[CJ~ Notice, too, that the congruence classes [0] and [I] serve as the zero element and multiplicative identity, respectively, whereas [—a] is the additive
inverse of [a] in Z". When no confusion is likely, we shall leave off the brackets from the elements of Zn, thereby making no genuine distinction
6
FIRST COURSE IN RINGS AND IDEALS
between a congruence class and its smallest nonnegative representative; under this convention, Z,I = {0, l, , n — 1}. It is perhaps worth com
menting that, since Z1 = 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 11. If R is a ring, then for any a, b, c e R
1) 0a = a0 = 0,
2) a(b) = (a)b = (ab), 3) (—a)(—b) = ab, and 4)a(b—c)=ab—ac,(bc)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
0a=(0+0)a=0a+0a. Thus, by the cancellation law for the additive group (R, + ), we have 0a = 0.
In a like manner, one obtains a0 = 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)) = a0 = 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 sé {0}, there exists some nonzero element a e R. If 0 and 1 were equal, it would follow that a = a1 = a0 = 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 l 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 79 as R, then a is called a left (right) zero divisor in R if there exists some b 7E 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 = nln2 in Z (0 < n1, n2 < n), then the product n1,,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 qé 0, implies b = c. Proof Suppose that R is without zero divisors and let ab = ac, a 7E 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 + 0. We then
have ab = a0, whence by cancellation b = 0.
Similarly, b 3!: 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 bestknown example of an integral domain is the ring of integers; hence the choice of terminology. 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
8
FIRST COURSE IN RINGS AND IDEALS
in this case the term “integral domain” would merely indicate a commutative ring without zero divisors. We change direction somewhat to deal with the situation where a subset of a ring again constitutes a ring. Formally speaking,
Deﬁnition 14. Let (R, + , ) be a ring and S E R be a nonempty subset of R. If the system (S, +, ) is itself a ring (using the induced operations), then (S, +, ) is said to be a subring of (R, +, ). This deﬁnition is adequate, but unwieldy, since all the aspects of the deﬁnition of a ring must be checked in determining whether a given subset is a subring. In seeking a simpler criterion, notice that (S, +, ) is a subring of (R, +, ) provided that (S, +) is a subgroup of (R, +), (S, ) is a subsemigroup of (R, ), and the two distributive laws are satisﬁed in S. But the distributive and associative laws hold automatically for elements of S as a consequence of their validity in R. Since these laws are inherited from R, there is no necessity of requiring them in the deﬁnition of a subring. Taking our cue from these remarks, a subring could just as well be deﬁned as follows. The system (S, +, ) forms a subring of the ring (R, +, ) if and only if 1) S is a nonempty subset of R, 2) (S, +) is a subgroup of (R, +), and 3) the set S is closed under multiplication. To add the ﬁnal touch, even this deﬁnition can be improved upon; for the reader versed in group theory will recall that (S, +) is a subgroup of the group (R, +) provided that a — b e S whenever a, b e S. By these observations we are led to a set of closure conditions which make it some
what easier to verify that a particular subset is actually a subring. Theorem 1—3. Let R be a ring and z 7E S E R. Then, S is a subring of R if and only if
1) a, b e S imply a — b e S
(closure under differences),
2) a, b e S imply ab 6 S
(closure under multiplication).
If S is a subring of the ring R, then the zero element of S is that of R and, moreover, the additive inverse of an element of the subring S is the same as its inverse as a member of R. Veriﬁcation of these assertions is left as an exercise.
Example 16. Every ring R has two obvious subrings, namely, the set {0}, consisting only of the zero element, and R itself. These two subrings are usually referred to as the trivial subrings of R; all other subrings (if any exist) are called nontrivial. We shall use the term proper subring to mean a subring which is different from R. ~
INTRODUCTORY CONCEPTS
9
Example 17. The set Z, of even integers forms a subring of the ring Z of integers, for Zn — 2m = 2(n — m)eZ,,
(2n)(2m) = 2(2nm) e 2,. 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
centR = {aeRlar = raforallreR}. 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 14. 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 — b 6 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' 7E 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 al’ 7E a. It is clear that
(al’)l’ = a(1’l’) = 01', or (al’ — a)1’ = 0. Since neither al’ — 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:
(tab) + (ed) = (a + c,b + d), (a,~b)(c, d) = (ac, M). A routine calculation will Show that Z x {0} = {(a, 0)a e Z} forms a subring with identity element (1, O). This obviously differs from the identity of the entire ring R, which turns out to be the ordered pair (1, 1). By our previous remarks,(1, 0) must be azero divisor inR;in fact,(l, 0)(0, 1) = (0, 0), where (0, 0) serves as the zero element of R.
If R is an arbitrary ring and n a positive integer, then the nth power a" of an element a e R is deﬁned by the inductive conditions a1 = a and a" = a"' 1a. From this the usual laws of exponents follow at once: a"a"‘ = a”+"',
(a")"' = a""'
(n, me 2+).
To establish these rules, ﬁx m and proceed by induction on n. Observe also that if two elements a, b e 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 a° = l, the symbol a" now has a welldeﬁned
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: la=a
and
na=(n—1)a+a,
when
n>1.
If it iS also agreed to let 0a = 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,g
(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 a certain sum of elements of R.
11
However, when there is an identity for
multiplication, it is possible to represent na as a product of two ring elements, namely, na = (n1)a. To proceed further with our limited program, we must ﬁrst frame a deﬁnition. Deﬁnition 15. 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, thering P(X) of subsets of a ﬁxed set X is of characteristic 2, since
2A=AAA=(A—A)u(A—A)=¢ for every subset A E 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 15.
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 charR = n > 0,thenna = 0for everyae Rand so, in particular, n1 = 0. Were ml = 0, where 0 < m < n, it would necessarily follow that
ma = m(la) = (ml)a = 0a = 0 for every element a 6R.
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 9E as R will then possess a ﬁnite additive order m, with m s n. (Recall that for an element
12
FIRST COURSE IN RINGS AND IDEALS
a aé 0 of the group (R, +) to have order m means that ma = 0 and ka + 0 if 0 < k < m.) But the relation 0 = ma = (m1)a implies that m1 = 0, for R is assumed to be free of zero divisors. We therefore conclude from the theorem that n S 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 m1 = 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 7E a e R and 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 16. The characteristic of an integral domain is either zero or a prime number. Proof. Let R be of positive characteristic n and assume that n is not a prime. Then, n has a nontrivial factorization n = n1n2, with 1 < n1, n2 < n. It follows that
0 = n1 = (n1n2)1 = (n1n2)12 = (n11)(n21)By supposition, R is without zero divisors, so that either n11 = 0 or n21 = 0. Since both n1 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 Z1 of integral multiples of the identity; stated symbolically,
21 = {nllneZ}. From the relations
n1 — m1 = (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 cyc group (21, +) is simply the characteristic of the given ring R. When R happens to be an integral domain, then 21 is a subdomain of R (that is, Z1 is also an integral domain with respect to the operations in R). In fact, Z1 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
l3
wherep is a prime, then we are able to deduce considerably more: each nonzero element of 21 is invertible. Before establishing this, ﬁrst observe that the set Z1, regarded as an additive cyclic group of order p, consists of , p — 1. Now p distinct elements, namely, the p sums n1, where n = 0, l, let n1 be any nonzero element ofZ1 (0 < n < p). Since n and p are relatively prime, there exist integers r and s for which rp + sn = 1. But then 1 = (rp + sn)l = r(p1) + (sl)(n1). As pl = 0, we obtain the equation 1 = (sl)(nl), so that s1 serves as the multiplicative inverse of n1 in 21. The value of these remarks will have to await further developments (in particular, see Chapter 4). PROBLEMS 1. Verify that the zero element of a ring R is unique, as is the additive inverse of each element a e R.
2. 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 + b = ab
for all a, b e R) must be the trivial ring R = {0}. 4. In a ring R with identity, establish each of the following: a) the identity element for multiplication is unique, b) if a e R has a multiplicative inverse, then a" 1 is unique, c) if the element a is invertible, then so also is —a,
d) 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 ofzero. For a, b e R, verify that a) ab = lifandonlyifba = 1, b) ifaz = 1,theneithera = lora = l. 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 + hr = a' + (:)a"'1b +
+ G)a""b" +
where n =
k is the usual binomial coeﬂicient.
n!
k!(n — k)!
(,.:,)ab"‘1 + b",
14
FIRST COURSE IN RINGS AND IDEALS
An element a of a ring R is said to be idempotent if a2 = a and nilpotent if a’l = 0 for some n e Z +. Show that a) a nonzero idempotent element cannot be nilpotent,
b) every nonzero nilpotent element is a zero divisor in R. Given that R is an integral domain, prove that a) the only nilpotent element is the zero element of R, b) the multiplicative identity is the only nonzero idempotent element. 10. If a is a nilpotent element of R, a ring with identity, establish that 1 + a is
invertible in R. [Hint: (l + a)‘1 = l — a + a2 +
+ (— l)”'1a"'1, where
a’I = 0.] 11. A Boolean ring is a ring with identity every element of which is idempotent. Prove that any Boolean ring R is commutative. [Hint: First show that a = —a for every 11 e R.] 12. Suppose the ring R contains an element a such that (l) a is idempotent and (2) a is not a zero divisor of R. Deduoe that a serves as a multiplicative identity for R. l3. Let S be a nonempty subset of the ﬁnite ring R. Prove that S is a subring of R if and only if S is closed under both the operations of addition and multiplication. 14. Assume that R is a ring and a e R. If C(a) denotes the set of all elements which commute with a,
C(a) = {reRar = ra},
show that C(a) is a subring of R. Also, verify the equality cent R = ﬂux C(a). 15. Given a ring R, prove that a) if S, is an arbitrary (indexed) collection of subrings of R, then their intersection n S, is also a subring of R; b) for a nonempty subset Tof R, the set (T) = n{STE S;SisasubringofR} is the smallest (in the sense of inclusion) subring of R to contain T; (T) is called the subring generated by T.
16. Let S be a subring of R, a ring with identity. For an arbitrary element a 9.5 S, the subring generated by the set S u {a} is represented by (S, a). If ae cent R, establish that
= {r0 + rla +
+ r,a"neZ+;r,eS}.
17. LetRbeanarbitraryringandneZ+. Ifthesetsnisdeﬁnedby
S,l = {aeRn"a = 0for somek > 0}, determine whether S,I is a subring of R. 18. Establish the following assertions concerning. the characteristic of a ring R:
PROBLEMS
15
a) if there exists an integer k such that ka = 0 for all a e R, then k is divisible by char R;
b) ifchar R > 0, then char S 3 char R for any subring S ofR; c) if R is an integral domain and S is a subdomain of R, then char S = char R. 19. Let R be a ring with a ﬁnite number of elements, say a,, 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 ni (i = l, 2, , n). Suppose that R is a ring with identity such that char R = n > 0. If n is not prime, show that R has divisors of zero. 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 a2 = a, then (ara — ar)2 = (ara — ra)2 = 0 for all r e R.] . Assume that R is a ring with the property that a2 + a 6 cent R for every element a in R. Prove that R is necessarily a commutative ring [Hint : Utilize the expression
£3
(a + b)2 + (a + b) to show ﬁrst that ab + ba lies in the center for all a, b e R.] 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 fo g of two functions ﬂ 9 e R are deﬁned by the usual rules
(f + (1)06) = f(x) + 906), (f° 006) = f(906))
(x 6 G)
Show that the resulting system (R, +, 0) forms a ring. At the same time determine the invertible elements of R.
.M'
. Let (G, ) be a ﬁnite group (written multiplicatively), say with elements x1, x2, . .. , x_, and let R be an arbitrary ring. Consider the set R(G) of all formal sums
l
rixi
(r, e R).
1
Two such expressions are regarded as equal if they have the same coeﬂicients. Addition and multiplication can be deﬁned in R(G) by taking .
l
rixi + Z sixi = 2 ('1 + 50x: 1 i=1 i=1 and
where x1m= M
(The meaning of the lastwritten sum is that the summation is to be extended over all subscripts j and k for which xix, = x,.) Prove that, with respect to these operations, R(G) constitutes a ring, the socalled 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 of subrings with a stronger type of multiplicative closure, namely, closure under multiplication by an arbitrary ring element. Deﬁnition 21. A subring I of the ring R is said to be a twosided ideal ofR ifand only ifreR and a6] imply both me! and anal. 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 twosided ideal may be reformulated as follows. Deﬁnition 2—2. Let I be a nonempty subset of a ring R. Then I is a twosided ideal of R if and only if 1) a,beIimplya — beI, and 2) r e R and a e I imply both products ra, ar e 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 twosided ideals.
CONVENTION In what follows, let us agree that the term “ideal”, unmodiﬁed, will always mean twosided ideal.
Before proceeding further, we pause to examine this concept by means of several speciﬁc examples. l6
IDEALS AND THEIR OPERATIONS
17
Example 2—1. For each integer a e Z, let (a) represent the set consisting of all integral multiples of a; that is,
(a) = {nan e Z}. The following relations conﬁrm (a) to be an ideal of the ring of integers: na — ma = (n — m)a, m(na)=(mn)a,
n,neZ.
In particular, since (2) = 2,, the ring of even integers forms an ideal of Z.
Notice, too, that (0) = {0} and (1) = 2..) Example 22. Another illustration is furnished by map (X, R), the ring of mappings from the set X into the ring R (see Example 1—4). For a ﬁxed element x e X, we denote by I, the set of all mappings which take on the value 0 at x:
1, = U6 map(X, R)f(x) = 0}Now, choose f, g e I, and h e map(X, R). From the deﬁnition of the ring operations in map (X, R),
(fg)(X) = f(X)g(X) = 00 = 0, while
(fh)(X) = f(x)h(X) = 0h(x) = 0, and, in a similar manner, (hf)(x) = 0. Thus, f — g, fh and hf all belong to 1,, which implies that 1,, is an ideal. More generally, if S is any nonempty subset of X, then
I = {femap(X,R)f(x) = Ofor allxeS} comprises an ideal of map(X, R). Since I = ﬂxeslx, we have a situation where the intersection of ideals is once again an ideal. (Theorem 2—2 shows that this is no accident.) Before presenting our next example, we derive a fact which, despite its apparent simplicity, will be frequently applied in the sequel. Theorem 21. If I is a proper (right, left, twosided) ideal of a ring R with identity, then no element of I possesses a multiplicative inverse; thatis,I n R* =z.
Proof Let I be an ideal of R and suppose that there is some member a + 0
of I such that a‘ 1 exists in R. (The theorem is trivial when I = {0}.) Since I is closed under multiplication by arbitrary ring elements, it follows that l = a' 1a e I. By the same reasoning, I contains r = r1 for every r in R;
18
FIRST COURSE IN RINGS AND IDEALS
that is, R E I, whence the equality I = R. This contradicts the hypothesis that I is a proper subset of R. Notice that, en route, we have also established
Corollary. In a ring with identity, no proper (right, left, twosided) ideal contains the identity element. Example 2—3. This example is given to show that the ring Mn(R#) of n x n matrices over the real numbers has no nontrivial ideals. As a notational device, let us deﬁne EU to bethe n x n matrix having 1 as its ijth entry and zeroes elsewhere. Now, suppose that I 7E {0} is any ideal of the ring Mn(R#).
Then I must contain some nonzero matrix (a,_,), with, say,
rsth entry a" + 0. Since I is a twosided ideal, the product Err (by) (aij )Ess
is a member of I, where the matrix (by) 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
13,, = 13,113,153,
(i,j = 1,2, ...,n)
implies that all n2 of the matrices EIU are contained in I. The clinching point is that the identity matrix (6“) can be written as
(ab = E11 + E22 +
+ E..."
which leads to the conclusion that (60) e I and, appealing to the above corollary, that I = M”(R#). In other words, Mn(R#) possesses no nonzero proper ideals, as asserted. ‘ As a matter of deﬁnition, let us call a ring R 7E {0} simple if R has no
twosided ideals other than {0} and R. In the light of Example 2—4, the matrix ring Mn(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 22. Let {1,} be an arbitrary collection of (right, left, twosided) ideals of the ring R, where i ranges over some index set. Then n I, is also a (right, left, twosided) ideal of R.
Proof. We give the proof for the case in which the ideals are twosided. First, observe that the intersection n Ii is nonempty, for each of the ideals Ii must contain the zero element of the ring. Suppose that the elements a, b e n Ii and r e R. Then a and b are members of Ii, where ivaries over the indexing set. Inasmuch as I, is assumed to be an ideal of R, it follows that a — b, ar and ra all lie in the set 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 1,, making n I1 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{IS E I; IisanidealofR}. 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 inclusion S E (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 E I, then necessarily (S) E 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 onesided ideals generated by S. ' If S consists of a ﬁnite number of elements, say a1, a2, ..., an, then the
ideal which they generate is customarily denoted by (a1, a2, , 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, twosided) 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 (re R), as well as the elements na (n an integer), and, hence, includes their sum ar + na. (As usual, the notation na represents the nfold 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 = a0 + la. These remarks make it clear that
(a), = {ar + nalreR; neZ}. 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(n1)=a(r+nl)=ar’, where r’ = r + n1 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 = {arlreR}. Similar remarks apply, of course, to the principal left ideal (a), 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 diﬂiculty, however, is that this ideal need not contain a itself.
With regard to the twosided ideal (a) generated by a, the situation is more complicated. Certam 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 2 riasi, where ri, s, e R, is also required to be in (a). The reader will experience no diﬁiculty in showing that the principal ideal generated by a is given by
(a) = {na + ra + as + Z riasilr, s, ri, si e R; n e Z}. ﬁnite
In case R happens to have an identity, this description of (a) reduces to the
set of all ﬁnite sums Z riasi.
A particularly important type of ring is a principal ideal ring, which we now deﬁne. Deﬁnition 23. 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 fumishes an example of such rings. Theorem 23. 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 idea] {0} is the principal ideal generated by 0. Suppose then that I does not consist of the zero element alone. Now, if m e 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) E I. To establish the inclusion I E (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 S r < n. Since k and qn are both members of I, it follows that r = k — qn e I. If r > 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 twosided ideals.) Given a ﬁnite number of ideals 11, 12, , In of the ring R, one deﬁnes their sum in the
natural way:
I1 + I2 +
+ 1,, = {a1 + a2 +
+ a,a,.eI,}.
IDEALS AND THEIR OPERATIONS
21
Then I, + I, + + In is likewise an ideal ofR and is the smallest ideal of R which contains every I,; phrased in another way, 11 + 12 + + I,I
is the ideal generated by the union I1 u I2 u
u I". In the special case
of two ideals I and J, our deﬁnition reduces to
I + J = {a + baeI;beJ}. More generally, let {Ii} be an arbitrary indexed collection of ideals of R. The sum of this collection may be denoted by Z Ii and is the ideal of R whose members are all possible ﬁnite sums of elements from the various idealsI,:
21= {zala em
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 Z Ii could be given by
Eli = {2 a,ai e 1,; all but a ﬁnite number of the a, are 0}, where it is understood that 2 represents an arbitrary sum with one or more terms. Just as n Ii can be interpreted as the largest ideal of R contained in every 1,, the sum 21,. supplies the dual notion of the smallest ideal containing every I,. HR = I1 + I2 + + I", then each element xe R can be expressed in the form x = a1 + a2 + + an, where a, lies in 1,. There is no guarantee, however, that this representation of x is unique. To ensure that
every member of R is uniquely expressible as a sum of elements from the ideals Ii, an auxiliary deﬁnition is required.
Deﬁnition 24. Let II, 12, , In be ideals ofthe ring R. We call R the internal direct sum ofIl, 12, , 1,,and writeR = I1 69 12 ® EB I", provided that a)R=I1 +12+ +I,,,and
b) Ii n (I1 +
+ Ii.l + IH1 +
+ I”) = {0} for each i.
As was heralded by our remarks, we are now in a position to prove Theorem 24.
Let 11, 12,
, I,, be ideals of the ring R.
Then the
following statements are equivalent: 1) R is the internal direct sum of 11, 12, , I,I 2) Each element x of R is uniquely expressible in the form + a,, where aieIr x = a1 + a2 + 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 = 11 63 12. Suppose further .that an element x e R has two representations x=al+b1=a2+b2
(aiEII,bIEIZ).
22
FIRST COURSE IN RINGS AND IDEALS
Then a1 — a2 = b2 — b1. But the lefthand side of this last equation lies in I1, while the righthand side is in 12, so that both sides belong to
I1 n I2 = {0}. Itfollowsthata1 — a2 = b2 — b1 = 0,0ra1 = a2,b1 = b2. In other words, x is uniquely representable as a sum a + b, a e I 1, b e I2. Conversely, assume that assertion (2) holds and that the element x 5 I1 0 12. We may then express x in two different ways as the sum of an element in I1 and an element in 1,; namely, x = x + 0 (here e1 and 0612) and x = 0 + x (here 0611 and x612). The uniqueness assumption of (2) implies that x = 0, in consequence of which I1 n I2 = {0}; hence, R = I 1 e 12. This completes the proof of the theorem. We now come to a less elementary, but extremely useful, notion; namely,
the product of ideals. Once again, assume that I and J are two ideals of the ring R. To be consistent with our earlier deﬁnition of the sum I + J, we should deﬁne the product IJ to be the collection of all simple products ab, where a e I and b G J. Unfortunately, the resulting set fails to form an
ideal. (Why?) To counter this diﬂiculty, we instead take the elements of II to be all possible ﬁnite sums of simple products; stated explicitly,
1.] = {Zaibilaieh bieJ}. ﬁnite
With this deﬁnition IJ indeed becomes an ideal of R. For, suppose that x,yeIJ and reRgthen,
x = alb1 + azb2 +
+ a,b_,,,
y = a’lb’1 + a’zb’2 +
+ aﬁnbjn,
where the a, and a; are in I, and the bi and b; are in J. From this we obtain x _ y = albl +
+ anb,l + (—a’1)b’1 +
rx = (”Obi + (ra2)b2 +
+ (—a;n)b;,.,
+ (ra,.
Now, the elements —a§ and rat necessarily lie in I, so that x — y and rx e U ; likewise, xr e 1.], making IJ an ideal of R. In point of fact, IJ is
just the ideal generated by the set of all products ab, a e I, b e J. There is no diﬂiculty in extending the above remarks to any ﬁnite number of ideals 11, 12, , In of the ring R. A moment’s thought shows that the product 1112 I,‘ is the ideal consisting of ﬁnite sums of terms of the form ala2 an, with a, in 1,. (It is perhaps appropriate to point out that, because of the associative law for multiplication in R, the notation
1112 In is unambiguous.) A special case immediately presents itself: namely, the situation where all the ideals are alike, say equal to the ideal I. Here, we see that I" is the set of ﬁnite sums of products of n elements
from I : I” = {age analz
“i. aikeI}.
IDEALS AND THEIR OPERATIONS
23
In this connection, it is important to observe that
12122132021"2forms a decreasing chain of ideals.
Remark. If I is a right ideal and S a nonempty subset of the ring R, then ﬁnite
forms a right idea] of R. In particular, if S = {a}, then a1 (a notation we prefer to {a}I) is given by
a1 = {arlr e I}. Analogous statements can be made when I is a left ideal of R, but not, of course, a twosided ideal.
The last idealtheoretic operation which we wish to consider is that of the quotient (or residual), deﬁned below.
Deﬁnition 25. 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 :,J), consists of all elements a e R such that aJ g I (Ja S I). In the event R is a commutative ring, we simply write I :J.
It is by no means obvious that the set
I:,J = {aeRlaJ E I} actually forms an ideal of R, whenever I and J are ideals. To verify this,
suppose that the elements a, b e I :, J and r e R. For any x e J, we clearly have (a — b)x = ax — be, 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] 2, J.
Likewise, the relations raJ 2 r1 9 I and
arJ g (1.] E I imply that ra, ar e 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 25. The following relations hold for ideals in a ring R (capital letters indicate ideals of R):
1) (n 1.02.1 = “(1:91), 2) 1:,2Ji= n(I :rJi),
3) I:,(JK) = (I:,IQ:,J.
24
FIRST COURSE IN RINGS AND IDEALS
Proof Concerning (l), we have
(nI,):,J = {aeRIaJ E nIi} = {aeRaJ E Ii for all i} = n{aeRaJ S 1,} = n(I,:,J). With an eye to proving (2), notice that the inclusion J, E 2 J, implies a(2 J,) S Iifand only ifaJ, E Ifor alli;thus,
I :,2 Ji = {aeRa(£ J,) E I} = {aeRIaJi E I for all i} = n(I:,Ji). Conﬁrmation of the ﬁnal assertion follows from
I:,(JK) = {aeRa(JK) g I} = {aeR(aJ)K E I} = {aeRaJ E 1:,K} = (I:,K):,J. Remark. Similar results hold for left quotients; the sole difference being that, instead of (3), one now has I 2, (JK) = (I :, J) :, 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 = {fiIfERJm} = 0}, J = {fi2 + ni2feR;f(0) = 0;neZ}, 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 6 J, while $2 é 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 of fe R and n e Z, with f(0) = 0. In consequence,
fi2 = 6 — n)i2, implying that f(x) = % — n 7E 0 for every 0 36 xeR"; 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”; in view of this, a’ is
IDEALS AND THEIR OPERATIONS
25
often referred to as the pseudoinverse of a. In the commutative case, the equation aa’a = a may, of course, be written as aza’ = a.
The result which we have in mind now follows. Theorem 26. 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 a6 1, then aa’a = a for some a’ in R.
Setting b = a’aa’, the
element b belongs to I and has the property that aba = a(a’aa’)a = (aa’a)a’a = aa’a = a. Our aim is to show that whenever a e J E I and r e R, then both ar and ra lie in J. We already know that ar e 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, ar 6 J. A symmetric argument conﬁrms that ra 6 J. Although Deﬁnition 2—6 appears to have a somewhat artiﬁcial air, we might remark that the set ofall 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 R and 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 onetoone as a map on the underlying sets is called an isomorphism. We emphasize that the + and  occurring on the lefthand sides of the equations in Deﬁnition 2—7 are those of R, whereas the + and  occurring on the righthand 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. Iff is a homomorphism of R into R’, then the image f(R) of R under f 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 homR will be used in place of hom(R, R). (Some authors prefer to write end R, for endomorphism, in place of homR; both notations have a certain suggestive power and it reduces to a matter of personal preference.) A knowledge of a few simpleminded examples will help to ﬁx ideas. Example 2—4. Let R and R’ be arbitrary rings andf: R —> R’ be the function which sends each element of R to the zero element of R’. Then,
f(a + b) = 0 = 0 + 0 =f(a) +f(b), f(ab) = 0 = 00 =f(a)f(b)
(a, b e R),
so thatfis a homomorphic mapping. This particular mapping, the socalled 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. That fis a homomorphism follows directly from the deﬁnition of the operations in 2,:
ﬁt! + b) = [a + b] = [a] +.. [b] =f(a) +..f(b), f(ab) = [017] = [a]'..[b] = f(a)°..f(’1)Example 2—6. In the ring map(X, R), deﬁne 17,, to be the function which assigns to each f 6 map (X, R) its value at a ﬁxed element a e X; in other
words, 1:“(f) = f(a). Then T“ is a homomorphism from map (X, R) into R, known as the evaluation homomorphism at a. We need only observe that
Ta(f+ g) = (f+ g)(a) =f(a) + 9(a) = T..(f) + Tutti), Mfg) = (fg)(a) = f(0)901) = t..(f)r..(g)We now list some of the structural features preserved under homomorphisms. Theorem 2—7. Let f be a homomorphism from the ring R into the ring R’. Then the following hold:
1) f(0) = 0, 2) f(—a) = —f(a) for all aeR. If, in addition, R and R’ are both rings with identity and f(R) = R’, then
3) f(1) = l, 4) f(a' 1) = f(a)‘ 1 for each invertible element a e R.
IDEALS AND THEIR OPERATIONS
27
Proof. ,From f(0) = f(O + 0) = f(O) + f(0), we obtain f(0) = o. The {act thatf(a) +f(—a) =f(a + (—a)) =f(0) = 0 yieldsf(—a) = —f(a). As regards (3), let the elementa e Rsatisfyﬂa) = 1;then,f(1) = f(a)f(1) =
f(al) = f(a) = 1. Finally, the equationf(a)f(a'1) =f(aa‘1) =f(l) = 1 shows that f(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(0)1 = f(a) = f(01) = f(a)f(1) 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(1) is the identity for the homomorphic image f(R). The element f(1) need not serve as an identity for the entire ring R’ and, indeed, it may very well happen
that f(l) + 1. We also observe, in passing, that, by virtue of statement (2),
f((1  b) =f(a) +f(b) =f(a) f(b). In short, any ring homomorphism preserves diﬂerences 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, f(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 partof the theorem, recall that, by deﬁnition,
the imagef(S) = {f(a)a e S}. Now, suppose thatf(a) andf(b) are arbitrary elements off(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)  f0?) = f(a  b) 6f(S) and
f(a)f(b) = f(ab) 6f(S)According to Theorem 1—3, these are suﬂicient 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 Rf(a) e 5’}. Thus, ifa, b ef‘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)€S' and
f(ab) = f(a)f(b) 6 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 ofreplacing the term “subring” in Theorem 2—8 by “ideal”. It is not diﬂicult 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 subringf‘1(1’) is an ideal of R. For instance, suppose that a e f '1(I’), 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 in f ‘ 1(1’). Likewise, ar e f ‘ 1(I’), which helps to make f ”(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) e f(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’) is an ideal of R. 2) Iff(R) = R', then for each ideal I of R, the subringf(1) is an ideal of R’. To go still further, we need to introduce a new idea.
Deﬁnition 28. 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: {aeRf(a) = 0}. Theorem 2—7 indicates that ker f 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 29. 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 ker f = f ' 1(0), the conclusion follows from the last corollary.
IDEALS AND THEIR OPERATIONS
29
The kernel of a homomorphism may be viewed as a measure of the extent to which the homomorphism fails to be onetoone (hence, fails to
be an isomorphism). In more concrete terms, we have Theorem 210. A homomorphism f from a ring R into a ring R’ is an
isomorphism if and only if ker f = {0}. Proof. First, iff is a onetoone function and f(a) = 0 = f(0), then a = O,
whence ker f = {0}. On the other hand, suppose that the kernel consists exactly of 0. Iff(a) = f(b), then
f(a  b) =f(a) f(b) = 0, which meansthata — b ekerf. Sincekerf = {O},wemust havea — b = O, or a = b, making f a onetoone function.
Two rings R and R’ are said to be isomorphic, denoted by R r: R’, if there exists an isomorphism from the ring R onto the ring R’. Although this deﬁnition is unsymmetric in that it makes mention of a function from one particular ring to another, let us remark that iff: R —> R’ is a onetoone, onto, homomorphic mapping, the function f ‘1 : R’ —> R also enjoys these properties. We may therefore ignore the apparent lack of symmetry and merely speak of two rings R and R’ as being isomorphic without specifying one ring as isomorphic to the other; notationally, this situation is recognized by writing either R 2 R’ or R’ 2 R. Isomorphic rings are indistinguishable from the structural point of view, even though they may differ in the notation for and nature of their elements and operations. Two such rings, although not in general formally identical, are the same for all purposes; the underlying feature is the existence of a mapping which transports the algebraic structure of one ring to the other. In practice, we shall often identify isomorphic rings without explicit mention. This seems to be a natural place to insert an example. Example 27. Consider an arbitrary ring R with identity and the mapping f: Z —> R given by f(n) = n1. (At the risk of being repetitious, let us again emphasize that n1 means the nfold sum of 1.) A simple computation shows that f, so deﬁned, is a homomorphism from the ring Z of integers into the ring R: f(n + m) = (n + m)1= n1 + ml =f(n) +f(m) and
f(nm} = (""01 = n(ml) = ("1) (ml) = f(n)f(m)Since ker f constitutes an ideal of Z, a principal ideal ring, it follows that
kerf: {neZInl = 0} = (p) for some nonnegative integer p. A moment’s reﬂection should convince the reader that the integer p is just the characteristic of R. In particular,
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 z 21, where l 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 if the ideals of R were in a onetoone 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 E J E I + kerf, then f(I) = f(J). The quickest way to see this is to notice that
f(I) SfU) EfU + kerf) =f(1) +f(kerf) =f(1), 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 ker f E I. In either event, it follows that I E J E I + ker f = I and, in
consequence, I = J. The ﬁrst of the restrictions just cited has the eﬂect of making the function f onetoone, in which case R and R’ are isomorphic rings (and it then comes as no surprise to ﬁnd their ideals in onetoone 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 ofR such that kerf E I, then I = f‘1(f(I)).
Proof. Suppose that the element a e f‘1(f(I)), so that f(a) e f(I). Then f(a) = f(r) for some choice of r in I. As a result, we will havef(a — r) = 0, or, what amounts to the same thing, a — r e ker f E I. This implies that
aeI, yielding the inclusion f ‘1(f(1)) E 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 onetoone correspon
dence between those ideals I of R such that ker f E I and the set of all ideals I’ of R’; speciﬁcally, I’ is given by I’ = f(1). 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 ker f E I,
IDEALS AND THEIR OPERATIONS
31
such that f(1) = I’. To accomplish this, it is suﬂicient to take I = f " 1(I’). By the corollary to Theorem 2—8, f ‘ 1(I’) certainly forms an ideal of R and, since 0 e I’,
keff = f"(0) E f‘ 1(1')lnasmuch as the function f is assumed to be an onto map, it also follows
thatfm= f(f 1(1'))) = 1'
Next, we argue that this correspondence lS onetoone. To make things more speciﬁc, let ideals I and J of R be given, where ker f E I, kerf E J, and satisfying f(I) = f(J). From the elementary lemma just established, we see that
I =f_1(f(1))=f'1(f(1))= JOne ﬁnds in this way that the correspondence I H f(I), where kerf E I, is indeed onetoone, completing the proof. Before announcing our next result, another deﬁnition is necessary.
Deﬁnition 29. A ring R is said to be imbedded in a ring R’ if there exists some subring S’ of R’ such that R 2 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)reR; neZ}. If addition and multiplication are deﬁned by
(a,n) + (b,m) = (a + b,n + m), (a, "Nb: 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)(0,1) = (a0 + 1a + n0, n1) = (a, n), and, similarly,
(0, l)(a, 71) = (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 T form (a, 0). Since
((1,0)  (b0) = (a  17,0), (a, 0)(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 f a homomorphism from I onto R’, a ring with identity. If I E 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(M) = f(a)f(u) = f(a)1 = f(a), showing that g actually extends the original function f. 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 + bu)
= f(0“) + f(bu) = g(a) + 9(1))
IDEALS AND THEIR OPERATIONS
33
As a preliminary step to demonstrating that g also preserves multiplication, notice that
f((abluz) = f(abu)f(u) = f(aim)From this we are able to conclude that
9(ab) = f(abu) = f(“bu”) = f((au)(bu)) = f(au)f(bu) = 9(a)g(b)The crucial third equality is justiﬁed by the fact that ue cent R, hence, commutes with b. As regards the uniqueness assertion, let us assume that there is another homomorphic extension offto the set R; call it h. Since fand 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(u) = h(a)h(u) = h(au) = f(014) = 9(0) 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 R1, R2, , R,I 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 ntuples (a,, a2,
, a,), with a, e Ri.
One can easily convert R into a ring by
performing the ring operations componentwise; in other words, if (a1, a2, , an) and (b1, b2, , bu) are two elements of R, simply deﬁne (a1! a2: "',an) + (b1, b2! "'abn) = (a1 + b11a2 + 1’29 "aan + b")
and (a1, a2,
, an)(b1, b2,
, b") = (albl, azbz,
, aubn).
The ring so obtained is called the external direct sum of R1, R2,
and is conveniently written R = R1 4— R2 i
, R,l
—i 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. , n), then the If R is the external direct sum of the rings R, (i = l, 2, individual R, need not be subrings, or even subsets, of R. However, there is
an ideal of R which is the isomorphic image of Ri. A straightforward calculation will convince the reader that the set
1,. = {(0,
,0, a, o,
, 0)a,e RE}
FIRST COURSE IN RINGS AND IDEALS
(that is, the set consisting of all ntuples with zeroes in all places but the ith) forms an ideal of R naturally isomorphic to Ri under the mapping which sends (0,
, 0, a,, 0,
, 0) to the element a,. Since
(a,,a,, ...,a,) = (a,,o,o, ...,0) + (o,a,,o, ...,0) +
+ (0,0, ...,o,a,),
it should also be clear that every member of R is uniquely representable as a sum of elements from the ideals 1,. Taking note of Theorem 2—4, this
means that R is the internal direct sum of the ideals I, and so
R1+R24L~+R,=I,@I,emel,
(R,=I,).
In summary, the external direct sum R of the rings R1, R2,
, R,l is also
the internal direct sum of the ideals I1, 12, , I,. and, for each i, R, and Ii are isomorphic. In 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 enotation. 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 6 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 6 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 — relr e 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),wherereeIandr — reeJ. SinceI n J = {0},thisis the only way r can be expressed as a sum of elements of I and J. (A moment’s
thought shows that ifae I n J, saya = r — re, thena = ae = (r — re)e =
r(e — e2) = 0.) It is also true that the ideal I = 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 214. If the ring R is a direct summand in every extension ring containing it as an idea], 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 2 R x {0}, where, as is easily veriﬁed, Rx {0} constitutes an ideal of R’. We may
._..s.. .1...a
34
PROBLEMS
35
therefore regard R as being an ideal of the ring R’. Our hypothesis now comes into play and asserts that R’ = R 69 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 lastwritten 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, 0) (e, — 1) = (re — r, 0) will consequently be in both R and 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 I. If I is a right ideal and J a lelt ideal of the ring R such that I n J = {0}, prove thatab=0forallael,beJ. 2. Given an ideal I of the ring R, deﬁne the set C(I) by
C(I) = {reRra — are] forall aeR}. 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 anideal of R. b) If{I‘}(i = 1,2, ...)isacollectionofidealsoftheringRsuchthatI1 E 12 E s I,I E , prove that u], is also an ideal ofR. 4. Consider the ring M, (R) of n x n matrices over R, a ring with identity. A square matrix (an) is said to be upper triangular if ail. = 0 for i > j and strictly upper
triangular if al.1 = 0 for i 2 j. Let T;(R) and Tf,(R) denote the sets of all upper triangular and strictly upper triangular matrices in M,(R), respectively. Prove each of the following: a) 72(R) and T,’,(R) are both subrings of M_(R).
b) T;(R) is an ideal of the ring 1:,(R). c) A matrix (an) e I,(R) is invertible in 7;,(R) if and only if all. is invertible in R
for i = 1, 2,
, n. [Hint: Induct on the order n]
d) Any matrix (aij) e T:(R) is nilpotent; in particular, (“i = 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 + ralieI, 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) + (7n) = (n, "1) = (d), where d is the greatest common divisor of n and m.
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
[J = {2 aibiaie I; b‘eJ}, where 2 represents a ﬁnite sum of one or more terms. Establish that H is a twosided ideal of R and, whenever I and J are themselves twosided, that U s I n J. If S is any given nonempty subset of the ring R, then
ann,S = {reRIar = 0 for all aeS} is called the right annihilator of S (in R); similarly,
ann,S = {reRIra = 0for all aeS} is the left annihilator of S. When R is a commutative ring, we simply speak of the annihilator of S and use the notation an S. Prove the assertions below: a) ann,S (ann,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 S is an ideal of R, then ann,S and ann,S are both ideals of R.
d) When R has an identity element, ann,R = ann,R = {0}. . Let 11,12, ...,I,I be ideals of the ring R with R = 11 + I2 + + I”. Show that this sum is direct ifand only ifa1 + a2 + + a,I = 0, with a,eIi, implies that each a, = 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 E X, P(Y) and P(X — Y) are both principal ideals of P(X), ‘
withP(X)=P(Y)€BP(X Y)11. Suppose that R is a commutative ring with identity and that the element a e R is an idempotent diﬂ'erent from 0 or 1. Prove that R is the direct sum ofthe principal ideals (a) and (l — a). ' 12. LetI,J andeideals of the ring R. Prove that a)I(J+K)=IJ+IK, (I+J)K=IK+JK; b)if12J,then1n(J+K)=J+(InK). 13.
Establish that in the ring 2, ifI = (12) and J = (21), then I+J=(3),
InJ=(84),
IJ=(252),
I:J=(4),
J:I=(7).
[Hint: In general, (a):(b) = (c), where c = a/gcd (a, b).] 14. Given ideals I and J of the ring R, verify that a) 0:,1 = ann,I, and 0:,1 = ann,I (notation as in Problem 8); b) 12,] (1:,J) is the largest ideal of R with the property that (I:,J)J .C. I (J(I:,J) E I). 15. Let I, J and K be ideals ofR, a commutative ring with identity. Prove the following assertions:
PROBLEMS a) b) c) d)
37
If! E J, then [2K 9 JzKand K:I 2 KzJ. 1:1“l = (I:J"):J = (I:J):J' for any neZ+. I:J= Rifandonlyif] S I. [:1 = I:(I + J).
1‘. If! is a right ideal ofR, a ring with identity, show that I:,R = {a e RRa E I} is the largest twosided ideal of R contained in 1.
I7. Given that f is a homomorphism from the ring R onto the ring R’, prove that
a) f(cent R) g 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)) = (f(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 ﬂ,(x) = am"1 is an automorphism of R.
19. Let f be a homomorphism from the ring R into itself and S be the set of elements that are left ﬁxed by f; in symbols,
S = {aeRf(a) = a}. Establish that S forms a subring of R. If f is a homomorphism from the ring R into the ring R’, where R has positive characteristic, verify that charf(R) 5 char R. 21. Let fbe 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(1 + J) =f(I) +f(J); h) K”) =f(I)f(J); c) f(I n J) Efﬂ) nf(J),withequalityifeitherI 2 kerforJ 2 kerf; d) f(I:J) Ef(I):f(J), with equality if] 2 kerf Show that the relation R 2 R’ is an equivalence relation on any set of rings Let R be an arbitrary ring. For each ﬁxed element a e R, deﬁne the leftmultiplication function 1;: R —> R by taking 7:,(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 mapping f(a) = 1; determines a homomorphism of R onto the ring I}; d) the kernel off is the ideal ann,R; e) if for each 0 79 a e R, there exists some b e R such that ab 99 0, then R 2 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 212. Establish that
a) Rx {0} isan ideal ofZ; b) Z 2 {0} x Z; c) if a is an idempotent element of R, then the pair (—a, l) is idempotent in R x Z, while (a, 0) is a zero divisor.
FIRST COURSE IN RINGS AND IDEALS
25. Suppose that R is a ring of characteristic n. If addition and multiplication are
A “mummyad
38
deﬁnedinRxZ, = {(x,a)xeR;a,,} by 0"“) + (Ysb) = (x + yaa +ub):
(x, @051?) = (xy + ay + bx,a'..b), prove that R xZn is an extension ring of R of characteristic n. Also show that R x 2,, has an identity element. . Let R = R1 9 R2 69
6 R, be the (external) direct sum ofa ﬁnite number of V
rings R,. with identity (i = 1, 2, , n). a) For ﬁxed i, deﬁne the mapping 7t,: R —» R, as follows: if a = (a1, a2, ', a_), where aie Rj, then ni(a) = a... Prove that n, is a homomorphism from the ring R onto R,. b) Show that every ideal I ofR is of the form I = I‘ 6 12 EB G) I”, with I, an ideal of Ri. [Hint: Take I, = 1r,(I). If bi e 1,, then there exists some (b1, , bi, ..., bn)eI. It follows that (bl, ...,bi, ...,bn)(0, ..., 1, ,0) = (0, ...,bi, , 0) e 1.] 27. A nonempty subset A of a ring R is termed an adeal of R if (i) a,beA implya+ beA, (ii) reR and aeA imply both areA and raeA. Prove that a) An adeal A ofR is an ideal ofR iffor each aeA there is an integer n + 0, depending upon a, such that na 6 aR + Ra. (This condition is satisﬁed, in particular, if R has a multiplicative identity.) b) Whenever R is a commutative ring, the condition in part (a) is a necessary as
well as sufﬁcient condition for an adeal to be an ideal. [Hint. For any a e R,
the set A = {naln e Z+} + aR is an adeal ofR; hence, an ideal ofR.] . 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 Mn(I) is an ideal of the matrix ring Mu(R). b) Every ideal of Mn(R) is of the form Mna), where I is an ideal of R. [Hint: Let
F,1(a) denote the matrix in MI (R) having a as its ijth entry and zeroes elsewhere. For any ideal .1! in MI(R), let I be the set of elements in R which appear as
entries for the matrices in .11. Given any a e I, say a is the rsth entry of a matrix A e .11, it follows that lib(a) = Fi,(l)AF_U(1) 6 JL] c) If R is a simple ring, then so is the matrix ring M,,(R). . 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.) IfR contains no divisors ofzero, prove that multiplication is commutative. [Hint: Given 0 + a e R, consider the subring S generated by a For arbitrary b e R, ab = r e S, so that ar = ra.]
THREE
THE CLASSICAL ISOMORPHISM THEOREMS
In this chapter we shall discuss a number of signiﬁcant results having to do with the relationship between homomorphisms and quotient rings (which we shall shortly deﬁne). Of these results perhaps the most crucial is Theorem 37, commonly known as the Fundamental Homomorphism Theorem for Rings. The importance of this result would be diﬁicult to overemphasize, for it stands as the cornerstone upon which much of the succeeding theory rests. The notion of an ideal carries with it a natural equivalence relation. For, given an ideal I of the ring R, it is a routine matter to check that the relation deﬁned by a E b if and only if a — b e I is actually an equivalence relation on R. As such, this relation induces a partition ofR into equivalence classes, the exact nature of which is determined below.Theorem 31. If I is an ideal of the ring R, then the equivalence class of b e R for the relation E is the set
b + I = {b + ilieI}. Prooﬁ Let [b] = {xeRx E b}. Ifa = b + iis any member ofb + I, then a — b = is I.
By deﬁnition of E, this implies that a e [b], and so
b + I E [b]. On the other hand, if x e [b], we must have x — b = i for some i in I, whence x = b + ieb + I. Thus, the inclusion [b] E b + I also holds and equality follows.
The usual practice is to speak of any set of the form b + I as a coset of I in R, and to refer to the element b as a representative of b + I. For future reference we next list some of the basic properties of cosets; these are wellknown facts about equivalence classes (see Appendix A) translated into the present notation. Theorem 32. If I is an ideal of the ring R and a, b e R, then each of
the following is true: 1) a + I = Iifandonlyifael. 2) a + I = b + Iifandonlyifa — be]. 3) Either a + I = b + I or else a + I and b + I are disjoint. 39
40
FIRST COURSE IN RINGS AND IDEALS
l
.3
Given an ideal I of the ring R, let us employ the symbol R/I to denote I the collection of all cosets of I in R; that is,
R/I = {a + IlaeR}. The set R/I can be endowed with the structure of a ring in a natural way; all we need do is deﬁne addition and multiplication as follows: (a+I)+(b+I)=(a+b)+I, (a+I)(b+I)=ab+I. One is faced with the usual problem of showing that these operations are actually welldeﬁned, so that the sum and product of the two cosets a + I and b + I do not depend on their particular representatives a and b. To this end, suppose that a+I=a’+I
and
b+I=b’+I.
Then a — a’ = i1 and b — b’ = i2 for some i1, i2 61. From this we conclude that (a+b)—(a’+b’)=(a—a’)+(b—b’) = i1 + i261, which, by Theorem 3—2, indicates that (a + b) + I = (a’ + b’) + I. The net result is that (a + I) + (b + I) = (a’ + I) + (b' + I). With regard to the multiplication of cosets, we observe that '
ab — a’b’
a(b  b’) + (a — a’)b’ ai2 + ilb’ e I,
Since both the products ai2 and ilb’ must be in I. The implication, of course, is that ab + I = a’b’ + I; hence, our deﬁnition of multiplication in R/I is
meaningful. The veriﬁcation that R/I, under the operations deﬁned above, forms a
ring is easy and the details are left to the reader. To assure completeness, we simply state Theorem 33. If I is an ideal of the ring R, then R/I is also a ring, known as the quotient ring (or factor ring) of R by I. In Theorem 2—9 we saw that certain ideals occur as kernels of homo
morphisms. Let us now demonstrate that every ideal does indeed arise in this manner.
Theorem 3—4. Let I be an ideal of the ring R. Then the mapping nat,: R —> R/I deﬁned by nat,(a) = a + I
is a homomorphism of R onto the quotient ring R/I ; the kernel of nat, is precisely the set I.
THE CLASSICAL ISOMORPHISM THEOREMS
41
Prooﬁ The fact that nat, is a homomorphism follows directly from the manner in which the operations are deﬁned in the quotient ring: nat,(a+b)=a+b+l=(a+I)+(b+I) = nat,(a) + nat,(b); nat,(ab) = ab + I = (a + I)(b + I) nat,(a) nat, (b). That nat, carries R onto R/I is all but obvious; indeed, every element of
R/I is a coset a + I, with a ER, and so by deﬁnition nat,(a) = a + I. Inasmuch as the coset I = 0 + I serves as the zero element for the
ring R/I, we necessarily have
ker (nat,) = {aeR nat,(a) = I} ={aeRa+I=I}=I. The last equality was achieved by invoking Theorem 3—2. It is customary to speak of the function nat,, which assigns to each element of R the coset in R/I of which it is the representative, as the natural,
or canonical, mapping of R onto the quotient ring R/I. When there is no danger of confusion, we shall omit the subscript I in writing this mapping. There are two standard techniques for investigating the structure of a particular ring. One method calls for ﬁnding all the ideals of the ring, in the hope of gaining information about the ring through its local structure. The other approach is to determine all homomorphisms from the given ring into a simpler ring; the idea here is that the homomorphic images will tend to reﬂect some of the algebraic properties of the original ring. (The reader is warned to proceed with some care, since, for example, it is quite possible
for multiplication to be commutative in the image ring, without the given ring being commutative.) Although these lines of attack aim in different directions, Theorems 2—9 and 3—4 show that for all practical purposes these are the same; every ideal determines a homomorphism, and every homo
morphism determines an ideal.
Example 3—1. A simple illustration to keep in mind when working with quotient rings is provided by the ring Z of integers and the principal ideal (n), where n is a positive integer. The cosets of (n) in Z take the form
a + (n) = {a + akEZ}, from which it is clear that the cosets of (n) are precisely the congruence classes modulo n. What we earlier described as the operations for congruence classes in Z" can now be viewed as coset operations in Z/(n):
(a+(n))+(b+(n))=a+b+(n), (a + (n))(b + (n)) = ab + (n).
FIRST COURSE IN RINGS AND IDEALS
42
In short, the ring Zn of integers modulo n could just as well be interpreted as the quotient ring of Z by (n). As regards the incidence of ideals in a quotient ring, it should be noted that the Correspondence Theorem applies, in particular, to the case in which we start with an ideal I of the ring R and take the homomorphism f to be the natural mapping natI: R —> R/I. Since ker (nat,) = I, the conclusion of the Correspondence Theorem is modiﬁed slightly. Theorem 35. Let I be an ideal of the ring R. Then there is a onetoone correspondence between those ideals J of R such that I E J and the set of all ideals J’ of the quotient ring R/I ; speciﬁcally, J’ is given by I
=
1131!].
Viewed otherwise, Theorem 3—5 asserts that the ideals of R/I have the form J/I, where J is an ideal of R containing I. In this context, J/I and
nat,J are both used to designate the set {a + I a 6 J}. By way ofan application ofthese ideas, consider the following statement: The ring 2,, of integers modulo n has exactly one ideal for each positive divisor m of n, and no other ideals. In the ﬁrst place, since Z" = Z/(n), Theorem 3—5 tells us that there is a onetoonc correspondence between those ideals of the ring Z which contain (n) and the set of ideals of Zn. But the ideals of Z are just the principal ideals (m), where m is a nonnegative integer. The outcome is that there is a onetoone correspondence between the ideals of Z,l and those ideals (m) of Z such that (m) 2 (n); this last inclusion occurs if and only if m divides n.
Theorem 36. (Factorization of Homomorphisms). Let f be a homomorphism of the ring R onto the ring R’, and I be an ideal of R such that I E kerf. Then there exists a unique homomorphismf: R/I —> R’
with the property thatf = fa natt. Proof. To start, we deﬁne a function]? R/I —> R’, called the induced mapping, by taking
7(0 + I) =f(a)
(Mill)
The ﬁrst question to be raised is whether or notfis actually welldeﬁned. That is to say, we must establish that this function has values which depend
only upon the cosets of I and in no way on their particular representatives. In orderto seethis, let usassumea + I = b + I. Then a — b e I E kerf.
This means that
ﬁt!) =f(a  b + 17) =f(a — b) .+f(b) =f(b) and, by the manner in which 7 was deﬁned, that 7(a + I) = ﬂb + I). Hence, the function fis constant on the cosets of I, as we wished to demonstrate.
THE CLASSICAL ISOMORPHISM THEOREMS
43
A routine computation, involving the deﬁnition of the operations in R/I, conﬁrms that fis indeed a homomorphism: 7((a+I)+(b+I))=f(a+b+I)=f(a+b)
=f(a) +f(b) =70! + I) +70? + I); and, likewise,
Na + 007 + 1)) =7(ab + I) =f(ab) =f(a)f(b) =7(a + 07(1) + I). In this connection, notice that for each element a e R,
H“) = 7(0 + I) = f(nat,(a)) = (7°Ilat1)(a) whence the equality f = f0 natI. It only remains to show that this factorization is unique. Suppose also thatf = g o nat, for some other function g: R/I —» R’. But then
7(a + I) =f(a) = (g°nat1)(a) = 9(a + I) for all a in R, and so g = f The induced mappingfis thus the only function from the quotient ring R/I into R’ satisfying the equation f = To nat,. Corollary. The induced mapping fis an isomorphism if and only if kerf C I. Proof What 1s required here 1s an explicit description of the kernel of f to wit
kerf: {a + Ilﬂa + I) = 0}
= {a+If(a)=0}
= {a + I aekerf} = nat,(kerf). With reference to Theorem 2—10, a necessary and sufﬁcient condition for
fto be an isomorphism is that kerf: I. In the present setting, this amounts to the demand that nat,(kerf) = I, which in turn is equivalent r to the inclusion kerf E I.
In View of the equality f = f0 natl, the conclusion of Theorem 3—6 is sometimes expressed by saying that the homomorphism f can be factored through the quotient ring R/I or, alternatively, that f can be factored by nat,. What we have proved, in a technical sense, is that there exists one and only one function f which makes the following diagram of maps commutative:
R—f—> R’
&\/f R/I .(Spcaking informally, a “mapping diagram” is commutative if, whenever there are two sequences of arrows in the diagram leading from one ring to
44
FIRST COURSE IN RINGS AND IDEALS
another, the composition of mappings along these paths produces the same function.) A rather simple observation, with farreaching implications, is that whenever I = ker )2 so that both the Factorization Theorem and its Corollary are applicable, f induces a mapping f under which R/I and R’ are isomorphic rings. We summarize all this in the following theorem, a result which will be invoked on many occasions in the sequel.
Theorem 37. (Fundamental Homomorphism Theorem). If f is a homomorphism from the ring R onto the ring R’, then R/ker f 2 R’. Theorem 3—7 states that the images of R under homomorphisms can be duplicated (up to isomorphism) by quotient rings of R; to put it another way, every homomorphism of R is “essentially” a natural mapping. Thus, the problem of determination of all homomorphic images of a ring has been reduced to the determination of its quotient rings. Let us use Theorem 3—7 to prove that any homomorphism onto the ring of integers is uniquely determined by its kernel. As the starting point of our endeavor, we establish a lemma which is of independent interest.
Lemma. The only nontrivial homomorphism from the ring Z of integers into itself is the identity map iz. Proof Because each positive integer n may be written as n = l + 1 + + 1 (n summands), the operationpreserving nature of f implies that f(n) = nf(1). On the other hand, if n is an arbitrary negative integer, then — n e Z+ and so
f(n) =f(—(—n)) = —f(n) = —(n)f(l) = "f(l). Plainly,f(0) = 0 = 0f(1). The upshot is thatf(n) = nf(1) for every n in Z. Because f is not identically zero, we must have f(1) = l; to see that this is so, simply apply the cancellation law to the relation f(m) = f(m1) = f(m)f(1), where f(m) 7E 0. One ﬁnds in this way that f(n) = n = iz(n) for all n e Z, making f the identity map on Z. Corollary. There is at most one homomorphism under which an arbitrary ring R is isomorphic to the ring Z. Proof. Suppose that the rings R and Z are isomorphic under two functions f, g: R —> Z. Then the composition fo g'1 is a homomorphic mapping from the ring Z onto itself. Knowing this, the lemma just proved implies
thatfog'1 = iz,orf= y. We now have the necessary information to prove the following result. Theorem 38. Any homomorphism from an arbitrary ring R onto the ring Z of integers is uniquely determined by its kernel.
THE CLASSICAL ISOMORPHISM THEOREMS
45
Proof Let f and g be two homomorphisms from the ring R onto Z with the property that kerf = ker g. Our aim, of course, is to show thatfand g must be the same function. Now, by Theorem 3—7, the quotient rings R/ker f and R/ker g are both isomorphic to the ring of integers via the induced mappings fand g, respectively. The assumption that f and g have a common kernel, when combined with the preceding corollary, forces ,7 = a. It follows at once from the factorizations f=.7° natlnerf’
g = gonatkerg
that the functions fand g are themselves identical. The next two theorems are somewhat deeper results than usual and require the full force of our accumulated machinery. They comprise what are often called the First and Second Isomorphism Theorems, and have important applications in the sequel. (The reader is cautioned that there seems to be no universally accepted numbering for these theorems.) Theorem 39. Let f be a homomorphism of the ring R onto the ring R’ and let I be an ideal of R. If ker f s I, then R/I 2 R’/f(I). Prooﬁ Before becoming involved in the details of the proof, let us remark that the corollary to Theorem 2—8 implies that f(I) is an ideal of the ring R’; thus, it is meaningful to speak of the quotient ring R’/f(I). Let us now deﬁne the function g: R —> R’/f(I) by g = nat[(1) o f, where natﬂn: R’ —> R’/f(I) is the usual natural mapping. Thus, g merely assigns to each element a e R the coset f(a) + f(I) in R’/f(I). Since the functions f and natw) are both onto homomorphisms, their composition carries R homomorphically onto the quotient ring R’/f(I). The crux of the argument is to show that ker g = I, for then the desired conclusion would be an immediate consequence of the Fundamental Homomorphism Theorem. Since the zero element of R’ /f(I) is just the coset f(I), the kernel of g consists of those members of R which are mapped by g ontof(I):
ker g = {a e R 9(a) = f(I)}  {a e R N) + {(1) = m} {a e R f(a)6f(1)} = f“‘(f(I))The hypothesis that ker f g I allows us to appeal to the lemma preceding
Theorem 2—11, from which it may be concluded that I = f '1(f(1)). But then I = ker g, completing the argument.
When applying this result, it is sometimes preferable to start with an arbitrary ideal in R’ and utilize inverse images rather than direct images. The theorem can then be reformulated in the following way.
FIRST COURSE IN RINGS AND IDEALS
46
Corollary 1. Let f be a homomorphism from the ring R onto the ring R'. HP is any ideal ofR’, then R/f‘1(1’) z R’/I’. Proof. In compliance with the corollary to Theorem 2—8, f ‘ 1(1’) forms an ideal of R. Furthermore, kerf E f ' ‘(I’), so that Theorem 3—9 leads directly to the isomorphism
R/f “(1') 2 R'/f(f'1(1’)) = R'/ ’Another special case, itself of interest, is the following.
Corollary 2.
Let I and J be two ideals of the ring R, with J E I. Then
I/J is an ideal of R/J and (R/J)/(I/J) 2 R/I.
Proof. AS we know, if I is an ideal of the ring R andfis any homomorphism of R, then f(I) constitutes an ideal of the image f(R). In the setting at hand, take f to be the natural mapping natJ: R —> R/J ; then I/J = nat, 1 forms an ideal of the quotient ring R’ = R/J. Since ker (natJ) = J E I, Theorem
3—9 implies that R/I is isomorphic to (R/J)/(I/J) under the induced mapping a where g = natm o natJ. The diagram displayed below may be of some help in visualizing the situation described by the last corollary: R
ml,
—> R/I
mt,
g
R/J :57 (R/ /(I/J) By virtue of our assumptions, there exists a (necessarily unique) isomorphism g : R/I —> (R/J)/(I/J) such that E o nat, = natm o natJ.
Let us now take up the second of our general isomorphism theorems. Theorem 3—10. If I and J are ideals of the ring R, then I/(I n J) z (I + J)/J. Proof. Reasoning as in Theorem 3—9, we seek a homomorphism f from I (regarded as a ring) onto the quotient ring (I + J)/J such that ker f = I n J. Our candidate for this function fis deﬁned by declaring that f(a) = a + J, a e I.
A trivial, but useful, observation is that I E I + J, whence f can
be obtained by composing the injection map i:3 I —> I + J with the natural mapping natJ: I + J —> (I + J)/J. To be quite explicit, f = nat, o i, or, in diagrammatic language,
I
"
I+J “3':
J'
(I + J)/’
THE CLASSICAL ISOMORPHISM THEOREMS
47
From this factorization, it is easy to see that f is a homomorphism with f") = (I + J)/J. To conﬁrm that the kernel offis precisely the set I n J, notice that the coset J serves as the zero element of (I + J)/J, and so
kerf= {aeIf(a) = J}
= {aeIa + J = J} = {aeIaeJ} = InJ.
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, twosided) 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
I" = {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 a1, a2, ..., ale I
(distinct or not), the product ala2 an = 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/I 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/I are nil rings and that a e R. Then there exists some positive integer n for which the coset
(a+I)"=a"+I=I, signifying that the element a” e I. Inasmuch as I is a nil ideal, (a")"' = a" = 0 for some m e Z +. This implies that a is nilpotent as a member of R and, in consequence, R is a nil ring. The remainder of the proof is left to the
reader’s care. Lemma. If N1 and N2 are two nil (nilpotent) ideals of the ring R, then their sum N1 + N2 is likewise a nil (nilpotent) ideal.
Proof. With reference to Theorem 3—10, we have (N1 + N2)/N1 : Nz/(N1 n N2). The righthand side (hence, the lefthand side) of this equation is a nil ring, being the homomorphic image of the nil ideal N2.
48
FIRST COURSE IN RINGS AND IDEALS
Since (N1 + N2)/N1 and N1 are both nil, it follows from the previous lemma that N1 + N2 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 311. The sum 2 N of all the nil ideals N of the ring R is a. nil ideal. Proof. If the element a e 2 N,, then, by deﬁnition, a lies in some ﬁnite sum
of nil ideals of R; say, aENl + N2 + + N”, where each Nk is nil. By virtue of the last corollary, the sum N1 + N2 + + N,. must be a nil ideal; hence, the element a is nilpotent. This argument shows that Z 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 Z N,, the sum of all nil ideals. But 2 N1 is itself a nil ideal, making N nil. Example 3—2. For examples of nilpotent ideals, let us turn to the rings Zp", where p is a ﬁxed prime and n > 1. By virtue of the remarks on page 42, ZI," has exactly one ideal for each positive divisor of p" and no other ideals; these are simply the principal ideals (pk) = p"Z,,n (0 S k S n). For 0 < k s n, we have
(p“)”=(p”")=(0)={0}, so that each proper ideal of ZI," is nilpotent. Before leaving this chapter, we should present an example to show that, in general, nil and nilpotent are diﬂ‘erent concepts.
Example 33. For a ﬁxed prime p, let S be the collection of sequences a = {a,,} with the property that the nth term a,' e ZV. (n 2 1). S can be made into a ring by performing the operations ofaddition and multiplication term by term: {an} + {bu} = {an + bu},
{an}{bn} = {anbn}'
The reader will ﬁnd that the zero element of this ring is just the sequence formed by the zero elements of the various a and the negative of {an} is {—an}. Now, consider the set R of all sequences in S which become zero
PROBLEMS
49
alter 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 nonnilpotcnt nil ideal. Let us denote by I the set of sequences in R whose nth term belongs to the principal ideal in Z1,. generated by p; in other words, the sequence a e I if and only if it is of the form a = (prl, prz,
, pr", 0, 0, ...)
(rt eZpk).
A routine calculation conﬁrms that I is an ideal of R. Since each term of
a is nilpotcnt in the appropriate ring, it follows that
a” = 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) as I for which a” =ﬁ O. For instance, deﬁne
a = {aﬁbytakingak = pifk = 1,2, ...,n +1anda,‘ = Oifk > n +1; that is, a = (p,...,p,p,0,...)
with
n +1p’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 Zr“, the sequence a'' 7E 0, implying that 1" 9E {0}. As this argument holds for any n 6 2+, the ideal I cannot be nilpotcnt.
We shall return to these ideas at the appropriate place in the sequel, at which time their importance will become clear.
PROBLEMS 1. Let a be an equivalence relation on the ring R. We say that a is compatible (with the ring operations) if and only if a E b implies a + c a b + c, ac E bc, ca 5 ob for all a, b, c e R. Prove that there is a onetoone 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,' = {nala e R} and J,l = {a 6 RIM = 0} are both ideals ofR; b) char (R/In) divides n;
c) if char R 7E 0, then char R divides n char (R/Jl).
50
FIRST COURSE IN RINGS AND IDEALS
. Let I be an ideal of the ring R. Establish each of the following: a) R/I has no divisors of zero if and only if ab 6 I implies that either a or b belongs to I.
b) R/I is commutative if and only if ab — ba 6 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 — aeIand ea — aeIforallainR. d) Whenever R is a commutative ring with identity, then so is the quotient ring R/I. . Let R be a commutative ring with identity and let N denote the set of all nilpotent elements in R. Verify that a) The set N forms an ideal of R. [Hint: If a" = b"I = 0 for integers n and m, consider (a — b)""'.] b) The quotient ring R/N has no nonzero nilpotent elements. . Prove the following generalization of the Factorization Theorem: Let f1 and f2 be homomorphisms from the ring R onto the rings R1 and R2, respectively. If kerf1 E kerfz, then there exists a unique homomorphism 7: R1 —> R2 satisfying f2 = 7° f1. [Hint: Mimic the argument of Theorem 3—6; that is, for any element f1(a) e Rn deﬁne f(f1(a)) = 1301).] . Let I be an ideal of the ring R. Assume further that J and K are two subrings of RwithI E L] S K. Show that
a) J s K ifand only ifnat,J E nat b) nat,(J n K) = natIJ n nat,K. . 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 I of R satisfying 1 c J c R; b) if R is a principal ideal ring, then so is the quotient ring R/I ; in particular, Z.
is a principal ideal ring for each n e Z+. [Hint: Problem 17, Chapter 2.] . a) Given a homomorphism f from the ring R onto the ring R’, show that
{f '1(b)b 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 integers are the rings Z", n > 0, and {0}. . Suppose that S is 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 R/I. [Hint: Utilize the mapping f(a) = a + I, where a e S.] 10. A commutator in a ring R is deﬁned to be any element ofthe 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 ifand 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/I is commutative if and only if
[R, R] g 1. 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] E kerf;
b) f = fa nat R], where fis the induced mapping; c) if kerf E [R, R], then R/[R, R] 2 R'/[ ’, R’]. 12. a) Suppose that I1 and [2 are ideals of the ring R for which R = I1 EB 12. Prove that R/Il z 12, and R/I2 z 11. b) Let R be the direct sum of the rings RE (i = 1, 2,, n). If I‘ is an ideal of Ri and! = II GB 1, GB 69 I,,showthat
R/1 z (RI/I1) 6 (Rz/Iz) 6
$ (R./1..)
[Hint. Find the kernel of the homomorphism f: R —> 2‘. 69 (Ri/Ii) that sends
a = (a1, a2, ...,a,,) tof(a) = (a1 + Iva2 + 12, ...,a,l + In).]
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(1)a) Show that h is a welldeﬁned isomorphism onto R’/f(I); hence, R/I 1: R’/f(I). b) Establish that h is the unique mapping that makes the diagram below commutative:
R LR' =f(R) R/I T R'/f(1) 14. Given integers m, n e Z +, establish that
a) ifm divides n, then Zn/(m)/(n) z z"; b) if m and n are relatively prime, then Z”, 2 Z,” 6 Z”.
15. If I is an ideal of the ring R, prove that the matrix ring MAR/1) is isomorphic to M,,(R)/M,I (I). [Hint: Consider the mapping f: Mu(R) —> MAR/I) deﬁned by f«“u» = (“u + 0]
.
16. Let R be a ring without divisors of zero. Imbed R in the ring R’ = RxZ, as described in Theorem 2—12. (The case R = Z, illustrates that R’ may contain zero divisors even though R does not.) Assuming that I denotes the left annihilator
of R in R’,
I = {aeR’Iar = Ofor all reR}, 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 to R. [Hint: Utilize Problem 9.]
FOUR 7‘
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 ﬂ 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 — {O} is a commutative group under the multiplication of F (the identity of this group will be written as l).

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 a0 = 0 = 0a for any a e F, all the members of F commute under multiplication and not merely the nonzero elements. Similarly, the relation 10 = 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 41. Here are some of the more standard illustrations of ﬁelds:
the set Q of all rational numbers, the set F = {a + bﬂla, b e 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 42. Consider the set C = R“ xR" of ordered pairs of real numbers. To turn C into a ﬁeld, we deﬁne addition and multiplication by
(a,b) + (c,d) = (a + c,b + d),
(a, b)(c, d) = (ac — bd, ad + be)The reader may verify without diﬂiculty 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 + 0, so that a1 + b2 > 0; thus,
L
—_b
2 + b2 ’ a2 + b2
exists in C and has the property that
(ab) a —b \={a2+b2a(—b)+ab\= (1, 0). ’ \a2+b2’a2+b2] \a2+b2’ a2+b2) This shows that each nonzero member of C has an inverse under multi
plication, 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, 0)a e R#}, it follows that R’“ z R"E 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, l) is such that (0, 1)2 = (0, 1)(0, 1) = (— l, 0). Introducing the symbol i as an abbreviation for (0, l), we have
(a, b) = (a, 0) + (b, 0)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 43. For an illustration of a division ring which is not a ﬁeld, we turn to the ring of (Hamilton’s) real quaternions. To introduce this ring, let the set H consist of all ordered 4—tuples of real numbers: H = {(a, b, c, d)a, b, c, d e R#}. Addition and multiplication of the elements of H are deﬁned by the rules (a, b, c, d) + (a’, b’, c’, d’) = (a + a', b + b’, c + c’, d + d’), (a, b, c, d)(a’, b’, c’, d’) = (aa’ — bb’ — cc’ — dd’, ab’ + ba’ + cd’ — dc’, ‘ ac’ — bd’ + ca’ + db’, ad’ + bc’ — cb’ + 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
l = (1, 0, 0, 0),
i = (0, 1, 0, 0), j = (0, 0, 0, 1, 0),
k = (0, 0, 0, l).
The elements 1, i, j, k have a number of distinctive properties; speciﬁcally,
I is the multiplicative identity of H and
i2=j2=k2= —1, ij=k, jk=i,
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, 19,0, d) = (a, 0, 0, 0)1 + (b, 0, 0, 0)i + (c, 0, 0, 0)i + (d, 0, 0, 0)k Since the subring {(r, 0, 0, 0)r 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 + cj + dka,b,c,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 fourdimensional 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, c, 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: i=a—bi—cj—dk.
INTEGRAL DOMAINS AND FIELDS
55
It is easily veriﬁed that the product
q§=¢jq=a2—(bi+cj+dk)2=a2+b2+c2+d2+0, thus exhibiting that q has the multiplicative inverse
q'1 = (a2 + b2 + c2 + d2)'1¢j. Incidentally, the totality of all members of H of the form (a, b, 0, 0) =
a + bi, the special quatemions, constitute a subring isomorphic to C; as substitutes, one might also consider the set of all elements (a, 0, b, 0) or all
elements (a, 0, 0, b). In this light, the real quatemions may be viewed as a suitable generalization of the complex numbers. The following theorem shows that any ﬁeld is without divisors of zero, and consequently a system in which the cancellation law for multiplication holds. ' Theorem 41. Every ﬁeld F is an integral domain. Prooﬂ Since every ﬁeld is a commutative ring with identity, we need only prove that F contains no zero divisors. To this purpose, suppose a, b e F, with ab = 0. If the element a 9E 0, then it must possess a multiplicative inverse a" 1 e F. But then the hypothesis that ab = 0 yields
0 = a‘10 = a‘1(ab) = 1b ='b, as desired.
There obviously exist integral domains which are not ﬁelds; a prime example is the ring Z of integers. However, an integral domain having only a ﬁnite number of elements must necessarily be a ﬁeld. Theorem 42. Any integral domain R with only a ﬁnite number of ideals is a ﬁeld. Proof Let a be any nonzero element of R. Consider the set of principal ideals (a"), where n e Z + :
(a" = {r a"r e R}. Since R has only a ﬁnite number of distinct ideals, it follows that (a"') = (a”) for certain positive integers m, n with m < n.
Now, a“, as an element of
(a"), must lie in (a"). This being so, there exists some 76R for which a“ = F a". By use of the cancellation law,
1 = ranm = (ram1y; Because multiplication is commutative, we therefore have a‘ 1 = 7 anm1 This argument shows that every nonzero element of R is invertible; hence,
R forms a ﬁeld.
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 adapt: to a variety of situations in which the underlying ring is ﬁnite. The reasoning proceeds as follows. Suppose that al, a2, , a, are tll members of the integral domain R. For a ﬁxed nonzero element a e R, we
consider the n products aal, aaz, , aau. These products are all distinct, for if aai = aaj, the cancellation law (valid in any integral domain) would
yield a, = aj. It follows that each element of R must be of the form on, for some choice of i. In particular, there exists some a, e R such that a5, = 1. From the commutativity of multiplication, we infer that a" 1 = 6,, 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 commutao tivity 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 2,. It has been previously shown that, for each positive integer n, ZI comprises a commutative ring with identity. A reasonable question is: F0! 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 43. A nonzero element [a] eZ,l is invertible in the ring Zn if and only if a and n are relatively prime integers (in the sense that god(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] = [W] +.. ["8] = [ar] +.. [0] = [a] '. [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
element of 2,; say, with inverse [b]. We thus have [ab] = [a] ,, [b] = [1]. so that there exists an integer k for which ab — l = kn. ab + n(—k) = 1; hence, a and n are relatively prime integers.
But then
Corollary. The zero divisors of Z,I are precisely the nonzero elements
of 2,, which are not invertible. Proof. Naturally, no zero divisor of Zn can possess a multiplicative inverse.
On the other hand, suppose that [a] 3!: [0] is not invertible in Z”, so that
INTEGRAL DOMAINS AND FIELDS
57
MM. n)’ = d, where l < d < n. Then, a = rd and n = sd for suitable nonzero integers r and s. This leads to
[a] 3. [S] = [GS] = ["13] = ["1] = [0]Since the deﬁning properties of 5 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 44. The ring Z,' of integers modulo n is a ﬁeld if and only if n is a prime number. If n is composite, then 2,, is not an integral domain
and the zero divisors of Z,I are those nonzero elements [a] for which
gcd(a, n) + 1.
Every ﬁeld necessarily has at least two elements (1 being diﬁ‘erent from 0); Theorem 4—4 indicates that there is a ﬁeld having this minimum number as its number of elements, viz. Z2. , 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/kerf 2 F. But ker f = (n) for some positive integer n, since Z is a principal ideal domain. (In this connection, observe that n + 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 Zn 2 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 Zn, forms a ﬁeld, n must be a prime number. A useful counting function is the socalled Euler phifunction (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—3, ¢(n) may also be characterized as the number of positive integers < n which are relatively prime to n. For instance, 45(6) = 2, 415(9) = 6, and ¢(12) = 4; it should be equally clear that whenever p is a prime number, then ¢(p) = p — 1. Lemma. If G,l is the subset of Z,I deﬁned by
G" = {[a] e Z,,a is relatively prime to n}, then (Gu, ,,) forms a ﬁnite group of order ¢(n).
Proof In the light of the preceding remarks, (Gn, ~,,) 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 45. (EulerFermat). If n is a positive integer and a is relatively prime to n, then a”) E 1 (mod n). Proof. The congruence class [a] can be viewed as an element of the multiplicative group (6", ,,). Since this group has order d>(n), it follows that
[a]4’(") = [1] or, equivalently, a”) E 1 (mod. n). (Recall that if G is a ﬁnite group of order k, then x" = 1 for all x e G.)
There is an interesting relationship between ﬁelds and the lack of ideals; what we shall show is that ﬁelds have as trivial an ideal structure as possible. Theorem 4—6. Let R be a commutative ring with identity. Then R is a ﬁeld if and only if R has no nontrivial ideals. Proof. Assume ﬁrst that R is a ﬁeld. We wish to show that the trivial ideals
{0} and R are its only ideals. Let us suppose to the contrary that there exists some nontrivial ideal I of R. By our assumption, the subset I is such
that I aé {0} and I 7E R. This means that there exists some nonzero element a e I. Since R is taken to be a ﬁeld, a has a multiplicative inverse a‘ 1 present in R. By the deﬁnition of ideal, we thus obtain 1 = a‘ 1 a e I, which in turn
implies that I = R, contradicting our choice of I. Conversely, suppose that the ring R has no nontrivial ideals. Given a nonzero element a e R, consider the principal ideal (a) generated by a:
(a) = {rare R}. Now, (a) cannot be the zero ideal, inasmuch as a = a 1 e (a), with a 7E 0. It follows from the hypothesis that the only other possibility is that (a) = R.
In particular, since 1 e (a), there exists an element 7' e R for which 7 a = l. Multiplication is commutative, so that 7' = a‘l. Therefore, each nonzero
element of R is multiplicatively invertible and we are done. In View of this last result, the ring Z of integers fails to be a ﬁeld since it contains the nontrivial ideal 2,. Theorem 4—6 is useful in revealing the nature of homomorphism between ﬁelds. We exploit it to prove Theorem 4—7. Let f be a homomorphism from the ﬁeld F into the ﬁeld F’. Then eitherfis the trivial homomorphism or elsefis oneto—one. Proof. The proof consists of noticing that since ker f is an ideal of the ﬁeld
F, either ker f = {0} or else ker f = F. The condition ker f = {0} implies that f is a onetoone function.
On the other hand, if it happens that
ker f = F, then each element of F is carried onto 0; that is to say, f is the
trivial homomorphism. Corollary. Any homomorphism of a ﬁeld F onto itself is an auto
morphism of F.
INTEGRAL DOMAINS AND FIELDS
59
Any ring with identity which is a subring of a ﬁeld must of necessity be an integral domain. Turning the situation around, one might ask whether each integral domain can be considered (apart from isomorphism) as a subring of some ﬁeld. More formally: Can a given integral domain be Embedded in a ﬁeld? In the ﬁnite case there is plainly no diﬁiculty, since any ﬁnite integral domain already forms a ﬁeld. Our concern with this question arises from the natural desire to solve the linear equation ax = b, where a aé 0. A major drawback to the notion of an integral domain is that it does not always furnish a solution within the system. (Of course, any such solution would have to be unique, since (m1 = b = ax2 implies that x1 = x2 by the cancellation law.) It hardly seems necessary to point out that when the integral domain happens to be a ﬁeld, the equation ax = b (a 7E 0) is always solvable, for one need only
take x = a'1b. We begin our discussion of this problem with an obvious deﬁnition. Deﬁnition 42. By a subﬁeld of a ﬁeld F is meant any subring F’ of F
which is itself a ﬁeld.
For example, the ring Q of rational numbers is a subﬁeld of the real
ﬁeld R“ ; the same is true of the ﬁeld F = {a + bﬂa, b e Q}.
Surely, the set F’ will be a subﬁeld of the ﬁeld F provided that (1) F’ is a subgroup of the additive group of F and that (2) F’ — {0} is a subgroup of the multiplicative group F — {0}. Recalling our minimal set of conditions for determining subgroups (see page 8), it follows that F’ will be a subﬁeld
of F if and only if the following requirements are met: I) F’ is a nonempty subset of F containing at least one nonzero element, 2) a,beF’implya— beF’,and
3) a, b e F’, with b 7E 0, imply ab”1 eF’. The coming theorem furnishes a clue to the nature of the ﬁeld in which we wish to imbed a given integral domain.
Theorem 4—8. Let the integral domain R be a subring of the ﬁeld F. If the set F’ is deﬁned by
F’ ={ab‘1a,beR;b aé 0}, then F’ forms a subﬁeld of F with R S F’; in fact, F’ is the smallest
(in the sense of inclusion) subﬁeld of F containing R.
Proof Notice ﬁrst that l = 11‘ 1 e F’, so that F’ 7E {0}. Now consider two arbitrary elements x, y of F’. With reference to the deﬁnition of F’, we then
have
x = ab'l,
y = ed—1
FIRST COURSE IN RINGS AND IDEALS
60
' for a suitable choice of a, b, c, deR, where b aé 0, d 7E 0. computation shows that
A simple
x — y = (ad — bc)(bd)'1 e F’. Also, if y is nonzero (that is, whenever c 79 0), we conclude that
xy'1 = (ad)(cb)‘l e F’. By virtue of the remarks following Deﬁnition 4—2, this is sufﬁcient to establish that the set F’ is a subﬁeld of F. Furthermore,
a = a1 = al‘leF’
for each a in R, implying that R E F'. Finally, any subﬁeld of F which contains R necessarily includes all products ab‘l, with a, 0 at: b e R, and, hence, contains F’.
Theorem 4—8 began with an integral domain already imbedded in a ﬁeld. In the general case it becomes necessary to construct the imbedding ﬁeld. Since the expression ab" 1 may not always exist, one must now work with ordered pairs (a, b), where b + 0. Our thinking is that (a, b) will play a role analogous to ab" 1 in the foregoing theorem. Actually, the proposed construction is not just conﬁned to integral domains, but will apply to a much wider class of rings; it will irnbed any commutative ring R that contains a (nonempty) set of elements that are not zero divisors in a ring Qc,(R), which may be described as follows.
Deﬁnition 43. Let R be a ring with at least one nonzerodivisor. A classical ring ofquotients of R is any ring c(R) satisfying the conditions 1)
R
E
c(R)r
2) every element of Q,,(R) has the form ab‘l, where a, b e R and b is a nonzero—divisor of R, and
3) every nonzerodivisor of R is invertible in Qc,(R). Convention. An element a e R is termed a nonzerodivisor if ar aé 0, and
ra 9E 0 for all 0 7E r e R; in particular, the phrase “nonzerodivisor” excludes the zero element. As a starting point, let S denote the set ofall elements ofR, a commutative ring, which are nonzerodivisors; we wﬂl assume hereafter that S 79 Q. Needless to say, if there happens to be an identity element 1 available, then 16 S. Notice, too, that the set S is closed under multiplication. For, suppose that the elements 51, s2 6 S and (slsz)a = 0. Then s1(s2a) =7 0 and, since s1 is not a divisor of zero, it follows that sza = 0; this in turn
implies that a = 0. Therefore, the product sls2 is not a zero divisor of R, ' whence sls2 e S.
INTEGRAL DOMAINS AND FIELDS
61
Now consider the set of ordered pairs
RxS= {(a,s)aeR, seS}. A relation ~ may be introduced in R x S by taking (a, 5) ~ (b, r) if and only if ar = bs. (We have in mind the previous theorem, where ab‘1 = cal‘1 if and only if ad = be.) It is not difﬁcult to verify that the relation ~, thus deﬁned, is an equivalence relation in R x S. The transitive property is perhaps the least obvious. To see this, assume that (a, 3) ~ (b, r) and (b, r) ~ (c, it), so that ar = Sb,
bi: = rc.
Now multiply the ﬁrst equation by t and the second by s to get art = sbt,
sbt = src.
Putting these relations together, we obtain atr = scr. Since r is not a zero divisor, the cancellation law gives us at = sc, which is exactly the condition that (a, 5) ~ (c, 1:). Next, we label those elements of R x S which are equivalent to the pair (a, s) by the symbol a/s; in other words, “/8 = {(b, 7') (a9 S) ~ (179 7)}
= {(b, r) ar = sb}.
The collection of all equivalence classes a/s relative to ~ will be denoted by Qc(R)3
c(R) = {a/sla e R; s e S}. From Theorem A—l, we know that the elements of c(R) constitute a partition of the set RxS. That is, the ordered pairs in Rx S fall into disjoint classes (called formalfractions), with each class consisting of equivalent pairs, and nonequivalent pairs belonging to diﬂ'erent classes. Further, two such classes (1/5 and b/r are identical if and only if ar = sb; in particular, all fractions of the form as/s, with s e S, are equal. ' With these remarks in mind, let us introduce the operations of addition
and multiplication required to make c(R) into a ring. means of the formulas
We do this by
a/s + b/r = (ar + sb)/sr, (a/s)(b/r) = ab/sr. Notice, incidentally, that since the set S is closed under multiplication, the righthand sides of the deﬁning equations are meaningful. As usual, our ﬁrst task is to justify that these operations are welldeﬁned;
62
FIRST COURSE IN RINGS AND IDEALS
that iS to say, it iS necessary to Show that the sum and product are independent of the particular elements of R used in their deﬁnition. Let us present the argument for addition in detail. Suppose, then, that a/s = a’/s’ and b/r = b’/r’ ; we must Show that (or + sb)/sr = (a’r’ + s’b’)/s’r’. From what is given, it follows at once that h
as’ = sa’,
br’ = rb’.
These equations imply (or + sb)(s’r’) — (a’r’ + S’b’)(sr) = (as’ — sa’)(rr’) + (br’ — rb’)(ss’) = 0(rr’) + 0(SS’) = 0. By the deﬁnition of equality of equivalence classes, this amounts to saying that (or + Sb)/sr = (a’r’ + s’b’)/s’r’, which proves addition to be welldeﬁned. In much the same way, one can establish that
ab/sr = a'b’ s’r’. The next lemma reveals the algebraic nature of c(R) under these operations. Lemma. The system c(R) forms a commutative ring with identity. Proof. It is an entirely straightforward matter to conﬁrm that Qc,(R) is'a
commutative ring. We leave the reader to make the necessary veriﬁcations at his leisure, and merely point out that 0/3 serves as the zero element, while — a/s is the negative of a/s. That the equivalence class s’/s’, where S’ iS any ﬁxed nonzerodivisor of R, constitutes the multiplicative identity is evidenced by the following computation : (a/s)(s’/s’) = as’/ss’ = a/s for arbitrary a/s in Qc,(R), since (as’)s = (ss’)a. Loosely speaking, common factors belonging to S may be .cancelled in a fraction as’/ss’. This proves part of the theorem. below. Theorem 49. Any commutative ring R with at least one nonzerodivisor possesses a classical ring of quotients. Proof. We begin by establishing that the ring c(R) contains a subring isomorphic to R. For this, consider the subset K of Qﬂ(R) consisting of all
INTEGRAL DOMAINS AND FIELDS
63
elements of the form aso/so, where so is a ﬁxed nonzerodivisor of R (recall that the equivalence class as'o/s0 depends only upon a, not upon the choice of so):
K = {use/so e armlae R}. The reader can easily check that K is a subring of Qc,(R). An obvious (onto) mapping f: R —r K is deﬁned by taking f(a) = aso/so. Since the condition
aso/so = bso/so implies that asé = bsfJ or, after cancelling, that a = b, f will be a onetoone function. Furthermore, it has the property of preserving both addition and multiplication:
f(0 + b) = (a + b)So/So = aso/So + bSo/So = f(a) + fU?) f(ab) = (ab)So/So = (41195033 = (“so/So)(bSo/so) = f(a)f(b)In this way, R can be isomorphically embedded in Qc,(R). By identifying R with K, we may henceforth regard R as actually being contained in c(R). In practice, one simply replaces the fraction aso/so e K by the corresponding element a e R. We proceed to show that all the elements of S are invertible in Qc1(R). Any nonzerodivisor s e S has, after identiﬁcation, the form sso/so. Now, the equivalence class so/sso is also a member of c(R) (note the crucial use of the closure of S under multiplication) and satisﬁes the equation (sso/so)(so/sso) = 533/533 = 50/50
Since so/so plays the role of the identity element for Qc,(R), we see at once
that (sso/s0)'l = so/sso. All that remains to complete the proof is to verify that each member 0/3 of Qc.(R) can be written as as' 1. It should be clear that “/3 = (“so/SoXSo/Sso) = (“So/so)(sso/So)_1
Replacing aso/s0 by a and sso/so by s, the displayed equation assumes the more familiar form a/s = as‘ 1. The point is this: the set Q,,(R) may now be interpreted as consisting of all quotients as‘ 1, where a e R, s e S. Thus, Qc,(R) satisﬁes Deﬁnition 4—3 in its entirety, thereby becoming a classical ring of quotients of R. Two comments are in order. In the ﬁrst place, given any element s e R which is a nonzerodivisor, it follows that (sso/so)(a/s) = “530/350 = “so/So
(50 ES)
ldentifying sso/so with s and aso/s0 with a, we conclude from this that the equation sx = a always possesses a solution in c(R), namely, x = a/s = as“ ‘. Second, notice that in Qc.(R) multiplicative inverses exist not only for
64
FIRST COURSE IN‘ RINGS AND IDEALS
members of S but for all elements of Q,.(R) which can be represented in the form r/s, where r, s are both nonzerodivisors; in fact,
(r/s)(s/r) = rs/sr = 30/30.
When the ring R is an integral domain, we may take the set S ofnonzerodivisors as consisting of all the elements of R which are not zero. The last remark of the preceding paragraph then leads to the following important theorem. Theorem 410. For any integral domain R, the system QC,(R) forms a ﬁeld, customarily known as theﬁeld of quotients of R.
Since an integral domain is (isomorphic to) a subring of its ﬁeld of quotients, we also obtain
Corollary. A ring is an integral domain if and only if it is a subring of a ﬁeld. It should be pointed out that the hypothesis of commutativity is essential to this last theorem; indeed, there exist noncommutative rings without
divisors of zero that cannot be imbedded in any division ring. The ﬁeld of quotients constructed from the integral domain Z is, of course, the rational number ﬁeld Q. Another fact of interest is that the ﬁeld of qubtients is the smallest ﬁeld in which an integral domain R can be imbedded, in the sense that any ﬁeld in which R is imbeddable contains a subﬁeld isomorphic to Qc,(R) (Problem 20). The existence theorem for the classical ring of quotients can be supplemented by the following result, which shows that it is essentially unique. Theorem 4—11. Let R and R’ be two commutative rings, each containing at least one nonzerodevisor. Then, any isomorphism ofR onto R' has a unique extension to an isomorphism of Qc,(R) onto c(R’).
Proof. To begin with, each member of Qc,(R) may be written in the form ab ‘ 1, where a, b e R and b is a nonzerodivisor in R. Given an isomorphism
(b: R —> R’, the element ¢(b) will be a nonzerodivisor of R’, so that (Mb)—1 is present in Q,,(R’). Suppose that (15 admits an extension to an isomorphism (I): c(R) —’ Qc1(R’). Since a = (ab1)b, we would then have
(15(0) = (a) = (I’(ab")(b) = (9(ab'1)¢(b), which, as a result, yields @(ab‘l) = ¢(a)¢(b)'1. Thus, (I) is completely determined by the eﬁect of ¢ on R and so determined uniquely, if it exists at all. These remarks suggest that, in attempting to extend (15, we should consider the assignment:
(Nab—1) = 4J(a)¢(b)'ll
for all
ab" 6 c(R)'
INTEGRAL DOMAINS AND FIELDS
65
For a veriﬁcation that (I) is a welldeﬁned function, let ab‘1 = cai'1 in Q¢.(R); that is to say, ad = be in R. Then. the equation ¢(a)¢(d) = ¢(b)¢(c) holds in R’ E Qc.(R’) or, viewed otherwise, 91>(a)(b)_1 = ¢(c)¢(d)'1. But this means @(ab‘l) = (Med1), so that (I) does not depend on how an element in Qc,(R) is expressed as a quotient. One veriﬁes routinely that (I), as deﬁned above, is a homomorphism of
Q..(R) into c(R’). This homomorphism certainly extends 1, is maximal if and only if n is a prime. Suppose that (n) is a maximal ideal of Z. If the integer n is not prime,
then n = nlnz, where 1 < n1 < n2 < n. This implies that the ideals (in!) and (n2) are such that
(n) C ("1) C Z,
(n) C ("2) C Z.
contrary to the maximality of (n). For the opposite direction, assume now that the integer n is prime. If the principal ideal (n) is not maximal in Z, then either (n) = Z or else there exists some proper ideal (m) satisfying (n) c (m) 1; this is equally untenable, since n is prime, not composite. At any rate, we conclude that (n) must be a maximal ideal.
Example 52. For an illustration of the practicality of Theorem 5—1, we take R = mapR’“, a commutative ring with identity (Example 4, Chapter 1). Consider the set M of all functions which vanish at 0:
M = {fe Rf(0) = o}. Evidently, M forms an ideal of the ring R; we contend that it is actually a maximal ideal. Indeed, iff 9f M and i is the identity map on R’“ (that is, i(x) = x), one may easily check that (i2 + f2)(x) 7E 0 for each xeR’. Hence, the function i2 + f2 is an invertible element of R.
Since
R 2 (M, j) 2 (i, f), with i2 +f2 e (i, f), this implies that (M, _f)=R; in consequence, M is a maximal ideal of R. (Here (i, f) denotes the ideal
generated by i and f; that is, (i, f) = {ri + sf Ir, s e R}.) Our immediate goal is to obtain a general result assuring the existence of suitably. many maximal ideals. As will be seen presently, the crucial step in the proof depends on Zorn’s Lemma (see Appendix B), an exceedingly powerful tool which is almost indispensable in modern mathematics. Zorn’s Lemma (traditionally called a lemma, but in fact an equivalent form of the Axiom of Choice) asserts:
Zorn’s lemma. If (S, s) is a partially ordered set with the property that every chain in S has an upper bound in S, then S possesses at least one maximal element. Clearly, some partial orderings are more useful than others in applications of Zorn’s Lemma. In our later investigations we shall frequently take S to be a family of subsets of a given set and the partial ordering to be the usual inclusion relation; an upper bound of any chain of elements would
MAXIMAL, PRIME, AND PRIMARY IDEALS
simply be their settheoretic union.
73
For this particular setting, Zorn’s
Lemma may be formulated as follows: Let .1! be a nonempty family of subsets of some ﬁxed nonempty set with the property that for each chain g in .21, the union u ‘6’ also belongs to .91. Then .32! contains a set which is maximal in the sense that it is not properly contained in any member of .21. Because this may be the reader’s ﬁrst contact with Zorn’s Lemma, we proceed in somewhat leisurely fashion to establish Theorem 5—2. If the ring R is ﬁnitely generated, then each proper ideal of R is contained in a maximal ideal. Prooﬂ Let I be any proper ideal of R, a ﬁnitely generated ring; say, R = (a1, a2, , an). We deﬁne a family of ideals of R by taking
.521: {JI E J; J is a proper ideal of R}. This family is obviously nonempty, for I itself belongs to .21.
Now, consider an arbitrary chain {1,} of ideals in .51. Our aim, of course, is to establish that u I, is again a member of .521. To this purpose, let the elements a, b e u Ii and r e R.
Then there exist indices i and j for which
as 1,, be II. As the collection {1,} forms a chain, either I, S I1 or else II s Ii. For deﬁniteness, suppose that I, E Ij, so that both a, b e Ij. But Ij is an ideal of R; hence, the difference a — b t S u 1,. Also, the
products ar and ra e I, E u Ii. All of this shows u Ii to be an ideal of R. Next, we must verify that u Ii is a proper ideal of R. Suppose, to the contrary, that U Ii = R = (a1, a2, , an). Then, each generator ak would belong to some idea] Iik of the chain {1,}. There being only ﬁnitely many I... one contains all others, call it I,“ Thus, a1, a2, ..., a,' all lie in this one
I,.. In consequence, Ii, = R, which is clearly impossible. Finally, notice that I E u I,, whence the union u 1,6 .51. Therefore, on the basis of Zorn’s Lemma, the family 421 contains a
maximal element M. It follows directly from the deﬁnition of .52! that M is a proper ideal of the ring R with I E M. We assert that M is in fact a maximal ideal. To see this, suppose that J is any ideal of R for which M c: J E R. Since M is a maximal element of the family at, J cannot belong to .21. Accordingly, the ideal J must be improper, which is to say that J = R. We thus conclude that M is a maximal ideal of R, completing the
proof. The signiﬁcant point, of course, is that this theorem asserts the existence of certain maximal ideals, but gives no clue as to hOW actually to ﬁnd them.
The chief virtue of Theorem 5—2 is that it leads immediately to the following celebrated result.
74
FIRST COURSE IN RINGS AND IDEALS
Theorem 53. (KrullZorn). In a ring R with identity each proper idd is contained in a maximal ideal.
Proof. An appeal to Theorem 5—2 is legitimate, since R = (1). Corollary. An element of a commutative ring R with identity is invertible if and only if it belongs to no maximal ideal of R. Although maximal ideals were deﬁned for arbitrary rings, we shall abandon a degree of generality and for the time being limit our discussion almost exclusively to commutative rings with identity. A ring of this kind is, of course, much easier to handle than one which is not commutative.
Another advantage stems from the fact that each ideal, other than the ring itself, will be contained in a maximal ideal. Thus, until further notice, we
shall assume that all given rings are commutative with identity, even when this is not explicitly mentioned. To be sure, a good deal of the subsequent material could be presented without this additional restriction. The KrullZorn Theorem has many important applications throughout ideal theory. For the moment, we content ourselves with giving an elementary proof of a somewhat special result; although the fact involved is rather interesting, there will be no occasion to make use of it.
Theorem 54. In a ring R having exactly one maximal ideal M, the only idempotents are 0 and 1. Proof Assume that the theorem is false; that is, suppose that there exists an idempotent a e R with a 7E 0, l. The relation a2 = a implies a(1 — a) = 0, so that a and 1 — a are both zero divisors. Hence, by Problem 4(d), Chapter 1, neither the element a nor 1 — a is invertible in R. But this means that the
principal ideals (a) and (1 — a) are both proper ideals of the ring R. As such, they must be contained in M, the sole maximal of R. Accordingly, the elements a and 1 — a lie in M, whence 1=a+(1——a)eM. This leads at once to the contradiction that M = R.
Although more elementary proofs are possible, Theorem 5—4 can be
used to show that a ﬁeld has no idempotents except 0 and 1. A full justiﬁcation of this statement consists of ﬁrst establishing that the zero ideal is the only maximal ideal in a ﬁeld. We now come to a characterization of maximal ideals in terms of their quotient rings. Theorem 55.
Let I be a proper ideal of the ring R. Then I is a maximal
ideal if and only if the quotient ring R/I is a ﬁeld. Proof. To begin, let I be a maximal ideal of R. Since R is a commutative
ring with identity, the quotient ring R/I also has these properties. Thus,
MAXIMAL, PRIME, AND PRIMARY IDEALS
75
to prove that R/I is a ﬁeld, it sufﬁces to show that each nonzero element of
R]! has a multiplicative inverse. Now, if the coset a + I + I, then a ¢ I. By virtue of the fact that I is a maximal ideal, the ideal (I, a) generated by l and a must be the whole ring R:
R = (La) = {i + raieI,reR}. That is to say, every element of R is expressible in the form i + ra, where is I and re R. The identity element 1, in particular, may be written as
l = 7+ Fa for suitable choice of is I, ieR.
But then, the difference
I  Fa e I. This obviously implies that
1+1=m+1=0+nm+n which asserts that 7 + I = (a + I)“. Hence, R/I is a ﬁeld. For the opposite direction, we suppose that R/I is a ﬁeld and J is any
ideal of R for which I c J E R. The argument consists of showing that J = R, for then I will be a maximal ideal. Since I is a proper subset of J, there exists an element a 6 J with a ¢ I. Consequently, the coset a + I 7E I, the zero element of R/I. Since R/I is assumed to be a ﬁeld, a + I must have an inverse under multiplication,
(a+I)(b+I)=ab+I=1+I, for some coset b + I e R/I. It then follows that 1 — ab 5 I f(P) deﬁnes a onetoone correspondence between the G of prime (primary) ideals of R which contain ker f and the set of all pl"(primary) ideals of R’.
18. IfM is a maximal ideal ofthe ring R and n 6 2+, show that the quotient ring R/M' has exactly one proper prime ideal. [Hint: Problem l7(c).] 19 a) For any ideal I of R, prove that I and J? are contained in precisely the in: ' prime ideals of R. b) Using part (a), deduce that whenever I is a prime ideal of R, then I = ﬂ . Let f be a homomorphism from the ring R onto the ring R’. Prove that a) ifIisanideal owithI 2 kerf, theaU) =f(JT), b) if I’ is an ideal of R’, then t/f‘IU') = f“(ﬁ’) 21. Verify that the intersection of semiprime ideals of the ring R is again a semiprinn ideal ofR.
. If I is an ideal of the ring R, prove that ﬂ is the smallest (in the settheoretic new)
3d
semiprime ideal of R which contains I. Establish that every divisor of zero in the ring Zp. (p a prime, n > 0) is nilpotd.
. Let I, J, and Q be ideals of the ring R, with Q primary. Prove the follow'u
statements: a) if] $ JQHhen the quotient Qzl = Q; h) ifIJ g QandI $ JQJhenJQ Q; c) ifIJ E Q and the idealJis ﬁnitely generated, then eitherI S Q or elseJ" E Q
for some n 6 2+. [Hint: If] S); Q, each generator of J is in JQ] . Assume that I is an ideal of the ring R. If ﬂ is a maximal ideal of R, show tht I is primary. [Hint: Mimic the argument of the corollary to Theorem 5—14.]
. Let R be an integral domain and P be a prime ideal of R. Consider R,, the I": of quotients of R relative to the complement of P:
R, = {ab'1 6 Q°,(R)a e R; b (E P}. Prove that the ring R, (which is known as the localization ofR at the prime ideal n has exactly one maximal ideal, namely, I = {ab‘1 6 c(R)a e P; b ¢ P}. . AringRissaidtobealocalringifithasauniquemaximalideal. IfRisaloal ring with M as its maximal ideal, show that any element a ¢ M is invertible in ll.
. A subring R of a ﬁeld F is said to be a valuation ring of F if for each nonm element a e F at least one of a or a" 1 belongs to R. Assuming that R is a valuatia ring of F, prove the following: a) R contains all the idempotent elements of F. b) R is a local ring, with unique maximal ideal M = {a e Rla‘l ¢ R}.
[Remark a'l denotes the inverse of a in F.] c) For any two elements a, b e R, either aR g bR or bR 2 aR. [Him Eitlﬂ
ab“1 6 R or ba‘1 6 R.]
PROBLEMS
89
d) lilisaproperideal ofRandbeR  I,thenI E bR. 1!. For a ﬁxed prime p, consider the subset of rational numbers deﬁned by
= {a/be QIP l' b}Show that a) V, is a valuation ring of Q;
h) the unique maximal ideal of V is M: {a/b e Qp fb, but pa}; c) the ﬁeld V,,/Mp _ Z. [Hint: Let the homomorphism f V —> Z be deﬁned
by HM?) = [a] [b]
]
3. a) Let 1,, 1,, , I,l be arbitrary ideals of the ring R and P be a prime ideal of R. If I II2 I” E P, establish that Ii g P for at least one value of i. [Hint: If a,.] Ii $ P for all i, choose ai e I. — P and consider the element a = ala2 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. 3]. Let R be an integral domain with the property that every proper ideal is the product of maximal ideals. Prove that a) [l' 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 76 {0}, pick 0 aé aeM. Then M,Mz Mn = (a) E M for suitable maximal ideals M,; hence, M = Mi for
some i.] b) If I, J, M are ideals ofR, with M maximal, then [M = JM implies I = J.
32. a) If I is an ideal of the ring R such that I E (a), show that there exists an ideal J of R for which a] = I. [Hint: Take J = (I z (a)).] b) Prove that if a principal ideal (a) of the ring R properly contains a prime ideal
P, then P g {“3 (a‘). n= 1
33. Let I be a primary ideal of the ring R. Prove that I has exactly one minimal prime
ideal, namely, JI— [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 Fundsmental 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 = 222335. 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 61. If a + 0 and b are elements of the ring R, then a is said to divide b, in symbols ab, provided that there exists some céR such that b = ac. In case a does not divide b, we shall write a l b.
Other language for the divisibility property ab 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 7E 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 61. Let the elements a, b, c e R. Then,
1) a0,1a, ala; 90
DIVISIBILITY THEORY IN INTEGRAL DOMAINS
91
2) al ﬁnd only if a is invertible; 3) ifalb, then acbc; 4) ifab and bc, then ale;
5) ifca and cb, then c(ax + by) for every x, y e R. Division of elements in a ring R is closely related to ideal inclusion: ab if and only if (b) E (a). Indeed, ab means that b = ac for some c e R; thus, b e (a), so that (b) E (a). Conversely, if (b) g (a), then there exists an element c in R for which b = ac,
implying that ab. Questions concerning divisibility are complicated somewhat by the presence of invertible elements. For, if 14 has a multiplicative inverse, any element of a e R can be expressed in the form a = a(uu' 1), so that both ua and u"a. An extreme situation occurs in the case of ﬁelds, where every nonzero element divides every other element.
0n the other hand, in the
ring Z, of even integers, the element 2 has no divisors at all. In order to overcome the diﬁiculty that is produced by invertible elements, we introduce the following deﬁnition.
Deﬁnition 62. Two elements a, b e R are 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 0 ~ 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 61. In the case of the ring Z, the only associates of an integer n e Z are in, since i1 are the only invertible elements.
Example 62. Consider the domain Z(i) of Gaussian integers, a subdomain of the complex number ﬁeld, whose elements form the set
Z(i) = {a+ bia,beZ;i2 = —1}. Here, the only invertible elements are i 1 and ii. For, supposea + bi e Z(i) hasamultiplicativeinversec + di. Then,wemusthave(a + bi)(c + di) = 1, so that (a — bi)(c — di) = 1. Therefore, 1 = (a + bi)(c + di)(a — bi)(c — di)
= (a2 + b2)(c2 + (12). 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 i1, ii. In con
92
FIRST COURSE IN RINGS AND IDEALS
sequence, the class of associates determined by any Gaussian integer a + II consists of exactly four members: a+bi,—a—bi,—b+ai,b—ai.
Since associated elements are rather closely related, it is not surprising that they have similar properties; for instance: Theorem 62. 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 ab and ba, 3) (a) = (b)Proof. To prove the equivalence of (1) and (2), suppose that a = bu, what
u is an invertible element; then, also, b = w”, so that both ab and ba. Going in the opposite direction, if ab, we can write b = ax for some x e R;
while, from bla, it follows that a = by with y e R. Therefore, b = (by)x = b(yx). Since b 7E 0, the cancellation law implies that 1 = yx. Hence, yin 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 63. Let a1, a2, , a,' be nonzero elements of the ring'R. An element d e R is a greatest common divisor of a1, a2, ..., a,I if it possesses the properties
1) dai for i = 1, 2,
, n (d is a common divisor),
2) cla, for i = 1, 2,
, n implies that cd.
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 a1, a2,
, all e R
can possess two diﬂerent 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 dd’ as well as d’d; according to Theorem 6—2, this implies that d and d' are associates. Thus, the greatest common divisor of a1, a2,
, a,I is unique, whenever it exists, up to arbitrary
invertible factors. We shall ﬁnd it convenient to denote any greatest common divisor of a1, a2,
, a,I by gcd (a1, a2,
, 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.
DIVISIBILITY THEORY IN INTEGRAL DOMAINS
93
Theorem 6—3.
Let a1, a2,
Then a1, a2, form
, a, have a greatest common divisor d, expressible in the
, a,I be nonzero elements of the ring R.
d = rla1 + rza2 + if and only if the ideal (a1, a2,
lies in the ideal (a1, a2,
(rl. e R),
, an) is principal.
Proof Suppose that d = gcd (a1, a2,
the form d = rlal + rza2 +
+ r,,a,l
, a”) exists and can be written in
+ ruan, with rieR. Then the element d
, an), which implies that (d) E (a1, a2,
, an).
To obtain the reverse inclusion, observe that since d = gcd (a1, a2,
, an),
each a, is a multiple of d; say, ai = xid, where xi 6 R. Thus, for an arbitrary
member yla1 + yza2 + ylal + y2a2 +
+ yuan of the ideal (a1, a2, + yuan = (ylxl + y2x2 +
This fact shows that (al, a1,
, an), we must have + ynxn)de(d)'
, an) E (d), and equality follows.
For the converse, let (a1, a2, (“1’ a2,
, an) be a principal ideal of R:
, an) = (d)
(d E R)
Our aim, of course, is to prove that d = gcd (a1, a2, , an). Since each a.e(d), there exist elements b, in R for which ai = bid, whence dai for i = l, 2, , n. It remains only to establish that any common divisor c of the al also divides d. Now, ai = sic for suitable s, e R. As an element of (0,. a1, , an), d must have the form d = rla1 + r2a2 + + rnan, with
r‘ in R. This means that d = (r151 + r252 +
+ rusu)c,
which is to say that cd. Thus, d is a greatest common divisor of a1, a2,
, an
and has the desired representation. Corollary. Any ﬁnite set ofnonzero elements a1, a2, . . . , an ofa principal ideal ring R has a greatest common divisor; in fact, gcd (a1, a2, ...,a,,) = rla1 + rza2 +
for suitable choice of r1, r2,
When (a1, a2,
+ rnan
, r" e R.
, a,,) = R, the elements a1, a2,
, a,l must have a
common divisor which is an invertible element of R; in this case, we say
that al, a1,
, a,l are relatively prime and shall write gcd (a1, a2,
, an) = 1.
, an are nonzero elements of a principal ideal ring R, then If a,, a2, the corollary to Theorem 6—3 tells us that al, a2, , an are relatively prime
if and only if there exist r1, r2, rla1 + rza2 +
, r,' e R such that
+ r,,a,I = l
(Bezout’s Identity).
One of the most useful applications of Bezout’s Identity is the following (it also serves to motivate our coming deﬁnition of a prime element).
94
FIRST COURSE IN RINGS AND IDEALS
”‘
Theorem 64. Let a, b, c be elements of the principal ideal ring R. I
cab, with a and c relatively prime, then c. Proof. Since a and c are relatively prime, so that god (a, c) = 1, there ex'ﬂ elements r, S e R satisfying 1 = ra + sc; hence, b=1b=rab+scb.
As cab and cc, Theorem 6—l(5) guarantees that c(rab + scb), or rather,
clb. Dual to the notion of greatest common divisor there is the idea of 0 least common multiple, deﬁned below.
Deﬁnition 64. Let a1, a2, , a,I be nonzero elements of a ring R. An element d e R is a least common multiple of a1, a2, , a" if
1) aid for i = 1, 2,
, n (d is a common multiple),
2) aic for i = 1, 2,, n implies dc. In brief, an element d e R is a least common multiple of a1, a2, ...,a, if it is a common multiple of a1, a2,
, an 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 a1, a2, , an, then dld’ and d’d ; hence, d and d’ are associates. We hereafter adopt the standard notation lcm (a1, a2, , an) to represent any least common multiple of a1, a2, , a...
The next result is a useful companion to Theorem 6—3.
Theorem 65.
Let a1, a2,
Then a1, a2,
, an have a least common multiple if and only if the ideal
, a,. be nonzero elements of the ring R.
n (a,) is principal. Proof We begin by assuming that d = lcm (a1, a2,
, an) exists. Then
the element d lies in each of the principal ideals (ai), for i = 1,2, ...,n,
whence in the intersection n (ai). This means that (d) S n (a,). On the other hand, any element r e n (ai) is a common multiple of each of the a,. But d is a least common multiple, so that dr, or, equivalently, r e (d). This leads to the inclusion n (at) E (d) and the subsequent equality. Going in the opposite direction, suppose that the intersection n (a,) in a principal ideal of R, say n (a,) = (a). Since (a) S (at), it follows that aila for every i, making a a common multiple of a1, a2, , a". Given any other common multiple b of a1, a2,
, an, the condition a,b implies that
(b) E (oi) for each value of i. As a result, (b) E n (a,) = (a) and so ab. Our argument establishes that a = 1cm (a1, a2, , an), completing the 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 65. A ring R is said to have the gcdproperty (lcmproperty) 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 gcdproperty if and only if every ﬁnitely generated ideal of R is principal. Likewise, Theorem 6—5 tells us that R possesses the lcmproperty 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 lcmproperty. 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 critioal stage of the discussion, let us ﬁrst establish a lemma. Lemma. Let a,, a2, domain R.
I) If lcm (a,, a2,
, a,. and r be nonzero elements of an integral
, an) exists, then lcm (ra,, raz,
lcm(ra1, raz,
2) . If gcd (ral, raz,
, a”).
, ran) = r lcm (a1, a2,
, ran) exists, then gcd (a1, a1,
gcd(ra1, raz,
, ran) also exists and
, an) also exists and
, ran) = r gcd (a1, a2,
, an).
Prooﬁ First, assume that d = lcm (a1, a2, ,an) exists. Then aild for each value of i, whence raird. Now, let d’ be any common multiple of
ra,, raz,
, ran. Then rld’, say d’ = rs, where s e R.
It follows that ails
for every i and so ds. As a result, rdrs or rdd’. But this means that lcm (ral, raz,
, ran) exists and equals rd = r lcm (a1, a2,
, a”).
As regards the second assertion, suppose that c = gcd (ral, raz,
, ran)
exists. Then rc; hence, c = rt for suitable te R. Since clrai, we have tai for every 1‘, signifying that t is a common divisor of the at. Now, consider an arbitrary common divisor t’ of a1, a2, , an. Then rt’ ra, for i = l, 2, ,n
and therefore rt’c. But c = rt, so that rt’ rt or t’lt. The implication is that god (a,, a2,
, an) exists and equals t. This proves what we wanted:
god (ra1, raz,
, ran) = c = rt = rgcd (a1, a2,
, a”).
Remark. It is entirely possible for gcd (a1, a2, , an) to exist without the , ran); this accounts for the lack of symmetry existence of god (ral, raz, in the statement of the above lemma. (See Example 6—4.)
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FIRST COURSE IN RINGS AND IDEALS
Although the coming theorem is somewhat specialized in character, tin information it contains is frequently useful. Theorem 66. Let a1, a2, , an and b1, b2, , b, be nonzero elemII = a,,b,l = x. of an integral domain R such that alb1 = azbz =
1) If 1cm (a1, a2,
, a") exists, then gcd (b1, b2,
, b") also exists and
satisﬁes 1cm (a1, a2,
, (1,.) god (b1, b2,
, b") = x.
, ran) exists for all 0 aé r e R, then 1cm (b1, b2, ..., 2) If gcd (ral, raz, bu) also exists and satisﬁes gcd (a1, a2, Proof.
, an) 1cm (b1, b2,
, bl) = x.
For a proof of statement (1), set a = 1cm (a1, a2,
aila for i = 1, 2,
,au).
Then
, n, say a = riai. From the relation x = aibi, we see
that ab, = (riai)b, = rix and so xlabi. On the other hand, consider any
divisor y of the abi. Then yai(ab,.)a,, or yaixa, making xa a common multiple ofya,, ya,, , ya". According to the lemma, lcm (yal, yaz, , ya.) exists and equals ya. Thus, by the deﬁnition of least common multiple, we
conclude that yaxa, whence yx. To recapitulate, we have shown that xab, for each i and whenever yab,, then ylx. This simply asserts that x = gcd(ab1, abz, = a gcd (b1, b2,
, ab") , b”) = 1cm (a1, a2,
, an) gcd (b1, b2,
, 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 (a1, a2, , an) exists but, more generally, the existence of ged(ra1, raz,
, ran) for all r 75 0.
Our next result is rather striking in that it tells us that, at least for integral
domains, the gcdproperty implies the lcmproperty, and conversely. Theorem 67. An integral domain R has the gcdproperty if and only if R has the lcmproperty.
Proof. Let b1, b2, , b" be nonzero elements of R and suppose that R b,l and (1,, = b1 bk.1 possesses the lcmproperty. Taking x = blb2 bk+1
b,' for k = l, 2,
, n, we may appeal to the ﬁrst part of Theorem
6—6 to conclude that god (bpbz, , b") exists; hence, the gcdproperty 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 1cm (b1, b2,
, bu) 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 pab implies that either pa or else pb. 2) A nonzero element q e R is said to be irreducible (or nonfactorizable) if and only if q is not invertible and in every factorization q = bc 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 ala2 a,' is divisible by a prime p, then p must divide at least one of the factors ai (i = l, 2,
, n).
Lemma. In an integral domain R, any prime element p is irreducible. Prooﬁ Suppose that p = ab for some a, b e R. Since p is prime, either pa or plb; 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 68. 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 re 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. Thin there would exist elements s, t e R for which 1 = sp + ta. Then,
b=bl=bsp+bta=bsp+pct=p(bs+ct),
which implies that plb. On the other hand, if r is invertible in R, then d = r‘lp e (p), whence (d) E (p). It follows that the element a e (p) and, in consequence, pa. At any rate, if pab, 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 69. Let R be a principal ideal domain. If {1"}, n e Z+, is any inﬁnite sequence of ideals of R satisfying
IlglzgnCI C1,“;then there exists an integer m such that I,I = I", for all n > m. Proof It is an easy matter to verify that I = 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 idea] In. For n > m, it then follows that I=(a)SImEI,,EI;
hence, I,I = 1",, 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 pe 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) =19 R is prime. Proof To begin, we suppose that p is an irreducible element of R and that (a) is any principal ideal for which (p) C (a) E 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) S (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. 0n 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 l = rb; but this results in the contradiction that b is invertible. Therefore, a ¢(p), yielding the proper inclusion (p) Q(ﬂ) deﬁned by f(a) = a is an isomorphism. To study divisibility properties of Q(ﬁ), it is convenient to make use of the concept of the norm of an element (an analog of the absolute value notion in Z):
Deﬁnition 69. For each elementa = a + bﬁ in Q(ﬂ), the norm N(a) of a is simply the product of a and its conjugate a:
N(oc) = and = (a + bﬁ)(a — bﬁ) = a2 — bzn. Some properties of the norm function which follow easily from the
deﬁnition are listed below. Lemma. For all a, ,B e Q(ﬂ), the following hold: 1) Ma) = Oifandonlyifa = 0;
2) NW?) = N(a)N(ﬂ);
3) N0) = 1; Prooﬁ Given a = a + bﬂ in Q(ﬁ), N(a) = a2 — bzn = 0 if and only if both a = b = O (that is, a = 0); otherwise, we would contradict the choice of n as a squarefree integer. Since the mapping f(a) = a is an isomorphism, N is a multiplicative function in the sense that
Mot/3) = 06/355 = 061%? = will? = N(a)N(ﬁ’) for all a, ,6 e QWE).
.
The proof of assertion (3) follows from the fact that
N0) = N0’) = N(1)N(l) = N0)”, whence N(l) = 1. Although Q(ﬁ) has been labeled as a ﬁeld, we actually have not proved this to be the case; it is high time to remedy this situation.
106
FIRST COURSE IN RINGS AND IDEALS
Theorem 616. For each squarefree integer n, the system QM?) form
a ﬁeld; in fact, QQ/ﬁ) is a subﬁeld of C.
Proof. The reader may easily verify that Q(ﬁ) is a commutative ring with identity. It remains only to establish that each nonzero element of Quin) has a multiplicative inverse in Q(ﬁ). Now, if 0 7E a e Q(\/E), then the element ,3 = a/N(oc) evidently lies in Q(ﬂ); furthermore, the product
(#9 = a(5t/N(a)) = N(a)/N(a) = 1, so that [2’ serves as the inverse of a.
Contained in each quadratic ﬁeld Q(ﬁ) is the integral domain Z(ﬂ) = {a + bﬂla, be Z}.
Since Z(ﬁ) is closed under conjugation, the norm function enables us to get a clear view of the sets of invertible and irreducible elements in these domains. The multiplicative property of the norm, for instance, transfers any factorization a = ﬂy of an element a(ﬂ) into a factorization N(a) = N(ﬂ)N(y) of the integer N(a). This is particularly helpful in proving Lemma. For any a e Z(ﬁ), the following hold: 1) N(a) = 11 if and only if a is invertible in Z(ﬂ); 2) if N(a) = ip, where p is a prime number, then a is an irreducible
element of Z(ﬁ). Proof As regards (1), observe that if N(ot) = l_1, then (15; = l_l ; thus all, which is to say that a is invertible. To prove the converse, let a be an invertible element of Z(ﬂ), so that aﬂ = 1 for some [3 in Z(ﬁ). Then,
Noam/5') = N(aﬁ’) = N(l) = 1. Since N(a) and N(15’) are both integers, this implies that N(a) = :1. Next, suppose that a has the property that N(oc) = i p, where p is e prime number. As N(0:) 7E 0, 1, the element a is neither 0 nor invertible in
Z(ﬂ). If a = ﬂy is a factorization of a in Z(ﬂ), then N(/>’)N(y) = Mon) = in
from which it follows that one of N(,8) or N(y) must have the value 1:1. From the ﬁrst part of the lemma, we may thus conclude that either [3 or 7 is invertible in Z(ﬁ), while the other is an associate of a. Accordingly, a
is an irreducible element of Z(ﬁ). Example 63. Let us ﬁnd all invertible elements in Z(i) = Z(JTIL the domain of Gausian integers, by ﬁnding those members a of Z(i) for which N(oc) = 1 (in this setting, the norm assumes nonnegative values). If a = a + bieZ(i) and N(a) = 1, then a2 + b2 =1, with a, beZ. This
DIVISIBILITY THEORY IN INTEGRAL DOMAINS
107
equation is possible only if a = i1 and b = 0, or a = 0 and b = :1.
Hence, the only choice for invertible elements in Z(i) are i 1 and i i.
Perhaps the most obvious approach to the question of unique factorization in the quadratic domains Z(ﬁ) is to try to show that they are Euclidean
domains (3 natural candidate for the Euclidean valuation is 6(a) = N(a) . In the coming theorem, we shall do precisely this for the domains Z(
— l),
Z(J——2), Z(ﬁ), and Z(ﬂ). Although there are other Euclidean quadratic domains, our attention is restricted to these few for which the division
algorithm is easily established.
Theorem 617. Each of the domains Z(ﬁ), where n = — 1, —2, 2, 3, is Euclidean; hence, is a unique factorization domain.
Prooﬁ The strategy employed in the proof is to show that the function 6
deﬁned on Z(ﬁ) by 6(a) = N(a) is a Euclidean valuation for n = — 1, —2, 2. 3. We clearly have 6(a) = 0 if and only if a = 0, so that 6(a) 2 1 for all a aé 0. Since both the norm and its absolute value are multiplicative, condition (2) of Deﬁnition 6—8 is always satisﬁed:
5(a/5') = (WW/5’) 2 5(a)'1 = 5(a) whenever a, B + 0 are in Z(ﬁ). Since ,8 at: 0, the product aﬂ‘l e Q(ﬁ) and so may be written in the
form aﬁ‘ ‘ = a + bﬁ, with a, b e Q. Select integers x and y (the nearest integers to a and b) such that
a — x] g 1/2,
lb — y] S 1/2.
Now, set a = x + y\/r_l. Then a e Z(Jri) and norm formula (valid also in aﬁ” shows that
ww*—m=ww—n+wwm
=w—W—w—w.
But the manner in which x and y were chosen imply that
—n/4 s (a — x)2 — n(b — y)2 S 1/4, 0 S (a — x)2 — n(b — y)2 S 1/4 + (—n)1/4,
ifn > 0, ifn < 0.
In terms of the function 6 this means 6(a)?“ — 0) = (a — x)’ — "(b — y)2 ’),N(Y)EZ+,
which in turn yields N0?) = N(y) = 3. Hence, if [5' = a + bJ:_’5, we ﬁnd that we must solve the equation a2 + 5b2 = 3 for integers a and b; but
this equation obviously has no solutionsinZ (b aé Oimplies that a2 + 5b2 2 5, and if b = 0, then a2 = 3). Thus, we have exhibited two genuinely different factorizations of the element 9 into irreducibles, so that unique factorization
does not hold in Z(\/——5).
Notice further that the common divisors of 9 and 3(2 + ,/——5) are 1, 3, and 2 + \/——5. None of these latter elements is divisible by the others, so that gcd (9, 3(2 + ‘/——5)) fails to exist (in particular, Z(\/——5) does not have the gcdproperty). 0n the other hand, the greatest common divisor
of 3 and 2 + F3 does exist and, in fact, gcd (3, 2 + f—_5) = 1. It follows that only the righthand side of the formula
gcd (33, 3(2 + ,/——5)) = 3gcd (3,2 + ,/——5) is deﬁned in Z(J——5), thereby illustrating the remark on page 95. This example has the additional feature of showing that the concepts
PROBLEMS
109
of an irreducible element and of a nonzero prime do not always coincide
in an arbitrary integral domain. Speciﬁcally, we have (2 + J — 5)33, but (2 + ,/—5)1’3, so that 2 + ,/—5 cannot be a prime element of Z(,/—5).
PROBLEMS I. 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 = (l — e + a)e.] b) If for some n e Z+ the elements a’' and e are associates, then a" and e are associates for all m 2 n.
2. Given that I, J, and K are ideals of a principal ideal domain R, derive the following relations: a) I” = (a) and J = (b), then I] = (ab); in particular, I" = (a').
b) [(1 n K) = U n IK. c) l+(JnK)=(I+J)n(I+K). d)ln(J+K)=(InJ)+(InK). e) I] = InJifandonlyifI + J = RforallnonzeroI,J. 3. Suppose that R = R1 63 R2 69 $ R”, where each R‘ is a principal ideal ring. Verify that R is also a principal ideal ring. 4. Let R be an integral domain having the godproperty. Assuming that equality holds to within associates, prove that, for nonzero a, b, c e R,
a) god (a, god (b, c)) = god (god (a, b), C)b) god (a, 1) = 1c) d) e) 1)
god (ca, ch) = c god (a, b); in particular, god (c, cb) = 1. il'gcd(a,b) = land gcd(a,c) = l,thengod(a,bc) = l. if god (a, b) = l, ac and bc, then able. god (a, b) 1cm (a, b) = ab. [Hint: Theorem 6—6.]
5. If R is an integral domain having the godproperty, show that a nonzero element of R is prime if and only if it is irreducible. i. In a principal ideal domain R, establish that the primary ideals are the two trivial ideals and ideals of the form (V), where p is a prime element of R and n 6 2+. [Hint: If I is primary, then ﬂ = (p) for some prime element p. Choose n e Z+
such that I g (p‘), but I $ (p"“), and show that I = (p").] 7. If R is a principal ideal doman, we deﬁne the length 1(a) for each nonzero a e R as follows: if a is invertible, then 1(a) = 0; otherwise, 3(a) is the number of primes (not necessarily distinct) in any factorization of a. Prove the following assertions: a) the length of a is welldeﬁned;
b) if ab, then 2(a) 5 Mb); c) ifab and 1(a) = Mb), then ba; d) if a J{ b and b l a, then there exist nonzero p, q s R such that
Mpa + qb) $ min Ma). 1101)}
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FIRST COURSE IN RINGS AND IDEALS
8. Let a be a nonzero element of the principal ideal domain R. If a has lend“ prove that there are at most 2'I ideals containing a. 9. Verify that any two nonzero elements of a unique factorization domain poml
greatest common divisor. [Hint: If a = p'ilpﬁz pi" and b = p'l'p',’ m” (h irreducible), then god (a, b) = pi‘p’z’ pl', where ji = min (k,, Ii).] 10. Let R be an integral domain. Prove that R is a unique factorization domain «at only if every nontrivial principal ideal of R is the product of a ﬁnite number N maximal principal ideals and these ideals are unique up to a permutation of 01th.
11. Show that 6(a) = a is not a Euclidean valuation on the domain Q. 12. Assuming that R is a Euclidean domain with valuation 6, prove the statemuts below: a) For nonzero a, b e R, if ab and 6(a) = 6(b), then a and b are associates. [Hm
Show that bla] b) For nonzero a, b e R, 6(ab) > 6(a) if and only if b is not an invertible claim [Hint: Use the division algorithm to write a = q(ab) + r.] c) If n is any integer such that 6(1) + n 2 0, then the function 6’: R  {0} 0 2 deﬁned by 6’(a) = 6(a) + n is also a Euclidean valuation on R.
13. For each ideal I in Z(i), the domain of Gaussian integers, establish that the quoe'ﬂ ring Z(i)/I is ﬁnite. [Hint: Write I = (a) and use the division algorithm on (1 Id
any .8 e 20).] 14. Let R be a Euclidean domain with valuation 6. a) Determine whether the set I = {a e R6(a) > 6(1)} u {0} is an ideal of R. b) Assuming that the set F = {a e_ R¢$(a) = 1} u {0} is closed under addiﬁtl. verify that F forms a ﬁeld. 15. a) Prove that if n1 and n2 are squarefree integers and n1 ){ n2, then the quadnti;
ﬁeld Q(\/n—1) is not isomorphic to Q(\/n_2).
b) For each squarefree integer n, determine all the subﬁelds of the quadnl': ﬁeld Q(ﬁ). 16. Establish the following assertions (where n is a squarefree integer): a) For n < — 1, the only invertible elements of the quadratic domain Z(ﬁ) In + 1.
b) For n > 1, Z(\/I_I) has inﬁnitely many invertible elements. [Hint: If a1, 1:1 is a solution of the equation a2 — nb2 = i1, conclude that a,” b, is ah I
solution, where a, + bk\/— = (a1 + b1 n)“, keZ+.] c) The invertible elements of Z(ﬂ) are precisely the elements of the form i(1 + ﬂ)“, n 6 2+. [Hint: If u is any positive invertible element of zuﬁ,
then (1 + ﬂ)" 3 u < (1 + ﬁr“ for some neZ+;hence,
lsql+ﬁ)'" 0; it is further evident that no pi(x) = q(x) for any invertible element u (otherwise, the polynomial obtained on dividing f(x) by q,(x) will have unique factorization; this implies that f(x) can also be factored uniquely). Let a, b be the leading coeﬂicients of p1(x), q1(x), respectively, and deﬁne
906) = af(x)  bp1(X)x"""‘qz(x)
t1,06)
On one hand, we have
900 = ap1(x)pz(x)
= p1(x)(ap2(X)
h(x) — bp1(x)x"""qz(x)
prov) — bx"""qz(x)
q.(x)
(h(x)),
and, on the other hand,
90¢) = aq;(X)qz(X) (h(x)  bp1(X)X"'"'qz(x) = (M106)  bP1(x)x"_"')qz(x) t1,06)
(h(x)
Now, either g(x) = 0, which forces aq1(x) = bp1(x)x"'"', or else deg g(x) < degf(x). In the latter event, g(x) must possess a unique factorization into , q,(x) and p1(x). The net result of irreducibles, some of which are q2(x),
this is that p1(x)g(x), but p1(x) If q..(x) for i > 1, so that p1(x)(aq1(x) _ bp1(x)x"_"'),
and therefore p1(x)aq1(x). In either of the two cases considered, we are able to conclude that p1(x) divides the product aq1(x); this being so, aq,(x) = p1(x)h(x) for some polynomial h(x) e R[x]. Since R is taken to be a unique factorization domain, a has a unique factorization as a product of irreducible elements of R—hence, of R[x]—say, a = c1c2 ck, where each
c, is irreducible in R[x]. (The only factorizations of a as an element of R[x] are those it had as an element of R.) Arguing from the representation €102
9:410“) = P1(x)h(x)a
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FIRST COURSE IN RINGS AND IDEALS
with p1(x) an irreducible, it follows that each c, and, in consequence, tb element a divides h(x). But, then, t“h(x) = P1(x)ahi(x)
for some h1(x) in R[x] or, upon canceling, q1(x) = p1(x)h1(x); in other
words, p1(x)q1(x). Using the irreducibility of q1(x) as a member of R[x], p1(x) must be an associate of q1(x). However, this conﬂicts with our original assumptions. Thus, we see that R[x] is indeed a unique factorization domain. Remark. For many years, it was an open question as to whether a power series ring over a unique factorization domain is again a unique factorization domain; a negative answer was established not long ago by Samuel [55]. To this we might add, on the positive side, that one can prove that the ring of formal power series over a principal ideal domain does in fact comprise a unique factorization domain (a not altogether trivial task).
Coming back to the corollary to Theorem 7—10, there is an interesting converse which deserves mention: namely, if R is an integral domain such that the polynomial ring R[x] forms a principal ideal domain, then R is necessarily a ﬁeld. In verifying this, the main point to be proved is that any nonzero element a e R is invertible in R. By virtue of our hypothesis, the ideal generated by x and a must be principal; for instance,
(36, a) = (f(xl),
0 75 f(x) '5 R[x]
Since both x, a e (f(x)), it follows that
a = g(x)f(x), and x = h(x)f(x) for suitable g(x), h(x) in R[x]. The ﬁrst of these relations signiﬁes that degf(x) = 0, say f(x) = a0, and as a result deg h(x) = 1, say h(x) = b0 + b1x(b1 75 0). We thus obtain x = ao(b0 + blx). But this means that the product aob1 = 1,thereby making ao (or, equivalently, f(x)) an invertible element of R. The implication is that the ideal (x, a) is the entire ring R[x].
It is therefore possible to write the identity element in the form 1 = xk1(x) + ak2(x),
with the two polynomials k1(x), k2(x)eR[x]. This can only happen if aco = 1, where co 9E 0 is the constant term of k2(x). In consequence, the element a has a multiplicative inverse in R, which settles the whole affair.
At the heart of all the interesting questions on factorization in R[x] lies the idea of an irreducible polynomial, which we formulate in a rather general way as follows: Deﬁnition 7—4. Let R be an integral domain. A nonconstant polynomial f(x) e R[x] is said to be irreducible over R, or is an irreducible
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polynomial in R[x], iff(x) cannot be expressed as the product of two polynomials (in R[x]) of positive degree. Otherwise, f(x) is termed
reducible in R[x]. In the case of the principal ideal domain F[x], where F is a ﬁeld, the irreducible polynomials are precisely the irreducible elements of F[x] (recall that the invertible elements of the polynomial ring F[x] are just the nonzero constant polynomials); by Theorem 5—9, these coincide with the prime
elements of F[x].
Of the equivalent notions, irreducible polynomial,
irreducible element, and prime element, the term “irreducible polynomial” is the one customarily preferred for F[x]. Perhaps we should emphasize that Deﬁnition 7—4 applies only to polynomials of positive degree; the constant polynomials are neither reducible nor irreducible. Thus, the factorization theory of F[x] concerns only polynomials of degree 2 1. The dependence of Deﬁnition 7—4 upon the polynomial domain R[x] is essential. It may very well happen that a given polynomial is irreducible when viewed as an element of one domain, yet reducible in another. One
such example is the polynomial x2 + 1; it is irreducible in R#[x], but reducible in both C[x], where x2 + 1 = (x + i)(x — i), and Zz[x], where x’ + l = (x + 1)(x + 1). Thus, to ask merely whether a polynomial is
irreducible, without specifying the coeﬂicient ring involved, is incomplete and meaningless. More often than not, it is a formidable task to decide when a given polynomial is irreducible over a speciﬁc ring. If F is a ﬁnite ﬁeld, say one
of the ﬁelds of integers modulo a prime, we may actually examine all of the possible roots. To cite a simple illustration, the polynomial f(x) = x3 + x + 1 is irreducible in Zz[x]. If there are any factors of this polynomial, at least one must be linear. But the only possible roots forf(x) are 0 and 1, yet f(0) = f(1) = 1 ye 0, showing that no roots exist in 22.
Example 73. Any linear polynomial ax + b, a 79 0, is irreducible in R[x], where R is an integral domain. Indeed, since the degree of a product of two polynomials is the sum of the degree of the factors, it follows that a representation ax + b = g(x)h(X), 90‘): “35) e R[x], with l s deg g(x), 1 s deg h(x) is impossible. reducible polynomial has degree at least 2.
This signiﬁes that every
Example 7—4. The polynomial x2 — 2 is irreducible in Q[x], where Q as usual is the ﬁeld of rational numbers. Otherwise, we would have
x2—2=(ax+b)(cx+d) = (ac)x2 + (ad + bc)x + bd,
FIRST COURSE IN RINGS AND IDEALS
128
with the coefﬁcients a, b, c, d 6 Q. Accordingly,
ac=l,
ad+bc=0,
bd=—2,
whence c = l/a, d = —2/b. Substituting in the relation ad + bc = 0," obtain

o = —2a/b + b/a = (2.:2 + b2)/ab. Thus, —2a2 + b2 = 0, or (b/a)2 = 2, which is impossible because J2 in not a rational number. Although irreducible in Q[x], the polynomial x2 — 2
isnonethelessreducibleinR’“[x];inthiscase,x2 — 2 = (x — J2)(x + J5) and both factors are in R#[x]. For ease of reference more than to present new concepts, let us summarize in the next theorem some of the results of previous chapters (speciﬁ
cally, Theorems 5—5 and 6—7) as applied to the principal ideal domain F[x]. Theorem 712. If F is a ﬁeld, the following statements are equivalent: 1) f(x) is an irreducible polynomial in F[x] ;
2) the principal ideal (f(x)) is a maximal (prime) ideal of F[x];
3) the quotient ring F[x]/(f(x)) forms a ﬁeld. The theorem on prime factorization of polynomials is Stated now. (Unique Factorization in F[x]). Each polynomial Theorem 713. f(x) e F[x] of positive degree is the product of a nonzero element of F and irreducible monic polynomials of F[x]. Apart from the order of the factors, this factorization is unique.
Sufﬁce it to say, Theorem 7—13 can be made more explicit for particular polynomial domains. When we deal with polynomials over the complex numbers, the crucial tool is the Fundamental Theorem of Algebra. Theorem 714. (The Fundamental Theorem of Algebra). Let C be the ﬁeld of complex numbers. If f(x) e C[x] iS a polynomial of positive degree, then f(x) has at least one root in C. Although many proofs ofthe result are available, none is strictly algebraic in nature; thus, we shall assume the validity of Theorem 7—14 without proof. The reader will experience little difﬁculty, however, in establishing the
following corollary. Corollary 1. Iff(x)_e C[x] is a polynomial of degree n > 0, then f(x)
can be expressed in C[x] as a product of n (not necessarily distinct) linear factors.
.
Another way of stating the corollary above is that the only irreducible polynomials in C[x] are the linear polynomials. Directing our attention
POLYNOMIAL RINGS
129
now to the real ﬁeld, we can obtain the form of the prime factorization in
R'[x] (bear in mind that polynomials with coeﬂicients from R‘“ are polynomials in C[x] and therefore have roots in C).
Corollary 2. If f(x)eR*[x] is of positive degree, then f(x) can be factored into linear and irreducible quadratic factors.
Proof: Since f(x) also belongs to C[x], f(x) factors in C[x] into a product
of linear polynomials x — ck, eke C.
If ckeR’“, then x — ckeR#[x].
Otherwise, c,k = a + bi, wherea, b e R’“ and b + 0. But the cOmplex roots of real polynomials occur in conjugate pairs (Problem 7—11), so that 7:, = a — bi is also a root off(x). Thus,
(x — ck) (x — Ek) = x2 — 2ax + (a2 + b2)eR#[x] is a factor of f(x). The quadratic polynomial x2 — 2ax + (a2. + b2) is irreducible in R#[x], since any factorization in R#[x] is also valid in C[x] and (x — ck) (x — 2,) is its unique factorization in C[x]. An interesting remark, to be recorded without proof, is that if F is a ﬁnite ﬁeld, the polynomial ring F[x] contains irreducible polynomials of every degree (see Theorem 9—10). This may be a convenient place to introduce the notion of a primitive polynomial. Deﬁnition 75. Let R be a unique factorization domain. The content of a nonconstant polynomial f(x) = a0 + alx + + a x” e R[x], denoted by the symbol contf(x) is deﬁned to be the greatest common divisor of its coefﬁcients: contf(x) = n (a0: a1:  a an)'
We call f(x) a primitive polynomial if contf(x) = Viewed otherwise, Deﬁnition 7—5 asserts that a polynomial f(x) e R[x] is primitive if and only if there is no irreducible element of R which, divides all of its coeﬂicients. In this connection, it may be noted that in the domain [Ix] of polynomials with coefﬁcients from a ﬁeld F, every nonconstant polynomial is primitive (indeed, there are no primes in F). The reader should also take care to remember that the notion of greatest common divisor and, in consequence, the content of a polynomial is not determined uniquely, but determined only to within associates. Given a polynomialf(x) e R[x] of positive degree, it is possible to write f(x) = cfl(x), where c e R and f1(x) is primitive; simply let 0 = cont f(x). To a certain extent this reduces the question of factorization in R[x] (at least, when R is a unique factorization domain) to that of primitive polynomials. By way of speciﬁc illustrations, we observe that f(x) = 3x3 —
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FIRST COURSE IN RINGS AND IDEALS
4x + 35 is a primitive polynomial in Z[x], while g(x) = 12x2 + 6x  3 3(4x2 + 2x — l) is not a primitive element of the same, Since g(x) has content 3.
Here is another new concept: Suppose that I is a (proper) ideal of R.n commutative ring with identity. There is an obvious mapping v: R[x] 4 (R/I)[x] ; for any polynomialf(x) e R[x] Simply apply nat, to the coefficients off(x), so that
v(f(x)) = (a0 + I) + (a1 + I)x +
or, more brieﬂy, v(f(x)) = 2(nattak)x".
+ (a, + I)x",
The reader will encounter no
difﬁculty in verifying that v, deﬁned in this way, is a homomorphism of R[x] onto (R/I)[x], the socalled reduction homomorphism modulo 1. The
polynomial v(f(x)) is said to be the reduction off(x) modulo 1. Although it might seem to be rather special, the reduction homomorphism will serve us in good stead on several occasions. We make immediate us of it to characterize primitive polynomials. Theorem 715. Let R be a unique factorization domain and let f(x) a
a0 + alx +
+ a,x"eR[x], with degf(x) > 0.
Then f(x) is a
primitive polynomial in R[x] if and only if, for each prime element p e R, the reduction off(x) modulo the principal ideal (p) is nonzero. Proof By deﬁnition, the reduction off(x) modulo (p) is
v(f(x)) = (ao + (12)) + (a1 + (p))x +
+ (a. + (p))x"
Thus, to say that v(f(x)) = 0 for some prime p e R is equivalent to asserting that ak e (p), or rather, plak for all k. But the latter condition signiﬁes tlnt cont f(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 716. (Gauss’s Lemma). Let R be a unique factorization domain. Iff(x), g(x) are both primitive polynomials in R[x], then their product f(x)g(x) is also primitive in R[x]. Proof? Given a prime element p e R, (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 off(x)g(x) cannot be the zero polynomial:
v(f(X)g(JC)) = V(f(X))V(g(x)) 7E 0The assertion of the theorem is now a direct consequence of our last result.
POLYNOMIAL RINGS
131
Corollary. If R is a unique factorization domain and f(x), g(x) e R[x],
then
cont (f(x)g(x)) = contf(x) cont g(x).
Proof. As noted earlier, we can write f(x) = af1(x), g(x) = bgl(x), where a = contf(x), b = cont g(x) and where f1(x), gl(x) are primitive in R[x]. Therefore, f(x)g(x) = abf1(x)g1(x). According to the theorem, the product f.(x)g,(x) is a primitive polynomial of R[x]. This entails that the content
off(x)g(x) is simply ab, or, what amounts to the same thing, cont f(x) cont g(x). Any unique factorization domain R, being an integral domain, possesses a ﬁeld of quotients K = Qc1(R) and we may consider the ring of polynomials R[x] as imbedded in the polynomial ring K[x]. The next theorem deals with the relation between the irreducibility of a polynomial in R[x] as compared to its irreducibility when considered as an element of the larger
ring K[x]. (The classic example of this situation is, of course, the polynomial domain Z[x] E Q[x].) Before concentrating our efforts on this relationship, we require a preliminary lemma. Lemma. Let R be a unique factorization domain, with ﬁeld of quotients K. Given a nonconstant polynomial f(x) e K[x], there exist (nonzero) elements a, b e R and a primitive polynomial f1(x) in R[x] such that
f(X) = 017—113(36)Furthermore, f1(x) is unique up to invertible elements of R as factors. Prooﬁ inasmuch as K is the ﬁeld of quotients of R,f(x) can be written in the
f(x) = (001761) + (01b1— 1)x +
+ (aubn 1)x”,
where a,, b, e R and bi at: 0. Take b to be any common multiple of the bi; for instance, b = bob1 bu. Then b + 0 and, since the coefﬁcients of bf(x) all lie in R, we have bf(x) = g(x) e R[x]. Accordingly,
f(x) = 51906) = ab'1f1(x), where fl(x) E R[x] is a primitive polynomial and a = cont g(x).
We
emphasize that f1(x) is of the same degree as f(x), so cannot be invertible
in R[x]. As for uniqueness, suppose that f(x) = ab'1f1(x) = cd'1f2(x) are two representations that satisfy the conditions of the theorem. Then, “(#100 = b6f2(x).
Since f,(x) and f2(x) are both primitive, the corollary to Gauss’s Lemma implies that we must have ad = ubc for some invertible element u e R.
In consequence, f1(x) = uf2(x), showing that f1(x) is unique to within invertible factors in R.
FIRST COURSE IN RINGS AND IDEALS
132
Theorem 717. Let R be a unique factorization domain, with ﬁeld if quotients K. Iff(x) e R[x] is an irreducible primitive polynomial, tllll
it is also irreducible as an element of K[x]. Proof. Assume to the contrary that f(x) is reducible over K. The; 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,
906) = ab'lglbd,
h(X) = cd'lhdx),
with a, b, c, d e R and g1(x), h1(x) primitive in R[x]. Thus,
bdfOC) = acgl(x)h1(X)Now, Gauss’s Lemma asserts that the product g1(x)hl(x) is a primitive polynomial in R[x], whencef(x) and gl(x)h1(x) differ by an invertible elem of R:
f(X) = 14910011106)Since deg gl(x) = deg g(x) > 0, deg h1(x) = deg h(x) > 0, the outcome is a nontrivial factorization off(x) in R[x], contrary to hypothesis. There is an obvious converse to Theorem 7—17, viz.: if the primitive polynomial f(x)eR[x] is irreducible as an element of K[x], it is also irreducible in R[x]. This is justiﬁed by the fact that R[x] (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] if and only iff(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 coeﬂiciu). The generalization which we have in mind is formulated below. Theorem 718. Let R be an integral domain and the nonconstant polynomialf(x) = a0 + alx + + a,,x'l e R[x]. Suppose thatthere exists a prime ideal P of R such that
l)a,,¢P,
2) akePfor03k F’; the reader will recall that this is deﬁned by taking 4),f(x) = f(r). As before, the image of F[x] under (I), is repre
sented by the symbol F[r]:
F[r] = {f(r)f(x)6F[x]}The set F[r] forms an integral domain (being a subring of the ﬁeld F’) and therefore has a ﬁeld of quotients K = c(F[r]) in F’. It is apparent
that F u {r} g F[r] E K. But F(r) is the smallest subﬁeld of F’ to contain both F and r, whence F(r) E K. On the other hand, any subring of F’ which contains F and r will necessarily contain the elements of F[r]; in particular, F[r] E F(r). Since F(r) is a subﬁeld of F’, it must also contain all the quotients of elements in F[r]. Thus, K E F(r) and equality follows. This leads to the more constructive description of F(r) as F(r) = Qc,(F[r]). The key to classifying simple extensions is the nature of the kernel of the substitution homomorphism (bear in mind that ker :15, consists of all
polynomials in F[x] having r as a root).
Theorem 719. Let F’ be an extension ﬁeld of the ﬁeld F and let r e F’. Then either
I) F(r) z c(F[x]), or else 2) F(r) z F[x]/(f(x) ) for some monic irreducible polynomial f(x) e F[x] such that f(r) = 0; this polynomial is uniquely determined. Proof. By the Fundamental Homomorphism Theorem, we know that
F[r] z F[x]/ker 4),. As F[r] is a subring of a ﬁeld, it must be an integral domain. Hence, ker deg h(x) = degf(x) + deg k(x) 2 degf(x) = n, a manifestly false conclusion. Thus, the polynomial h(x) = 0, which forces = "_ 1 = 0. The proof that the n elements the coefﬁcients c0 = c1 = 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.
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FIRST COURSE IN RINGS AND IDEALS
Corollary 1. If r e F’ 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 I polynomial of degree less than n: F(r) = {a0 + alr +
+ an_1r"'1a,,eF}.
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 726. 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 reF’ 2 F be algebraic and [F'zF] ﬁnite.
Then
F’ = F(r) if and only if [F(r): F] = [F’zF]. Proof By the lastwritten 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 710. Consider the element r = J? + i e C 2 Q, C as usual being the complex number ﬁeld. Then r2 = l + Zﬂi, so that (r2  l)2 = —8 or r4 — 2r2 + 9 = 0. Thus, risaroot of the polynomial
f(x) = x4 — 2x2 + 9eQ[x] and, hence, is an algebraic element over Q. Now, f(x) has the irreducible factorization over C,
f(x)=(x—ﬁ+i)(x—ﬂ—D(x+\/§+i)(x+\/§TD, which indicates that f(x) has no linear or quadratic factors in Q[x]. Therefore, f(x) is irreducible as a member of Q[x] 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 fourdimensional vector space over Q, with basis
1,
r=ﬁ+i,
‘
r2=1+2ﬂi,
r3=—,/§+5i.
At the same time r is a root ofthe polynomial x2 — Zﬁx + 3 e R*[x], with x2 — Zﬂx + 3 irreducible over R’“; thus, r is of degree 2 over R’. Example 71]. For a second illustration, we turn to the extension ﬁeld
Q(ﬂ, ﬂ). The elements ﬂ and J3 are clearly algebraic over Q, being roots
of the polynomials x2 —, 2, x2 — 3 e Q[x], respectively. Our contention is
POLYNOMIAL RINGS
143
that the ﬁeld Q(J§, J3) is actually a simple algebraic extension of Q; in
fact, Q(J§, J3) = Q(J§ + J5), with J5 + J? algebraic over Q. Since the element J? + J? belongs to Q(JZ J3), we certainly have
Q(J§ + J3) S Q(J§, J3). As regards the reverse inclusion, a simple computation shows that
zﬁaﬂwWJM+ﬂ is a member of Q(J§ + J3), and therefore so is J5. But then,
ﬁﬁﬂ+ﬂ—ﬂ
also lies in Q(J§ + J3). This leads to the inclusion Q(JZ J3) E Q(J§ + J3) and the asserted equality. To see that r = J5 + J3 is an algebraic element over Q, notice that
r2 = 5 + 2J3, (r2 — 5)2 = 24, and, hence, the polynomial f(x) = x4 — 10x2 + 1 has r as a root. One may verify that f(x) is irreducible in Q[x], making it the minimum polynomial of r. Perhaps the quickest way to see this is as follows. Let F’ = Q(J§); then [F'zQ] = 2, with basis {1, J5},
and [F’(Ji):F’] = 2, with basis {1, J3}. From Theorem 7—24, it follows that [F’(Jj): Q] = 4andabasis for F’(J§) overQis given by {1,J§,J§,J3}. But
FM) = mm = QMZ ﬂ) = Q(«/5 + ﬂ),
and we know that the dimension of Q(J§ + J3) is equal to the degree of
the minimum polynomial of r = J5 + J3. Incidentally, there are ﬁve ﬁelds between Q and Q(J§, J3), namely,
Q, g(ﬁ), g(ﬂ), Q(J6), and Q(J§, J3). Taking stock of Steinitz’s theorem (page 140), it should come as no surprise that Q(JZ 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 polynomial f(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 of F’[x], has a root. (If degf(x) = 1, 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 of F 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) 2 F[x]/(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.
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FIRST COURSE IN RINGS AND IDEALS
Theorem 727. (Kronecker). If f(x) is an irreducible polynomial 'F[x], then there is an extension ﬁeld of F 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 of F, consider the natural mapping nat,: F[x] —> F’. According to Theorem 4—7, either the restriction nat,F is the trivial homomorphism or else nat, (F) forms a ﬁeld isomorphic to F, where as usual
nat,(F) = {a + Ia e F}. The ﬁrst possibility is immediately excluded by the fact that nat,(l) = 1 + I 7E I, which is the zero element of F’. 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 + alx +
+ aux”, then, from the
deﬁnitions of coset addition and multiplication,
(ao+I)+(a1+I)(x+I)++(a.+I)(x+I)" =ao+a1x++a,,x"+I=f(x)+I=0+I. I Ifwe now identify an element a,‘ e F with the coset at + I which it determines in F’ (the fact that F is isomorphic to nat,(F) permits this), we obtain ao+a1(x+I)++a,,(x+I)"=0,
which is equivalent to asserting that f(x + I) = 0. In other words, the coset x + I = 1x + 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] 1s of positive degree, then there exists an extension ﬁeld of F containing a root off(x) To go back to Theorem 7—27 for a moment, let us take a closer look at
the nature of the cosets of I = (f(x)) in F[x], with the aim of expressing the extension ﬁeld F’ = F[x]/I 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
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145
has degree less than that off(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 deg f(x) = n > 1 (for instance, f(x) = a0 + alx +
then the extension ﬁeld F’ may be described by
F’ = {b0 + blx +
+ anx"),
+ b,,_,x"‘1 + IbkeF}.
Identifying bk + I with the element bk, we see as before that a typical coset can be uniquely represented in the form b0
+
b1(x +
I) +
"'
+
bu_1(x
+
Ir—l.
As a ﬁnal simpliﬁcation, let us replace x + I by some new symbol 1, so that the elements of F’ become polynomials in A: F,
=
{b0
+
b1]. +
"'
+
bn_1l"_lbh€F}.
Observe that since A = x + I is a root off()6) in F’, calculations are carried out with the aid of the relation a0 + all + + all" = 0. The last paragraph serves to bring out the point that F’ is a ﬁnite extension of F with basis {1, l, 2.2, , 11"1}; in particular, we infer that [F’zF] = n = degf(x).
To recapitulate: if f(x) e F[x] is an irreducible polynomial over F, then there exists a ﬁnite extension F’ of F, 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 712. Consider Z2, the ﬁeld of integers modulo 2, and the polynomial f(x) = x3 + x + 1 e Zz[x]. Since neither of the elements 0 and 1
is a root of x3 + x + 1, f(x) must be irreducible in Zz[x]. Theorem 7—27 thus guarantees the existence of an extension of 22,
speciﬁcally, the ﬁeld Z2[x]/(f(x) ), in which the given polynomial has a root. Denoting this root by A, the discussion above tells us that
Z,[x]/(f(x)) = {a + b). + c12a, b,ceZz}
= {0, 1,1,1 + 1,12,1+ 1.2,). + 12,1+ A + 112}, where, of course, 13 + 11 + 1 = 0. As an example of operating in this ﬁeld, let us calculate the inverse of l + J. + 12. Before starting, observe that by using the relations
A3=—(/1+1)=/1+1,
A4=AZ+A
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FIRST COURSE IN RINGS AND IDEALS
(our coefﬁcients come from Z2, where —l = 1), the degree of any product can be kept less than 3. Now, the problem is to determine elements a, b, c e Z2 for which
(1+ 1 + 1% + b}. + c112) = 1. Carrying out the multiplication and substituting for A3, 14in terms of l,
A, and 12, we obtain
(a+b+c)+a/1+(a+b)).2=1. This yields the system of linear equations a+b+c=1,
a=0,
a+b=0,
with solution a = b = 0, c = 1; therefore, (1 +1 + f)—1 = 12.
It is worth noting that x3 + x + 1 factors completely into linal’ factors in Z2[x]/(f(x)) and has the three roots 2., 12, and A + 2.2: x3 + x+1=(x— A)(x — lz)(x—(). + 12)).
Example 713. The quadratic polynomial x2 + l is irreducible in R'[x]. For, if x2 + 1 were reducible, it would be of the form
x2+1
(ax+b)(cx+d) = acx2 + (ad + bc)x + bd,
where a, b, c, d e R’“. It follows at once that ac = bd = landad + be = 0. Therefore, bc = —(ad), and
1 = (aC)(bd) = (ad)(b6) = (ad)2, or rather, (ad)2 = — l, which is impossible. In this instance, the extension ﬁeld R”‘[x]/(x2 + 1) is described by
R”‘[x]/(x2 + 1) = {a + blla,beR#; 12 + 1 = 0}. Performing the usual operations for polynomials, we see that (a+b).)+(c+dl)=(a+c)+(b+d)l and
(a + bl)(c + d1) = (ac — bd) + (ad + bc)). + d.2 + l) = (ac — bd) + (ad + bdM. 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 (I): R"‘[x]/(x2 + l) —> C given by (a + bl) = a + bi. Thil
provides an elegant way of constructing C from R’“.
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147
Before proceeding further, two comments are in order. First, Example
712 shows that there exist ﬁnite ﬁelds other than the ﬁelds ZI, of integers modulo a prime p. The fact that the ﬁeld of this example has 23 = 8 elements is typical of the general situation: if F is a ﬁnite ﬁeld, then F contains 11" elements, where the prime p is the characteristic of F (Theorem 9—7). In the second place, the construction of Theorem 7—27 yields an exten
sion of the ﬁeld F in which a given (nonconstant) polynomial f(x) e F[x] . splits oﬂ' one linear factor. By repeated application of this procedure, we can build up an extension F' of F in which f(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 off(x) (technically speaking, the polynomial splits completely in F’[x]). We present this result in the form of an existence theorem. ' Theorem 728. Iff(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 727, there is an extension ﬁeld K of F in which
a(x) and, hence, f(x), has a root r1; speciﬁcally, the ﬁeld K = F[x] /(g(x) ) Thus, f(x) can be written in K[x] as f(x) = (x — r1)h(x), where deg h(x) = n — 1. By our induction assumption, there is an extension F’ of K in which Mx) splits completely; say h(x) = a(x — r2)(x — r3) (x — r"), with r, e F’, a 4: 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 of F in which f(x) has n (not necessarily distinct) roots.
Example 714. To illustrate this situation, let us look at the polynomial
f(x) = (x2 — 2)(x2 — 3) over the ﬁeld Q of rational numbers.
From
Example 7—4, x2 — 2 (and by similar reasoning, x2 — 3) is already known to be irreducible in Q[x]. So we begin by extending Q to the ﬁeld F1, where
Fl = Q[x]/(x2 — 2) = {a + b).a,beQ;/12 — 2 = 0}; and obtain the factorization
f(x) = (x — l)(x + 10062  3) = (x — ﬂ)(x + ﬁxxz — 3).
(As 1.1 = 2, one customarily identiﬁes A with ﬂ.)
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FIRST COURSE IN RINGS AND IDEALS
However, f(x) does not split completely, since the polynomial x’  3
remains irreducible in F1[_x]. For, suppose to the contrary that x2  3 III a root in F1; say c + d\/2, with c, d 6 Q. Substituting, we ﬁnd that (c2 + 2d2 — 3) + 2cdﬂ = 0, or, what amounts to the same thing, c2+2d2—3=0,
cd=0.
The latter equation implies that either c = 0 or d = 2. But neither c not d can be zero, since this would mean that d2 = 3/2 or c2 = 3, which is clearly impossible. Accordingly,x2 — 3 does not Split In F1[x]. In order to factorf(x) into linear factors, it becomes necessary to extend the coeﬂicient ﬁeld further. We therefore construct a second extension P, where
F2 = F1[x]/(x2 — 3) = {a + ﬂﬂla,ﬂEFl;/t2 — 3 = 0}. The elements of F2 can be expressed alternatively in the form
(a+bﬂ)+(c+d\/§)\/3=a+b\/§+c\/3+d\/3, where, of course, the coefficients a, b, c, d all lie in Q. It follows without
difﬁculty that the original polynomial now factors in F2[x] as
f(x) = (x  1X30 + 11)(x  £006 + I!)
= (x — Jinx + ﬂux — J3)(x + J3).
Let a ﬁeld F be given and consider a nonconstant polynomialf(x) 5 ﬁx]. An extension ﬁeld F’ of F is said to be a splittingﬁeldforf(x) over F provided that f(x) can be factored completely into linear factors in F’[x], but not so factored over any proper subﬁeld of F’ 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:
f(x) = 6106 — r1)(x — r2)
(x  7.)
(nEF’)
To obtain a splitting ﬁeld for f(x), we need only consider the family {FJ of all extension ﬁelds Fi in which f(x) can be decomposed as a product of linear factors (Theorem 7—27 guarantees the existence of such extensions); then 0 Fl. 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. Let f(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) z 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) 2 F[x]/(f(x)) via an isomorphism 0 with the property that 4), = 0 o natmx». (As usual, ¢,:F[x] —> K is the substitution homomorphism induced by r.) Regarding the last statement of the lemma, we necessarily have
7 = ¢r(X) = (0° natmxnxx) = 0(x + (f(10))The chief value of this lemma is that it leads almost immediately to the following theorem. Theorem 729.
(Isomorphism Extension Theorem).
Let a be an iso
morphism from the ﬁeldF onto theﬁeld F'. Also, letf(x) = a0 + alx + + a,,x'I be an irreducible polynomial in F[x] and f’(y) = 6(ao) + 0(a1)y + + a(a,,)y" be the corresponding polynomial in F’[y]. Then, f'(y) is likewise irreducible. Furthermore, if r is a root of f(x) in some extension ﬁeld of F and r’ is a root off’(y) in some extension ﬁeld of F’, then a can be extended to an isomorphism (I) of F(r) onto F’(r’) with (I)(r) = r’. Proof: Let us ﬁrst extend a to a mapping 6 between the polynomial rings
F[x] and F’[y] by setting adx) =
6(bo
+
b1x
+
“'
+
bnx") =
”(bo) +
db1)y +
"'
+
o'(bn)y.I
for any polynomial g(x) = b0 + blx + + bnx'l e F[x] We bequeath to the reader the task of supplying the necessary details that 6— is an isomorphism of F[x] onto F’[y]. It is important to notice that for any polynomial g(x) in F[x], an element a e F is a root of g(x) if and only if 0(a) is a root of &g(x). Indeed, if, as before, g(x) = b0 + blx + + bnx", then, upon evaluating Eg(x) at 0(a),
(ag(x»(a(a)) = «(120) + 0(b1)o(a) + = «(b0 + bla +
+ «(boner
+ bua”)
= 0(g(a)), from which our assertion follows. In particular, we infer that the polynomials g(x) and 6g(x) are simultaneously reducible or irreducible in F[x]
and F’[y], respectively. This being so, f’(y) = 6f(x) is irreducible in F’[y].
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FIRST COURSE IN RINGS AND IDEALS
 By the foregoing lemma, we know that there exist isomorphism ac: F(r) —» F[x]/(f(x)) and ﬂ: F’(r’) > F'[y]/(f’(y)), with
W) = x + (f(x»,
50’) = y + (f’(.v))
MoreOVer, it is an easy matter to show that there is also an isomorphism tol'
F[x]/(f(x)) onto F’[y]/(f’(y)) deﬁned by
r(g(x) + (f(x))) = 69(x) + (f’(y))
(g(x)e F[x]).
Observe particularly that ‘L’ carries the coset x + (f(x)) onto y + (f'(y)). We contend that F(r) : F’(r’) under the composition of maps
(I) = I!" o r o a, where (I): F(r) —> F’(r’); this situation is portrayed in the diagram below: F(r) —(p—> F’(r’)
a
,3" 1
F[X]/(f(x)) i» F’[y]/(f’(y)) Certainly, (I) is an isomorphism ofF(r) onto F’(r’), for the individual mapping a, r, ,8" 1 are themselves isomorphisms. If a is an arbitrary element of F, then
(Ma) = 09“ ° T)(a(a)) = (5'1 ° T)(a + (f(x))) = [3‘1(0(a) + (f'00)) = 0(a), whence (I) is actually an extension of a to all of F(r). Finally, we point out that
(F(r) = (1?"1 ° T)(a(r)) = (19" ° r)(x + (f(x)) = ﬂ'1(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# ; let f(x) e R*[x] he the irreducible polynomial f(x) = x2 + 1, so that f’(y) = y2 + I (recall that the identity map is the only isomorphism of R” onto itself). Finally, choose r = iand r’ = —i. Theorem 7—29 then asserts that R*(i) z R*(—i) under anisomorphism which carriesionto —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 a be an isomorphism of the ﬁeld F onto the ﬂu“ F’. Let f(x) = a0 + alx + + a,x”eF[x] and f’(y) = duo) +
PROBLEMS
151
0(a1)y + + a(a,,)y" be the corresponding polynomial in F’[y]. If K is a splitting ﬁeld off(x) and K’ a splitting ﬁeld off’(y), then a can be extended to an isomorphism (I) of K onto K’. Proof Our argument will be by induction on the number n of roots off(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 off(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 a 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’01) 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 2 1). Iff(x) e F[x] is a polynomial having n roots outside of F, then not all of the irreducible factors off(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; (all it r. Similarly, one of the roots of the polynomial f’(y), say r’, is a root of g’(y) in K’. By Theorem 7—29, 0 can be extended to an isomorphism a’
between the ﬁelds F(r) and F’(r’). Now, K is a splitting ﬁeld off(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 off(x) lying outside of F(r) is less than n, the induction hypothesis permits us to extend a" (itself an extension of a) to an isomorphism (I) of K’ onto K. This completes the induction step and the proof of the theorem as well, for a 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 (I) such that the restric
tion (DF is the identity mapping. Prooﬂ This is an immediate consequence of the theorem on taking F = F’ and a to be the identity isomorphism iF. PROBLEMS I. If R is a commutative ring with identity, prove that
a) The set I = {f(x)eR[[x]]ordf(x) > 0} u {0} forms an ideal of the ring R[[x]]; in fact, I = (x).
FIRST COURSE IN RINGS AND IDEALS
152
b) The ideal 1" consists of all power series having order 2n, together with 0. C)
nnEZ+ II! = {o}'
2. For any ﬁeld F, consider the set F(x) consisting of all expressions of the in w
z akx" = a_,,x"' + a_,,+1x"'+1 +
+ a_1x'l + a0 + alx + azx’ +
k= n
where all the a,‘ e F and n 2 0 varies. If addition and multiplication are deﬁned in the obvious way, F(1:) bacon I ring, known as the ring of extended (formal) power series over F. Show that 17(3)
is in fact the ﬁeld of quotients of the domain F[[x]]. [Hint: Given neZ” c(F[[x]]) must contain x‘”.] 3. Let R be a commutative ring with identity. If R is a local ring, prove that III 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) = a0 + alx + + 11.x" is a zero divisor in R[x] then there exists an element 0 + r e R such that rf(x) 0. [Hmt Assume tint f(x)g(x) = 0. Use the polynomials akg(x) to obtain 0 99 h(x)eR[x], with deg h(x) < deg f(x), satisfying h(x)f(x) = 0.] 5. If R is a commutative ring with identity, verify that the polynomial l + as '
invertible in R[x] if and only if the element a is nilpotent in R. [Hint: Prohlan in. 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) charR = 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]. [Hinu Induct on in degree of polynomials in I[x].] 7. Establish the following assertions concerning the polynomial ring Z[x]: a) The ideal
(x) = {alx + azx2 +
+ a,x"a.,eZ;n 21}
is a prime ideal of Z[x], but not a maximal ideal. Incidentally, (x) is maxi“
in F[x], where F is a ﬁeld.
b) Z[x] is not a principal ideal domain. [Hint: Consider (x, 2), the (maximh ideal of polynomials with even constant terms] c) The primary ideal (x, 4) is not the power of any prime ideal of Z[x]. [113: (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, 'l
M[x] a maximal ideal of R[x]?
PROBLEMS
153
9. Consider the polynomial domain F[x], where F is a ﬁeld, and a ﬁxed element r e F. Show that the set of all polynomials having r as a root,
M, = {f(x)€F[XJIf(r) = 0}, forms a maximal ideal of F[x], with F[x]/M, : F. [Hint: M, = ker 45,, where ¢, : F[x] —> F is the substitution homomorphism induced by r.] IO. Regarding the ring ofExample 8, Chapter 1, show that the polynomial (a, 0)::2 e R[x]
has inﬁnitely many roots in R[x]. II. Given f(x) = a0 + alx +
+ a,x"e C[x], deﬁne the polynomialﬂx) by
f(x) = 6., + 51x +
+ aux“,
where E,‘ denotes the usual complex conjugate of a," Verify that
a) r e C is a root off(x) if and only if F is a root off(x). [Hintﬂﬁ = 7(7)] b) Iff(x) e R#[x] E CDC] and r is a complex root off(x), then 7 is also a root
off(x).
12. Let R be a commutative ring with identity and'let f(x) e R[x]. The function 7': R » R deﬁned by taking f(r) = f(r) for every r e R is called the polynomial
junction induced by f(x). Assuming that PR denotes the set of all polynomial functions induced by elements of R[x], prove that a) PR forms a subring of map R, known as the ring of polynomial functions on R;
b) the mapping a: R[x] —» PR given by a(f(x)) = 7' is a homomorphism of R[x] onto PR;
e) if the element r e R is ﬁxed and I, = {76 PR7(r) = 0}, then Ir is an ideal of PR. 13. a) When R is an integral domain, show that distinct polynomials in R[x] induce distinct polynomial functions (in other words, the mapping a: R[x] —> PR is onetoone) if and only if R has an inﬁnite number of elements. b) Give an example of two distinct polynomials which induce the same polynomial function. Id. Let R be a commutative ring with identity anddeﬁne the function 6: R[x] —> R[x], the socalled derivative function, as follows:
If
f(x) = a0 + alx +
+ a,#eR[x],
then
6f(x) = a1 + 2a2x +
+ naux'“.
For any f(x), g(x) e R[x] and any r e R, establish that
a) 5(f(x) + 906)) = 6f(x) + 59(36)b) 6(rf(x)) = r5f(x). c) 6(f(x)g(x)) = 6f(x)g(x) + f(x)6g(x). [Hint: Induct on the number of terms off(x).] 15. Suppose that R is a commutative ring with identity and let r e R be a root of the nonzero polynomial f(x) e R[x]. We call r a multiple root off(x) provided that
f(X) = (x — r)”g(X)
(n > 1),
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FIRST COURSE IN RINGS AND IDEALS
where g(x) e R[x] is a polynomial such that g(r) + 0.
Prove that an elm
r e R is a multiple root off(x) if and only if r is a root of both f(x) and 6f(x).
16. Let F be a ﬁeld and f(x) e F[x] be a polynomial of degree 2 or 3. Deduce tilt f(x) is irreducible in F[x] if and only if it has no root in F. Give an example whit shows that this result need not hold if degf(x) 2 4. 17 Prove that if a polynomial f(x) e F[x] (F a ﬁeld of characteristic 0) is irreducible. then all of its roots in any ﬁeld containing F must be distinct. [Hint: First show that god (f(x), 6f(x)) = 1.] 18. Given that I is a proper ideal of R, a commutative ring with identity, establish the assertions below: a) If v: R[x] —» (R/I)[x] is the reduction homomorphism modulo the ideal I,
then ker v = I[x]; hence, R[x]/I[x] z (R/I)[x]. b) If the polynomial f(x) e R[x] is such that v(f(x)) is irreducible in (R/l)[x]. then f(x) is irreducible in R[x]. c) The polynomial f(x) = x3 — x2 + 1 is irreducible in Z[x]. [Hint: Redum
the coeﬂicients modulo 2.] 19. Let R be a unique factorization domain. Show that any nonconstant divisor of I primitive polynomial in R[x] is again primitive.
. Utilize Gauss’s Lemma to give an alternative proof of the fact that if R is a unique factorization domain, then so is the polynomial ring R[x]. [Hinn For 0 39 f(x) e R[x], induct on deg f(x); if deg f(x) > 0, write f(x) = cf,(x), where c e R and f1(x) is primitive; if f,(x) is reducible, apply induction to its factors] 21. Apply the Eisenstein Criterion to establish that the following polynomials are
irreducibleinQ[x]:f(x) = x2 + l,g(x) = x2 — x + 1,andh(x) = 2x5 — 6x3 + 9x2 — 15. [Hint: Consider f(x + 1), g(—x).] 22. Let R be a unique factorization domain and K its ﬁeld of quotients. Assume that ab" 1 e K (where a and b are relatively prime) is a nonzero root of the polynomial
f(x) = a0 + alx +
+ aux” e R[x]. Verify that aa0 and ba,,.
. Prove the following assertions concerning the polynomial ring Z[x, y]: a) The ideals (x), (x, y) and (2, x, y) are all prime in Z[x, y], but only the last is maximal.
b) (x, y) = J(x2, y) = J(x’,xy,y1)c) The ideal (x", xy, yz) is primary in Z[x, y] for any integer k e Z+. d) If I = (x2, xy), then ﬂ is a prime ideal, but I is not primary. [Hint: f = (x).] . Consider the polynomial domain F[x, y], where F forms a ﬁeld.
a) Show that (x2, xy, yz) is not a principal ideal of F[x, y]. b) Establish the isomorphism F[x, y]/(x + y) z F[x]. . Let the element r be algebraic over the ﬁeld F and let f(x) e F[x] be a monic polynomial such that f(r) = 0. Prove that f(x) is the minimum polynomial of r over F if and only iff(x) is irreduciblein F[x]
. Assuming that F’ is a ﬁnite extension of the ﬁeld F, verify each of the statements below:
PROBLEMS
155
a) When [F’ :F] is prime, F’ is a simple extension of F; in fact, F’ = F(r) for every element re F’ — F.
b) Iff(x) e F[x] is an irreducible polynomial whose degree is relatively prime to [F’: F] then f(x) has no roots in F’. c) If r e F’ is algebraic of degree n, then each element of F(r) has as its degree an integer dividing rt
d) Given ﬁelds K, (i = 1, 2) such that F’ 2 K, 2 F, with [Kl] and [K2s relatively prime integers, necessarily Kl n K2 = F.
27. Show that the following extension ﬁelds of Q are simple extensions and determine
their respective degrees: Q(J§, ﬂ), QUE i), Q(ﬂ, 3/3).
3. a) Prove that the extension ﬁeld F’ = F(rl, r2,
, r"), where each element r is
algebraic over F, forms a ﬁnite extension of F. [Hint: If F, = F,_1(r,), then F. = F’and [F’: F] = II[F,+1.:F].] b) If F” is an algebraic extension of F’ and F’ is an algebraic extension of F, show
that F” is an algebraic extension of F. [Hint: Each r e F” is a root of some polynomial f(x) = a0 + alx + + a,x"eF’[x]; consider the extension ﬁelds K = F(ao, a1,
, a.) and K’ = K(r); r is algebraic over K’.]
3. Let F’ be an extension of the ﬁeld F. Prove that the set of all elements in F’ which
are algebraic over F constitute a subﬁeld of F’; applied to the case where F’ = C and F = Q, this yields the ﬁeld of algebraic numbers. [Hint: If r, s are algebraic over F, [F(r, s) :F] is ﬁnite; hence, F(r, s) is an algebraic extension of F] II. a) Granting that f(x) = x2 + x + 2 is an irreducible polynomial in Za[x],
construct the multiplication table for the ﬁeld Zs[x]/(f(x)).
1)) Show that the polynomial f(x) = x3 + x2 + 1eZZ[x] factors into linear factors in Zz[x]/(f(x)) by actually ﬁnding the factorization.
31. If n 9!: l is a (nonzero) squarefree integer, verify that Q[x]/(x2 — n) forms a ﬁeld isomorphic to the quadratic ﬁeld
em) = {’a + bﬁla, be Q}32. Describe the splitting ﬁelds of the following polynomials: a) x3 — 3eQ[x], b) x2 + x + leZS[x],
c) x‘ + 2x2 + 1eR*[x], d) (x2 — 2)(x2 + 1)eQ[x]. 33. Let r be a root of the polynomial f(x) = x3 — x + l e Q[x]. Find the inverse
of l — 2r + 3r2 in Q(r). 34. Let f(x) e F[x] be an irreducible polynomial and r, s be two roots off(x) in some splitting ﬁeld. Show that F[r] z F[s], by a unique isomorphism that leaves every element of F ﬁxed and takes r into s. 35. Suppose that F’ is the splitting ﬁeld for the polynomial f(x) e F[x]; say f(x) = a(x — r1)(x — r2)
(x — r”)
(rieF’, a 3/: 0).
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FIRST COURSE IN RINGS AND IDEALS
Prove that F’ = F(r,,r2, ...,r_).
As a particular illustration, establish M
Q(\/2 J3) is the splitting ﬁeld of (x2  2)(x2 — 3) e Q[x].
36. Let f(x) e Z [x] be an irreducible polynomial of degree n, p a prime. Verify U
the ﬁeld F= Z [x]/(f(20) contains p'I elements.
37. If F’ is a splitting ﬁeld of a polynomial of degree 71 over F, show that [F’ :F] 5 ll 38. A ﬁeld F’ is said to be algebraically closed if F’ has no proper algebraic extension. Assuming that F' is an algebraic extension of F, prove the equivalence of lb following statements: a) F’ 18 algebraically closed. b) Every irreducible polynomial in F’[x] is linear. c) Every polynomial in F[x] splits in F’. (For a proof that every ﬁeld has an algebraic extension which is algebraically closed, the reader is referred to [23].)
EIGHT
CERTAIN RADICALS OF A. RING We touched earlier on the radical concept by brieﬂy considering the notion of the nil radical of an ideal. There are a number of other radicals in circulation; several of the more prominent ones are introduced in this chapter. These various formulations are not, in general, equivalent to one
another and this has given rise to certain confusion and ambiguity in the use of the term. (Indeed, whenever the reader encounters the word “radical”
by itself, he should take some pains to discover just what is meant by it.) By way of removing some of this confusion, qualifying adjectives are given to the different types of radicals which appear here. In addition to indicating the importance of these new radicals in the structure theory, we will be concerned with the nature of the inclusion relations between them and the circumstances under which various radicals coincide. The reader is again reminded that, in the absence of any statement to the contrary, the term
“ring” will always mean a commutative ring with identity. It appears in order to deﬁne one of the radicals around which our interest centers. Deﬁnition 8—]. The Jacobson radical of a ring R, denoted by rad R, is
the set
rad R = n {MIM is a maximal ideal of R}. If rad R = {0}, then R is said to be a ring without Jacobson radical or, more brieﬂy, R is a semisimple ring. The Jacobson radical always exists, since we know by Theorem 5—2 that
any commutative ring with identity contains at least one maximal ideal. It is also immediately obvious from the deﬁnition that rad R forms an ideal of R which is contained in each maximal ideal. To ﬁx ideas, let us determine the Jacobson radical in several concrete
rings. Example 81. The ring Z of integers is a semisimple ring. For, according 157
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FIRST COURSE IN RINGS AND IDEALS
to Example 5—1, the maximal ideals of Z are precisely the principal ideals generated by the prime numbers; thus,
rad R = n {(p)p is a prime number}. Since no nonzero integer is divisible by every prime, we see at once thlt
rad R = {0}. Example 82. A more penetrating illustration is furnished by the ring R = map (X, F), where X is an arbitrary set and F a ﬁeld. For any element x e X, consider the function 1,,f = f(x) which assigns to each function in R its value at x. It is easily checked that t, is a homomorphism of R into F; since R contains all the constant functions, this homomorphism actually maps onto the ﬁeld F. Thus, by Problem 10, Chapter 5, its kernel is the maximal ideal
M, = {fe Rf(x) = 0}. Because rad R E n M, = {fe Rf(x) = 0 for all x e X} = {0}, it follows that R must be a semisimple ring.
Example 83. For a ﬁnal example, we turn to the ring R[[x]] of form! power series.
Here, there is a onetoone correspondence between the
maximal ideals M of R and 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 1: (Theorem 7—4). Thus,
rad R[[x]] = n {M’IM’ is a maximal ideal of R[[x]]} = n {(M, x)M is a maximal ideal of R} = (n M, x) = (rad R, x).
In particular, if R is taken to be a ﬁeld F, we have rad .F[[x]] = (x), the principal ideal generated by x. Our ﬁrst theorem establishes a basic connection between the Jacobson
radical and invertibility of ring elements.
Theorem 81. Let I be an ideal of the ring R. Then I E rad R if and only if each element of the coset 1 + I has an inverse in R. Proof. We begin by assuming that I E rad R and that there is some element a e I for which 1 + a is not invertible. Our object, of course, is
to derive a contradiction. By the corollary to Theorem 5—3, the element 1 + a must belong to some maximal ideal M of the ring R. Since a e rad R, a is also contained in M, and therefore 1 = (1 + a) — a lies in M. But this means that M = R, which is clearly impossible.
CERTAIN RADICALS OF A RING
159
To prove the converse, suppose that each member of l + I has a multiplicative inverse in R, but I 5; rad R. By deﬁnition of the Jacobson radical, there will exist a maximal ideal M of R with I $ M. Now, if a is
any element of I which is not in M, the maximality of M implies that the ideal (M, a) = R. Knowing this, the identity element 1 can be expressed in the form 1 = m + ra for suitable choice of m e M and r e R. But then, m = 1 — me 1 + I, so that m possesses an inverse. The conclusion is
untenable, since no proper ideal contains an invertible element. The form which this result takes when I is the principal ideal generated by a e rad R furnishes a characterization of the Jacobson radical in terms of elements rather than ideals. Although actually a corollary to the theorem just proved, it is important enough to be singled out as a theorem. Theorem 82. In any ring R, an element a e rad R if and only if 1 — ra is invertible for each r e R. This theorem adapts itself to many uses. instructive applications are presented below.
Three fairly short and
Corollary 1. An element a is invertible in the ring R if and only if the coset a + rad R is invertible in the quotient ring R/rad R. Proof Assume that the coset a + rad R has an inverse in R/rad R, so that
for some b e R.
(a + rad R)(b + rad R)= 1+ rad R Then 1 — ab lies in rad R. We now appeal to Theorem
8—2, with r = l, to conclude that the product ab = 1 — 1(1 — ab) is invertible; this, in turn, forces the element a to have an inverse in R. The
other direction of the corollary is all but obvious. Corollary 2. The only idempotent element in rad R is 0. Proof. Let the element a e rad R with a2 = a. Taking r = 1 in the prewding theorem, we see that 1 — a has an inverse in R; say (1 — a)b = 1, where b e R. This leads immediately to
a=a(1—a)b=(a—a2)b=0, which completes the proof. Corollary 3. Every nil ideal of R is contained in rad R. Proof Let N be a nil ideal of R and suppose that a e N. For every r e R, we then have me N, so that the product ra is nilpotent. Problem 10, Chapter 1, therefore implies that 1 — ra is invertible in R. This shows that the element a lies in rad R, from which it follows that N g rad R.
Although the Jacobson radical of a ring R is not necessarily a nil ideal, very little restriction on R forces it to be nil. We shall see subsequently
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FIRST COURSE IN RINGS AND IDEALS
that, if every ideal of R is ﬁnitely generated, then rad R is not only nil bu nilpotent. This is a convenient place to also point out that a homomorphic imp of a semisimple ring need not be semisimple. An explicit example of this situation can easily be obtained from the ring Z of integers. While Z form a ring without a Jacobson radical, its homomorphic image Z, (p a prime; n > 1) contains the nil ideal (p); appealing to Corollary 3 above, we see that ZI," cannot be semisimple.
Example 8—4. Consider F[[x]], the ring of formal power series over a ﬁeld F.
From the lemma on page 116, it is known that an element f(x) 
I a0 + alx +
+ aux" +
has an inverse in F[[x]] if and only ifthe
constant term a0 aé 0. This observation (in conjunction with Theorem 8—2) implies that if g(x) = bo + blx + + bnx" + , then
rad F[[x]] = {f(x)1 — f(x)g(x) is invertible for all g(x) e F[[x]]} ={f(x)1— aobo + o for all boeF} = {f(X)Iao = 0} = (x)Thus, we have a second proof of the fact that the Jacobson radical of F[[xI] is the principal ideal generated by x. We next prove several results bearing on the Jacobson radical ofquotient rings. The ﬁrst of these provides a convenient method for manufacturing semisimple rings; its proof utilizes both implications of the last theorem. Theorem 83. For any ring R, the quotient ring R/rad R is semisimple; that is, rad(R/rad R) = {0}. Proof. Before becoming involved in details, let us remark that since rad R
constitutes an ideal of R, we may certainly form the quotient ring R/rad R. To simplify notation somewhat, we will temporarily denote rad R by I. Suppose that the coset a + I e rad (R/I). Our strategy is to Show that the element a e I, for then a + I = I, which would imply that rad (R/I) consists of only the zero element ofR/I. Since a + I is a member ofrad (R/I), Theorem 8—2 asserts that (l+I)—(r+I)(a+I)=l—ra+I
is invertible in R/I for each choice of r e R. Accordingly, there exists a coset b + I (depending, of course, on both r and a) such that (l—ra+I)(b+I)=l+I. This is plainly equivalent to requiring l— (b — rab)eI = radR.
CERTAIN RADICALS OF A RING
161
Ayin appealing to Theorem 8—2, we conclude that the element b—rab=1—l(1—b+rab) has an inverse c in R. But then (1 — ra)(bc) = (b — rab)c = 1, so that l — ra possesses a multiplicative inverse in R. As this argument holds for every r e R, it follows that a e rad R = I, as desired.
Continuing this theme, let us express the Jacobson radical of the quotient ring R/I as a function of the radical of R. Theorem 8—4. If I is an ideal of the ring R, then
1) rad (R/I) 2 Elli—+1, and, 2) whenever I E rad R, rad (R/I) = (rad R)/I. Proof Perhaps the quickest way to establish the ﬁrst assertion is by means of the Correspondence Theorem; using this, one has
rad (R/I)
n {M’IM’ is a maximal ideal of R/I} n {nat,MM is a maximal ideal ofR with I S M} rad R + I 2 nat, (radR + I) = —I ,
which is the ﬁrst part of our theorem (the crucial step requires the inclusion
ﬂmMM 2 I + radR).
With an eye to proving (2), notice that whenever I g rad R, then rad (R/I) 2
radR +1 2 (rad R)/I. I
Thus, we need only to show the inclusion (rad R)/I 2 rad (R/I). To this purpose, choose the coset a + I erad(R/I) and let M be an arbitrary maximal ideal of R. Since I E rad R g M, the image nat,M = M/I must be a maximal ideal of the quotient ring R/I (Problem 3, Chapter 5). But then,
a + Ierad(R/I) S M/I,
forcing the element a to lie in M. As this holds for every maximal ideal of R, it follows that a e rad R and so a + I e (rad R)/I. All in all, we have proved that rad (R/I) E (rad R)/I, which, combined with our earlier inclusion, leads to (2). Armed with Theorem 8—4, we are in a position to establish:
FIRST COURSE 1N RINGS AND IDEALS
162
Theorem 85. For any ring R, rad R is the smallest ideal I of R such that the quotient ring R/I is semisimple (in other words, if R/I is I semisimple ring, then rad R g I).
Proof. From Theorem 83, it is already known that R/rad R is without Jacobson radical.
Now, assume that I is any ideal of R for which the
associated quotient ring R/I is semisimple. Using part (1) of the preceding theorem, we can then deduce the equality (I + rad R)/I = I. This in turn leads to the inclusion rad R E I, which is what we sought to prove. This may be a good place to mention two theorems concerning the number of maximal ideals in a ring; these are of a rather special character, but typify the results that can be obtained.
Theorem 8—6. Let R be a principal ideal domain. Then, R is semisimple if and only if R is either a ﬁeld or has an inﬁnite number of maximal ideals. Proof. Let {pi} be the set of prime elements of R. According to Theorem 6—10, the maximal ideals of R are simply the principal ideals (pi). It follows that an element a e rad R if and only if a is divisible by each prime p,. if R has an inﬁnite set of maximal ideals, then a = 0, since every nonzero
noninvertible element of R is uniquely representable as a ﬁnite product of primes.
On the other hand, if R contains only a ﬁnite number of primer
171, p2,
, p", we have
radR = ﬂ (pa = (12112212.) aé {0}, so that R cannot be semisimple. Finally, observe that if the set' {pi} is empty, then each nonzero element of R is invertible and R is a ﬁeld (in which case
rad R = {0}). Corollary. The ring Z of integers is semisimple.
Theorem 87. Let {Mi}, ie I, be the set of maximal ideals of the ring R.
If, for each i, there exists an element aie Mi such that 1 — aie
rad R — Mi, then {Mi} is a ﬁnite set.
Proof. Suppose that the index set J is inﬁnite. Then there exists a wellordering s of I under which I has no last e1 ment. (See Appendix A for terminology.) For each ief, we deﬁne I, a“ ,4 M}. Then {1,} forms a chain of proper ideals of R. By hypothesis, we can select an element a, e M, such that 1 — a, e I, — Mi. Now the ideal I = U I, is also aproper ideal of R, since 1 9.6 I. By our choice of the 1,, I is not contained in any maximal ideal of R. Indeed, suppose that there does exist an index i for which I C Mi; then, 1— aisligl g M,,
CERTAIN RADICALS OF A RING
163
yielding the contradiction 1 e M,. But it is known that every proper ideal of R is contained in a maximal ideal of R (Theorem 5—2). From this ' contradiction we conclude that .f must be ﬁnite. Let us now turn to a consideration of another radical which plays an essential role in ring theory, to. wit, the prime radical. Its deﬁnition may also be framed in terms of the intersection of certain ideals. Deﬁnition 82. The prime radical of a ring R, denoted by Rad R (in contrast with rad R), is the set
Rad R = n {PP is a prime ideal of R}. If Rad R = {0}, we say that the ring R is without prime radical or has zero prime radical. Theorem 5—7, together with Deﬁnition 8—1, shows that the prime radical exists, forms an ideal of R, and satisﬁes the inclusion Rad R E rad R. It
is useful to keep in mind that, for any integral domain, the zero ideal is a
prime ideal; for these rings, Rad R = {0}. In particular, the ring F[[x]] of formal power series over a ﬁeld F has zero prime radical but, as we already know, a nontrivial Jacobson radical.
Perhaps the most striking result of the present chapter is that the prime radical, although seemingly quite different, is actually equal to the nil radical of a ring. The lemma below provides the key to establishing this assertion. Lemma. Let I be an ideal of the ring R. Further, assume that the subset S E R is closed under multiplication and disjoint from I. Then there exists an ideal P which is maximal in the set of ideals which contain I and do not meet S; any such ideal is necessarily prime. Prooﬂ Consider the family .97; of all ideals J of R such that I E J and J n S = d). This family is not empty since I itself satisﬁes the indicated conditions. Our immediate aim is to show that for any chain of ideals {Ji} in 9', their union u Ji also belongs to 3". It has already been established in Theorem 5—2 that the union of a chain of ideals is again an ideal; more
over, since I ; Ji for each i, we certainly have I g u Ji. Finally, observe that (uJi)nS= u(J,nS)= u¢=¢. The crux of the matter is that Zom’s Lemma can now be applied to infer that .97 has a maximal element P; this is the ideal that we want. By deﬁnition, P is maximal in the set of ideals which contain I but do not meet S. To settle the whole aliair there remains simply to show that P is a prime ideal. For this purpose, assume that the product ab 6 P but that a at P and b ¢ P. Since it is strictly larger than P, the ideal (P, a) must
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contain some element r of S; similarly, we can ﬁnd an element s e S end! that s e (P, b). This means that rs e (P, a)(P, b) E '(P, ab) 9 P. As S is hypothesized to be closed under multiplication, the product rs also lies in S. But this obviously contradicts the fact that P n S = a. Our argument therefore shows that either a or b is a member of P, which proves
that P is a prime ideal. Remark. The ideal P need not be a maximal ideal of R, in the usual meaning of the term, but only maximal with respect to exclusion of the set S. To put it another way, if J is any ideal of the ring R which properly contains P,
then J must contain elements of S. Two special cases of this general setting are particularly noteworthy:
S = {1} and I = {0}. In the event S = {1}, the ideal P mentioned in the lemma is actually a maximal ideal (in the usual idealtheoretic sense); consequently, we have a somewhat different proof of the facts that (i) every proper ideal is contained in a maximal ideal and (ii) each maximal ideal is prime. The case where I is the zero ideal is the subject of the following corollary, a result which will be utilized on several occasions in the sequel. Corollary. Let S be a subset of the ring R which is closed under multiplication and does not contain 0. Then there exists an ideal maximal in the set of ideals disjoint from S; any such ideal is prime. As it stands, the preceding lemma is just the opening wedge; we can exploit it rather effectively by now proving
Theorem 88. The intersection of all prime ideals of R which contain a given ideal I is precisely the nil radical of 1:
ﬂ = n {PP 2 I; Pisaprime ideal}. Proof Ifthe element (1 ¢ ﬂ, then the set S = {a’ln e Z+} does not intersect I. Since S is closed under multiplication, the preceding lemma insures the existence of some prime ideal P which contains 1, but not a; that is, a does not belong to the intersection of prime ideals containing I. This establishes the inclusion
n {PIP a I; P is a prime ideal} E ,[I— The reverse inclusion follows readily upon noting that if there exists a prime ideal which contains I but not a, then a 9! ﬂ, since no power of a belongs to P.
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165
As with the case of the Jacobson radical, the prime radical may be
characterized by its elements; this is brought out by a result promised earlier.
Corollary. The prime radical of a ring R coincides with the nil radical of R; that is, Rad R is simply the ideal of all nilpotent elements of R. Proof The assertion is all but obvious upon taking I '= {0} in Theorem
H. An immediate consequence of this last corollary is the potentially powerful statement: every nil ideal of R is contained in the prime radical, not simply contained in the larger Jacobson radical (Corollary 3 to Theorem 8—2). Example 85. For an illustration of Theorem 8—8, let us fall back on the
ring Z of integers. In this setting, the nontrivial prime ideals are the principal ideals (p), where p is a prime number. Given n > 1, the ideal (n) E (p) if and only if p divides n; this being so, m = ﬂ (Pi)pilll
Thus, if we assume that n has the prime power factorization it follows that
n = p’i‘p'é’ "14" (n) = (m) n (pz) 0
(1962+), n (p,) = (mp2
1),)
Let us go back to Theorem 8—8 for a moment. Another of its advantages is that it permits a rather simple characterization of semiprime ideals. (The reader is reminded that we deﬁned an ideal I to be semiprime provided
that I = ﬂ).
Theorem 89. An ideal I of the ring R is a semiprime ideal if and only if I is an intersection of prime ideals of R. Proof The proof is left to the reader; it should offer no difﬁculties. Corollary. The prime radical Rad R is a semiprime ideal which is contained in every semiprime ideal of R. Before pressing on, we should also prove the prime radical counterpart of Theorem 8—3. Theorem 810. For any ring R, the quotient ring R/Rad R is without prime radical. Proof For clarity of exposition, set I = Rad R. Suppose that a + I is any nilpotent element of R/I. Then, for some positive integer n, (a+I)"=a"+I=I,
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so that a" e I. But I consists of all nilpotent elements of R. Thus, we must
have (a")"' = 0 for suitably chosen me Z +; this is simply the statement that a e I, and, hence, a + I is the zero element of R/I.
Our argument
implies that the quotient ring R/I has no nonzero nilpotent elements, which is to say that Rad (R/I) = {0}. To round out the picture, two theorems are stated without proof; it will
be observed that these take the same form as the corresponding result established for the Jacobson radical (Theorems 8—4 and 8—5). Theorem 811. If I is an ideal of the ring R, then
1) Rad (R/I) 2
RadR + I
, and,
I 2) whenever I E Rad R, Rad (R/I) = (Rad R)/I.
Theorem 812. For any ring R, Rad R is the smallest ideal I of R such that the quotient ring R/I is without prime radical. A problem exerting a natural appeal is that of describing the prime
radical of the polynomial ring R[x] in terms of the prime radical of R. As a starting point, let us ﬁrst prove a lemma which is of interest for various parts of ring theory. Lemma.
A polynomial f(x) = a0 + alx +
+ aux" is invertiblein
R[x] if and only if a0 is invertible in R and all the other coeﬂicients a1, a2,
, a,l are nilpotent elements of R.
Proof. If a0 has an inverse in R and a1, a2, , a,I are all nilpotent, then the polynomial f(x) = a0 + alx + + aux" is the sum of an invertible element and a nilpotent element. Hence, f(x) must itself be an invertible
element of R[x] (Problem 5, Chapter 7). Going in the other direction, assume that the polynomial f(x) = a0 + alx + + anx"e R[x] possesses an inverse. That a0 is then
invertible in R should be obvious. For any prime ideal P of R, P[x] is a prime ideal of R[x] and the quotient ring R[x]/P[x] z (R/P)[x]. Thus, the homomorphic image off(x) in (R/P)[x],
(a0 + P) + (a1 + P)x +
+ (a,l + P)x'I
must have an inverse. Since R/P is an integral domain, the invertible elements in (R/P)[x] are nonzero constant polynomials. This implies that a1+P=a2+P=...¥an+P=P;
hence, the elements a1, a2,
, a, all lie in P. As this statement holds for
every prime ideal of R, it follows that al, a2, to Theorem 8—8, the elements a1, a2,
, a,I € Rad R. By the corollary
, a,I must therefore be nilpotent.
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167
Having dealt with these preliminaries we are now ready to prove
Theorem 813. For any ring R, rad R[x] = Rad R[x]. Proof: It is enough to establish the inclusion rad R[x] S Rad R[x]. If + aux"e rad R[x], Theorem 8—2 the polynomial f(x) = a0 + alx + tells us that
1+ xf(x) = l + aox +
+ anx"+1
must be invertible in R[x]. Hence, by the above lemma, the coefﬁcients , an are all nilpotent elements of R. For a suﬂiciently large power, a0, a1,
f(x) will then be nilpotent in R[x] and thus be in Rad R[x]. The assertion of Theorem 8—13 can be improved upon. For the reader will have little diﬂiculty in now convincing himself that
Rad R[x] = (Rad R)[x], where (Rad R)[x] denotes the ring of polynomials in x with coefﬁcients
from Rad R. In fact, the inclusion Rad R[x] E (Rad R)[x] is implicit in the foregoing proof; the opposite inclusion requires the corollary to Theorem 88 By virtue of the displayed equation, we have
rad F[x] = (Rad F)[x] = {0} for any ﬁeld F. That is to say, the polynomial ring F[x] constitutes a semisimple ring. Suppose for the moment that I is an ideal of the ring R with I E Rad R. Given an idempotent element e 7E 0 in R, we know that the coset e + I will be idempotent in R/I. What is not so obvious is that e + I 7E I ; this follows from the fact that Rad R contains no nonzero idempotents (Corollary to Theorem 8—2). We are mainly concerned with the converse here: If u + I is a nonzero idempotent of the quotient ring R/I, does there exist an idempotent e e R for which e + I = u + I? Before becoming involved in this discussion, let us give a general deﬁnition.
Deﬁnition 8—3. Let I be an arbitrary ideal of the ring R. We say that the idempotents of R/I can be raised or lifted into R in case every idempotent element of R/I is of the form e + I, where e is idempotent in R. Deﬁnition 8—3 means just this: the idempotents of R/I can be lifted if for each element ue R such that u2 — us I there exists some element e2 = e e R with e — u e I. Although it is surely too much to expect the lifting of idempotents to take place for every I, we shall see that this situation does occur whenever I is a nil ideal (or, equivalently, whenever I E Rad R). Let us begin with a lemma, important in itself.
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Lemma. If e and e’ are two idempotent elements of the ring R such that e — e’ e Rad R, then e = e’.
Proof Inasmuch as the product (e — e’) (1 — (e + e’)) = 0, it is enough to show that l — (e + e’) is an invertible element of R. Now, one may write 1 — (e + e’) in the form l—(e+e’)=(l—2e)+(e—e’),
where(1 — 2e)2 = l — 4e + 4e2 = 1. Theimplicationisthatl — (e + e’). being the sum of a nilpotent element and an invertible element, is necessarily
invertible in R (Problem 10, Chapter 1). Corollary. Let I, J be ideals of the ring R with I E J g Rad R. If the idempotents of R/J can be lifted into R, then so can the idempotents of R/I. Proof. Suppose that u + I is any idempotent of R/I. Since I S J, it follows that u + J is an idempotent element of R/J ; u2 — u e I E J. By assumption, there must exist some e2 = e in R such that e + J = u + J, whence
e — ueJ. But then (e+I)—(u+I)=e—u+IeJ/IS(RadR)/I=Rad(R/I).
Applying the lemma to the quotient ring R/I, we conclude that the eoset u + I = e + I and so the idempotents of R/I can be lifted. The key to showing that the idempotents of R/Rad R are liftable is the circumstance that certain quadratic equations have a solution in the prime radical of R. Theorem 814. For any ring R, the idempotents of R/Rad R can be lifted into R. Proof. Let u + Rad R be an idempotent element of R/RadR, so that u2 — u = r e Rad R. The problem is to ﬁnd an idempotent e e R with e — u e Rad R or, putting it another way, to obtain a solution a of the
equation (u + a)2 = u + a, with a e Rad R.

We ﬁrst set a = x(1 — 2a), where x is yet to be determined. Now, the requirement that the element u + a = u + x(l — 214) be idempotent in equivalent to the equation
(x2 — x)(1 + 4r) + r = 0. By the quadratic formula, this has a formal solution
x .. l 1 _ ; ‘2
./1 + 4;
= 1/2(2r — cw + (era — m).
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169
Since r is nilpotent (being a member of Rad R), the displayed series will terminate in a ﬁnite number of steps; the result is a perfectly meaningful polynomial in r with integral coefﬁcients. Thus, the desired idempotent is e = u + x(1 — 2a), where x e Rad R. Corollary. For any nil ideal I of R, the idempotents of R/I can be lifted. Proof Because I s Rad R, an appeal to the last lemma (with J = Rad R)
is legitimate. Let us deﬁne a ring R to be primary whenever the zero ideal is a primary ideal of R. This readily translates into a statement involving the elements of R: R is a primary ring if and only if every zero divisor of R is nilpotent.
Integral domains are examples of primary rings. In general, primary rings can be obtained by constructing quotient rings R/Q, where Q is a primary
ideal of R. As an application of the preceding ideas, we shall characterize such rings in terms of minimal prime ideals. (A prime ideal is said to be a minimal prime ideal if it is minimal in the set of prime ideals; in a commutative ring with identity, such ideals are necessarily proper.) The crucial step in the proof is the corollary on page 164. Theorem 815. A ring R is a primary ring if and only if R has a minimal prime ideal which contains all zero divisors. Proof. For the ﬁrst half of the proof, let R be a primary ring. Then the set of zero divisors of R, along with zero, coincides with the ideal N of nilpotent
elements, and N will be prime. Being equal to the prime radical of R, N is necessarily contained in every prime ideal of R; that is to say, N is a minimal prime ideal. The converse is less obvious; in fact, it is easiest to prove the contra
positive form of the converse. Suppose, then, that R has a minimal prime ideal P which contains all zero divisors and let a e R. be any nonnilpotent element. We deﬁne the set S by
S = {ra"r¢P;n 2 0}. S is easily seen to be closed under multiplication and 1 e S. Notice, particularly, that the zero element does not lie in S, for, otherwise, we would have
M" = 0 with a" % 0; this implies that r is a zero divisor and therefore a member of P. We now appeal to the corollary on page 164 to infer that the complement of S contains a prime ideal P’. Since P’ E R — S S P, with P being a minimal prime ideal, it follows that R — S = P. But a e S, whence a ¢ P, so that a cannot be a zero divisor of R. In other words, every
zero divisor of R is nilpotent, which completes the proof.
For the sake of reﬁnement, let us temporarily drop the assumption that all rings must have a multiplicative identity (commutativity could also be
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FIRST COURSE IN RINGS AND IDEALS
abandoned, but this seems unnecessarily elaborate for the purposes in mind). To make things more speciﬁc, we pose the problem of constructing a radical which will agree with the Jacobson radical when an identity element is available. Of course, it is always possible to imbed a ring in a ring with identity, but the imbedding is often unnatural and distorts essential features of the given ring. The direct approach of considering the intersection of all maximal ideal: is not very eﬁ‘ective, because one no longer knows that such ideals exist (Theorem 5—2, our basic existence theorem for maximal ideals, clearly requires the presence of an identity). A more useful clue is provided by Theorem 8—2, which asserts that an element a e rad R if and only if 1 — m is invertible for every choice of r in R, or, to put it somewhat diﬂ'erently, the principal ideal (I — ra) = R for all r e R. This latter condition can be
written as {x — raxlx e R} = R for each r in R, and is meaningful in the absence of a multiplicative identity. It thus would appear that Theorem 8—2 constitutes a hopeful starting point to the solution of the problem before  us. Needless to say, it will be necessary to introduce concepts capable of replacing the notions of an invertible element and maximal ideal which were so essential to our earlier work. One begins by associating with each element a e R the set
1,, = {ax — xxeR}. A moment’s thought shows I, to be an ideal of R. Now, it may very well happen that I, = R; in this event, we shall say that a is a quasiregular element. There is another way of looking at quasiregularity:
Theorem 8—16. An element a of the ring R is quasiregular if and only if there exists some b e R such that
a + b — ab = 0. The element b satisfying this equation is called a quasiinverse of a. Proof Suppose that a iS quasiregular, so that I, = R. Since as 1,, we must have a = ab — b for suitable b in R, whence a + b — ab = 0. On the other hand, if there exists some element b e R satisfying a + b — ab = 0, then as 1,. Thus, for any 1' in R, are 1,. By virtue ofthe deﬁnition of 1,, we also have ar — r e 1,, which implies that
r = ar — (ar — r)eI,. This means R E In, or rather R = I“, and a is quasiregular. Corollary. An element a GR is quasiregular if and only if a e I“.  Here are some consequences: Every nilpotent element of R is quasiregular. Indeed, if a" = 0, a straightforward calculation will establish that
CERTAIN RADICALS OF A RING
171
= —a  a2 — — a"‘1 is a quasiinverse of a. Notice also that zero is the only idempotent which is quasiregular. For, if a2 = a and a + b — ab = 0 for some b in R, then we have
a=a2+ab—'ab=a(a+b—ab)=0. One of the most useful tools in handling the concept of quasiregularity is the socalled “circle operation” of Perlis [54]. Given a, b e R, we deﬁne a o b by a o b = a + b — ab.
With this notation, Theorem 8—16 may be rephrased so as'to assert that an element a e R is quasiregular if and only if there exists some second element beR for which do b = 0. It is a simple matter to verify that the pair (R, o) is a semigroup with identity element 0; in particular, one infers from this that quasiinverses are unique, whenever they exist. An even stronger result is that the quasiregular elements of R form a group with respect to the circle operation. Lastly, let us call attention to the fact that if R possesses a multiplicative identity 1, then (1—a)(l—b)=l—aob. Accordingly, a is quasiregular if and only if 1 — a is an invertible element
of R. (Tying this idea more closely to Theorem 8—2, we see that the product ra is quasiregular for every r e R if and only if 1 — ra is invertible for all r in R.) Example 86. Consider the assertion: if every element of a commutative ring R is quasiregular, with exactly one exception, then R must be a ﬁeld. To see this, let us take the element e to be the one exception; certam
e =/= 0, since 0 is quasiregular. Now, e2 o a = e o (—eo a) 7E 0 for each aeR, from which we infer that e2 = e.
Observe also that if e o a 79 e, then there would exist some
element b e R such that (e o a) o b = 0. Associating, we obtain e o (a o b) = 0 and so e is a quasiregular element, a contradiction. Accordingly, we must
have e o a = e for every choice of a e R or, upon expanding, ae = a for all in R; this implies that e acts as a multiplicative identity for R. Finally, given an element x 7E 0, we write x = e — a, with aeR. Then, since a aé e,
x(e—b)=(e—a)(e—b)=e—aob=e for suitable b e R. In other words, every nonzero element of R is invertible,
conﬁrming R to be a ﬁeld. As heralded by our earlier remarks, we now deﬁne the Jradical (for Jacobson, naturally enough) of a ring R to consist of those elements a for
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which 1,, = R for every r e R; recasting this in terms of the notion of qualiregularity: Deﬁnition 84. The Jradical J(R) of a ring R, with or without an identity, is the set
J(R) = {a e Rar is quasiregular for all r e R}. If J(R) = {0}, then R is said to be a Jsemisimple ring. To reinforce these ideas, let us consider several examples.
Example 87. The ring Z, of even integers is JSemisimple. For, suppou that the integer n e J(Ze ). Then, in particular, n2 is quasiregular. But Ill equation
71’ + x — n2x = 0,
or, equivalently, n2 = (n2 — 1)x, has no solution among the even integer: unless n = 0. This implies that J(Ze) = {0}. Example 88. Any commutative regular ring R is Jsemisimple. Indeed, given a e J(R), there is some a’ in R such that aza' = a. Now, aa’ must he quasiregular, so we can ﬁnd an element x e R satisfying aa’+x—aa’x=0.
Multiplying this equation by a and using the fact that aza’ = 0, we deduc
that a = 0, whence J(R) = {0}. Example 8—9. Consider the ring F[[x]] of formal power series over the
ﬁeld F. As we know, F[[x]] is a commutative ring with identity in which an element f(x) = Z akx" is invertible if and only if a0 7E 0. If f(x) belongs to the principal ideal (x), thenf(x) has zero constant term ; hence, (I — f(x))" exists in F[[x]] Taking f(x)’ = (1 — f(x))' 1, we see that f(x) o f(x)’ = 0.
Thus, every member of(x) is quasiregular, which implies that (x) E J(F[[xﬂ). On the other hand, any element not in (x) is invertible and therefore cannot be in the Jradical. (In general, if a e J(R) has a multiplicative inverse, then 1 = aa‘1 is quasiregular; but zero is the only quasiregular idempotent,
so that 1 = 0, a contradiction.) The implication is that J(F[[x]]) E (x) and equality follows:
J(F[[x]]) = (x) = rad F[[x]]. Turning once again to generalities, let us Show that any element a e J(R) is itself quasiregular. Since ar is quasiregular for each choice of r in R, it follows that a2 in particular will be quasiregular. Therefore, we can obtain an element b e R for which a2 o b = 0. But a Simple computation shows that
ao((—a)ob)= (ao(—a))ob= azob =0
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173
One ﬁnds in this way that the element a is quasiregular with quasiinverse (a) o b.
It is by no means apparent from Deﬁnition 8—3 that J(R) forms an ideal of R; our next concern is to establish that this is actually the case.
Theorem 817. For any ring R, the Jradical J(R) is an ideal of R. Proaﬁ Suppose that the element a e J(R), so that for any choice of x e R, ax is quasiregular. If re R, then certainly (ar)y = a(ry) must be quasiregular for all y in R, and therefore ar e J(R). It remains to show that whenever a, b e J(R), then the difference a — b lies in J(R). Given x, u, v e R, a fairly routine calculation establishes the identity (a — b)xo(uov) = a(x — ux)ov + (—bxou) — (—bxou)v. Taking stock of the fact that a, b belong to J(R), we can select an element u such that —(bx) o u = b(—x) o u = 0 and a second element 1: for which dx — ax) o v = 0; in consequence, (a — b)x o (u o v) = 0. This being the case, (a — b)x is quasiregular for every x e R, whence a — b e J(R). Thus, the Jradical satisﬁes the deﬁning conditions for an ideal of R. As one would expect, there are many theorems concerning the Jradical which are completely analagous to theorems stated in terms of the Jacobson radical.
Although it would be tedious to prove all of these results, the
following deserves to be carried through. Theorem 818. For any ring R, the quotient ring R/J(R) is Jsemisimple.
Prooﬁ Take a + J(R) to be an arbitrary element of J(R/J(R)). Then (a + J(R))(x + J(R)) = ax + J(R) is quasiregular for each xeR. Accordingly, there exists some coset y + J(R) in R/J(R), depending on both a and x, for which (ax + J(R))o (y + J(R)) = J(R). But this implies that the element ax o y lies in J(R), and, hence, is quasiregular as a member of R; say (ax o y) o z = 0, where z e R. It follows from the associativity of e that ax is itself quasiregular in R, with quasiinverse y o 2. Since this holds for every x e R, the element a belongs to J(R), and we have a + J(R) = J(R), the zero of the quotient ring R/J(R). It turns out that the class of ideals which must replace the maximal ideals are precisely those ideals whose quotient rings possess a multiplicative identity. Deﬁnition 85. An ideal I of the ring R is called modular (or regular, in the older terminology) if and only if there exists an element e e R such
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174
that ae — a e I for every a in R. Such an element e is said to be an identity for R relative to I, or modulo I.
In passing, we Should remark that whenever R has an identity element 1, then 1 can be taken as the element e of Deﬁnition 8—5, and all ideals of
R are modular. Notice, too, that if e is an identity for R relative to I, then the same is true for the elements e + i, where is I, and e" (n e Z).
By a modular maximal ideal, we shall mean a maximal (hence, proper) ideal which is also modular. Paralleling the proof of Theorem 5—5, it can be Shown without too much diﬁiculty that a proper ideal M of R is a modular maximal ideal if and only if the quotient ring R/M forms a ﬁeld. The existence of suitably many modular maximal ideals is assured by the following:
Theorem 819. Each proper modular ideal of the ring R is contained in a modular maximal ideal of R. Proof. Let I be a proper modular ideal of R and e be an identity element for R relative to I. We consider the family .52! of all proper ideals of R which contain 1; because I itself is such an ideal, .2! is certainly nonempty. It is important to observe that the element e lies outside each ideal I of .11. Indeed, if e did belong to J, we would then have ae e J for all a in R. By virtue of the fact that I is modular, ae — a e I E J, from which it follows
that
a = ae — (ae — a)eJ.
One ﬁnds in this way that J = R, a ﬂat contradiction, inasmuch as J is a
proper ideal of R by deﬁnition of .21. Now, let {Ji} be any chain of ideals from .31. Then the settheoretic union L) Ji forms an ideal 'of R containing 1. Since e¢ uJ,, this ideal is proper, whence L) J, e .51. Thus, Zom’s Lemma asserts the existence of a maximal ideal M of R with I g M. Any such ideal will be modular, because
ae — aeI g MforeachelementaeR. This theorem has a number of important consequences (which we list as corollaries) having to do with quasiregularity. Corollary 1. If the element a e R is not quasiregular, then there exists a modular maximal ideal M of R such that a ¢ M. Proof Since a is not quasiregular, I, = {ra — rr e R} forms a proper ideal of R. Moreover, Ia is modular, with the element a as an identity for . R modulo I4. Knowing this, it follows from the theorem that there exists a
modular maximal ideal M of R containing I, and excluding a.
Corollary 2. If J(R) + R, then the ring R contains modular maximal
ideals; in fact, for any a ¢ J(R), there exists a modular maximal ideal M with a ¢ M.
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175
Proof If the element a 9! J(R), then there is some x e R such that ax is not quasiregular. Corollary 1 asserts the existence of a modular maximal ideal of R which excludes ax and, in consequence, does not contain (1. Corollary 3. An element a e R is quasiregular if and only if, for each modular maximal ideal M of R, there exists an element b such that a o b e M.
Proof The indicated condition is clearly necessary, for it sufﬁces to take the quasiinverse of a as the element b. Suppose now that the condition is satisﬁed, but that a does not possess a quasiinverse. Then the modular
ideal I, = {or — rr e R} will be contained in some modular maximal ideal M of R. By assumption, we can ﬁnd an element b in R for which a o b e M. But ab — b e In, whence a = aob + (ab — b)eM. It follows that are M for arbitrary r in R, and, consequently, that
r = or — (ar — r) e M. Therefore, M = R, which is impossible. In the presence of an identity element, the Jacobson radical rad R is
the intersection of all the maximal ideals of a ring R. One would rightly suspect that there is a similar characterization of the Jradical in terms of modular maximal ideals (the sole difference being that, in the present
setting, we must impose the demand that J(R) does not exhaust the ring R). Theorem 820. If R is a ring such that J(R) 7E R, then
J(R) = n {MM is a modular maximal ideal of R}. Proof As so often happens, one inclusion will be quite straightforward and easy, and the other will be deeper and more complicated. In the ﬁrst place, suppose that the element a lies in every modular maximal ideal of R, but that a ¢ J(R). Using Corollary 2, we could then ﬁnd a modular maximal ideal M for which a é M, a contradiction; consequently, a e J(R). Going in the other direction, take a in J(R). We wish to show that a e M, where M is any modular maximal ideal of R. Assume for the moment that a at M. Owing to the fact that M is maximal, the ideal generated by M and a must be the whole ring R; therefore (in the absence of an identity), R = {i + ra + naieM,reR,neZ}. Now, let e be an identity element for R modulo M. Then there exist suitable is M, r e R and an integer n for which e=i+ra+na.
As J(R) forms an ideal of R, the sum ra + na 6 J(R), so that e — ie J(R).
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FIRST COURSE IN RINGS AND IDEALS
This implies that e — i is quasiregular, say with quasiinverse x. From ill equation (e — i) o x = 0, together with the modularity of M, we obtain
e=i—ix+(xe—x)eM. In consequence, M = R, and an obvious contradiction ensues. Thus, a e M, from which one concludes that J(R) is contained in the intersection of the modular maximal ideals of R, completing the proof.
The hypothesis that J(R) 7E R is certainly fulﬁlled whenever the ring R possesses a multiplicative identity 1. Speciﬁcally, the element 1 itself in not quasiregular, whence 1 ¢ J(R); in fact, if 1 + b — 1b = 0 for some b in R, we would have 1 = 0, a contradiction. When an identity element is available, all ideals of R are automatically modular. In this situation, the
Jradical will coincide with the Jacobson radical of R: J(R) = rad R. If J(R) = R, then the ring R may contain maximal ideals, but no such ideal can be modular. Indeed, suppose that I is any modular ideal of R, R, with e acting as an identity for R modulo I. By supposition, the element e e J(R), so that e has a quasiinverse e’. The modularity of I then yields e = e’e — e’ 61, which implies that I = R. Accordingly, the ring R possesses no proper modular ideals and, in particular, no modular maximal ideals. However, the possibility of the existence of maximal ideals in R is not excluded. The following theorem provides a convenient result with which to claw this chapter. Theorem 8.21. A ring R can be imbedded in a ring R’ with identity such that J(R) = rad R’. Proof. If R already has an identity, we simply take R’ = R. Otherwise, we imbed R in the ring R’ = R x Z in the standard way (see Theorem 2—12 for details). Then R, or more precisely, its isomorphic image R x {0}, is an ideal of R’ and R’/R 2 Z; thus, R’/R is semisimple. This being so, it follows from Theorem 8—5 that rad R’ g R. Since R is an ideal of R’, we also have J(R) = J(R’) n R = rad R’ n R (Problem 26). But rad R’ E R. which implies that J(R) = rad R’.
PROBLEMS Unless indicated to the contrary, all rings areassumed to becommutativewith identity.
1. Describe the Jacobson radical of the ring Z, of integers modulo n. [Hint: Considu the prime factorization of n.] In particular, show that Z,I is semisimple if and only if n is a squarefree integer.
2. Prove that F[x], the ring of polynomials in x over a ﬁeld F, is semisimple.
PROBLEMS
177
3. Prove that rad R is the largest (in the settheoretic sense) ideal I of R such that l + a is invertible for all a e I.
4. a) Let the ring R have the property that all zero divisors lie in rad R. If (a) = (b), show that the elements a and b must be associates. ' b) Verify that if the element a e rad R and ax = x for some x e R, then x = 0.
5. Provethatapowerseriesf(x) = a0 + alx +
+4v1,,x'I +
belongstoradR[[x]]
if and only if its constant term a0 belongs to radR.
6. Prove the following assertions concerning semisimple rings: a) A ring R is semisimple if and only if a 7E 0 implies that there exists some element r e R for which 1 — ra is not invertible. b) Every commutative regular ring is semisimple. c) Suppose that {1,} is a family of ideals of R such that R/Ii is semisimple for each i and n Ii = {0}. Then R itselfis a semisimple ring. [Hint: Theorem 8—5.]
e R, is the direct sum of a ﬁnite number of rings R, 7. If R = R1 6 R2 Q (i = 1,2, ...,n), prove that radR=radRl EBradRze
aradRu.
.. Establish that the conditions below are equivalent: a) the ring R has exactly one maximal ideal (that is, R is a local ring); b) radR is a maximal ideal ofR;
c) the set of noninvertible elements of R coincides with rad R; d) the set of noninvertible elements of R form an ideal; e) the sum of two noninvertible elements of R is again noninvertible; f) for each element r e R, either r or 1 — r is invertible.
9. Let R be a principal ideal domain. If the element a e R has the prime factorization
a = p’j‘p’é2 (plpz
pf', prove that the Jacobson radical of the quotient ring R/(a) is
p,)/(a). [Hint: Themaximalideals ofR containing (a)are (pl), (172),
, (p,).]
10. Letf be a homomorphism from the ring R onto the ring R’. Show thatf(rad R) S rad R’ and, whenever kerf E rad R, then rad R = f‘1(rad R’); do the same for
Rad R. 11. Prove: An ideal I of R is semiprime if and only if a2 e I implies that a e I.
12. Establish that an ideal I of R contains a prime ideal if and only if for each n, ala, a,l = 0 implies that a,‘ e I for some k. [Hint: The set S = {b1b2 "' bulbk¢I; n 21}
is closed under multiplication and 0 ¢ 8.] 13. Show that the prime radical of a ring R contains the sum of all nilpotent ideals of R. 14. Establish the equivalence of the statements below: a) {0} is the only nilpotent ideal of R;
b) R is without prime radical; that is, Rad R = {0}; c) for any ideals I and J of R, U = {0} implies that I n J = {0}.
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FIRST COURSE IN RINGS AND IDEALS
15. A multiplicatively closed subset S of the ring R is said to be saturated if dies implies that both a e S and b e S. Prove that a) Sis a saturated multiplicatively closed subset of R if and only if its complemt
R— S is a union of prime ideals; b) the set of nonzerodivisors of R is a saturated multiplicatively closed subs! (hence, the set of zero divisors of R, along with zero, is a union of prime ideals)
16. a) Prove that Rad R is the maximal nil ideal of R (maximal among the set of nil ideals); this property is often taken as the deﬁnition of the prime radical of R. b) If R has no nonzero nil ideals, deduce that the polynomial ring R[x] is semisimple. c) Let charR = n > 0. Prove that if R is without prime radical, then n is a squarefree integer. [Hintz Assume that n = pzq for some prime p; then there exists an element a e R such that pqa 99 0, but (pqa)2 = 0.] 17. Prove that the following statements are equivalent: a) R has a unique proper prime ideal; b) R is a local ring with rad R = Rad R; 0) every noninvertible element of R is nilpotent;
d) R is a primary ring and every noninvertible element of R is either a zero divisor or zero. 18. Supply a proof of Theorems 8—11 and 8—12. 19. A ring R is termed a Hilbert ring if each proper prime ideal of R is an intersection of maximal ideals. Prove that: a) R is a Hilbert ring if and only if for every proper ideal I of R, rad (R/I) = Rad (R/I). b) Any homomorphic image of a Hilbert ring is again a Hilbert ring. c) If the polynomial ring R[x] is a Hilbert ring, then R is one also. [Hint: Utilin (b) and the fact that R[x]/(x) : R . If I is an ideal of the ring R, prove each of the following statements: a) The nil radical of I is the intersection of all the minimal prime ideals of I. b) Rad R is the intersection of all the minimal prime ideals of R.
c) The union of all the minimal prime ideals of R is the set
S = {reeaeRadR,for somea¢RadR}. d) If I is a primary ideal of R, then ﬂ is the only minimal prime ideal of I.
[Hint. Problem 19, Chapter 5.] In Problems 21—30, the ring R need not possess an identity element. 21. Consider the ring P(X) of subsets of some (nonempty) set X. Show that in this setting the circle operation reduces to the union operation and determine the quasiregular elements. 22. a) Prove that if the element a e R has the property that a“ is quasiregular for some n 6 2+, then a itselft be quasiregular. [Hint. a" = a o( —2;; id]
PROBLEMS
179
b) Verify that the annihilator of a Jsemisimple ring R is zero; in other words,
annR = {0}.
'
n. Iff is a homomorphism from the ring R onto the ring R’, establish the inclusion f(J(R)) E J(R’); also show that if kerf S J(R), then J(R) = f "(J(R’)). 24. Prove each of the statements below: a) J(R) contains every nil ideal of the ring R. b) J(R) is a semiprime ideal of R. [Hint: Theorem 5—11.] c) For any ring R, Rad R E J(R). 25. If we deﬁne R = {Zn/(2m + l)n, m e Z}, then R forms a commutative ring under ordinary addition and multiplication. Show that J(R)‘ = R, while Rad R = {0}. 2n
—2n
H' t.o — = 0. [m 2; thus, p is an odd prime. We take for F the ﬁeld Zp. Then, from above, every nonzero
element of Zp is a root of the polynomial x"—1 — 1 e 2,, [x]:
x"'1 — 1 E (x — l)(x — 2)
(x — (p — 1)) (mod p).
Putting x E 0 (mod p) in the justwritten equation, it follows that
—1 a (—1)(—2) (—(p — 1))=(—1)P—1(p — 1)! (modp). As p — 1 is even, this leads directly to Wilson’s Theorem. We now settle the question of the existence of ﬁnite ﬁelds.
Theorem 98. For any prime number p and positive integer n, there exists a ﬁeld (unique up to isomorphism) with exactly p" elements.
TWO CLASSIC THEOREMS
189
Proof Consider the splitting ﬁeld F’ of the polynomial f(x) = x"" — x in
Zp[x].
Since F’ contains 2?, it has ﬁnite characteristic p.
Now, the
derivative off(x) is 6f(x) = p"x""'l — l = —1.
By virtue of Problem 15, Chapter 7, this means that the polynomial has no repeated roots in F’. Let the subset F E F’ consist of the p” distinct roots off(x) in F’:
F = {aeF’Ia"" = a}. Clearly, the elements 0 and 1 lie in F. If a, b e F, with b 3E 0, then we have
(a — b)"" = a’" — b""= a —‘ b,
(ab1r" = aver)" = ab" (again using the fact char F’ = p), so that both a — b, ab‘1 6 F.
Con
sequently, the set F constitutes a subﬁeld of F’ and, thus, a ﬁeld with p"
elements. From the corollary of the last theorem, we infer that F = F’ and F’ is the desired ﬁeld. Uniqueness follows from the result that any two ﬁelds having p" elements are isomorphic (Corollary 2). From this, it is a short step to
Corollary. For any ﬁnite ﬁeld F and positive integer n, there exists an extension ﬁeld of F of degree n.
Proof Suppose that F has (1 = p’" elements, where p is the characteristic of F and m = [F : Z1,]. By the theorem just proved, there exists a ﬁeld F’
with p’“ = q" elements, namely, the splitting ﬁeld of x"""' — x over ZF.
We contend that F’ is actually an extension of F. Indeed, every element of F is a root of the polynomial xl’" — x; this fact, together with the relation
p" = "‘p'"("' 1), implies that for any a e F ' apm" = grub—1): aP'"("_2) = ... = apm = a.
Thus, each element of F is also a root of the polynomial x"""' — x, which is to say that F g F’. Finally, observe that
mn = [F’zZp] = [F’zF][F:Zp] = [F’zF]m, whence [F’: F] = n, completing the proof. We next examine the multiplicative structure of a ﬁnite ﬁeld. The ﬁniteness assumption leads to a particularly simple description: the nonzero elements comprise a cyclic group under multiplication. Theorem 99. The multiplicative group of a ﬁnite ﬁeld is cyclic.
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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:
h = are? 4;", where the q, are distinct primes and ri 6 2+.
For i = 1, 2,
, m,’ net
hi = h/qi. Now, there exists a nonzero element a, e F which is not a root
of the polynomial x"i — 1 e F[x] ; for this polynomial has at most hi distinct roots in F and h, < h, the number of nonzero elements of F. Next, take where
bi = ali'lsi,
and deﬁne b = blb2
Si = q?
(i = l, 2, ... m)
bm. We certainly have b? = a]: = 1’
so that the order bi must divide S, = qf‘. On the other hand, if
b?” = 1,
then
a? = alil/qi = bin1: 1,
contrary to our original choice of the element a,. The implication is that b, has order qlf‘. To settle the whole affair, we will show that the element b is of order It.
In the contrary case, the order of b must be a proper divisor of h (since b’. = l, the order of b certainly divides h) and therefore divides at least one
of the integers hi (i = l, 2,
, m), say h. We then have
1: bill = biilbgl “'bﬂ.
If2 S i s m, then qﬂhl, which implies that b?‘ = 1 and so b’;‘ = 1. This means that q',1 (the order of b1) divides h, 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 of F*. 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 Zp. 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 of F can be written as 0, 1, b, b2, , b”'"2 for some b e F*; in other words, the ﬁeld F = Zp (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. Theorem 910. Let F be a ﬁnite ﬁeld. For each positive integer 71, there exists an irreducible polynomial f(x) e F[x] with deg f(x) = n. Proof Suppose that F’ is an extension of F 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’zF] = 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 21,. 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
Z, [x]/(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 (Wedderbum’s Theorem).
Our
approach is founded on a treatment by Herstein [43]. Although. this is perhaps the most elementary, other proofs of Wedderbum’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 Wedderbum’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 a?" = a for some m > 0. Then there exists an x e R for which 1) xax'1 7E a, 2) xax'1 e ZF(a), the extension ﬁeld obtained by adjoining a to Zp.
Proof. Let ZI, be the prime subﬁeld of R. Since a?" — 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’ﬂ elements for some n 6 2+. Furthermore,
each r e Zp (a) satisﬁes r’" = r.
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FIRST COURSE IN RINGS AND IDEALS
Now, deﬁne the functions —> R by setting f(x) = xa — ax for all x in R. Using induction, it is not diﬂicult to Show that the composite k
Z (1)‘(':‘)aixa"'i f"(X) = i=0
(k 2 1)
When k = p, the foregoing equation reduces simply to f”(x) = xa" — a’x, because p(‘,’) for 0 < i < p (recall also that char R = p). Another routine induction argument extends this to
f""(x) = xa’" — a’"x. But a’" = a, whence f""(x) = xa — ax = f(x) for all xeR, which is equivalent to asserting thatfP" = f. For each element r e Z"(a), consider the function T, on R deﬁned by 7;(x) = rx. Our contention is that f commutes with all such 7;. The reasoning proceeds as follows: Being a ﬁeld, Zp(a) is commutative, so that, if x e R,
(f ° 7;)(x) = f(H) = (DC)a  a(r36) = W!  fax = r(xa — ax) = (71°f)(x)This, in short, means that f o T; = T, o ffor every r in Zp(a). From the corollary on page 188, the polynomial yp" — yeZ,[y] factors completely in Zp(a); in other words, we have yp"_y=
H (y_r)’ reZ,(a)
or, what amounts to the same thing,
H (yr)y""—y=y Ofrezpul) This formal identity requires only that y commute with all elements r e ZI,(a). Taking stock of the fact that fo ’1; = T, o]: as well as the relation fP" = f, we thereby obtain ‘
0=f” —f=f° n (f— 7;). Oahrezpos)
(In essence, one applies the substitution homomorphism 45, to the ring of polynomials whose coeﬂicients are homomorphisms on (R, +).) If, for every r + 0 in Zp(a), it happens that (f — T,)(x) = 0 implies x = 0, then the lastwritten 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 + reZp(a) and some element x aé 0 in R for which (f — 1;)(x) = 0; that is to say, xa — ax = rx and so
xax‘1 = r + ap(a). Since r 7E 0, certainly the product xax' 1 + a.
TWO CLASSIC THEQREMS
193
Corollary. In the lemma, xax‘ 1 = a" 7E a for some integer k 6 2+.
Proof. Since a!” 1 = 1, the element a has ﬁnite order as a member of the multiplicative group R*. Let s be the order of a. Then, in the ﬁeld Z”(a), each of the s elements 1, a, a2,
y e Zp[x].
1, a,
, a"1 is a root of the polynomial ys —
This polynomial can possess at most s roots in Zp(a) and
, as‘ 1 are all distinct. But xax‘ 1 e Zp(a) and clearly (xax'1)’ = xa‘x'1 = xx‘1 = 1.
In consequence, xax'1 = a" for some k, with2 S k S s — 1.
To cope with the problem at hand, we shall also need the following: Lemma 2. If F is a ﬁnite ﬁeld and 0 + a e F, then there exist elements
a,beuch thata = a2 + b2. Proof We ﬁrst dispense with the case where char F = 2. In this special situation, F has 2" elements and any element of F satisﬁes the equation x2" = x. Thus, every nonzero member a of F is a square and, in particular,
a = a2" = (arm)? The lemma is thereby established on taking a = on?"1 and b = 0. Now, if the characteristic of F is an odd prime p, then F will contain
p" elements. Let f be the mapping of F“ into itself deﬁned by f(x) = x2 (as usual, F* denotes the multiplicative group of F). Then f is a group homomorphism, with
kerf= {xeF“‘x2 = 1} = {1, —1}. Since char F 7E 2, 1 and — 1 are necessarily distinct. This implies that, for
each [I ef(F*), there exist exactly two elements a1, a2 in F* with a} = ocﬁ = ,8 ; in fact, a2 = —a1. To put it another way, for each pair of elements a, and —al in F*, we get one element which is a square. Hence, half the elements
of F* will be squares, call these ,61, [12, , ,8“, where the integer k = (p'. — 1)/2. Given 0 7E a e F, assume that a is not a square and consider the set S = {a — ﬁili = l,2,...,k}. If it turns out that a — B, is not a square for any value of i, the set S (which contains k distinct elements) must coincide with the k nonsquares of F*. But then a will lie in S, yielding a = a — ﬂ, for some choice of i; whence ﬂ, = 0, an obvious contradiction. This being the case, we conclude that
a — ,6, = ﬂj for suitable integers i and j, or a = ,8, + ﬁj. Thus, a is the sum of two squares in R and the requisite equation holds.
Corollary. If F is a ﬁnite ﬁeld and 0 7E a e F, then there exist elements
a,binFsuchthatl + a2 — ocb2 = 0.
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FIRST COURSE IN RINGS AND IDEALS
After this preparation, we now undertake the task ofproving the theorem which serves as the focal point of the present chapter.
Theorem 911. (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 ab" = b"a, but ab # ba, then b" 6 cent R. For, consider the centralizer of b" in R:
C(b") = {xeRbk = bkx}. It follows without difﬁculty that C(b") comprises a division ring (a division
subring of R). If C(b") 7E R, then by our hypothesis C(b") would necessarily be commutative. But a, b both lie in C(b") and these elements clearly do not commute. This entails that C(b") = R, which is scarcely more than a restatement that b" 6 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 = {meZ+for somecécentR, cmecent R} is not empty. Pick the integer n to be minimal in S.
Then there exists an
element a ¢ cent R such that a” 6 cent R. We assert that n is a prime number. Indeed,weren = n1n2,with1 < n1, n2 < mitwouldfollowthata’" ¢cent R,
yet (a'”)"2 = a" 6 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 R and an integer k such
that xax‘1 = " 7E a. At the outset, observe that xzax'2 = Jc(xax'1)x‘1 = xa"x‘1 = (xax‘1)" = a”, so, by induction, x”— lax(P 1) = aw 1. Since we know that n is prime, the Little Fermat Theorem (Problem 10, Chapter 4) tells us that there exists an integer u satisfying k" 1 = 1 + an. Therefore, a""'I = a”'"' = M” = ra,
where r = (a")" 6 cent R. Setting b = x”'1, one gets bab‘1 = ra. Now, the element x ¢ cent R, as xax'l 9E a, so that b cannot belong to cent R by the minimal nature of n. From the observation at the beginning of the theorem, we thus conclude that ba =,£ ab. The implication of all this is that r 7E 1. On the other hand, Since r and a” both lie in the center of R,
r"a" = (ra)" = (bab'1)" = ba"b'1 = a",
TWO CLASSIC THEOREMS
195
whence 1" = 1. Because n is a prime; the order of r (as a member of R“) must be n. Finally, it is worth noticing that
b’l = f'b" = (rb)" = (a’lba)" = a‘lb"a, from which we derive ab" = b"a. Again in the light of our opening remarks, since a commutes with b'' but not with b, necessarily b" 6 cent R.
We now'assert that whenever an element y of R satisﬁes y’I = 1, then it must be of the form y = r‘, where 0 s i S n — 1. Indeed, the extension ﬁeld (cent R) (y) (although awkward, the notation conveys the point) contains at most n roots of the polynomial z" — 1. But, since r is of prime order n, the elements 1, r, r2, , r"‘1 comprise n distinct roots of z” — 1 in this ﬁeld. These remarks should make it plain that y = ri for some i. In passing, we might also observe that, because y 6 cent R, (cent R) (y) = cent R. With reference to Theorem 9—9, inasmuch as cent R constitutes a ﬁnite
ﬁeld, its multiplicative group of nonzero elements must be cyclic; say with generator s. Accordingly, a” = s], b" = s' for suitable integers j and 1. Furthermore, n divides neither j nor 1. To see this, suppose that j = nk; then, a" = sj = 5"", whence a"(s"‘)'I = 1. As the element s lies in cent R, we would have (as—k)" = 1. But the preceding paragraph then yields as"‘ = r' for some integer i, or a = ris“ 6 cent R, which is impossible. In a similar fashion, one is able to establish that n does not divide I. We now
set c = a', d = b}. Then, c”=a"'=sﬂ=b"j=d".
This relation, in conjunction with ba = rab, leads to cd = tdc, where the
element t = f” 6 cent R. A fact which will not detainpus long is that t 7E 1. In the contrary case, r'jl = 1, which implies that nlil; since n is a prime number, either nli or nll, resulting in a contradiction. One can deduce a little more, namely, that
t" = (rﬁ)n = (r")‘j' = 1. Pausing for a moment to tidy up, let us point out that the reasoning so far has succeeded in producing two elements c, d e R with the following properties: 1) c'I = d" = aecentR, 2) cd = tdc, with te cent R,
3) tsé 1,butt’I = 1.
From these relations, we may easily compute (c‘ld)". In this connection, notice that
(c‘ld)2 = c'1(dc‘1)d = c'1(tc'1d)d = tc'zdz,
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FIRST COURSE IN RINGS AND IDEALS
while a similar approach leads to (c‘ld)3 = t1+2c'3d3. A straightforward induction argument, which we bequeath to the reader, extends this to (Cld)n = t1+2+~~+(n1)c—udn = tl+2++(nl) = tn(n1)/2_
The ﬁnal stage of the proof achieves the longsought contradiction. We consider two cases in turn: n is an odd prime and n = 2. If n > 2, then (n — l)/2 is an integer and so tu(n1)/2 = (tn)(u1)/2 ___ l,
which implies that (c'ld)’I = 1. Being a solution of the equation y" = 1,
it follows from earlier reasoning that c’ld = r" 6 cent R for some choice of i. But then d ‘ 1c = (c' ld)‘ 1 6 cent R and so (using (2) above), t = c"1tc = (dc—1d'1)c = d(d'1c)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, t2 = land, of course, 1: 7E 1, whencet = —1. Then,
cd = —dc 79 dc; consequently, the characteristic of R is different from 2. Applying Lemma 2 to the ﬁeld cent R, we can ﬁnd elements x, (i = l, 2) in cent R satisfying
l+xf—ax§=0
(a=c2=d2).
Armed with this, a direct computation shows that
(c + dx1 + cdx2)2 = c2(l + xi — axg) = 0, which, because R is a division ring, leads to c + dx1 + cdx2 = 0. To clinch matters, since char R at: 2,
0 7E 2c2 = c(c + dxl + cdxz) + (c + dx1 + cdx2)c = 0, an absurdity. This contradiction ﬁnally completes the proofofWedderburn’s Theorem. We next proceed to take up a class of rings introduced by Jacobson. Deﬁnition 91. A ring R with identity is called a Jring if, for each xeR, there exists an integer n(x) > 1 (depending on x) such that
x"(’" = x. Our immediate goal is to prove that every Jring 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 'I‘HEOREMS
197
Theorem 9—12. Let R be a Jring. 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 7E 2. Consider any elementa 7E Oin R. By hypothesis,
there exist integers h, k > 1 for which a” = a, (2a)" = 2a.
Setting
q = (h — 1)(k — 1) + 1 > 1, it follows that both a“ = a and (2a)“ = 2a. From this, we obtain (2‘ — 2)a = 0, with 24 — 2 + 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 Zp be the prime subﬁeld of R.
Since a’I = a, the element a is
algebraic over Z1, and, hence, the extension Zp(a) constitutes a ﬁnite ﬁeld; say with p" elements. In particular, a itself lies in Zp(a), so that a?" = a. If we now assume that a ¢ cent R, then all the hypothesis of Lemma 1 will be satisﬁed; thus, there exists an element b e R and integer k > 1 satisfying bab'1 = a" 7E a. Similar reasoning applied to the extension ﬁeld Zp(b) indicates that W" = b for some integer m > 1. At this point we turn our attention to the set of ﬁnite sums
“$2”:20 ’ijab'lruezr} It should be apparent that W is a ﬁnite set which is closed under addition. Since the relation db = ba allows us to bring the a’s and b’s together in a product, Wis 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 = a" 75 a. Having arrived at a suitable contradiction, we infer that the choice of a at 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 Jring. Then every right ideal I of R is a twosided ideal of R. Proof. To begin with, we assert that R can possess no nonzero nilpotent elements Indeed, if x 7E 0, the condition 36""): x necessarily implies that x"I 7E 0 for all m > 1. Now, suppose that e is any idempotent element of R; then, for any x e R,
(xe — exe)2 = (ex — exe)’ =
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FIRST COURSE IN RINGS AND IDEALS
so that xe — exe = 0 = ex — exe.
Therefore, ex = exe = xe, in con
sequence of which e 5 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"'I is an idempotent element of R: (an—1)2 = “Zn2 = “nan2 = aanZ ____ all—1.
Hence, a"'1 6 cent R and so, for any r in R, ra = ra”‘1a = a”‘1ra = a(a"‘2ra) = ar’, where r’ = a”‘2ra. Since at“ e I, this shows that ra e I also, making I a twosided ideal of R.
Lemma 2. in rad R.
Let R be a Jring. For all a, b e 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 twosided 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""" = x. Thus, we are thrown back to a situation where Theorem 912
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, b in R, or, equivalently, ab — ba 6 M. As this last relation holds for every maximal ideal of R, it follows that ab — ba 6 rad R. With these preliminaries established, we now have the constituent pieces to prove
Theorem 913. (Jacobson). If R is a Jring, 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 onlyidempotent belongingto rad R;hence,theelemente = 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 6 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 subﬁeld 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, let S 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 coeﬂicients 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, a“) = a, where n(a) is the number of elements in S. 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 depending on x such that x"(") — x 6 cent R. Then R is commutative. We have noted that in a J—ring some positive power of every element
lies in the center. This provides another path along which to proceed to commutativity.
Theorem 915. (Herstein). Let R be a ring with the property that for each x e R there exists a positive integer n(x) depending on x such that x""" 6 cent R. If R contains no nonzero nil ideals, then R is commutative.
PROBLEMS l. a) If the Boolean ring R has at least three elements, prove that every nonzero element except the identity is a zero divisor of R. b) Verify that the idempotent elements of any commutative ring with identity of characteristic 2 form a Boolean subring. 2. Show that any ring R (not necessarily with identity) in which each element is
idempotent can be imbedded in a Boolean ring. [Hint: Let R’ = Rs and mimic the argument of Theorem 2—12.]
3. a) Let R be a commutative ring with identity and S the set of all idempotent elements of R. Deﬁne a new sum of a and b in S by taking a+’b=a+b—2ab.
200
FIRST COURSE IN RINGS AND IDEALS
With +’ as the addition, prove that S constitutes a Boolean ring, known a the idempotent Boolean ring of R. b) In particular, obtain the idempotent Boolean ring of Z1 2.
c) Show that the idempotent Boolean ring of any integral domain is isomorphic to 2,. 4. Establish the statements below: a) Up to isomorphism, the only simple Boolean ring is Zz. b) If R is a Boolean ring, then R = Q,,(R).
c) A ring R is a Boolean ring if and only if R is commutative with identity and ab(a + b) = 0for all a, beR. ‘ 5. In any Boolean ring R, an order relation 5 may be introduced by taking a S b if and only if ab = a. Given that the elements a, b, c, d all lie in R, conﬁrm the
following orderproperties: a) a S a,0 S a s 1 foreveryaeR; b) a S bandb S cimplya s c; c) a S bandb s aimplya= b; d) a s bandc S dimplyacs bd; e) bc = Oimpliesac = 0(c at: 0)ifandonlyifa s b. 6. a) Let I be a nonempty subset of the Boolean ring R. Show that I is an ideal of R ifand only if i) a,beIimplya + beI,and ii) aeIand reR withr S aimply rel.
b) Verify that the set I, = {re Rr s a} forms an ideal of R. 7. By an atom of a Boolean ring R is meant an element a 36 0 such that r s a implies either r = a or r = 0. Prove that a) the ideal I, is maximal if and only if 1 — a is an atom of R (see Problem 6 for
the deﬁnition of I“); b) any maximal ideal contains all the atoms of R, except at most one. [Hint: Use Theorem 9—1.] 8. Let R .be a Boolean ring. For any nonzero element a e R, show that there exists a
maximal ideal M of R such that a ¢ M from this, deduce that R is semisimple. [Hint: Apply Zorn’s Lemma to the family of all ideals of R which contain 1 — a,
but not a.] 9. For any nonzero element a e R, R a Boolean ring, deﬁne the set S, by
S,I = {MM is a maximal ideal ofR; a¢M}. Establish the following properties of the sets S“:
a) S, aé ﬁ whenever a =/= 0.
b) S,” = SaA Sb. c) S,,, = S“ n Sb. d) S.I = Sbﬂand only ifa = b. [Hint:a¢Mifandonlyif1— aeM.] 10. With reference to Problem 9, prove that if
X = {MM isa maximal ideal ofR},
PROBLEMS
201
then the ring R is isomorphic to a subring of the ring (P(X), A, n). [Hint: Consider the mapping f: R 9 P(X) deﬁned by f(a) = Sr] 11. Assume that a and b are elements of the Boolean ring R with a $ b. Deduce the existence of a maximal ideal M ofR such that a ¢ M, but b e M.
12. Prove that any proper ideal of a Boolean ring R is a semiprime ideal. 13. Let I be a proper ideal of a Boolean ring R and deﬁne the set I’ by
1’ ={1— aIaEI}. Show that I U I’ is the smallest subring of R in which I is a maximal ideal. 14. Suppose that S is a subring of the Boolean ring R. Prove that any homomorphism f from S onto the ﬁeld Z2 can be extended to all of R. [Hint: Use Theorem 2—6; ker f is contained in a maximal ideal M, where R/M z 22.]
15. Prove that a commutative ring R is regular if and only if every principal ideal of R is a direct summand of R.
16. Let R be a regular ring. Establish that R has no nonzero nilpotent elements if and only if for each a e R there exists an element a’ in R such that a = a’a’. [Hint: If R has no nonzero nilpotent elements, then aa’ being idempotent lies in
cent R.] 17. Show that if R is a regular ring, then cent R is also regular. [Hint. Given that a 6 cent R, then aa’a = a for some a’ e R; show that axa = a, where x = a(a’)2
belongs to cent R.]
'
18. Assuming that R is a regular ring with identity, prove the statements below: a) if 0 and 1 are the only idempotent elements of R, then R is a division ring (this holds, in particular, if R has no divisors of zero);
b) if R is of positive characteristic n, then n is a squarefree integer; c) R has no nonzero nilpotent (twosided) ideals; d) for every right ideal I of R, I = ann,(ann,1). [Hint: If I = eR, where e is idempotent, then annJ = R(l — e).] e) If R has no nonzero nilpotent elements, then aR = Ra for every a e R. [Hint: Choose a’ e R such that aa’a = a; since the idempotent a’a 6 cent R, ar = ar(a’a) = sa for any r e R.]
19. If R = R1 EB R2 63 R, (i = 1,2,
63 R, is the direct sum ofa ﬁnite number of regular rings
, n), show thatR isalso regular.
. Verify that the ring L(V) of linear mappings of an ndimensional vector space V into itself forms a regular ring; in this setting, ring multiplication is taken to be functional composition. [Hint: Starting with 0 aé fe L(V), a basis {xv , x,,}
for kerfand a basis {xv {f(l),
, x,"
,f(x,,)} to a basis {yv
, n} for V, extend the linearly independent set ,y,,f(x,,+1),
,f(x,,)} for V. Given any k
elements 21, ,zke Kdeﬁnef’eMV)byf’(yi) = ziforl s i s k,f’(f(xi)) = x, for k + 1 s i S n] 21. Prove that an integer n > 1 is prime if and only if (n — 1)! + 1 is divisible by n.
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FIRST COURSE IN RINGS AND IDEALS
22. Show that GF(p”) is (isomorphic to) a subﬁeld of GF(p") if and only if mln; in fact, if mn, then there is exactly one subﬁeld with p” elements. [Hint: In case mn,
use the fact that a — 1a" — 1 for k > 1 to conclude that x’" — xIxP' — x.] 23. Establish the following assertions:
a) given that an irreducible polynomial f(x)e Z,[x], then f(x)x"" — x if and only if degf(x)n; b) if an irreducible polynomial f(x) e Z, [x] has a root in GF(pP), then f(x) splits completely in GF(p");
c) x’" — x is the product of all the irreducible monic polynomials f(x) e Z,[x] such that deg f(x)n. 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 of F* will generate F as a vector space over K.]
27. Let F be a ﬁnite ﬁeld with p” elements. Prove that the mapping 0,: F > F deﬁned by taking ap(a) = a” is an automorphism, the socalled Frobenius automorphism of F; furthermore, a; = i F.
28. 3) 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 if f is a homomorphism from a ring R onto a division ring, then kerfforms a maximal ideal ofR. 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) = {reea = 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 qll 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 equal 2.
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 9!: 0, the linear mapping Tax = ax is onetoone; hence, onto R.]
33. a) Prove that every ﬁnite ﬁeld is a Jring. b) More generally, establish that a ﬁeld F is a Jring if and only if F 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 Jrings 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) > 1 for which a"“" = a, then R must be commutative. [Hint:Since any idempotenteisincent R,the subringS = eR = Re has e for an identity and, hence, is commutative by Theorem 9—13; then (xy — yx)e = 0. for all x, yeR; now use the fact that e = (xy — yx)"'1 is
idempotent] 35. A ring R is called an Hring if for every x e R there exists an integer n(x) > 1
such that x”) — x e cent R. Assuming that R is an Hring, prove the following assertions: a) Any homomorphic image of R is again an Hring.
b) For each x e R, there exist arbitrarily large n for which it“ — x 6 cent R. c) All the idempotent and nilpotent elements of R lie in cent R. d) If a e R is a zero divisor, then there exists some nonzero c 6 cent R such that ac = 0. [Hint.Ifab = 0,withb + 0,thenc = b“ — becentRandac = 0;
if c = 0, look at the idempotent d = b"‘l.]
TEN
DIRECT SUMS OF RINGS
If {Ri} is an indexed family of rings (not necessarily distinct), it is reasonable to ask whether there is some promising way to use the rings R, to build up new rings. Towards this end, we now introduce the notion of the complete
direct sum of a set of rings (the term “direct product” is also employed in the literature). Although complete direct sums can be rather complicated, there is a special class of such sums that are more manageable; namely, the subdirect sums. We gain much and lose little by soon turning in this direction. The question as to whether a ring R is isomorphic to a subdirect sum of rings of some speciﬁed kind will be shown to be equivalent to the problem of ascertaining whether certain types of ideals of R have zero intersection. In the absence of any statement to the contrary, we shall restrict ourselves to commutative rings with identity. We begin our material by framing the deﬁnition of a complete direct sum. Deﬁnition 101. Let {Rt} be a family of rings indexed by some set I. The complete direct sum of the rings Ri, denoted by Z 69 Ri, consists of all functions a deﬁned on the index set J subject to the condition that for each element ie I the functional value a(i) lies in Ri:
2 Q R, = {aa:.9' —> URE; a(i)eR,}. The rings R, are called the component rings of the sum 2 ED RI. and, more speciﬁcally, we say that R, is the ith component.
Addition and multiplication may be introduced in the set 2 EB R, by means of the corresponding operations in the individual components; writing this as a formula, we have
(a + b)(i) = a(i) + b(i), (ab)(i) = a(1)b(i)
for all ie J.
It follows without diﬂiculty from the ring axioms in each component that the resulting system comprises a ring. The zero element of Z G Ri is the 204
DIRECT sums OF RINGS
205
function 0: J —> u R, deﬁned by taking 0(i) = 06 R, for every index i; similarly, the negative —a of a function a e 2 GB R, is given by the rule
(a)(i) = —a(i). At this point, we should make several remarks. For one thing, in deﬁning the complete direct sum 2 63 Rf, the component rings were not required to be distinct; some, or even all, of these rings may coincide.
A
case of particular interest occurs when R, = R for every value of i 6 J. Under these circumstances, 2 69 R, becomes the set of all functions deﬁned
on J and having values in the given ring R; in short, 2 69 R, = map (J, R). Secondly, if i runs over a ﬁnite index set J (there is no loss in assuming that J = {1, 2,, n}), then the situation is even simpler than it ﬁrst
appears. When this happens, the ring R = 2 e R, may be interpreted as consisting of all ordered ntuples (a1, a2, , an), where the element a, e R,. Addition and multiplication are still to be carried out componentwise; that is to say, (a1, a2, ...,a”) + (b1, b2, ...,b”) = (a1 + b1, ‘12 + b2, ...,an + b”), (01, a2, u. , an)(b1, b2, u. , b”) = (01b1, (12b2, H. , dub")
, an) e R with the Now, let us deﬁne Ii to be the set of all h—tuples (a1, a2, property that a,‘ = 0 for k + i. It is easily checked that I, constitutes an ideal of R, which is isomorphic to the ring R, under the assignment
a, —» (0, ...,o,a,.,o,
, 0)
(a,éR,.).
Furthermore, every element of R has a unique representation in the form (a1, a2,
, an) = (a1,0,
,0) + (0, a2,
, 0) +
+ (0,
, 0, an).
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 Ii. 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 213:1 ® R, or R1 9 R2 Q Q R". We might also mention in passing that if J is the positive integers, then 2 @ Ri may be viewed as the set of all inﬁnite sequences (a1, a2,
, an, ...) such
that ai e R, for each ie 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, is the subring of Z (43 Ri 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 1N RINGS AND IDEALS
exceptions.” It would not be too far removed from traditional connotations
to represent the discrete direct sum of the rings Ri by 2‘ ® Ri:
2‘ GD R, = {a e Z 6 R,a(i) = 0 for all but a ﬁnite number of i}. Again, if the index set I is taken to be ﬁnite, say I = {1, 2, , n}, then the stipulation “for almost all i” is redundant and may be dropped from the
description of 2" 6 RE; in this latter setting,
Zd®Ra=R1®[email protected]'€BR.Another special subring of the complete direct sum 2 e R, which is worthy of consideration is the socalled 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 xi: 2 GB R, —> Ri by the equation
75:01) = a0‘)One can verify that n, is a homomorphism of 2 $ R, onto the ring R" called the ith component projection. If S is any subring of Z 69 R,, the restriction 72iS deﬁnes a homomorphism of S into Ri and, hence, onto a
subring 75i (S) of RE. The case of principal interest is that in which 1t,(S) = Ri 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 102. A subring S of the complete direct sum 2 69 R, is said to be a subdirect sum of the rings Ri, written S = 2‘ 69 R,, if the induced projection niSz S —> R, is an onto mapping for each i. The subdirect sum is nontrivial if none of the mappings 1t,S is onetoone (hence, S is not isomorphic to any Ri). In eﬂect, a subring S E 2 $ 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 suﬂicient conditions upon a ring R will enable us to write it (up to isomorphism) as the subdirect sum of more tractable rings Ri? 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 —> z 63 R, such that, for each i, n, o f is a homomorphism of R onto Ri. Proof. Given an isomorphism f of R onto a subdirect sum 2" EB R, of the rings K, the composition 7:, of: R —> R, deﬁnes a homomorphism of R into
DIRECT SUMS 0F RINGS
207
R, (n, itself being a homomorphism). Since 2‘ ® R, is a subdirect sum, 7ri of actually carries R onto Rt. On the other hand, if there happens to exist an isomorphism f satisfying the indicated conditions, then we certainly have
R :f(R) = 23912,. 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]. A ring R is isomorphic to a subdirect sum of rings R,if and only if R contains a collection of ideals {Ii} such that R/Ii 2 R,
and n Ii = {0}. Proof. To start, we assume that R z 2‘ GD Ri. Then there exists an isomorphism f: R —> Z“ 6 R, such that the “natural” homomorphisms 7n n f: R —> R1. are all onto mappings. Using the Fundamental Homomorphism Theorem, this implies that R/Ii 2R1, where I, = ker(7z, of). Note further that
kerf= {reRf(r) = 0} = {reR(7z,of)(r) = 0for all i} = nI,. Since f is a onetoone function, kerf = {0}, from which it follows that n Ii = 0 . Goiigg}in the other direction, suppose that we are given a set of ideals
{Ii} ofR with R/I, 2 R, and n1, = {0}. Deﬁneafunctionf: R —> 2 Q R,by requiringf(a) to be such that its ith projection 7t,( (f(a)) = a + Ii. (I‘he essential point here is that any element of 2 ® R, is completely determined by its projections.) Then R is isomorphic by means off to a subring of the
direct sum 2 EB Ri. To see thatfis oneto—one, for instance, simply observe that
kerf = {a e R(7z, of)(a) = I, for all i} = {aeRa + 1,. = I, for all i} = n I. = {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/Ii if and only if R contains a collection of ideals {Ii} such that
n II. = {0}. Furthermore, the subdirect sum is nontrivial if and only if II. + {0} for all i.
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FIRST COURSE IN RINGS AND IDEALS
If a ring R is isomorphic to a subdirect sum 2’ 6 Ri of rings Ri, it is convenient to speak of 2‘ G R, as being a representation of R (as a subdirect sum of the rings R1). 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.
Theorem 102. A ring R is isomorphic to a subdirect sum of ﬁelds if and only if R is semisimple. Proof. A ring R is semisimple if and only ifthe intersection of all its maximal ideals Mi 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/Mi, 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 103. 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 103
implies the following: a (commutative) ring R with no nonzero nilpotent elements is isomorphic to a subdirect sum of ﬁelds.
Example 101. 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 ensures that it is isomorphic to a subdirect sum of the rings Z/(p) = Z1,, where p is a prime number:
Z 2 2‘ EB 21,, ppnme
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 Zp” since the
intersection of the ideals (p2) is also the zero ideal: Z ’2
ES
6 Zp2.
p prime
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 or RINGS
209
Example 10—2. For another application of Theorem 10—1, consider the ring map R‘“ of realvalued functions on R"E . As we know, each of the ideals
M,,={femapR*f(x)=0},
xeR’“
is maximal in map R’“. Since ﬂxen M,‘ = {0}, it follows that map R’“ 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 104. 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 + R
such that I + J = R. Proof. Let I 7E {0} be an ideal of R, where R is isomorphic to a subdirect sum of ﬁelds. Then R contains a collection {Mi} of maximal ideals with
n Mi = {0}. Since I is nonzero, this entails that I $ Mi 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 + a e R, so that the principal idea]
(a) 96 {0} (there is no loss in supposing also that (a) 7/: 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 E M and (ii) a ¢ M. To see that M is actually a maximal ideal of R, consider any ideal K with M C K E R. Then, by the maximal nature of M, the element a e K; hence, R = (a) + J E (a) + K E 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 105. 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 I S, = {a,a2,
, a", ...}
is closed under multiplication and does not contain 0. Thus, there exists a prime ideal P, of R, with P, n S, = Z (corollary on page 164). We assert that R z 2‘ @ (R/P,), where the summation ranges over all the nonnilpotent elements of R.
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FIRST COURSE IN RINGS AND IDEALS
Clearly, I = n P,l comprises an ideal of R and is not nil by hypothesis.
If I 7E {0}, we can select some nonnilpotent element b e I. But then I E P“ 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/P,. 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 I1, 12,
, I,I be a ﬁnite set of (nontrivial) ideals of
the ring R. If], + I]. = Rwheneveri 7E j, then R/n I, z z e (R/I,). Proof To start, we deﬁne a mapping f: R —> Z q; (R/Ii) by
f(x) = (x + 1,,x + 1,, ...,x+ In). The reader can painlessly supply a proof that f is a homomorphism with ker f = n Ii. Our problem is to Show that, under this homomorphism, any element (x1 + 11, x2 + 12, ...,x,, + 1,.) of the complete direct sum 2 EB (R/Ii) appears as the image of some element in R; the stated result then hinges upon an application of the Fundamental Homomorphism Theorem. Fix the index j for the moment. Using the fact that Ii + Ij = R whenever i 7E j, there exist elements a, e Ia, bi e Ij with ai + b‘ = 1. This ensures that the product
rj = a1a2
aJ_1aj+1
as 01... 1
J
Furthermore, since 1 — ait, the coset ai + I]. = 1 + I]. for alli 7E j, whence rj + [1. =1 + If. Now, pick arbitrary elements x, e R (i = 1, 2,
, n); our contention is
that f(X) = (x1 + 11, x2 + 12’ "axn + In),
where x = Z rixi. To see this, observe that we may write x + Ij as iﬂ But rit for i 9E j, while r, + Ij = l + IJ, so the displayed equation reduces to x + Ij = x, + I1.0 = 1, 2,
, n). This substantiates the claim
that f is actually an onto mapping, leading to the isomorphism R/ﬂ Ii 2 2 $ (R/Ii)'
Careful scrutiny of the above argument shows that we have proved a subresult of independent interest; namely,
DIRECT SUMS or RINGS
Corollary. Let I 1, 12,
211
, I,I be a ﬁnite set of ideals of the ring R with
the property that I, + I, = R whenever i + j. Given any n elements x1, x2, , x,l eR, there exists some xeR such that x — xie I, for i = 1,2, ...,n.
This corollary may be applied to the ring Z of integers and to the principal ideals (m1), (m2), , (m), where the integers mi 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 AD. 250): Theorem 107. (Chinese Remainder Theorem). Let m1, m2, , m,l be positive integers such that gcd (mi, mj) = 1 for i 7E j. If a1, a2, , a,l are any n integers, then the system of congruences
x E a1 (mod m1),
x E a2 (mod m2),
,
x E a,l (mod m")
admits a simultaneous solution. Furthermore, this solution is unique modulo m = mlm2 m The hypothesis in Theorem 10—6 is conveniently expressed in terms of the following: a ﬁnite set of ideals I1, 12, , I,l of a ring R is said to be pairwise comaximal (or pairwise relatively prime, in the older terminology) ifI, 7E R and 1,. + I]. = R fori 7E j; when n. = 2, we simply term I1 and I2 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 1, 7E Ij for i aé j, as well as 1, 7E {0} for all i. If, in the representation of a ring R as a subdirect sum of the rings R,, the “natural” homomorphism of R onto R, 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 2 R, by way of some mapping other than the “natural” homomorphism of R onto Ri.) 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 103. A ring R is said to be subdirectly irreducible if, in any representation of R as a subdirect sum of the rings R,, at least one of the
associated homomorphisms of R onto R, 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.
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FIRST COURSE IN RINGS AND IDEALS
The importance of subdirectly irreducible rings is demonstrated by the following representation theorem due to Birkhoﬂ'. Theorem 10—8. (Birkhoff). Every ring R is isomorphic to a subdirect sum of subdirectly irreducible rings. Proof. For each element a 3e 0 of R, Zom’s Lemma can be used to select an ideal I, 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 deﬁnition of 1,, implies that if I is any ideal of R with the property
that I, C I, then a e I. We should also point out that the intersection of the ideals I, (where a runs over all nonzero elements of R) is zero. Indeed, if b e ﬂﬁeo I, with b 79 0, then b must, in particular, lie in the ideal 1,;
this contradicts the fact that 1,, was originally chosen so as to exclude the element b; hence, ﬂ“,0 I, = {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/I, is itself subdirectly irreducible or, more to the point, that the intersection of all the
nonzero ideals of R/I, is nonzero. By the Correspondence Theorem, it sufﬁces to establish that the intersection of all the ideals of R properly containing Iu again contains Ia as a proper subset. In light of the maximality of 1,, the element a must belong to all such ideals; therefore, their inter
section contains a and, hence, contains 1, properly. The implication is that the coset a + I, is nonzero and lies in every nonzero ideal of R/Ia. 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 {II is a nonzero ideal of R}. I We observe that R" is a minimal ideal of R which is contained in each nonzero ideal of R; for this reason, R" is frequently called the minimal ideal of R. When R" 96 {0}, it is not hard to see that RV 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 RV aé {0}. A deﬁnition deserves a theorem, so we oblige with the following:
Theorem 10—9. (McCoy). If R is a ring for which Rv 7E {0}, then 1) arm RV is a maximal ideal of R; 2) arm R" consists of all zero divisors of R, plus zero; 3) whenever R is without prime radical, R forms a ﬁeld.
DIRECT SUMS 0F RINGS
213
Proof. First, suppose that the element r ¢ ann RV. Then ra + 0 for some choice of a in R". Since ra lies in R", it will serve as a generator for R" ;
that is, RV = (ra). Thus, we can ﬁnd an element 5 e R satisfying a = ras, whence the product (1 — rs)a = 0. It follows that 1 — rs e ann R", in consequence of which 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) 7E {0} and, since RV is contained in every nonzero ideal of R, ann (r) 2 R". This last inclusion simply asserts that
re ann (ann (r)) 9 arm R", so that ann R" 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 aé {0} (by Problem 14, Chapter 8, {0} is the only nilpotent ideal of R). Thus, there exists some element r e R" for which rR" 7E {0}. The implication of this fact is that R" $ ann RV. Inasmuch as Rv is minimal in the set of nonzero ideals of R, we conclude at once that ann Rv = {0}. The rest follows from (1): {O} is a maximal ideal of R and 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 Z2 (Theorem 9—2). There is a corollary to Theorem 10—9 that will be useful later on. Corollary. If R V 79 {0}, then the annihilator of the set of zero divisors of R is precisely R". Proof. With reference to the theorem, it is enough to prove that ann (ann R") = R". Since one always has R" g ann(annR"), let us concentrate on the reverse inclusion. If a is any nonzero element of arm (ann RV), then R" E (a) and, hence, 0 79 ar e R" for some choice of r é ann Rv (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 1 — rs 6 arm R".
This means that a(l — rs) = 0 and so a = (ar)seRV. ann (ann R") g R", which completes the argument.
It follows that
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 = 2,,» of integers modulo a power of a prime; in this setting, one has
RV = (1)”) 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.
.
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FIRST COURSE IN RINGS AND IDEALS
PROBLEMS
In the set of problems below, all rings are assumed to be commutative with identity. 1. Prove each of the following assertions regarding the complete direct sum 2 e R,: a) If a and b are elements of Z 9 R, such that 7:1(a) = 7r.(b) for each index i, then a = b. b) If an element ri e R, is given for each i, then there exists a unique a e 2 6 R, satisfying 7t.(a) = n.
c) If R, 7E Q for all i, then the ith projection 7:, maps 2 EB Ri onto R,. 2. Prove that an arbitrary function f from a ring R into the complete direct sum
2 63 R, of the rings R, is a homomorphism if and only if the composition 7:, of: R —> R, is itself a homomorphism for each value of i.
3. Consider the complete direct sum 2 $ R,. For a ﬁxed index i, deﬁne the sets
Ii = {aez ® Ria(i) = 0forj + i}, J, = {aez ea R,a(i) = 0}. Verify that I. and J, are both ideals of the ring 2 63 R, and that Z$Ri=li®JP
4. Prove that a ring R is isomorphic to a subdirect sum of rings Ri if and only if, for each i, there exists a homomorphism g, of R onto R. such that if 0 aé r e R, then 9.0) at: 0 for at least one value of i. [Hint: Assume that the stated condition holds. For ﬁxed r e R, deﬁne f, e Z 63 R. by f,(i) = g,(r). Now, consider the mapping
s —> Z 63 R. in whichf(r) = f,.] 5. Establish that each of the given rings has a representation as a subdirect sum of the rings R, (i = 1, 2, 3, ...): a) Z; R, = Zp., where p is a ﬁxed prime. b) Z; R, = Zw where p, is an odd prime. c) 2,; R, = Ze/(2'). (In the situation considered, (2') = {2‘r + 2"nlreZ,; neZ} = 2'2.) d) 2,; R, = Ze/(pi), where p, is an odd prime. 6. Suppose that R is isomorphic to a subdirect sum of the rings R. under the homomorphism ﬁ We say that the subdirect sum is irredundant if, for each index j, the
mapping h}: R —> 2w 6 Ri deﬁned by hj(r) = f(rug,J QB R, is not one~to~one (that is, ker hj aé {0}. Prove the equivalence of the following statements: a) the subdirect sum 2‘ e R, is irredundant; b) ker(1rj of) $ ﬂiﬁ ker (7:, of) for each indexj; c) there exists a collection of ideals {1,} of R such that (l) R, 2 R/Ii, (2) n I, = {0}, and (3) Hi.” I. at: {0} for each index j.
7. Given that thcringR = {(a, b)a, b e Z; a — b e Z,},showthatRisanirredundant subdirect sum (but not the direct sum) of two copies of Z.
PROBLEMS
215
. Prove that an irredundant subdirect sum of a ﬁnite number of simple rings is their direct sum.
. 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 (ii)
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 ideals 1,. are allowed. [Hint: Consider the ring Z and the ideals Ii = (12,), where pi is the ith prime]
, a,I be a ﬁnite set of nonzero elements of the principal ideal domain R such that ai and aj are relatively prime for i 79 j. If a = 1cm (a1, a2, , an),
12. a) Let a1, a2,
show that R/(a) z 2 ® (R/(ai)). b) Prove that if the integer n > 1 has the prime factorization n = pf‘pé"
pfr,
then Z,l 2 z 6 Zp'h
l3. Let II, 12, , 1,, be a ﬁnite set of ideals of the ring R. Prove that a) the ideals Ii are pairwise comaximal if and only if their nil radicals J7) are pairwise comaximal; b) if the ideals Ii are pairwise comaximal, then their product 1112 ~I,l = 11 n I2 n
n In.
[Hint: Use induction on 71. Notice that I,I is comaximal with I1 0 since
n In“
R=R"=l'I(I,,+Ii)EI,.+(nIi)gR forlSiSn—IJ 14. Assume that the ring R is subdirectly irreducible. Establish that there exists an element 0 at: r e R with f(r) = 0 for every homomorphism f on R which is not onetoone.
15. Prove that any subdirectly irreducible ring has characteristic zero or a power of a prime. In particular, conclude that Z,I is subdirectly irreducible if and only if n is a power of a prime. 16. If R is a subdirectly irreducible ring, show that 0 and 1 are the only idempotents of R. [Hint: For an idempotent eeR, consider the principal ideals (e) and
(1 — e).] 17. 3.) Verify that any subdirectly irreducible Boolean ring is a ﬁeld. b) Prove that a ring R is a Boolean ring if and only if R is isomorphic to a subdirect sum of ﬁelds 22. [Hint: Theorem 10—9 and part (a).]
216
FIRST COURSE IN RINGS AND IDEALS
18. Prove that a ring R is subdirectly irreducible if and only if R contains an element rv with the following two properties: i) the principal ideal 0") has nonzero intersection with every nonzero ideal of R; ii) arm 0") is a maximal ideal of R. [Hint: Assume the conditions and let a + 0; from (a) n 0") + {0}, deduce that
(M s (a)] 19. Prove that the idempotent Boolean ring of Z, is isomorphic to the Boolean ring of 2" elements, where k is the number of distinct prime divisors of n. [Hint: Show
that Z,I has exactly 2" idempotents or that x2 E x (mod n) has 2" solutions; for k > 1 use the Chinese Remainder Theorem]
ELEVEN
RINGS WITH CHAIN CONDITIONS
In pursuit of the deeper results of ideal theory, it will be necessary to limit ourselves somewhat and hereafter study special classes of rings. Noetherian rings, which we are about to introduce, are particularly versatile.
These
satisfy a certain ﬁniteness condition, namely, that every ideal of the ring should be ﬁnitely generated. As will be seen presently, an equivalent formulation of the Noetherian requirement is that the ideals of the ring satisfy the socalled ascending chain condition. From this idea, we are led in a natural way to consider a number of results relevant to rings with descending chain condition for ideals. Our investigation culminates in a structure theorem for semisimple Artinian rings which dates back to Wedderburn. (By a ring, we shall continue to mean a commutative ring with identity.) The following deﬁnition serves as a convenient starting point. Deﬁnition 111. A ring R satisﬁes the ascending chain condition for ideals if, given any sequence of ideals 11, I2, of R with
IlglzgnEI
C... "—
there exists an integer n (depending on the sequence) such that Im = I,l for all m 2 n. Deﬁnition 11—1 amounts to saying that every inﬁnite ascending chain of ideals of R must “break off” at some point; that is, equality must hold
beyond some index. In the case of noncommutative rings, it should be apparent how to deﬁne the ascending chain condition for left ideals or for right ideals. We illustrate this idea with several exan‘1ples.
Example 111. In a trivial sense (being simple rings), every ﬁeld and the ring Mn (F) of matrices over a ﬁeld F satisfy the ascending chain condition. So also do the rings Z,,, for they have only a ﬁnite number of ideals. 217
218
FIRST COURSE IN RINGS AND IDEALS
Example 11—2.
In the ring of integers, the inclusion (n) g (m) implies that
m divides n. Since a nonzero integer can have only a ﬁnite number of distinct divisors, the ring Z evidently satisﬁes Deﬁnition 111.
Example 113. As a more interesting example, let us show that the ascending chain condition is satisﬁed by any principal ideal ring R. For this purpose, consider an increasing sequence of ideals of R,
IlgIzgnglug It is easily checked that the set theoretic union I = u I,l is also an ideal of R. Moreover, since R is principal, we must have I = (a) for suitable a e R. Now, the element a lies 1n one of the ideals of the union, say the ideal I. For m > n, it then follows that
I=(a)SIn.C_I,,,EI, whence I", = I", as desired. Example 114. To provide an illustration of a ring in which the ascending chain condition fails to hold, let R denote the collection of all ﬁnite subsets
of Z +. Then (R, A, n) is a commutative ring without identity (in fact, R . is an ideal of the ring of sets P(Z+)). If 1,, = {1, 2, , n}, then the reader may verify that
P(Il) C P02) C P03)C forms an increasing chain of ideals of R which terminates at no point. Our ﬁrst theorem establishes several equivalent formulations of the ascending chain condition. Before presenting this, we make one deﬁnition. Deﬁnition 112. The maximum condition (for ideals) is said to hold in a ring R if every nonempty set of ideals of R, partially ordered by inclusion, has at least one maximal element (that is, an ideal which is not properly contained in any other ideal of the set). We make immediate use of this idea in proving
Theorem 111. The following statements concerning the ideals of a ring R are equivalent: 1) R satisﬁes the ascending chain condition for ideals. 2) The maximum condition holds in R. 3) Every ideal of R is ﬁnitely generated.
Proof. With an eye to proving that the ascending chain condition implies statement (2), let .9’ be a nonempty collection of ideals of R. We shall suppose that .57 has no maximal element and derive a contradiction. Since .9’ is not empty, pick an ideal I1 6 .9’. By assumption, I1 cannot be maximal
RINGS WITH CHAIN CONDITIONS
219
in .9 ; hence, I1 is properly contained in some ideal I2 e .9’. Likewise, I2
is not maximal, so there exists an ideal 13 in .9’ with 12 c I3. Continuing in this fashion, we obtain an inﬁnite ascending chain of ideals of R,
Ilc 12C I3 c , all ofwhose inclusions are proper ; this violates the ascending chain condition. We now assume that the maximum condition holds and let I be any
ideal of R.
If I = {0}, then I is generated by one element, namely, 0.
Otherwise, choose a nonzero element al 6].
Either the principal ideal
(a1) = I and we are through, or else there is an element a2 6 I which does
not lie in ((11); then, (al) C (a1, a2) E 1. Again, if (a1,'a2)aé I, there exists some a3 in I such that (al, a2) C (a1, a2, a3). This reasoning leads to an ascending chain of ideals of R: (“1) C (01, a2) C (a1, “2: as) C
'
The maximum condition assures us that the above set of ideals possesses a , an), , an). Were I 76 (a1, a2, maximal element, say the ideal (a1, a2, , an); accordingly, the ideal we could then ﬁnd some a e I with a ¢ (a1, a2, , an) would properly contain (a1, a2, (a, a1, Thus, I is generated by the n elements a1, a2,
, a”), Which is impossible. , an.
The proof of the theorem is completed by showing that (l) is a consequence of (3). For this, assume that we have an ascending chain of ideals of R,
IlgIzsuglngand let I = u Ii.
Then I is an ideal of R which, by hypothesis, must be
ﬁnitely generated; suppose, for instance, that I = (a1, a2,
, a,).
Now,
each generator ak is an element of some ideal Iik of the given chain. Choosing n to be the largest of the indices ik, it follows that all the ak lie in the ideal In. But then, for m 2 n, I : (a1,a2,...,a,) E I,I E I," E 1; hence, I", = In. Our argument shows that every ascending chain of ideals
R terminates at some point. Rings satisfying any one of the three equivalent conditions of Theorem 11—1 (hence, all three conditions) are called Noetherian rings, in honor of Emmy Noether, who ﬁrst initiated their study. The fact that, when dealing
with Noetherian rings, we can restrict our attention to ﬁnitely generated ideals is of great advantage; the next two results should amply illustrate this. Theorem 112. If I is an ideal of the Noetherian ring R, then I contains some power of its nil radical; that is, (JIY' E I for some n e Z +.
220
FIRST COURSE IN RINGS AND IDEALS
Proof. In view of Theorem 11—1, JI— is a ﬁnitely generated ideal, say
ﬂ = (a1, a2,
, am). Since each ai 6 ﬂ, there exist positive integers n,
for which a2“ 6 I. Taken = n1 + n2 +
+ nm. Now,agenerating system
for NT)" is provided by the products a'{‘ a? n = k1 + k2 + "‘ + km'
air; where kieZ and
But,if
k1+k2+'+k,,.=n1+n2++nm,
then we must have k, 2 n, for some index i (i = 1, 2,
that a2“ 6 I, hence that the element a’{‘ a?of (ﬂy lie in I, it follows that (ﬂy E I.
, m). This implies
of: e I. Since all the generators
Corollary. Let Q be a primary ideal of the Noetherian ring R and I and J be ideals with 1.1 E Q. Then either I E Q or else (\/J)’l E Q for some n 6 2+. Proof. Taking stock of Problem 24(c), Chapter 5, the condition IJ E Q implies that either I g Q or there exists a positive integer m for which .1” E Q. Since R iS Noetherian, we also have (ﬁr E J for some k e Z +. This being so,
(#7)“ E J'" E Q, which is what had to be proven. The Hilbert Basis Theorem asserts that if R is a Noetherian ring (commutative with identity), then the polynomial ring R[x] inherits this property. Since any principal ideal domain and, in particular, any ﬁeld, is Noetherian, Hilbert’s Theorem provides us with a rather extensive class of Noetherian rings. The proof is somewhat demanding, but the result so elegant, that we hope all readers will work through the details. Theorem 113.
(Hilbert Basis Theorem). If R is a Noetherian ring,
then the polynomial ring R[x] is also Noetherian. Proof. Let I be an arbitrary nonzero ideal of R[x]. To prove that R[x] is Noetherian, it is enough to Show that I is ﬁnitely generated. For each integer k 2 0, we ﬁrst consider the set 1,, consisting of zero and those elements r e R which appear as the leading (nonzero) coefﬁcient of some polynomial of degree k lying in I:
1,, = {reRlao + alx +
+ rxkeI} u {0}.
It is easily checked that Ik forms an ideal of the ring R with I,‘ Q [n+1(The second assertion follows from the fact that if r e I» then r occurs as the leading coefﬁcient of x“ 1 when the corresponding polynomial is multiplied by x; hence, r e IH 1.) Since we are assuming that the ascending chain
RINGS WITH CHAIN CONDITIONS
221
condition holds in R, there exists an integer n such that 1,, = In for all k 2 n. Moreover, each of the ideals Ii (i = 1, 2,
, n) has a ﬁnite basis, say (i = 0,1,..., n),
I, = (at1,0229 ...,aim
where aU is the leading coeﬂicient of ﬁj(x), a polynomial of degree i in I. + m” We now set ourselves to the prime task of proving that the mo + polynomials fij(x) generate I. , fun") is ﬁnitely generated and, , fur, , f0“), The ideal J = (f01, by our choice of the I’D(x), must be contained in I. To obtain the reverse inclusion and thereby complete the proof, consider an arbitrary polynomial f(x) e I, say, of degree r:
f(x) = bo + blx +
+ b,_1x"1 + bx'.
The argument proceeds by induction on r. If r = 0, then we have f(x) = b0 6 Io E J and nothing needs to be proven. Next, assume inductively that any polynomial of degree r — l lying in I also belongs to the ideal generated by the ﬁj(x).
When r > n, the leading coefﬁcient b e I, = In and one may write b = anlc1 + auzc2 +
+ amﬂc"l
for suitable choice of c, e R. Then the polynomial f1(x) = f(x) _ x'_"(cif;u(x) + 6213.206) +
+ cmnfilmnoc»
belongs to I and has degree less than r; indeed, the coefﬁcient of x' in this
polynomial is b — i: Cram = 0. (Notice particularly that f1(x) differs from f(x) by an element of J.) At this point, the inductive assumption can be applied to f1(x) to conclude that f1(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 beI,, we can always ﬁnd elements d1,d2, ...,d,,,r in R such that the
polynomial
f2(x) = f(x)  ((1111106) + 4213206) +
'
+ dmfnulx»
is an element of I with degree r — l 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.
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FIRST COURSE IN RINGS AND IDEALS
Corollary. If R is a Noetherian ring, then so is the polynomial ring , x". , x,] in a ﬁnite number of indeterminants x1, x2, R[x1, x2, We recall that an ideal I is nilpotent provided that there exists an integer
n for which I'. = {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 114. (Levitsky). In a Noetherian ring R, the prime radical 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 N1 be an arbitrary nilpotent ideal of R, say N’1‘ = {0} ; assume
further that Ni = {0}. Then (N + N1?” = {0}, so that the idealN + N1 is nilpotent. From the inclusion N E N + N1 and the maximal property of N, it follows that N = N + N1. One is then left with N1 9 N, which settles the point. Now every nilpotent ideal must also be nil and thus N E 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 a" + N = (a + N)” = N for some neZ+, implying that a” e N. Because N is a nil ideal, there exists a positive integer m for which (a")"‘ = 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) E N, by the maximality of N. The rest should be clear: Since a e (a) E 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 of R possessing a quotient ring without prime radical (Theorem 8—12). Therefore, Rad R E 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
(36)” = 06") 75 {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,
Ilalzaualu 2»
3
there exists an integer n such that 1,, = I"+1 = [n+2 = As in Theorem 11—1, this deﬁnition leads to Theorem 11—5. 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). 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 Artinian 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 115. The statement of the Hilbert Basis Theorem is no longer true if Artinian is subtituted for Noetherian. For example, if F is any ﬁeld,
then
(x) = (x2) = (x3) 2 is a strictly descending chain of principal ideals of F[x]
descending chain condition fails to be satisﬁed in F[x].
Thus, the
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FIRST COURSE IN RINGS AND IDEALS
Example 116. Consider the ring R = map R” of realvalued functions on R‘“. Given an arbitrary real number r > 0, we deﬁne
I, = {feRf(x) = Ofor —r S x s r}. Then I, is an ideal of R and it is not difﬁt to seethat ” C Is C 12 C II c 11/2 c 11/3 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
Ir is properly contained in the maximal ideal M = { f6 RI f(0) = 0}. Example 117. We next give an example of an Artinian ring which is not Noetherian. For this purpose, let p be a ﬁxed prime. Cousider 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:
2(1)”) = {m/p"0 s m < p”; mez; n = 0, 1,2, ...}. We make Z(p°°) 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 ofZ(p°°) and choose k to be the smallest positive integer such that for some a, a/p" ¢ I; we implicitly assume that a
and p are relatively prime. Then I must contain all the elements 0, l/pk‘l,
2/p"'1, ..., (p"'1 — 1)/p"‘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 2 k and, of course, b and p are relatively prime. One can then ﬁnd integers
r, s for which rb + Sp = 1.
Since both the rational numbers (reduced
modulo 1)
(rp"")b/p" = WP"
and
W"1 = Sp/p"
lie in I, it follows that (rb + sp)/p" = l/p" also belongs to I, contradicting the minimality of k. Thus, the ideal I is ﬁnite and is given by
I = {0,l/Pk'1,2/P"'l, .,(P"'1  1)/P"_1}Representing I by the symbol 1*. 1, we conclude that the only ideals of Z(p°°) are those which appear in the chain {0} C I1 C 12 C ... c 1": ... c 2(1)”).
Therefore, Z(p°°) possesses an inﬁnite (strictly) ascending chain of ideals, but any descending chain is of ﬁnite length.
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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 116. 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 9 [’2 E E I; E of ideals of
R’. Put It =f‘1(1;‘), for k = 1,2,
Then II E I2 9.
S In E
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", = In
for all m 2 n. Taking stock of the fact that f is an onto mapping, we have f(1*) = 1;. Hence, 1;, = 1;, whenever m 2 n, so that the original chain also stabilizes at some point. Letting f be 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). V 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) JEK, (2) JnI=KnI, and (3) J/I=K/I, thenJ = K.
Proof. Evidently, we need only establish the inclusion K E J. To this purpose, select any member k of K. On the basis of (3), there exists an element j 6 J for which j + I = k + I, which signiﬁes that k — j = i for some choice of i in I. But, since J E K, the diﬂerence k — j also lies in the
ideal K. Using condition (2), we thus ﬁnd that i=k—jeInK=JnI,
and, in consequence, k = i + jeJ. This fact is enough to enable us to prove a partial converse of the last corollary. Theorem 117. 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 J1 E J2 E
E J,l E
be any ascending chain of
ideals of R. From this, we may construct a chain of ideals of I,
JlnIEJaEHEJnnIEu,
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FIRST COURSE IN RINGS AND IDEALS
as well as a chain of ideals of the quotient ring R/I, Jl/I E Jz/I E
E Jn/I E m,
where
Jk/I = nat,J,,.
Since we are told that I and R/I are both Noetherian, each of these chains becomes stationary from some point on; say after r and s steps, respectively. Now, take the integer n to be the larger of r and s, so that
JmnI=JnnI
and
J,,,/I=J,,/I
for all m 2 n. This being the case, an invocation of the lemma is permissible;
it follows that J,,l = J, whenever m 2 n, whence R comprises a Noetherian ring.
Artinian rings are generally more restrictive than Noetherian rings; for instance, the only integral domains which satisfy the descending chain condition are ﬁelds (this is not to suggest, however, that Artinian rings are
without interest).
Theorem 118. Any Artinian domain R (integral domain and Artinian ring) is a ﬁeld. Proof It obviously suﬂices to show that each nonzero element of R has a multiplicative inverse. Thus, suppose that a + 0 in R and consider the descending chain of ideals
(a) 2 (a2) 2 (a3) 2 By the descending chain condition, this chain must be of ﬁnite length, say
(a) 2 (a2) 2
2 (an = (aw) = (W) = .
Then there exists an element r e R such that a" = ra"+ 1; using the cancellation law, it follows that l = ra, which proves our assertion.
Corollary. An integral domain with only a ﬁnite number of ideals is a ﬁeld. With the aid of this result, we can now prove that in the presence of the descending chain condition the Jacobson radical and prime radical coincide.
Theorem 119. If R is an Artinian ring, then every proper prime ideal of R is a maximal ideal. Proof Suppose that I is a proper prime ideal of R. Then, the quotient ring R/I forms an integral domain which satisﬁes the descending chain condition. because R does.
It follows from Theorem 11—8 that R/I must be a ﬁeld,
whence I is a maximal ideal of R. Corollary. In any Artinian ring R, rad R = Rad R.
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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 {Pi} of distinct proper prime ideals of R. We would then be able to form a descending chain of ideals P1
2P1P22P1P2P32
"’.
Since R is Artinian, there exists a positive integer n for which P1P2‘”Pu = P1P2’”PuPn+1
It follows from this that P1P2 P E Pu“, whence Pk E P"+1 for some k s n. But Pk is a maximal ideal of R, so that we must have Pk = Pu“, contrary to the fact that the Pi are distinct. These observations are summarized as Theorem 1110. 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 1111. 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 E rad R and I2 = I. Our contention is that I = {0}.
Assume for the moment that I 7E {0} and consider the family of all ideals J of R such that (i) J E I and (ii) .11 7E {0}. This collection is not empty since it contains I and, hence, it admits a minimal member K. By
(ii), KI 7E {0}, so that al 7E {0} for some nonzero element a e K. Thus, (aI)I = a12 = a1 + {0}, with al E K E I; hence, a1 = 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 E rad R, which implies that 1 — 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 7E {0}.
This contradiction signiﬁes that
I = (rad R)’I = {0}, as asserted. With little additional effort we can learn a good deal more about nilpotent ideals in rings with the descending chain condition.
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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 coincides 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, RV 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 RV, 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 RV; thus, there exists an element x in R
such that ax = r". The only other fact which we will require is that the annihilator of the set of zero divisors of R is precisely the ideal Rv = 0"). Theorem 11—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
(a) '2 (a2) 2 (a3) 2 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"(1 — ra) = 0, with r e R. Inasmuch as a" + 0, the expression in parentheses is a zero divisor of R and, hence, lies in ann Rv by
Theorem 10—9. Thus, foranynonzeroelementx e R",we have x(1 — ra) = 0. But xa = 0, since a also belongs to arm R", 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 ofR which is not nilpotent. Then all the powers a2, a3, ... , a”, are zero divisors and, of course, none is zero.
Thus, for every power a",
there exists an element x,I such that axl = a2x2= ... = a"); = ... =rV +0.
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229
Consider the ideals 1,, = ann ((1‘); clearly, we have 11E12§"'EI"E...
Now, x,l is not in I”, but a(a"x,,) = ar" = 0, since a is a zero divisor of R.
Therefore, x” lies in I“I 1 and the 1,, 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 b are both zero divisors ofR. 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, 06 R such that
xu = yv 7E 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 1113. Let I be an ideal of the ring R with I E rad R. R is a local ring if and only if R/I is a local ring.
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 = l + r with r e I. Since the ideal I S 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 noninvertible elements of R. The reasoning of the previous paragraph shows that the cosets a + I and b + I lack inverses in R/I. Since the quotient ring R/I constitutes a local ring, their sum (a + I) + (b + I) = a + b + I is again a noninvertible 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 E radR. Then R is a local ring if and only if R/D is a local ring.
Theorem 1114. 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 E rad R. Then R forms a local ring. 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 = cx with ceR. But then x = by = box, or x(1 — bc) = 0, which implies that l — bc lies in D S 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 1 — 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 E 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 1115. (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 diﬂ‘erent 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 M1, M2, , M,l of R such that n M, = {0}, but the ideals Ii = ﬂkﬂMk aé {0} foreveryi. InasmuchasMiismaximal,R =' I, + Mi; moreover, I, n M, = {0}, so that this sum is actually direct. Hence, the
RINGS WITH CHAIN CONDITIONS
231
natural mapping natMi: R > R/Mi induces an isomorphism Ii z R/Mi, making I, a ﬁeld. By virtue of the relation R = Ii + Mi, one can write, for each element a e R: a = x, + yi,
x,eI,,y,eM,
(i = 1,2, ...,n).
Given any'integer k between 1 and n,
a— i=(a—x,,) ieMk, i=1
i+k
sincea — xk = ykeMk, while xieli g Mk fori + k. Thus,
a — ienM,‘ = {0} and so a = 2 xi. R = 11 + I2 +
This implies that the ring R may be represented as + In;that the foregoing sum is direct follows from the
fact that 2,.“ Ii E Mk, whence Ik n(i;k1i)g I,‘ 0 MI, = {0}. It is of interest to compare Theorems 10—2 and 11—15.
We have
exchanged the subdirect sum part of Theorem 10—2 for a direct sum (in fact, a ﬁnite direct sum) in Theorem 11—15; however, Theorem 11—15 was
obtained at the cost of an additional hypothesis: the ring must be Artinian. We offer a second proof of Wedderburn’s Theorem, the relative merits
of which can be weighed by the reader; although this second proof is clearly less complicated than the ﬁrst, it nonetheless relies more heavily on the
results of the previous chapter. Second proofof Theorem 11—15. Let Ml, M2,
, M,‘ be the maximal ideals
of R (there is no harm in assuming that {0} is not a maximal ideal, for otherwise the theorem follows trivially). The maximality of these ideals implies that M, + MJ = R whenever i 7E j. Thus, by Theorem 10—5, it follows that
R = R/n M, 2 2 @ (R/Mi)
(complete direct sum),
where each of the quotient rings R/Mi is a ﬁeld (i = 1, 2,
, n). But, in
the ﬁnite case, the complete direct sum coincides with the usual direct sum.
Notice that, in carrying out the above argument, we have proved a subresult which is interesting in its. own right : If a ring R has a ﬁnite number
of maximal ideals Mi with zero intersection, then R z 2 69 (R/Mi). There is a corollary to Theorem 11—15 which is worthy of emphasis. Corollary. Any semisimple Artinian ring is Noetherian.
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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. . Prove that every homomorphismfof a Noetherian ring R onto itself is necessarily
onetoone. [Hint: Consider the ascending chain {0} E kerf S kerf2 S ideals of R.]
of
. a) If I is an ideal of the Noetherian ring R, show that ﬁ/I forms a nilpotent ideal of R/I. b) Let I and J be two ideals of the Noetherian ring R. Establish that I" E J for
some integer n e Z+ if and only if ﬂ 9 ﬂ
. Prove that every ideal of a Noetherian ring R contains a product of prime ideals. [Hint: If not, let S be the set of those ideals of R which do not contain a product of prime ideals and apply the maximum condition] . 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]] is Noetherian if and only if R is a Noetherian ring.
[Hint: Mimic Hilbert’s Theorem, now using elements of
lowest order when deﬁning the ideals 1*] c) Let R’ be an extension ring of the Noetherian ring R. For a ﬁxed element r e R’, show that the ring
RD] = {f(r)f(x) e R[x]} is Noetherian.
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. . Let R be a ring with at least one nonzerodivisor. Prove that if R is Noetherian, then its classical ring ofquotients Qc.(R)isalso Noetherian. [Hint:IfJl E J2 E is an ascending chain of ideals of QC,(R), then, by Problem 29, Chapter 4,
I1 E I2 E
forms an ascending chain of ideals of R, where I, = Jk n R.]
PROBLEMS
233
. Let p be a ﬁxed prime number and put
Q, = {m/rlmez; 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 Q,). Establish that a) 2 forms an ideal of the resulting ring Q,; h) the quotient ring Qp/Z is isomorphic to Z(p°°);
10. a) Prove that a ﬁnite direct sum 2 69 Ri is Noetherian (Artinian) if and only if each of the component rings Ri is Noetherian (Artinian). [Hint: If n = 2, say
R = R1 9 R2, then R/R1 2 R2.] b) Let Rbearing having aﬁnite number ofideals I1, 12, , I, such that n I, = {0}. If each of the quotient rings R/Ii is Noetherian (Artinian), show that R is also Noetherian (Artinian). ll. In an Artinian ring R, prove that the zero ideal is a product of maximal ideals. [Hint: rad R = M1 n M2 n n M", where each M, is maximal; use Theorem
11—11 and Problem 13, Chapter 10, to conclude {0} = (rad R)“ = Mf for some integer k.]
M:
12. Establish that an Artinian ring R is isomorphic to a ﬁnite direct sum of Artinian local rings. [Hint: From Problem 11 and the fact that the ideals Mf are comaximal,
we have Mf n
n Mt = Mf
M: = {0}. By Theorem 10—1, R 2 2 e R/MQ‘.
Now use Problem 5—18.] 13. Prove that any semisimple Artinian ring possesses only a ﬁnite number of ideals. [Hint: Assume that R admits that decomposition R = F1 EB F2 $ 63 F”, F, a ﬁeld; ifI is an ideal ofR, then I = II $ 69 I,I with I, an ideal ofF,.] 14. a) Let I be a nontrivial minimal ideal of the Artinian ring R.
Show that the
annihilator arm I forms a prime and, hence, maximal, ideal of R.
[Hint: If
a ¢ arm I, at! = I.] b) Assume that I is a nonzero ideal of the ring R (no chain conditions). If P is a maximal member of the collection {arm (x)0 aé x e I), deduce that P is a prime ideal. [Hint: Let ab 6 P = ann (r), with a g! P; then, P E (P, b) E ann (ar).] 15. A ring R is termed divisible if every nonzerodivisor 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 II n I2 + {0} for any two nonzero ideals of R, then R is a local ring. [Hint: Show that the sum of two noninvertible elements 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 22. [Hint. See the remark following the second proof of Theorem 11—15.] b) Prove that a ﬁnite Boolean ring has 2'l 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 crosssection 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’i‘p’;2
pf' is a factorization of the positive integer n into
distinct primes p,, then an integer m is divisible by n if and only if m is divisible by each pi“; in the notation of principal ideals, this amounts to asserting that
(n) = (1"?) n (1";2 n where each of the (p?) is a primary ideal of Z.
0 (pt), 
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 0N NOETHERIAN RINGS
23S
arbitrary Noetherian ring. A convenient vehicle for this discussion is the notion of an irreducible ideal. Deﬁnition 12]. 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 P is always irreducible. For, suppose that there exist ideals I and J of R satisfying PCJ. PCI, P=InJ, We can then select elements a e I — P and b G 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 (nonprime) 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 M2 = (x2, xy, y2) must be primary (see Example 7—8); however, M2 has the following representation as an intersection of proper ideals of F[x, y]:
M2 = (M2, x) n (M2, 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 .97 be the family ofall ideals of R which are not ﬁnite intersections of irreducible ideals. If it happens that 37 7E Q, then Theorem 11—1 asserts the existence of an ideal I which is maximal in the set .97 (this is where the Noetherian hypothesis enters). Then any ideal of R properly containing I must be a ﬁnite intersection of irreducible ideals. Since I e .97, I is not itself irreducible. Thus, we can write I = J n K, where J and K are ideals of
R strictly containing 1. The maximal nature of I implies that .I and K both are ﬁnite intersections of irreducible ideals; hence, I is one also. But
this clearly contradicts the fact that I e .9". Accordingly, the set 9" 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 6 I, b ¢ I and no power of a belongs to I. Now,
I:(a) E I:(az) ;
g I:(a") ;
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FIRST COURSE IN RINGS AND IDEALS
forms an ascending chain of ideals of R; indeed, if rd" 6 I, certainly ra'“r 1 e I. Because R iS taken to be Noetherian, we can therefore ﬁnd an integer k
for which I: (a") = I: (akH). The bulk of our argument consists of showing that I can be expressed as
I=amnaw Evidently, each ideal on the righthand side of this equation contains 1, so that I g (I, a") n (I, b). To obtain the opposite inclusion, select an arbitrary
x e (I, a") n (I, b). Then, x=i+ra"=i’+r’b for suitably chosen elements 1', i’ e I and r, r’ e R. Consequently, the product
wk“ = (i’ — i)a + r’(ab)eI, which in turn implies that reI: (a"+1) = I: (ak). But this signiﬁes that ra" e I and so the element x = i + ra" lies in I, as we wished to Show. To complete the proof, observe that I C (I, a"), for our hypothesis assures us that a"¢ 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. These results may now be put into the form of a decomposition theorem, the socalled 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.
Let us call a representation of an ideal I in the form I = ﬂ, Q,, where each Q, is a primary ideal, a primary representation of I; the individual QII are said to be the primary components of the representation, while J5, 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 122. A primary representation I = ﬂ}; 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, ﬂ,,,, Q, 7E ﬂ Q, for anyj = 1, 2,, n.
2) J5, 75 JEfori #1. If an ideal I admits a ﬁnite primary representation, say I = ﬂ','=1 Q,, then enough of the Q,’s can be omitted to yield an irredundant representation. To make this precise, we ﬁrst let Q’i be the intersection of all those primary components which have the same associated prime ideal; that is,
FURTHER RESULTS ON NOETHERIAN RINGS
if x/Qi, = x/Q, =
= x/Q}, simply take Q; = Qi, n Q1.2 n
237
n Q».
By the corollary to Theorem 5—13, the ideal Q; is itself primary with
ﬂ = \/Q—,k and, ofcourse, we have I = ﬂiQQ. 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 Q; 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 = ﬂ';=1 Qi. Then. a prime ideal
P of R contains I if and only if P contains some \/Qi.
Proof The if part is trivial: JQ, E P implies that I E Q, S P. Conversely, assume that there is no ﬁt which is contained in P.
For each
index i, we can then choose an element a, e JQ, with a, at P. There also
exist suitable integers k, such that a2“ E Q. Setting a = ai‘m’;2 follows that a e (12:, Q, = I E P. Now, the product
aﬁ", it
are? at") e P with a1 ¢ P and so, by the deﬁnition ofprime ideal, (a?    aﬁ") e P. Repeating this argument, we ﬁnally obtain at" 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 123. 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 = (15;, Q, of an ideal I as a ﬁnite intersection of primary ideals Qi, 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
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FIRST COURSE IN RINGS AND IDEALS
\/—: 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 (x2, xy). It is easy to see that (x2, xy) consists of those polynomials which have x as a factor and which do not possess linear terms. Now, the
nonzero elements of (x2, xy, yz) are precisely the polynomials each of whose terms are of degree 22; hence the intersection (x2, xy, yz) n (x) contains the zero polynomial together with all polynomials of degree 22 which have x as a factor. Thus, we have
(x2, xy) = (x2, xy, yz) n 06)Besides this irredundant representation, there is yet another:
(x2, xy) = (x2, y) n ()0 To derive the relation above, notice that for a polynomial to lie in (x2, xy), it is sufﬁcient to require that the polynomial be divisible by x and that the
coeﬂiCient of y be zero. As has been seen in Example 7—8, (x2, xy, yz) and (x2, 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 124. Suppose that an ideal I of the ring R has a ﬁnite irredundant primary representation, say I = ﬂy: 1 Q, and let P be any
prime ideal of R. Then P = JQ for some i if and only if there exists an element a «at I such that P = , /I: (a). Proof To start with, assume that P = JQ, 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 = ﬂ,‘ Q,‘ implies that there exists an element a 6 ﬂ”, Qk, but a ¢ I. For any such element a, we must have
'I:(a) 9 J5, ; ./I:(a). The ﬁrst inclusion is justiﬁed by the fact that, since a(I : (a)) E I E Q, with
a ¢ Qt, necessarily I z (a) S JG. To see the second inclusion, simply note that aQi E I, whence Q, E I: (a).
FURTHER RESULTS 0N NOETHERIAN RINGS
239
Next, suppose that the product bc 6 I: (a), but b 9% J5. Then, a(bc) e I E 5,. Since Qi is primary and b¢ J5“ it follows that ac E Q. Also, ac e (a) E ﬂ”, Qk, which gives ac 6 ﬂ,‘ 5,, = I, forcing the element c to lie in I: (a). c e I : (a).
In other words, bceI: (a) with b¢ J5, implies that
Thus, all the conditions of Theorem 5—14 are satisﬁed and we
may conclude that P = J5 = JI : (a). Going in the other direction, let the elementa ¢ I besuch thatP = J1 : (a). With reference to Theorem 2—5,
Me) = ((3 Qt):(a) = n (we) and, the intersection being ﬁnite, P = H, , /Q;~3 (a). Observe that if a e 5,, then JQi: (a) = Qi: (a) = R. On the other hand, if a¢ Qi, reason as in the ﬁrst part of the proof with I now replaced by the primary ideal Q; for these Qi, we then obtain 5,: (a) = J5i. In consequence, P is the intersection of some of the Q; let us say that P =
VQil n
Qiz n
” n
Qir'
Knowing this, the proof is easily completed; for, by Problem 30(a), Chapter 5, P must contain one of the ideals x/n and is obviously contained by
it, whence P = JQik. What we are really after is the corollary below. Corollary. Let I be an ideal of the Noetherian ring R. Suppose that I = 51 n n Q,l = Q’1 n n Q; are two ﬁnite irredundant primary representations of I. Then. n = m and the associated prime ideals of these two representations are equal (that is, with a suitable renumber
ing of the indices J5, = J5; for 1 s i s n = m). Proof. The theorem provides an intrinsic characterization of the associated prime ideals in terms of I alone. Example 122. In the polynomial ring F[x, y], the ideal (x2, xy) has the irredundant primary representation
(x2, xy) = (x2, xy, y") n (X) regardless of the choice of n > 1. The corresponding nil radicals (that is, the associated prime ideals) are (x, y) and (x). Thus, in any irredundant representation
(x2, xy) = Ql 0 Q2 by primary ideals Q1 and 52, we must have VQI = (x, y),
‘VQZ = (X).
Of these, only (x) is a minimal prime of (x2, xy). Now, it so happens that
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FIRST COURSE IN RINGS AND IDEALS
the nil radical JI of an ideal I in a Noetherian ring is the intersection of the minimal primes of I; granting this fact, one ﬁnds that
\/(x2, xy) = (x)Incidentally, our example has the added advantage of showing that J7 can be prime without the ideal I being primary. 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 = 02;, Q,. Then I is semiprime (that is,
I = JI) if and only if each Q, is a prime ideal of R. Proof. We begin by assuming that all the Q, in the given primary representation of I are prime ideals. If the element a e JI, then a" e I for some positive integer n; hence, a” 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 JI E 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,
I=ﬁ=¢m=ga
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 = ﬂiﬁ , /Q,. But then
I = (we; {19:2 I, 1+}
tat]
which means that I = ﬂiﬁ Qt. This, however, contradicts the hypothesis that the given representation of I is irredundant.
Next, ﬁx the integer j and let a be any element of the ideal J6. Since
ﬂiﬂJ6, 7E ﬂ J6“ we can ﬁnd some b e ﬂiﬁJ6iwith b ¢ Qj. Then the product ab 6 ﬂ J6 = I E Q, whence a is a member of 61.. Therefore, J6j E Qj, yielding the subsequent equality Qj = J6,” This implies that Qj is a prime ideal, which was what had to be proved. 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 0N NOETHERIAN RINGS
241
Theorem 126. (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 .9; of ideals of R which are not ﬁnitely generated is nonempty. Appealing to Zorn’s lemma, .9?" must contain a maximal element, call it I. By virtue ofour hypothesis, I cannot be a prime ideal of 4 Consequently, there exist elements a, b of R which are not in I such t, at their product ab e I. Now, both the ideals (I, b) and I : (b) properly contain I ; in particular, a e I : (b). By the maximal nature ofI in .97, these ideals are ﬁnitely generated. For deﬁniteness, let us suppose that
(I, b) = (c1, c2,
, c")
and
I: (b) = ((1,, d2,
Then c, = a, + bri, where aieI and rieR(i = 1,2, (I, b) = (01, 02,
, dm).
, n), so that
a an: b)‘
Next, consider the ideal J generated by the elements a, and bdj; in other words, the ideal J = (a1,
, an, bdl,
, bdm).
Since M, e I for every j, the inclusion J E I certainly holds. What is not so obvious is that I E J. To see this, let x be an arbitrary member of I ; because x e (I, b), it may be written in the form (yi’yeR)'
x = alyl + H. + anyn + by
As each ai 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 2i 6 R such that y = (11121 + dzz2 +
+ dmzm,
leading directly to x = aly + = aly1 +
+ any" + b(d121 + + any,I + (bd1)z1 +
+ dmzm) + (bdm)z,,,eJ.
The equality I = J now follows and so one concludes that I itself is ﬁnitely generated, an impossibility since I e 3'". 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 (I, b) and I: (b) are both ﬁnitely generated,
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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)"eIEM for some positive integer n. This implies that l — ee 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 inter
section 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 U = I, then there exists an element r 6 J such that (1 — r)I = {0}.
Proof. Suppose that I is generated by the elements a1, a2,
denote the ideal (ai, “H1,
, an. Let I,
, an) and put I”+1 = {0}. By induction on i,
we shall prove the existence of an element r, 6 J such that (1 — ri)I E Ii (i = 1, 2,
, n + 1); in particular, r”+1 will be the element mentioned in
the statement of the theorem. When i = 1, the ideal I1 = I and one can Simply take r1 = 0. Using the induction hypothesis that (1 — r,)I E Ii for some r, e J, together with the fact that I E U, we have
(1 — r,)I ; (1 — r,)IJ ; 1,J. Since each generator a, lies in I, it follows that (1 — r,)a, e I,J and so (1 — rt)“: = gibmak
(bad)
In consequence,
(1 — 7': — bii)ai =
2": bikakEIHl' k=i+1
FURTHER RESULTS 0N NOETHERIAN RINGS
243
It now suﬂices to take] — n+1 = (l — r,)(1 — r, — bu); clearly, r,+1 eJ and, as a little computation will show, (1 — ri+l)I = (1 _ r;)(l — r, _ bu”
E (1 — 7i _ buﬂ: = (1 — ri “ bii)(aiali+1) E Ii+1' 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 note—
worthy result of Nakayama. Theorem 127.
(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 E M, leading to the contradiction that l e M. Accordingly, I — 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.
1'38
Theorem 128 Let I be a proper ideal of the Noetherian ring R. Then
I" = {reR(1 — a)r = 0 for some aeI}.
Proof For ease of notation, let S denote the righthand side of the indicated equation. If the element r e S, so that (1 — a)r = 0 for suitable a e I, we would necessarily have r=ar=a2r=~=a"r= The implication of these relations is that r belongs to I” for every integer n and, hence, r 6 m, I"; in other words, S s ﬂu I". The opposite inclusion is less obvious. To obtain this, put J = ﬂu I" and consider the irredundant primary representation of the ideal IJ : L] = ﬂ Qi:
(Qi Primal?)
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FIRST COURSE IN RINGS AND IDEALS
We wish to establish theequality IJ = J (once this has been accomplished the previous lemma can be applied). Since I.I E J, it will be enough to Show that J E Q for each index i. Now, U E Q, so, by the corollary to Theorem 11—2, either J g Q, or else (ﬂy 9 Qi for some n e Z +. But if
(ﬁr 9 Qt: then J g I" .9 WI)“; Q,. In any event, J E Q, for each value of i, whence J E IJ and equality follows. From the lemma just proved, there exists an element a e I such that (1 — a)J = {0}. But this amounts to asserting that J = ﬂ, PI is contained in the set S and thereby completes the proof. ' This leads almost immediately to an important theorem of Krull. Theorem 129. (Krull Intersection Theorem). Let I be a proper ideal of
the Noetherian ring R. Then (13;, I” = {0} if and only if no element ofthe set 1 — I = {1 — aaeI} is a zero divisor in R. There are, of course, a number of interesting consequences of this last result. Some of these are the content of the corollaries below. Corollary 1. If I is a proper ideal of the Noetherian domain R (in particular, if R is a principal ideal domain), then H :11 I” = {0}. Corollary 2. In any Noetherian ring R, ﬂz‘; 1 (rad R)” = {0}. Proof Take I = radR in Krull’s Theorem. By Theorem 8—2, every element of 1 — rad R is invertible and thus cannot be a zero divisor. Corollary 3. In a Noetherian domain R, any prime principal ideal (a) is a minimal prime ideal of R. Proof. Suppose to the contrary that there exists a prime ideal P of R
satisfying {0} C P C (a).
Since the element a¢ P, the condition rae P
implies that re P; hence, P = aP.
Utilizing Corollary 1, we therefore
conclude that
P=aP=a2P= E ﬂ(a”)= ﬂ(a)”={0}, which is impossible. ideal of R.
This line of reasoning makes (a) a minimal prime
Given a zero divisor r = 1 — a, with a e I, the element 1 — r evidently
belongs to the ideal I. In the light of this, the Intersection Theorem is often phrased in a slightly different form: Let I be an ideal of the Noetherian
ring R; then,
:21 I" = {0} if and only if there is no zero divisor r of R
such that 1 — r e I.
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245
We conclude the present phase of our investigation by showing that, for any ideal I of a Noetherian ring, Off: 1 I'' is equal to the intersection of certain primary components of the zero ideal. This result, which may be regarded as a reﬁnement of the Krull Intersection Theorem, is due to
Northcott [53]. Theorem 12—10.
Let I be an ideal of the Noetherian ring R and let
{0} = Q1 n Q2 n
n Q,' be an irredundant primary representation
of {0}. Assume further that J6i n (1 .— 1) 7E Q for m + l S i S n, but not for 1 S i S m Then,
(511"=QQ~nQ... Proof According to Theorem 12—8, it is enough to show that the set
S = {reR(l — a)r = 0for someaeI} = Q1 n Q2 0
n 6",.
Suppose that the element r e S, so that xr = 0 e Q, for suitable x e 1 — I.
For 1 S i S m, our hypothesis implies that x ¢ J6,, whence r e 6,; this establishes the inclusion S E Q1 0 Q2 0 n 6",. Now, let y be an arbitrary member of the intersection Q1 n Q2 n n 6",. When the integer i 2 m + 1, it is possible to choose an element
a, 6 J6, n (l — I). For k suﬂiciently large, we will then have y(am+l ”‘ an)keQ1 ﬂ ... n Q»: n
that is to say, y(a,,,+1
D Q» = {0},
an)" = 0. Because 1 — I forms a multiplicatively
closed subset of R, the product am+1 ye S. Our argument gives Q1 0 Q2 0 equality follows.
a,, e 1 — I and so, by deﬁnition, n 6,, E S, fromwhich thedesired
One could rephrase Theorem 12—10 to read as follows. Given an ideal I of the Noetherian ring R, ﬂf=1 I” is the intersection of those primary
components Q, of {0} for which J6, n (1 — I) = 9’. We next intend to prove the following result on local rings: if a local
ring R has principal maximal ideal M, with 0;: 1 M" = {0}, then R must be Noetherian. Such rings have an extremely simple ideal theory in that every nontrivial ideal is a power of the maximal ideal. The obvious example
of this situation is the power series ring F[[x]], where F is a ﬁeld; by
Theorem 7—3, we know that F[[x]] forms a local ring with maximal ideal (x), in consequence of which 30:13:)" = 0310‘") = {0}
Before we consider the stated result, a lemma is required. Lemma. Let R be a local ring whose maximal ideal M is principal. Then,
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FIRST COURSE IN RINGS AND IDEALS
1) for nonzero a, b eM, (a) = (b) if and only if a = bu, where u is invertible in R; 2) if M = (p) 7/: {0}, then p is an irreducible element of R; 3) if 0 aé a = q"u, with q 6 M and u invertible, then this factorization is essentially unique (that is, the integer n is uniquely determined by q). Proof. If (a) = (b), then a = br and b = as for suitable r, s e R. Accordingly, a = asr or a(1 — sr) = 0. Suppose that one of the elements r or s lies in M, so that the product sr e M. We observe that 1 — sr ¢ M, for other
wise 1 e M. This means that l — sr is an invertible element of R (recall that M consists of all the noninvertible elements of R). Then the relation a(1 — sr) = 0 yields a = 0, a contradiction. Thus, neither r nor s belongs to M, which signiﬁes that they are both invertible. The converse should be obvious. Concerning (2), let p = ab. If the element b is not invertible, then part (1) forces M = (p) c (a). The maximality of M then ensures that (a) = R, whence a has an inverse in R. To see the ﬁnal assertion, assume that a = q"u = qmv, with m > n. Then q”u(1 — q’"'"vu‘ 1) = 0. By the argument of the ﬁrst paragraph, this relation implies that a = 0. Theorem 1211. Let R be a local ring with principal maximal ideal M = (p). Then every element 0 79 a e M has a factorization in the form a = pmu, where u is invertible, if and only if ﬂ,°,°=1 (p”) = {0}.
Proof. Assume that the intersection ﬂ” (p‘) = {0}.
If 0 7E a e M, then
a = pr for some r in R. If it happens that r is not an invertible element, then r e M; thus, we can write r = ps or, upon substituting, a = p23. This process must eventually terminate, for otherwiseawould lie in ﬂ” (1)") = {0}. For the converse, suppose that there exists some nonzero element a in
ﬂ, (p"). In particular, a e (p) = M, so that a = p"'u for some invertible u. For any integer k > m, we then have
(p"') E (a) E (17’) 9 (pm), whence the equality (p"‘ = (p"). The lemma now tells us that p’" = p"v, where v is an invertible element of R. This means that a = pkvu, with vu invertible, contradicting the last assertion of the lemma.
Corollary 1. Let R be a local ring with principal maximal ideal M = (p).
Assume further that (13;, 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 E 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 pku e I.
It then follows that I 9 (pk). On the other hand, Since p"u e I, so does p" = (pku)u‘1; this implies that (p") E I and equality follows.
FURTHER RESULTS 0N NOETHERIAN RINGS
247
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 JordanH'o'lder 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 moduletheoretic prerequisites.
Our discussion is not entirely self
contained 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.
Deﬁnition 12—3. Let R be a ring with identity. By a left module over R (or a left Rmodule), 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,
2) (rs)a = r(sa), 3) r(a + b) = ra + rb, 4) 1a = a, for all r, s e R and a, b e M. The parallel notion of a right Rmodule 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 m is often called the module product of r and a. In effect, Deﬁnition 12—3 states that a left Rmodule 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 Rmodule M”. We pause to look at some typical examples of modules.
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Example 123.
If R = F, where F is any ﬁeld, then a left R module is
simply a vector space over F. Example 124. Every commutative group (G, +) can be considered as a left Zmodule in a natural way. For, given an integer n and element a e G, na has a welldeﬁned 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 Rmodule. Indeed, the deﬁnition of left ideal insures that the ring product ra e I for every r e R and a e I. AS a special case, any ring R with identity can be viewed as a left (or right) Rmodule over itself. Example 12—6. Consider the set homG 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, + , 0) constitutes a ring with identity, where 0 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 fa=f(a) (fehomG,aeG). 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) 0a = r0 = 0, (ii) r(—a) = (—r)a = —(ra), forallreRandaeM. 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 “Rmodule” will always mean “left Rmodule”; it Should be apparent that the entire discussion applies equally well to right Rmodules. Of course, when R is a commutative ring, any left Rmodule can be turned into a right Rmodule simply by
putting ar = ra. Modules over commutative rings are essentially twosided 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 Rsubmodule 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 124. A nonempty subset N of the Rmodule 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, b e N, then the difference a — b e N. Every Rmodule 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 Rmodule M simple, if M 7E {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 Fsubmodule 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 Rsubmodules. The concept of a quotient structure carries over to modules as expeCted. To be more concrete, let N be a submodule of a given Rmodule 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. This amounts to showing thatwhenevera + N = a’ + N, then r(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’)eN. 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 Rmodule (the socalled 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
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FIRST COURSE IN RINGS AND IDEALS
quotient structure were welldeﬁned. Let us emphasize that, in the case of modules, no such distinguished subsystem need be deﬁned; for each sub
module of an Rmodule M, we can construct a quotient module of M. The counterpart of the Correspondence Theorem remains valid for modules and will be relevant to our discussion; we take the opportunity to record this as Theorem 1212. Let N be a submodule of the Rmodule M. Then there is a onetoone inclusion preserving correspondence between the submodules of M/N and those submodules of M which contain N. The notions of homomorphism and isomorphism can be deﬁned for modules in the obvious way. Deﬁnition 12—5. Given two Rmodules M and N, a mappingf: M —> N is called a module homomorphism or merely an (R) homomorphism if
1) fis a group homomorphism from (M, +) into (N, +); 2) f(ra) = rf(a) for all re R and a GM.
When f is onetoone and onto N, it is called an (R) isomorphism; one then says that M and N are (R) isomorphic and writes M 2 N. Example 127.
If R = F, where F is an arbitrary ﬁeld, the Rhomomor
phisms are just the linear mappings (linear transformations) from M to N. Example 12—8. Let N be a submodule of the Rmodule M. The function natN : M —> M/N which assigns to each element a e M its coset a + N is an Rhomomorphism; for, by deﬁnition,
natN(ra) = ra + N = r(a + N) = rnatN(a). As in the ringtheoretic case, we shall call natN the natural mapping of M onto the quotient module M/N. With the above deﬁnitions in View, the reader will experience no diﬂiculty in proving the appropriate results. These are set out in the following omnibus theorem. Theorem 1213. Let M and N be two Rmodules and' f: M > N be an Rhomomorphism from M into N. Then, 1) the kernel off, kerf = {a e M f(a) = 0}, is a submodule of M ; 2) the image of M under 12 f(M) = {f(a)a e M}, is a submodule of N; 3) f is a onetoone function if and only if ker f = {0};
4) M/kerf=f(M)At this point, it should come as no surprise that such ideas as the ascending (descending) chain condition and the maximum (minimum)
FURTHER RESULTS ON NOETHERIAN RINGS
251
condition can also be applied to Rmodules, the sole difference being that, in our earlier deﬁnitions, the term “idea ” must now be replaced by the word
“submodule”. Adapting the argument of Theorem 11—2, it is a simple matter to show that an Rmodule 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 1214. 1) If the Rmodule M satisﬁes the ascending (descending) chain condition, then so does every homomorphic image of M. 2) Let N be a submodule of the Rmodule M. Then M satisﬁes the ascending (descending) chain condition if and only if N and M/N both satisfy it. For the most part, the stated results are merely a translation of Theorems 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 Rmodule M is meant a (ﬁnite) chain of R
submodules running from M to {0}: M=M02M,22Mn_12M,,={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=N02N122Nm_12Nm={0} is said to be a refinement of :Mn={0} M=M02M12m2Mnl —
, n} into provided that there exists a onetoone function f from {0, l, Mi every that saying to , m} such that M, = N1’“). This amounts {0, 1, termed is series normal a of reﬁnement A Nj. the of one as must appear proper if the reﬁnement contains a submodule not in the original series. A normal series which admits no proper reﬁnement is called a composition series. We summarize this in the following deﬁnition.
Deﬁnition 126. A composition series for an Rmodule M is a normal series (without repetitions) M=MODMID"' DMlIIDMII={0}
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FIRST COURSE IN RINGS AND IDEALS
such that the quotient modules Mi/MH 1 are all simple; in other words, the inclusions Mi 2 N 2 MH 1, where N is a submodule of M,, imply that either N = M, or N = MH1. The number of submodules in a composition series is called the length of the series. Two normal series for the Rmodule M, M=M03M13~DM,,_1=M,,={0}
and M = No 3 N1 3. "'
.
D Nm_1 D N"! = {0}
are termed equivalent if they have the same length (n = m) and there exists a onetoone correspondence f between their indices such that
Mi/Mi+l 3 Nf(i)/Nf(i+1)' Expressed diﬁerently, two normal series are equivalent if their associated quotient modules are pairwise isomorphic in some order. Using these deﬁnitions, the classic JordanHolder Theorem asserts that any two composition series for an Rmodule M are equivalent and therefore have the same length; this common value is called the length of M and denoted by 1(M). In eﬂ‘ect, an Rmodule has essentially one composition series. We omit the proof, but the details can be found, for example, in the
admirable book by Northcott [28]. Example 129. In the Zmodule 224, the normal series
224 3 (2) D (12) 3 {0} is not a composition series, since it may be reﬁned by inserting either of the submodules (4) or (6). (In the situation considered, the notation (n) stands for the cyclic subgroup generated by n.) On the other hand,
224 3 (2) 3 (4) D (8) 3 {0} and
224 3 (3) D (6) D (12) D {0} both form composition series for Z24. One way to verify this is to check the orders of the subgroups involved. For instance, to insert a submodule between (2) and (4) there would have to exist a subgroup of Z24 of order n, 6 < n < 12, such that n divides 12 and is itself divisible by 6; clearly, no
such subgroup exists. To go still further we need a criterion for the existence of composition series. Theorem 1215. An Rmodule M has a composition series if and only if M satisﬁes both chain conditions for submodules.
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Proof Suppose that both chain conditions and, hence, the maximum and minimum conditions, hold in M. Applying the maximum condition to the set of submodules different from M, we can select a maximal submodule
M1 C M. Now, either M1 = {0} and the proof halts, or there exists a submodule M2 of M1 which is maximal with respect to being proper. Continuing in this way, we get a strictly decreasing chain of Rsubmodules of M: M=MODMIDM23'”.
By virtue of the descending chain condition such a chain must eventually terminate; thus, M,l = {0} for some integer n and a composition series for M is obtained. As regards the converse, we proceed by induction on the length l(M)
of M. If l(M) = 1, then M D. {0} is a composition series; hence, M is a simple module and both chain conditions hold trivially.
Next, assume
inductively that all Rmodules of length n — 1 satisfy the two chain conditions and let l(M) = n. Given any composition series for M, say 1)M=MODMID..
DMn={0},
then, upon setting N, = M,/M,,_,, the chain 2)N=NoD N1 3
3Nu—1={0}
will form a normal chain for the quotient module N = M/M,,_1. Using the ﬁrst isomorphism theorem for modules (Problem 26, Chapter 12), Ni/Ni+1 = (Mi/Mn1)/(Mi+l/Mnl) '3' Mi/Mi+1'
This implies that the module Ni/NH 1 is Simple and so the chain (2) actually comprises a composition series for M/Mrl. As a result, we are able to conclude that l(M/Mn_l) = n — 1. By our induction assumption, the quotient module M/M,,_ 1 must satisfy both the ascending and descending chain conditions. Since Mn_ 1 is a simple Rmodule, an appeal to Theorem 12—14 is legitimate; we thus deduce that M itself satisﬁes both chain
conditions for submodules. It is hardly necessary to point out that the concepts of normal series and composition series apply equally well to the ideals (the Rsubmodules) of a ring R. In what follows, whenever we speak of a composition series for a ring R, we shall mean a composition series for R considered as a module over itself. We are now in a position, having assembled the necessary mathematical machinery, to attack Hopkin’s Theorem. To set the stage for our presentation, part of the argument is separated off as two lemmas. Hereafter, R will denote a commutative ring with identity.
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Lemma 1. Suppose that in the ring R the zero ideal is a product of
maximal ideals, say {0} = M1M2 (MIMz
Mn. If
Mi—1)/(M1M2
Mi)
viewed asavector space over R/M,is ﬁnite dimensional fori = 1, 2,
, n,
then R has a composition series. Proof. We ﬁrst set N, = Ml M, for i = 1,2, ..., n. Observe that the quotient Rmodule Ni_ 1/Ni can be regarded as a vector space over the ﬁeld R/Mi. Its elements are simply the cosets x + Ni, with x e Ni_1, and scalar multiplication is deﬁned (on the right) by (x + N,)(r + M,) = xr + N,
(reR).
Since N,_ 1/Ni is annihilated by Mi, this deﬁnition makes sense; in fact, if x — x’ 6 N,, where x, x’ lie in Ni_1, and r — r’ 6 Mi, we necessarily have
xr — x’r’ = x(r' — r’) + (x — x’)r'eN,_1M, + N, 9 Ni. Let us now consider the descending chain
R=N02N12N222N,,={0} of Rsubmodules. It is well known that any ﬁnite dimensional vector space admits a composition series [33]. This being so, our hypothesis guarantees that NI_1/N, has a composition series as an R/Mimodule and, hence, as an Rmodule. (Let us stress that, by virtue of the deﬁnition of scalar multiplication, the R/Misubmodules of Ni_1/N,. are identical with the Rsub— modules of Ni_1/Ni.) Using Theorem 12—12, a composition series can therefore be inserted between Nl._1 and N,. By putting all these series together, we obtain a composition series for R itself.
Lemma 2. If in the ring R, {0} = MlM2
M", where the M, are
maximal ideals, then either chain condition implies the other. Proof. Again, let N, = Ml M, for i = 1, 2, , n and consider the quotient module Ni_l/Ni as a vector space over R/Mi. Now, Ni_1/Ni forms an Rsubmodule of R/Ni which, in turn, is a homomorphic image of R; it follows that if either chain condition on ideals holds in R, then Ni_ l/N, must satisfy the corresponding chain condition on Rsubmodules (hence, on R/Misubspaces). But, in a vector space, either chain condition implies that the space is ﬁnite dimensional [33]. From Lemma 1, the ring R thereby admits a composition series and so, with the aid of Theorem 12—15, we
conclude that both chain conditions hold in R. Here now is the main result of this chapter; our proof follows the lines of [36].
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255
Theorem 1216. (AkizukiHopkins). A ring R is Artinian if and only if R is Noetherian and every proper prime ideal of R is maximal.
Proof. We ﬁrst suppose that R satisﬁes the ascending chain condition and that every proper prime ideal of R is maximal. By Problem 4, Chapter 11, every ideal of R contains a product of prime ideals (the Noetherian hypothesis ensures this). In particular, {0} must be a product of prime, and therefore maximal, ideals.
That R is Artinian follows immediately from
Lemma 2. Conversely, let us now assume that R satisﬁes the descending chain
condition. If P is any proper prime ideal of R, the quotient ring R/P also satisﬁes this chain condition and, of course, is an integral domain. Appealing to Theorem 11—8, we see that R/P is in fact a ﬁeld, whence P forms a maximal ideal.
To prove that R is Noetherian, it is again enough to establish that {0} is a product of prime (hence, maximal) ideals. We assert ﬁrst that for every proper ideal I of R there exists a prime ideal P for which I C I : P. To see this, deﬁne the family .97 by
33" = {JIJisanidealofR;I:J 7E R}. .7 surely is not empty, because R is a member of 97. Use the minimum condition to select an ideal J’ which is minimal in this collection. Then, P = I: J’ forms a prime ideal of R. If not, there would exist elements a, b
not in P such that their product ab 6 P. Therefore, P CP:(a) C R. That is to say, I : J’ C I : J’a 9E R, whence J’a C J’. Since this contradicts the minimal nature of J’ in .97, P must indeed be a prime ideal. Now, the quotient ideal I : P 2 I; inasmuch as I : P 2 J, which is not contained in I, it follows that I C I : P and our assertion is proved. (Incidentally, this
argument proves the existence of prime ideals in R.) For the ﬁnal stage of the proof, let K be minimal in the set of those ideals of R which are products of prime ideals. If K + {0}, then the ideal I = 0: K is different from R, for 1 ¢ I. By the last paragraph, there is a prime ideal P of R such that I C I: P; in other words, 0: K C 0: KP. This implies
that the ideal KP C K and contradicts the minimality of K. In consequence,
K = {0}, so that {0} is a product of prime (maximal) ideals. Lemma 2 now completes our task. Corollary. Any commutative Artinian ring with identity is Noetherian. Having come this far, it might be useful to prove Fitting’s Lemma, a result which requires both chain conditions on submodules. First, we pause to establish a fact of independent interest.
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Lemma. Let M be an Rmodule satisfying both chain conditions and let f: M —> M be an Rhomomorphism. Then f is a onetoone function if and only iff maps onto M. Proof. To start, suppose that f is onetoone and consider the chain of Rsubmodules
M 2f(M) 2f2(M) 2 Since M satisﬁes the descending chain condition, this chain will terminate
after a ﬁnite number of steps, say n steps; then f"(M) = f”+1(M). Given an arbitrary x e M, f"(x) = f"“(y) for suitable y in M. As f is assumed to be a onetoone function, f" also enjoys this property, whence x = f(y). The implication is that M = f(M) and so f maps onto M. Next, letfcarry the set M onto itself. Notice that we have the following ascending chain of Rsubmodules:
{0} E kerfg kerf2 E By hypothesis, there exists an integer m for which ker f"' = kerf'"+1. Select any x e M with f(x) = 0. Inasmuch as fmaps onto M, so also must f"‘. Thus, it is possible to choose an element y e M such that f"'(y) = x.
But thenf"'“(y) = f(x) = 0, implying that y e kerf"+1 = ker f"'. Accordingly, x = f"'(y) = 0 and, hence, ker f = {0}. This makes f a onetoone function and we are done. The result which we have in mind is stated below.
Theorem 1217. (Fitting’s Lemma). Let the Rmodule M satisfy both chain conditions. Given an Rhomomorphism f: M —> M, there exists some n e Z + such that M =f"(M) ® kerf". Proof As observed in the proof of the lemma, we have two chains of Rsubmodules at our disposal: M2f(M) 2f2(]ll) 2 ...,
{0} Skerfg kerf2 g Because both chain conditions hold, each of these chains ultimately stops,
for instance, after r and s steps, respectively. The theorem now follows on taking n to be the larger of r and s. For, suppose that x ef"(M) n kerf"; then f"(y) = x for some y e M, while f"(x) = 0. Therefore, f2"(y) = f”(x) = 0, so that y lies in kerf2" = ker f”. But this means that x = f"(y) = 0, whence the intersection f”(M) n kerf" = {0}. Now, pick any element x e M. Since f"(x) ef"(M) = f2"(M), there
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257
exists some y e M withf"(x) = f2"(y). Thus,'f"(x — f"(y)) = 0, or, equivalently, x — f"(y) e kerf". As a result,
x =f"(y) + (x — f"(y))ef"(M) + kerf", which forces R = f”(M) + ker f". As we noted earlier, any ﬁnite dimensional vector space satisﬁes both
chain conditions (on subspaces). This being the case, Fitting’s Lemma can be interpreted in vector space terminology as asserting Corollary. Let V be a ﬁnite dimensional vector space and let f: V —> V be a linear transformation. Then V = W1 GB WZ, where W1 and W2 are both invariant subspaces under f, f W1 is nonsingular, and f[W2 is nilpotent. Proof. Take W1 = f"(V) and W2 = kerf", as indicated above. By the lemma to Theorem 12—17, the restriction f Wl being an onto mapping is also onetoone; hence, a vector space isomorphism (to put it another way, f  W1 is a nonsingular transformation). PROBLEMS
Unless indicated to the contrary, all rings considered are assumed to be commutative with identity. 1. Let I be a semiprime ideal of the ring R. Prove that I is a prime ideal if and only if it is irreducible. [Hint: If I is irreducible, but not prime, then there exist elements
a,b¢I with abeI; argue that \/I—2 (La) 0 (Lb) D 1.] 2. a) In the polynomial ring F[x], where F is a ﬁeld, show that the ideal (xz, 2x, 4) is primary, but reducible. [Hint. (x2, 2x, 4) = (x2, 2) n (x, 4).] b) Express the ideal (x2, xy, 3) as an intersection of primary ideals in F[x, y].
3. Let R be a Noetherian ring, and I and J two ideals ofR with J E I. If I = ﬂ}; 1 Q, is an irredundant primary representation of I, establish that a) I/J = ﬂg=1(Q,/J) is an irredundant primary representation of the ideal 1/] in R/J ;
b) ,/(Qi/J) = JE/J are the associated prime ideals of 1/]. 4. Find an irredundant primary representation'for the ideal (x2, 2xy) in F[x, y], F a ﬁeld; determine the associated prime ideals of (x2, 2xy), as well as its minimal
primes. [Hint: (x2, 2xy) = (x2, xy, y’) n (x) n (x2, 2x, 4).] 5. Let I be an ideal of the Noetherian ring R. Without recourse to Problem 20, Chapter 8, prove the statements below:
3.) J7 is the intersection of the minimal prime ideals of I. [Hint: If I has the irredundant primary representation I = ﬂ}; 1 Q, then J— : 0L1 JEN]
b) The set of nilpotent elements of R is the intersection of the minimal prime ideals of R.
258
FIRST COURSE IN RINGS AND IDEALS
c) JI is a prime ideal if and only if I has a single minimal prime. (1) If P is a minimal prime ideal of I, then the primary component corresponding to P is the same for all irredundant primary representations of I. 6. Let I be an ideal of a Noetherian ring R in which every nontrivial prime ideal is maximal. Show that I is a product of primary ideals. [Hint: If I = ﬂ, Q, then the ideals \/—Q_i are pairwise comaximal when nontrivial; now use Problem 13,
Chapter 10.] . Let R be a Noetherianring, and I and J two ideals of R with I aé R. Prove that
I :J = I if and only if J is contained in no associated prime ideal of I. [Hint: Assume that I = ﬂ; Q, Q primary. If J $ @ for all i, then, by Problem 24(a),
Chapter 5,Qi:J = Q... Conversely, 1etI:J = I. IN E E, then J” S NE)” E Q, for some n; whence I = IzJ" = ﬂi(Qi:J") = ﬂi+k(Q,:J") 2 ﬂiﬂ Qi 2 I.] . Given that R is a Noetherian ring, prove that a) An element a belongs to some associated prime ideal of the ideal I if and only if there exists some b ¢ I for which ab 6 I. [Hint: Apply Problem 7 to the ideal
J = (a).] b) The set of all zero divisors of R together with zero is the union of the associated
prime ideals of {0}. [Hint: Part (a) with I = {0}.] c) An ideal I of R consists entirely of zero divisors (along with 0) if and only if I
is contained in some associated prime ideal of {0}.
[Hint: Part (b) and
Theorem 5—16.]
9
Let I be an ideal of the ring R. An element a e R is said to be related to I if there
exists some r ¢ I such that ar e I. Prove each of the assertions below: a) An element a e R is related to I if and only if the quotient ideal I :(a) 7E I. b) An element a e R is related to I if and only if the coset a + I is either zero or a divisor of zero in R/I. c) Every element of the nil radical ﬂ is related to I. d) If R is Noetherian and I = (11;. Q. is an irredundant primary representation
of I, then an element aeR is related to I if and only if ae UH/E. [Hint: Problem 8(a)] 10. Assume that R is a principal ideal ring with zero prime radical. Deduce that the zero ideal is the intersection of a ﬁnite number of prime ideals. ll. Given that I is an ideal of the Noetherian ring R, establish the following: a) If] E radR, then @2111" = {0}.
b) H?“ (I + (rad R)") = I. [Hint: Apply part (a) to R/I.] c) I” + radR = R, then I = R. [Hint: R = R" =
3°=1(I + radR)" E
:21
(I + (rad R)”) = 1.] 12. Let R be a Noetherian local ring with maximal ideal M.
a) Verify that the intersection “:21 M“ = {0}. b) If I is any ideal of R for whichM = I + M2, prove thatI = M. [Hint: From M=I+M2=I+M(I+M2)=I+M3=m, deduce that M: {1:1 (I + M") = I.]
PROBLEMS
259
13. a) Derive the Krull Intersection Theorem from Theorem 12—10. [Hint: Problem
8(b).]
b) Show that the settheoretic condition JE n (1 — I) = ¢ appearing in
Theorem 12—10 is equivalent to requiring that m + I 95 R. 14. Let I be a proper ideal of the integral domain R. Assume further that, for any ideal J of R, there exists an integer k for which I" n J S U (when R is a Noetherian domain, every idea” has this property [33]). Prove that the intersection 3°: 1 I" = {0}. [Hint: Takea e 0,211 PI and consider the principal ideal J = (a).] 15. Suppose that R is a local ring whose maximal ideal M is principal, say M = (p). If M is a nil ideal of R, prove that a) M is a nilpotent ideal of R. . b) For any proper ideal I of R, I = ann (arm I). [Hint: By Theorem 12—11,
I = (pk) for some integer k.] 16. Let R be a ring possessing an ideal M which is both maximal and nil. Verify that R is a local ring with unique maximal ideal M. [Hint. If the element a 9% M, show that a is invertible by expanding (ab — 1)“.] In Problems 17—26, the term Rmodule means left Rmodule.
17. Prove the following statements concerning submodules of the Rmodules M : a) A nonempty subset N E M forms a submodule of M if and only if (i) x, y e N imply x + ye N and (ii) xeN, r ER imply that rx 6 N.
b) If S is a subring of the ring R, then every Rsubmodule of M can also be regarded as an Ssubmodule.
c) If] is an ideal of R and x a ﬁxed element of M, then the set N,‘ = {rxx e 1} forms a submodule of M.
18. a) Verify that the submodule [S] of the Rmodule M generated by a nonempty subset S _C. M consists of all ﬁnite Rlinear combinations of elements of S; that is,
[S] = {Z rixilri e R, x, e S}. b) Let f," g: M —> N be two Rhomomorphisms of the Rmodule M into the Rmodule N. Iff(x) = g(x) for every x in a nonempty subset S E M, show that
f and 9 agree on the submodule [S]. 19. An element x of an Rmodule M is said to be a torsion element if there exists some r aé 0 in R for which rx = 0. Show that the set Tof torsion elements of M forms a submodule of M and that the quotient module M/T is torsionfree (in other words, M/T has no nonzero torsion elements). 20. Let f: M —> N be an Rhomomorphism of the simple Rmodule M into the Rmodule N. Establish thatf(M) is a simple submodule of N and thatfis onetoone whenever f(M) aé {0}. 21. Let M 1, M2, , M,l be submodules of the Rmodule M. We call M the (internal) direct sum obMz, ,M,I and writeM = M1 69 M2 6 e M, if
i) M = M1 + M2 +
+ M, = {x1 + x2 +
+ x,,x,,eM,,}, and
260
FIRST COURSE IN RINGS AND IDEALS
ii) M1“ (M: + + Mk—i + Mk+1 + + Mn) = {0} for all k. Prove that M is the direct sum of M1, M2, , M,I ifand only if each x e M can be expressed uniquely as a ﬁnite sum
x = x1 + x2 +
+ x.
(xkeMk).
. Suppose that M is an Rmodule with submodules M1,M2, ...,M,I such that
M = M1 Q M2 69 N = N1 + N2 +
63 M". For eachk, let Nkbea submodule ofM, and set + N,,.Verifythat
3) N=N1$N2$'“$Nn;
b) M/N I: M1/N1 e Mz/N2 69
(B Mu/Nu, as Rmodules.
. Let M be an Rmodule. Prove the assertions below:
a) The set A(M) = {reRrx = 0 for all xeM} is an ideal ofR, known as the annihilator of M. b) M becomes an R/A(M)—module on deﬁning the module product by (r + A(M))x = rx, where r e R, x e M. c) Viewed as an R/A(M)module, M has zero annihilator. d) A nonempty subset N E M is an Rsubmodule of M if and only if it is an R/A(M)submodule of M. e) The length of M as an Rmodule is the same as its length when considered as an R/A(M)module.
. Given an Rmodule M 7E {0}, establish that a) M is a simple module if and only if Rx = M for each 0 + x e M; here the set
Rx = {rxlreR}; b) if N1, N; are submodules of M, with N1 simple, and ifN1 (1 N2 + {0}, then N1 g N,.
. Derive the Second Isomorphism Theorem for Modules: If N1 and N2 are 'two submodules of the Rmodule M, then N1/(N1 n N2) z (N1 + N2)/N2. [Hint:
Mimic the argument of Theorem 310.] Derive the First Isomorphism Theorem for Modules: If N 1 and N 2 are two submodules of an Rmodule M with N1 9 N2, then N,jN1 is a submodule of M/Nl and (M/N1)/(Nz/N1) '3 M/Nz
[Hint: Mimic the argument of Theorem 3—9.] 27. An Rmodule is said to be indecomposable if it is not the direct sum of two nonzero submodules. Let M satisfy both chain conditions on submodules and let f be an Rhomomorphism of M into itself. Prove that M is indecomposable if and only if f is either nilpotent or an automorphism. [Hint: Use Fitting’s Lemma]
Let R be a commutative ring (not necessarily with identity) and let homRR be the set of all Rhomomorphisms of the additive group of R into itself. For each a e R, deﬁne 1;: R —> R by setting 72(x) = ax. If TR denotes the set of all such functions, prove the following: a) homRR forms a ring with identity, where multiplication is taken to be functional composition;
PROBLEMS
261
b) TR is an ideal of homRR; c) the mapping f(a) = fl:I determines a (ring) homomorphism of R onto T.;
d) if for each 0 + a e R, there exists an element b e R such that ab 55 0, then R z TR (hence, R can be imbedded as an ideal in a ring with identity); e) whenever R has a multiplicative identity, then I} = homR R.
THIRTEEN
SOME NONCOMMUTATIVE THEORY
This, our concluding chapter, is designed primarily for the reader who wishes to know something about noncommutative ideal theory. It is not our intention to treat this subject in any exhaustive manner; rather, we have
concentrated on those major results which could be ﬁtted into a concise development and which do not require many specialized preliminaries (even
with this restraint, some of the theorems are fairly sophisticated). Particular effort is devoted to proving the farreaching Wedderburn Structure Theorems for nilsemisimple rings satisfying the descending chain condition on right ideals. These and other related results make intriguing use of the general
theory of idempotents, as developed in the present chapter. From this point onward, R will denote a ring with identity element, not necessarily commutative (for most of our work the assumption of an
identity is not really essential). In the previous chapters, considerable progress was made after imposing a chain condition on the ideals of the ring; this was an entirely natural procedure and it is equally expedient to do so here. To have a concise statement, we shall call a ring R right Artinian if it satisﬁes the descending chain condition on right ideals. This chain Condition admits the usual equivalent formulation: every nonempty set of right ideals of R possesses a minimal member. An important theorem of Brauer, which requires only the hypothesis that R be right Artinian, is that each nonnilpotent right ideal of R contains an idempotent element. We choose to begin our discussion with a proof of this result. Theorem 13—1. (Brauer). In a right Artinian ring R, every nonnilpotent right ideal I contains a nonzero idempotent element. Proof. The collection of nonnilpotent right ideals of R which are contained in I is not empty, for I itself is such an ideal. By the minimum condition on right ideals (equivalent to the assumed chain condition), there exists a minimal member I1 of this collection. In particular, any right ideal of R properly included in II must be nilpotent. Since Ii forms a nonnilpotent
right ideal contained in I1, it follows that If = 11. 262
SOME NONCOMMUTATIVE THEORY
263
Now, consider the family of all right ideals J of R with the properties
(i) .111 7E {0} and (ii) J E I 1. Such ideals certainly exist, for we have just seen that I1 satisﬁes the indicated conditions. From among these ideals, a minimal one can be obtained, call it J1. Using (i), there exists an element
u 7E 0 in J1 such that 1411 7E {0}. Since 1411 is a right ideal of R contained in I 1, with (“11)11 = "1% = “Ii 7E {0}, the minimality of J1 implies that M1 = J1. As a result, it is possible to ﬁnd an element a 6 II E I for which ua = u. Hence,
u=ua=ua2=m or u = ua” for all n.
The conclusion is that a is not nilpotent and, in
consequence, I cannot be a nil right ideal. The key to constructing the required idempotent is to consider the right annihilator of u in I 1, deﬁned by
A(u) = {reIllur = 0}. Notice that A(u) is a right idea] of R which is properly contained in I1, since uI1 = J1 79 {0}. By the minimality of I1, A(u) must be a nilpotent right ideal. Inasmuch as the product u(a2 — a) = 0, a2 — a lies in A(u), and, hence, is a nilpotent element of R. '
For the ﬁnal stage of the proof, we propose to show that there exists a polynomial f(x), with integral coefﬁcients, such that e = af(a) is a nonzero idempotent in I. To this purpose, suppose that (a2 — a)" = 0. Then, upon expanding, one obtains a" = a”“g(a) for a suitable polynomial g(x) e Z[x]. It follows that a" = a(a"g(a)) = a"+2g(a)2. Continuing, this process eventually leads to a" = a2"g(a)". If we now set e = a”g(a)", then surely e belongs to I, and furthermore satisﬁes the equation
62 = a2”g(a)2” = (a2"g(a)”)g(a)" = a"g(a)" = e. Were e = 0, this would mean that a" = a2"g(a)" = a"e = 0, a palpable contradiction (a being nonnilpotent from the ﬁrst paragraph). Therefore, e serves as the desired nonzero idempotent in I, which proves the theorem. The attentive reader will have noticed that by proving Brauer’s Theorem we have actually obtained a criterion for a right ideal to be nilpotent.
Corollary. Let R be a right Artinian ring. Then a right ideal I of R is nilpotent if and only if every element of I is nilpotent (that is to say, I is a nil ideal). Proof. Necessity follows from the deﬁnition of nilpotent ideal. That the stated condition is also suﬂicient is a direct consequence of the theorem and the observation that a nonzero idempotent cannot be nilpotent.
264
FIRST COURSE IN RINGS AND. IDEALS
Before progressing further, we need the following fact about nilpotent ideals, important in itself. Lemma. Let R be a ring which has no nilpotent twosided ideals, except the zero ideal. Then R possesses no nonzero nilpotent right (left) ideals. Proof. We take I to be any nilpotent right idea] of R, say I" = {0}. Since I is a right ideal, so also is R1; at the same time, R being a left ideal implies
that RI forms one too. More simply put, the set RI comprises a twosided ideal of R. Now, (RI)'l can be written as
(RI)” = R(IR)(IR) (IR)I = R(IR)”‘1I ; RI"‘11 = R1” = {0}, so that RI is a nilpotent ideal in its own right. If RI 7E {0}, a contradiction obviously ensues. Hence, we necessarily have R1 = {0} E I, making I a twosided ideal of R. Because I is nilpotent, our hypothesis guarantees
that I = {0}, which proves the contention. This lemma prompts us to make a deﬁnition. A ring R will be called nilsemisimple when it has no nilpotent ideals different from zero. By what was just proved, every nilsemisimple ring contains no nilpotent onesided
ideals, other than {0}. A word of caution: Many authors apply the term “semisimple” (standing alone) to any ring R such that (i) R satisﬁes the descending chain condition
on right ideals and (ii) R has no nonzero nilpotent ideals. The use of this nomenclature is justiﬁed by the fact that every such ring is the direct sum of ﬁnitely many simple rings (Theorem 13—3). Unfortunately, the term would cause difﬁculty in the present text, where semisimple has another
meaning.
'
We now restrict the scope of our discussion to nilsemisimple right Artinian rings. Rings satisfying these hypotheses turn out to be of great importance in the noncommutative theory and the rest of the section centers around their study. Let us also abandon, for the present, the assumption
that all rings under consideration must possess a multiplicative identity. (It will be demonstrated shortly that any nilsemisimple right Artinian ring actually does have an identity element.) The coming theorem shows that idempotent elements occur as an unavoidable part of our theory; in fact, every right ideal is principal, with an idempotent generator.
Theorem 132. Let R be a nilsemisimple right Artinian ring.
Then
any nonzero right ideal I of R is generated by an idempotent element, that is, I = eR for some idempotent e in R.
SOME NONCOMMUTATIVE THEORY
265
Prooﬁ By Brauer’s Theorem, I contains nonzero idempotent elements. For each such idempotent e, we obtain a right ideal
A(e) = {eIex = 0} of R. Use the minimum condition to select an idempotent 0 7E e e I such that A(e) is minimal in this collection. If A(e) aé {0}, then it has at least one nonzero idempotent, say e1.
Next, set e2 = e + e1 — ele.
Then
e2 6 I and is itself an idempotent element, since ee1 = 0 (e1 being a member
of A(e)). Furthermore, ee2 = e2 = e, which signiﬁes that A(ez) E A(e). The preceding inclusion is necessarily proper, for ee1 = 0, while ezel = ee1 + e} — e1(ee1) = ei = e1 99 0; in particular, we conclude that e2 9E 0. This leads to a contradiction to the minimal nature of A(e), thereby forcing A(e) = {0}. Now, for any element x e I, the product e(x — ex) = 0 and so x — ex 6 A(e) = {0}. It follows that x = ex for all x in I, and consequently, I = eI. But then I = eI E eR E I,
which yields the subsequent equality I = eR. Remark. Notice that the idempotent e acts as a left identity for the right ideal I = eR. Indeed, if x e I, then x = ey for some y e R; therefore,
ex=e2y=ey=x. The foregoing theorem allows us to gather more detailed information concerning the idempotents of R. Corollary 1. Let R be a nilsemisimple right Artinian ring. If I is any nonzero twosided ideal of R, then I = eR = Re for some unique idempotent e aé 0 lying in the center of R. Proof By the theorem, we already know that I is idempotent generated as a right ideal; for the sake of argument, suppose that I = eR, e2 = e 9E 0. Now consider the set
J = {x — xele}. Then J is a left ideal of R, with J2 S J] = J(eR) = {0}. Since R contains no nilpotent left ideals other than the zero ideal, it follows that J = {0}. As a result, we must have x = xe for all x in I, or, what amounts to the
same thing, I = Ie. Reasoning as in the theorem, this entails that I = Re. To conﬁrm that e 6 cent R, simply observe that for each choice of r e R the elements er and re both belong to I; therefore
re = e(re) = (er)e = er.
Finally, if e’ is any other idempotent generator of I, then e = ee’ = e’.
266
FIRST COURSE IN RINGS AND IDEALS
Another way of phrasing Corollary 1 is to say that any twosided ideal of R has a multiplicative identity, namely, the generating idempotent. In the light of the fact that the entire ring R is itself an ideal, we can deduce the fairly remarkable result that R must possess an identity. Corollary 2. A nilsemisimple right Artinian ring has an identity element. We continue a little further in this vein by proving
Lemma. LetRbeanilsemisimple right Artinian ring and] = eR = Re be an ideal of R, e an idempotent. Then any right (left, twosided) ideal of I is also a right (left, twosided) ideal of R. Proof. Suppose that J is an arbitrary right idea] of I, considered as a ring. Since J 9 Re, each element a 6 J can be written in the form a = re, with r e R; but then a = re = (re)e = aeeJe,
leading to the equality J = Je. Knowing this, one ﬁnds that JR = (Je)R = J(eR) = .11 E J, which makes J a right ideal of R. This last lemma is considerably deeper than it ﬁrst appears. For most purposes, its value lies in the corollary below. Corollary. Let R be a nilsemisimple right Artinian ring. rings,
Viewed as
1) each ideal of R is itself a nilsemisimple right Artinian ring, and 2) any minimal ideal of R is a simple ring.
This preparation brings us to a profound result, the First Wedderburn Structure Theorem.
As in the commutative case, the ultimate aim is to
characterize those rings under consideration as a direct sum of certain rings of known type. Theorem 133. (Wedderburn). Let R be a nilsemisimple right Artinian ring. Then R is the (ﬁnite) direct sum of its minimal twosided ideals, each of which is a simple right Artinian ring. Proof. Since the minimum condition on right ideals holds in K, it is possible
to ﬁnd a minimal twosided ideal I1 7E {0} (simply apply the condition to
the collection of all nonzero twosided ideals of R). With reference to Theorem 13—2, we know that I1 = elR = Rel, e1 being a suitably chosen idempotent in the center of R. Then 1 — el 6 cent R, from which it follows
that J1 = (l — e1)R forms an ideal of R. Now, any element x e R can be written as x = elx + (1 — e1)x, whence the relation R = I1 + J1 holds.
SOME NONCOMMUTATIVE THEORY
267
To see that this sum is actually direct, select an arbitrary x in I 1 n J1. On the one hand, x = (1 — e1)r, so that elx = 0, and, on the other hand,
x = els, implying that elx = eﬁs = els = x; thus, the element x = 0, or
equivalently, 11 n J1 = {0}. In consequence, R = I1 6 J1 is the direct sum of the ideals I1 and J1. Furthermore, the ideal I1 is simple when regarded as a ring (Corollary 2 above). The heart of our argument is the observation that the ideal J1, being an ideal of a nilsemisimple right Artinian ring, inherits these properties (as a ring). Therefore, if J1 7E {O}, the technique of the preceding paragraph may be repeated with J1 now replacing R. This yields the decomposition J1 = I2 (B J2, with J2 an ideal contained in J1. Repeating the process, we arrive at
R=h®5®m®h®h where each Ii = eiR is a simple, idempotentgenerated, minimal ideal of R. Since J1 2 J2 2 J3 2 ~, the descending chain condition on right ideals
implies that J,l = {0} for some n. That is to say, at some point R is exhibited as the direct sum
[email protected]@[email protected] To complete the proof, it remains only to show that the Ii include all the minimal twosided ideals of R. Pursuing this aim, let I 79 {0} be any minimal ideal of R. Since R admits the direct sum decomposition R = I1 ® 12 69
ED 1", we thus have
I = R1 = 111691219
91,1.
Now, each I,1 is an ideal of R which is contained in Ii. By the minimality of 1,, either 1,1 = {0} or else 1,1 = I. However, if it happened that 1,1 = {0} for every i (i = 1, 2, ..., n), then we would necessarily have
I = {0}, which is nonsense. The implication is that 1,1 = I, for some choice of i. But then I, = 1,1 g I and so the minimal nature of I forces 1,. = I, as asserted. This reasoning also allows us to conclude that the direct sum decomposition of R is unique, up to the order of occurrence of the summands. Our assertions are now veriﬁed. Since the ideals I, = eiR (i = 1, 2,, n) are the only (twosided) minimal ideals of R, we conclude that any nilsemisimple right Artinian ring R has a ﬁnite number of minimal ideals. This observation can be sharpened to a statement regarding the number of ideals of R, minimal or otherwise. 
Corollary. A nilsemisimple right Artinian ring .R has 2'I ideals for some n e Z +.
268
FIRST COURSE IN RINGS AND IDEALS
Q I", where each Ii
Proof. According to the theorem, R = I1 Q I2 Q
forms a minimal ideal of R. If I is an arbitrary ideal of R, then = IR = III 6 121 $ "‘ Q III.
As before, the ideal 1,1 is contained in I“ so that either 1,1 = {0} or [£1 = 1,. In other words, we can express I as I = I“ + Ii; + "' + Ilk’
where {in i2, , ik} is a set of distinct integers between 1 and n. It follows that there are exactly 2" ideals in R, namely, the ideals I“ + 1:2 + + Iik. In the foregoing structure theorem, we obtained a decomposition R = elR Q e2R Q
Q enR,
where each ei is a nonzero idempotent element of R. Let us next Show that the ei form an orthogonal set of idempotents, in the sense that ei ej = 0 whenever i + j. This depends on the observation that the intersection of two simple ideals is a twosided ideal and so must be zero. In the case at hand, we have
(e,R)(ejR) E e,R n ejR = {0},
fori 3E j,
which, of course, gives e, ej = 0. Now, let 1 e R be written as
l = elrl + ezr2 +
+ e,,r,,
(rieR).
Multiplying this equation by en it follows that e, = efr, = em; hence, the identity element can be expressed more succinctly as l=e1+e2++e,,. Remark. Theorem 13—3 could be used to establish that any nilsemisimple right Artinian ring R necessarily has an identity element, viz., the idempotent + en. (In the absence of an identity, the notation e = e1 + e2 + (1 — e1)R occurring in the structure theorem must be interpreted as meaning the set {r — e1rr e R}.) The reasoning proceeds along the following lines. Since R = elR Q ezR Q Q enR, each element re R can be represented asr = elr1 + l2 + + e,,r,l for suitable rl. in R. Thus, the equation er = (e1 + e2 + + e,,)(e1r1 + ezrz + + enrn) = eir1 + e§r2 +
+ eﬁru
= elr1 + ezr2 +
+ e,,r,, = r
holds, making e a left identity for R. On the other hand, consider the left
ideal I = {r — relreR}. Because R = eR, we have (r — re)R = (r — re)eR = r(e — e2)R = {0},
SOME NONCOMMUTATIVE THEORY
269
from which it follows that IR = {0}. Thus, I forms a twosided ideal of R,
with 12 2 IR = {0}. But R is hypothesized to be a nilsemisimple ring, whence I = {0}. The implication is that e also serves as a right identity for R and so a twosided identity. We pause to summarize what has been proved so far.
Theorem 134. For any nilsemisimple right Artinian ring R, there exists a decomposition R=Il®12®'[email protected]ﬂ
into minimal twosided ideals. The ideals I1, 12, by orthogonal idempotents; that is,
Ii = eiR,
Where
, I, are generated
eiej = {3i :2: :3 ,
and these are such that 1 = el + e2 + + e,,. Furthermore, the direct sum decomposition is unique apart from the order of the summands. Knowing that any nilsemisimple right Artinian ring can be represented as a direct sum of simple right Artinian rings, our problem is thus reduced to determining a satisfactory structure theory for simple rings in which the descending chain condition on right ideals holds. It will be found in due
course that such rings are isomorphic to the ring of all linear transformations on a suitably deﬁned vector space. For the present, we content ourselves with the observation that any simple ring R for which R2 7E {0} is automatically nilsemisimple. Indeed, if I is any nilpotent twosided ideal of R,
then either I = {O} or I = R.
Inasmuch as I is nilpotent, the latter
possibility implies that R" = {0} for some n; but this is ruled out by the
fact that R2 = R (since R2 is an ideal with R2 7E {0}, necessarily R2 = R). Hence, the ideal I = {0}, as required. To set the stage for our principal theorem, we next draw attention to certain relations between the structure of eR and that of eRe. One result which we have in mind is the following. Theorem 135. Let R be a nilsemisimple ring (no chain conditions) and let e 36 0 be an idempotent element of R. "Then eR is a minimal right idea] of R if and only if eRe is a division ring. Proof. Before embarking on the proof proper, we note that the set eRe forms a nonzero subring of R with identity element e. Suppose ﬁrst that eR is a minimal right ideal of R. To show that eRe comprises a division ring, it is enough to ﬁndaninverse for each nonzero element. If 0 9E ereeeRe,
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FIRST COURSE IN RINGS AND IDEALS
then ereR is a nonzero right ideal of R contained in eR.
Since eR is a
minimal right idea], we must have ereR = eR. Therefore, (ere)(eRe) = (ereR)e = eRe.
This relation implies that we can ﬁnd an element xeeRe for which (ere)x = e. Thus, every nonzero element in eRe has a right inverse with respect to the identity e. In particular, there is some y e eRe satisfying x(ere)y = e; but then e = x(ere)y = xe(ere)y = x[(ere)x](ere)y
= x(ere)[x(ere)y] = x(ere)e = x(ere). This enables us to conclude that each right inverse is also a left inverse, yielding the desired outcome. As regards the converse, assume that eRe constitutes a division ring and let I 79 {0} be any right ideal of R contained in eR. This gives eI = 1.
Notice also that Ie at: {O}; in the contrary case, I2 E IeR = {0}, which conﬂicts with our hypothesis that R has no nonzero nilpotent ideals. Accordingly, there exists an element r e I such that re 7E 0 and, since I = eI, we must have ere = re 79 0. Because eRe is taken to be a division ring, ere possesses an inverse s e eRe. But, ere e I; hence, e = (ere)s e I. This forces
eR E I and the equality I = eR follows. The above theorem is evidentally true with “right” replaced by “left” throughout; this symmetry allows us to add Corollary. Let R be a nilsemisimple ring and let 0 + e e R be idempotent. Then eR is a minimal right ideal if and only if Re is a minimal left ideal.
It is reasonable to ask whether the statement of Theorem 13—5 could be improved upon by the stipulation of a chain condition. In pursuit of an answer, we make the following deﬁnition.
An idempotent 0 7E e e R is
called primitive if e is not the sum of two orthogonal nonzero idempotents of R; that is, it is not possible to write e = u + v, where u2 = u 7E 0,
vz=vaé0anduv=vu=0. We can characterize when an idempotent element of R is primitive in terms of the idempotents of the ring eRe. To be precise:
Lemma. An idempotent 0 aé e e R is primitive if and only if e is the only nonzero idempotent in the ring eRe. Proof. Let e be primitive and assume that ere is idempotent for some r e R.
Then the element e — ere is also idempotent: (e — ere)2 = e2 — ezre — ere2 + (ere)2 =e—ere—ere+ere=e—ere.
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SOME NONCOMMUTATIVE THEORY
At the same time, ere(e — ere) = (e — ere)ere = 0.
Thus, we have
e = ere + (e — ere), where both ere and e — ere are idempotent and
orthogonal. From the primitivity of e, it may .be concluded that one of these is zero; that is, either ere = 0 or ere = e. Conversely, if e is not primitive, then we may write e = u + v, where
uandvarenonzero orthogonalidempotents. Hence,u 7E eand eu = ue = 14, which implies that the element u = eue is in eRe.
We put the ﬁnishing touches on our theory of idempotent generated minimal ideals with Theorem 136. Let R be a nilsemisimple right Artinian ring. Then an idempotent 0 7E e e R is primitive if and only if eR is a minimal right ideal of R. Proof. We begin by supposing that eR is not a minimal right ideal. Then eR properly contains a right ideal I 3E {0} of R, which is of the form I = uR, u a nonzero idempotent. Since u e eR, it follows that u = er for some re R, whence u = e2r = eu. Now, set U = ue and w = e — ue. An
easy calculation shows that e = v + w, where v and w are orthogonal idempotents :
v2 = (ue)2 = u(eu)e = uze = ue = v, w2=(e—ue)2=e2—eue—ue2+(ue)2=e—ue—ue+ue=w,
vw
ue(e — ue) = ue2 — (ue)2 = ue — ue = 0,
wv=(e—ue)ue=eue—(ue)2=ue—ue=0. It is also important to observe that v and w are both nonzero.
Indeed,
were v = 0, we would obtain the contradiction
u=u2=u(eu)=(ue)u=vu=0. 0n the other hand, suppose that w = 0, so that e = ue.
Then the right
ideal I = uR will contain the element e; this implies that I = eR, which is
impossible. Having thus represented e as the sum of two orthogonal nonzero idempotents, we infer that e is not primitive. Going in the opposite direction, we assume that the idempotent e is not primitive. Then e can be expressed as e = u + v, a sum of orthogonal nonzero idempotents.
euR E eR.
Now, u = eu = ue, whence the right ideal uR =
This inclusion is proper, since the idempotent e e eR, while
e ¢ uR. In fact, if e = urforasuitableelementrof R, thenv = ve = vur = 0,
an obvious contradiction. Therefore, {0} 7E uR C eR, so that eR cannot be a minimal right ideal of R. This completes the proof.
The two preceding theorems, taken together, yield a single result:
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FIRST COURSE IN RINGS ANDVIDEALS
Theorem 137. In a nilsemisimple right Artinian ring R, an idempotent e 75 0 is primitive if and only if eRe forms a division ring.
We continue our development after a few preparatory remarks about modules. Let M, N be nonzero right Rmodules, where R is any ring. (The reader is reminded that, whenever there is reference to a ring R, it is tacitly assumed that R possesses an identity element.) We shall hereafter use the notation homR(M, N) to denote the set of all Rhomomorphisms from M into N; in other words, the set of all mappings f: M —> N such that
(x,yeM;r6R)
f(x + y) = f(x) + f(y), f(xr) = f(X)r,
Under the usual pointwise addition of functions, homR (M, N) forms a
subgroup of the commutative group hom (M, N). Equally important is the observation that hom,R (M, M) turns out to be a subring (with identity) of the ring hom (M, M), ring multiplication being composition of mappings. In fact, for arbitrary f, g e homR(M, M), x e M and r e R,
(f  g)(xr) = f(M) — 9(xr)
= f(x)r  9007 = (f(x)  9(X))r = (f  g)(x)r, whence f — g e homR(M, M). It is just as routine that
(f° g)(xr) = f(9(xr)) = f(906)?) = f(9(x))r = (f° g)(X)r, and so fo 9 also lies in homR(M, M). One calls homR (M, M) the ring of Rendomorphisms of M, or sometimes the centralizer of the Rmodule M. The reason for this latter choice of terminology is that the elements of homR (M, M) are precisely those additive endomorphisms f of M which commute with the right multiplications determined by elements of R. Indeed, with every element r e R there is associated a mapping T; of M into itself given by 71x = xr (as an immediate consequence of its deﬁnition, T, e homR(M, M)). Then the condition f(xr) = f(x)r can be reformulated as
(f° 7306) = (71°f)(x)
(xeM),
or, equivalently, as f o T, '= T, of. It is natural to ask whether homR(M, M) can be turned into an Rmodule. In answering this question, let us suppose for the moment that R is a commutative ring and that f belongs to homR (M, M). Given a ﬁxed r in R, deﬁneﬂ by means of
U006) = f(X)r
(x e M).
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273
With this deﬁnition, one ﬁnds that homR(M, M) (regarded merely as a commutative group) becomes a right Rmodule. Clearly, ﬂ is an additive homomorphism from M into M. Moreover. for any 5 e R,
(ﬁ)(XS) = f(xS)r = f(308' = f(W8 = (fr)(x)SHere, rs = sr, because R is taken to be commutative.
Thus, fr actually
does form an Rendomorphism. To recapitulate: in the commutative case, homR (M, M) is both a ring and an Rmodule, with ring multiplication and module multiplication linked by the relation
(f°g)r =f°(gr) = (fr)°g
(ﬂgehomn(M,M); reR)
In particular, when M = R, a clear distinction must be made between the
Rhomomorphisms of R (considered as module over itself) and its homomorphisms as a ring. Before turning to the next theorem, it is necessary to recall that any right ideal I of the ring R is a right Rmodule; the module product being given by the ordinary ring multiplication xr, where x e I and r e R. Theorem 138. Let R be a simple ring with minimal right ideal I = eR, e 7E 0 an idempotent. Then homR (I, I) z eRe, viewed as rings. Proof. As remarked above, we consider I as a right Rmodule.
Thus,
fe homR(I, 1) implies that
f(a + b) =f(a) + f(b), f(ar) =f(a)r
(a, beI; rel?)
The ﬁrst task is to characterize these Rendomorphisms in some convenient way. We handle this problem as follows. Letfe homR(I, I) and the element a e I = eR, say a = er. .Then f(a) = f(er) = f(e)r, so that the function f is completely determined once the value f(e) is known. Since e is an idempotent of R, we have f(e) = f(e2) = f(e)e. But f(e)e eR, whence f(e) = es for some choice of s in R, and so f(e) = f(e)e = ese. It readily follows that
f(a) = f(e)r = (ese)r = (ese2)r = (ese)er = (ese)a. In brief, f acts on the elements of I as left multiplication by ese, where f(e) = ese. We shall utilize this information presently. Now, deﬁne a mapping ¢zhomR(I, I) —> eRe by means of the rule ¢(f) = f(e). It is immediately apparent that
¢(f + 9) = (f + g)(e) =f(e) + 9(6) = ¢(f) + ¢(y)If f is caused by a left multiplication by the element ese and g is a left multiplication by ete, then (I) enjoys the further property
¢(f ° 9) = (f ° g)(e) = 119(8)) = f(ete) = esete = (ese)(ete) = f(e)g(e) = ¢(f)¢(g).
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FIRST COURSE IN RINGS AND IDEALS
This permits us to conclude that d), at the very least, is a homomorphism from the ring homRU, I) into the ring eRe. We next proceed to show that ker 4) = {0}. If ¢(f) = 0, then f(e) = 0, which can be applied to yield
N!) = f(er) = f(e)r = 0 for any a e I = eR. This of course means that f = 0, making d) a onetoone function. Finally, we select an arbitrary x in eRe (for instance, x = ere) and deﬁne the endomorphism h e homR(I, I) by setting h(a) = xa = (ere)a, where a e I. Then we will have ¢(h) = h(e) = (ere)e = ere = x,
so that 43 maps onto eRe. The proof that 11) serves as a ring isomorphism between hOmR(I, I) and eRe is now complete. In combination with Theorem 13—5, one obtains as a corollary Corollary. If R is a simple ring with minimal right ideal I, then homR (I, 1) forms a division ring.
The foregoing corollary is a special case of a much wider theorem due to Schur. In accordance with the terminology of Section 12, we shall call an Rmodule simple provided it is nonzero and has no nontrivial submodules. Theorem 139. (Schur’s Lemma). If M is a simple Rmodule, then homR (M, M) forms a division ring. Proof. What we must prove is that any nonzero element f e homR (M, M)
has an inverse in homR(M, M). Since the image f(M) is a nonzero submodule of M, it follows that f(M) = M (M being a simple module by hypothesis) and therefore f maps onto M. In addition, f is necessarily a onetoone function. For ker f is certainly a submodule of M and cannot equal M, because f 79 0; the implication is that kerf = {0}. All told, the inverse f "1 : M > M exists. The reader is left the task of verifying that
f ' 1 is actually an Rhomomorphism. This might be the proper place to characterize all simple Rmodules of a given ring R. To this end, observe that if I is a right ideal of R, then
we can certainly form the quotient group R/I ; that is, R/I is the additive group whose elements are the cosets x + I, x e R, with addition deﬁned by
(x+I)+(y+I)=x+y+I. We can, however, do more than this. For, setting
(x+1)r=xr+I
(reR),
R/I carries a welldeﬁned structure as a right Rmodule. Indeed, the veri
ﬁcation that the module axioms are satisﬁed presents no problem and will
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275
therefore be omitted. The reader may also conﬁrm that the submodules of R/I are of the form J/I, where J is a right ideal of R and J 2 I. These remarks lead us to a precise identiﬁcation of all simple Rmodules. Theorem 1310. A right Rmodule M is simple if and only if it is Risomorphic to a quotient module R/I for some maximal right ideal I of R. Proof Assume that M is a simple Rmodule. For a ﬁxed nonzero element
x e M, deﬁne the right ideal I by I = {a e Rxa = 0}. To see that I is maximal, take J to be any right ideal of R satisfying I C J E R. Now, the set xJ forms a nonzero submodule of M. From the supposition that M is simple, it follows that ad = M = xR. (Actually, any element 0 7E y e M is cyclic in the sense that yR = M.) Thus, for arbitrary r e R, there exists some b e J such that xb = xr. But then, x(r — b) = 0, giving r — b e I E J; this says that r 6 J, whence the equality J = R, and I is a maximal right ideal of R. Next, consider the module homomorphism f: R > M in which f(a) = xa. The elements of the kernel offare precisely the elements of I. Furthermore, the condition xR = M assures us that f maps onto M and so, invoking Theorem 12—13, there is induced an Risomorphism R/I 2 M. The other direction of the theorem relies on the fact that R/I is a simple right Rmodule if and only if I is a maximal right ideal of R; the reader may supply the necessary details. If M is a right Rmodule, then the set
A(M) = {r e RMr = {0}} is called the annihilator of the module M. As is easily veriﬁed, A(M) forms a twosided ideal of R. We shall refer to M as a faithful Rmodule or speak of R as acting faithfully on M whenever A(M) = {0}; that is to say, whenever Mr = {0} implies that r = 0. Since the ring R is assumed to possess an identity element, R is evidently faithful as a module over itself. To round out our studies, let us introduce a notion of radical which is
meaningful for general rings and which agrees with the previously deﬁned Jacobson radical in the event that the ring is commutative. The deﬁnition we give is phrased in terms of annihilators of modules. Deﬁnition 131. The radical of the ring R is the set J(R) = n A(M), where the intersection varies over all simple right Rmodules M; if there are no simple right Rmodules, put J(R) = R.
In short, the radical of a ring R may be described as the annihilator of all simple right Rmodules. Notice, too, that J(R), being the intersection of twosided ideals of R, itself is an ideal of R. A ﬁnal point to which attention should be drawn is that if R happens to have a faithful simple right R
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FIRST COURSE IN RINGS AND IDEALS
module, then J(R) = {0}. (This observation is made in preparation for Theorem 13—13.) The relevance of J(R) to some of our earlier investigations is clariﬁed by the next theorem. What one can prove is Theorem 1311.~ In a ring R with identity, J(R) = n {II is a maximal right ideal of R}. Proof If the element reJ(R), then r annihilates every simple right Rmodule. In particular, for any maximal right idea] I of R, the quotient module R/I forms a simple Rmodule (we remark that maximal right ideals exist by Zorn’s Lemma). Hence, (R/I)r = {0} or, in other terms, Rr 9 I. Since R possesses an identity element, it may be inferred that r = 1r 6 1. Thus, r lies in every maximal right ideal of R and (with selfexplanatory notation) J(R) E n I. For proof of the opposite inclusion, let reK = n I, where I runs through all maximal right ideals of R. We contend that the right ideal (I — r)R = R. In the contrary case, (1 — r)R would be contained in some
maximal right ideal I’ ofR; from re K S 1' follows 1 = (1 — r) + re 1’, which is impossible. The next step is to consider any simple right Rmodule
M. If Mr 7E {0}, then xr qé 0 for a suitable element x of M. Since xK is a submodule of M, this yields xK = M. Accordingly, there exists an element 5 e K for which xs = x. As was pointed out a moment ago, 5 e K secures that (l — s)R = R; hence, (I — s)t = —s for some t in R. This
relation gives 0=x(s+t—st)=xs+xt—xst=x+xt—xt=x,
leaving us with a contradiction. The implication is that Mr = {0} for every simple Rmodule M. Thus, the element r e J(R), which settles the proof. Remark. For the reader’s guidance, it needs to be stated that, when R lacks
an identity element, the analog of Theorem 13—11 asserts that J(R) is the intersection of the modular maximal right ideals of R. (Of course, in a ring with identity, all ideals are trivially modular.) Furthermore, the intersection of the modular maximal right ideals of R always coincides with the intersection of all modular maximal left ideals [10]. In order to extend Schur’s Lemma, we introduce a further concept,
which, to begin with, may appear unrelated. Let (G, +) be a commutative group having more than one element and let E(G) denote the collection of endomorphisms of G. (Remember that by an endomorphism of G is meant a homomorphism of G into itself.) Under the operations of pointwise addition and functional composition, E(G) forms a ring with identity. Any
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277
subring S of the ring E(G) is called a primitive ring of endomorphisms of G if for all x, y e G, with x + 0, there exists some fe S such that f(x) = y. As regards notation, we shall hereafter write C(S) fer the centralizer of S
in E(G); that is, C(S) consists of those endomorphisms of G which commute, under composition, with every member of S. It is readily veriﬁed that C(S) is a subring of E(G) containing the identity endomorphism. We are now ready to prove the following theorem. Theorem 1312. Let G be a commutative group with more than one element. If S is a primitive ring of endomorphisms of G, then the centralizer C(S) forms a division ring. Proof Given that 0 7E fe C(S), let the element x e G be such thatf(x) 7E 0. For each y e G, the primitivity of S assures us that there exists an endo
morphism g e S satisfying g(f(x)) = y; hence,
f(9(x)) = g(f(x)) = yAs this holds for arbitrary y in G, we conclude that f maps G onto itself.
Now, assume thatf(x1) = f(x2), Where x1, x2 6 G, so thatf(x1 — x2) = 0. If the element x1 — x2 7!: 0, then for any y e G we can choose some h in S with h(x1 — x2) = y. (Once again, this is possible by the primitivity of S.) But then,
ﬂy) = f01061  362)) = h(f(x1  x2)) = W) = 0, which means thatf = 0, an obvious contradiction. In other words, we must
have x1 — x2 = 0, or, rather, x1 = x2, ensuring that the mapping f is a onetoone function. As a result, there exists a welldeﬁned multiplicative inverse f '1: G —> G, which is also an endomorphism of G. For any 9 e S,
f_l°(.¢l°f)°f_l =f_1°(f°g)°f_1, leaving us with f ‘1 o g = g of ‘1; therefore, f"1 e C(S) and all is proved. Note that this theorem actually does generalize the one due to Schur. For, let M be any simple right Rmodule. In the ring E(M) of endomorphisms of the additive group of M, consider the subring S of right multiplication functions induced by elements of R:
S = {7;reR; T,(x) = xrforxeM}. As was seen earlier, the oentralizer ofS in E(M) is exactly the ring homR (M, M) of Rendomorphisms of M. We contend further that S is a primitive ring of endomorphisms of the module M.
Indeed, if x, y e M, where x + 0,
then the set {T,xr e R} forms a nonzero submodule of M and so must be M itself; this guarantees that T,x = y for some choice of r e R. As the conditions of Theorem 13—12 hold, we ﬁnd that homR (M, M) = C(S) is a division ring.
FIRST COURSE IN RINGS AND IDEALS
278
It is appropriate to call attention here to a very special class of rings which are called primitive rings. A primitive ring is, by deﬁnition, isomorphic to a primitive ring of endomorphisms of some commutative group. That any division ring D is primitive should be clear. (The required primitive ring of endomorphisms is just the ring of right multiplication functions on the additive group of D.) One frequently classiﬁes rings according to the modules which they admit; in this scheme, the primitive rings are perhaps the simplest type, for these admit a faithful simple module. Theorem 1313. A ring R is primitive if and only if it has a faithful simple Rmodule. Proof. Let M be any faithful simple Rmodule. Since M is simple, we have M = xR for each 0 aé x e M. The force of this observation is that the ring of right multiplications T,, r e R, forms a primitive ring of (group) endomorphisms of M. Furthermore, the supposition that M is faithful implies that T, = 0 if and only if r = 0. The mapping p: r —> T, thus determines a ringisomorphism of R onto a primitive ring of endomorphisms of M, giving rise to the conclusion that R is a primitive ring.
On the other hand, suppose that R is isomorphic by means of the mapping r —> r’ to a primitive ring R’ of endomorphisms of a commutative group M.
Deﬁne a module structure on M by setting xr = r’(x), where
x e M, r e R. It follows at once that Mr = {0} if and only if r’(x) = 0 for all xeM; that is to say, if and only if the endomorphism r’ = 0, or, equivalently, r = 0.
Thus, R acts faithfully on M as a right Rmodule.
To see that the module in question is simple, notice that the primitivity of
R’ implies that M = {r’(x)r' e R’} for each nonzero x e M; this in its turn tells us that M = xR for any 0 7E xe M, which makes M a simple Rmodule.
Note that as an immediate consequence of the theorem we get
Corollary. For any primitive ring R, the radical J(R) = {0}. Now that Theorem 13—13 is available, let us make immediate use of it
to prove that a commutative primitive ring R is a ﬁeld. (Theorem 13—12 could just as well be employed.) To set this fact in evidence, suppose that 0 7E a e R and that M is any faithful simple right Rmodule. Then there exists some x e M such that xa + 0 and, as a result, xaR = M. It is there
fore possible to select an element b in R satisfying xab = x. The commutativity of R now implies that, for every r e R, xr(ab) = x(ab)r = xr. But xR = M, whence Mab = M; in other terms, M(ab — 1) = {0}. Since M is a faithful Rmodule, this gives ab = 1, so that a is an invertible element of R.
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279
It might be of interest to present a somewhat different proof of the fact that a commutative primitive ring R forms a ﬁeld. As before, let M be any faithful simple right Rmodule. Theorem 13—10 asserts that M : R/I, viewed as Rmodules, for some maximal (twosided) ideal I of R. Now, if . reIandaeR,then (a+I)r=ar+I=I, which says that the element r e A(R/I). As a result, we deduce that
I ; A(R/I) = A(M) = {0}. Since {0} is a maximal ideal of the ring R, it follows that R must be a ﬁeld. In this connection, we should observe that every simple ring R (hence, any ﬁeld) is primitive. For, if I is a maximal right ideal of R, then R/I is a simple right Rmodule. Its annihilator A(R/I) forms a twosided ideal of R, which is necessarily proper since 1¢A(R/I). By hypothesis, we thus have A(R/I) = {0}, so that R/I is a faithful simple Rmodule; this assures that R is a primitive ring. The theory of primitive rings is extensive but somewhat specialized from our present point of view. We shall therefore break off the discussion in order to turn to more important matters. Let M 9E {0} be a simple right Rmodule, where R is a given ring with identity, and consider the division ring D = homR(M, M) of all Rendomorphisms of M. For any fe D, we can deﬁne a (left) module product fx by fx = f(x), where x e M. With this deﬁnition M becomes in a natural way a left Dmodule (that is, a vector space over the division ring D); the veriﬁcation is routine. Bearing this in mind, we next present a technical lemma, the value of
which will become clear as we proceed. Lemma. For any ﬁnite set x1, x2, , x,' in M, linearly independent with respect to D, there exists an element a e R, such that
xla =
= x,,_1a = 0,
xna 75 0.
Proof The proof argues by induction of n. To get the induction started, take n = l; in this case, we have simply to establish that l 7E {0}.
Suppose that it happened that l = {0}. Since x1 aé 0, the set
N = {x e MxR = {0}} would form a nonzero Rsubmodule of M. Taking stock of the hypothesis
that M is simple, we infer that N = M. It then follows that MR = {0}, which is deﬁnitely impossible when R has an identity.
Now, assume inductively that n > 1 is arbitrary and that the assertion of the lemma is already proved for n — 1. Put
I ={aeRlx1a =
= u_2a = 0}.
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FIRST COURSE IN RINGS AND IDEALS
Then I is a right ideal of R and so xn_1I constitutes an Rsubmodule of M.
By the induction hypothesis, xrll + {0}, which implies that x” 1I = M (M, of course, being simple). We wish to establish the existence of some element a e I with xn_1a = 0, while xua 7E 0. Let us assume that this is not the case; in other words, for all aeI, whenever xrla = 0, then necessarily xna = 0. To derive a contradiction from this last sentence, we
tentatively deﬁne a mappinge —> M by setting f(xu1a) = xna
(“ED
and proceed to show that fis an Rendomorphism of M. First, one should check to see that f is welldeﬁned. If xn_1a = xn_1b for two elements a, b e I, then xn_ 1(a — b) = 0. But the assumption made above assures us that, in this event, xn(a — b) = 0, whence xna = xnb, producing a welldeﬁned function. The following Simple calculation conﬁrms that f is an endomorphism of M: f(xu1a + xn—lb) =f(xnl(a + b»
aa+w = xna + xnb =f(xn1a) +f(xn1b)'
Using the fact that I is a right ideal of R, we note further that if a e I and r e R, then f(xn—1ar) = xu(ar) = (xna)r = f(xnla)r'
By these considerations, f becomes an Rendomorphism of M; that is, a
member of the division ring D. This means that the module product fxr 1a = xua for all a e I. Thus, for each a in I, we have
3‘1“ =
= xn—za = (fxnl _ x».)“ = 0
Again, from our induction supposition that the lemma holds for n — 1 elements, it follows that
x1, x2, waxn—za fxnl _ xu
(0 #feD)
must be linearly dependent over D. As a result, the elements x1, x2,
, x"
are also dependent and we arrive at the desired contradiction. Armed with this rather intricate machinery, we can now derive the fundamental structure theorem for simple right Artinian rings (the socalled Second Wedderburn Structure Theorem). From the many ways of proving this result, we select a moduletheoretic approach essentially due to
Henderson [40].
SOME NONCOMMUTATIVE THEORY
281
Theorem 1314. (WedderburnArtin). Let R be a simple right Artinian ring with identity. Then R I: homD (I, I) (thought of as rings) for some ﬁnite dimensional vector space I over a division ring D. Proof. By virtue of the minimum condition, as applied to the collection of all nonzero right ideals of R, there exists a minimal right ideal I of R; furthermore, I is of the form I = eR, e a nonzero idempotent. The next
step is to observe that, since (eRe)eR = eR, the ideal I = eR can be viewed as a left vector space over the division ring D = eRe. (Theorem 13—5 afﬁrms that the set eRe comprises a division ring.) Now, let Rﬂ denote the set of all right multiplications ’12: I —> I determined by elements of R; that is,
Rn = {'1;aeR; ﬁx = xa fore}. There is no special difﬁculty in verifying that R" is a subring of the endomorphism ring homD(I, I). At the same time, the correspondence a —> ’1; sets up in a natural way a nonzero homomorphism of R onto Rn, whose kernel is a twosided ideal of R. Inasmuch as R is assumed to be simple, this kernel is {0}, whence the isomorphism R 2 Rﬂ follows. The main contention of our proof is that R" = homD(I, I). To settle
‘this point, note ﬁrst that the set
ReR = {Z aiebilai, bi e R}
(here 2 represents an arbitrary ﬁnite sum) constitutes a twosided ideal of R different from zero, since 0 sé e = lel e ReR.
Hence, we must have
ReR = R. In particular, the identity 1 e ReR, so that it is possible to select elements ai, bie R satisfying 1 = Zaiebi. Now, choose any Dendomorphism fe homD(I, I) and any element x = ere I. A straightforward computation gives
f(x) = W1) = f(er 2 aieb.) = f(Z eraieb.) = Zf(eraiebi)
= Z eraief(ebi)
(since eraie e D)
= er 2 aief(ebi) = x Z aief(3171)From this formula, it appears that f(x) = Tsx, where the element 3 (which does not depend on x) is given by s = Z a,ef(eb,.). Therefore, fe R", conﬁrming that R" = homD(I, I). Putting our remarks together, we obtain the isomorphism R 2 RI, = homD(I, I).
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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 x1, x2, of linearly
independent elements. That is, for each integer n, the set {xv x2,
, ll}
is independent with respect to D. Given an integer n, deﬁne
In = {aeRla =
= xna = 0}.
Then 11 2 I2 2 2 I,l 2 forms a descending chain of right ideals of R. Since the ideal I = eR can be regarded as a simple (right) Rmodule, a direct appeal to the last lemma is permissible. Thus, there exists some element a e R for which xla =
= xn_1a = 0,
xna 75 0
leading to the conclusion that In is properly contained in 1,,_ 1. The point which we wish to make is that I1 3 I2 3 D I,' D is a strictly descending chain, in violation of the assumption that R is a right Artinian. Accordingly, the dimension dimDI < 00 and this ﬁnally ends the proof of Theorem 13—14.
If V is an ndimensional vector space over the division ring D, then the ring homD(V, V) is wellknown to be isomorphic to the ring Mn(D) of n x n matrices over D. To spell out some details, let Vhave the basis {x1, x2, , xn}. Iffe homD(V, V), each of these basis elements will be mapped by f into a uniquely determined linear combination of x1, x2,
, x". In other words,
there exist elements an. e D, uniquely deﬁned by f: such that f(x) can be expressed in the form f(x,) = Z": ailx,
(i = 1,2, ...,n).
(Observe that the summation is over the ﬁrst index.) Thus, to each endomorphism fe homD(I/; V) there corresponds a unique n x n matrix (a,,) with entries from D. There is no problem in showing that the map f —> (aij) yields a oneto
one function from homD(V, V) onto the matrix ring Mn (D). Indeed, starting with an arbitrary (a,J.) 6 Mn (D), one deﬁnesaDendomorphismfe homD(V, V) by ﬁrst setting f(x) = 22;, ailx, and then extending linearly to all of V; it is evident that (aij) is precisely the matrix identiﬁed with the resulting endomorphism. If (an) and (bu) are the matrix representations of two elements of homD(I/; V), say f and g, then
(f + g)(x,) = f(x,) + g(x,) = Z aijx. + 2 b.,x. = 2(a + bum and
PROBLEMS
283
(f° g)(xj) = 41:21 bik) = 1:11 bkjf(xk)
= kgl bkj i; aikx‘ =
i= =1 i1(1:i
dub”)
These relations make clear that
f+ g " (at) + by) = (aij) + (by), and, by deﬁnition of the product of two matrices, that f° g "( 2": aikbkj) = (aij)(bij)
The conclusion is that the mapping which associates with eachfe homD(V, V) its matrix representation (av) (relative to the ﬁxed basis) induces a ring isomorphism
Mmmmzmm. On the strength of these remarks, our various results may be collected to give a description of nilsemisimple right Artinian rings in terms of matrix rings. Theorem 1315. Let R be a nilsemisimple right Artinian ring. Then there exist division rings D, and suitable integers ni (i = 1, 2, such that
R z MM(D1) GB Mn2(D2) Q
, r)
GB Mnr(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 [15] 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 diﬁiculties.
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 twosided ideal I of R, show that the subring eIe = I n (eRe).
2. If R is a right Artinian ring, prove that whenever there exists two elements a, b e R
with ab = 1, then be = 1. [Hint: Consider the descending chain bR 2 b2R 2 2 b"R 2
of right ideals of R.]
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FIRST COURSE IN RINGS AND IDEALS
3. Given a right Artinian ring R, establish that a) The sum N of all nilpotent right ideals of R is again a nilpotent right ideal. [Hint: If N is not nilpot‘ent, it contains a nonzero idempotent by Theorem 13—1.] b) N forms a twosided ideal of R. [Hint: The ideal RN is such that (RN)’l s RN",
whence RN S N.] c) The ideal N contains any nilpotent left ideal I of R. [Hint: I S RN E N.]
. Let I be a minimal right ideal of the ring R. Show that either 12 = {0} or I2 = I ; in the second case, deduce that I = eR = eI for some idempotent e + 0 in I. [Hint: See the proofs of Theorems 13—1 and 13—2; ﬁrst, establish the existence of
an element a e I for which a1 = 1.] . Assume that I and J are two minimal right ideals of the ring R which are isomorphic as right R—modules. If I2 = I, prove that a) Any Risomorphism f: I —> J is of the form f(x) = ax for some a 6 J. b) The product JI = J. [Hint: By part (a), J = a1, where a eJ.] . For nonzero idempotents e, u of the ring R, prove that a) eR = uRifandonlyifeu = uandue = e. b) eR and uR are isomorphic as Rmodules if and only if there exist x, y e R such that xy = u and yx = e. [Hint: Since xe = ux and ya = ey, the function f(er) = x(er) = u(xr) deﬁnes an Risomorphism f: eR —> uR, with f "1(us) =
you) = e.] c) eR z uR, as Rmodules, if and only if Re 2 Ru.
. Let eR and uR be two minimal right ideals of the ring R, where e and u are nonzero idempotents of R. Show that eR and uR are Risomorphic if and only if their product uReR + {0}. [Hint. If we aé 0 for some r e R, deﬁne the Risomorphism f: eR —> uR by f(es) = (ure)s.] . Let R be a nilsemisimple ring without identity. Ifthe element a e R and aR = {0},
establish that a = 0.
A(R)2 ; A(R)R = {0}.]
[Hint: The ideal A(R) = {reRrR = {0}} satisﬁes
. Establish the statements below:
a) A right 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 idea] I of the ring R is a direct summand of R if and only if I2 qé {0}. 10. Prove that a right Artinian ring R is nilsemisimple if and only if I2 aé {0} for each minimal right ideal I of R. 11. Assuming that R is a nilsemisimple right Artinian ring, verify the following ' 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 onetoone correspondence between the set of all idempotent elements e 6 cent R and the set of twosided ideals of R. ' c) For any twosided ideal I of R, ann,I = ann,I and so R = I 6 ann I.
(1) 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 ﬁnite direct sum of right ideals Ii + {0}, say R = 11 9 I2 $ 6 In. If 1 = el + e, + + e", where eieIi, prove that the elements ei form a set of orthogonal idempotents and that Ii = eiR (i = 1,2, , n). b) If the ideals Ii of part (a) are all twosided, show that e, 6 cent R and so serves as an identity element for 1,. 13. a) Prove that an idempotent e 5!: 0 of the ring R is primitive if and only if R ‘ contains no idempotent u =ﬁ e such that eu = ue = u. b) Establish that any idempotent element e 3!: 0 of a nilsemisimple right Artinian ring R is the sum of a ﬁnite number of orthogonal primitive idempotents.  [Hint: There exists a minimal right ideal I g eR. Write eR = I EB J, where
the right ideal J g I. Now, either J = {0}, or else e = e1 + 22, with e1 6 I a primitive idempotent. If e2 6 J is not primitive, repeat this process as applied to J = ezR.] 14. a) If M 7E {0} is a right Rmodule, verify that M becomes a left homR(M, M)module on deﬁning the module product fx by
fx =f(x)
(fehomR(M,M);xeM).
b) Let M and N be right Rmodules which are Risomorphic under the mapping azM —> N. Show that homR(M, M) a homR(N, N), as rings, by means of the
function that carries fe homR(M, M) to a o fo a". 15. Let F be a ﬁeld and M2(F) denote the ring of 2 x 2 matrices over F. Prove that 3) The matrices e1 = ((1) g and e2 = 3 (1)) are orthogonal idempotents; b) It = eiMz(F)(i = 1, 2)isaminima1rightidealos(F),withM2(F) = 11 69 12; c) eiM2(F)ei : 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 nilsemisimple.
[Hint: Problems 19 and 20, Chapter 9.] 18. Let M aé {0} be a simple right Rmodule 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 x1, x2, , x,I e M which are linearly independent with respect to D and arbitrary y1, y2, , y” e M, there exists some element a e R (equivalently, some Dendomorphism I}, e R“) such that xka = y,‘ for k = 1, 2,
, n. [Hint: From the lemma preceding Theorem 13—14, it is possible
to choose elements a; e R for which =0forj+i
x’a‘7é0forj=i' Let rt 6 R be an element such that xiairi = yi. Now, consider a = 2L1 airp]
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FIRST COURSE IN RINGS AND IDEALS
19. Given a right Rmodule M, set M* = homR(M, R) Prove the statements below: a) M“ can be made into a left Rmodule (known as the dual module of M) by deﬁning the module product If as
(If)(x) = ,106)
(r6R,xeM).
b) When R is a division ring, so that M* forms a vector space over R, then
[M*:R] = [MzR]. [Hint:Ifx1,xz,...,xnisabasisforMJhenthenfunctions f5“ eM* (i = l, 2,, n) prescribed by lfori = ‘
ﬁx!) = ‘5‘1‘ ={0 for i +5
serve as a basis for M*.] c) M and M** are naturally isomorphic as right Rmodules, where M“ = (M*)* = homR(M*, R). [Hint: For each x e M, deﬁne 32 e M** by 55(g) = g(x), g e M*; then M 2 M“ under the Risomorphism that sends x to 3.]
. Let M be a right Rmodule and I a right ideal of R. Prove that
a) The set MI = {Z x,r,x,.eM, rieI}, where 2 is an arbitrary ﬁnite sum, constitutes a submodule of M. b) If M and I are both simple (as Rmodules) with M1 aé {0}, then M 2 I. [Hint: Since x1 = M for some xeM, an Risomorphism f: I —> M can be given by f(a) = xa, where a e I.] 21. Let R be a nilsemisimple right Artinian ring. Verify that, up to Risomorphism, there exist only a ﬁnite number of simple right Rmodules. [Hint: R is a direct sum R = I1 G 12 e e In of ﬁnitely many twoSided ideals. If M =)é {0} is any simple right Rmodule, use Problem 20(b) to conclude that M 2 Ii for some i.] 22. Prove that the ring R is Simple if and only if every Simple right Rmodule is faithful. [Hint: If R has nontrivial twosided ideals, it possesses a maximal one I by Zorn’s Lemma. Let J be any maximal right idea] of R with J 2 I and obtain a contradiction by considering the simple right Rmodule A(R/J) 2 1.] . Prove each of the following assertions: a) The radical J(R) = ﬂ (I :,R), where the intersection runs over all maximal right
ideals I of R. b) Any nonzero ideal ofa primitive ring ofendomorphisms 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 I such that
the quotient ideal (I :,R) = {0}. 24. An ideal I of the ring R is said to be a primitive ideal if R/I is a primitive ring. Establish that the radical J(R) can be represented as J(R) = H P, where the intersection is taken over all primitive ideals P of R. [Hint: If P = A(M), where M is a simple Rmodule, then M is a faithful simple module over the ring R/P.]
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 ofparticular 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 settheoretic language as Deﬁnition Al. A (binary) relation R in a nonempty set S 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 Ybe a given mapping. Take for ~ the relation a ~ b if and only iff(a) = f(b); then ~ is an equivalence relation in X, called the equivalence relation associated with the mappingf. More generally, if .9; is an arbitrary family of functions from X into Y; an equivalence relation
can be introduced in X by interpreting a ~ b to mean f(a) = f(b) for every fe 3". (The underlying feature in the latter case is that any intersection of equivalence relations in X is again an equivalence relation, for N = ﬂfef {(a, b)f(a) = f(b)})
One is often led to conclude, incorrectly, that the reﬂexive property is redundant in Deﬁnition A—2. The argument proceeds like this: if a ~ b, then the symmetric property implies that b ~ a; from a ~ b and b ~ a, together with the transitivity of ~ , it follows that a ~ a. Thus, there appears to be no necessity for the reﬂexive condition at all. The ﬂaw in this reasoning lies, of course, in the fact that, for some element a e S, there may not exist
any b in S such that a ~ b. As a result, we would not have a ~ a for every member of S, as the reﬂexive property requires. Any equivalence relation ~ determines a separation of the set S into a collection of subsets of a kind which we now describe. For each a e S,
let [a] denote the subset of S consisting of all elements which are equivalent
to a:
[a] = {beSb ~ a}.
This set [a] is referred to as the equivalence class determined by a. (The reader should realize that, in general, the equivalence class [a] is the same
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289
as the class [a’] for many elements a’ e S.) As a notational device, let us henceforth use the symbol S/~ to represent the set of all equivalence classes of the relation ~ ; that is,
S/~ = {[a]ae 5}Some of the basic properties of equivalence classes are listed in the theorem below. Theorem A—l. Let ~.be an equivalence relation in the set S. Then, 1) for each a e S, the class [a] 75 Q ;
2) if be [a], then [a] = [b]; in other words, any element of an equivalence class determines that class;
3) foralla,beS, [a] = [b]ifand onlyifa ~ b; 4) for all a, b e S, either [a] n [b] = Q or [a] = [b];
5) U [a] = S. was
Proof. Clearly, the element as [a], for a ~ a. To prove the second assertion, let b e [a], so that b ~ a. Now, suppose that x e [a], which means x ~ a. Using the symmetric and transitive properties of ~, we thus
obtain x ~ b, whence x e [b]. Since x is an arbitrary member of [a], this establishes the inclusion [a] s [b]. A similar argument yields the reverse inclusion and equality follows. As regards (3), ﬁrst assume that [a] = [b]; then a_e [a] = [b] and so a ~ b. Conversley, if we let a ~ b, then the
element a e [b]; hence, [a] = [b] from (2). To derive (3), suppose that [a] and [b] have an element in common,
say, c e'[a] n [b]. Statement (2) then informs us that [a] = [c] = [b]. In brief, if [a] n [b] 75 Q, then we must have [a] = [b]. Finally, since each class [a] E S, the inclusion u {[a]a e S} E S certamly holds. For the opposite inclusion, it is enough to show that each element a e S belongs
to some equivalence class; but this is no problem, for a e [a]. As evidenced by Example A—4, any mapping determines an equivalence relation in its domain. The following corollary indicates that every equivalence relation arises in this manner; that is to say, each equivalence
relation is the associated equivalence relation of some function. Corollary. Let ~ be an equivalence relation in the set S. Then there exists a set T and a mapping 9: S —> T such that a ~ b if and only if
g(a) = 907)Proof. Simply take T = 8/ ~ and g: S —> T to be the mapping deﬁned by g(a) = [a]; in other words, send each element of S onto the (necessarily unique) equivalence class to which it belongs. By the foregoing theorem, a ~ b if and only if [a] = [b], or equivalently, g(a) = g(b).
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FIRST COURSE IN RINGS AND IDEALS
Example A—5. This example is given to illustrate that any mapping can be written as the composition of a onetoone function and an onto function. Letf: X —u Ybe an arbitrary mapping and consider the equivalence relation ~ associated with f. If the element a e X, then we have
[a] = {b E Xf(b) = f(a)} = f'1(f(a)). In effect, the equivalence classes for the relation ~ are just the inverse
images off ‘1(y), where y e f(X) E X Now, deﬁne the functionj—"z X/~ —> Yby the rulef([a]) = f(a). Since [a] = [b] if and only iff(a) = f(b),fis welldeﬁned. Observe that whereas the original function f may not have been onetoone, f happens to be onetoone; indeed, f([a]) = f([b]) implies that f(a) = f(b), whence [a] = [b]. At this point, we introduce the onto function g: X —> X/ ~ by
setting 9(a) = [a].
Then f(a) = f([a]) = f(g(a)) = (f o g)(a) for all a in
X, in consequence of which f = f0 g. This achieves our stated aim. We next connect the notion of an equivalence relation in S with that of a partition of S. Deﬁnition A3. By a partition of the set S is meant a family 9’ of subsets . of S with the properties
1) Q 915 9', 2) for any A, B e 9’, either A = B or A n B = Q (pairwise disjoint),
3) u g» = s. Expressed otherwise, a partition of S is a collection 9’ of nonempty subsets of S such that every element ofS belongs to one and only one member of 9’. The set Z of integers, for instance, can be partitioned into the subsets
of odd and even integers; another partition of Z might consist of the sets of positive integers, negative integers and {0}. Theorem A—1 may be viewed as asserting that each equivalence relation ~ on a set S yields a partition of S, namely, the partition S/~ into the equivalence classes for ~. (In this connection, notice that, for the equiva
lence relation ofequality, the corresponding classes contain only one element each; hence, the resulting partition is the ﬁnest possible.) We now reverse the situation and show that a given partition of S induces an equivalence relation in S. But ﬁrst a preliminary lemma is required. Lemma. Two equivalence relations ~ and ~’ in the set S are the same if and only if S/~ and S/~’ are the same.
Proof. If ~ and ~’ are the same, then surely S/~ = S/~’. So, suppose that ~ and ~’ are distinct equivalence relations in S. Then there exists a pair of elements a, b e S which are equivalent under one of the relations,
RELATIONS
291
but not under the other; say (1 ~ b, but not a ~’ b. By Theorem A—l, there is an equivalence class in S/ ~ containing both a and b, while no such class appears in S/~’. Accordingly, S/~ and S/~’ differ. Theorem A2. If 9 is a partition of the set S, then there is a unique equivalence relation in S whose equivalence classes are precisely the members of 9. Proof. Given a, b e S, we write a ~ b if and only if a and b both belong to
the same subset in 9. (The fact that 9 partitions S guarantees that each element of S lies in exactly one member of 9.) The reader may easily check that the relation ~, deﬁned in this way, is indeed an equivalence relation in S.
Let us prove that the partition 9 has the form S/ ~. If the subset P e 9, then a e P for some a in S. Now, the element b e 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 9, it follows that 9 E S/~. On the other hand, let [a] be an arbitrary equivalence class and P be the partition set in 9 to which the element a belongs. By similar reasoning, we conclude that [a] = P; hence, S/~ E 9. Thus, the set of equivalence classes for ~ coincides with the partition 9. The uniqueness assertion is an immediate consequence of the lemma. To summarize, there is a natural onetoone 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 socalled 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 A4. 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 S to represent a partial order relation, writing a, S b in place of aRb; the foregoing axioms then read: (1) a S a, (2) if a S b and b S a, then a = b, and (3) if a S b and b S c, then a S c. As a linguistic convention, let us also agree to say (depending on the circumstance) that “a precedes b” or
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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 S b and a 7E b. By a partially ordered set is meant a pair (S, S ) consisting of a set S and a partial order relation S 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 S is partially ordered by S. 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 S. Two elements a and b of S such that either a S b or b S 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 AS. A partial order S in a set S is termed total (sometimes simple, or linear) if any two elements of S are comparable; that is, a S b or b S a for any two elements a and b of S. A partially ordered set (S,S) whose relation S 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 S (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 S 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 realvalued functions deﬁned on a nonempty set X. Iff S g is interpreted to mean f(x) S g(x) for all x e X, then S partially, but not totally, orders S. Example A9. For a ﬁnal illustration, consider the set P(X) of all subsets of a set X. The relation A S B if and only if A E 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 S B nor B S A holds. As regards terminology, any family of sets ordered in this manner will be said to be ordered by inclusion.
Let (A, S) and (B, S) be two partially ordered sets (when there is no danger of confusion, we write S for the partial orders in both A and B).
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A mappingf: A —> B is said to be orderpreserving or an orderhomomorphism if for all a, b e A, a S b implies f(a) S f(b) in B. A onetoone orderhomomorphism f of the set A onto B whose inverse is also an orderhomomorphism (from B onto A) is an orderisomorphism. If such a function exists, we say that the two partially ordered sets (A,S ) and (B, S) are orderisomorphic. When the partial order is the primary object of interest and the nature of the elements plays no essential role, orderisomorphic 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 S. Then A is orderisomorphic to a family of subsets of A, partially ordered by inclusion.
Proof For each a 6A, let I, = {x e Ax S a}. It is not hard to verify that the mapping f: A —> P(A) deﬁned by f(a) = I, is an orderhomomorphism of A into P(A). Indeed, if a S b, then the condition x S a implies x S b and therefore Ia E ID, or, equivalently, f(a) E f(b). To see that f is onetoone, suppose a, b e A are such that f(a) = f(b). Then the element a e I, = 1b, and, hence, a S b by deﬁnition of Ib; likewise, b S a,
from which it follows that a = b. Finally, the inverse f '1 is also orderpreserving. For, if the inclusion 1,, E Ib holds, then a e 1,, and so a S b. These calculations make it clear that A is orderisomorphic to a certain set of subsets of P(A).
Corollary. For no set A is A orderisomorphic to P(A).
Proof. We argue that iff: A —> P(A) is any orderhomomorphism from A into P(A), then f cannot map onto P(A). For purposes of contradiction, assume that f does carry A onto P(A). Deﬁne B = {a e Aa ¢ f(a)} and
B* = {ceAc S a for some aeB}. By supposition, the set B* = f(b) for some element b e A. If b 9*. B*, then, according to the deﬁnition of B, b e B E B*, a contradiction. Hence,
b e B* and so b S a for some a in B. From the orderpreserving character of f, B* = f(b) E f(a). But then, a e B E B* E f(a). The implication is that a 93 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 S . An element x e S is said to be a minimal (maximal) element of S if a e S anda Sx(x Sa)implya = x.
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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 A10. The simplest illustration of a partially ordered set without minimal or maximal elements is furnished by the set R”e with the ordering S in the usual sense.
Example A—ll. In the collection P(X) — {Q} of all nonempty subsets of a nonempty set S (partially ordered by settheoretic inclusion), the minimal elements are those subsets consisting of a single element. Example A12. Consider the set S of all integers greater than 1 and the partial order S deﬁned by a S 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.
Let S be a set partially ordered by the relation S. Deﬁnition A—7. An element x e S is called the ﬁrst (last) element of S if x S a (a S 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, S) has two ﬁrst elements, say x and y; then, x S y and y S x, so that x = y by the anti
symmetric 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 S and let A be a subset of S.
An element xeS is said to be a lower (upper)
bound for A ifx S a(a S x) for all aeA.
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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 E R# with respect to s ) 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 .2! E P(X) is any set containing u .21, while a lower bound is any set contained in n .521.
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 Zom’s Lemma. Our presentation is descriptive and most of the facts are merely stated. The reader who is not content with this bird’seye 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, S) is said to be wellordered if every nonempty subset A g S has a ﬁrst element (“with respect to S ” being understood). The set Z+ is wellordered by the usual S ; each nonempty subset has a ﬁrst element, namely, the integer of smallest magnitude in the set. Notice that any wellordered set (S, S ) is in fact totally ordered. For, each subset {a, b} g S must have a ﬁrst element. According as the ﬁrst element is a or b, we see that a S b or b S a, whence the two elements
a and b are comparable. Going in the other direction, any total ordering of a ﬁnite set is a wellordering of that set. Let it also be remarked that a subset of a wellordered set is again wellordered (by the restriction of the ordering). Example B—l. Consider the Cartesian product S = Z + xZ+. We partially order S as follows: if (a, b) and (a’, b’) are ordered pairs of positive integers, (a, b) S (a’, b’) means that (1) a < a’ (in the usual sense) or (2)
a = a’ and b S b’. (This is called the lexicographic order of Z + xZ+, because of its resemblance to the way words are arranged in a dictionary.) For instance, (4, 8) S (5, 2), while (3, 5) S (3, 9). To conﬁrm that S is a wellordering of S, let @ 7E A E S and deﬁne B = {a+(a, b)eA}. Since A is a nonempty subset of Z +, it has a ﬁrst element, call it a0. Now, let C = {b eZ+(a0, b)eA}. Again, the wellordering of Z + under S guarantees that C has a ﬁrst element, say be. We leave it to the reader to convince himself that the pair (a0, b0) serves as the ﬁrst element of A, thereby making S a wellordered set relative to S . 70‘
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A fundamental axiom of set theory, which has a surprising variety of logically equivalent formulations, is the socalled Well—Ordering Theorem
of Zermelo (1904). The designation “theorem” notwithstanding, we take this to be an axiom (assumed and unproven) of our system. We state: Zermelo’s Theorem. Any set S can be wellordered; that is, there is a
partial ordering s for S such that (S, s) is a wellordered 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 wellorders an uncountable set. Moreover, the promised wellordering may bear no relation to any other ordering that the given set may already possess; the wellordering of R’“, for instance, cannot coincide with its customary ordering. Zermelo based the “proof” of his classical WellOrdering 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 ‘6 be a (nonempty) collection of nonempty sets. A function f: ‘6 —> u (K is called a choice function for ‘6 if f(A) e A for every set A in ‘6.
Informally, a choice function f can be thought of as “selecting” from each set A e ‘6 a certain representative element f(A) of that set. As a simple illustration, there are two distinct choice functions f, and f2 for the family
of nonempty supsets of {1, 2}:
f1({1,2}) = 1, f2({1, 2}) = 2,
'
MW) = 1, f2({1}) = 1,
f1({2}) = 2, f2({2}) = 2
The question arises whether this selection process can actually be carried out when g has inﬁnitely many members. The possibility of making such choices is handled by the axiom mentioned above: Axiom of Choice. Every collection ‘6 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 is! Xi is nonempty (it should be clear that the elements of X Xi are precisely the choice functions for {Xi}). For another common phrasing, which again expresses the idea of selection, let ‘6 be a collection of disjoint, nonempty sets. The axiom of choice, as we have stated it, is equivalent to
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asserting the existence of a set S with the property that A n S contains exactly one element, for each A in ‘6. Granting Zermelo’s Theorem, it is clear that a choice function f can be deﬁned for any collection ‘6 of nonempty sets: having wellordered u ‘6, 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 ofchoice, so that these are in reality two equivalent principles (although seemingly quite diﬂ‘erent). Although the axiom of choice may strike the reader as being intuitively obvious, the soundness 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 nonconstructive criterion for the existence of maximal elements is given by the socalled “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.) Zorn’s Lemma. Let S be a nonempty set partially ordered by S . Suppose that every subset A E S which is totally ordered by S has an upper bound (in S). Then S possesses at least one maximal element. Zom’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 practise, let us prove what is sometimes known as Hausdorff’s Theorem (recall that by a chain is meant a totally ordered set):
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Theorem 3]. 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 ‘6 of all chains of a partially ordered set (S, s );