Open Collections

UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

Some results in the theory of radicals of associative rings Heinicke, Allan George 1968

Your browser doesn't seem to have a PDF viewer, please download the PDF to view this item.

Item Metadata

Download

Media
831-UBC_1969_A1 H45.pdf [ 8.64MB ]
Metadata
JSON: 831-1.0080525.json
JSON-LD: 831-1.0080525-ld.json
RDF/XML (Pretty): 831-1.0080525-rdf.xml
RDF/JSON: 831-1.0080525-rdf.json
Turtle: 831-1.0080525-turtle.txt
N-Triples: 831-1.0080525-rdf-ntriples.txt
Original Record: 831-1.0080525-source.json
Full Text
831-1.0080525-fulltext.txt
Citation
831-1.0080525.ris

Full Text

SOME RESULTS IN THE THEORY OF RADICALS  OF ASSOCIATIVE RINGS . by ALLAN GEORGE HEINTCKE B . S c , U n i v e r s i t y of -Manitoba, 1961 M. Sc., U n i v e r s i t y of Manitoba, 1965 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY / ' • • i n the Department of MATHEMATICS We accept t h i s thesis, as conforming to.the r e q u i r e d standard The U n i v e r s i t y o f B r i t i s h Columbia? December 1968 In p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the r e q u i r e m e n t s f o r an advanced degree a t the U n i v e r s i t y of B r i t i s h Columbia, I a g r e e t h a t the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e and Study. I f u r t h e r a g r e e t h a t p e r m i s s i o n f o r e x t e n s i v e c o p y i n g of t h i s t h e s i s f o r s c h o l a r l y purposes may be g r a n t e d by the Head o f my Department or by h i s r e p r e s e n t a t i v e s . I t i s u n d e r s t o o d t h a t c o p y i n g or p u b l i c a t i o n of t h i s thes,is f o r f i n a n c i a l g a i n s h a l l not be a l l o w e d w i t h o u t my w r i t t e n p e r m i s s i o n . Department o f The U n i v e r s i t y o f B r i t i s h Columbia Vancouver 8, Canada ABSTRACT Supervisor: N.J. Divinsky Several aspects of the theory of r a d i c a l classes i n associative r i n g theory are investigated. In Chapter three, the Andrunakievic-Rjabuhin construction of r a d i c a l s by means of annihilators.of modules i s employed to define several r a d i c a l properties. One of these i s shown to be the "weak r a d i c a l " of Koh and Mewborn. The relations between these r a d i c a l s / t h e i r properties and some of t h e i r applications to the study of c l a s s i c a l quotient rings are investigated. / In Chapter four, the ideals of a r i n g K of the form R(K), for a hereditary r a d i c a l , R, are studied. A closure operation on the l a t t i c e of ideals i s introduced, and the "closed" ideals are p r e c i s e l y the ideals of thi s type. It i s proved that the ascending and descending chain conditions on the closed ideals of a rin g imply that the rin g has only a ' f i n i t e number of closed ideals. In Chapter f i v e , f i n i t e subdirect sums of rings are studied. The properties of hereditary r a d i c a l s and of the various structure spaces, i n a s i t u a t i o n where one has a f i n i t e subdirect sum of rings, are investigated. i i i . TABLE OF CONTENTS page J INTRODUCTION • 1 CHAPTER 1. GENERAL RADICAL THEORY 1-1 R a d i c a l p r o p e r t i e s and r a d i c a l c l a s se s . . . 5 1-2 The lower r a d i c a l c o n s t r u c t i o n . . . . . . ' 6 1-5 The upper r a d i c a l c o n s t r u c t i o n M O 1-4 H e r e d i t a r y r a d i c a l s 12 1- 5 S u p e r n l l p o t e n t , SP, ' and dua l r a d i c a l s i 4 CHAPTER 2 RADICALS AND MODULES 2- 1 D e f i n i n g r a d i c a l s by c l a s se s of modules .; 18 2- 2 I n t e r s e c t i o n s of one-s ided i d e a l s 28 CHAPTER 3 THE UNIFORM, RATIONAL, AND WEAK RADICALS 3- 1 E s s e n t i a l and r a t i o n a l ex tens ions ' 3 1 3-2 The u n i f o r m , r a t i o n a l , , and weak r a d i c a l s . 33 3-3 M a t r i x r i n g s and the r a d i c a l s U , U * , and W - 41 1 2 3 3-4 R e l a t i o n s between the c l a s s e s L , £ , I . . . . 49 3-5 Rings w i t h the ascending cha in c o n d i t i o n . . . . . . 56 3-6 Q u a s i - i n j e c t i v e modules . 66 • 3-7 Dens i ty theorems f o r the r a d i c a l s - H^ and W .. : 80 3-8 Prime r i n g s w i t h zero s i n g u l a r i d e a l and a un i form r i g h t i d e a l .• ' 8 8 i v . CHAPTER 4 HEREDITARY RADICAL IDEALS OF A RING 4 - 1 An equiva lence r e l a t i o n f o r r i n g s . . . . - 9 6 4 - 2 A c lo su re ^operation . . . ' 1 0 6 4 - 3 M i n i m a l ' c l o s e d i d e a l s of a r i n g 1 1 2 4 - 4 Rings w i t h cha in c o n d i t i o n s on c lo sed i d e a l s . . . 1 1 5 4- 5 Rings determined by t h e i r atoms 1 2 2 CHAPTER 5 RINGS WITH CHAIN CONDITIONS ON CHARACTERISTIC IDEALS 5 - 1 C h a r a c t e r i s t i c i d e a l s !. . 1 3 4 5 - 2 Some s t r u c t u r e theorems 1 3 7 5 - 3 F i n i t e s u b d i r e c t sums o f r i n g s . . . . 1 4 9 5- 4 H e r e d i t a r y r a d i c a l s of f i n i t e s u b d i r e c t sums of r i n g s .• . 1 5 4 5 - 5 F i n i t e s u b d i r e c t sums and s t r u c t u r e spaces . . . . 1 5 8 BIBLIOGRAPHY . 1 6 9 v.. ACKNOWLEDGEMENTS The' author wishes t o express h i s thanks t o h i s re s e a r c h s u p e r v i s o r , Dr. N.J. D i v i n s k y , f o r encouragement and advice tendered d u r i n g the e x e c u t i o n of the r e s e a r c h h e r e i n d e s c r i b e d . . ; The author a l s o f i s h e s to,.thank Dr. T. Anderson f o r many h e l p f u l c o n v e r s a t i o n s . T h e • f i n a n c i a l support of the H.R.'MacMillan f a m i l y , and of the N a t i o n a l Research C o u n c i l o f Canada i s g r a t e f u l l y . , acknowledged. . ) INTRODUCTION In t h i s t h e s i s , by a r i n g , we mean an a s s o c i a t i v e r i n g , not n e c e s s a r i l y p o s s e s s i n g a u n i t y element. When K i s a r i n g , a K-module'will be a r i g h t K-module, and w i l l u s u a l l y be denoted by M or, when there i s danger of ambiguity, by Mg. "Of course, a r i n g K may be regarded as a r i g h t K-module, and, i n t h i s case, .the submodules are J u s t the r i g h t i d e a l s of K. :-; I f f: L-^  -• Lg i s a 'homomorphism of r i n g s ( r e s p . of r i g h t K-modules), we s a y . t h a t f i s a monomorphism i f f(x).> = f ( y ) i m p l i e s x = y, and we say t h a t f i s a sur s u r j e c t i o n (or a s u r j e c t i v e mapping) i f , f o r every element w of L 2 there i s an element x i n such t h a t f ( x ) = w. We'are mainly concerned, i n t h i s t h e s i s , with v a r i o u s aspects of r a d i c a l theory. In Chapter one, the d e f i n i t i o n s and b a s i c p r o p e r t i e s of r a d i c a l s are d i s c u s s e d . And through-out the t h e s i s , p r o o f s of r e s u l t s are seldom g i v e n i f the r e s u l t i s proved i n the book .by D i v i n s k y ( 6 ) . However, i n such cases, e x p l i c i t r e f e r e n c e s are g i v e n t o d i r e c t the reader to a p r o o f . I t i s w e l l known t h a t the Jacobson r a d i c a l - o f a r i n g K can be d e s c r i b e d i n terms of r i g h t K-modules. "This r e s u l t was g e n e r a l i z e d by A n d r u n a k i e v i c and Rjabuhin, who showed tha.t any r a d i c a l p r o p e r t y can be d e s c r i b e d i n terms of modules. The f - i r s t p a r t of Chapter two i s devoted t o an e x p o s i t i o n of 2. these r e s u l t s . The work In Chapter two then turns t o a g e n e r a l i z a t i o n (Theorem 2 . 2 . 1 ) of the well-known r e s u l t t h a t the Jacobson r a d i c a l of a r i n g K i s the i n t e r s e c t i o n of the maximal modular r i g h t i d e a l s of K. In r e c e n t y e a r s , r i n g theory, from the p o i n t of view of r a d i c a l theory, has di v e r g e d g r e a t l y from the other d i r e c t i o n s , t h a t have been taken. An attempt t o b r i d g e t h i s gap i s made i n Chapter t h r e e . I n - t h i s chapter, the techniques d i s c u s s e d i n Chapter two are a p p l i e d t o d e f i n e a number of new r a d i c a l p r o p e r t i e s . The r e l a t i o n s between these r a d i c a l s and some of the more c l a s s i c a l r a d i c a l s are d i s c u s s e d , and two r e s u l t s (Theorems 3 -7 .1 and 3 . 7 . 2 ) g i v e n g e n e r a l i z a t i o n s of the Jacobson d e n s i t y theorem. A l s o i t - i s shown t h a t one of the r a d i c a l s d i s c u s s e d here c o i n c i d e s . w i t h the "weak r a d i c a l " of Koh and Mewborn ,(2). ' ' ' : . - • Chapter f o u r marks a r e t u r n t o g e n e r a l r a d i c a l theory'. A c l o s u r e o p e r a t i o n on the l a t t i c e of two-sided i d e a l s of a r i n g K i s i n t r o d u c e d ' f o r which the " c l o s e d " i d e a l s of K are those i d e a l s of the form H(K),'- where H i s a h e r e d i -t a r y r a d i c a l . In TheoreW . , .4.4.3 i t i s shown t h a t a r i n g K has the ascending and descending c h a i n c o n d i t i o n s f o r c l o s e d i d e a l s i f and only i f K possesses a f i n i t e number of c l o s e d i d e a l s . Some a t t e n t i o n i s a l s o g i v e n t o the case where a r i n g i s determined by i t s minimal c l o s e d i d e a l s . In the f i r s t p a r t of. Chapter f i v e , a s l i g h t g e n e r a l i z a t i o n of some r e s u l t s of A n d r u n a k i e v i c ( l ) i s gi v e n . T h i s l e a d s t o r e p r e s e n t a t i o n s of c e r t a i n r i n g s as f i n i t e s u b d i r e c t sums of prime r i n g s . In the l a t t e r p a r t o f Chapter f i v e , f i n i t e s u b d i r e c t sums i n g e n e r a l are examined. L e t a r i n g K be a ""/ • n s u b d i r e c t sum of the-Tings K,, K0,... K , and l e t S = © K.. 1 d n i = i 1 The r e l a t i o n s between H(K) and H(S), where H i s a h e r e d i t a r y r a d i c a l , and the r e l a t i o n s between the v a r i o u s s t r u c t u r e spaces of K and of S are i n v e s t i g a t e d (Theorems 5.4.5 and 5 - 5 .1 ) . " V Throughout the t h e s i s , d e f i n i t i o n s are-, i n d i c a t e d by u n d e r l i n i n g of the term b e i n g d e f i n e d . Other t h i n g s , such as theorems, c o r o l l a r i e s , and examples are numbered by three i n t e g e r s . For example, Example 5*3-3 i s found i n Chapter f i v e , § 3 j immediately f o l l o w i n g Lemma 5 -3 -2 . Most o f the n o t a t i o n used i s standard. A few exceptions are the f o l l o w i n g . I f K i s a r i n g , and S a subset of K, | S > K i s the r i g h t , i d e a l o f K generated by S , and <S> K ( o r simply < S > i f there i s no danger o f ambiguity) denotes the two-sided i d e a l o f K generated by S . A l s o , suppose t h a t F' i s a s e t whose members are a l l • subsets o f a g i v e n s e t S. I f F i s v o i d , d e f i n e |~| F to-be S, and otherwise d e f i n e [] F to be D X. In e i t h e r • xeF case-we a l s o d e f i n e M F t o be u X.' Thus, f o r example, . XeF the well-known c h a r a c t e r i z a t i o n of the Jacobson r a d i c a l o f a r i n g K as the i n t e r s e c t i o n of a l l the modular maximal r i g h t . . . 4 . i d e a l s of K takes the form J(K) = [~[ F> where f = {I: I i s a modular maximal r i g h t i d e a l of K}. I f P i s v o i d , then (both i n f a c t and i n t h i s n o t a t i o n ) J(K) = K. CHAPTER ONE GENERAL RADICAL THEORY 1.1 R a d i c a l P r o p e r t i e s and R a d i c a l C l a s s e s V A c l a s s R of a s s o c i a t i v e r i n g s i s c a l l e d a r a d i c a l  c l a s s i f the f o l l o w i n g c o n d i t i o n s are s a t i s f i e d : (A) Any homomorphic image of a member of R ' i s a l s o In R, (B) Any r i n g K has an i d e a l R(K)-.-which, as a r i n g , Is i n R, and which i s maximum among the i d e a l s of K which are i n R. 1 . (C) For any r i n g K, K/R(K) ha.s only one i d e a l i n the Clas s R, namely the i d e a l 0. ( I n other words, R(K/R(K)) = 0.) As i s w e l l known, the c l a s s e s J , c o n s i s t i n g of a l l -r i g h t q u a s i - r e g u l a r r i n g s , and N, c o n s i s t i n g of a l l n i l r i n g s , are r a d i c a l c l a s s e s . On the other hand, the c l a s s o f a l l n i l p o t e n t r i n g s i s not a r a d i c a l c l a s s . For other examples o f . r a d i c a l c l a s s e s , we r e f e r t o D i v i n s k y (6). A l s o , f o r some of the elementary p r o p e r t i e s _ of r a d i c a l c l a s s e s , we r e f e r t o Chapter one of the same r e f e r e n c e . . Many authors r e f e r t o a " r a d i c a l p r o p e r t y " ; r a t h e r than t o a " r a d i c a l c l a s s " . I f P i s a p r o p e r t y of r i n g s , then P i s a r a d i c a l p r o p e r t y i f f the c l a s s S of a l l r i n g s with p r o p e r t y P i s a r a d i c a l c l a s s . Conversely, i f R i s a r a d i c a l c l a s s , then the p r o p e r t y "belonging t o R" i s a r a d i c a l p r o p e r t y . For example, " n i l " i s a r a d i c a l p r o p e r t y , while " n i l p o t e n t " i s not. I f R i s a r a d i c a l c l a s s , and i f K i s a r i n g which belongs t o R, then R(K) = K, and we c a l l K an R - r a d i c a l  r i n g . I f , on the other hand, R(K) = 0, then we say t h a t K i s R-semisimple. For some of the more f a m i l i a r r a d i c a l c l a s s e s , R-semisimple i s sometimes g i v e n another name. • For example, i f J i s the Jacobson r a d i c a l c l a s s , c o n s i s t i n g of a l l r i g h t q u a s i - r e g u l a r r i n g s , a r i n g K f o r which J(K) = 0 i s sometimes r e f e r r e d • ' to as a s e m i p r i m i t i v e r i n g . A g ain, I f B i s the Baer Lower R a d i c a l (see D i v i n s k y ( 6 ) , sec. 3.3) . , then the B semisimple r i n g s are c a l l e d semiprime r i n g s . > -1 . 2 The Lower Ra d i c a l " ' C o n s t r u c t i o n Given a c l a s s of r i n g s , M, i t - i s r e a s o n a b l e ' t o ask, • i f M i s not i t s e l f a r a d i c a l c l a s s , whether M i s contained i n a r a d i c a l c l a s s , arid, ' i f so, whether there i s a minimal or minimum r a d i c a l c l a s s c o n t a i n i n g M. The answer t o the f i r s t q u e s t i o n i s "yes", f o r the c l a s s of a l l r i n g s i s c l e a r l y a r a d i c a l c l a s s . ' ' . , -The answer t o the second q u e s t i o n i s a l s o "yes". We s h a l l show t h a t , g i v e n any c l a s s M of r i n g s , there i s a minimum r a d i c a l c l a s s c o n t a i n i n g M. T h i s w i l l be denoted S (M), and w i l l be c a l l e d the. lower r a d i c a l c l a s s with o x ^ r e s p e c t t o ' M.• The proof of the E x i s t e n c e of S (M) i s due t o Kurosh ( 1 6 ) . The p r e s e n t c o n s t r u c t i o n i s -due t o S u l i n s k i , Anderson, and D i v i n s k y ( 6 ) . The c o n s t r u c t i o n i s achieved as f o l l o w s . Suppose we are g i v e n a c l a s s of r i n g s M. Define t o be the c l a s s of a l l r i n g s which are homomorphic images of members of M. Given any o r d i n a l a > 1 , i f has been d e f i n e d f o r a l l o r d i n a l s B < a, ,then d e f i n e t o be the c l a s s of. a l l r i n g s K f o r which every non-zero homomorphic image o f K has a non-zero i d e a l i n M^, f o r some 8 < a. I t i s e a s i l y seen t h a t , f o r each o r d i n a l a, M i s h o m o r p h i c a l l y c l o s e d ( i . e . each homomorphic image of a member of M"a i s i n M Q), and a l s o t h a t , i f a and y & T e o r d i n a l s , where a < y, then, M c M . Define S (M) >= 11 M , the u n i o n b e i n g taken over a — y 0 a a l l o r d i n a l s . Before p r o v i n g t h a t SQ(.M) i s indeed the minimum r a d i c a l c l a s s c o n t a i n i n g M; we quote the f o l l o w i n g lemma, whose pr o o f i s g i v e n i n D i v i n s k y (6). p.4. Lemma 1 . 2 . 1 A" c l a s s R of r i n g s i s a r a d i c a l c l a s s i f and only i f i t s a t i s f i e s the f o l l o w i n g c o n d i t i o n s : (A) Any homomorphic image of a member,of R i s i n R (D) I f K i s a r i n g such t h a t every non-zero homomorphic 8 . image has "a non-zero i d e a l i n R, then K i s i t s e l f a member of R. With t h i s lemma, we can now prove the f o l l o w i n g theorem. Theorem 1.2.2 S (M) i s a r a d i c a l c l a s s which c o n t a i n s M. Furthermore, i f R i s any r a d i c a l c l a s s c o n t a i n i n g M, R co n t a i n s S (M). ' o v ' • Proof; C l e a r l y we have M c So(M)... Furthermore, s i n c e each M Q i s homomorphically c l o s e d , S Q(M) has c o n d i t i o n (A). Suppose K i s a r i n g f o r which every non-zero homomorphic image has a non-zero i d e a l i n S Q(M). ,• For each;, i d e a l I of K f o r which I/K, the f a c t o r r i n g K/I has a non-zero i d e a l i n . S (M). For each such i d e a l I , we choose an o r d i n a l o v a T such t h a t K/I has a non-zero i d e a l i n M . Since the c o l l e c t i o n of a l l i d e a l s of K Is a proper s e t , there i s an o r d i n a l p such t h a t fJ > ct-j. f o r a l l I. Now l e t K' be any non-zero homomorphic image of K. Then K' = K/I f o r -some i d e a l I of- K. Since K/I has a non-zero i d e a l i n -M c M A, the image,, of t h i s i d e a l under the isomorphism i s a Ct-j- p ^ ) non-zero i d e a l of K' which i s a l s o i n M Q. Thus we see t h a t P any non-zero homomorphic image of K has a non-zero i d e a l -i n Mft, which g i v e s K e M R . c S (M). We have shown t h a t S Q(M) s a t i s f i e s c o n d i t i o n (D), and thus i s a r a d i c a l c l a s s . Now suppose t h a t R i s a r a d i c a l c l a s s , and t h a t M c R. Then, s i n c e R i s homomorphically c l o s e d , cr R. Suppose t h a t a i s an o r d i n a l , and suppose t h a t , f o r a l l P < a, we have t h a t cr R. Then, from the d e f i n i t i o n of M , and from the f a c t t h a t R has c o n d i t i o n (D), we o b t a i n CL a t once t h a t M cr R. Thus S (M) cr R. "This proves the theorem. a — o v •' — r S I t i s perhaps worthwhile to note t h a t S (M) was shown t o be a r a d i c a l c l a s s without knowing whether or not the a s s o c i a t e d sequence of M 's terminated i n t h e ; s e n s e t h a t , f o r a some o r d i n a l a, ML- = M ,, (from which i t f o l l o w s t h a t ' a a+1 v M a = MB-' f o T a 1 1 P 2 a ) • I n S u l i n s k i , Anderson, and D i v i n s k y , (24), i t was shown t h a t t h i s c o n s t r u c t i o n does indeed t e r m i n a t e , and t h a t i t terminates a t m , the f i r s t i n f i n i t e o r d i n a l . Thus, no matter what c l a s s M i s taken, i t i s t r u e t h a t S (M) = M . Of course, i t may be t r u e t h a t we have S (M) = M, , o v ' wQ o v ' k' where k i s a f i n i t e o r d i n a l , f o r some choices of the c l a s s • M. In p a r t i c u l a r , i f M i s ' i t s e l f a r a d i c a l c l a s s , then '~ S (M) = = M. For M equal t o the c l a s s of a l l n i l p o t e n t r i n g s , denote S (M) by B. ' T h i s ' i s the well-known Baer Lower r a d i c a l . In S u l i n s k i , Anderson, and D i v i n s k y (24), i t was' shown t h a t B = M 0. •' • 1 0 . The author has shown (He i n i c k e ( 1 0 ) ) t h a t ' t h e r e e x i s t s a c l a s s M f o r which the bound (joQ i s a t t a i n e d . That, i s t o say, there i s a c l a s s M f o r which S (M) / M"k f o r any f i n i t e o r d i n a l k. 1 . 5 The Upper R a d i c a l C o n s t r u c t i o n A q u e s t i o n which i s " d u a l " t o the one r a i s e d In §1.2 i s : when can a g i v e n c l a s s of r i n g s be the c l a s s o f R-semisimple r i n g s .for some r a d i c a l c l a s s R, and, i f the g i v e n c l a s s C i s not I t s e l f a semisimple c l a s s , i s there a maximum r a d i c a l c l a s s R, f o r which any member of C , i s R-semisimple? In Chapter one of D i v i n s k y (6) i t i s shown th a t a c l a s s C i s the c l a s s of.R-semisimple r i n g s f o r a r a d i c a l c l a s s . R i f and only i f C s a t i s f i e s both: • (E) Every non-zero i d e a l of a member of C can be homomorphically mapped onto a non-zero member of C, and • • (F) I f K i s a r i n g f o r which e v e r y non-zero i d e a l can be homomorphically mapped onto a non-zero member of C, then K i s i t s e l f i n C. • Furthermore, i t i s a l s o shown t h a t , i f C i s a c l a s s -of r i n g s with ( E ) , then the c l a s s "C, d e f i n e d as the c l a s s of a l l r i n g s K f o r which every non-zero i d e a l can be mapped onto a non-zero member of C, has both (E) and ( F ) , and i s .therefore the c l a s s of R-semisimple r i n g s for. some r a d i c a l p r o p e r t y R. It. t u r n s out th a t R c o n s i s t s of a l l r i n g s - K 1 1 . which cannot he mapped,onto a non-zero member of CL A l s o , R i s the l a r g e s t r a d i c a l c l a s s S f o r which every member of C i s S-semisimple. The c l a s s R, d e f i n e d i n t h i s manner, i s the upper r a d i c a l c l a s s with r e s p e c t t o C. I t might be p o i n t e d out t h a t , g i v e n any c l a s s C of r i n g s , i t ' i s p o s s i b l e t o enlarge i t t o o b t a i n a c l a s s C which has (E) i n such a way t h a t the upper r a d i c a l w i t h r e s p e c t to C' i s , i n f a c t , the l a r g e s t r a d i c a l c l a s s S f o r which a l l members of C are S-semisimple. For theorem 47 o f D i v i n s k y (6) shows t h a t i f I i s an i d e a l of K, then, f o r any r a d i c a l c l a s s R, R ( l ) i s a l s o an i d e a l of K. An immediate c o r o l l a r y -is t h a t i f K i s R-semisimple, and i f I i s an i d e a l of K, then I i s a l s o R-semisimple. Thus, f o r any r a d i c a l c l a s s R, the c l a s s of R-semisimple - r i n g s i s c l o s e d under the t a k i n g of i d e a l s . • (Henceforth, a c l a s s c l o s e d under the t a k i n g of i d e a l s w i l l be c a l l e d a h e r e d i t a r y  c l a s s . ) C l e a r l y , any h e r e d i t a r y c l a s s has ( E ) . I f K i s a r i n g , and i f S ' i s ' a s u b r i n g , we say. t h a t S i s a c c e s s i b l e t o K i f there i s a f i n i t e c h a i n S <S S1 <3 S 2 . . . <J S n = K, (where A 4 B means t h a t A i s an i d e a l of B). ' I f C i s any c l a s s of r i n g s , d e f i n e C-' t o be the c l a s s of a l l r i n g s isomorphic t o an a c c e s s i b l e s u b r i n g of a member of C. I t i s simple t o v e r i f y t h a t C i s a h e r e d i t a r y c l a s s c o n t a i n i n g C. I f R i s the. upper r a d i c a l w i t h r e s p e c t t o C , then any member of C i s R-semisimple. A l s o , ; i f S i s a r a d i c a l 12 . class for which every member.of C i s S-semisimple, i t follows that any accessible subring of a member i s also 5- semisimple, and thus every member of C i s S-semisimple. Therefore, • S cr R. For thi s reason we are j u s t i f i e d i n r e f e r r i n g to R as the upper r a d i c a l with respect to C. 1.4 Hereditary Radicals . Many, of the•familiar r a d i c a l classes, for example J, N, and B, are known to be hereditary classes. However, not a l l r a d i c a l classes are hereditary (see Divinsky ( 6 ) , p. 1 0 ) . It i s known, (see Divinsky, ( 6 ) , p. 1 2 5 ) , that a '. r a d i c a l class R i s hereditary i f and only i f , f o r any ide a l of a r i n g K, R(I) = I n R(K). The following i s due to Hoffman and Le a v i t t ( 1 2 ) . Theorem 1 .4 .1 If M i s a hereditary c l a s s , then S (M) i s a hereditary r a d i c a l class. -Proof; It s u f f i c e s to show that each M a is- hereditary.. Let K e M1 and l e t - I be an i d e a l of K. Then there i s a homomorphism . 8; from a member L of, M' onto. K. Now 6 - 1 ( l ) i s an i d e a l of L, and, since M i s hereditary, e ~ 1 ( l ) i s i n M. Then I = 8 ( 0 _ 1 ( l ) ) . i s i n - and i s hereditary. ' ~ Let a be an ordinal > 1, and suppose that M^ i s hereditary f o r a l l 8 < a. Let K e MQ, and l e t I be an • 1 5 -') i d e a l of K, and l e t i|i be any non-zero homomorphism of I. If. J i s the k e r n e l o f , then = I / J . We s h a l l show-t h a t I / J has a non-zero i d e a l i n M f o r some X < a. A. L e t U. be an i d e a l o f K which i s maximal with r e s p e c t to h a v i n g I fl U c J . (Such a U e x i s t s by Zorn's lemma.) I f U = ;K, then I c J , and i s a zero map, c o n t r a r y t o our assumption. Thus, U / K. Since K (and t h e r e f o r e K/U) i s i n M , K/U has.a non-zero i d e a l W/U i n M f o r some X < a. W i s s t r i c t l y l a r g e r than U, and/hence W n I <j£. J . Th e r e f o r e , (W fl I + J ) / J i s a non-zero i d e a l of I / J . We show t h a t (W fl I -!- J ) / J i s i n M^. I f W n I c U, then W n I c tJ fl I c ' J , a c o n t r a d i c t i o n . T h e r e f o r e , W fl I $z U, and (W fl I + U)/U i s a non-zero i d e a l of -W/U'€ M^. Since i s h e r e d i t a r y , (W fl I + U)/U i s i n M^. T h e r e f o r e , (W n l)/(W n I fl II) i s i n M^. Since W fl I fl U c W n J , we can map " (W n I)/(W fl I fl U) homo-m o r p h i c a l l y onto (W fl I)/(W n J ) , and the l a t t e r i s then i n M, . Since (W fl I + J ) / J = (W n l)/(W fl I fl- J) = A. (W n I)/(W n J ) r we have- t h a t (W n I + J ) / J i s i n M , as A. d e s i r e d . We have shown t h a t M-j* i s h e r e d i t a r y , and a l s o t h a t i f i s h e r e d i t a r y f o r a l l o r d i n a l s 1 < X < a, then' M a i s ' h e r e d i t a r y . It- f o l l o w s by t r a n s f i n i t e i n d u c t i o n t h a t , f o r each o r d i n a l a _>•!,.-M i s h e r e d i t a r y . T h e r e f o r e S (M) 14. i s hereditary. Q.E.D. The.converse of thi s theorem i s not true. In Michler (20) i t i s shown that the class of weakly regular rings i s an example of a non-hereditary class whose lower radical, i s hereditary. 1.5 Supernilpotent, SP, and Dual Radicals In many parts of r i n g theory, prime rings and prime ideals play an important r o l e . An,ideal I of a r i n g K i s a prime i d e a l i f I ^ K and,"' i f a.Kb e I, then either a e I or b e I. This l a t t e r condition i s well known to be equivalent to: i f A and B are both r i g h t (or both l e f t , or both two-sided) ideals of K, and i f AB c I, then either A c I or B e I. A r i n g K i s a prime r i n g i f and only i f 0 i s a prime, i d e a l . . I t i s well known that the class of prime rings i s a hereditary--'class, and one can then form the upper r a d i c a l with respect to the class of prime rings. A well-known r e s u l t of L e v i t z k i (see chapter 3 of Divinsky (5)) says that t h i s upper r a d i c a l i s the same as the Baer Lower r a d i c a l class. This,, i t w i l l be r e c a l l e d , i s the lower r a d i c a l , with ., respect to the class of nilpotent rings. Furthermore, L e v i t z k i showed that, f o r any r i n g K, B(K) i s the i n t e r s e c t i o n of a l l the prime ideals of K. -This s'ort of a r e s u l t , where the R-radical of an ar b i t r a r y r i n g i s the in t e r s e c t i o n of certa i n ideals i s quite common i n r a d i c a l theory. This occurs, for example with the 1 5 . s o - c a l l e d S P - r a d i c a l s , ( c a l l e d s p e c i a l r a d i c a l s i n the l i t e r a t u r e . ) A c l a s s of r i n g s i s s a i d to be ,a s p e c i a l c l a s s i f : (X) Every member i s a prime r i n g . (Y) The c l a s s , toge ther w i t h the one-element r i n g 0, . i s a h e r e d i t a r y c l a s s . (Z) I f A i s i n the c l a s s , and i f A i s an i d e a l i n K , then K / A * i s i n the c l a s s , where t A* = {x 6 K: Ax = xA = 0}. Since a s p e c i a l c l a s s toge ther w i t h 0, i s h e r e d i t a r y , the upper r a d i c a l w i t h respec t to a s p e c i a l c l a s s C i s the c l a s s of a l l r i n g s which cannot be mapped onto a non-zero member of C. I t was shown by Andrunak i ev i c ( l ) (see a l s o D i v i n s k y ( 6 ) , chapter 7) t h a t . s u c h an upper r a d i c a l i s h e r e d i -t a r y . Since any s p e c i a l c l a s s c o n s i s t s of prime r i n g s , such^ an upper r a d i c a l c l a s s must c o n t a i n B , the Baer Lower r a d i c a l . A r a d i c a l t ha t con ta ins B . .and i s h e r e d i t a r y i s c a l l e d a s u p e r n i l p o t e n t r a d i c a l . A n d r u n a k i e v i c a l s o proved t h a t , f o r any r i n g K , i f R i s the upper r a d i c a l w i t h r espec t to a s p e c i a l c l a s s C, then R(K-) = ]~| { I : I 4 K , and R / I e C}. This l a s t , statement i m p l i e s tha t an R-semisimple r i n g i s a subdirect- sum of r i n g s from the s p e c i a l c l a s s C. We s h a l l c a l l an u p p e r , r a d i c a l w i t h respec t to a s p e c i a l c l a s s an S P - r a d i c a l c l a s s . ( I n the l i t e r a t u r e , - t h e s e r a d i c a l s are c a l l e d " s p e c i a l r a d i c a l s " . We p r e f e r to a v o i d t h i s 1 te rm, because to use i t would .necess i ta te t a l k i n g about " s p e c i a l 16 . r a d i c a l c l a s s e s " which are not s p e c i a l c l a s s e s . ) A number of ques t ions can.be asked about S P - r a d i c a l s . Since every S P - r a d i c a l c l a s s i s s u p e r n i l p o t e n t , the most obvious ques t ion i s : i s the converse t rue? R e c e n t l y , R jabuh in (22) c la imed to have, shown tha t the converse i s not t r u e . The author has not had access to t h i s r e s u l t , and • cannot v e r i f y or d i sp rove i t . > - . A ques t i on more e a s i l y answered i s : can d i s t i n c t s p e c i a l c l a s se s def ine the same S P - r a d i c a l ? I t i s known (see D i v i n s k y ( 6 ) , chapter 7) t ha t the c l a s s of r i g h t p r i m i t i v e r i n g s i s a s p e c i a l c l a s s which g ives r i s e to the Jacobson r a d i c a l . S ince the Jacobson r a d i c a l i s r i g h t - l e f t symmetric , the c lass , of l e f t p r i m i t i v e r i n g s i s a l s o a s p e c i a l c l a s s whose upper r a d i c a l i s the Jacobson r a d i c a l . Bergman (4) has shown tha t these are d i s t i n c t s p e c i a l c l a s s e s . A r i n g i s s u b d i r e c t l y i r r e d u c i b l e i f the i n t e r s e c t i o n of the non-zero i d e a l s i s non-zero . In a s u b d i r e c t l y i r r e d u c i b l e r i n g K , t h i s i n t e r s e c t i o n i s c a l l e d the hear t of K. The hea r t H of a s u b d i r e c t l y i r r e d u c i b l e r i n g is, a unique minimal 2 2 two-s ided i d e a l , and e i t h e r H = H , or H = 0. I n the former case the hear t i s - . s a i d to be idempotent. I t i s known 2 -t ha t i f H = H , then H i s , a s imple r i n g . A l s o , •& s u b d i r e c t l y I r r e d u c i b l e r i n g w i t h an idempotent-hear t i s a prime r i n g . (The proofs f o r these a s s e r t i o n s a n d f o r the'' ones tha t f o l l o w , can be found i n chapter 7 of D i v i n s k y ( 6 ) . ) 17 . Suppose tha t P. i s a .proper ty df r i n g s which i s i n v a r i a n t under isomorphisms. That i s , i f S = T, then S has P i f and only i f T has P. Then i t i s t rue that, the c l a s s of a l l s u b d i r e c t l y i r r e d u c i b l e r i n g s which have idempotent h e a r t s , and whose hear t s have P , i s a s p e c i a l c l a s s . I f we denote such a s p e c i a l c l a s s by C, then the upper r a d i c a l w i t h r e s p e c t . t o C i s c a l l e d a dua l r a d i c a l • • Le t R be any s u p e r n i l p o t e n t r a d i c a l c l a s s . Define C, to be the c l a s s of a l l prime R-semisimple r i n g s , and i • '> • def ine Cg to be . the c l a s s of a l l s u b d i r e c t l y I r r e d u c i b l e r i n g s w i t h idempotent h e a r t s , whose hear t s are R-semis imple . These are both s p e c i a l c l a s s e s , and £ C^. I f the correspond i n g upper r a d i c a l s are denoted by R , cor responding to Cg, and R , cor responding to C, , then the f o l l o w i n g statements S -J-are t rue : (1) R c R o c R -v ' — s — cp - . -(2) R i s the s m a l l e s t s p e c i a l r a d i c a l c o n t a i n i n g R. (3) R i s the s m a l l e s t dua l r a d i c a l c o n t a i n i n g R. ^ cp These r e s u l t s are due to Andrunak iev i c ( l ) . Proofs may a l s o be found i n D i v i n s k y ( 6 ) . ;, 18. CHAPTER TWO , RADICALS AND MODULES 2.1 D e f i n i n g R a d i c a l s by Classes of Modules I t i s . w e l l known th a t there 'are two ways to describe the Jacobson r a d i c a l . The f i r s t method could be considered, as " i n t e r n a l " or "element-wise". In this* approach, the Jacobson r a d i c a l c l a s s is. the c l a s s of a l l , r i g h t q u a s i - r e g u l a r r i n g s . (A r i n g i s r i g h t q u a s i - r e g u l a r I f , f o r , each x there i s a. y such t h a t x + y - xy = 0 .) The second method,'an " e x t e r n a l " method, concentrates on the r i g h t p r i m i t i v e r i n g s . A r i n g i s s a i d to'be r i g h t p r i m i t i v e i f there i s a simple r i g h t K - module M such that (0:M) - [x e K: Mx = 0} = 0. Suppose, f o r any r i n g K, we define to be the cl a s s of simple r i g h t K -modules M f o r which MK / 0. ( I n the sequel, we s h a l l c a l l such a module an i r r e d u c i b l e module.) Then, as i s w e l l known, the Jacobson r a d i c a l of any r i n g K can be w r i t t e n J(K)'= fl. {(0':M) : M e £. }. In g e n e r a l , i f A and B are subsets or a r i g h t K-module M, we define (A:B) = [k e K : Bk c A}. I f 'A i s a submodule of M, then' (A:B) i s a r i g h t i d e a l o f . K. I f B Is a l s o a submodule of M, then (A:B) i s a two-sided i d e a l M of K. A l s o , i f I i s a l e f t i d e a l of K, we denote by I the set of a l l m e M such that - ml = 0. This i s a submodule 19-o f M. Suppose tha t M i s an, i r r e d u c i b l e r i g h t K - m o d u l e . • M Then, for any non-zero m i n M, mK ^  0. (For i f mK = 0 , m e K , and K = M, whence MK = 0 , which i s f a l s e . ) Since M i s s i m p l e , MK = M. We then have a homomorphism (Q(k) = mk) 6 : K -• M of K - m o d u l e s which is . s u r j e c t i v e . Any i r r e d u c i b l e , r i g h t K - module i s the re fore isomorphic to a module of the form K / I , where I i s a r i g h t i d e a l of K. Such a r i g h t . i d e a l must be a maximal r i g h t i d e a l , s ince M i s s i m p l e , and, as i s w e l l known, must a l s o be a modular r i g h t i d e a l . (A r ight- i d e a l T ' of K i s modular i f and only if", there i s an element e i n K, , e not i n I , such tha t {ex-x : x. e K} c I . I f K has a l e f t u n i t y element, then every r i g h t i d e a l i s modular . ) I f M i s any r i g h t K - m o d u l e , then (,0:M) = I"! {(0:m) : m e M} . I f M i s i r r e d u c i b l e , each (0 :m) , where m 0, i s a maximal modular r i g h t i d e a l . We then o b t a i n the '" r e s u l t tha t , , f o r any r i n g K , J (K) = [\ ((0:M.) : M e S^} = f l {(0:m) ; 0 / m e M, M e S K ) - Th i s shows tha t J (K) can be expressed 'as an i n t e r s e c t i o n of maximal modular r i g h t i d e a l s . On the other hand, i f I i s a maximal modular r i g h t i d e a l , then. K / I i s an- i r r e d u c i b l e r i g h t K-module. I t i s w e l l known tha t ( 0 : K / l ) c I . - S i n c e J (K) c . ( 0 : K / l ) , we have tha t J (K) c I . There fo re , J (K) c J~| {'1:1 i s a maximal modular r i g h t i d e a l of K } . Combining t h i s w i t h the r e s u l t 20. above, we can conclude t h a t J ( k ) = ]~| £ I : I i s a maximal modular r i g h t I d e a l o f X ] . I n t h i s c h a p t e r , we s h a l l see how to- g e n e r a l i z e t h i s t o c e r t a i n o t h e r r a d i c a l s . Suppose t h a t cp: K -* K' i s a r i n g homomorphism. I f M' i s a r i g h t K' - module, t h e n we can g i v e M' a r i g h t K -module s t r u c t u r e by d e f i n i n g rn'*k = .m' .<p(k), f o r a l l m' i n M' and . k i n . K. A l s o , i f cp . ' i s a s u r j e c t i o n , t h e n any r i g h t K-' module M f o r which " ( 0 : M ) ; 3 k'er(cp') can be g i v e n a r i g h t K'- module s t r u c t u r e as f o l l o w s : i f m e M and k' e K' , choose k i n cp" (k' ) and d e f i n e m*k' = m«k. T h i s i s w e l l d e f i n e d , s i n c e i f cp{k-^) = cp(k^) = k', t h e n k l ~ k 2 e k e r ( c p ) — ( 0 : M ) ^ a n d m-k-^  = m«k 2. I t i s e a s i l y v e r i f i e d t h a t , i f cp i s a s u r j e c t i o n , t h e n , i n each o f the two cases d e s c r i b e d above, the l a t t i c e s o f submodules o f the o r i g i n a l and o f the i n d u c e d module a r e I s o m o r p h i c . A l s o , the Induced module i s t r i v i a l i f and o n l y i f the o r i g i n a l module i s t r i v i a l . I n 1964, A n d r u n a k i e v i c and R j a b u h i n ( 2 ) , c o n s i d e r e d the f o l l o w i n g s i t u a t i o n . Suppose t h a t t o e v e r y r i n g K t h e r e i s a s s i g n e d a ( p o s s i b l y empty) c l a s s of r i g h t K -modules s a t i s f y i n g t h e . f o l l o w i n g c o n d i t i o n s : ( P . 0 ) I f M';e ST^, • t h e n MK / 0. ( P . l ) I f B 0 K, and M e "E K /g* t h e n M, w i t h the i n d u c e d K - module s t r u c t u r e d e s c r i b e d above, i s i n t-rr , 21. (P. 2) I f -M e.. S K , and B <3 K such tha t B cz ( 0 : M ) , then . M, as a K / B - module, i s i n ^YL/B' (P. 3) I f , ' 0 •= ]-|{(o':M) : M e S K } , and i f B i s a non-zero i d e a l of K , then 2g ^ 0. (P .4) I f , for ' each non-zero i d e a l B of K , Eg / 0, then 0 = nt(O'M) : M e 2 ^ . Theorem 2 .1 .1 ( l ) I f , f o r each r i n g K there i s ass igned a c l a s s Zlr o f r i g h t K - m o d u l e s such that) P.O - P .3 are s a t i s f i e d , then the c l a s s S, c o n s i s t i n g of a l l r i n g s K f o r which' 2 R = 0, i s a r a d i c a l c l a s s . • (2) Under the assumptions of ( l ) , c o n d i t i o n P .4 is'" s a t i s f i e d i f and on ly i f , f o r each r i n g K, •S(K) = ni (0 :M) : M € EK3." Proof : ( l ) (This p roof Is due to I . H e n t z e l ( 1 0 ) ) . For each r i n g K, def ine k e r ( 2 , K ) = |~|{(0:M) : M e 2 K ) • I f k e r ( 2 , K ) = 0 , we; c a l l the c l a s s 2 K f a i t h f u l . A l s o , de f ine L(2) to be the c l a s s of r i n g s K f o r which t h e r e " e x i s t s an M e 2 R such tha t (0:M) = 0._ .(A K - m o d u l e ; M f o r which (0:M) = 0 w i l l be c a l l e d a f a i t h f u l module. Thus, i f 2^ has a f a i t h f u l module, the c l a s s £ ' i s f a i t h f u l . ) 2 2 . R e c a l l , from § 1 . 3 , t ha t i f a c l a s s C of rings-s a t i s f i e s : (E) Every non-zero i d e a l of a member of C can be mapped onto a non-zero member of C then the upper r a d i c a l w i t h respec t to C c o n s i s t s of a l l r i n g s which cannot be mapped onto a non-zero member of C. We s h a l l show tha t the cl,ass L ( s ) s a t i s f i e s ( E ) . Suppose tha t K i s i n L ( s ) , and tha t B / 0 i s an i d e a l of K. Since K ' i s i n L ( s ) , E'K- i s c l e a r l y f a i t h f u l , and so , by P . 3 , there i s ; an , M i n Eg. Le t I = (0:M) <! B. Then I / B, and by P. 2 , M, as a B / I - module, i s i n Z-Q/J-An easy c a l c u l a t i o n shows t h a t , i n the r i n g B / I , ( ^ : M ^ B / I = ® ' Thus B / I i s a non-zero member of L ( s ) . Prom the d i s c u s s i o n i n § 1 . 3 , the c l a s s S, c o n s i s t i n g of a l l r i n g s which cannot be mapped onto a non-zero member of L ( s ) i s a r a d i c a l c l a s s . I f K i s a r i n g such "that £ / 0, and i f M i s i n E ^ , the same argument as was used f o r B i n the p rev ious paragraph shows tha t ' K/(0 :M) i s i n L ( s ) . Conver se ly , i f 1 0 K , and i f K / I e L ( S ) , then there i s a r i g h t K / I module M i n ^YL/V F r o m p < 1 w e n a v e tha t M, w i t h the induced K module s t r u c t u r e i s i n S^. Thus we have tha t S T, / 0 i f and on ly i f K can be mapped onto a non-zero K member of L ( 2 ) , o r , e q u i v a l e n t l y , the upper r a d i c a l c l a s s w i t h respect to L(S) c o n s i s t s of a l l r i n g s K f o r which £ K = 0. 23. I f S(K) d\ k e r ( E , K ) , t hen , f o r some M e 2 R , S(K) i ( 0 :M) . Denot ing (0:M) by B , we have tha t ( S(K ) + B ) / B i s a non-zero i d e a l op K / B € L ' (E) . Thus, by c o n d i t i o n ( E ) , (S(K)+B)/B can be mapped to a non-zero member of L ( E ) . Since. ( S(K ) + B ) / B = S ( K ) / ( S(K)n B ) , and s ince the l a t t e r i s a homomorphic image of S ( K ) , we have tha t S(K) can be mapped onto a non-zero member.of. L ( E ) . Th i s homomorphic image T of S(K) then has £ T / 0. But T, be ing a homomorphic image.of an S - r a d i c a l r i n g , must be S - r a d i c a l , and thus E T = P C Th i s , c o n t r a d i c t i o n shows tha t S(K) c k e r ( E , K ) . (2) Suppose now tha t P. 4 h o l d s . K / S ( K ) ' i s S-semis imple , and ( D i v i n s k y (.6), page 125) so i s every- i d e a l of K / S ( K ) . Thus, f o r every non-zero i d e a l B of K / S ( K ) , B i s not S - r a d i c a l , and so Eg / 0. By, P. 4, k e r ( £ , K / S ( K ) ) = 0. For each M i n ^YL/S(K)} M C ^ N B E & I V E N A K - module s t r u c t u r e , and, as a K - module, M i s i n E^-. An easy c a l c u l a t i o n shows tha t ( 0 : M ) R ' = {k e K : M 6(k) = 0} = 8 _ 1 ( ( 0 : M ) K / S ( K ) ^ where 8 i s the n a t u r a l homomorph'ism of K to K / S ( K ) . . S ince 0 = k e r ( E , K / S ( K ) ) , we o b t a i n S(K) - 8~ 1 (0) = 0 " X ( PI -C(0:M) : M e % / s ( K ) 3 ) = F l C(0:M) M i s a K module induced from some M' e ^ K / S ( K ) ^ ' S i n c e the l a t t e r i n t e r s e c t i o n con ta ins k e r ( £ , K ) , we have S ( K ) =) k e r ( E , K ) , and thus the two are e q u a l . . 24. Conversely, suppose t h a t , f o r every r i n g K, 'S(K) = k e r ( S , K ) . If• the h y p o t h e s i s of .-P.4.. i s s a t i s f i e d , no non-zero i d e a l o f . K can be S-radical?, and so , 0 = S(K) = ker(£,K). T h i s completes the p r o o f of ( 2 ) . Q.E.D. • I t c o u l d be p o i n t e d out t h a t .any r a d i c a l c l a s s can be d e s c r i b e d i n t h i s f a s h i o n . To be s p e c i f i c , A n d r u n a k i e v i c and Rjabuhin a l s o proved t h a t , i f R i s -any r a d i c a l c l a s s , then the assignment t o a r i n g K the c l a s s of a l l n o n t r i v i a l r i g h t K-modules' M f o r which ,/K/(0:M) i s R-semisimple g i v e s elasges of modul©s -sat isfying F.O = P , 4 , and the r a d i c a l c l a s s S of the theorem i s the same as the c l a s s R. Me give two examples t o show how some f a m i l i a r r a d i c a l s are d e f i n e d i n terms of modules. Example 2 . 1 . 2 The Jacobson R a d i c a l For each r i n g K, l e t 2^ be the c l a s s of a l l i r r e d u c i -b l e r i g h t K-modules. I t i s e a s i l y seen t h a t P.O, P . l , and P.2 are s a t i s f i e d . Suppose t h a t ker(2,K) = 0 , and t h a t B / 0 Is an i d e a l of K. Then f o r some M i n £ K , B c£ (0:M), and MB ^ 0 . Since B . i s an i d e a l , MB i s . a K-submodule, _ and so MB = M. We can a l s o c o n s i d e r M . as a Br-module. We show t h a t , as a B-module, M i s simple. I f m e M, and m / 0 , then mB / 0 , f o r otherwise, s i n c e B i s a two-sided i d e a l , the s e t {n e M : nB = 0} would be a. non-zero submodule, and thus would be a l l of M. T h i s would gi v e MB = 0 , a c o n t r a d i c t i o n . .Now mB i s a non-zero K-25 . submodule of M, and so i s a l l of M. Thus M, a6 a B module, i s simple and n o n - t r i v i a l - i . e i M i s i n Eg. T h i s shows P. 3 i s v a l i d . Suppose t h a t K i s a r i n g f o r which Eg ^ 0 whenever B ' i s a non-zero i d e a l of K. I f T = k e r ( E , K) / 0 , then there i s a simple n o n - t r i v i a l T- module M. Then-there i s a t e T such t h a t Mt / 0. L e t m e'M be- such t h a t mt / 0. Then, ( u s i n g the same argument as was used i n the p r e c e d i n g paragraph) mtT / C]> and so M(tK) / 0. m(tK) i s c l e a r l y a T-submodule of M. We can g i v e i t the s t r u c t u r e of a K-module by d e f i n i n g m(tk')*k' to be M(tkk'). Furthermore, i f x i s i n T, then m(tk)*x = m(tk)x, and so any K- submodule of m(tK), with t h i s m u l t i p l i c a t i o n , i s a T-submodule of the o r i g i n a l , T-module M. T h i s shows t h a t M(tK) i s a simple K-module. A l s o , m(tK) i s not a t r i v i a l K-module, f o r , i f m(tK)*K = 0 , then m(tT)T = 0. Since m(tT) / 0 , {n e M: nT = 0} i s non-zero, i s a submodule, and so must b e , a l l of M. T h i s would g i v e MT = 0 , a c o n t r a d i c t i o n . We have shown t h a t the K-module m(tK) i s i n £ K - Since-, T = k e r ( E , K ) , T c ( 0 : m ( t K ) ) , and 0 = m(tK)*T •= m(tK)T. The argument above shows t h a t t h i s l e a d s t o a c o n t r a d i c t i o n . T h e r e f o r e T = 0 , . and P.k i s proved. i The corre s p o n d i n g r a d i c a l property,- as d e s c r i b e d i n . Theorem 2 . 1 . 1 , i s the upper r a d i c a l p r o p e r t y with, r e s p e c t t o the c l a s s of a l l r i n g s -K with a f a i t h f u l simple n o n t r i v i a l 26 . module. ' These are the r i g h t p r i m i t i v e r i n g s , and the r a d i c a l p r o p e r t y i s the Jacobson r a d i c a l p r o p e r t y . Example 2 . 1 . 3 The Baer Lower R a d i c a l As was p o i n t e d out i n Chapter one, the Baer Lower r a d i c a l i s a l s o the upper r a d i c a l with r e s p e c t t o the c l a s s of a l l prime r i n g s . We show how t h i s r a d i c a l can be d e s c r i b e d i n terms of modules. (See A n d r u n a k i e v i c and Rjabuhin ( l ) . ) A r i g h t K - module i s prime i f and only i f : MK ^ 0 , and i f m e M and B <J K are such t h a t mB = 0 , then m = 0 or MB = 0 . Lemma 2 . 1 . 4 A r i g h t K-module i s prime i f and only i f MK y 0 , and f o r every non-zero submodule N of M,, we have (0:N) = (0 Proof: I f M i s a prime r i g h t K--module, then MK / 0 . L e t N be a non-zero submodule of M. C l e a r l y (0:N) D , ( 0:M). A l s o , N(0:N) = 0, . and (0:N) i s an i d e a l o f K. Since N / 0 , we have M(0:N) = "6 , i . e . ... (0:N) c (0:M). Conversely, suppose t h a t M "is a module s a t i s f y i n g the c o n d i t i o n , and suppose t h a t m i n M and B. 4 K are such t h a t mB = 0 . Then mKB = 0 . I f mK 0 , i t i s a non-zero submodule, and we have B e (0:mK) = (0:M), and MB.= 0 . I f mK = 0 , then m = 0 , f o r otherwise [n e M : nK = 0} i s a non-zero submodule N such t h a t NK = 0 . T h i s would g i v e MK = 0 , - which i s f a l s e . T h i s proves the.lemma. Q.E.D. 27 . For each r i n g K, d e f i n e Eg to be the c l a s s of a l l prime r i g h t K-modules. I t i s easy t o v e r i f y t h a t P.O, P . l , and'P. 2 are s a t i s f i e d . I f ker(£,K) = 0 , and i f B . i s a non-zero i d e a l of K, then there i s a member M of Eg such t h a t MB / 0 . We show t h a t M i s a l s o a prime B module. L e t I < B and m e M s a t i s f y ml = 0 . Then mBI = 0 , and, s i n c e mB i s a K submodule of - M, e i t h e r MI = 0 (by the lemma), or mB = 0 . Since MB •/ 0 , the l a t t e r s i t u a t i o n i m p l i e s t h a t m = 0 . Thus M i s i n Eg, and P.3 i s proved. We now prove P.4.. Suppose t h a t , f o r each non-zero i d e a l B of K, Eg / 0. I f T = ker(£,K) / 0 , then there i s a prime T - module •M. Proceeding as we d i d i n ' t h e p r e v i o u s example, we can f i n d m i n M and t i n T such t h a t mt j?;' 0 . Then, s i n c e M i s prime and MT / 0 , m(tT) / 0 , and so m(tK) / 0 . As i n the. p r e v i o u s example, we can g i v e • m(tK) a r i g h t K-module s t r u c t u r e by d e f i n i n g m(tk)*k' t o be m(tkk'). We show t h a t t h i s K-module i s prime. S i n c e , as a T module, M i s prime, we have m(tT)T / 0 , and so m(tK)*K / 0 . I f B <1 K, and 0 = m(tk)*B = m(tkB), then m(tkTB) = 0 . Now, TB < B, and, s i n c e M i s a prime B-module, e i t h e r mtk = 0 or M(TB) = 0 . In the l a t t e r case, m(tKB) = 0 = m(tK)*B. Thus 'm(tK) € £ R , and (0: m(tK) ) 3 T = ker(E,K) - i . e . m(tK)*T = 0 . Repeating the same argument t h a t we. have used b e f o r e , the primeness of M as a T module leads t o a c o n t r a d i c t i o n . T h e r e f o r e , T = 0 , as desired." / 28. We have shown t h a t , i f , t o each r i n g K we a s s i g n the c l a s s L^. of prime r i g h t K modules, we have the c o n d i t i o n s P.O - P.4 s a t i s f i e d . The corres p o n d i n g r a d i c a l i s the upper r a d i c a l with r e s p e c t t o the c l a s s L ( Z ) of a l l r i n g s with a prime f a i t h f u l r i g h t module. The next lemma shows t h a t L ( E ) i s the c l a s s of a l l prime r i n g s . Lemma 2.1.5 K. has a prime f a i t h f u l r i g h t module M i f and only i f K i s a prime r i n g . Proof: L e t M be a prime f a i t h f u l r i g h t K-module, and suppose t h a t aKb =0, where a and b are i n K. I f a ^ 0, then Ma / 0, and MaK / 0, but MaKb = 0. MaK i s a sub-module, and t h e r e f o r e b e (0:MaK) '= (Q:M). ( T h i s l a s t e q u a l i t y come from Lemma 2.1.4.) Thus Mb = 0, and b = 0, s i n c e M i s f a i t h f u l . Conversely, suppose t h a t K . i s a prime r i n g . I t i s s t r a i g h t f o r w a r d t o v e r i f y t h a t • K, con s i d e r e d as a r i g h t module over i t s e l f , i s prime and f a i t h f u l . Q.E.D. T h i s lemma shows t h a t the r a d i c a l a s s o c i a t e d with the c l a s s e s o f prime K modules ( f o r a l l r i n g s K) i s the upper r a d i c a l w i t h r e s p e c t ' t o the c l a s s of a l l prime r i n g s - i . e . the Baer Lower r a d i c a l . 2.2 I n t e r s e c t i o n s , of One-sided I d e a l s In an e a r l i e r paper than the one c i t e d i n the pr e v i o u s s e c t i o n , A n d r u n a k i e v i c and Rjabuhin (2) showed t h a t any 2 9 . S P - r a d i c a l can be d e s c r i b e d as i n the p r e v i o u s s e c t i o n i n such a way t h a t a l l members of 2^ -, f o r a l l K, are prime modules. Prime modules, t h e r e f o r e , occur i n many s i t u a t i o n s . In c o n t r a s t t o the s i t u a t i o n f o r the Jacobson r a d i c a l , i t i s not c l e a r f o r s p e c i a l r a d i c a l s i n g e n e r a l , l e t alone f o r a r b i t r a r y r a d i c a l s , whether or not the r a d i c a l o f a r i n g K can be expressed i n terms of one-sided i d e a l s . Of course, f o r the Jacobson r a d i c a l case* J(K) i s always the i n t e r s e c t i o n of the maximal modular r i g h t i d e a l s . We give next a g e n e r a l i -zation of th i s result.. ' 7 In the s e q u e l , the n o t a t i o n I < K means I i s a • r t r i g h t i d e a l of K. , We c o n s i d e r c l a s s e s 2 K of K-modules which s a t i s f y P.O - P.4 and a l s o the a d d i t i o n c o n d i t i o n (P. 5) Every non-zero submodule of a member of 2^ - i s a l s o a member of' 2^. Th i s c o n d i t i o n i s s a t i s f i e d , f o r example,if 2^ . i s the c l a s s of i r r e d u c i b l e r i g h t K-modules, or i f 2^ i s the c l a s s of prime r i g h t K-modules. Theorem 2 . 2 . 1 Suppose t h a t , f o r each r i n g K there i s ass i g n e d a class. 2T, of r i g h t K-modules' such P.O- - P. 5 K „ , are s a t i s f i e d . For any r i n g K, l e t s(K) ='•••{I: I < K and ' ' r t K/I e 2^}. Then s(K) has a Subset J such t h a t , i f S i s \ 30. the r a d i c a l d e f i n e d by the c l a s s e s Eg (as i n theorem 2 . 1 . 1 ) , then S(K) = TJ{I: I-« K a n d 1 e J3 • r t Furthermore, i f each member of s(K) i s a modular r i g h t i d e a l , then we may take J t o be a l l of s ( K ) , and thus S(K) = (1(1: 1 < K a n d K / I € 2 K } . r t * Proof: L e t m / 0 be a member of M, where M e Eg. Then mK y 0 , f o r otherwise N = { x e M : x K = 0 } would be a non-zero submodule of M, and,, by P.5> N would be i n Eg. However NK = 0 , and t h i s c o n t r a d i c t s P.O. .) Th e r e f o r e mK / 0 , and t h i s i s (by P.5) a member Eg. Since mK = K/(0:m), we have (0:m) i s i n s ( K ) . Now S(K) = fl (0:M), by theorem 2 . 2 . 1 , so we have MeEg - 1 S(K) = 0. (0:M) = n n { ( ° * m ) * 0 / m e M} . Thus, i f we MeEg MeEg . -set J = {I: I < K f o r which 3 MeEKJ, and 3 m / 0 / m e M r t . such t h a t • I = (0:m)}, we have J c s(K) and S(K) = f~| J-Suppose now t h a t each member o f s(K) i s modular. That i s , suppose t h a t f o r each I i n s(K) there i s an element e^. i n K such t h a t e^x - x e I f o r each, x i n K. L e t m = e I + I e K/I. Then (0:m) = {k e K : e'jk e I) = [k.e K : k e I) = 1 . Thus I i s i n J and J = s ( K ) . ..Q.E.D. 31. CHAPTER THREE • THE UNIFORM, RATIONAL, AND WEAK RADICALS In t h i s chapter, we s h a l l d e f i n e , by means of modules, three r a d i c a l c l a s s e s , and some of the p r o p e r t i e s of these r a d i c a l s w i l l be i n v e s t i g a t e d . " The b a s i c concepts used here are u s u a l l y a p p l i e d i n the study of q u o t i e n t r i n g s , as found i n the papers of Utumi, Johnson, and G o l d i e . .A good r e f e r e n c e which c o n t a i n s most of the b a s i c r e s u l t s i s F a i t h ( 7 ) . 3.1 E s s e n t i a l and. R a t i o n a l Extensions A r i g h t K-module M i s an e s s e n t i a l e x t e n s i o n of a submodule N i f T D N )4 0 whenever T i s a non-zero submodule of M. Under these circumstances, we a l s o say t h a t N i s an e s s e n t i a l ,-submodule of M. We say t h a t M i s a r a t i o n a l e x t e n s i o n of a -submodule N i f , whenever we have N c T c M, T a submodule of M, and f:T -* M a module homomorphism such t h a t f(N) = 0, then f =' 0. F o l l o w i n g F a i t h ( l ) , we w r i t e M v N i f M i s an e s s e n t i a l e x t e n s i o n of N, and M v N i f M i s a r a t i o n a l e x t e n s i o n of N. Lemma 3.1.1 I f M v N, ' then M v N. . Proof: I f T i s a submodule of M such t h a t T.D N = 0, then d e f i n e f: N '© T - M by "f(.t) = t , f o r t e T, and 32. f(n) = 0 f o r n e N. . Then f(N) = 0 , and f must therefore . be the zero map - that i s , T = 0 , Q.E.D. Lemma 3 . 1 . 2 M v N i f and only i f , whenever x and y are i n M, with y ^ 0 , there i s a k e K and an integer n such that xk + nx e N and yk + ny ^  0 . Proof: See Faith ( 7 ) , page 58 , f o r a proof of thi s r e s u l t . I f M i s a r i g h t K-module, an element m. of M i s said to be a singular element i f (0:m) i s an ess e n t i a l submodule ( i . e . r i g h t ideal) !iof K. I f m i s a singular element, and i f x € K, then mx i s , a l s o singular. For-l e t 0 'ft I < K. I f x l = 0 , then mxl = 0 , and I c (0:mx). r t I f x i yi 0 , then (0:m) fl x l / 0 , and there i s an i / 0 i n I such that mxi = 0 . In either case, I D (0:mx) / 0 , for a l l 0V 4 I <1 K, and so mx i s singular. Clearly a r t f i n i t e i n t e r s e c t i o n of e s s e n t i a l r i g h t ideals i s e s s e n t i a l , so, i f x and y are .singular elements, (0:x-y) £ (0:x) fl ( 0:y), and x-y i s also singular. We see therefore that the set 1 of singular elements of a module forms a submodule. This i s known as the singular submodule of M, and i s denoted by Z(M). C l e a r l y , i f N i s a submodule of M, then Z(N) = N fl Z(M). I f M = K, • as a r i g h t K - module, the -singular submodule i s a two-sided i d e a l of'• K, as i s c a l l e d the righ t singular i d e a l of K. This i s usually, denoted Z r(K). The l e f t singular i d e a l i s defined i n the obvious way. The 3 3 . r i g h t and l e f t singular ideals of a r i n g need not be the. same. (See, for example, Small ( 2 5 ) . ) The next lemma provides a p a r t i a l converse to Lemma 3 . 1 . 1 . Lemma. 3 . 1 . 3 If Z(M) = 0 , and M v N, then M v N. Proof: Let x and y ^ 0 be i n M. Since N i s es s e n t i a l i n M, i t follows that (N:x) i s es s e n t i a l i n K. ( I f I i s a r i g h t i d e a l of K, either x i = 0 c-N, and I c (N:x), or x l ^  0 , i n which case x i fl N ^  0 and therefore I n (N:x) ^ 0 . ) . Since Z(M) = 0 , y | Z(M), and (0:y) i s not e s s e n t i a l , whence (N:x) ( 0:y). There i s therefore an element k of K i n (N:x) but not i n ( 0:y). Thus xk e N, and yk - f - Q . ' j The rest follows from Lemma 3 . 1 . 2 . Q.E.D. 3 . 2 The Uniform, Rational, and Weak Radicals A module i s c a l l e d uniform i f i t Is an e s s e n t i a l exten-sion of every non-zero submodule. I f , i n addition, M i s a ra t i o n a l extension of every non-zero submodule, we s h a l l say that M i s r a t i o n a l l y uniform. We say that two modules M and M' are subisomorphic i f there are monomorphisms f:M -• M' and g:M' M. A module M which i s subisomorphic to every non-zero submodule w i l l be c a l l e d a homogeneous module. x Lemma 3 . 2 . 1 A homogeneous module which i s n o n - t r i v i a l i s prime. Proof: Let N. be a non-zero submodule of a homogeneous -34. module M. Then there-Is a monomorphism f:M-»-N. Since f i s one-to-one, i t follows that (0:M) = (0:f(M) )3 and t h i s contains (0:N). Since N c M., (0:M) c (0:N), and so (0:N) = (0.:M). ' By Lemma, 2 . 1 . 4 , M i s therefore a prime module. Q.E.D. We now define., by using these classes of modules, three r a d i c a l properties. Two of them are new, and we s h a l l show that the t h i r d coincides with the weak r a d i c a l of Koh and Mewborn (17 ) . • For each r i n g K, we .assign the following classes of r i g h t K modules: , . j Sg i s the class of a l l prime, uniform r i g h t K-modules. p ; S K . i s the class of a l l prime, r a t i o n a l l y uniform r i g h t K -modules. 3 ' Eg i s the class of a l l r a t i o n a l l y uniform, homogeneous ri g h t K - modules. Theorem 3 . 2 . 2 The properties P.O - P.5 are s a t i s f i e d 1 2 3 • for each of the classes S , S , and S . (See Chapter two). Proof: It i s e a s i l y v e r i f i e d that the properties of uniformity, r a t i o n a l uniformity, and homogeneity, as well as primeness, are inherited by non-zero submodules, and so P.0-, P.5 are seen to be true. The discussion of Example 2 . 1 . 3 showed that the property of prim'eness i s preserved under the induced . . 35-module s t r u c t u r e s a r i s i n g i n . P . l and P.2. U n i f o r m i t y i s a l s o p r e s e r v e d s i n c e i t i s a p r o p e r t y of the l a t t i c e of submodules of a module, and the K- and K/B -modules o c c u r i n g i n P . l and P.2 have isomorphic l a t t i c e s of submodules. The other p r o p e r t i e s , r a t i o n a l i t y and homogeniety, are d e f i n e d i n terms of module homomorphisms from submodules of M . t o submodules o f M. I t i s e a s i l y v e r i f i e d t h a t these are p r e s e r v e d under the o p e r a t i o n s of P . l and P.2. I t 1 p 3 f o l l o w s then t h a t each of the c l a s s e s 2 , 2 , and 2 s a t i s f y P . l and P. 2 . We now e s t a b l i s h P.3. Suppose t h a t 0 = n C ( 0 : M ) : M e'Sg}, where i i s 1 , 2 , or 3 , and l e t B / 0 be an i d e a l of K. Then there i s an M i n 2 ^ such t h a t MB / 6 . The d i s c u s s i o n , i n Example 2 . 1 . 3 showed t h a t M, as a B-module, Is prime." I f M i s u n i f o r m as a- K-module, and i f U and V are two non-zero B-submodules of M, then ( u s i n g the f a c t t h a t M i s prime)' UB and VB are non-zero. These are K-submodules of M, a n d ' t h e r e f o r e have a non-zero i n t e r s e c t i o n . Thus U (1 V ^ 03 and M i s a uniform B-module. Suppose t h a t M, as a K-module, i s a r a t i o n a l e x t e n s i o n of every non-zero submodule, and suppose t h a t T and N are B-submodules o f ' M r.O / ' T c N, and suppose t h a t f:N - M i s 'a B-homomorphism such t h a t f ( T ) = 0 . Then TB and NB are non-zero ( s i n c e B is' prime) and the r e s t r i c t i o n f'~ 3 6 . of f to NB i s a B-homomorphism for which f'(TB) = 0. We claim f' • i s a K-homomorphism. Any element of NB i s of the form x = E ^ i ViPx3 w h e r e v i e N> a n d b i e B* F o r k e K, f ^ y . b ^ k = ( f ' ( y i ) b i )k = f ^ y . ^ b . k ) = f'ty.b.k)., and i t follows that f'(x)k = f'(xk). Since M i s a r a t i o n a l extension of the K - submodule TB, and since f'(TB) = 0, we must have f = 0, and therefore f(NB) =0. For n e N, f(n)B = 0. The primeness of M, and the fa c t that MB ^ 0, give us that f(n) =. 0, and so f i s the zero map. This shows that M^, . as a B module, i s a r a t i o n a l extension of T, where T i s any non-zero submodule. - I F i n a l l y , suppose that M i s subisomorphic to every non-zero K-submodule of M, and l e t N be any non-zero B-submodule. Then' NB i s a non-zero K-submodule, and NB i s subisomorphic to M. The k-monomorphism g:M-» NB c N, and the embedding map of N into M are B-monomorphlsms, and so M and N are subisomorphic. This completes the ' 1 2 3 proof that each of the classes E , £ , and z s a t i s f y P. 3 Now we e s t a b l i s h P.k. Let . i be 1, 2, or 3, and suppose that Eg / 0 f o r any non-zero id e a l B of K, but 0 7^  n { ( ° : M ) : M 6 2g} = T. 'Then,there exists a prime T-module M i n E^. As i n Example 2.1.3, there i s m; i n .M, t i n T such that m(tK) ^ 0> and t h i s has a K-module structure 37. (defined by ( m(.tk) )*k' :=' m(tkk') ) for which N = m(tK) is a prime K-module. If M •is uniform as a T module, since K-submodules of N are also T-submodules, N is also uniform as a K-module. Any non-trivial K homomorphism from a K-submodule of N to N is-a T-homomorphism from a T-submodule of M into M, and must therefore have zero kernel i f - M, as, a T-module, is a rational extension of every non-zero submodule. Thus this latter condition on the T-module M is passed on to the K-submodule N. , 3 i Finally, suppose that M € 2 ^ , and let' L be a K-submodule of N. Then L is a T-submodule of M, and there is a T-monomorphism f: M -• L. We show that the restriction f of f to m(tK) is indeed a K-homomorphism. For any k^ and k' in K, ^- we have f(m(tk))*k / = (. f(m)*tk)*k' (since f is a T-homomorphism) = ( f(m)*tk)*k' (since f(m) is in L c N, and, in T the multiplications • and * agree) = f(m)*(tk'k') (since L is a K-moduleJ ; = f(m*tkk') (since tkk' e T, and f is, a T-homomorphism) . • \ = f((m*tk)*k' ) (from the definition of the multiplication *) The map f, restricted to N, i s therefore a K-homomorphism, 38. and i t i s one-to-one. This map, together with the imbedding of L into N, are the maps which guarantee that N and L • 3' are subisomorphic, and N e 2j£. Q.E.D. The previous theorem assures us that each of the 1 2 3 classes 2 , 2 , and 2 define r a d i c a l properties as described i n Theorem 2 . 1 . 1 , and that, f o r any ri n g K, the "radical Is n { ( ° : M ) ' M E 2 K 3 > "the 2 being 2 , 2 , or 2^, as the case may be. We-shall c a l l these r a d i c a l s the uniform r a d i c a l , the r a t i o n a l r a d i c a l , and the weak r a d i c a l , respectively, and we denote them by U, U*, and W. Thus U i s the upper r a d i c a l with respect to the class C y of a l l rings K with a faithful", uniform., prime r i g h t K-module; U* is the upper r a d i c a l with respect to the class of a l l rings K with a f a i t h f u l , r a t i o n a l l y uniform and prime r i g h t K-modulej and W i s the upper -radical with respect to the class Cfi of a l l rings K with a f a i t h f u l , r a t i o n a l l y uniform, homogeneous r i g h t K-module. Since C w c c C y, we have U cz U* c W. Prom Lemma 2.1 .5a ™Q see that any member of Cy i s a prime ri n g . On the other hand, any ri g h t primitive r i n g (a.ring with a..simple n o n - t r i v i a l r i g h t module which l i s f a i t h f u l ) i s c l e a r l y i n _ C^. Therefore we have B c U G U* c W c J , where B and J are the Baer Lower and Jacobson r a d i c a l classes. 39. Theorem 3.2.3 I f K i s a commutative r i n g , then B(K) = U(K) = U*(K) = ¥(K). Proof: From Example 2.1.3, we have that B(K) =-[1(1 < K:K/I is a prime ring}. Therefore, i f we can show that any commu-tat i v e prime r i n g i s i n C^, then for a commutative prime ri n g K, we w i l l have that W(K) i s contained i n each, i d e a l ' '.I f o r which ,K/I: i s prime, -,and then W ( K ) c B ( K ) . Let K be a commutative prime r i n g . Then K i s a domain (xy =0 implies x = 0 or y = 0). K, as a ri g h t K module i s c l e a r l y uniform, f o r i f U and V are -non-zero submodules ( i . e . i d e a l s ) , then 0 / U V = V U c r U n V . Since K has no zero d i v i s o r s , Z r(K) =0, and, by Lemma 3.1.3 we have that K i s r a t i o n a l l y uniform/ I f U i s . a non-zero submodule, and i f . 0 ,/ u e U, then. L y : K U, where L u(k) = uk, i s a K-monomorphism, and U i s homogeneous. K i s e a s i l y seen to be a faithful.K-module, and so the r i n g K i s i n C w. Q.E.D. Corollary 3.2.3 The ra d i c a l s J and W are d i s t i n c t . Proof: The subring S of the ra t i o n a l s consisting of numbers of the form (even integer)/(odd integer) i s a commutative prime r i n g which i s also r i g h t quasi-regular. Then W(S) = 0, and J(S) = S. Q.E.D. .; Recall from §1.5, that a class C of rings i s c a l l e d a. s p e c i a l class i f i t s a t i s f i e s : ... (X) Every member i s prime.. . , ; (Y) C u {0} i s hereditary. -(Z) I f A c C, and A <3 K, then K/A* i s i n C, where A* = {x e K:xA = Ax = 0}. The upper r a d i c a l s with respect to the s p e c i a l classes are c a l l e d the SP-radicals. ' Theorem 3 . 2 . 4 Each of the classes Cy, and are sp e c i a l classes, and therefore U, U* and W are a l l SP-radicals. '. • ,. • . • . . • j. Proof; The proof i s the same i n a l l three cases. Let 1 2 3 Eg denote one of Eg, Eg, or Eg, and l e t C be the corresponding class Cy, Cy*, or C^. We already know that any member of C i s a prime ri;ng. Suppose that K 6 C and 0 V B < K. Let M be a • f a i t h f u l member of Eg. - M can be regarded as a B module, and {b e B : Mb ±* 0} = 0. The proof that M, as a B module, i s i n Eg, proceeds exactly as i n the v e r i f i c a t i o n of property P.3 i n Theorem 3 . 2 . 2 . This proves that B i s i n C and condition (Y) i s established. Suppose that A\"is i n C, and that A <J K. In order to prove that condition (Z) i s true, we must show that K/A* i s i n C, where A* = (k e K : kA = Ak =0}. We show that A* i s indeed an i d e a l of K. I f x and y are i n A*, then x-y i s also. Suppose.that x e A* and .k e K. 4 l . Then A(xk) = 0. Also (xk)A c xA = Oj and thus xk e A*, and A* i s a r i g h t i d e a l of K. S i m i l a r l y A* i s a l e f t i d e a l . Now A i s i n C, so there i s a f a i t h f u l member M of E^. Proceeding exactly as i n the v e r i f i c a t i o n of P.4, there are m i n M, and a i n A such that m(aK) / 0. Then N = m(aK) has a K module structure defined by m(ak)*k' = m(akk')> and (as i n the proof of P.4) t h i s i s a member of E R . What i s (0:N R)? Since a K c A, c l e a r l y we have A* c (0:N R). Suppose now that x 6 (0:N). Then 0 = m(aK) (AxA). Since m(aK) i s an A submodule of M, since AxA < A, and since M i s prime, M(AxA) = 0. But M i s also f a i t h f u l , and so AxA = 0. Thus (Ax) - (xA) = 0 and since A i s a prime r i n g , xA = Ax = 0 i . e . x € A*. Thus A* = (0:N K). Then, by P. 2, N € E R /, A* and (°:N K /, A*) 3.3 Matrix Rings and the Radicals U, U*, and W For many r a d i c a l s R, i t i s true that R(K n) = R ( K ) n > where S denotes the r i n g of n x n matrices with entries n i n a r i n g S. However, t h i s i s not true f o r a l l r a d i c a l s -not even f o r a l l SP-radicals. For example, the generalized n i l r a d i c a l N , defined as the upper r a d i c a l with respect S to the class of a l l rings with no zero d i v i s o r s , does not have t h i s property. A f i e l d F i s N semisimple, but the r i n g \ 42. of 2 x 2 matrices over P i s a simple ring with zero divisors, and so is radical. We shall prove now that the radicals U, U* and W do have this property. Recall that U, U* and W are a i l SP-radicals, and (see Divinsky, (6) ), any SP-radical i s hereditary. Lemma 3.5.1 For a hereditary radical R, i n order to prove that R(K n) = R(K)^... for a l l rings . K, . i t is sufficient to prove i t for a l l rings which have a unity element. Proof: It i s well known that i f S i s a ring with unity, there i s a one-to one correspondence between the two-sided ideals of S and the two-sided ideals of S . An ideal I n of S corresponds to I n , and an ideal B of S n corres-ponds to the set of a l l members of S which are an entry of some member of B. This correspondence is an isomorphism of the lattices of two sided ideals of S and of Sn. It is equally well-known that any ring K can be embedded as an ideal in a ring K with unity. If R Is any hereditary radical,, then R(K) = K fl R(K). Under the lat t i c e isomorphism, we get R ( K ) n = K n n R ( ^ ) n * T h e r e f ° r e -i f a l l rings S with unity satisfy R(S n) = R(S) n, for any ring K we have R(K) n « K n ft R(K) n = K n n R(Kn) = R(K ft), the last equality being due to the fact that R i s hereditary, and K n 4 K n» Q.E.D. I 43. In the sequel, 1^. w i l l denote the matrix which, when mu l t i p l i e d on the r i g h t to a matrix M has the e f f e c t of interchanging the i ' t h and j ' t h columns of M. Thus hi = <rst>> w h e r e r i i " r j j - °> r i J - r J i - X> r s s - 1 f o r s / i , s / j , and r t » 0 f o r a l l other s and t. Also, f o r any element k of a r i n g K, Ej, .j(k) w i l l denote the matrix which has k i n the p o s i t i o n , and zeros i n a l l other p o s i t i o n s . Lemma 3 . 3 . 2 Let C represent any of the classes C u, Cy*, or C^. I f K i s a r i n g with unity, and i f K i s i n C, then K i s i n C. n Proof; Let M be a f a i t h f u l member of the class E K (= where i = l , 2 , or 3 ) . Define H to be the Cartesian product of n copies of M. With componentwise addition and the obvious m u l t i p l i c a t i o n of an n-tuple (m^nig,. * ,mn) on the r i g h t by an n x n matrix from the r i n g K n, M becomes a r i g h t K^-module. Since M i s i n Ej^, M i s a prime, module. We show that M i s also prime. Let N be a non-zero submodule of M, and l e t 0 / x = (x-^Xg,... ,x n) € U. Suppose, say, that Xj / 0. Then 1? also contains a n d "this n-tuple has i t s f i r s t entry non-zero. Let 1 < K n, and l e t B be the 44 set of members of K which are entries of members of B. Then, as was mentioned earlier, B A K, and Bfi = B". If w = (w1,w2,...,wn) sa t i s f i e s . xB = 0, where w^  ^  0, say, then B" = 0. For l e t y • (y s t ) € 13. Then 0 = " ^ ^ ^ ^ ( " ^ J y ^ t l ^ 1 ^ = (Wjrygt,0,....,0) for a l l r i n K. Thus w j K y g t = 0 i n M, which i s a prime f a i t h f u l K-module, and w^  ^  0. We must have y ^  = 0, and y = 0. This shows that "M i s prime and f a i t h f u l . Now suppose that ' M is uniform.' Let TJ and V be non-zero submodules of M, and l e t 1 u = (u1,u2>...,un) and v = (v^Vg,... ,v n) be non-zero i n TJ and V respectively. The remarks i n the previous paragraph allow us to assume that and v.^  are not zero, and we so assume. Let k and k' be elements of K such that u^k = v.jk' ^  0. Then uE i ; L(k) = vE 1 1 ( k / ) = (ujkjO,... ,0) 0, and this is i n U n V. - Therefore "H i s uniform, provided M is. Now we consider the case where M Is rationally uniform. Let N be a non-zero submodule in M, and suppose that w = ( wi*»»»* w n) ^  0 is i n N. As we have seen, we can take w1 0. Let x = (x^Xg,... ,x n) and y = (y^yg* • • »y n) 0 be any members of M. Suppose that y^ ^  0. Since M Is a rational extension of w.jK, there is a k i n K such that s 45. x^k € w^K, and y^k / 0. (This comes'from Lemma 3.1.3). Suppose that x^k = w^k'. Then x E t l ( k ) =» (x tk,0,0,...,0) = (w.jk', 0,0,0) - wE 1 1(k / ) 6 N, and y E t l ( k ) = (y tk,0,... ,0) / 0. By Lemma 3.1.2, M i s a r a t i o n a l extension of N. Thus, i f M i s r a t i o n a l l y uniform, so i s M. Suppose that M i s homogeneous, and l e t N be a non-zero submodule of M. I f x = (x 1,x 2,...,x n) i s i n N, so i s x E t l ( l ) = (x.^,0,... ,0). I f we define N to be {meM : x = {x^9..,,x ) e N such that = m}, then N i s a submodule of. M which contains a l l elements of M appearing as entries of members of N. Consider {(x 1,...,x n) € "M: x ± e N, i=l,2>... ,n}. This set N', we have seen, contains N. Conversely, i f (x^,...',x ) eW9 each x^ i s the f i r s t entry of some w.^  i n 1?, and w 1 E 1 1 ( l ) + ... + w n E ^ n ( l ) - (xl90909...90) + (0,x 2,0,...,0) + ... + (0,0,...,x f i) = x. Therefore x € N, and N = !'. M i s subisomorphic to N, and so there i s a K-monomorphism f:M-» N. Define 7:M -» N by f(x^,...,x n) =T (; f ( x 1 ) , . . . , f (x n) ). This i s one-to-one, and i s a Kn~homo-morphism. Also, f(M) c N' » N. This shows that M and N are subisomorphic. Q.E.D. The converse of t h i s lemma i s also true. 46. Lemma 3.3."5 I f C represents any one of the classes Cy, C u # , or C w then, i f K i s a r i n g with unity, and i f . K e C, then K € C. n 1 2 Proof: Let M be a f a i t h f u l member of E K ( = E R , E R , n n n or E^ as the case .--may be), and define a K-module structure K n * on M be defining m*k to be mE 1 1(k). Since M i s f a i t h f u l as a K • module, M*K = 0. Choose m such that n m*K 0, and l e t M' = m*K = {mE^k): k € K).. This w i l l be. a member of E K, and w i l l be f a i t h f u l . To show primeness, suppose that k' annihilates a non-zero submodule N of M'. Then there i s a k i n K such that m*k / 0, m*k € N, and, m*k*K*k' = 0. For any * - < * i j > i n V x - s i , j = i V x i J } ' a n d ^ n ( k ) x E i : L ( k ' ) m E 1 1 ( k ) E 1 1 ( x 1 1 ) E 1 1 ( k ' ) = m*k*x 1 1*k / = 0. Since m*k « mE i ; L(k) and since M, as a Kft module i s f a i t h f u l and prime, E n ( k ' ) e'(0: mE i : L(k)K n) - (0:M) « 0, whence k' = o . Thus every non-zero submodule of M' i s f a i t h f u l , and M' i s prime and f a i t h f u l . Suppose now that M i s a uniform K n module, and l e t U' and V' be non-zero submodules of M'. Let u «=• m*k and v = m*k' be non-zero and i n U' and V' respectively. Then uK n vK ^ 0. Let x • and 47 y = (y^j) he such that ux = vy. ^ 0. Writing x as 2 i , j = l E i j ( x i j ^ ' w e s®e u x 13 m E n ( k ) x = m( E i j ( k x i j ^ I f i t were true that, f o r t=i,2,...,n, u x E ^ l ) = 0, we would have 0 = E ^ i u x E t i ( 1 ) E i t ^ 1 ^ = u x * w n i c n i s f a l s e . Therefore, f o r some t , u x E t l ( l ) = m E ^ k x ^ ) =/ 0. Computation shows that u * x l t = u x E t l ( l ) , and that •v*y l t = v y E ^ ( l ) . Therefore u * x l t = v*y 1 < f c 0, and U' 0 V 0. M' i s shown to be uniform. Assume that M i s a r a t i o n a l extension of every non-zero submodule, and l e t N' be a non-zero'submodule of M'. Suppose that x = m*k^ and y = m*kg ft 0 are i n M', and l e t z = m*r be a non-zero member of N'. M i s a r a t i o n a l extension of zK n ft 0, and, by Lemma 3.1.2, there i s an a = (a^j) such that xa e zK n and ya ft 0. Now, ya = m( E j ^ i E l j ^ k 2 a l j ^ ^' and, as i n the previous paragraph, y a E t l ( l ) = m E 1 1 ( k 2 a l t ) ft 0 f o r some t. Therefore y * a i t = m E 1 1 ( k 2 a l t ) ^ 0. We also know that xa € zK n. Suppose that-xa as zw, where w = ( w i j ) * Then x * a i t ~ x E l l ^ a l t ^ ~ m E 1 1 ( k 1 ) a E t l ( l ) = x a E t l ' ( l ) = z w E t l ( l ) = m E ^ r ) w E t l ( l ) = mE i : L(rw l t) = m E 1 1 ( r ) E 1 1 ( w l t ) * Z * M \ % € Z * K £ N ' • Therefore y * a l t ft 0 and "K*8--^ e N' showing that M' i s a r a t i o n a l extension of N'. 48. /. Suppose now that M i s homogeneous, and l e t N' / 0 be a submodule of M'. Let z = m*k / 0 be i n N'. Then zK i s a non-zero submodule of the K module M, and there n n i s a K n homomorphism f:M -• zK n. The r e s t r i c t i o n g of f to M' i s one-to-one and i s a K-homomorphism. I t remains only to show that g(M') c N'. Rec a l l that M' = m*K. Let f(m) = zx e zK n5 where x » ( x ^ j ) . For-any m' i n M', m' « m*r « mE i ; L(r), and g(m') =. f(mE ] L 1(r)) = z x E 1 1 ( r ) = m E 1 1 ( k ) x E 1 1 ( r ) = m E 1 1 ( k x 1 1 r ) = m E 1 1 ( k ) E n ( x ] L l r ) = z*(x11r) € z*K c N'. The proof i s complete. Q.E.D. Theorem 3.3.4 I f K i s a r i n g with unity, and i f R i s any one of the r a d i c a l s U> U*, or W, then R(K n) = R ( K ) n > Proof; By theorem 2.1.1, f o r any r i n g K, R(K) = fl{(OsM): M e E K ) . From properties P . l and P.2, i t follows that an i d e a l I of K i s (0;M) f o r some M e E K i f and only i f K/I i s i n C. Also, i f K has a unity, the l a t t i c e s of two-sided ideals of K and of K n are isomorphic, and, for-I < K, K n A n = (KA)n " that i s , they are isomorphic rings. I t follows then that K/I e C i f and only i f K n / I n e C. Therefore R(K).= H{I < K;K/I e C} = f l C l « K: K n / I n e C}. Using the l a t t i c e isomorphism, t h i s gives '< 49-R(K) n - Cn £ 1 4 K: K n / I n € C } ] n = ntln <l K n: K n / I n € C}, ' and t h i s l a s t i n t e r s e c t i o n i s j u s t R ( K n ) » Q.E.D. Corollary 3 * 3 * 5 I f K i s any r i n g , and i f R i s any one of U, U*, or W, then R(K n) = R(K) n. Proof: Apply Lemma 3 . 3 . 1 and Theorem 3 . 3 . 4 . Q.E.D. 1 2 3 3 . 4 Relations Between the classes £ , E , E In order to c l a r i f y the re l a t i o n s between these classes, and between the corresponding r a d i c a l s , we introduce two new IT r a d i c a l s . For any r i n g K, define ££ to be the class of a l l n o n - t r i v i a l homogeneous r i g h t K-modules. R e c a l l , from H K Lemma 3 . 2 . 1 , that every member of £„ i s a prime module. TJTT "1 IT Also, f o r any- r i n g K, we define E K to be £ R fl £ R . An examination of the proofs of Theorems 3 . 2 . 2 and 3 . 2 . 4 w i l l show that these classes s a t i s f y P.O - P.5* and that the r a d i c a l s they define are SP-radicals. We s h a l l c a l l these H r a d i c a l s the homogeneous r a d i c a l , corresponding to £ , which which w i l l be denoted by H, and the uniformly homogeneous r a d i c a l , denoted by H U, corresponding to E ^ . From the "5 • ' "5 2 H d e f i n i t i o n of £ , we see that E ^ , = £ R n E £ f o r any r i n g K. The r e l a t i o n between these classes can be expressed i n the following diagram (where the arrows represent i n c l u s i o n s ) . 50 . I r r e d u c i b l e K-modules F i g . 3 . 4 . 1 Theorem 3 . 4 . 1 Any member of C^, the c l a s s of r i n g s K H with a f a i t h f u l member of ET/., has no non-zero n i l r i g h t is. i d e a l s and no non-zero n i l l e f t i d e a l s . Proof: Suppose' t h a t K e C^, and l e t L be a non-zero n i l l e f t i d e a l . I f a e L, then, f o r any x i n K,(xa) i s i n L, and ( x a ) n = 0 f o r some. n. Then ( ? a x ) n + 1 = 0 , and aK 'is a n i l r i g h t i d e a l . Since K i s a prime r i n g , by Lemma 2 . 1 . 5 , aK / 0 . T h e r e f o r e , i f there i s a non-zero n i l l e f t i d e a l i n K, there i s a l s o a non-zero n i l r i g h t i d e a l . We denote such a r i g h t i d e a l by T. Let M be a f a i t h f u l member of 2 ^ . Then MT / 0 , and so mT / 0 f o r some m i n M. Now mT i s a K-submodule of M, and so there i s a K-monomorphism f:M -• mT ( s i n c e M, i s homogeneous). Suppose t h a t f(m) = mt. By i n d u c t i o n we o b t a i n f (m) = mt f o r every p o s i t i v e i n t e g e r n. The n element t i s n i l p o t e n t , and so f °(m) = 0 f o r some n- . 51 Since f i s one-to-one,' we obtain m =?. 0, a contradiction. Therefore there cannot be any non-zero n i l r i g h t or l e f t ideals i n K. Q.E.D. Corollary 3.4.2 The n i l r a d i c a l N i s contained i n H. Proof; For any r i n g K, l e t M € E^, and l e t A = (0:M). Then (N(K)+A)/A i s a n i l i d e a l of K/A,* a member of C R, and therefore N(K)+A = A, and N(K) c A. Therefore N(K) c f|C(0:M) : M e E^} = H(K). Q.E.D. • Recal l that the generalized n i l r a d i c a l N g i s the upper r a d i c a l with respect to the class of a l l rings with no zero d i v i s o r s . -Lemma 3.4.3 A r i n g with no zero d i v i s o r s i s i n C H. Proof: Consider K, a r i n g with no zero d i v i s o r s , as a rig h t module over i t s e l f . For any non-zero submodule ( r i g h t ideal) T, l e t t ^  0 be i n T. The map L t:K - T, where L^(k) = tk, i s a K-monomorphism, which, together with the _ natural imbedding of T into K, gives a pai r of maps which shows K and T are subisomorphic. Therefore K i s i n E^, and i s f a i t h f u l . Q.E.D.' Theorem 3.4*4 The r a d i c a l H i s contained i n N .. 52. Proof: From § 1 . 3 and the d e f i n i t i o n of H, an H-radical ri n g K i s one which cannot be homomorphically mapped onto a non-zero member of C^, and, by the previous lemma, such a r i n g cannot be mapped onto a non-zero ri n g with no zero d i v i s o r s . K i s therefore N g r a d i c a l . Q.E.D. The r e l a t i o n between the yvarious r a d i c a l s i s i l l u s t r a t e d i n the following diagram. ( C f . Divinsky, s ( 6 ), page 1 5 6 . ) U U * g Fig. 3 . 4 . 2 By considering the singular submodule Z(M) of a module-M, we can obtain more information about the rel a t i o n s between these module classes. If M i s a r i g h t K-module, and i f A <3 K, A c (0:M), then the additive abelian group of M can be given a K/A-module structure as described i n the property P . l . Suppose we denote these modules by M K . and M j ^ . It makes sense, i f we consider these two modules as having the same underlying set, to compare Z(UK) and Z ( M K / A ) . ; 55. Lemma 3.4.5 I f M i s a r i g h t K-module, and i f A 4 K, A c ( 0 : M ) , then Z(M R /, A) c Z ^ ) . ' Proof; For m i n M, {k 6 K:mk = 0) = (0:m) K i s a r i g h t i d e a l of K which contains A, and which, under the natural homomorphism K -• K/A maps to £k e K/A : mk = 0} = iQ:m^K./At I f m e Z(M Ky A), then (0''m)jz/A i s a n e s s e n t i a l r i g h t i d e a l of K/A. Let I be a non-zero r i g h t i d e a l of K. I f I n A ft 0, then I D (0:m) R ft 0. I f I n A = 0, then (I+A)/A i s . a non-zero r i g h t i d e a i ^ i n K/A, and (I+A)/A D (0:m)Ky-A ft 0. This implies (I+A) fl (0:ni)g ^  A. Let x = i+a be i n (I+A) n (0:m) K but not i n A. Then i and i - x - a i s i n (0:m) K + A = (0:m) K and i e I. This proves that (0:m) K i s e s s e n t i a l i n K, and m € Z(M R). Q.E.D. Theorem 3.4.6 Let K be a r i n g , and l e t M be a r i g h t ,2 'K K-module such that M € E K , and such that M ft Z(M). Then (1) M has a submodule N / 0 such that N e E R . Either N c Z(M) or Z(M) = 0. (2) I f K i s semiprime, then Z(M) = 0. (3) I f K i s a prime r i n g , then Z r(K) = Z 1(K) = 0. Proof: ( l ) Let M. and K s a t i s f y the assumptions of the theorem, and l e t (0:M) ** A. Prom Lemma 2.1.5* K/A i s a 54. prime r i n g , and so B(K), the Baer Lower r a d i c a l of K, i s contained i n A. Let "K" » K/B(K). Using property P.2, . M can be considered a K" module, and, as such, M i s i n Eg. From the previous lemma, we have that Z(Mg) c Z(Mg) ft M, and so there i s an m i n M such that (0:m)-g i s not an e s s e n t i a l r i g h t i d e a l of "K. There i s , therefore, a non-zero r i g h t i d e a l I of K such that I fl (0:m)^ «= 0. I t follows that the K-module homomorphism f:K -» M, where f(k) = mk, i s one-to-one on I. I i s therefore isomorphic to a K"-submodule of Mg. We show that' I (and hence f ( l ) ) i s •2 2 i n E^. From P. 5 , f ( l ) i s i n Eg, and thus so i s I. Suppose that V i s a non-zero K-submodule of I - that i s , V i s a r i g h t i d e a l of K contained i n I. Since K i s semiprime, V 2 ft 0, and so, f o r some v i n V, v i ft 0. Since I i s a r a t i o n a l extension of every non-zero submodule, the K-module homomorphism of I to V which maps i to v i must be one-to-one. Therefore V and I are subisomorphic. Since V i s any submodule of I, t h i s shows that I 6 E^. The K-submodule f ( I ) i s then i n and, by property P.1, f ( l ) , as a K-module, i s i n Eg. This i s the desired module N. I f Z(N) ft 0 then N i s subisomorphic to Z(N), and i t follows that N • Z(N) c Z(M). I f , on the other hand, Z(N) a 0, then 0' - Z(N) = N fl Z(M), and, since M i s uniform, Z(M) •» 0. 55. (2) Suppose now that K i s semiprime. In the notation of the previous paragraph, K = K. In order to show that Z(M),« 0 i t i s s u f f i c i e n t to show that 2(N) = 0, that i s , to show Z(I). = 0. But i f u e Z ( l ) , then (0:u) i s e s s e n t i a l i n K, and (0:u) n I and so there i s i € I i / 0, such that u i = 0. Then the map 1^:1 -• I> where L u ( i ) = u i , has a non-zero kernel, and, since I i s r a t i o n a l over t h i s kernel, 1^ = 0, and u l = 0. Thus Z(I) c {x e K : x l = 0} = 2 ' l a ( l ) . Then Z(I) c la(l)»I = 0. Since K i s semiprime, Z(I) » 0. (3) Suppose that K i s a prime r i n g . I f T i s any two-sided i d e a l , and i f I i s any r i g h t i d e a l , then 0 / IT c I D T . This shows that any two-sided i d e a l must meet i n a non-zero fashion any r i g h t i d e a l . Now Z r(K) i s a two-sided i d e a l , and Z ( l ) = i n Z r(K) - 0. Therefore Z r(K) = 0. In order to show that Z 1(K) = 0 also, we f i r s t show that, f o r any non-zero u i n I, (0:u) i s a maximal r i g h t annihi-l a t o r . Suppose b € K, and (0:b) p (0:u). I f J i s a non-zero r i g h t i d e a l of K, either J fl (0:u) i s zero or i t i s non-zero. In the l a t t e r case, J n (0:b) / 0, while i n the former case, .^0:u) + J £ (0:u). Since K/(0:u) = uK c I, K/(0:u) i s uniform, and i t follows that ( (0:u) + J) n (0:b) ^ (0:u). Suppose that a € (0:u) and j € J are su'ch that a + j =» x i s i n (0:b) but i s not i n (0:u). Then j / 0, and j - x-a i s i n (0:b) n J . Thus, again i n t h i s 56. case J n (0:b) / 0. Therefore (0:b) i s e s s e n t i a l as a ri g h t i d e a l of K and b e Z r(K) = 0. This shows that (0:u) i s a maximal r i g h t a n n i h i l a t o r i n K. We choose u i n I, u / 0. I f Z 1(K) / 0, then, since K i s prime, there i s an s i n Z 1(K) such that usu / 0. Then 0 / su e Z 1(K), and so la(su) i s an es s e n t i a l l e f t i d e a l , and la(su) 0 Ku /•(). Thus there i s a k i n K such that ., ku / 0, but kusu = 0. Then su € (0:ku) = (0:u). (The l a s t equality comes from that f a c t t h a t (0 : u ) I s a maximal r i g h t a n n i h i l a t o r , (Oiku) s ( 0 : u ) , and ku / 0.) Thus usu = 0, a contradiction. Therefore Z 1(K) = 0. Q.E.D. Corollary 3.4.7 I f M € E^> and,if Z(M) = 0, the con-clusions of the theorem .hold f o r M and K. Proof; M i s uniform, and so, by Lemma 3.1.3* M i s a ' 2 r a t i o n a l extension of every non-zero submodule. Thus M e Eg. We then apply the theorem. Q.E.D. 3.5 Rings With the Ascending Chain Condition Let P be a property pf submodules of a r i g h t K-module M. We say that M s a t i s f i e s the ascending chain  condition f o r P-submodules i f , f o r every increasing sequence Nn c N 0 c ... c N,... of submodules which a i l have property 5 7 . P, there i s an integer t such that N^ = N t + i f o r a 1 1 i ^ 0. This i s equivalent to the statement: every non-empty set of submodules, each having property P, has a maximal ( r e l a t i v e to inclusion) member. If P ' i s merely the property of being a submodule of M, and i f M has the ascending chain condition f o r P-submodules, we say that M i s Noetherian. I f K, as a ri g h t K-module over i t s e l f , i s Noetherian, we say that K i s a r : i-g nt Noetherian r i n g . A K-module i s f i n i t e l y generated i f there i s a f i n i t e set x 1, x 2> • • • x n °f elements from M such that the smallest submodule of M containing x^, Xg, .. ( which we denote by |x 1,x 2,...,x n> ) i s M i t s e l f . I t i s well known that a module i s Noetherian i f and only i f every submodule i s f i n i t e l y generated. A submodule N of a r i g h t K-module M i s c a l l e d a complement submodule i f there i s a submodule A of M such that N i s maximal among the submodules T s a t i s f y i n g T n A = 0. Also we say that N i s a complement of A. By Zorn's Lemma, i t follows that every submodule has -a complement N. Also, A i s e s s e n t i a l i n M i f and only i f 0 i s a complement f o r A. The ascending chain condition f o r comple-ment submodules of a r i g h t K-module w i l l be denoted by max.-rc The next few re s u l t s are due to Goldie (8). Theorem 3 . 5 . 1 A r i g h t K-module has max-rc i f and only i f M 58. contains no i n f i n i t e d i r e c t sums of non-zero submodules. Proof: Let M have max-rc, and suppose that V = B. i s a d i r e c t sum of submodules of M. Let T Q be a comple-ment of V, and suppose we have defined T^, 0 £ i _< n, such that, f or i >. 1> T i i s a complement of B J J A N D T i - T i - 1 + B i * T n n 2 j - n + l B j ~ 9' a n d h e n c e t h e s u m T n + Bn + 1 1 5 d i r e C t > a n d <Tn + W n *J-n+2 B3 = ° ' B y Zorn's Lemma, there i s a submodule T n + 1 which i s maximal among the submodules X s a t i s f y i n g X fl ^ j = n + 2 B j = °> x - T n + Bn+1* It i s clear that ^ n + ^ i s a complement of $ j _ m + 2 B i * T n e sequence T Q c- c£ ... c- T n of complement submodules can thus be extended to a sequence t Q ^  T l ^  T2 •** ^  T n ^  Tn+1' where T n + 1 i s also a complement submodule. I t follows by induction that there i s an i n f i n i t e s t r i c t l y increasing sequence of complement submodules, contrary to our assumption. Therefore max-rc implies that there are no i n f i n i t e d i r e c t sums of submodules. Conversely, suppose there was a s t r i c t l y increasing sequence T i ^ T 2 ^ ' " ^  " ' o f c o m P l e m e n ' f c submodules, and suppose that T^ i s a complement of , A^. Now B^ = T, ,, n A. / 0, and B„ n L1}"} B, cz A 0 T *''0 f o r each n. i - f l i • n i = l 1 — n n It follows that the sum B i i s a d i r e c t sum. Therefore, 59-i f there are no i n f i n i t e d i r e c t sums, M must have max-rc. Q.E.D. Lemma 3 . 5 . 2 Let M be a ri g h t K-module with max-rc. Then every submodule contains a uniform submodule. Proof: Suppose that N i s a submodule which contains no uniform submodules. Then N i s i t s e l f not uniform, and N contains non-zero submodules and such that N.^  D N- = 0, and the sum + N'^ i s d i r e c t . Suppose that, f o r an integer n, we have a d i r e c t sum N, + N 0 + ... N + N' of non-zero l d n n submodules of N. Since N' i s not uniform, there submodules n N and N'.. of whose i n t e r s e c t i o n Is zero. I t n+l n+l n^ follows that the sum Nn + N 0 + ... N + N , + N' , i s 1 2 n n+l n+l d i r e c t . I t follows that we can construct a sequence f N^,Ng,... of submodules f o r which the sum i s an i n f i n i t e d i r e c t sum. By the previous theorem, t h i s contradicts max-rc. Q.E.D. I f M i s a r i g h t K-module with max-rc, the previous lemma assures us that M contains uniform submodules. I f we l e t {U : a e A ) be the set of non-zero uniform submodules-of M, an a p p l i c a t i o n of Zorn's Lemma guarantees that there i s a subset F of A maximal among the subsets X of A f o r which U~ i s a d i r e c t sum. This F i s a f i n i t e set,, by Theorem 3.5.1. Suppose that £ U a > cieF} = {U^Ug,.. .U n). Then V « ^  • Ug • ... • U n i s e s s e n t i a l , f o r i f V fl A = 0, 'Go. f o r some submodule A / 0, then A contains a uniform sub-module W / 0. I f W = U. , where \ e A, then G = F u {X } 3 F , and £ UTJ i s d i r e c t . This contradicts ° * . . . the maximality i f F . These remarks show,that, i f M has max-rc, there i s a family U^Ug*... *U n of uniform submodules such that V = i s a d i r e c t sum, and i s e s s e n t i a l i n M. Remark The integer n appearing i n the previous paragraph i s an invariant of the module, c a l l e d the dimension of M. It i s the maximum possible length of a d i r e c t sum of submodules of M. (See Goldie' ( 8 ) , Chapter three.) Neither the invariance of n, nor the f a c t that there are no d i r e c t sums of more than n submodules w i l l be needed f o r our purposes. Theorem 3 .5 .3 Let K , be a semiprime r i n g with a family {¥ : a e A} of uniform r i g h t ideals such that £ W i s cteA an e s s e n t i a l r i g h t i d e a l of K.' Then U(K) = 0. Proof: Any uniform r i g h t i d e a l W of a semiprime r i n g K i s a prime r i g h t K-module (and hence W e £ ^ ) . For i f V / 0 i s a submodule of W, and i f k e (0:V), then Vk = 0, VKk = 0, and V c la(Kk) = {x:xKk =0}. Now Kk i s a l e f t i d e a l , and so la(Kk) i s e a s i l y seen to be a two-sided i d e a l of K. I f WkK / 0, then T = V f l WkK / 0, since W i s uniform, but 61. T c V'WkK c la(Kk)«KkK = 0. Since K i s semiprime, T must be zero, and so WkK = 0, and Wk c la ( K ) . But la(K) = 0, so Wk = 0. Thus, i f V i s a non-zero submodule of W, then (0:V) c (0:W). The other i n c l u s i o n i s always true, so (0:V) = (0:W). Also WK ^ 0, since W2 ^ 0. By Lemma 2.1.4, W i s thus a prime K-module. If T = Z W ,is e s s e n t i a l i n K, where each W i s uniform, and hence i n E T,, we have U(K) c fl (0:W ) c (0:T). a€A The proof w i l l be complete i f we can show that (0:T) = 0. But T i s e s s e n t i a l , so i f (0:T) ^  0, then (0:T) fl T j4 0, and t h i s i n t e r s e c t i o n i s a r i g h t i d e a l whose square i s zero. Since K i s semiprime, t h i s i s a contradiction. Therefore (0:T) = 0. Q.E.D. Corollary 5.5.4 I f K i s a r i n g f o r which K/B(K) has max-rc, then U(K) = B(K). Proof; I f K = K/B(K) has max-rc, then as we have seen, there i s a f i n i t e family of uniform r i g h t ideals whose sum i s an e s s e n t i a l r i g h t i d e a l of K. By the theorem, 'U(K) = J0, which implies U(K) c B(K). The other i n c l u s i o n i s always true, so we have U(K) = B(K). Q.E.D. Theorem J>.5.5 I f K i s a semiprime r i n g with max-rc and Z r(K) = 0, then W(K) = 0. 62. Proof: The condition max-rc guarantees that there is a family of uniform right ideals Wg, ..., w"n such that T = Si"1Wi is essential. The argument i n Theorem 3.5.3 shows that each Wi is in E^. Since Z(W±) = W± n Z r(K) = 0, Lemma 3.1-3 guarantees that each Wi is a rational extension of every p non-zero submodule, and so Wi € £ K. If V is a submodule o of Wi (i.e. a right ideal of K contained i n W.^ ), V / 0. and so there i s a v in V such that vW^  / 0. Since W^  is rational over a l l submodules, the map Ly:W^ -• V, where L (w) = vw, must have kernel zero and hence is one-to-one. v / Thus W^  and V are subisomorphic, and, since V was any non-zero submodule, W^  is homogeneous. This shows that each ' -z. n W. £ s£. Thus W(K) c n (0:¥.) c (0:T), and, as in Theorem 1 A ~ i=l 1 ~ 3-5.3, (0:T) = 0. Q.E.D. Corollary 3.5.6 If K is a right Noetherian ring, then B(K) = W(K) and this is a nilpotent ideal. Proof: It is known that, in a right Noetherian ring K, B(K) is nilpotent. The ring K = K/B(K) is also right Noetherian, and hence has max-rc. We now show that Z r(K) = 0. For x in K, (Orx 1) c ( 0:x i + 1). Since K is right Noetherian, for each x there 63. > i s an n such that (0:x n) = ( 0 : x n + 1 ) . Suppose that x i s i n Z r(K). Then, i f n i s as before, and i f x n ^ 0, x1^. ^ 0 (since K i s semiprime), and x 1 ^ fl (0:x) ^  0, (since x i s i n Z r ( K ) ) . Therefore there i s a k i n K such that x"k ^ 0 but xx0!"; = 0, and (0:x n) ^ ( 0 : x n + 1 ) . This contradiction shows that, f o r x i n Z r(K), x n = 0, and so Z r(K) i s a n i l i d e a l . By L e v i t z k i ' s theorem, (see Faith ( l ) , Chapter ten) Z r(K) i s nilpotent, and thus Z r(K) = 0. We can now apply Theorem 3.5.5 to conclude that W(K) =* 0, and so W(K) c B(K). Since we always have W(K) => B(K), we are done. Q.E.D. Remark I t i s well known (see, f o r example, Faith (7), Chapter nine, Theorem 7) that a semiprime r i n g K with max-rc and Z r(K) = 0 has a c l a s s i c a l r i g h t quotient r i n g Q which i s semiprime and r i g h t A r t i n i a n . Conversely, i t i s also true that i f K has a semiprime r i g h t A r t i n i a n r i g h t c l a s s i c a l quotient r i n g Q, then K i s semiprime with max-rc and Z r(K) = 0. Thus the conditions i n Theorem 3.5.5 are f a m i l i a r ones i n r i n g theory. The concept of homogeneous modules i s related to the concept o f basic modules, as introduced by Goldie i n ( 9 ) . For Noetherian modules over r i g h t Noetherian modules, Goldie defined a.basic module M to be one which i s homogeneous and 64. and has Z(M) = 0. By Lemma3.5.2, a Noetherian module M contains a uniform submodule N. I f M i s also: homogeneous, M i s isomorphic to a submodule of N, and thus M i s i t s e l f uniform. Also Z(M) = 0 and M uniform imply (Lemma 3 . 1 . 5 ) that M i s a r a t i o n a l extension of each non-zero submodule. 3 ' Thus a basic module, by Goldie's d e f i n i t i o n , i s i n Eg. For our purposes, we s h a l l define a basic module to be one which i s uniform, homogeneous, and which has Z(M) = 0. This i s equivalent to saying a basic module i s a member of Eg with singular submodule zero. The next r e s u l t i s due to Goldie (9) . Theorem 5 . 5 . 7 Let K be a r i g h t Noetherian r i n g , and l e t M be any r i g h t K-module. Either Z(M) i s e s s e n t i a l i n M or M contains a basic module. Proof: I f Z(M) i s not e s s e n t i a l i n M, there- i s a submodule N of M f o r which Z(N) = N n Z(M) = 0. Since K i s r i g h t Noetherian, there i s a non-zero submodule N' of N such that (0:N') i s a maximal member of {(0:X): X i s a non-zero submodule of N}. Then N'K f4 0» f o r N'K = 0 implies N' c N fl Z(M) = 0,- which i s f a l s e . Also, i f T i s a non-zero submodule of N', then (0:T) 3 (0:N'), and the maximality of (0:N') gives (0:T) = (0:T) = (0:N'). N' i s therefore a prime K-module, by Lemma 2.1 .4. For any n ft 0 i n N', nK ft o (elsewise n e N n Z(M) = 0), and nK i s a Noetherian 65. right K-module which, by Lemma 3.5»2, must contain a non-zero uniform submodule M'. Thus M' is prime, uniform, and has Z(M') c Z(N) = 0. By Corollary 3.4.7 M' contains a non-zero submodule M" i n Eg. This submodule is basic. Q.EvD. In the reference (9), Goldie showed that, for the- cases he was considering, the uniform right ideals of a ring serve as examples of basic modules. One can show that, in a sense, these are the only examples. This is the importance of the next theorem. Theorem 5.5.8 (1) A basic K-module M i s isomorphic, as a K-module, to a uniform right ideal of K. (2) If K is semiprime and i f Z r(K) = 0, any uniform right- ideal of K i s a basic K-module. Proof; (1) If M is a basic K-module, Z(M) = 0, and for m / 0 in My (0;m) is not essential i n K. Therefore there is a right ideal I of K such that i n (0:m) =0. The K-module homomorphism f:K K -* M, where f(k) = mk is one-to-one on I, so I i s isomorphic to a submodule of M. The homogeneity of M guarantees that there is a K-monomorphism of M into f ( l ) . Composing this mapping with the inverse of the restriction of f to I gives a K-monomorphism g from M into K. M is basic, and therefore uniform, so g(M) i s a uniform right ideal of K. 6 6 . ( 2 ) If K is semiprime, and Z r(K) = 0 , any uniform right ideal I satisfies Z(l) = 0 . Proceeding just as i n • Theorem 3 . 5 . 5 , I is i n Z^, and so I is indeed basic. Q.E.D. 3 . 6 Quasi-injective Modules > In this section, we introduce the concept of a quasi-injective module, and sketch some of the results known about them. Most of the results of this section are due to Johnson and Wong and their proofs may be found i n Faith (7). Later we shall employ these results to obtain generalizations of the Jacobson density theorem. A right K-module E is said to be injective i f , given a K-module M and a submodule N, and given a K-homomorphism f:N -» E, there is a K-homomorphism g:M -» E such that the restriction of g to N is f. It is well known that, given any right K-module M, there is an injective K-module E(M) and a K-monomorphism i:M -• E(M) such that E(M) is an essential extension of i(M). Furthermore, i f there i s another injective K-module E'(M), and a K-monomorphism i':M-» E'(M) such that E'(M) is an essential extension of i'(M), i t i s known that there i s a K-isomorphism f:E(M) -• E'(M) such that f i(m) = i'(m) for a l l m in M. Therefore the pair (E(M), i) i s , in a sense, unique up to an isomorphism which respects the imbedding of M. The pair (E(M), i) i s called the injective h u l l of M. 6 7 . A right K-module Q is quasi-injective i f , whenever we have N a submodule of Q, and f:N -» Q a K-homomorphism, there is a K-homomorphism g:Q - Q whose restriction to N is f. Clearly any infective module is quasi-injective, and any simple module is quasi-injective. However, injectives and simple modules do not exhaust the class of quasi-injective modules. For example, l e t K be the ring of integers, and l e t M be the K-module (i . e . , an abelian group) of integers modulo 4 . It is well known that, for this choice of K, the K-injectives are the divisible, abelian groups, so M is neither simple nor injective. The only submodules of M are M, 0 , and { 0 , 2"}. Clearly, to show M is quasi-injective, i t is sufficient to show that any homomorphism from N (= {"2, C7}) to M can be extended to an endomorphism of M. Let f be such a homomorphism. • Then 2f(2') = f ( 1 ) = f (0) = C7, and so f(e>) = "2" or 0 . In the former case, l e t g(T) - T , and i n the latter case l e t g(T) = 0 , where i = 0 , 1 , 2 , or 3 . This gives the desired extension, so M is quasi-injective. Theorem 3 . 6 . 1 Let M be any right K-module, and l e t (E(M), i) be the injective h u l l of M. (If we identify M ~ and i(M), we can regard M as .being a submodule, and an essential submodule, of E(M).) If we define H = HomK( E(M),E(M) ), then, writing the operators of H on the l e f t , HM i s a quasi-injective K-module which is an essential extension of M. M i s i t s e l f quasi-injective i f 68. and only i f M = HM. Proof; Clearly HM is a K-submodule of E(M), and i t contains M. Since E(M) is an essential extension of M, i t follows that HM is also an essential extension of M. Suppose that N i s a submodule of HM, and that f:N -» HM is a K-homomorphism. We can regard f as mapping N to E(M). From the inj e c t i v i t y of E(M), there is an h i n H such that h(n) = f(n) for a l l n i n N. If w e HM, w = h-j^m^), where b^ e H and m^ ' e M. Then h(w) = S i = l h h i ( m i ^ 6 ™ > a n d s o £ ^ Therefore the rest r i c -tion h of h to < HM is an endomorphism of HM which extends' f. This proves that HM i s quasi-injective. Clearly, i f M = HM, then M is quasi-injective. Conversely, suppose that M is quasi-injective, and l e t h e H. If h = 0, then E(M) = h _ 1(M). If h ^ 0, then h(E(M)) ft o , h(E(M)) n M ft 0 (since E(M) is an essential extension of M), and so h"1(M) ft o , and W = M n h - 1(M) ft 0. The restriction of h to W is a map g:W -» M which, since M is quasi-injective, can be extended to a map g;M -• M. By the i n j e c t i v i t y of E(M) there is h in H such that h(m) = g(m) for a l l m i n M. Consider (h-h)M. If this were non-zero, then (h-h)M D M ft o , and there are m and m' i n M such that m = h(m') - h(m') ft 0. Then h(m/) = E(m' ) - m =s Km') - m e M, and so m' € W. But this gives 69. g(m') = h(m'), and we obtain h(m/) = g(ra/) = g(m') = h(m'), and m = 0, a contradiction. Therefore (h-h)M = 0. Then, for m i n M, h(m) = ,h(m) = g(m) e M, and so h(M) € M, and HM c M. Since M c HM, we have HM = M. Q.E.D. In the future the quasi-injective module HM obtained from the injective h u l l E(M) of M w i l l be denoted Q(M). Remark Using the same notation as in the theorem, for any h in H, we have h(HM) c HM. Therefore the restriction of h to HM is in A = Hom( Q(M),Q(M) ). It follows then that we also have Q(M) = AM. Lemma 3 . 6 . 2 Any complement submodule of a quasi-injective module M i s a direct summand of M. Proof: Let M be quasi-injective, and l e t N be a complement submodule of M. Suppose that N i s a complement of L. Then N i s maximal among the submodules X of M such that X n L = 0. Consider the map gQ:N + L -» N defined by g Q(n + a) = n, where n e N and a e L. We claim that this can be extended to a homomorphism f:M -• N. This is sufficient to prove the theorem, for then M = N © (l-f)M, -for any m in M may be written m = f(m) + (m - f(m)), and i f n = x - f(x) is in the intersection of these submodules, f(x) e N, ff(x) = g Qf(x) = f(x), and we have n = f(n) = f(x) - ff(x) = f ( x ) - f(x) =0. Consider the set of a l l ordered pairs (Y,t), where 7 0 . Y i s a submodule of M containing N S> L, and t i s a K-homomorphism from Y into N extending g Q. We p a r t i a l l y order t h i s set by saying (Y-^t^) ^. (Ygjtg) i f and only i f Y l - Y2> a n d \(y) = t 2 ^ y ^ f o r a 1 1 y i n Y2* Z t i s e a s i l y v e r i f i e d that t h i s p a r t i a l ordering i s inductive, and so we can apply Zorn's Lemma to obtain a maximal extension (W,g), r e l a t i v e to t h i s ordering, of g Q. Then W _> N + L, g:W N i s a K-homomorphism which i s an extension of g Q . We claim that W = M, and g i s the desired map. For suppose that W jt M. Since M i s qu a s i - i n j e c t i v e , the map g:W -» N c M can be extended to a homomorphism f:M M, and f(M) £ N. (For f(M) c N implies that (M,f) V, (W,g), contradicting the maximality of (W,g)). Then f(M) + N £ N, and so (F(M) + N) n L ^ 0 . Let f(m) + n = a j£ 0 , where a i s i n L. I f m e W, then f(m) = g(m) i s i n N, . and a i s i n L n N = 0 , which i s f a l s e . / Thus m i W. I f we set T = f - 1 ( N + L ) , we see that m i s i n T, and that T 3 W. Therefore T p W. I f we l e t f' be the r e s t r i c t i o n of f to T, i t i s e a s i l y v e r i f i e d that (T, g Q f ' ) ^ (W,g), contradicting the maximality of (W,g). Thus W = M, and g:M -* N extends g Q. Q.E.D. A ri g h t K-module i s said to be indecomposable i f and only i f M cannot be written as A $ B, where A and B \ i are non-zero submodules. 7 1 . Corollary 5 . 6 . 5 A quasi-injective module M is uniform i f and only i f i t is indecomposable. Proof; Clearly any uniform module is indecomposable. Conversely, i f M is not uniform, there are submodules S and T, both non-zero, such that S n T = 0. Zorn's Lemma can be applied to obtain a complement S' of S containing T. By the theorem, S' is a direct summand of M, and S' / M, S' / 0. This shows that. M is hot indecomposable. Q.E.D. Theorem 5.6.4 If M is any right K-module, the following are equivalent: (1) M is uniform ( (2) Q(M) is uniform (5) Q(M) is indecomposable. Proof: ( l ) , together with the fact that Q(M) is an essential extension of M implies^ (,2), and (2 ) clearly, implies ( 1 ) . Since Q(M) is quasi-injective, (2) and (5 ) are equivalent by Corollary 5 . 6 . 5 . Q.E.D. A ring is said to be regular i f , for every x there is a y such that x = xyx. If x = xyx, then e = xy i s idempotent, and |x > = | e >. (Recall that | w > means the submodule - in this case, the right ideal - generated by w.) Consequently, in a regular ring, every principal right ideal is generated ,by an idempotent. Conversely, i f every principal right ideal in a ring K is generated by-ah 72. idempotent, then, for any x, |x> = |e>, and e = xy for some y i n K. But x = ew, for some w, and ex = eew = . ew = x. Therefore x = ex = xyx, and K is a regular ring. If y is a l e f t quasi-regular element in a ring K with unity, then x + y - xy = 0, and (l-x)(l-y) = 1, so 1-y has a l e f t inverse. Conversely, i f K i s a ring with unity, and i f 1-y is an element with a l e f t inverse z, then writing x = 1-z, we have (l-x)(l-y) = z(l-y) = 1. This implies that x + y - xy = 0, and y is l e f t quasi-regular. A unit of a ring K with unity is an element with a (two-sided) inverse. If K is a ring for which the non-units form an ideal I, we c a l l K a local ring. If K is a local ring, no element of J(K) can he a unit (for i f i t were otherwise J(K) would contain 1, and 1 is clearly not right or l e f t quasi-regular), and so J(K) c I. Conversely, i f x e I, then 1-x \ I (for otherwise 1 = 1-x + x would be i n I, and hence a non-unit, which is absurd) and so (1-x)" 1 exists. As above, this implies that x is l e f t quasi-regular, and I is a l e f t quasi-regular l e f t ideal, that is I c J(K). Consequently, i n a local ring K, the set of non-invertible elements of K is precisely J(K). It follows at once that, in this case, K/J(K) is a division ring. It i s a well known theorem, known as Schur's Lemma, that, i f M i s a simple right K-module, then HomK(M,M) i s 73. a division ring. Simple modules are clearly indecomposable quasi-injectives, so the second part of the following theorem provides a generalization of Schur's Lemma. Theorem 3 . 6 . 5 Suppose that Q i s a quasi-injective "right K-module. Denote A = HomK(Q,Q). Then J ( A ) = {OCA: Ker(ct) is essential i n Q} and . A / J ( A ) i s a regular ring. Further-more, A is a local ring i f and only i f Q is indecomposable, and, i n this case, J(A) = {aeA: ker(a) / 0}. Proof: Let I *= {a€A: ker(a) is essential i n Q}. If X € A , le t L be a maximal member of the set of a l l submodules X of Q which satisfy X n ker (x) = 0. Then L $ ker (x) is essential i n Q. For i f T / 0' is a submodule of Q either T c L or T + L ^ L, i n which case (T+L) fl ker(X) / 0. In either case i t follows that T n (ker(x) © L) / 0. Define f: XL - Q by setting f( Xx) = x. If xx = Xy» where x and y are in L, then x-y e L PI ker (x) =0, so f is well defined on L. • Since Q is quasi-injective, f can be extended to 8:Q-» Q. A simple calculation shows that (X - XSX) (L & ker (x) ) = 0, so X - X6X e I. Since ker(a - B) 3 ker(a) fl ker(8), and since ker(Xa) 3 ker(a) i t follows that I i s a l e f t ideal of A . If ael, since ker(l-a) n ker(a) = 0, we have ker(l-a) = 0. If T = (l-a)Q, then the map g:T - Q, where g( (l-a)q) = q, is well defined and can be extended to a map 6€A. Then 6(l-a) = 1, the identity map on Q, so (1-a) has a l e f t 74. inverse, and a i s l e f t quasi-regular. Therefore I i s a l e f t quasi-regular l e f t i d e a l , and I c J ( A ) . Suppose now that X e J ( A ) . Then, as above, there i s a @ e A such that X - X0X e I. Since XQ € J ( A ) , XQ. i s l e f t and r i g h t quasi-regular, and ( l - X S ) " 1 e x i s t s . Since I i s a l e f t i d e a l , x = (1-X0)" 1 (l-XS)X = ( l - x e ) " 1 (X-X0X) e I. Therefore J ( A ) C I , and J ( A ) = I. We have proved that J ( A ) = I , and that A / J ( A ) i s a regular ring. Suppose now that X + J ( A ) i s an idempotent element of A / J ( A ) . Then x 2 - X .e J ( A ) , and X = ker(x 2-X) i s es s e n t i a l i n Q. Now we claim that the sum XX + (l-x)X i s d i r e c t . For i f Xx = x' - Xx'. f o r x, x' i n X, then 2 2 2 X x = (X-X )x' = 0 . Also, (X -X)x = 0 so we have Xx = 0 . Consequently, i f Q i s indecomposable (and hence uniform, by Corollary 3.6.3) and i f x + J ( A ) i s idempotent i n A / J ( A ) , then eit h e r XX = 0 , and X e J ( A ) (since X i s e s s e n t i a l ) , or (l-x)X = 0 , and (1-x) e J ( A ) . Thus the indecomposability of Q implies that A / J ( A ) has no idempotents other than 0 and 1. Since A / J ( A ) i s regular, given any x / 0 i n • A / J ( A ) , there i s a y such that xyx = x. Therefore xy and yx are non-zero idempotents and so these are both 1, showing that A / J ( A ) i s a d i v i s i o n r i n g . Now, i f a i s a non-unit i n A , then, f o r some p € A , we have a-a&a -( l - a p ) a e J ( A ) . Since A / J ( A ) i s a d i v i s i o n r i n g , t h i s implies that either a or (1-ap) i s i n J ( A ) . The l a t t e r 7 5 . i m p l i e s t h a t 1 - (1-aB) = aB i s i n v e r ^ t i b l e , so a would be i n v e r t i b l e , which i s f a l s e . T h e r e f o r e , i f a i s a non-u n i t i n A , we have a e J ( A ) . .Since any member of J ( A ) . i s . a non-unit, the non-units form an i d e a l and A i s a l o c a l r i n g , p r o v i d e d Q i s indecomposable. Conversely, assume t h a t A i s a l o c a l r i n g . I f Q i s not indecomposable, we can w r i t e Q = S I T where S and T are non-zero. The map p:Q Q, where p(s+t) = s, i s an idempotent element,.of A , and so p ( l - p ) = 0., Since A / J ( A ) i s a d i v i s i o n r i n g , e i t h e r p or 1-p i s i n J ( A ) and e i t h e r k er(p) = T or k e r ( l - p ) = S i s e s s e n t i a l i n Q. Since both S' and T are not e s s e n t i a l , Q must be indecom-posable. • " I f Q i s indecomposable, and•therefore u n i f o r m , (by C o r o l l a r y 3 .6.3) , . any non-zero submodule of Q, i s e s s e n t i a l , ^ and so i t f o l l o w s from the f i r s t p a r t of the theorem t h a t J ( A ) = {a: ker(a) 0}. Q.E.D.-. Theorem 3 .6.6 I f M i s a prime module, then Q(M) i s prime,' and (0:M) = (0,:Q(M)). Proof: R e c a l l t h a t Q(M) = HM, where, H = Hom K(E(M),E(M)), and where E(M) i s the" i n j e c t i v e h u l l ) o f , M. Suppose t h a t T i s a non-zero submodule o f Q(M). T fl M / 0 , s i n c e Q(M) i s an e s s e n t i a l e x t e n s i o n of M. From the primeness of M, • we have (0:M) = (0: T fl M) .p_ (0:T) 3 (0:Q(M)). However, ' ' , ; • 76. i t i s c l e a r from'the f a c t t h a t Q(.M) = HM t h a t ' (0:Q(M)) => (0:M). There f o r e (0:T) = (0:Q(M)) = ( 0:M), and Q(M) i s prime by Lemma 2 . 1 . 4 . Q.E.D. ' For a r i g h t K-module M, and N a- subset o f M, l e t N^ = {keK: Nk = 0 } , and f o r T a subset of' K, l e t T M = {meM: mT = 0 } . ( T M ) K i s denoted Tm and ( N K ) M i s denoted N ™ . We note f o r f u t u r e r e f e r e n c e t h a t , i f M i s a prime M K-module, then K = 0 . Theorem 5 . 6 . 7 L e t Q be a q u a s i - i n j e c t i v e r i g h t K-module, and l e t A = Hom K(Q,Q). Then Q has a l e f t A - , r i g h t K-bimodule s t r u c t u r e . I f N . i s any A-submodule s a t i s f y i n g 1 ^ . = N, then, f o r any x i n Q, ( N + A X ) K Q = N + A * . Proof: L e t A = N? and B = x K . Then A Q = N. ...Simple ) computation shows' B'= x K = ( A X ) K . . " C l e a r l y N + A X c ( N + A X ) K < ^ , and so i t s u f f i c e s to prove ( N + A X ) K Q C m+AX. .Now ' ( N + A X ) K = l ^ n ( A X ) K = A n B, so we only need show (A n B ) ^ c ' N + A X . Suppose y e (AflB)^, and c o n s i d e r 8:xA -» yA d e f i n e d by S(xa) =• ya. I f xa = xa', K where a and a' • are''in A, then a-a', e A n x •= A n B. Since ; y•• e ( A f l B ) ^ , y(a-a'). = 0 , and the mapping'is w e l l -d e f i n e d . T h i s i s a K-homomorphism from xA to yA c Q, and the q u a s i - i n j e c t i v i t y o f Q i m p l i e s t h a t 6 .may be extended. 77. to X:Q -» Q. Then X x a = ya f o r a l l a i n A, so (Xx-y) e (X-x-yj e A Q = N, and y e N + A X . Q.E.D. C o r o l l a r y 3 . 6 . 8 Suppose t h a t Q Is q u a s i - i n j e c t i v e , and that. K^ .' = 0. I f x.^,. x2,..,',x are any elements from Q, then ( Z . ^ A x ± ) K Q = Z ^ A X . . Proof: T a k i n g 1 N = 0 i n the theorem, since' 0 K ^ = K^ = 0, \ we have ( A x ^ ) ^ = A X ^ . The r e s t f o l l o w s 'by .the obvious i n d u c t i o n . Q.E.D. We s h a l l say t h a t the elements x^,x 2,...,x n of Q are A - i n d e p e n d e n t I f , f o r each i , x. £ £ ^ AX ..• Theorem - 5 . 6 . 9 ( D e n s i t y theorem f o r q u a s i - i n j e c t i v e modules)' I f Q i s a'' q u a s i - i n j e c t i v e r i g h t K-module such t h a t K^ = 0, and i f x ^ , X g , . . . , x n are A - i n d e p e n d e n t elements of Q (where' A = HomK(Q,Q) ), then there are r i g h t i d e a l s A n . A „ , . . . a A i n K such t h a t x.A. / 0, x.A. = 0 - "for j / i . Furthermore, i f y-^Yg* • • • ^ y n a r e elements of Q such t h a t y. e x.A. f o r each i , then there e x i s t s k i n ^ 1 1 x ' K such t h a t . x J s = y i , f o r i == 1,2,... ,n. Proof: L e t -A. = '(2 A.- A X . ) K . ' T h i s i s a r i g h t i d e a l i n K, : j / i 1 7 8 . and, by C o r o l l a r y 3 . 6 . 8 , A.^ = 2 ^ A X . , which does not c o n t a i n x.. Th e r e f o r e x.A. 4 0 , but x.A. =.0 f o r j / i . l i i I j Suppose now t h a t y^ i s i n xj,Aj_3 i = 1*2,...,n, and suppose y: = x.a., where a. e A.. Then k = 2 - N N a. w i l l s a t i s f y °i i i l l i = l l ^ x^k = y f o r a l l i . Q.E.D. Remark I f M i s a f a i t h f u l , i r r e d u c i b l e r i g h t K-module, then M i s simple and t h e r e f o r e q u a s i - i n j e c t i v e . A l s o , K1^ = 0 , and A i s a d i v i s i o n r i n g . Furthermore, i f x x_,,x„,...,x ,. and A..,... ,A are as i n the theorem, then I3 23 3 n3 I3 3 n • 3 the s i m p l i c i t y of M give s = M ' Thus, i n t h i s case, i f x ^,x 2,...,x n are A - i n d e p e n d e n t , and i f y - j ^ . . . * y n are a r b i t r a r y , there i s a k i n K such t h a t x^k = y^ for. i = l , 2 , . . . , n . T h i s i s the c l a s s i c a l Jacobson D e n s i t y Theorem. Remark I f M i s a prime r i g h t K-module, Q(M) i s q u a s i -i n j e c t i v e and prime ( T h e o r e m ' 3 . 6 . 6 ) , and so K ^ M ^ = 0 , and Theorem 3 . 6 . 9 i s a p p l i c a b l e t o Q(M). We s h a l l make use of t h i s f a c t l a t e r . In Theorems 3 . 6 . 4 and 3 - 6 . 6 , we have seen t h a t the p r o p e r t i e s of primeness and u n i f o r m i t y are t r a n s f e r r e d , from M t o Q(M), and from •'• Q(M) to " M. We now show t h a t t h i s i s ' a l s o t rue f o r the p r o p e r t y o f b e i n g a r a t i o n a l e x t e n s i o n of every non-zero submodule.. 7 9 . Theorem 3 . 6 . 1 0 L e t M be a r i g h t K-module, E ( M ) the i n j e c t i v e h u l l of M, H = Hom^(E(.M), E(M)), and Q = HM. The f o l l o w i n g are e q u i v a l e n t ; (1) Q i s a r a t i o n a l e x t e n s i o n o f every non-zero . submodule. , (2) M i s a r a t i o n a l e x t e n s i o n of every non-zero submodule. > (3) A = Hom K(Q, Q) i s a d i v i s i o n r i n g . Proof: The p r o p e r t y o f being., a r a t i o n a l e x t e n s i o n of every non-zero submodule i s i n h e r i t e d by non-zero submodules, so, ( l ) i m p l i e s (2) . ' We show (2) i m p l i e s (3),. Since r a t i o n a l extensions are e s s e n t i a l e x t e n s i o n s , (2) i m p l i e s t h a t M i s uniform. By Theorems 3 . 6 . 4 and, 3 . 6 . 5 , A i s a l o c a l r i n g , and J ( A ) = {a: ker(a) / 0}. We w i l l prove t h a t J ( A ) = 0. . L e t a e J ( A ) , and l e t L = k e r ( a ) . F i r s t of a l l . M c L. For otherwise, a(M) / 0 , M f l a(M) / 0 , and so W = {m e M: a(m) e M} . has a(W) / 0. Now the. r e s t r i c t i o n . f of a t o W has k e r n e l ' L f l - M / 0 ( s i n c e M i s e s s e n t i a l . i n Q). Since M i s a r a t i o n a l , e x t e n s i o n o f L)(] M, we have 0 = f(W),= a(W) a c o n t r a d i c t i o n . T h i s shows t h a t a e J ( A ) i m p l i e s a(M) = 0... Recall-now t h a t Q = A M . (See the remark f o l l o w i n g Theorem 3 . 6 . 1 . ) v Thus any , q in- Q may be w r i t t e n q = S I N 1 \t\> and. a(<l) = a\±(m±). Since a e J ( A ) , a\- e J ( A ) , a\.(m.) = 0 , and so a(q) = Q. Thus J ( A ) = "0, 80. and A i s a d i v i s i o n r i n g . • \ F i n a l l y vie show (5) i m p l i e s ' ( l ) . Suppose 0 / X c T, where X and T are submodules of Q, and suppose f:T -* Q i f a K-homomorphism wi t h f(X) = 0 . Q Is q u a s i - i n j e c t i v e , and so f can be. extended to a member X of A . Now i f X ^ 0, X i s i n v e r t i b l e , since A i s a d i v i s i o n r i n g , and so X i s one-to-one. But f(X) = x(X) = 0. ' Therefore X = 0, and so f.(T) = 0. This, shows that ,-Q i s a r a t i o n a l extension of the submodule X. Since X was an a r b i t r a r y non-zero submodule of Q, we are) done. ' Q.E.D. 5.7 Density Theorems f o r the R a d i c a l s and • ¥ In t h i s s e c t i o n , we prove some r e s u l t s which g e n e r a l i z e the well-known'Jacobson Density Theorem. A l s o we w i l l prove th a t the r a d i c a l ¥ c o i n c i d e s with the "weak r a d i c a l " of Koh and Mewborn ( 1 7 ) . We s h a l l say a r i n g ' K i s an L.W . - t r a n s i t i v e r i n g i f and, only i f : . (1) There i s a l o c a l r i n g A and a l e f t A-module V such t h a t K acts f a i t h f u l l y as a subring of Horn (V,V).. That I s , V has a l e f t A - , r i g h t K-bimodule structure', and i s f a i t h f u l . (2) V contains a K-submodule M which i s uniform, and for-which A M -'-'V. 8 l . (3) I f x^,X£,...,x i s any s e t o f A - i n d e p e n d e n t elements o f V, and i f y-^yg*''' J ^ n ^ s a n y s e ^ of elements from M (note - not from anywhere i n . V) t h e n t h e r e e x i s t k i n K and an i n v e r t i b l e element A. i n A such t h a t x^k = \y^3 f o r i =. 1,2,... 3 n . Furthermore we s h a l l - say t h a t a r i n g K i s ¥-transitive i f K i s L . W . - t r a n s i t i v e , and i f the l o c a l r i n g A i s a d i v i s i o n r i n g . Theorem 5 .7 .1 • A: r i n g • K i s an L . W . - t r a n s i t i v e r i n g i f and o n l y i f K has a f a i t h f u l module M i n Thus an H U-s e m i s i m p l e ring., i s - a s u b d i r e c t sum o f L . W . - t r a n s i t i v e r i n g s . P r o o f : Suppose f i r s t t h a t M' e S g H , and t h a t (0:M) = 0 . L e t V = Q(M) and A = Hom K(V,V). S i n c e . M i s u n i f o r m , Theorems 3 . 6 . 4 and 3 . 6 . 5 t e l l us t h a t A i s a l o c a l r i n g . V, as a r i g h t K-module, i f f a i t h f u l by Theorem 3 . 6 . 6 and the f a c t t h a t . M i s prime and f a i t h f u l , and the remark f o l l o w i n g Theorem 3 . 6 .1 says t h a t A M = V. Thus ( l ) and (2) o f the d e f i n i t i o n o f L . W . - t r a n s i t i v e , r i n g s . a r e s a t i s f i e d . Suppose now t h a t x-^x^,... ,x^ a r e A - i n d e p e n d e n t elements o f -V, and y^, y^3 . . . 3 y n a r e any elements o f M. S i n c e V i s a prime K-module (Theorem 3 . 6 . 6 ) , we can a p p l y Theorem 3 . 6 . 9 t o f i n d r i g h t " i d e a l s A^, i = l , 2 , . . . , n , such t h a t x i A i ^ °> b u t x i A - i = 0 f o r i / j . S i n c e M i s 32. u n i f o r m , V i s un i fo rm.by Theorem 3 . 6 . 4 , and so n P = M n ( fl x . A . ) / 0. M i s homogeneous, so there i s a 1=1 1 1 K-monomorphism f : M -* P wh ich , by ,the q u a s i - i n j e c t i v i t y of V, can be extended to a member. \ ' of A . I f k e r ( x ) / 0 , then 0 / M f] k e r ( x ) c k e r ( f ) , which i s f a l s e . Therefore k e r ( x ) Is z e r o , and X i s a monomorphism. By Theorem 3 . 6 . 5 , X i J ( A ) and so x - 1 e x i s t s . Now X ( y i ) = f(y±) E P £ f o r each i , so , b y . t h e second p a r t of Theorem 3 . 6 . 9 there i s an element k of K . such tha t x^k = Xy^ f o r i = 1 , 2 , . . . K is t he re fo re L . W . - t r a n s i t i v e . • ' Conversely suppose tha t K ' i s L . W . - t r a n s i t i v e , and suppose t h a t ' V", A , and )M s a t i s f y , the c o n d i t i o n s . 'We choose m / 0 i n M, and l e t I = {k e K: mk = 0} . Consider the r i g h t K-module K / I . Th i s Is i somorphic to mK c M, and-so K / I i s un i fo rm. We show tha t K / I i s homogeneous. I f X / I i s a non-zero submodule o f K / I , "where X i s a r i g h t . i d e a l o f K , X ^ I , then mX / 0. Suppose x e X i s such tha t mx / 0 . By c o n d i t i o n ( 3 ) , there are k i n K . and. X i n A , X I n v e r t i b l e , such tha t mxk = xm. Consider the K-module homomorphism f :K-» X / I , where f ( s ) = xks + I . The k e r n e l o f f i s { s :xks e, 1} = {s: mxks = 0} . Since X i s i n v e r t i b l e , t h i s k e r n e l i s p r e c i s e l y {s: ms = 0} = I . Then f induces a K-module monomorphism f : K / I -• X / I , and' so K / I and • X / I are subisomorphic . This shows tha t K / I 8 3 . i s homogeneous. I f s e. (0:K/I), /then Ks c I, and mKs = 0. For any y i n , M there are 0 . i n A , 0 i n v e r t -i b l e , and 'r i n K such t h a t mr = 0 y . Then 8ys.= mrs = 0, and so ys = 0. Thus i f s e (0:K/I), Ms = 0, . and so Vs = A M S = Oo Since K • acts, f a i t h f u l l y , on V, t h i s i m p l i e s s = 0. Thus K/I i s a f a i t h f u l r i g h t K-module which i s homogenous and t h e r e f o r e ) p r i m e (Lemma 3 . 2 . 1 ) . . K/I i s t h e r e f o r e u n i f o r m , prime, homogeneous and f a i t h f u l , ; a n d i s t h e r e f o r e a UH f a i t h f u l member of STjr . -' IV Now, i f K Is an . semisimple r i n g , K i s a ' TIH s u b d i r e c t sum of r i n g s K w i t h a f a i t h f u l member of 2^ . Q.E.D. . a Theorem 5 . 7 « 2 A r i n g K i s ¥-transitive i f and only i f K 3 has a f a i t h f u l member of 2^. . A l s o , a W-semisimple r i n g i s ^ a s u b d i r e c t sum of W - t r a n s i t i v e r i n g s . 3 Proof: Suppose M i s a f a i t h f u l member of 2^. Then, as i n the p r e v i o u s theorem, we can show t h a t K a c t s i n an L.W.-t r a n s i t i v e f a s h i o n on Q,(M). In order to show .that' K i s W - t r a n s i t i v e , i t s u f f i c e s t o show t h a t A = Hom K(Q(M), Q(M)) i s a d i v i s i o n r i n g . But t h i s f o l l o w s immediately from the f a c t t h a t M i s a r a t i o n a l e x t e n s i o n of each non-zero sub-module and from Theorem 3 . 6 . 1 0 . • ' < : . ' Conversely, suppose t h a t K- i s W - t r a n s i t i v e , and t h a t 84. V, A , and M s a t i s f y ( l ) - (3) of the d e f i n i t i o n . As i n the p r e v i o u s theorem, we can choose m / 0 i n M and show UH t h a t K/I, where I = (0:m), i s a f a i t h f u l member of £ K . In order to complete the p r o o f i t i s s u f f i c i e n t t o prove t h a t K/I i s a r a t i o n a l e x t e n s i o n of every non-zero submodule. Suppose t h a t f : T / I ~» K/I i s a K-homomorphism with k e r n e l ' J / I , where J and T are r i g h t i d e a l s . of K, ' and I <± J c T. We w i l l show f = 0. I f f / 0, l e t f ( T / l ) = X/I / 0, where X ^ I, and X i s a r i g h t i d e a l of K. Then mX / 0, and there e x i s t r i n K, x i n X, and X. i n A , X i i n v e r t i b l e , such,that mxr = \m / 0. Since \ i s i n v e r t i b l e , (0:mxr) = (0:m) = L, and x r ^ I. But mxrl = Xml = 0, whence x r l c I. Now 0 / x r + I e X/I = f ( T / l ) , and so we can l e t x r + I = f ( t + I) f o r some t i n T, t not i n I. Then f ( t l + I) = [ f ( t + l ) ] I = ( x r + l ) l = x r l + 1 = 0 ( i n . K / I ) , so t l <= j . ( R e c a l l J / l i s the k e r n e l o f f ) . . E i t h e r mt and m are dependent or they are indepen-dent. I f mt and rn . are A-independent, there are s i n K and 8 i n A , 8 i n v e r t i b l e , such t h a t mts = 0 , ms = 0m / 0. Then t s e I c J , but s / I.. • •' -I f mt and m are A-dependent there i s a B i n A (a d i v i s i o n r i n g ) , such t h a t ,B (mt) = m. The (0:"mt) = (0:m) = I, and i t f o l l o w s t h a t t l c I. However, t J gd I, f o r t J ' cz I i m p l i e s mtJ = 0, whence mj = 0, and J c I, which i s f a l s e . T h u s . i t f o l l o w s t h a t ( t J + l ) % I and t h a t 8 5 . ( t J + I ) / I n J / I / 0 i n K/I ( r e c a l l K/I i s u n i f o r m ) . T h e r e f o r e Y = ( t J + l ) fl I, and mY / 0. As we have seen b e f o r e , there are y i n Y, d i n K, and y i n A such t h a t myd = -:ym / 0, and y' = yd s a t i s f i e s y ' l c I but y' / I. Now y' e Y, so we can w r i t e y' = t j + i e J , where t j i I. I f j e. I, then t j e t i c I, which i s f a l s e . A g a i n i n t h i s case we have an element s"(=j) such t h a t s / I, but t s e J . Now 0 = f ( t s + l ) = f ( t + l ) s = ( x r + l ) s = x r s + 1 , so xr s e I. But then s e (0:mxr) = (0:m) = I , • a c o n t r a d i c t i o n . T h i s shows 'that f = 0, and t h e r e f o r e K/I i s a r a t i o n a l e x t e n s i o n o f an a r b i t r a r y submodule J / I / 0. Now, i f K i s a W-semisimple r i n g , K i s a' subdirect. sum o f r i n g s K , where each K has a ' f a i t h f u l member of ° o r a 2 ^ .Q.E.D. -a A r i n g A i s a r i g h t order i n "a r i n g B i f A i s a s u b r i n g of B, i f every element of A which i s not a z e r o - d i v i s o r i n A i s i n v e r t i b l e i n B, and i f every element of B can be w r i t t e n as uv'V where u, v are i n A, and v i s not a z e r o - d i v i s o r i r i A. > . In (17), Koh and Mewborn d e f i n e d a r i n g t o be a weakly  t r a n s i t i v e r i n g i f : ( l ) K a c t s f a i t h f u l l y as a r i n g of l i n e a r t ransforma-t i o n s on a l e f t v e c t o r space V over a d i v i s i o n r i n g D, 86. ( 2 ) There i s a r i g h t order S I n .D, and a l e f t S-, r i g h t K-submodule M of < V which i s a bimodule, which i s u n i f o r m as a K-module, and ' which s a t i s f i e s V = DM, ( 5 ) I f x-^jXg,. ..-,x i s a f i n i t e D-independent s e t of elements from M, and i f y- j ^ y 2 , •»• * y n a r e any elements of M, then there are k i n K and s i n S such t h a t x.^ k = sy^, f o r i = 1 , 2} • • • ) n.»>' ^ i-• Furthermore, they d e f i n e d a r a d i c a l c l a s s , which we, i w i l l denote by W*, by d e f i n i n g W*(K) = 4 K: K/I i s . weakly t r a n s i t i v e } . '• We now show t h a t a r i n g a c t s i n a W - t r a n s i t i v e f a s h i o n on some v e c t o r space i f and only i f i t a c t s i n a weakly t r a n s i t i v e f a s h i o n on some (perhaps d i f f e r e n t ) v e c t o r space. I t i s an immediate consequence of t h i s f a c t , t h e - p r e v i o u s . theorem, and Theorem 2 . 1 . 1 t h a t W(K) = W*(K) f o r a l l r i n g s K. Theorem 3 . 7 . 5 - A r i n g K i s weakly t r a n s i t i v e i f and onl y I f i t has a f a i t h f u l member of 2j£. . -Proof: I f K' a c t s i n a weakly t r a n s i t i v e f a s h i o n on V, the same arguments as i n Theorems 5 . 7 . 1 and 5 . 7 . 2 give a f a i t h f u l member of 2^. Conversely,.suppose t h a t M i s a f a i t h f u l member of 5 ' -2~. As i n . t h e f i r s t p a r t of the p r o o f o f Theorem 5 . 7 . 2 , , 8 7 . A = Hom K(Q(M),Q(M)) i s a d i v i s i o n r i n g , and K a c t s i n a W - t r a n s i t i v e f a s h i o n on Q(M). We w i l l show t h a t K a c t s i n a weakly t r a n s i t i v e f a s h i o n on Q(M). Le t S = HomK(M,M).. Since M i s a r a t i o n a l e x t e n s i o n of every non-zero submodule, every member of S i s a mono-morphism, and S has no zero d i v i s o r s . The q u a s i - i n j e c t i v i t y of Q(M) guarantees t h a t each s i n S can be extended t o a X i n A . I f X-^ a n d ^2 a r e ^ w o e x - f : e n s : 3 - o n s ° ^ s ^- n S, then M c ker(X-£ - X 2 ) . By Theorem 3 . 6 . 5 , Xj^ - X2 e J ( A ) = 0 . Thus any element s i n S has a unique e x t e n s i o n "s i n . . A . In t h i s way we have induced a mapping from S t o A . which i s e a s i l y v e r i f i e d ' t o be a . r i n g monomorphism. We now show t h a t S = {~s: s e S} i s a r i g h t order i n A . L e t . X 6 A , X jt 0 . Then W = {m e M: \(m) e M} i s a non-zero--K-submodule of M, and \(W) c.M.' Since W and M are subisomorphic, there i s a module monomorphism f:M -» W. We can re g a r d f as a mapping from M to ,M, t h a t i s , as a member of S. Then X.°f: M - W - x(W) d M, .and Xf can a l s o be co n s i d e r e d as a member of S. Now Xf and Xf agree on M, and so Xf .=• Xf = s. Thus X = s ( f ) ~ . Since A i s a d i v i s i o n r i n g , any non-zero member of "3" c l e a r l y has • an i n v e r s e i n A , so "S i s a r i g h t order i n A . '• F i n a l l y suppose -that x^, x^,; • • • x n a r e A-independent members of M, ' and t h a t , y-^yg* • • • * y n are I n , M. In the pr o o f of Theorem - 3 . 7 . 1 * i t was shown t h a t , x i k '= Xyj_ f o r some k 88. i n K and X £ A , where X was an e x t e n s i o n o f a map f:M -• P c M. C l e a r l y , i n t h i s case, X = f . This completes the p r o o f . Q.E.D. 5.8 Prime Rings with Zero S i n g u l a r I d e a l and a Uniform R i g h t I d e a l In F a i t h ( 7 ) , page 129, the f o l l o w i n g problem i s posed: i f . K i s a.prime r i n g w i t h a u n i f o r m r i g h t i d e a l U and i f Z (K) = 0 then S = HomK(U,U) i s a r i g h t Ore domain ( t h a t i s , S i s a r i g h t order i n a d i v i s i o n r i n g ) . I f x^,x 2,...,x n are S-independent elements of U, and i f y l j y 2 * * * " * yn a r e i n U * does there e x i s t s / 0 i n S and k i n K such t h a t x^k = sy^, i = 1,2,..., n.' ? I f K and U are as above, then U, as a K-module, 5 i s a f a i t h f u l member of S^. T h i s f o l l o w s because the argu- -5 ment i n Theorem 5.5*5 shows t h a t U i s i n 2^, and s i n c e U ' (0:U) = 0 , and s i n c e K i s a prime r i n g , ' (0:U) = 0. In Theorem 5 . 7 . 5 i t was shown t h a t the c o n c l u s i o n of the problem i s v a l i d p r o v i d e d we regard U as a submodule of Q(U), and -choose our x 's t o be A - i n d e p e n d e n t ( A = Hom K(Q(U),Q(U)) not S-independent. : The f o l l o w i n g example shows t h a t t h i s i s the best t h a t can be done, and t h a t the co n j e c t u r e i n the form s t a t e d above i s f a l s e . . ' 8 9 . Example 5 . 8 . 1 L e t K be a r i n g with 1 and with no z e r o - d i v i s o r s , and f o r which K, as a r i g h t module over i t s e l f , i s uniform, but f o r which. K i s not u n i f o r m as a l e f t module over i t s e l f . Such a r i n g , or r a t h e r one with " r i g h t " and " l e f t " i n t e r -changed i s g i v e n i n Example 9> page 7 1 , of D i v i n s k y ( 6 ) . o Kg i s uniform, and, si n c e ; K has no zero - d i v i s o r s , Z (K) = 0. The r i n g , K . i s a right'"order i n a d i v i s i o n r i n g D, and Q(Kg) = D, as"a r i g h t K-module. K i s isomorphic t o Hom K(K,K), and the embedding of Hom K(K,K) i n t o HomD(D,D) = D i s j u s t the embedding of ' K, i n t o D. The r i n g D i s a d i v i s i o n r i n g , and any two elements of D are l e f t D-dependent. However, s i n c e K, as a l e f t K-module, i s not uniform, there are x and ,y non-zero i n K such that,- Kx n Ky - 0 , and x and y are l e f t S ( i . e . K) independent. I f the co n j e c t u r e were t r u e , there would exist- k' i n . K and s i n S ( i . e . K) such t h a t xk = sy and/ ys = sy, where sy ^ o. But then (x-y)k = 0 , and, s i n c e K has no zero ' d i v i s o r s , e i t h e r x = y or k = 0 , both of which are f a l s e . T h e r e f o r e the c o n j e c t u r e i n i t s o r i g i n a l form i s f a l s e . The f o l l o w i n g r e s u l t s show how G o l d i e • s Theorem on prime r i n g s may be p l a c e d i n a context s i m i l a r , t o t h a t o f the c l a s s i c a l d e n s i t y theorem f o r p r i m i t i v e r i n g s . 9 0 . Suppose t h a t M i s a f a i t h f u l member of 2 g , and t h a t Z(M) = 0 . .(That i s , suppose t h a t M i s a f a i t h f u l b a s i c module.) Then D = Homg( Q(M),Q(M) ) i s a d i v i s i o n r i n g , and K a c t s i n a W - t r a n s i t i v e f a s h i o n on D Q (Theorem 3 . 7 . 2 ) . The next' theorem shows what happens when i s f i n i t e -d i m e n s i o n a l . Theorem 3 . 8 . 2 L e t Mg.. be a f a i t h f u l b a s i c module, and suppos t h a t i s f i n i t e - d i m e n s i o n a l , . where Q = Q(M), and , D = Hom K(Q,Q).. Then K i s a r i g h t order i n the simple a r t i n i a n r i n g L = Hom D(Q,Q). Proof: The r i n g L i s the r i n g of a l l endomorphisms of a f i n i t e - d i m e n s i o n a l v e c t o r space over D, and so L i s indeed simple a r t i n i a n . We show f i r s t ' t h a t Kg i s e s s e n t i a l i n Lg. L e t X l , X 2 i ' * * * X n 1 ° e a n y ^ a s i s f o r ~ jfi> a n < ^ suppose t h a t s / 0 i s i n L. Then x^s / 0 f o r some i , which we may take t o n \. be 1. Since : Q., i s uniform, T = fl x.K i s ' n o n - z e r o and : 1=1 1 i s an e s s e n t i a l submodule of Qg. For each i , E^ = {keK: ( x i s ) k e T ) i s an e s s e n t i a l r i g h t i d e a l of K, and~so n • E = fl E. i s e s s e n t i a l i n K. Now x. sE / 0 ( otherwise . i = l . 1 1 x, s e Z(QtK) = 0 ) and so there i s an e i n E such t h a t 9 1 . se / 0 . • For each i , x ^ e e T, . and so, by Theorem 3 . 6 . 9 * there i s an element k i n K such t h a t x^k = x^se f o r i = l , 2 , . . . , n . Since x ^ , x 2,...x n i s a b a s i s f o r DQ, . k = se / 0. " Ther e f o r e sK fl K / 0 f o r any s / 0 i n L, and Kg i s an e s s e n t i a l submodule, of Lg. A l s o , K i s a s u b r i n g o f the l e f t and r i g h t a r t i n i a n r i n g L, and so K has the descending c h a i n c o n d i t i o n on both l e f t and r i g h t a n n i h i l a t o r s . Since K. i s a l s o prime, we can apply the re c e n t r e s u l t o f Johnson and Levy (14) t o conclude t h a t any e s s e n t i a l r i g h t i d e a l of K >contains a r e g u l a r element. U s i n g the n o t a t i o n of the l a s t paragraph, f o r s / 0 i n L, the I d e a l E has a r e g u l a r element c. As above, there i s an element k i n K such t h a t k = sc. The ..proof w i l l be complete when we show t h a t any r e g u l a r element of K has an. i n v e r s e i n L. Suppose t h a t k i s a r e g u l a r element i n K. -Then i f s i n L i s such t h a t ks = 0, then k( sK fl K) = 0, and so sK fl K = 0, which Implies t h a t s = 0, by the f i r s t paragraph.. Ther e f o r e a r e g u l a r element i n K has no r i g h t a n n i h i l a t o r (except 0) i n L. Since L i s a r t i n i a n , t h i s i s s u f f i c i e n t f o r t h i s element t o have an i n v e r s e • i n L. Q.E.D. C o r o l l a r y • 3 . 8 . 3 •'. For a prime r i n g K, the f o l l o w i n g are e q u i v a l e n t . • ( l ) K i s a r i g h t order i n a simple a r t i n i a n r i n g . ." 9 2 . (2 ) Z (K) = 0 , and K has. max-rc. (3) K has a f a i t h f u l b a s i c module M, and K s a t i s f i e s the descending c h a i n c o n d i t i o n on r i g h t a n n i h i l a t o r s of subsets of' M. (4) Z (K) = 0 , K has a uni f o r m r i g h t i d e a l U, and • : • K s a t i s f i e s the descending c h a i n c o n d i t i o n on r i g h t a n n i h i l a t o r s of U. • Proof: The equivalence o f ( l ) and (2 ) i s w e l l known. F u r t h e r -more, these c o n d i t i o n s imply t h a t K has a uniform r i g h t i d e a l U. Since K has an a r t i n i a n q u o t i e n t r i n g , and s i n c e the descending c h a i n c o n d i t i o n on r i g h t a n n i h i l a t o r s i s i n h e r i t e d by s u b r i n g s , (4) i s e a s i l y seen t o be t r u e . I f ( ; 4 ) i s s a t i s f i e d , then the u n i f o r m r i g h t i d e a l U i s a b a s i c module (Theorem 3 - 5 . 8 ) . A l s o U^. i s f a i t h f u l , s i n c e K i s a prime r i n g . T h e r e f o r e (3) i s i m p l i e d by ( 4 ) . Now ±fJ ( 3 ) : i s " • s a t i s f i e d , then Q(M) i s ' f i n i t e - d i m e n s i o n a l over the d i v i s i o n r i n g D = HomK( ' Q(M,), Q(M) ). For i f i t were otherwise, since' :Q(M) = DM, ' we would be abl e t o f i n d a sequence x - ^ X g , . . . of D-independent elements i n M. By the c h a i n c o n d i t i o n i n ( 3 ) * we o b t a i n { x ^ , X g , . . . * x n } K = {x^, X g , x n , x n | ] _ } K f ° r some. n. But then, by C o r o l l a r y n -3 . 6 . 8 , we o b t a i n 2 Dx. = 2 n -Dx., c o n t r a d i c t i n g the 1=1 1 i = l 1 D-independence o f the x.'s. 93. . Thus. Q(M) i s f i n i t e d i m e n s i o n a l , and ( l ) f o l l o w s from the p r e v i o u s theorem. Q.E.D. I t i s well-known t h a t a r i n g which a c t s . t r a n s i t i v e l y on a f i n i t e - d i m e n s i o n a l v e c t o r space must he the endomorphism r i n g o f t h a t v e c t o r space, and consequently i s a simple a r t i n i a n r i n g . . The next theorem i s the analogue of t h i s , r e s u l t f o r W - t r a n s i t i v e r i n g s . \', • Theorem 5 . 8 . 4 L e t K " he a W - t r a n s i t i v e r i n g , and suppose t h a t D"V i s f i n i t e d i m e n s i o n a l . Then K has a simple a r t i n i a n c l a s s i c a l r i g h t q u o t i e n t r i n g . ) Proof: We are assuming t h a t we have D, - QV ^ , and M^ - as i n the d e f i n i t i o n o f W - t r a n s i t i v i t y , and t h a t i s f i n i t e -d i m e n s i o n a l . We show f i r s t t h a t Z ( V K ) = 0. For i f v e Z ( V K ) , v / 0, then f o r any m i n M,- there, are k i n K and d i n D, d / 0, such t h a t , vk = dm. Then ,: (0:m) = (0:dm) = (0:vk), and the l a t t e r i s e s s e n t i a l s i n c e v e Z(v" K). Now V = DM, and t h i s i s f i n i t e - d i m e n s i o n a l , so there i s a basis, n rn-.,mQ,...,m c o n s i s t i n g of elements o f M. But then E = fl . . 1=1 n (0:m.) i s e s s e n t i a l , and so i s non-zero.- But VE = ( 2 Dm. )E 1 - i = l 1 and t h i s c o n t r a d i c t s the f a c t t h a t K a c t s f a i t h f u l l y on V. Hence Z ( V K ) = 0, 'and so Z(M K) = 0 . 9 4 . T h e r e f o r e M i s a un i f o r m K-module f o r which.-Z(M) = 0 . Since V = DM i s f a i t h f u l as a K-module, M i s a l s o f a i t h f u l . A l s o M i s a prime K-module. For suppose t h a t ml = 0 f o r .some m i n M and 1-4 K. By Theorem 3 . 7 - 2 , . K has some f a i t h f u l member.of 2j£, and so K i s a prime r i n g . T h e r e f o r e e i t h e r I = 0 , or I, b e i n g a' two-sided i d e a l i n a prime r i n g , i s an e s s e n t i a l r i g h t i d e a l , - a n d m e Z(M) = 0 . T h i s e s t a b l i s h e s the primeness o f M. -Thus M i s a faithful.-member of S2., and Z(M) = 0 . By Theorem 3 . 4 . 6 , M has a submodule N / 0 which i s a member of 2.g. A l s o . N K i s f a i t h f u l and Z(N)'= 0 . L e t V •= Q(N) and D' = HomK( V , V ). The l a t t e r i s a d i v i s i o n , r i n g , by Theorem 3 . 6 . 1 0 and Lemma 3 . 1 . 3 . We w i l l , now show t h a t dim(jy V')..• <_ dim(^V). Once t h i s has been e s t a b l i s h e d , we may apply Theorem 3.8 . 2 to conclude that. K i s a r i g h t order i n 'Hom^/(jy V'.,jy V ' ) , which i s a simple a r t i n i a n r i n g . , Suppose now t h a t d i m ^ V ) = n, but dim( D/V') > n. Then, s i n c e D'N = V', we can f i n d x 1 , x 2 , . . . * x n + 1 * a s e t of D'-independent elements o f N. However, s i n c e N c M c V, these elements are not D-independent. Without l o s s o f n g e n e r a l i t y , we may assume that.we may w r i t e x - = 2 d.x , •" • . N + ± .1=1 J J 95. where d ^ d g , . . . , d n are i n D. Then ( 0 : X n + i ) H n (°: xj)> n fl j = l from which i t f o l l o w s t h a t { x ^ x ^ j . . . ^ ^ ^ = { x ^ , X g , . . . ,x^, xn + ] _ } ^ " • We may then apply C o r o l l a r y 3.6.8 and conclude . n n-i-1 t h a t 2 D'x. = 2 D'x., c o n t r a d i c t i n g the D'-independence o f the X j ' s . Thus d i m ( D , V') : <_ d i m ( D V ) . Q.E.D... Remark: Under the c o n d i t i o n s of the theorem, K i s not, i n g e n e r a l , a r i g h t order i n ,Hom^(^V,^V). For example, l e t D be any d i v i s i o n r i n g which i s not commutative, and l e t C be the centre of D. L e t K = C,,/ D V K ' = D D C > a n d \ = c I t i s . simply v e r i f i e d tha,t K a c t s ¥-transitively on the ve c t o r space'' V, and t h a t V i s one-dimensional over D. But K i s the centre of a d i v i s i o n r i n g , and t h e r e f o r e Is a . f i e l d , and i s i t s own c l a s s i c a l r i g h t q u o t i e n t r i n g . A l s o , Hom D(V,V) = D, \tfhich i s . not commutative, so D / K = r i g h t q u o t i e n t r i n g of K. 96. . CHAPTER FOUR HEREDITARY RADICAL IDEALS OF A RING In t h i s chapter we focus a t t e n t i o n on the i d e a l s I of a r i n g K f o r which I = R(K) f o r some h e r e d i t a r y r a d i c a l R.. These, as we s h a l l see, w i l l be the c l o s e d i d e a l s f o r a c l o s u r e o p e r a t i o n d e f i n e d on the l a t t i c e of a l l i d e a l s of the r i n g K. The p r o p e r t i e s of t h i s c o l l e c t i o n o f c l o s e d i d e a l s w i l l be i n v e s t i g a t e d when c e r t a i n c o n d i t i o n s are imposed on K. 4 . 1 An Equivalence R e l a t i o n f o r Rings R e c a l l from Chapter one t h a t a r a d i c a l c l a s s i s s a i d to be h e r e d i t a r y . i f and'only i f any i d e a l o f a member of R' Is a l s o a member of R. A l s o r e c a l l t h a t a s u b r i n g T of a r i n g K i s c a l l e d ;an a c c e s s i b l e s u b r i n g o f K i f there i s a f i n i t e c h a i n , - " . •T = T Q. c T 1 c. T 2 . . .: c T.n = K ,. where T. i s an i d e a l o f T. ,',. I t f o l l o w s t h a t , i f R ' i s 1 • l + l ' a h e r e d i t a r y - r a d i c a l , and i f K e R, then any a c c e s s i b l e . ,. s u b r i n g of K. i s i n R. In Theorem 1.4 . 1 , we saw t h a t i f , M i s a h e r e d i t a r y c l a s s o f r i n g s which i s a l s o c l o s e d under" the t a k i n g of homomorphic images, then S Q ( M ) , the lower r a d i c a l c l a s s w i t h r e s p e c t t o M, i s a h e r e d i t a r y ' r a d i c a l c l a s s . Suppose 97. t h a t we are g i v e n any c l a s s M of r i n g s . Does there n e c e s s a r i l y e x i s t a s m a l l e s t h e r e d i t a r y r a d i c a l c l a s s c o n t a i n i n g M? T h i s q u e s t i o n i s answered a f f i r m a t i v e l y by the next theorem. Theorem 4.1.1- Suppose M Is an a r b i t r a r y c l a s s of r i n g s , and d e f i n e M* to be the c l a s s o f a l l homomorphic images of a c c e s s i b l e subrings of members of M. Then M* i s a homo-m o r p h i c a l l y . c l o s e d h e r e d i t a r y c l a s s , and- SQ(M*) i s the s m a l l e s t h e r e d i t a r y r a d i c a l c l a s s c o n t a i n i n g M. Proof: .M* i s c l e a r l y homomorphically c l o s e d . We show t h a t M* . i s h e r e d i t a r y . I f K e M*, then there i s ; a r i n g S i n M, a 'chain T = T, c.T- c .. . c T = S, where T. i s an i d e a l of- T^+^,' and there i s a homomorphism f:T -• K which maps T onto ' K. Now, i f I i s an i d e a l of K, then c • f - 1 ( l ) . =. .J i s an i d e a l of T, and hence J i s an a c c e s s i b l e s u b r i n g of S. A l s o , the r e s t r i c t i o n of f t o i s a r i n g homomorphism from J onto I. T h i s shows t h a t I e M*, and thus M* i s h e r e d i t a r y . By Theorem 1.4.1, S Q ( M * ) i s h e r e d i t a r y . I f R i s • any h e r e d i t a r y r a d i c a l c l a s s , and i f M e R, then, s i n c e R i s c l o s e d under t a k i n g of a c c e s s i b l e subrings and homomorphic images, M* c R. Prom the p r o p e r t i e s of the lower r a d i c a l , ~ we obtain. S Q ( M * ) c R. Q.E.D. For a g i v e n class 7).,M of r i n g s , we w i l l denote SQ(M*) 9 8 . by H(M). I f M c o n s i s t s of a s i n g l e r i n g K, we s h a l l denote- H(M') - by Hg. . . We d e f i n e an e q uivalence r e l a t i o n f o r r i n g s by s a y i n g the r i n g s K^ and Kg are h e r e d i t a r i l y e q u i v a l e n t i f and only i f Hg = Hg , a s i t u a t i o n which we denote by K-^  ~ Kg. T h i s i s e a s i l y seen t o be an e q uivalence r e l a t i o n . For example, any r i n g i s h e r e d i t a r i l y e q u i v a l e n t to a d i r e c t sum of copies of i t s e l f . For i f S = ® K , where ae K == K. f o r each a, then K . i s isomorphic to an i d e a l of S, and K e Hg. On the other hand, .. S i s a d i r e c t sum of'' c o pies of K, /-and so S e Hg. That t h i s guarantees . S ~ K f o l l o w s from the next r e s u l t . P r o p o s i t i o n 4.1.2 For any r i n g s K" and K', K ~ K' i f and o n l y i f K e Hg/ and.- K' e Hg. Proof: I f ; K ~ ,K', then K e Hg = Hg, and K' e Hg/ = Hg. Conversely,- i f K e Hg/, then, by Theorem 4.1.1 we have Hg c Hg/.. S i m i l a r l y Hg/ c Hg, and we o b t a i n Hg = Hg/, , -t h a t i s K ~ K'. :Q.E.D. P r o p o s i t i o n 4.1.3 I f K and K' are r i n g s , then K ~ K' i f and only i f , f o r every h e r e d i t a r y - r a d i c a l - R,. the c o n d i t i o n s R(K) = K and R(K') = K' are e q u i v a l e n t . 99. Proof: Suppose f i r s t t h a t K ~. K' . I f R is'.'a hereditary-r a d i c a l , and i f R(K) = K, then K e R, and Hg, = Hg c R, by Theorem 4.1.1. Thus K' e R, or R(K') = K'. S i m i l a r l y R(K') = K' i m p l i e s R(K) = K. Conversely, i f the c o n d i t i o n s R(K) = K and R(K') = K are e q u i v a l e n t f o r a l l h e r e d i t a r y radicals-•-R, then Hg/(K') K' i m p l i e s Hg, (K) <= K, or K e Hg/. S i m i l a r l y K' e Hg. By the p r e v i o u s p r o p o s i t i o n we have K ~ K'. Q.E.D. For any r i n g K, we s h a l l denote by E(K) the c l a s s , o f a l l r i n g s K' f o r which K K'. C l e a r l y , ' i f K' ~ K, then K' e Hg so we have a t once t h a t E(K) C Hg. The opposite i n c l u s i o n i s not always t r u e , f o r the one-element r i n g i s always i n Hg but, u n l e s s K = 0, 0 £\E(K). A s l i g h t l y l e s s t r i v i a l example i s the f o l l o w i n g . L e t K = C , the zero r i n g on the ( a d d i t i v e ) i n f i n i t e c y c l i c CO group, and l e t K' = Cg, the zero r i n g on the ( a d d i t i v e ) c y c l i c , group of order 2. Then K' ..'is a homomorphic image of K, and so K' e Hg. However, K' £ E(K). T h i s can be seen as f o l l o w s : the c l a s s R of a l l . r i n g s whose ( a d d i t i v e ) u n d e r l y i n g a b e l i a h group i s a t o r s i o n group i s e a s i l y seen to be a r a d i c a l c l a s s , and i s a h e r e d i t a r y r a d i c a l c l a s s . Now K' e R but K / R. Thus K/'K'\ I t i s of some I n t e r e s t t o ask the qu e s t i o n : when i s 100. •is E(K) U {0} equal t o the c l a s s Hg? The next few r e s u l t s l e a d up t o the answer t o t h i s q u e s t i o n . We s h a l l see t h a t E(K) U {0} = Hg. i f and only if:. K i s h e r e d i t a r i l y e q u i v a l e n t to a simple r i n g . Lemma 4 . 1 . 4 L e t R be. any r a d i c a l c l a s s (not n e c e s s a r i l y h e r e d i t a r y ) , and l e t K be a r i n g . Then': (1) I f B OK, and i f both B and K/B are i n R, then K i s i n R. . (2) I f {Bar: aeA) i s a f a m i l y o f i d e a l s of K, and i f B £ R f o r each a, then £ jB i s i n R. Proof: . (1) C l e a r l y R~(K) p_ B, and so K/R(K) i s a homo-morphic image of .K/B. Since K/B i s i n R, so must K/R(K) be i n R. T h e r e f o r e , 1 K/R(K) i s both R - r a d i c a l and R-semisimple, so K/R(K) = 0 , K = R(K), and K e R. (2) L e t S =. .2 B . Then, f o r . each a, "B e R, . . aeA B <I S, so B c R(S). T h e r e f o r e S = 2 B c R(S), so ae A S = R(S), and S e R. Q.E.D. The next lemma i s , i n essence, due to D i v i n s k y , Anderson, and S u l i n s k i . ( c f . D i v i n s k y ( 6 ) , Theorem 47 ) . Lemma 4 . 1 . 5 Let' R be any r a d i c a l c l a s s , and l e t K be . a r i n g . Suppose that,.' K 0 J M , and t h a t I e. R. ( l ) I f x e K, then I + x l i s an i d e a l of J , and 101. I + x i i s i n R. ( 2 ) The i d e a l of K generated by I, I.+ KI + IK + KIK = <I> K, i s i n R. Proof: I + x l i s c l o s e d under a d d i t i o n . . A l s o ( I + x l ) J c I+xI, and J ( I + x l ) c J I + (Jx.)T c I. The r e f o r e I+xI i s an i d e a l of J . Consider the map f : I . - ( I + x l ) / I , where f ( y ) =• xy+I. T h i s s a t i s f i e s f(y+y') = f ( y ) + f(-y')> .and f ( y y ' ) + I = (xy)y' + I = I ( s i n c e xy e J ) , w hile f ( y ) f ( y ' ) = (xy)(xy') + I = (xyx)y' + 1 = 1 . Theref o r e f i s a r i n g homomorphism and i s a s u r j e c t i o n . Since I e R, we have t h a t I and (I+x.l)/I are i n R, whence, by Lemma 4.1.4, ' I+xI i s i n R. ( 2 ) In a s i m i l a r f a s h i o n : one''-can show., t h a t I+Ix i s an i d e a l i n J and i s a mebaber of R. ; If...we r e p l a c e I by I+zI, f o r z e K, we a l s o o b t a i n . ( l + z l ) + ( l + z l ) x i s an i d e a l i n J which belongs t o R. Now we' can- w r i t e <I>^- as 1 + 2 (I+xI) + 2 (l+Iy) + 2 {(I+zI) + (I+zl)w}. By xeK yeK z',v/eK p a r t ( l ) t h i s i s a sum of i d e a l s of J which are a l l members of R. By Lemma 4.1 . 4 , <I> K i s i n R. Q.E.D. C o r o l l a r y 4.1 .6 L e t R be any r a d i c a l c l a s s , and l e t I be an a c c e s s i b l e s u b r i n g of a r i n g K. I f I e R,- then ; <I> K i s a l s o in. R. Proof: We have seen t h a t the statement i s true whenever I <I J <J K. Suppose t h a t we have proved the r e s u l t f o r a l l 102. a c c e s s i b l e subr ings I ' which s a t i s f y I ' <J J ' 2 4 J ' ^ < . . . <3 J ' ri_-^ = K , ' and suppose tha t we have I <J <} ... <j = K . Then l e t T = <I> T = I + J a I + I J , -I- J-,IJ-, . By the lemma J-*, 3 3 3 3 we have tha t T e R . Now, we have T < J-, < J,, < ... <J J = K , 3 4 n ^ and so , by our i n d u c t i o n assumpt ion, < ! - ^ K i s i n R » ' S ince I c T, we have <1>K £ < T > K * On,the other hand, i t i s e a s i l y seen tha t T c <I>Ka and so <T>K c <I> R . Therefore <T>K = <I> K i and. t h i s i n R . •' Q . E . D . Theorem 4 . 1 . 7 L e t M be a homomorphically c lo sed c l a s s of r i n g s . Then a r i n g K i s i n S Q ( M ) i f and only i f every non-zero homomorphic image of K has an a c c e s s i b l e sub r ing which i s ' a non-zero member of M. . \-Proof : We use the n o t a t i o n of §1 . .2 . . Suppose f i r s t tha t K i s i n S Q ( M ) , and l e t K ' / 0 be 'a homomorphic ima.ge o f K . Then K ' e S Q ( M ) , and, by the d e f i n i t i o n ' o f - S Q ( M ) , K ' e M Q f o r some o r d i n a l a. I f a / 1 , then K-' has a non-zero i d e a l ' K ' which belongs to M , • f o r some o r d i n a l a, < a.~ 1 D a 1 Suppose tha t we have found K ' 2 , K ' n , where K ' . <J K ' . . , and K ' . e M where a < a , < . . . ou. I f l •. l - l J j a. n n-1 1 J a n / 1, then K ' N has a non-zero i d e a l K ' N + ] _ which i s i n M L f o r some a . < a . Thus, i f K ' had no a c c e s s i b l e a ,, n-rl n 3 n-hl 103. subrings i n ( = M ) , we would be'able to f i n d an i n f i n i t e descending sequence of o r d i n a l numbers. Since the o r d i n a l s are w e l l ordered, t h i s - i s i m p o s s i b l e , and so K' must have an a c c e s s i b l e subring which i s a member of M. Conversely,' suppose the c o n d i t i o n holds. I f S Q ( M ) ( K ) / K , then K / S Q ( M ) ( K ) . = K' / 0 , and,'"'by our assumption, t h i s has a non-zero a c c e s s i b l e subring I which i s a member of S0(.M). By C o r o l l a r y 4 . 1 . 6 , <I>K/ i s ' a l s o i n S Q ( M ) . I f J i s the inverse image of < ^ > j ^ ' under the n a t u r a l homo-morphism from : K to K', Lemma- 4 .1 .4 guarantees that i J e S 0 ( M ) . This i s a c o n t r a d i t i o n since J ^  S Q ( ' M ) ( K ) . Therefore K = S Q ( M ) ( K ) , or K e S0(M)..- Q.E.D. Theorem 4.1.8 For any non-zero r i n g K , the f o l l o w i n g are' e q u i v a l e n t : ( l ) K ~ T whenever T i s a non-zero homomorphic image of an a c c e s s i b l e subring of K. •> (2) E ( K ) u {0} = HK. ; • (3) E ( K ) y {0} i s a h e r e d i t a r y r a d i c a l c l a s s . (4) E ( K ) U {0} i s h e r e d i t a r y and i s homomorphically closed. , ( 5 ) For any r i n g L, and f o r any h e r e d i t a r y r a d i c a l R, i f 0 ^ R(L) c H K(L), then R(L) = H.K(L). . ( i . e . H K(L) i s e i t h e r zero or a minimal non-zero 10 4 . h e r e d i t a r y r a d i c a l i d e a l o f L ) . ( 6 ) E ( K ) c o n t a i n s a s i m p l e r i n g . P r o o f : ( l ) i m p l i e s ( 2 ) . We a l r e a d y know t h a t E ( K ) U {0} c Hg. Suppose t h a t L i s a non-zero member of Hg. By Theorem 4 . 1 . 1 and the p r e v i o u s theorem, L has a non-zero a c c e s s i b l e s u b r i n g L' i n {K}*. . By d e f i n i t i o n o f {K} *, L' i s a homomorphic image o f an a c c e s s i b l e s u b r i n g o f K, and so ( l ) i m p l i e s L' ~ K. Now L' i s an a c c e s s i b l e s u b r i n g o f L and s o l / e H^ c Hg, and Hg = H L/ c H L c Hg. T h e r e f o r e K ~ L,. and thus"' Hg c E ( K ) U { 0 } . That ( 2 ) i m p l i e s ( 3 ) , and t h a t ( 3 ) i m p l i e s ( 4 ) , are b o t h o b v i o u s . ( 4 ) i m p l i e s ( l ) . Suppose t h a t we have K !> t> Kg ... t> K , and-a homomorphism f o f K n onto a non-zero r i n g .. T. Now K e E ( K ) , and by u s i n g (4) we s e e . t h a t K-^ , Kg, ... K^ and f i n a l l y T . a r e i n E ( K ) U ' { 0 } . S i n c e . T / 0 , we have • T e E ( K ) , or K ~ T. ( 2 ) i m p l i e s ( 5 ) . Suppose t h a t 'R i s a h e r e d i t a r y , r a d i c a l , and t h a t L i s a r i n g f o r which 0 / R(L) c Hg(L).-Then the r i n g T = R(L) i s i n Hg and t h e r e f o r e i s I n E ( K ) , s o ' Hg = H T. I n p a r t i c u l a r , H R ( L ) = H T ( L ) : . S i n c e T = R ( K ) , T e R, and H T c R. T h e r e f o r e H R ( L ) = H T ( L ) c R'(L), and we have R(L) = H K ( L ) . 105. (5) i m p l i e s ( 2 ) . Suppose tha t L i s a non-zero member o f Hg. Le t K ' be a r i n g i somorphic to K , and cons ide r the ( e x t e r n a l ) d i r e c t sum K ' & L . Since K ' and L are both i n Hg, we have 0 / L c H L (K' -$L) C Hg(K'&L) = K ' & L . ' By c o n d i t i o n ( 5 ) , K ' & L = H . L ( K ' $ L ) , which i m p l i e s K ' e H L . S ince K ' = K , we have K e and L e Hg, which g i v e s , by P r o p o s i t i o n 4 . 1 . 2 , K ~ L . (6) i m p l i e s ( 2 ) . Suppose tha t E(K) conta ins a s imple r i n g S. Then S c e r t a i n l y s a t i s f i e s ( l ) , and, by a. p rev ious p a r t of t h i s p r o o f , H g = E(S) U {0}'. Since Hg = Hg • and s ince E(S) = -E(K) , the r e s u l t f o l l o w s . (4) i m p l i e s ( 6 ) . We choose x / 0 i n K. By Zorn ' s Lemma, there i s an idea l . . U. of K maximal among those i d e a l s which do not c o n t a i n x . I t f o l l o w s then t h a t , i f H -i s the image o f the ( two-s ided) i d e a l generated by x under the n a t u r a l homomorphism ..of K onto . K / U , H i s the i n t e r -s e c t i o n of a l l non-zero i d e a l s o f K / ' J , and H / 0. By c p c o n d i t i o n (4)., H i s i n ^ E ( K ) . • I f H = H then H i s w e l l 2 2 known to be a s imple r i n g . I f H / H , t h e n ; H = 0 . I n t h i s case l e t h / . 0 be i n H . The a d d i t i v e subgroup T generated by h i s an i d e a l of H , and t h i s Is a c y c l i c a b e l i a n group. T can there fore be homomorphically mapped onto 0^ ( the zero r i n g on the a d d i t i v e c y c l i c group of order p) f o r some prime p . A g a i n u s i n g ( 4 ) , we have tha t T and C_ are i n E ( K ) . But C i s a s i m p l e r i n g . Q . E . D . 106. 4 .2 A Closure Opera t ion We in t roduce i n t h i s s e c t i o n a c lo su re o p e r a t i o n i n the l a t t i c e of a l l i d e a l s of a, r i n g K , and show tha t the " c l o s e d " i d e a l s are p r e c i s e l y those i d e a l s of the form R(K) f o r some h e r e d i t a r y r a d i c a l R. Some o f the p r o p e r t i e s of t h i s l a t t i c e are cons ide red . A mapping c of a l a t t i c e to i t s e l f i s c a l l e d a-c l o s u r e o p e r a t i o n i f i t s a t i s f i e s : C I . A <_ B i m p l i e s c(A) <_ c ( B ) . C 2. c ( c ( X ) ) ; = c(X) f o r a l l X i n the l a t t i c e . C 3. X.<_ c(X) f o r a l l X i n the l a t t i c e . The elements of the l a t t i c e f o r which X = c(X) are c a l l e d c lo sed elements ( r e l a t i v e to the c lo su re o p e r a t i o n i n q u e s t i o n ) . Theorem 4.2.1. : The mapping c of the l a t t i c e of two-s ided i d e a l s of a r i n g K i n t o i t s e l f , def ined -by c ( l ) = H-j-(K), i s a c l o su re o p e r a t i o n . The c lo sed i d e a l s of K are p r e c i s e l y those i d e a l s of the form R(K') f o r some h e r e d i t a r y r a d i c a l R. Proof : I f A , B are i d e a l s o f K, and i f A c B then A <J B , and so A e Eg whence R"A .c Kg . Therefore c (A) . = H A (K) c H B ( K ) = c ( B ) . . C l e a r l y C 3 i s s a t i s f i e d . - We now e s t a b l i s h C 2. I f . B = R(K) f o r some h e r e d i t a r y r a d i c a l R, then B e R, H c R, and so c(B) = H R(K)" c R(K) = B , and B = c ( B ) . -1 0 7 . Thus any i d e a l of the form R(K) i s c l o s e d . In p a r t i c u l a r , t h i s shows tha t C 2 i s s a t i s f i e d . ; F i n a l l y , i f B i s a c lo sed i d e a l , B = c(B) = E g ( K ) , where Hg i s , of course , a h e r e d i t a r y r a d i c a l . Q . E . D . In the sequel-, f o r B an i d e a l o f K , we s h a l l denote c(B) by B . Theorem 4.2.2 I f A and B are i d e a l s o f a r i n g K , . then A = B i f and on ly I f A ~ B.';, . P roof : , I f A = B":, then A <3 A = B = H B ( K ) , SO A e Eg . ' S i m i l a r l y , B e H ^ , and thus •. A ~ B , by P r o p o s i t i o n 4 . 1 . 2 . Conver se ly , i f - A ~ B then H A = 'Hg, and c l e a r l y A = B . Q.E.D, Theorem 4.2.3 Le t K be any r i n g . Then: ( 1 ) 0 i s a c lo sed I d e a l o f K. - • (2) The i n t e r s e c t i o n o f any set of c l o sed i d e a l s i s c l o s e d . (3) I f { A a : ,aeA} i s .a f a m i l y of i d e a l s of K , then • S" A = 2 A . aeA a aeA a Proof : Since the c l a s s c o n s i s t i n g o f t h e - s i n g l e r i n g 0 i s a h e r e d i t a r y r a d i c a l c l a s s , ( l ) i s s a t i s f i e d . Suppose tha t { C Q : aeA} i s a se t of c lo sed i d e a l s of K. S ince fl C c Cg f o r each 3, we have fl C c C a = C Q , aeA , . p aeA P P .108. and hence fl c fi ^g* Together w i t h p rope r ty C I , ' aeA peA ... P . t h i s i m p l i e s (2). Suppose now tha t [ A a : aeA] i s a set of i d e a l s o f K . Then A c J, f o r each ' a , and so • £ A cz £ A . On the a — a • 3 a — a aeA . aeA other hand, A c £ A f t and so A c £ AT v which g ives , l3eA P v a ~ p e A p £ A" c £ ~KZ . . A p p l y i n g the c l o s u r e - o p e r a t i o n to t h i s l a s t aeA a f 3 e A p i n e q u a l i t y e s t a b l i s h e s (3). Q . E . D . The se t of c l o sed i d e a l s of a r i n g K has a p a r t i a l : o r d e r i n g ( i n c l u s i n n ) and t h i s induces a l a t t i c e s t r u c t u r e on the se t of c l o sed i d e a l s of K. I f we denote these l a t t i c e opera t ions by A and y, then t h i s l a t t i c e i s a complete l a t t i c e , ' w h e r e , i f {C : aeA] i s a set of c lo sed i d e a l s , A C •' = n C , and v C ' = £ C . a a a a aeA aeA aeA aeA . - -A l a t t i c e i s s a i d t o be modular , i f , whenever a , b and c are i n the l a t t i c e , where b <_ a , we have a A (bye) = b v ( a A c ) . Many o f the- common l a t t i c e s i n a l g e b r a i c , systems, such as the l a t t i c e of i d e a l s i n a r i n g , are modular l a t t i c e s , and t h i s f a c t i s c r u c i a l i n p r o v i n g r e s u l t s as the Jordan-Hoelder Theorem. The f o l l o w i n g example shows tha t "the l a t t i c e of c l o sed i d e a l s of a r i n g need not be a modular l a t t i c e . In p a r t i c u l a r , t h i s example a l so , shows tha t the l a t t i c e opera t ions (y and A ) i n the l a t t i c e o f c l o s e d i d e a l s are 1 0 9 . not n e c e s s a r i l y the same as the l a t t i c e opera t ions (+ and fl) i n the l a t t i c e of a l l i d e a l s of a r i n g . That i s , the l a t t i c e of c lo sed i d e a l s i s not n e c e s s a r i l y a s u b l a t t i c e o f the l a t t i c e of a l l i d e a l s of a r i n g . Example 4 , 2 . 4 Let S be a s imple n o n - t r i v i a l Jacobson r a d i c a l r i n g of c h a r a c t e r i s t i c 2 . (Such r i n g s are known to e x i s t . See D i v i n s k y ( 6 ) , p . 1 1 2 , or Sas iada and Cohn ( 2 3 ) . ) We embed S i n t o a r i n g S' w i t h u n i t y . b y p u t t i n g a r i n g s t r u c t u r e on the C a r t e s i a n product S x (Z^ be ing the r i n g of i n t e -gers modulo 2) by d e f i n i n g a d d i t i o n componentwise and m u l t i -p l i c a t i o n by ( s , n ) ( s 1 , n 1 ) = (ss^ + ns^ + n^s, no^) . I t i s w e l l known tha t S' . has an a s s o c i a t i v e r i n g s t ruc tu re^ tha t ( 0 , 1 ) i s a u n i t y f o r S ' , and tha t { ( s , 0 ) : se'S} i s an i d e a l o f S' which i s i somorphic to S.. L e t Z be the r i n g of i n t e g e r s , and l e t K == Z&S' . We c l a i m , tha t Z , S, S ' , K and 0 are a l l c l o sed Idea l s of K. We have mentioned tha t the c l a s s T"; "of a l l r i n g s whose u n d e r l y i n g ( a d d i t i v e ) a b e l i a n groups are t o r s i o n groups i s a h e r e d i t a r y r a d i c a l c l a s s . (See § 4 . 1 . ) C l e a r l y f o r the r i n g K , T(K) = S ' . Since J (K) = S, we have tha t K , 0 , S and S' are a l l c lo sed i d e a l s . I f Z was not c l o sed i n -K, we would have Z / H ^ ( K ) , and the re fo re H Z ( K / Z ) = H Z ( S ' ) V 0 . By Theorem 4 . 1 . 7 and the d e f i n i t i o n of H 7 t h i s means tha t S' would have a 1 1 0 . non-zero a c c e s s i b l e sub r ing L which i s a homomorphic image of an a c c e s s i b l e sub r ing of .Z, and L . would be commutative. Since S i s the unique maximal i d e a l o f S'., and s ince S i s a s imple r i n g , we would, have to have L = S' or L = S, and thus S would be a commutative s imple r i n g . Such r i n g s are w e l l known to be e i t h e r f i e l d s or r i n g s f o r which the m u l t i p l i c a t i o n i s t r i v i a l . On the other hand S. i s a non-t r i v i a l Jacobson r a d i c a l r i n g , and. thus can be n e i t h e r a f i e l d nor a t r i v i a l r i n g . Therefore Z must be a c l o sed i d e a l o f K. Since K/(Z+S) = - S / / s ' = Z23 a n o m o m o r P h i c image of Z , we have K/(Z-i-S) e cz Hg+S " By Lemma. 4 . 1 . 4 , we then o b t a i n K e H2+S o r z + s = K . ^ e c a n n o w s e e that. . the l a t t i c e of c l o sed i d e a l s of K i s not modular , s ince S' n (S+Z) = S' n K = S' , . w h i l e '• S + (S'f lZ) = S+0 = S = S. A l s o , s ince S+Z / S+Z, the l a t t i c e of c lo sed i d e a l s i s not a s u b l a t t i c e of the l a t t i c e of a l l i d e a l s . Theorem 4.2.5 • Le t A and B be Idea l s of a r i n g K , and l e t B c A . Then (A/B) c ( A ) / B . _ . Proof : A / B i s a homomorphic image of A , so H^y^ c H ^ . Therefore (A/B)'== H ^ y ^ K / B ) - c H A ( K / B ) • = ' H A ( K ) / B = A / B . Q . E . D . C o r o l l a r y 4.2.6 ; I f A . and B ' are i d e a l s of K, where B c A , then , i f A i s c l o s e d , A / B i s a c l o sed i d e a l of K / B . 111. Proof: (A/B) c A/B = A/B. Q.E.D. The converse of the c o r o l l a r y i s f a l s e . The f o l l o w i n g example shows that closed i d e a l s are not preserved under extensions. We give an example of a r i n g K with i d e a l s A and • B, where A 3 B, such t h a t B i s a closed i d e a l of K, A/B Is a closed i d e a l of K/B, but • A i s not closed in. K. ' • Example 4.2.7 Let F be the f i e l d o f two elements, and l e t A = 2 2 Fx $ Fy, where x = 0 and y = y. Let K =, { ( a , f ) : aeA,feF} wi t h a d d i t i o n defined componentwise and m u l t i p l i c a t i o n defined by ( a , f ) ( a / , f / ) = (aa' + f a ' + f a , ft'). We can i d e n t i f y the elements, of A with {(a,0): aeA} c K, and thus regard A as a subring of K. Under t h i s i d e n t i f i c a t i o n , i t i s e a s i l y seen t h a t A and B = Fy are i d e a l s i n K. I t i s e q u a l l y easy to see t h a t K/B = { ( f x , g ) : f,g e F} w i t h a d d i t i o n defined componentwise and m u l t i p l i c a t i o n defined by (fx,g) (f'x,g') = ( (gf' +g' f )x,gg' ). This l a t t e r ring, has four elements, and' the only n o n - t r i v i a l • a c c e s s i b l e subring i s the one Isomorphic to A/B, ..and t h i s i s a n i l p o t e n t subring. I t f o l l o w s , since B has a u n i t y element and has two elements, that K/B has no non-zero accessible subrings which, are homomorphic images of a c c e s s i b l e subrings of B. By Theorem 4.1.7, H B(K/B) = 0, and so B = B. ' In K/B, the i d e a l A/B i s n i l p o t e n t , and i s , i n f a c t the Baer Lower R a d i c a l of K/B. -112. Therefore A / B i s c lo sed i n K / B . I t ' i s easy to see tha t , K/A ~ F = BeH^. Thus, by Lemma 4 . 1 . 4 , A = H^(K) = K , and A i s 'not c l o s e d . 4.3 Min ima l Closed Idea l s of a R ing We s h a l l c a l l a min ima l . (non-ze ro ) c lo sed i d e a l o f a r i n g an atom. Then Theorem 4 . 1 . 8 a s se r t s tha t E(K) con ta ins a s imple r i n g i f and only i f , f o r every r i n g L , H^(L) i s e i t h e r zero or an atom/ R e c a l l tha t B denotes the Baer Lower R a d i c a l . I t i s shown i n D i v i n s k y (6) (page 43) tha t B = S Q ( {C^} )» J-which i m p l i e s a t o n c e , / s i n c e B ' i s h e r e d i t a r y , t ha t B = CO Lemma 4 . 5 . 1 Le t K be a r i n g f o r which B(K) - i s non-zero and has no elements o f f i n i t e a d d i t i v e o rder . Then B(K) i s an atom of . K. y Proof : I f T i s any non-zero i d e a l o f K conta ined i n B ( K ) , then T e, B . By Theorem 4 . 1 . 7 , T then has an a c c e s s i b l e s u b r i n g T' which i s a non-zero homomorphic image o f an a c c e s s i b l e sub r ing of C . But B ( K ) , and the re fore T, has no elements of f i n i t e a d d i t i v e o rde r , and C i s i somorphic CO to e v e r y • s u b r i n g of i t s e l f . I t f o l l o w s tha t T' must be i somorphic to C . Since T' i s an a c c e s s i b l e sub r ing o f ' CO -T, T'eHVp, and we have B = H c = E^, c: H^, c.B-. . T h e r e f o r e c o , B(K) = H T ( K ) = T, and B(K) can not p r o p e r l y c o n t a i n a n y ' 113. non-zero c l o s e d i d e a l s . Q.E.D. For any r i n g K , and any prime p, l e t Fp(K) be the s e t of a l l elements of K whose a d d i t i v e order i s a power of p.- I t i s e a s i l y v e r i f i e d t h a t Fp(K) i s an i d e a l of K , and t h a t F P ( K / F P ( K ) ) = 0 . From t h i s i t f o l l o w s t h a t the c l a s s F of a l l r i n g s whose elements a l l have a d d i t i v e P . order a power of p i s a r a d i c a l c l a s s . C l e a r l y t h i s c l a s s i s h e r e d i t a r y . :, Theorem 4 . 3 . 2 I f K i s . a r i n g f o r which B ( K ) 4 6 , then K c o n t a i n s an atom. Proof: I f the a d d i t i v e group o f B ( K ) has no elements o f f i n i t e order, Lemma 4 . 3 . 1 says t h a t B ( K ) i s i t s e l f an atom. I f B ( K ) does have elements o f f i n i t e order, i t f o l l o w s from-standard arguments o f a b e l i a n group theory t h a t T = F ( B ( K ) ) / 0 . . . I f o r some prime p. Now T = F (B(K)) ="'F (K) fl B(K), and t h i s i s an i d e a l o f K . Also,- T i s i n B, and so, by Lemma 4 . 1 . 4 , -T has a non-zero a c c e s s i b l e s u b r i n g T' which i s a homomorphic image o f C . Since T a l s o i s i n F , so i s T', and so-, T' = C n f o r some n. But then T' P P has an i d e a l T" which i s isomorphic t o C , and T" i s a l s o an a c c e s s i b l e s u b r i n g o f K . By C o r o l l a r y 4 . 1 . 6 , <T">V e H-„•= H p and so H r (K) / 0 . By Theorem 4 . 1 . 8 , K T C p •• °P " . 114. and the f a c t that C i s a simple r i n g , HL, (K) i s an P •, P atom of K. Q.E.D. Theorem 4.5.5 Let I-''be a minimal i d e a l of a-ring K. Then T contains an atom. 2 Proof: I f I i s a minimal id e a l of K, either I = . I or 2 2 " - '-' ' I =0. I f 1 = 1 , i t i s well known that I i s a simple ring. Then, by Theorem 4.1.8, 1 = H_(K) ' i s an atom. 2' "' If I =0, consider the additive group of I. I f this has elements of f i n i t e order, then, for some prime p, I n F p(K) ft 0. By the minimality of I, I c F (K). • The same arguments as' i n the proof of the previous theorem guar-antees that- I has an accessible subring T" isomorphic to C . We then obtain H T„ c Hj., and( so T p_ H T / /(K) = H c (K) ft 0. By Theorem-4.1.8, H^ (K),. since i t Is--not zero, P ' ' P ": 2 i s an atom. On the other hand, i f I =0, and i f I has no elements of f i n i t e additive order, choose x ft 0 from I. Then the additive subgroup of I generated by x i s an ide a l of I, and i s isomorphic to Ceo. Therefore- B = H^ c Hj / . CO and B(K) <= H-j-(K) c " l . By Theorem. 4.5.2,' this contains an atom. Q.E.D. Remark I t i s not necessarily true that T i s i t s e l f an atom. 1 1 5 . For example, l e t F be the f i e l d of r a t i o n a l numbers, and p 2 A = Fx+Fy, where x = .0 , xy .= x = yx, y = y. I f we s e t K = A&C , Fx ' is' an i d e a l of K. . In f a c t , Fx i s a minimal P i d e a l of K. For i f L Us an ide ;al, of K contained i n A, L i s an i d e a l of A, and, s i n c e ' A has a u n i t y , L must be a subalgebra of A. Since Fx i s of dimension 1 over F, Fx i s a minimal subalgebra of A, and hence i s a minimal i d e a l of K. Now Fx c o n t a i n s a copy of C as an i d e a l , and so. 00 i t f o l l o w s t h a t .H p x = B, and Fx = B(K) = Fx ® C . T h i s i s not an atom, f o r i t p r o p e r l y c o n t a i n s = F P ( K ) . . . 4 . 4 Rings with Chain C o n d i t i o n s on Closed I d e a l s In t h i s s e c t i o n we s h a l l c o n s i d e r r i n g s which..have e i t h e r or both of the ascending and descending c h a i n c o n d i t i o n s on c l o s e d i d e a l s . Since the l a t t i c e of c l o s e d i d e a l s i s not modular, we have no reason to expect a r e s u l t analogous to the Jordan-Hoelder Theorem. On the other hand, a c l o s e d i d e a l C of a r i n g must c o n t a i n a l l homomorphic images of . C which are i d e a l s of • K, 'so d i s t i n c t c l o s e d i d e a l s must d i f f e r i n some, d e f i n i t e way from one another. In the presence of,' say,-both the ascending and descending c h a i n c o n d i t i o n s f o r c l o s e d i d e a l s , we might hope t h a t there would not be too many d i f f -e r e n t c l o s e d i d e a l s . We s h a l l see t h a t t h i s i s the case. Indeed, we. s h a l l prove t h a t the two c h a i n c o n d i t i o n s are necessary and s u f f i c i e n t c o n d i t i o n s to guarantee t h a t there are o n l y a f i n i t e number of c l o s e d i d e a l s i n the ring.. 116. F i r s t of a l l , we show t h a t the ascending c h a i n c o n d i t i o n on c l o s e d i d e a l s i s not a p r o p e r t y which i s pre s e r v e d under homomorphisms. Example 4 . 4 . 1 . We give an example of a r i n g K which has the ascend-i n g c h a i n c o n d i t i o n f o r c l o s e d i d e a l s , hut f o r which there i s a homomorphic image of K which does not have t h i s p r o p e r t y . L e t K = & K , where K... = Z, the r i n g o f i n t e g e r s , n=l n V and l e t A = © p K , where {p.: i=l,2,....} ( i s the sequence n=l 09 of prime numbers. Then A <J K, and K/A = ©• Z. . = B, where 1=1 p i m Z i s the r i n g o f i n t e g e r s module p.. L e t W = © Z : P i 1 m i = l P i m t h i s i s an i d e a l . o f B. For x e Wm, (^II p.^ ) x =0, and the same i s t r u e i f x i s any element o f - a homomorphic image o f an a c c e s s i b l e s u b r i n g of W . On the other hand, i f y e B/W , &> n m m and i f ( II p.)y = 0, i t f o l l o w s t h a t y = 0 ( s i n c e y i = l 1-m has a d d i t i v e order r e l a t i v e l y prime to H p . ) . There f o r e i = l 1 B/W has no non-zero a c c e s s i b l e subrings- which are homomorphic images of a c c e s s i b l e subrings o f -W . From Theorem 4 . 1 . 8 ° ° m we o b t a i n (B/W-) =0, and so W^  i s c l o s e d i n B. Since m 1 1 7 . I\L cj w"^ +^  we have t h a t B = K/A does not have the ascending c h a i n on c l o s e d i d e a l s . We now show t h a t K does have the ascending c h a i n c o n d i t i o n on c l o s e d i d e a l s . L e t C L , i = 1 , 2 , . . . be an ascending c h a i n of c l o s e d i d e a l s of K. For each j there i s an i such that""'0 = C'.K. c C . fl K.. For any i and j • 3 3- 3 c o n s i d e r the i ' t h p r o j e c t i o n TT^ of K onto Z, . and co n s i d e r the s et S = { T T ^ ( C . fl K^) : i and j are p o s i t i v e i n t e g e r s } . Since Z has the ascending c h a i n on a l l i d e a l s , there are i and j such t h a t -uf. \ c . fl K. ) i s a maximal member o u o '.1 > , i . i _ ' of S. We note t h a t , i f i s the embedding of Z i n t o K which takes Z onto- K. , the f a c t ; t h a t C. i s c l o s e d 1 J guarantees t h a t C. => 6 '1 TT . / (C . fl K., ) f o r a l l i , i ' , and - j . Consider n > j Q . I f C^ ^  Cj > there i s an element x i n C which does not belong t o C. . W r i t i n g x = x,+x 0+. . .+x, , where x. e K.,.' we must have x. / C. f o r some i . Novr has a u n i t y e^ ,.... and ' x^ = xe^ e C n > T h e r e f o r e x.. e C 'fl K . , / a n d 7r.(x.) e ir,(C fl K. ). We I n i 5 ; , • , iv. i ' i v n i ' have Q.TT. ("c. 'fl K. ) c C. 'fl K. c C " f i K., 'and so o "o o "o • TT. (C . n K. )- = TT.0 . TT. (C . fl K. ) c IT. (C fl K. ). By the O J ~ ^ i i O j„ I ' — i ^ n i ' • o °o • • o o °o o 118 . maximality of TT. (C. fl K . ) , we must have e q u a l i t y , and -^o J o . xo-thus 7r.(x.) e Tr. (C. f l K . ). But then x. = 0.TT.(X.) e 1 1 1 X i i I I I V I o o o Q.TT. ( C . fl K . ) c C . . T h i s i s a c o n t r a d i c t i o n . T h e r e f o r e 1 V J b V - Jo = C . f o r a l l n > j . T h i s proves t h a t K has 'the n J Q °o • * ascending, c h a i n c o n d i t i o n on c l o s e d i d e a l s . Lemma 4 . 4 . 2 L e t {K Q : aeA} be a f a m i l y of r i n g s indexed by some' s e t A , and suppose' t h a t each i s an i d e a l i n a r i n g T. ( T h i s i s no r e a l l i m i t a t i o n , f o r weicould take T = 6> K . ) Then K = S ( >j H ). aeA a £ A : . aeA a Proof: Since each '• i s h e r e d i t a r y , u H K i s a h e r e d i t a r y a aeA a homomorphically c l o s e d c l a s s , and so S Q ( u Hg ) i s a h e r e d i t a r y aeA a r a d i c a l c l a s s , by Theorem 1 . 4 . 1 . Since each K " e S n ( \j Ev ) , -• aeA a Lemma 4 . 1 . 4 . g i v e s . L K e S n ( u H~, ), -and so H K c aeA u aeA a 2 a aeA ' S Q ( U H ). aeA a - •'••' Now, l e t B be a member of S ( U-Hg ).•' By Theorem ° aeA a" 4 . 1 . 7 ^ every non-zero homomorphic image " B' of B has a non-zero a c c e s s i b l e s u b r i n g .¥' - i n ' u RK , and thus aeA a W'' e Hg for.some' a e A . We apply Theorem 4 . 1 . 7 a g a i n , and a -119. r e c a l l the d e f i n i t i o n o f Hg , and we. can conclude tha t ¥ ' has an a c c e s s i b l e s u b r i n g W" which i s a homomorphic image of an a c c e s s i b l e sub r ing Y of K' . S ince ; . K 0 £ K , a a aeA Y i s an a c c e s s i b l e sub r ing o f £ K . A l s o W" i s an C6£A a c c e s s i b l e sub r ing o f B . ' . Therefore B ' has a non-zero a c c e s s i b l e s u b r i n g ( ¥ " ) which i s \ a non-zero homomorphic Image of an a c c e s s i b l e sub r ing of £ K Q . A g a i n a p p l y i n g . OKA Theorem 4 . 1 . 7 and the d e f i n i t i o n o f H*,,- B e H K . Q . E . D . C o r o l l a r y 4 . 4 . 5 Under the same c o n d i t i o n s as i n the Lemma, . l e t V be a r i n g f o r which g ^ / 0. Then, f o r a€A A some aeA, Hg (V) / 0. a Proof : By the 'theorem, S Q ( u H K )(V) / 0. By Theorem ' aeA a 4 . 1 . 7 , V has an a c c e s s i b l e sub r ing ¥ which i s a non-zero member of U Hg and . ¥ €'. Hg. f o r some . a. By C o r o l l a r y aeA a a. 4 . 1 . 6 , • <¥> y e H g . , and so 0 / <W>R c Hg ( V ) . Q . E . D . C6 CL I f A and B are c lo sed i d e a l s o f a r i n g K , we s h a l l say tha t A covers B - i f A ^ B , but. there no c losed i d e a l s T f o r which A^T^B.:-Lemma 4 .4 .4 Le t K be a r i n g w i t h the ascending cha in 120. c o n d i t i o n f o r c l o sed i d e a l s . Then there are on ly a f i n i t e number o f c l o sed i d e a l s .which cover a .given c lo sed i d e a l . P roof : Le t B be a c lo sed i d e a l of K , and l e t (A : aeA} 3 L a J be the set o f c lo sed i d e a l s which cover. B . By'-the assumed cha in c o n d i t i o n there i s a f i n i t e subse t ; F Q o f A f o r which " H AT i s a maximal member o f {2 K~~:' G i s a f i n i t e subset o f A } . Suppose tha t F Q = {a^, a ^ , . . . , a^}. For each \£A, K^+sJ= 1Ka_ = sJI^K^ , -'and the re fore c 2 - ^ K ^ . This i m p l i e s tha t K / B e H . . F o r some i , by C o r o l l a r y X E . , K i = l a ^ 4 . 4 . 3 , we have ( K / B ) i s non-zero . But then 0 / I H„ ( K . / B ) = K . / B n H„ (K/B) = K. / B fl K„ / R . Th i s l a s t a. * V * ' " i 7 . i i e q u a l i t y i s due to the f a c t tha t B c K , ,and the f a c t t ha t I K i s c l o s e d . Therefore 0' / (K. fl K ) / B . Since K and K are both c l o s e d , so i s t h e i r i n t e r s e c t i o n . " However, a . 3 • s ince both K, and K . cover' B , we must have K, = K . Therefore A = F Q . Q . E . D . Theorem 4.4.5.- A r i n g K has both the ascending and descending 121. cha in c o n d i t i o n s on c lo sed i d e a l s i f and only i f there i s a t most a f i n i t e number of c lo sed i d e a l s . P roof : O b v i o u s l y , i f t he re i s on ly a f i n i t e number o f c l o sed i d e a l s , the cha in c o n d i t i o n s must h o l d . Suppose tha t both cha in c o n d i t i o n s h o l d . We assume there Is an i n f i n i t e number of c l o sed i d e a l s and o b t a i n a c o n t r a d i c t i o n . I f there i s an i n f i n i t e number of c l o sed i d e a l s , then P = { I : I i s a c l o sed i d e a l , and there are i n f i n i t e l y many c lo sed i d e a l s " J . such tha t J p 1} i s a non-vo id s e t . (For the id e a l 0 e P.) By the ascending chain condition, there i s a maximal member T Q of P. Le t L-^, L £ , . . . , L n be the ( f i n i t e ) se t o f c l o sed Idea l s which cover T Q . (The p rev ious Lemma guarantees tha t t h i s i s a f i n i t e s e t . ) Each L . i s riot i n P, so there i s j u s t a f i n i t e set L . . , • i • i > j j = 1,2,...,^ of c l o sed i d e a l s W s a t i s f y i n g W o. L ^ . We • c l a i m tha t any c lo sed i d e a l which p r o p e r l y conta ins T Q must n • N . be long to the f i n i t e se t U - U { L . • . } , and t h i s w i l l 1=1 j = l l j J c o n t r a d i c t . T Q e P. Le t J be a c l o s e d i d e a l of K such tha t J ^ T Q . By the descend ing•cha in c o n d i t i o n f o r c l o sed i d e a l s , there i s a c lo sed I d e a l L o f K m in ima l w i t h respec t to the p roper ty T Q Q L c J , C l e a r l y L covers T Q , and so"' L = L i f o r some i . But then J = L . . f o r some j . Q . E . D . 1 3 J 122. 4 .5 Rings Determined by T h e i r Atoms In module t heo ry , the-: concept o f a comple te ly r e d u c i b l e module i s w e l l e s t a b l i s h e d . (See, f o r example, Jacobson ( 1 3 ) , Chapter t h r e e . ) The analogous concept f o r r i n g s i s a l s o we l l -known. The main r e s u l t i n t h i s d i r e c t i o n i s the f o l l o w -i n g theorem, which i s due to B l a i r ( 5 ) . Theorem 4 . 5 . 1 For a g i v e n r i n g K , the f o l l o w i n g are e q u i v a l e n t : (1) K = 2 S , where each S i s a s imple r i n g and aeA an i d e a l of K. ' (2) K = $ S% where•each S i s a s imple r i n g and y e r Y " Y an i d e a l of K. •(3) For any i d e a l I of K , there i s an i d e a l J such t ha t I $ J = K. Proof : C l e a r l y (2) i m p l i e s ( l ) . ' We show tha t ( l ) i m p l i e s (2 ) . I f we choose a e A , the set {S } i s t r i v i a l l y an independent f a m i l y of i d e a l s , tha t i s . the sum of a l l the i d e a l s i n the se t i s a d i r e c t sum. By Z o r n ' s Lemma there i s a subset T of_ A which i s maximal w i t h r e spec t^ to the p roper ty tha t 2 S y e r Y i s a d i r e c t sum. • For any a e A , i f S Q(& S) = 0, then • • y e r •• " 2 S R i s d i r e c t , c o n t r a d i c t i n g the max ima l i ty o f r .-Peru {a} p , ; ' There fore , f o r any a e A , S^ n ( 9- S ) / 0. Since S is- a y e r . Y 123. s imple r i n g , the i n t e r s e c t i o n i s a l l of S .. There fo re , S c & S , and K = $ 'S . a - yer Y . Y E R Y (2) i m p l i e s ( 3 ) . ' Le t . I be an i d e a l of.. K. I f I = K , there i s no more to be done.. I f I 4 K , \ then , f o r some y £ r , I S^, and .1 P. S = 0,- s ince S^ i s s imp le . Therefore the c o l l e c t i o n of a l l subsets A o f •T f o r which I n ( ©• S ) = 0 i s not empty. Z o r n ' s Lemma can be a p p l i e d to g ive a subset A .of A maximal w i t h t h i s p r o p e r t y . Then the sum I + D, where D = 9- S , i s d i r e c t . I f 6 e A c I $ D 4 K , then gf I + D f o r some'' Y e r . ' 'S ince S^ i s s i m p l e , S^ Pi (I&D) = 0 , and we o b t a i n a c o n t r a d i c t i o n to the max ima l i ty o f A • Therefore I © D = K , as d e s i r e d . (3) i m p l i e s ( l ) . F i r s t of a l l , we show tha t K has minimal i d e a l s which are s imple rings."" Le t . x e K , x 4 0. By Z o r n ' s Lemma, there i s an i d e a l J o f K maximal w i t h r e spec t to not c o n t a i n i n g x . •  We c l a i m tha t J i s a maximal i d e a l o f K. I f . B 4 K , B ^ J , then x e B . There i s an _ i d e a l C of K such tha t B © C = K . Since the l a t t i c e of i d e a l s of K Is modular , J = J + (BflC) = B fl (J+C) . Now, i f J cz- J + C, then x i s i n both B and J+C, -and the re -fore x e J , which i s f a l s e . Therefore J =..J+C, or C c- J . But J c B , and the re fo re C c C n B = 0. Th i s shows tha t 124. B = K , and J i s the re fo re a maximal i d e a l . By c o n d i t i o n (3), we can w r i t e K = J © J ' . I f J ' i s not a s imple ring',--, i t has a .p roper i d e a l L ' , which i s an i d e a l of K , s ince j ' i s a d i r e c t summand of K. We then o b t a i n J c J © L ' c K , c o n t r a d i c t i n g the max ima l i ty of K. Le t S be the sum of a l l the minimal i d e a l s of K which are s imple r i n g s . Then, i f S K , K = S 'I S' , f o r some i d e a l S' / 0. Le t s' / 0 be i n S' , and l e t . J be an i d e a l of K maximal w i t h r e s p e c t to c o n t a i n i n g S and not c o n t a i n i n g s ' . The same proof as above shows! tha t J i s a maximal i d e a l of K. I f K = J ® . J ' , J ' i s then a minimal i d e a l of K , J ' i s a s imple r i n g , and J ' c S. But then J ' c J ' fl J = 0. This shows S = K , a n d ' ( l ) i s s a t i s f i e d . Q .E . C o r o l l a r y 4 . 5 . 2 • I f K s a t i s f i e s the c o n d i t i o n s of the theorem, any i d e a l I of K i s of the' form 0- S. f o r Proof : We have seen tha t K = I & D. A p p l y i n g the arguments used i n the p roof of the p rev ious theorem, we can w r i t e Therefore J ' i s a s imple r i n g and a min imal i d e a l of K. some subset O f A . K = D $ T where T = 9 S 8e 1 Then I = K/D = T = & S We remark a t t h i s p o i n t t h a t , In the p roo f of Theorem 4.5.1, the m o d u l a r i t y o f the l a t t i c e of two-s ided i d e a l s of K was used. We have seen ( i n Example 4.2.4) tha t the l a t t i c e of c lo sed i d e a l s of a r i n g i s not n e c e s s a r i l y a modular l a t t i c e . We s h a l l a t t e m p t ' t o see how much of Theorem 4.5.1 remains t rue when we d i s cus s the l a t t i c e of c losed i d e a l s , r a the r J than the l a t t i c e o f a l l i d e a l s . For the r e s t o f t h i s chap te r , { S a : £A} w i l l denote the f a m i l y of atoms of K , and S X (K) = L S a Theorem 4.5.5 Le t K be a r i n g , and l e t L be a c lo sed i d e a l o f K such t h a t L c S 1 ( K ) . Then there i s a f a m i l y [ S & : 6 e A } o f atoms such tha t S (K) = L © (©• Sj . 6 £ A Proof: . I f L = S ^ ( K ) , l e t /\ =';0. Otherwise , f o r some atom -Sa o f K , Sa <£ L . Then . S fl L ..Is a c lo sed i d e a l of K • . " / p r o p e r l y conta ined i n S , and the re fore S fl L = 0 . The - a a c o l l e c t i o n of a l l subsets T ' o f A f o r which [S^: y £ r } i s an independent f a m i l y , and f o r which L fl ( ©• S ) = 0 , i s -not •vyer Y • empty. Zo rn ' s Lemma can be a p p l i e d to give, a maximal such f a m i l y A . We c l a i m S, (K) = L 9- (©• S77. • ' • 6 6 A ' 0 5 " ~ ' * • I f t h i s e q u a l i t y were not. t r u e , then ...L •$' (Q> S ) must f a i l to" c o n t a i n , and hence must have zero i n t e r s e c t i o n w i t h , 126 . some atom S . But then S fl (L €> (©• S..) ) = 0,' and we 6 £ A 0 • • o b t a i n a c o n t r a d i c t i o n to the max ima l i ty of'- A . Q . E . D . C o r o l l a r y 4 . 5 . 4 I f K i s a r i n g , then (1) the f a m i l y o f a l l atoms I s , a n independent f a m i l y (2) f o r each c lo sed i d e a l L c ' S - ^ K ) , there Is a c lo sed i d e a l L ' o f K of the form & ST such 6£A 6 t ha t S 1 ( K ) = L ® L ' . P roof : We show (2) f i r s t . Under the present ' 'hypotheses, the statement o f the p r ev ious theorem becomes S, (K) = L ©• (©• S ) 6 e * o I f we l e t W = ©• S x we have S n (K) = L & W. We do not 6 e a o know tha t W i s n e c e s s a r i l y c l o s e d . However, i t f o l l o w s from Theorem 4 . 2 . 3 tha t we a l s o have S-^(K) = L ® W. We now e s t a b l i s h ( l ) . B y a p p l y i n g the p rev ious theorem. t a k i n g L - 0 , we o b t a i n ' ' S, (K) = S f o r some subset • v r^ Y r o f A . I t remains only to show tha t r = A . . For any cx 6 A , S c S, (K) = $ S~~ = R V V o ( K ) . I t f o l l o w s by an a p p l i c a t i o n o f C o r o l l a r y 4 . 4 . 3 . t h a t , fo r some yeT, 0 i H S ( S J = S a 0 H S ( K ) = S a n V S l n C e ' 3 a ^ % ^ both a toms , t h i s i m p l i e s tha t S Q = S , and a = y Q . E . D . 127. Remark In the case where K = S ^ ( K ) , part (2) of the i c o r o l l a r y says,'that for.: any closed i d e a l L of K there i s a closed. i d e a l L' of the .form ~9 ST" such that S, (K) = 6 € A K = L 9 L' . In a.n a r b i t r a r y r i n g there may' be more that one closed i d e a l L' ( f o r a given closed, i d e a l L) such that K = L 9 L' . For example, consider the r i n g K of Example 4.2.4. In t h i s r i n g , Z, S . and S' are closed I d e a l s , and K = Z 9 S = Z 9 S' , where S ^ S' . This" r i n g K does not s a t i s f y K = S- L(K). ""It would be i n t e r e s t i n g ' t o -know whether, i f K = S 1 ( K ) , .; one can have L, L' , and L" - c l o s e d , L' p L" , and K = L 9 L' = L & L" . In t h i s example, since S i s an atom, S i s c l e a r l y minimal among the closed i d e a l s 1 / such t h a t K = L 7 9 Z. -The same can be s a i d sometimes i n a more general s i t u a t i o n . Theorem 4 . 5 . 5 I f K i s a r i n g , L a closed i d e a l of K, and i f : [S^: aefi} ,is a f a m i l y of atoms--of K ' such that K = L CO (© S ), then f o r any closed i d e a l 1 / such t h a t K = L&I/ , then L ' o l B -. • In other words, ¥ S~ i s ' a - w e n 1 " wen.10 unique minimal closed 1/ such that K = L ® L' . Proof: Suppose th a t K = L 9 L' . "For each ..io e 0 , S^ c • -L ©• L' , and S e H T ffi,T/. By C o r o l l a r y 4 . 4 . 3 , e i t h e r 128, H T(S ) 4 0, o r • H T,(S ) 4 0.. B u t H T ( S ) = S n HT ( K ) = S fl L = 0, a n d s o 0 4 H T /(S ) = S n L' . S i n c e S i s uu • L ^ a) • uu uu a n a t o m , ' S c 1/ . T h u s © S c 1/ , a n d , s i n c e L' I s c l o s e d , ® S, c 1/ . Q . E . D . . . . T h e o r e m 4 . 5 . 6 T h e f o l l o w i n g a r e e q u i v a l e n t f o r a n y r i n g K : (1) K = S X ( K ) . (2) ( a ) F o r e a c h c l o s e d i d e a l -L t h e r e i s a c l o s e d i d e a l L' s u c h t h a t K = L &• L' . ( b ) E v e r y n o n - z e r o c l o s e d I d e a l c o n t a i n s a n a t o m . (3) ( a ) F o r e a c h c l o s e d i d e a l L / K t h e r e . i s a c l o s e d i d e a l 1/ 4 0" s u c h t h a t L fl 1/ =0 . ( b ) E v e r y c l o s e d i d e a l c o n t a i n s a n a t o m . P r o o f : . (1) i m p l i e s ( 2 ) . T h a t ( l ) i m p l i e s (2) ( a ) f o l l o w s f r o m C o r o l l a r y 4 . 5 . 4 . A l s o , we h a v e s e e n i n t h e s a m e c o r o l l a r y t h a t t h e s e t o f a t o m s i s a n i n d e p e n d e n t f a m i l y o f i d e a l s . S u p p o s e . L 4 0 i s a c l o s e d i d e a l , [ S : a e A } ' i s t h e f a m i l y a o f a t o m s o f K , a n d t h a t K = L © ( &• S„) . S i n c e L 4 0, 6e£ 0 _ . t h e r e i s a n a i n A b u t n o t i n A . B y C o r o l l a r y 4 . 4 . 3 , e i t h e r H r ( . S ) 4 0 o r , ' f o r s o m e 6, H q ( S ) / 0. - I n t h e l a t t e r c a s e , we w o u l d h a v e 0 / H c ( K ) n S = S . n S . S i n c e o c a o a o • , S c a n d S a r e b o t h a t o m s , t h i s , w o u l d i m p l y t h a t S = S„ , 6 a ^ J a s 129. which i s f a l s e . Therefore we have 0 4 H T ( S J = S n H T(K) = L u a i j S a fl L. Since S a i s an atom, t h i s gives Sa c L, as des i r e d . C l e a r l y (2) i m p l i e s ( 3 ) . To show that (3) i m p l i e s ( l ) , suppose that (3) ' h o l d s , but t h a t - K 4 S 1 ( K ) . By ( 3)(a) there i s a closed :ideal L' such t h a t S^(K) fl L' = 0 , and, by ( 3 ) ( b ) , L' contains an atom T. Then T c S 1(K) n T c S 1(K) n L' = 0 , a c o n t r a d i c t i o n . Therefore K = S 1 ( K ) . Q.E.D. Theorem 4.5.7 Let K be a r i n g , and {S Q: aeA) the f a m i l y " of atoms, of • K. Suppose a l s o t h a t K = S 1 ( K ) . Then'the l a t t i c e of closed i d e a l s of K i s modular i f and only i f , , f o r every closed i d e a l L of K' there i s a. subset A l of A such that . L = S> S . In t h i s case the l a t t i c e of closed X € A L . i d e a l s i s d i s t r i b u t i v e . (That i s , i f A,B,, and L are closed i d e a l s , then L fl (A+B) = LOA + LP.B .••) ' -Proof: Suppose f i r s t t h a t the l a t t i c e of closed i d e a l s i s modular, and l e t L be a closed i d e a l of K. By C o r o l l a r y 4.5.4, there Is a subset A of A such that L (~¥~ST) = K. • 6£A A l s o by the same c o r o l l a r y , the f a m i l y of a l l atoms i s inde-pendent. I f we set A-^  = A ~ A then we also, have TT& B7T © ( ©• S ) = ~W~S~ ' = K. From Theorem 4.5.5, we XeAr .6eA aeA a ' 130. o b t a i n &• S c L. I f we denote ©• S • by B, . then the \£A L • XeA L assumed modularity of the l a t t i c e of closed i d e a l s gives 6£A 6 as d e s i r e d . 5 Conversely, suppose that any closed i d e a l L i s of the form ~S S7~ f o r some A T <= A . We note f i r s t t h a t i f <H) and A are subsets of A , then- ( £ S ) + ( £ S ) = £ S •06 © d 6.£A y£AU©Y and ( £ S ) fl ( £ S ) = ".S S B. . . The f i r s t ' , e q u a l i t y i s ' Se © e • 6eA Pe@nA obvious. A l s o .. E S.'g. i s c l e a r l y contained i n both E . S 6o3HA < P 9 £ © and E S 0 , - By our assumption, we may w r i t e 6£A 9 - ' '.'. • . the closed i d e a l E ''Slfl n E S-. as E S f o r r c A . For-ge© 6£A y £r Y each y e r , S- c E S. , and, by C o r o l l a r y 4.4.3,. f o r Y 9e some 9 i n © , we• have 0 4 H 0 (S ) = Sn f] S . Since these S0 Y 9 Y are both atoms, S^ = S Q and' y=-@.' S i m i l a r l y , y' i s i n A. Therefore we have £ S fl < £ S c: £ S . . 9 e © B "• 6eA aeAPi© An immediate consequence of these remarks i s th a t i f L = E. S and L = £ Sg are closed i d e a l s i n K, and \ £ A L aeAr .. i f L c L, , then A t <= A T • — I Li — J-i^ 131. We can now show tha t the l a t t i c e of c lo sed i d e a l s o f K i s d i s t r i b u t i v e . The above remarks show tha t the mapping from subsets o f A to c lo sed i d e a l s , where A maps to £ St , Is a l a t t i c e homomorphism. Our assumption j u s t says 6 6 A 6 tha t t h i s map i s onto the l a t t i c e of c lo sed i d e a l s . . Since the l a t t i c e o f subsets of any set i s a d i s t r i b u t i v e l a t t i c e , and s ince d l s t r i b u t i v i t y i s preserved u n d e r l a t t i c e homomorphims, the r e s u l t f o l l o w s a t once. Q.E.D.' ' I n c o n c l u s i o n , we examine what occurs i f we impose a s t ronger c o n d i t i o n than j u s t K = S-^(K). i Theorem 4 . 5 . 8 Suppose tha t K i s a r i n g , {S a:aeA} the set of atoms o f K, and suppose tha t K = £ S . Then: aeA , (1) the f a m i l y of atoms i s independent, ( 2 ) f o r any c lo sed i d e a l L there i s a unique- c lo sed i d e a l L ' such tha t L © L ' = K (3) f o r each c l o s e d i d e a l L there i s a subset A T o f A such tha t L = £ S. . ( 4 ) the l a t t i c e o f c lo sed i d e a l s of K i s a d i s t r i b u -t i v e s u b l a t t i c e of the l a t t i c e of i d e a l s o f K. Proof : ( l ) i s t rue f o r any r i n g K , by C o r o l l a r y 4 . 5 . 4 . We. now prove - ( 2 ) . I f L "* i s a c lo sed i d e a l o f K , by C o r o l l a r y 4 . 5 . 4 , there i s a subset A of' A such tha t K = L © ( £ S..) • ' 6eA 132. L & ( 2 S ) . For each X S A - ^ A , from C o r o l l a r y 4.4 .3 we o b t a i n t ha t e i t h e r H T ( S . ) / 0 , o r , fo r some 6eA, H 0 (S. ) / 0. X J A . A. As we- have seen b e f o r e , the l a t t e r would imply tha t S f i = S^, and X = 6, which i s f a l s e . There fo re , f o r Xe/wA, 0 ^ H i / S X ^ and, as b e f o r e , We o b t a i n S c L . There fore , K =• © S = K X£A ( © S ) © ( 2 S ) c L © ( 2 S ) c K , and thus K = L © ( © S ) X£A~A X 6£A 6€A 6£A We - c l a i m tha t K = L © (~¥~S7).. For i f L fl ( ® S ) / 0 , 6eA • 6£A ° then 0 / H ^ g ( L ) , and, by C o r o l l a r y 4 . 4 . 3 / ' w e would have 5£A 5 Ho (L) / 0 f o r some" 6eA :. As we have seen p r e v i o u s l y , t h i s 6 would i m p l y S^ c L , which i s f a l s e . Thus we have K = L © I / , where L ' = © S_ i s a c l o s e d i d e a l . I f we a l s o had K = L©L", 56A 5 where L" i s a l s o c l o s e d , then , I / . = K / L = L" . S ince L ' and L" are i somorphic c lo sed i d e a l s , they are the same. Th i s proves ( 2 ) . U s i n g the n o t a t i o n of the p rev ious paragraph, ' i f L i s a c l o sed i d e a l , we have ,-K = ( © S •) © ( © S ) = L © ( © S ) , XSA-A X 6£A 5 • 6£A 5 where L 3 © S '.- I t f o l l o w s immediately tha t L-= © . S . '• X£A~A * : XCA~A X . . • ( I n g e n e r a l , i f K = U © -V = U ' © V , where- U c U ' , then •; U = U ' • ) This proves ( 3 ) . ' Now we show ( 4 ) . -Let L. and L ' be c lo sed i d e a l s -133-of K. In order to show tha t the l a t t i c e o f c lo sed i d e a l s of K i s a sub- la t t ice of the l a t t i c e of a l l i d e a l s of K , we must show tha t L-HL' = L + L ' . We may w r i t e L+T7" = £ S \eT Y by ( 3 ) . Then, f o r each yeT., j - i / ( 2y ^  ^s n o n - z e r ° . There fo re , b y . C o r o l l a r y 4 . 4 . 3 , e i t h e r H ^ S ) or H£, (S--)- i s . n o n - z e r o . Since S i s an atom, t h i s i m p l i e s tha t e i t h e r S c L or Y ; . Y -S c L ' and, i n e i t h e r case , S c L + 1/ . Therefore Y - Y ~ L+L7" c L + 1/ . The l a t t i c e of a l l i d e a l s of K i s a modular l a t t i c e , and so i s any s u b l a t t i c e . Theorem 4 . 5 - 7 can now be a p p l i e d to conclude tha t the l a t t i c e of c l o sed i d e a l s i s indeed d i s t r i b u t i v e . Q . E . D . •' 134. CHAPTER FIVE RINGS WITH CHAIN CONDITIONS ON CHARACTERISTIC IDEALS 5 . 1 C h a r a c t e r i s t i c Idea l s An i d e a l I of a r i n g K i s a c h a r a c t e r i s t i c i d e a l of K i f , 'whenever K i s an i d e a l of a r i n g S, I i s a l s o an i d e a l of S. C h a r a c t e r i s t i c i d e a l s are not uncommon i n a r i n g . C l e a r l y , i n a r ing- K, K and 0 are c h a r a c t e r i s t i c i d e a l s . I f K has a u n i t y element e, whenever K i s an i d e a l i n a r i n g S we can w r i t e S = eSe > ( l - e ) S e +. e S ( l - e ) + ( l - e ) S ( l However, eS and Se are i n K , and so eS = eSe = Se, which g ives e S ( l - e ) = ( l - e ) S e = 0 . A l s o we o b t a i n . eSe = K . -It f o l l o w s then tha t S = K © T, where T i s a two-s ided i d e a l of S, a s , of course , i s K. From t h i s we see tha t any i d e a l o f K i s an i d e a l o f S, which i m p l i e s tha t any i d e a l of X i s a c h a r a c t e r i s t i c i d e a l of K. From the Andrunak i ev i c Lemma ( D i v i n s k y ( 6 ) , page 107) we can see tha t any idempotent i d e a l I • of K i s ' a- charac-t e r i s t i c i d e a l of K. For i f I < K , and I 2 = I , then , whenever K <3 S, the i d e a l <I>3 of S generated by I . 3 3 ° s a t i s f i e s I = I c <I> S £ I , from which we see tha t 3 " I = <I> S i s an i d e a l o f S. The re levance of c h a r a c t e r i s t i c i d e a l s to the study 135-of r a d i c a l s comes from the f a c t t h a t , f o r any r a d i c a l c l a s s R, and f o r any r i n g K , R(K) ) i s a c h a r a c t e r i s t i c i d e a l of K. (See D i v i n s k y ( 6 ) , page 124. ) Another example of a c h a r a c t e r i s t i c i d e a l is- an i d e a l I of a r i n g K which i s such tha t K / I Is a semiprime r i n g . For i f K i s an i d e a l of a r i n g , S, then <I>g c: K , . .and,. by the A n d r u n a k i e v i c Lemma, <l>gJ £ I . S ince K / I i s semiprime, we o b t a i n <I> S £ I* and ttrus I = <I>S* an i d e a l o f S. Therefore I i s a c h a r a c t e r i s t i c i d e a l of K. As a f i n a l example of c h a r a c t e r i s t i c i d e a l s , we note t h a t , i n a semiprime r i n g K , f o r any i d e a l I o f K , I* = {xeK: Ix = 0} i's a two-s ided c h a r a c t e r i s t i c i d e a l of K. F i r s t of a l l , I* i s e a s i l y seen to be a r i g h t i d e a l o f K. I t i s a l e f t i d e a l . o f K because, i f x e I* and keK, then I (kx ) c Ix = 0... Here we have used the f a c t t ha t I i s • a two-s ided i d e a l o f K. . Since .. I I * = 0 , we have (1*1)^ = 0 , which g i v e s , s ince K i s semiprime, 1*1 = 0. I t f o l l o w s tha t I* '= {xeK: x l = Ix = 0} . Suppose now tha t K i s an • i d e a l i n a r i n g S. . . Then. I*S c K , and I ( I*S) = ( I I * ) S = 0 , so I*S c I* . ' A l s o , • SI* c K , and ' ( I S I * ) 2 = 0. Us ing the. semiprimeness of K , we have I SI* = 0 , or SI* c I * . Th i s proves tha t I* i s a c h a r a c t e r i s t i c i d e a l of K. C l e a r l y , i f K i s a semiprime r i n g and i f I i s an i d e a l of K, then .1 c I** . Since t a k i n g a n n i h i l a t o r s reverses i n c l u s i o n s , we have I* 3 However, i f we ••" 136. replace I by I * i n the equation I c I * * we see th a t I * c I * * * , and thus- I*. = I * * * . I t i s an immediate consequence of these remarks t h a t , i n a semiprime r i n g K, e i t h e r of the ascending or descending chain c o n d i t i o n s on i d e a l s of the form I * , where I i s an i d e a l of K, i m p l i e s the other : chain c o n d i t i o n , and a l s o that these c o n d i t i o n s are i m p l i e d by e i t h e r the ascending or descending chain condition'on c h a r a c t e r i s t i c i d e a l s . -An ideal, of the form I * , where I i s an i d e a l of K, w i l l be c a l l e d an a n h i h i l a t o r , i d e a l of K. Theorem 5 -1 .1 Let K be a r i n g , and l e t I .be a character-i s t i c i d e a l of K. I f K has the ascending (reap., descending) chain c o n d i t i o n on c h a r a c t e r i s i t i c i d e a l s , then K/I. a l s o has the ascending (resp. descending) chain c o n d i t i o n on character-i s t i c i d e a l s . Proof: Let C , n = 1 , 2 , . . . be an ascending chain of c h a r a c t e r i s t i c i d e a l s of K/I. I f we l e t T be .the inverse n image 'of C n under the n a t u r a l homomorphism f:K -* K/I, . then T i s an I d e a l of K, and T c T ,_. n ' n — n+1 We cl a i m that T n i s a c h a r a c t e r i s t i c i d e a l of K. For i f K <3 S, since .'I Is c h a r a c t e r i s t i c , I 0 S, and we can form the f a c t o r r i n g S/I. This has K/I as an i d e a l . Als-o, the n a t u r a l homomorphism6 f can be extended to g:S - S/I. Since C = T / I ' i s a c h a r a c t e r i s t i c i d e a l of K/f, 137. C n i s an i d e a l of S/I, and thus T?n = f " 1 ( C n ) = g"* 1(C n) i s an i d e a l i n S. This shows that T i s indeed a character-n i s t i c i d e a l of K. By the assumed chain condition, there Is an integer m such that T = T ,. for a l l i > 0. Therefore C = C ,. m m+i — m m+i fo r a l l i J> 0, and t h i s proves K/I has the ascending chain condition on c h a r a c t e r i s t i c i d e a l . • The descending chain condition case i s proved i n the same way. Q.E.D. 5.2 Some Structure Theorems In t h i s section, some re s u l t s of Andrunakievic (1) are applied to show that c e r t a i n rings can be described as f i n i t e subdirect sums of prime rings. These r e s u l t s of Andrunakievic were also noted by Levy (19). R e c a l l from Chapter one the d e f i n i t i o n s of supernil-potent, SP, and dual r a d i c a l s . I f R i s any supernilpotent r a d i c a l , i t was asserted i n Chapter one that we have R c R g c R^, where R g i s the upper r a d i c a l with respect to the class of prime R-semisimple rin g s , and R^ i s the upper r a d i c a l with respect to the class of a l l subdirectly i r r e -ducible . rings with R-semisimple hearts. The ra d i c a l s R and R^ are, respectively, the smallest SP-radical, and the smallest dual r a d i c a l , containing the r a d i c a l class R. Theorem 5.2.1 ( c . f . Andrunakievic ( l ) , Lemma 16) 138 Let A be a non-zero, i d e a l of a semiprime r i n g K. If K has the ACC on an n i h i l a t o r i d e a l s , then A contains a subring A' which is. a prime r i n g , and which i s an i d e a l of K. Proof: Suppose the statement i s f a l s e . Then A Is not i t s e l f a prime r i n g , and so A has non-zero ideals A^' and such that A^B.^ = 0. Let A][ = <\'>y? > and. B.^ '= <B 1 /> K . Since K i s semiprime, neither A 1 nor B^ i s zero, and, by the Andrunakievic Lemma,, A^ B.^  c A^ /B 1 / =0. Suppose that, f o r a p o s i t i v e integer n we have found A^, , i .= l,2,...,n, such that A 1 + 1 and B i + 1 are non-zero ideals of K contained i n A^ and A^B^ = 0. By our assumption, A n i s not a prime r i n g , and so i t contains' ideals (of A } U and V which are non-zero but which v n' s a t i s f y UV = 0. I f we l e t A n + 1 = <U>K5 and B n + 1 = <V>K5, then these are non-zero ideals of K (since K i s semiprime) > contained i n A . . Also A .B r t l. c UV = 0. Thus the sequences n n+1 n+1 — ^ of A^'s and B^'s, i = l,2,...,n, can be extended by adjoining A n + 1 and B n + 1 » By induction, there are i n f i n i t e sequences A^ and B^, i a- p o s i t i v e integer* such that, fo r each i , A^ +^ and B^ + 1 are non-zero ideals of K contained i n A^ and A i B i = 0 f o r a l l i . Since 3 A i + i > 139. we have A.* c A 1+1 By the ACC on an n i h l l a t o r i d e a l s , there B, i s an m such that A * = A, m+1 — ra+1 = A *, m 3 and therefore B, m 'm+1 0. The semiprimeness of K then gives B m = 0, a contradiction. Q.E. Lemma 5.2.2 ( c f . Andrunakievic ( l ) , Corollary 8) Let R be a supernilpotent r a d i c a l , and K an R-semisimple r i n g with the ACC on annihilator i d e a l s . Denote by the class of prime R-semisimple rings. Then K con-tains an i d e a l B such that B* ^  0, and K/B i s i n C^. Proof: Since R i s a supernilpotent r a d i c a l , K i s also a semiprime r i n g . By the previous theorem, there i s a non-zero i d e a l A of K which i s a prime r i n g . Prom Divinsky (6), Corollary 2 of Theorem 47 (page 125), A i s also R-semisimple, and so A i s i n C^. From Chapter one, ^ i s a sp e c i a l c l a s s , and so K/A* i s i n C^. Since 0 ^ A c A**, we see that B = A* i s an i d e a l of the desired form. Q.E.D. Theorem 5.2.3 ( c f . Andrunakievic ( l ) , Theorem 14) Let R be a supernilpotent r a d i c a l , and l e t K be an R-semisimple r i n g with ACC on ann i h i l a t o r i d e a l s . Then K i s a subdirect sum of a f i n i t e number of prime R-semisimple rings. 140. Proof; Consider £P : a€A} where, f o r each cfcA, P_, i s • • 1 CL CL an i d e a l of K such that P Q* 4 0 and K/P a i s a prime R-semisimple r i n g . By the previous lemma, t h i s i s a non-void family of ideals of K. Let A = D P . We show f i r s t aeA that A = 0. Prom Theorem 5.2.1, i f A / 0, A contains a non-zero i d e a l B of K which i s a prime r i n g . Proceeding as i n the proof of Lemma 5.2.2, we obtain the r e s u l t that B* e [P : O € A ) . CL Suppose that B* = P„ \p 'A* Since B c A, we have B* 3 A*, and hence B* 3 A+A*. ' Taking ann i h i l a t o r s gives B** = P„ * c (A+A*)* c A* n A**. 0 ~ o Then (P„ *) c A*A** = 0, and, since K i s semiprime, a o P * = 0. This contradiction of the d e f i n i t i o n of the P •s ao a shows A = 0. It remains only to show that A i s the i n t e r s e c t i o n o f a f i n i t e number of P a's. F i r s t of a l l , each P„ s a t i s f i e s P„ = P **. For P-3 a a a a i s a prime i d e a l of K, and 0 = P„*P~** <= P . Thus either CL CL "—— CL P Q* c P a (which would imply ( P a * ) 2 = 0, which i s f a l s e ) , or P ** c: P . Since P i s always contained i n P **, the CL CL CL CL equality follows. ihi. We now show that, for any f i n i t e subset F of A , '( n P j * * = n P R. We have n Pft = n (Pp**) = ( 2 Pp*)* 3 : peF p BeF p peF p BeF p SeF p ( fl P Q)**. The l a s t inequality comes from the e a s i l y v e r i f i e d f a c t that ( fl P R ) * p_ 2 ( P Q * ) J and the second equality comes PeF p BeF P from the fac t (also e a s i l y v e r i f i e d ) that, for ideals - A,, n n A „ , . . . . . A „ of K, fl A * = ( 2 A,)*. - Since ( n P R ) * * d ' n 1=1 1 1=1 1 BeF P always contains 0 P R, the desired equality follows. BeF p Since K i s semiprime, the remarks i n §5-1 show that K also has the DCC on ann i h i l a t o r i d e a l s . Also, for any f i n i t e subset F df A , fi P R i s i t s own "double anni-BeF p h i l a t o r " , and i s therefore an anni h i l a t o r i d e a l . By the DCC on an n i h i l a t o r ideals,-- there i s a f i n i t e subset F Q of A such that fi P R i s a minimal member of { n P : G i s a • BeF Q p yeG Y f i n i t e subset of A}. It follows then that, for each X i n A , P. P, ( n Pa) = n P«, or n P R C P . Therefore K • SeF 0 p peF 0 p SeF 0 p K n P r c A = o. SeF Q P We have found a f i n i t e subset {P R: PeF Q}, the i n t e r -section of whose members i s zero. The r i n g K i s therefore a subdirect sum of the f i n i t e number of rings (K/Pft: P£^o^3 and each of these rings' 1 i s a prime R-semisImple r i n g . . Q.E.D. 142 . Theorem 5 . 2 . 4 Let K be a ri n g with the ACC (resp. DCC) on c h a r a c t e r i s t i c i d e a l s , and ; l e t R be any supernilpotent • r a d i c a l class. Then R(K) = R 0(K), and K/R(K) i s a f i n i t e . subdirect sum of prime R-semisimple rings s a t i s f y i n g the same chain condition. Proof: As we have seen i n § 5 . 1 * R(K) i s a c h a r a c t e r i s t i c i d e a l of K, and so, by Theorem 5 . 1 . 1 , K/R(K) also has the ACC (resp. DCC) on c h a r a c t e r i s t i c i d e a l s . I t was also shown . i n § 5 « 1 that either of these chain conditions i n a semiprime r i n g implies "both the ACC and DCC for ann i h i l a t o r i d e a l s . Therefore K/R(K) s a t i s f i e s the hypotheses of Theorem 5 . 2 . 3 , and we can conclude that K/R(K.) i s a subdirect sum of prime R-semisimple rings K^,Kg,...,K n« Each i s a homomorphic .image of, K/R(K), and^ thus of K. Suppose K i = K / I ^ Then each 1^ i s a prime i d e a l of K, and i s thus,a c h a r a c t e r i s t i c i d e a l of K (see § 5 . 1 ) . By'.Theorem 5 - 1 . 1 , K./ has the ACC (resp. D.C.C.) on c h a r a c t e r i s t i c i d e a l s . Since each i s prime and R-semisimple, i t i s Rg.' semisimple. Since, a s " i s well known, for any r a d i c a l class ^ P, a subdirect sum of P-semisimple rings i s P-semisimple, we have that . K/R(K)' is,R-semisimple. Since R(K) C R_(K), t h i s implies R(K) = R_(K). Q.E.D. , . " Lemma 5 . 2 . 5 Let S be a prime r i n g with DCC on character-i s t i c i d e a l s . Then S - i s - a subdirectly i r r e d u c i b l e r i n g with 143. h e a r t H s a t i s f y i n g H = H. Proof: Since S, and t h e r e f o r e S n f o r a l l i n t e g e r s n, i s a c h a r a c t e r i s t i c i d e a l of S, the DCC on c h a r a c t e r i s i t c i d e a l s guarantees t h a t S n = S n + Z f o r some n. Since S i s prime S n / 0. Thus S n i s a non-zero idempotent i d e a l . We saw i n §5.1 t h a t any idempotent i d e a l i s c h a r a c t e r i s t i c . T h e r e f o r e we can apply the DCC on c h a r a c t e r i s i t c ' i d e a l s t o f i n d a minimal non-zero idempotent i d e a l I. Suppose t h a t I' i s any minimal non-zero idempotent i d e a l . Then i n I' i s a c h a r a c t e r i s t i c i d e a l which, i f i t -i s not z e r o , has some power of i t s e l f , say ( i n i ' ) m idempotent. But the m i n i m a l i t y of both I ahd I' must then g i v e I = ( i n i ' ) m =1'. On the other hand, we cannot have ih'I' = 0, f o r t h i s would imply I I ' =0,. c o n t r a d i c t i n g the f a c t t h a t S i s a prime \ r i n g . T h e r e f o r e we conclude t h a t there i s a unique non-zero minimal idempotent i d e a l I. Now - l e t B be any' non-zero i d e a l of S. Then SBS i s a c h a r a c t e r i s t i c i d e a l , and u s i n g "the DCC and the f a c t t h a t S i s prime we see t h a t some power of SBS i s idempotent. ' A g a i n u s i n g the DCC on c h a r a c t e r i s t i c i d e a l s , we can conclude t h a t there i s a minimal idempotent i d e a l of S contained i n SBS, and h e n c e - i n . B.-^  T h e r e f o r e • I c B. S i n c e . I i s an i d e a l of S contained i n every non-zero i d e a l o f - S, S i s indeed s u b d i r e c t l y i r r e d u c i b l e , and I, the h e a r t of S i s idempotent. Q.E.D. ' '• 144. Theorem 5 . 2 . 6 ' L e t K be a r i n g with DCC on c h a r a c t e r i s t i c i d e a l s , and l e t R be a s u p e r n i l p o t e n t r a d i c a l p r o p e r t y . Then R(K) = R (K) and K/R(K) i s a s u b d i r e c t sum of r i n g s . cp. K^, i = l , 2 , . . . , n , where each i s prime, R-semisimple, s u b d i r e c t l y i r r e d u c i b l e , and s a t i s f i e s the DCC on c h a r a c t e r i s t i c ' i d e a l s . Proof: By Theorem 5 . 2 ;4, K/R(K) can be- rep r e s e n t e d as a s u b d i r e c t sum of r i n g s K-^Kg,. .. ,K n, where each i s prime, R-semisimple, and s a t i s f i e s the DCC on c h a r a c t e r i s t i c ideals. By the previous Lemma, each K.^  is subdirectly i r r e d u c i b l e . L e t the h e a r t o f K. be H.. Since H. i s . i i I an i d e a l ' of the R-semisimple r i n g K^, i s a l s o R-semisimple. From the d e f i n i t i o n of R we see t h a t each K. i s a l s o cp i R - semi simple. K/R(K}' i s a s u b d i r e c t sum of R - s e m i s i m p l e " r i n g s , and i s t h e r e f o r e R^-semisimple.- Since R(K) c R (K), we o b t a i n the d e s i r e d e q u a l i t y . Q.E.D. In what f o l l o w s , L w i l l denote the L e v i t z k i . ( l o c a l l y , n i l p o t e n t ) r a d i c a l c l a s s , and B w i l l denote the Baer Lower-r a d i c a l . , Lemma 5 . 2 . 7 In a r i n g K, l e t {N^ ,: aeA} be the f a m i l y of a l l n i l p o t e n t i d e a l s o f K, and l e t T = ,E N . Then • Ct€A T i s a c h a r a c t e r i s t i c i d e a l . -145. • Proof: Suppose K is.-an i d e a l of a r i n g S, and co n s i d e r <N > c, the i d e a l i n S generated by N . T h i s i s an i d e a l of S. contained i n K, and s'o>< i s an i d e a l of K. By the 3 A n d r u n a k i e v i c Lemma, <N'V>C, c N . so <N > Q -. i s n i l p o t e n t , and <N > c c T. • Then <T> Q c S <N> C c T, so T = <T> Q, aeA or T i s an i d e a l o f S. Q.E.D. Theorem 5 . 2 . 8 L e t K be a r i n g with e i t h e r the ACC or the DCC on c h a r a c t e r i s t i c ideals.' : / Then B(K) = T, and t h i s i s n i l p o t e n t . ' ' • • Proof: I t i s w e l l known, t h a t the nilpotence' of T will-, imply t h a t T = B(K). I t i s s u f f i c i e n t , ' t h e r e f o r e , ' to' prove t h a t T i s n i l p o t e n t . . • Suppose f i r s t t h a t K has the DCC on c h a r a c t e r i s t i c i d e a l s . I f T . i s not n i l p o t e n t , some power o f . T, say T n , i s non-zero and idempotent ( s i n c e ' T ...is-' c h a r a c t e r i s t i c ) . Then I = T n s a t i s f i e s I 2 =1? = I. . By the DCC on character-i s t i c i d e a l s , there i s a minimal member of the s e t . o f a l l c h a r a c t e r i s t i c i d e a l s J • f o r which I J I 4 0. L e t t h i s minimal member of J Q . Then, f o r some x i n J Q , I x l 4 0. Now I x l i s a c h a r a c t e r i s t i c . i d e a l of K contained i n J Q , and 1 ( 1 x 1 ) 1 = I 2 x l 2 = 1 x 1 4 0. By the m i n i m a l i t y of J Q , 'j . n • • J 0 = I x l , and we nave x = L y 3_xz i f o r y 1 ,y 2,... , y n , 146. z l - ' * ' ' J , z n i n 1'' S I N C E ^ - T E B(K-) C L ( K ) , the s u b r i n g W of K generated by the y^'s> the z j _ ' s ' a n ^ x i s n i l p o t e n t of index, say . m. Repeated s u b s t i t u t i o n . f o r 'x _m times i n n the r i g h t hand s i d e of x = £ y±xz± &^ves x = 0 , a i = l • . " c o n t r a d i c t i o n of I x l / 0. In the DCC case, t h e r e f o r e , T i s n i l p o t e n t . Consider now the case where K has the ACC on c h a r a c t e r i s t i c i d e a l s . The s e t o f a l l n i l p o t e n t c h a r a c t e r i s t i c i d e a l s i s not empty, f o r i t cont a i n s the i d e a l 0 , and so i t c o n t a i n s a maximal member H. For any n i l p o t e n t i d e a l N of . K, . KNK i s a n i l p o t e n t c h a r a c t e r i s t i c i d e a l , and so i s H + KM.- By the maximality o f ' H, . H = H + KNK,' or KNK c H. Then T"5 c KTK = K( £ N )K c £ KN K c H, so aeA aeA • "5 ' " T , and t h e r e f o r e T, . i s n i l p o t e n t . C o r o l l a r y 5 . 2 . 9 I f K i s a' r i n g w i t h the DCC on cha r a c t e r - , i s t i c i d e a l s , L(K) = B(K), and t h i s i s n i l p o t e n t . Proof: From ,Theorem 5 . 2 . 8 , . B(K) i s n i l p o t e n t . From Theorem 5 . 2 . 6 , B(K) = B (K), and (see D i v i n s k y ( l ) , Theorem 67) -L(K) c 3 c p(K). Th e r e f o r e B(K) = L(K) = B (K), and t h i s i s n i l p o t e n t . Q.E.D. We would l i k e to have a converse to Theorem 5 . 2 . 4 or to-Theorem 5 - 2 . 6 . The problem i n o b t a i n i n g one i s t h a t , i f 147. K i s a r i n g , and A and B are c h a r a c t e r i s t i c i d e a l s . . where A ' D B, . we do not know.whether or not A/B i s a c h a r a c t e r i s t i c i d e a l o f K/B. I f t h i s q u e s t i o n could be answered a f f i r m a t i v e l y , ' the 'techniques nased i n the f o l l o w i n g • theorem could be used t o o b t a i n converses to Theorems 5.2.4 and 5 . 2 . 6 . Theorem 5»2.10 L e t R be a s u p e r n i l p o t e n t r a d i c a l p r o p e r t y . A r i n g K i s R-semisimple and has the ACC (resp . DCC) on a l l . two-sided i d e a l s i f and only i f i t i s a f i n i t e s u b d i r e c t sum of prime R=s8mi@lmple rings with the ACC (resp. DOG) on a l l two-sided i d e a l s . Proof: We s h a l l prove the r e s u l t i n the ACC case. The p r o o f i n the DCC case i s s i m i l a r . Suppose f i r s t t h a t K i s R-semisimple and has the ACC on a l l i d e a l s . . By Theorem 5.2.4, K i s a s u b d i r e c t .sum of-r i n g s K^,...,-K , and each K^ i s prime and R-semisimple. From the p r o p e r t i e s of s u b d i r e c t sums, each i s a homo-morphic image of •K. I t f o l l o w s t h a t a l s o has the ACC o n . a l l i d e a l s . Conversely suppose t h a t K i s a s u b d i r e c t sum of the prime R-semisimple r i n g s K-^Kg,...,K 3 where each Ky has the ACC on a l l i d e a l s . , Then K, b e i n g a s u b d i r e c t sum of R-semisimple r i n g s , Is R-semisimple. ' I t remains o n l y to show t h a t K, has the ACC on a l l 148. i d e a l s . Suppose t h a t I - ^ I g , . . . , ! are i d e a l s of K such n-t h a t 0.1^ = 0, and = K/I^. (Such i d e a l s e x i s t , s i n c e i = l - ' K i s a s u b d i r e c t sum of the K. ' s.) Le t {L^; t a p o s i t i v e integer}, be an ascending c h a i n of i d e a l s o f K. We' can use the ACC i n to f i n d , f o r j = l , 2 , . . . , n , an i n t e g e r N. such that- ( L ^ fl I Q D 1^ P. ... I. = ( L N > + r fl I 0 n . . . fl 1^)+. I j f o r a l l r > 0, where we denote K by I Q . I f we s e t N '= max{N^,N2,...,Nn}, then, f o r a l l r _> 0, we. have ^ . L N = L N + r + 11 -(2) L N n I-L + i 2 = L N + R n I-L +. i 2 (n) L N n i x n i 2 ' n . . . n i n - 1 + i n = L N + R n i x n . . . ' n Zn-1 + V /. We show th a t L ^ '= L N + R f o r a l l r, •.._> 0. For any x ; i n L.Tl„ there i s an \x-, i n L , T ' such t h a t s n = x - x, e I n N+r 1 N . 1 - / 1 1 •Then € L ^ + r H I-^ and there i s an x 2 i n L ^ fl - I-^ such t h a t s 2 = - x 2 € "^2* ^ s o > . x^ "' and x 2 , ' and hence S 2 a r e i n , LN+r n V ^ n d S O S2.. e LN+r' n h n. ^ : - -1 4 9 . C o n t i n u i n g i n t h i s manner,, we f i n d , f o r j = l , 2 , . . . , n elements x. E . L N n ^ n. . . . n 1 ^ and B. = s.^ - X j. e L ^ R n i x n ip n . . . n i . . ' We have . ' ' ' ' ' s l = x ~ x l S 2 = s l ~ X 2 -S 3 ~ S 2 ~ X 3 S T — S — X -n - 1 n - 2 n - 1 s n = s n _ 1 - x n , where e L N + r n I x n I 2 ... n I n = 0 and where each x. e. L,,. Then s , = x e L,T. Climbing back j N n - 1 n N ° . up, we see t h a t s n _ ] _ j s n 2'* * * >s2> a n c ^ f i n a l l y s^ are i n L^. But then x = s^ + x^ i s a l s o i n . L^. T h i s shows t h a t L , T L c L,T f o r a l l r > 0 . Since the L.'s are an a s c e n d i n g N+r — N — I 0 1 sequence, we have = ' L N + r - f o r a H r 2. 0 * - T h i s proves t h a t K has the ACC o n , a l l i d e a l s . Q.E.D. 5.5 F i n i t e S u b d i r e c t Sums of Rings J -..Suppose t h a t a r i n g K i s a s u b d i r e c t sum of• a f a m i l y of r i n g s {K : a€A}. More p r e c i s e l y , suppose t h a t K has a f a m i l y of i d e a l s ct£A} such t h a t , f o r each a, there 150 . i s an isomorphism 0 : K/I - K , and n I = 0, and aeA denote by i the monomorphism of K i n t o J T K , where — CL i ( x ) = (8 (x+I ) ) . F o l l o w i n g Levy ( 1 9 ) , we s h a l l say t h a t the s u b d i r e c t embedding i i s i r r e d u n d a n t i f each K i s . PI T V o A l s o necessary i n the sense t h a t , f o r each B S A , a € /^j-g} ' * we w i l l say t h a t K i s an i r r e d u n d a n t s u b d i r e c t sum of the V s - . Not a l l s u b d i r e c t sums are i r r e d u n d a n t . For example, i f K i s the r i n g of even i n t e g e r s , then K can be r e p r e s e n t e d as a s u b d i r e c t sum of the n i l p o t e n t r i n g s 2Z/2 nZ, n = 1,2,..., (where Z i s the r i n g of i n t e g e r s ) , and any f i n i t e ' n u m b e r of the r i n g s 2Z/2 nZ may be omitted. On the other hand, i f K i s a s u b d i r e c t sum of a f i n i t e number of r i n g s . K-^Kg,... ,K n, then K may be r e p r e -sented as an i r r e d u n d a n t s u b d i r e c t sum of some of the K^'s. For suppose the. corresponding - i d e a l s of K (such t h a t n K/I. = K.) are K. , i = l , 2 , . . . , n . . Then D I. = 0 , arid we 0 - 3. i i = l 1 • can f i n d a subset F of {l,2,...,n} minimal with r e s p e c t t o h a v i n g the p r o p e r t y t h a t 0. 1 = 0 , . , Then K i s an Y'GF y i r r e d u n d a n t s u b d i r e c t sum of {K^r Y e F ) • The next o b s e r v a t i o n i s due to Levy ( 1 9 ) « 151. Lemma 5.5.1 L e t i : K--» .TT K A be a s u b d i r e c t embedding. — 6 e A p For each a we i d e n t i f y K with the i d e a l { ( x Q ) Q • xQ=0 J a p p£A . p 6 f o r 3 / a} of • TT K Q* Then i i s an i r r e d u n d a n t s u b d i r e c t S e A P embedding i f and only i f , f o r each aeA., i ( K ) n K i s a non-zero i d e a l of TTo K Q . - ^ p S A p .^ Proof:. F i r s t o f ' a l l , i ( K ) n K ' i s an i d e a l of ' JT K • aeA For suppose x = - ( x ) , c ' e i ( K ) fl K . Then x. = 0 f o r a l l . ^ 6 o €A s a o 6 / a. Now suppose t h a t ( y * ) 6 e i s a n element of J J K . A aeA From the p r o p e r t i e s of s u b d i r e c t embeddings, there i s an element a i n K such t h a t the a'.th component o f i ( a ) ' i s y . ,'Then we see t h a t xy '= ( x A y . )- and t h a t x y = 0 . Cc 0 0 0 /\ 0 0 f o r 6 4 a. Since x e i ( K ) , l e t x = i ( b ) f o r some b e K. Simple computation shows t h a t .. xy = i ( b ) y = i ( b ) i ( a ) = i ( b a ) , and t h a t t h i s i s i n i ( K ) fl K . Ther e f o r e •(• i ( K ) n O J T K R c i ( K ) H K , and i ( K ) n K . i s a r i g h t i d e a l o f JT- Kft. In a s i m i l a r f a s h i o n , i ( K ) fl K Q i s a l e f t i d e a l . -B'eA ' -Suppose now that, the map i i s an i r r e d u n d a n t s u b d i r e c t embedding. Then, f o r each a, there i s a non-zero element a i n fi I R . Simple computation shows t h a t 0 / i ( a ) e i ( K ) fl Conversely, suppose t h a t , f o r each a, i ( K ) D K 4 0. 152. Then, f o r a g i v e n a, i f a i n K i s such t h a t 0 / i ( a ) e.-i ( K ) fl K , i t i s e a s i l y seen that' a e n I A . Q.E.D. A P E A P We have seen, i n Theorem 5 - 2 . 6 , t h a t , under c e r t a i n conditions,'we can express a r i n g K as a f i n i t e s u b d i r e c t sum of s u b d i r e c t l y i r r e d u c i b l e • r i n g s . As noted'above, K can then be expressed as an irred u n d a n t s u b d i r e c t sum of some of these r i n g s . Under these circumstances,, the f o l l o w i n g r e s u l t h o l d s . , n ; Lemma 5 . 5 . 2 L e t I: K -• © K. = s • be an irr e d u n d a n t 1=1 s u b d i r e c t embedding, and -^ L-et each be s u b d i r e c t l y . i r r e -d u c i b l e with h e a r t H. . n n : " Then we have H = © H. c © ( i ( K ) fl ;K. ) G I(K) C S, ' 1=1 1 i = l x i n -and H and © ( i ( K ) fl'K. ) are i d e a l s i n S, and hence i n 1=1 1 i ( K ) . Proof: By the pr e v i o u s lemma, f o r each i , i ( K ) fl i s a non-zero i d e a l o f S. I t i s t h e r e f o r e a non-zero i d e a l of K^, and t h e r e f o r e .contains H\. The r e s t f o l l o w s immediately. Q.E.D. ' . ' -Under the c o n d i t i o n s d e s c r i b e d i n the Lemma, i t i s n a t u r a l t o ask whether more can be s a i d about i'(K). For 153. example, i s i ( K ) an I d e a l of S, or i s . .... n i ( K ) = © ( i ( K ) fl K. ) ? The fo l lowing-example shows tha t i = l .. • ' . 1 • . • • n e i t h e r o f these statements need be t r u e . Example 5>3.5 Let L be any r i n g of c h a r a c t e r i s t i c p , and l e t L ' be .the r i n g { ( x , n ) : ' x e L , n e Z^ (the i n t ege r s modulo p) } wi th , a d d i t i o n def ined componentwise and m u l t i p l i c a t i o n def ined by ( x , n ) ( x ' , n ' ) = (xx ' H- 'n 'x + n x ' , n n ' ) . ( c f . Example 4 .2.4) Th is i s ' a r i n g w i t h u n i t y element (0,1) and which has a copy o f L as an i d e a l . Le t S be a s imple n o n - t r i v i a l Jacobson r a d i c a l r i n g o f c h a r a c t e r i s t i c p (see Example 4 . 2 . 4 ) , and form the r i n g T = S 1 / © S 2' , where S 1 =; Sg = S. A l s o , l e t ¥ = (S^ © S g ) ' . There i s a r i n g monomorphism i : ¥ - T, def ined . by i [ ( ( s 1 , ' s 2 ) , n ) ] = ( (s-^n) ^ ( s ^ n ) ) •.- Th i s map i , ' ' composed w i t h the p r o j e c t i o n s of T onto S^ 7 and S^ 7 g ives mappings from ¥ onto S^ 7 and S 2 ' , so i t f o l l o w s tha t i - Is a ' s u b d i r e c t embedding. • I t i s e a s i l y v e r i f i e d tha t S' i s s u b d i r e c t l y i r r e d u c -i b l e w i t h hear t S, and a l s o tha t S]_ © S 2 = ( ! ( ¥ ' ) n S 1 / ) © (!(¥') n S 2 ' ) p i ( ¥ ' ) . Since S ± = i ( ¥ ' ) n S f , and s i m i l a r l y f o r S c , Lemma 5 .3 .1 assures us tha t i i s an i r redundant 154. s u b d i r e c t embedding. A l s o , i ( W ) i s not an i d e a l of T, f o r , i f i t were, we would have i ( W ' ) must conta in-( ( 0 , 1 ) , (0 ,1 ) );( ( 0 , l ) , ( 0 , 0 ) . ) = ( ( 0 , 1 ) , ( 0 , 0 ) ) :. and t h i s i s not t r u e . ... .. -, 5.4 H e r e d i t a r y R a d i c a l s of F i n i t e Subd i rec t Sums of Rings I f H i s a r h e r e d i t a r y r a d i c a l c l a s s , a n d , i f I i s an i d e a l o f a r i n g S, then H ( l ) = I fl H(S) ( D i v i n s k y ( 6 ) , page 125) . Th is i s not i n genera l t rue i f I i s j u s t a sub r ing of S, not even I f I i s s u b d i r e c t l y embedded In to S.. For example, we have seen tha t the r i n g o f ,even i n t e g e r s , which i s J -semis imple (where J i s the Jacobson r a d i c a l ) i s a s u b d i r e c t sum of n i l p o t e n t r i n g s . I t i s e a s i l y v e r i f i e d t h a t , i f each K a i s a r i g h t q u a s i - r e g u l a r r i n g , then TT K i s a l s o r i g h t q u a s i - r e g u l a r . The even i n t e g e r s , t h e r e f o r e , can be s u b d i r e c t l y embedded i n t o TT (2Z/2 nZ) , - -, ... ' . n=I which i s Jacobson r a d i c a l . In t h i s s e c t i o n , we s h a l l see that, the s i t u a t i o n i s -d i f f e r e n t f o r f i n i t e s u b d i r e c t sums.. We s h a l l prove t h a t , n i f i : K -• © K . =. & •-, i s a s u b d i r e c t embedding, then , f o r j = l J : " - , every h e r e d i t a r y r a d i c a l H , H ( i ( K ) ) = i ( K ) fl H ( S ) . Note t h a t , i n Example 5 . 3 . 3 , we saw tha t i ( K ) heed not be an i d e a l of S. • . 155. n Lemma 5 .4 .1 I f S' = ©• K., and i f H i s any h e r e d i t a r y , J - l 3 -n r a d i c a l , then H(S) =' & H(K . ) . j = l J-Proof: Since K <J S f o r each j , H(K.) <I S, s i n c e H(K.) i s a c h a r a c t e r i s t i c i d e a l of K.. There f o r e we have 3 • " n n $ H(K.) c H(S). Now l e t x = E x . be i n . H(S), where j = l J . ~ j = l 3 x. e K. f o r each j , and c o n s i d e r the p r o j e c t i o n ' p.: S -* K.. 3 3 cJ .0 Under t h i s map we have x. = p .(x) e P-(H(S)) c H(K.). 3 3 3 3 n T h e r e f o r e .x e 2 H(K .). Q.E.D. j = l J n Lemma 5 . 4 . 2 L e t i : K -• S = ©• K. be a s u b d i r e c t embedding. . 0 = 1 J > Then i( H (K)) c i ( K ) . n H(S). Proof: L e t p. be the p r o j e c t i o n of S onto K.. From the 3 3 p r o p e r t i e s of s u b d i r e c t sums,- the composition p . i maps K 3 onto K. f o r each j , and so p.I(H(K)) i s an i d e a l of J 3 . _ K •, and i s t h e r e f o r e contained i n H(K'.). Then, f o r y e H(K), n '• n we have i ( y ) = 2 P-i(V) e 2 H(K.) = H(S). Since i ( H ( K ) ) c j = l J j = l '" i ( K ) , we have i ( H ( K ) ) c H(S) P, i ( K ) . Q.E.D. Lemma .5 .4 .5 Using the same n o t a t i o n as above, H(S) fl i("K) 156. i s a s u b d i r e c t sum of the H - r a d i c a l r i n g s p.(.H(S) fl i ( K ) ) , j == l j 2 j . . . • j n . Proof : H(S) fl i ( X ) i s an i d e a l of i ( K ) , so p . (H(S) n i ( K ) ) J i s an i d e a l of p . ( i ( K ) ) = K . . A l s o , p . (H(S) fl i ( 'K)) c J 3 , 3 p . ( H ( S ) ) = H ( K . ) * where t h i s l a s t e q u a l i t y f o l l o w s from 3 .3 Lemma 5 . 4 . 1 . Therefore p . (H(S) fl i ( K ) ) ^ i s indeed an H -r a d i c a l r i n g . ' ; ;./ •. n For any x i n H(S) fl i ( K ) , we can w r i t e x = Z p . ( x ) , J - l . 3 n and so we have H(S) fl i ( K ) c 9- p . (H(S) fl i ( K ) ) . I t i s " J=l 3 c l e a r t ha t the i n c l u s i o n mapping i s - a s u b d i r e c t embedding. Q . E . D . •Lemma 5 . 4 . 4 Le t P be a s u b d i r e c t sum of r i n g s P - ^ P g , . . . and suppose tha t each P . i s H - r a d i c a l , where H i s a h e r e d i t a r y r a d i c a l . Then P i s a l s o an H - r a d i c a l r i n g . P roof : We know tha t .there are i d e a l s 1^, I g , . . . , I n such tha t n . P . - P / I . , and fl I . = 0. We proceed by i n d u c t i o n on n. 3 . J ' j"=l 3 . When n = 1 , the r e s u l t i s t r i v i a l l y t r u e . Suppose tha t we know the r e s u l t i s t rue f o r n = k - 1 , and "consider n the case where n = k . I f we l e t ' T, = 1 - , , and : T0 = fl I . , "-• 3=2 J i t f o l l o w s tha t P / T 1 = P 1 , so P / T 1 i s H - r a d i c a l . . Also_ 1 5 7 . i t i s e a s i l y seen t h a t P/T 2 i s a s u b d i r e c t sum of the r i n g s ( P / T p ) / ( I , / T p ) ~ P/I-, j = 2,3,...,n, and, by our i n d u c t i o n assumption, P/T 2 d s . ^ H - r a d i c a l . A l s o T 1 f l Tg = 0 , so P i s a s u b d i r e c t sum of the r i n g s P/T^ and • P/Tg. I t i s t h e r e f o r e s u f f i c i e n t t o prove the r e s u l t when k = 2 . Since ^ f l T g •= G, T± = ( T 1 ) / ( T i n T 2 ) % (T 1+T 2 )A 2 '<> p / T 2 * Since. P/T 2 i s H - r a d i c a l , and s i n c e H i s h e r e d i t a r y , i s t h e r e f o r e an H - r a d i c a l ' r i n g . T h e r e f o r e c H(P), which implies H(P/T 1) = H(F)/T^. But H(P/T 1) = P/T^ and i t f o l l o w s t h a t H(P) = P, and P i s H - r a d i c a l . Q.E.D. v' n Theorem 5 .4.5 L e t i : K'~» S = © K. be -a s u b d i r e c t embedding. Then, f o r H a h e r e d i t a r y r a d i c a l , H ( i ( K ) ) = 1(K) f l H(S.). Proof: Since i i s a'monomorphism, we have H(i.(K))'= i ( H ( K ) ) , and t h i s i s contained i n i ( K ) f l H(S) by Lemma 5.4 . 2 . Also,. i ( K ) f l H(S). i s . a n i d e a l of i ( K ) , and, by Lemmas 5 - 4 . 3 and 5 . 4 . 4 , I(K) fl H(S) : i s an H - r a d i c a l r i n g . , There f o r e i ( K ) '0 H(S) c H ( i ( K ) ) . Q.E.D. n Remark: I t i s not n e c e s s a r i l y true t h a t • H( i ( K ) ) = © (H(K . ) f l i ( K ) ) . In Example 5 . 3 - 3 , l e t H be the r a d i c a l F . (See § 4 . 3 ) . Then F (i(W')) = i(W'), w hile (F p(S ]_') fl i(W')) © ( F p ( S 2 ' ) fi i(W')) = Q- s 2 , and .these are not equal. 158. 5.5 F i n i t e S u b d i r e c t Sums and S t r u c t u r e Spaces Let P be a p r o p e r t y of r i n g s such t h a t i f S has P, and i f S' i s isomorphic to S, then S' a l s o has P. For any r i n g K, l e t F(K:P) be- {I: I i s a prime i d e a l of I / K, and K/I has P}. For example, i f P i s the p r o p e r t y of being, a r i g h t p r i m i t i v e r i n g , F(K:P) i s the s e t of a l l i d e a l s I .of K f o r which K/P i s a r i g h t p r i m i t i v e r i n g . As i s w e l l known, a t l e a s t in. s p e c i a l cases, i t i s p o s s i b l e t o d e f i n e a topology on F(K:P) by- i n t r o d u c i n g a c l o s u r e o p e r a t i o n i n the following manner; i f F(K:P), -then clftt;) i s d e f i n e d to be {I e F(K:P): I => -WW) > where r]'Ut= fl U. T h i s c l o s u r e o p e r a t i o n s a t i s f i e s : .. CI 1: c l ( 0 ) = 0 CI 2: For \a F(K: P ) , % cz cl(1(). CI 3: For \ c F(K:P), c l cl(\) = cl(\). c l 4: For 'U. and I f contained i n F(K:P), cl(Uu'V) = c l ( U ) U clClf). I t i s a standard r e s u l t of topology t h a t , when we have a c l o s u r e o p e r a t i o n s a t i s f y i n g C l 1 - C l 4, the subsets 1A. which s a t i s f y 1L= - c l ( l \ ) form the c l o s e d s e t s f o r a topology. ( K e l l e y (15), Chapter one) We show t h a t C l 1 - C l 4 are s a t i s f i e d . -By d e f i n i t i o n , f o r  r\\j= 0, we s e t [ J U L ^ With t h i s convention, C l 1 -and' C l 2 are e a s i l y seen. From C l 2, i f ^ c F(K:P), we have c l ( \ ) c: , c l ' / c l ( ^ . Now, i f l e c l cl (1 \ ) , then 159. 1 3 f l = f l { J e.F ( K : P ) : J 3 []%}. I t f o l l o w s that' f l n c l ( U l ) - E z> f r o m which we see I e &(%) and CI 3 i s proved. I f 0^ and are subsets of F(K:P), and i f %cz\fi, then. f l U p. [~M and so cl ( U ) £ cl(0j) • Th e r e f o r e cl ( U ) U dCtf) c cl(^uiy) f o r a 1 1 subsets % . and iT of F(K:P). Now l e t I e cl ( ^ u U ) . Then :I 3 f l (UU*\f) = (FiU) H ( nif) E ( f l U ) ' (TTlT) • Since I' i s a prime i d e a l of K, we have e i t h e r I 3 f j l f and I e c l ( U ) , or I 3 f]U^ and I € cl ( U ) i In e i t h e r case I e cl(70 U c l f l f ) . T h i s e s t a b l i s h e s CI 4. Variou s p r o p e r t i e s of s t r u c t u r e spaces,. !and the r e l a t i o n s between t o p o l o g i c a l p r o p e r t i e s of the s t r u c t u r e space and the r i n g - t h e o r e t i c p r o p e r t i e s o f the r i n g have been i n v e s t i g a t e d by s e v e r a l authors. (See, f o r example, M i c h l e r ( 2 1 ) , and the r e f e r e n c e s c i t e d . t h e r e . ) Our p r e s e n t g o a l i s t o prove the f o l l o w i n g : . n ' . Theorem 5 .5 .1 L e t K .be a r i n g , and l e t i : K •-» S = 9 K. ' J'=l 3 be a s u b d i r e c t embedding. i f P i s a p r o p e r t y of r i n g s which respects' isomorphisms, then the map f : F(S:P) -» F(i(K-):P), where f ( l ) =.I fl i ( K ) , i s a continuous c l o s e d mapping onto F ( i ( K ) : P ) . Furthermore, f o r .1 e F(S:P),' S/I = i ( K ) / f ( l ) . . From these r e s u l t s i t , f o l l o w s t h a t ' F ( i ( K ) : P ) . (and t h e r e f o r e F(K:P) ) has the topology g i v e n by a q u o t i e n t ' t o p o l o g y f o r F(S:P). . 160. ¥e s h a l l break the p r o o f i n t o a number of st e p s . Lemma 5» 5» 2 (1) I f I 4 S i s a prime i d e a l o f S, then I n i ( K ) i s a prime i d e a l of i ( K ) , and - i ( K ) fl. I 4 i ( K ) . (2) S/I ~ i ( K ) / ( i ( K ) fl I ) , under the c o n d i t i o n s of ' CD- . / > ' ' . (3) I f I e F ( S : P ) , then I fl i(K-) e F ( i ( K ) : P ) . n Proof: C l e a r l y (3) " w i l l f o l l o w from (2). I f I / S = & K., :' V- ' J ' = 1 J then, f o r some: j ..we have K. <jt "L- For any - j 4 j we have • ° . . ^o ' ° 0 = K .K. ; c I, so i f I i s a prime i d e a l of • S, K. c: I f o r a l l j / j '. Then we see t h a t we can w r i t e I = (IDK. ) ® J o ( & K^). T h e r e f o r e S = K. + I.""' ']'o The map . p . i : K -» K. - i s onto. K. (where p . i s ' n the p r o j e c t i o n of . S onto K.), so t h a t , f o r :.x = 2 x.; d • ' •: ' J=l J .. i n S, where each x. i s i n K., there i s a k i n K such t h a t p. i ( k ) = x. . Since I p z K., i t f o l l o w s t h a t J o J o ^ j / j Q J x :e i ( K ) + I, and hence S = i ( K ) + I. I t i s a consequence of t h i s t h a t S/I = ( i ( K ) + I)/ I . = i ( K ) / ( i ( K ) fl I ) . From this- isomorphism, both ( l ) and (2) f o l l o w . Q.E.D. 161. Lemma 5 -5 -5 The mapping f:.F(.S:P) - F ( i ( K ) : P ) , where f ( l ) = I fl i ( K ) , i s a continuous mapping. i Proof: By Lemma 5 . 5 - 2 , the map f as d e f i n e d i s indeed a . map i n t o f ( i ( K ) : P ) . To show t h a t i t i s continuous,' i t s u f f i c e s t o show that- the i n v e r s e image of a . c l o s e d s e t i s c l o s e d . L e t C be a c l o s e d subset of F ( i ( K ) : P ) . Then f _ 1 ( C ) = {I e F(S:P): I fj i(K> p_ |lC}. I f I' e c l ( f _ 1 ( C ) ) , then. I' cont a i n s flf" 1(C)V' and I' n i ( K ) p_ ( n f _ 1 ( C ) ) f l i ( K ) ( n i) n I ( K ) = n i n I ( K ) = flc. Thus i'ni(K)3_nc, Ief* 1(C) I e f _ 1 ( C ) or f ( I ' ) = I' n i ( K ) e cl(C) = C. Thus I' e f _ 1 ( C ) , and f - 1 ( C ) i s c l o s e d . Q.E.D. n Lemma 5 . 5 . 4 L e t i : K -» S =' ©• K . be s u b d i r e c t embedding, 3=1 3 and l e t f : F(S:P) -• F ( i ( K ) : P ) be the continuous mapping which sends I t o I n i ( K ) . - For 1(Q) e F ( i ( K ) : P ) , f _ 1 ( Q ) i s a f i n i t e non-empty s e t . In p a r t i c u l a r , f i s a s.urjection. Proof: L e t D., j = l , 2 , . . . , n , be the k e r n e l of the map J p . i : K - . K - . , where p.' i s the j ' t h p r o j e c t i o n of - S onto 3 3 3 . ' K . . I t i s e a s i l y v e r i f i e d t h a t I(D.) = i(K-) f l ( 2 K ) . J J • m/j Suppose, t h a t Q i s a prime i d e a l i n K . Since 162. n . L\.Dp...D c. fl D = 0, • we have t h a t Q o D. f o r l e a s t one ... .. j = l 3 3 j . L e t T = { j : 1 <_ j < n and D. £ Q}. I f Q" e f _ 1 ( i ( Q ) ) , t h e n Q" n i ( K ) = i ( Q ) / i ( K ) , so Q" must f a i l t o c o n t a i n K. f o r some j . We show t h a t j i s i n T. J 0 J o J o Since Q" i s a prime i d e a l of S, and s i n c e K.K = 0 c Q", f o r j / j , i t f o l l o w s t h a t K. c Q", ' and d J Q o 0 . so -£ K c Q". Then f(Q") = i(Q) = Q" n i ( K ) p_ ( E K.) H J V J 0 J J / J 0 J i ( K ) = i ( D ), whence D. c Q and J ,e T. ,Thus, i f J o J o ° Q" e f""'"(i(Q)), there i s a unique g* i n T such t h a t Q" p_ S K ... T h e r e f o r e f _ 1 ( i ( Q ) ) = y { Q" e f _ 1 ( I ( Q)): j y j Q J teT. Q" 3 £ K,}1 " jVt J We know t h a t T i s a f i n i t e non-void s e t . To complete, the p r o o f , we w i l l show t h a t , f o r t i n T, {Q" e f - 1 ( i ( Q ) ) : Q" r> E K .} c o n s i s t s of the s i n g l e member-Q' =.p ti(Q) + E F i r s t of a l l , S/Q' = K t/pM(Q) = jVt. p M ( K ) / p M ( Q ) . Since i ) t = k e r ( p M ) c Q,; we have S/Q' ~ (K/D t)/(Q/D t) =• K/Q. The r e f o r e Q' i s indeed a, prime i d e a l of 'S i f Q M s a prime i d e a l of K, and Q' e F(S:P) . i f • Q e F(K:P). • \ l 6 j . Now we show t h a t Q' i s indeed i n f ~ ( i ( Q ) ) . ... For n . q e Q, we have i ( q ) = £ p . i ( q ) ' e (p.i(Q) + £ K.) n i ( K ) , J=l 3 z jVt J and thus I(Q) c Q' n i ( K ) . L e t x = i ( k ) e 0/ fl i ( K ) . We may w r i t e x = £ p . i ( k ) . Since x e Q' , there' i s a q i n •3=1 3 Q such t h a t p^'i(q) = p ^ i ( k ) . Then, i ( q - k ) = i ( q ) - i ( k ) e i ( K ) n ( E K.) = i ( D . ) c i ( Q ) . Then i ( k ) i s i n ' i ( Q ) , and jVt J . • thus i ( K ) fl Q' c i ( Q ) . We have, now shown t h a t f(Q') = i ( Q ) , as d e s i r e d . At t h i s p o i n t , we have shown that;, f o r any Q. i n F(K:P), f ( i ( Q ) ) i s not empty, and thus f maps onto F ( i ( K ) : P ) . ' .. V -Suppose now t h a t Q" e f _ 1 ( i ( Q ) ) , and Q" z>. £ K.. We s h a l l show Q" = Q' . Now f(Q") = Q" n i ( K ) = i ( Q ) , so -P t i ( Q ) c p^(Q / /) c Q" . ( T h i s l a s t i n e q u a l i t y comes from the f a c t "that Q" => . E K..) From >the d e f i n i t i o n of Q' , we have ~ z-A 3 n .. I f x = 2 x. i s " i n Q", where x. e K., then j = l J J . J . Q" -3 x - E x. = x.. -Since " i i s a s u b d i r e c t embedding, • JVt J t - • r . there i s a k i n K such t h a t p ^ i ( k ) = x^. Then i ( k ) = n £ p , i ( k ) e Q" +. £-K. = Q". Th e r e f o r e i ( k ) € i ( K ) n Q" = J = I . T j y t J i(Q) = i ( K ) fl Q'.c'Q'. - Since Q' ' a l s o c o n t a i n s 2 K., 1 6 4 . we o b t a i n x, as a'member of Q ' . But then x = x, + 2 x . i s a l s o i n 0/ ,. or Q" c Q' . Thus Q" = Q' , as d e s i r e d . Q.E. Lemma 5 . 5 . 5 Under the same assumptions as b e f o r e , f• • i s ' a c lo sed mapping.. P roof : As i n the p rev ious lemma, i f I i s a prime^ i d e a l of S ( I 4 S) there i s a unique j i n { l,2 , . . . n } such tha t I => E K . . I t follows.- tha t we can w r i t e I = ( I f lK. ) + E K . . . . I f we se t F . = [ I e F (S :P ) : I 3 2 K } i t i s easy j V j n . J ' '' " V j m n to v e r i f y tha t F(S':P) = U F . , and tha t t h i s i s a d i s j o i n t 3=1 3 u n i o n . From the d e f i n i t i o n of the c lo su re ope ra t ion i n the t opo logy , each F . i s a c lo sed set i n F ( S : P ) . I f C' i s • 3 ' n. any c lo sed subset of F ( S : P ) , we can w r i t e C = • \j (F. fl C ) , - j = l ' 3 n and f (C) = U f ( F . fl. C). Since C i s c l o s e d , so i s F . n C. 3=1 3 . . . - 3 In order to show tha t f i s a c l o s e d mapping, i t i s . therefore s u f f i c i e n t to show tha t each f ( F . fl C) i s c lo sed i f C is"" c l o s e d . ,./\ Le t ' i ( Q ) e , c i ( f ' (F ,nC) ) ' . Then i ( Q ) p _ fl ( Q"ni ( K ) ) p J Q"eF.DC - j ( E . K m ) fl i ( K ) = i ( D . . ) . From the p rev ious lemma, or r a t h e r V j the p roof of the lemma, we have tha t 1 Q' = P - i ( Q ) . + v ,- i :' m^j . 165. s a t i s f i e s f(Q') = i(Q)., We show th a t 0/ e c l ( F . fl C ) . n Let . x = £ x e n ( F . f l C ) . Since ' i i s a s u b d i r e c t • m=l ° embedding, there i s a k V i n - K' such t h a t p , i ( k ) = x.. . . J • J . Then i ( k ) = x. •+,. E p i ( k ) . Since .each ;Q* i n F. (1 C 3 ' m . •• / 3 contains E K . i t f o l l o w s that x. and then , i ( k ) are m/j m 3 . i n n ( P - n c ) * a n d s o l ( k ) e i(K) n (Tl ( F . n C ) . In other words, f o r Q" In F. fl C , i( ,k) e i ( K ) fl Q" = f ( Q " ) . Therefore 3 i ( k ) e n [ f ( Q ' ) - s Q* e C n F - ) . Since i(Q) e ,c l ( ?(F,P,C)) j 1 j we have i ( k ) i s i n i(Q),. so k e Q. Therefore x . = • 3 • n p i ( k ) e p i ( Q ) , and x = E x e p , i(Q) + E K = Q'. J J m=l 1 3 m/j m This shows. Q' p_ r~I( F- n c); o r Q' e c l ( F . f l C ) = F. fl C . Now i(Q) = f(Q') e f ( F . f) C ) , and so f ( F . D C ) J J , i s closed. Q.E.D. C o r o l l a r y 5*5-6 • The topology on F(K:P) has the topology of a quotient space of F(S:P) induced by the map f. Proof: We have seen t h a t the. map f: F(S:P) -» F ( i ( K ) : P ) i s continuous, c l o s e d , and s u r j e c t i v e . From-Kelley (15) (Theorem 3 - 8 , page 95) F ( i ( K ) : P ) _ has the quotient topology induced by f. Q.E.D. Remark: This completes the proof of Theorem 5 -5 -1 -166. We have seen t h a t the mapping f : F(S:P) - P(K:P) i s a c l o s e d mapping. The f o l l o w i n g example shows t h a t f need not.be an open mapping. Example 5*5.7 We c o n s i d e r the r i n g s W' and- T as d e f i n e d i n Example 5.3.3. R e c a l l t h a t , u s i n g the n o t a t i o n of the example, we had an irredunda.nt s u b d i r e c t embedding i : ( S ^ © S 2 ) ' -• S^' © S 2 ' . L e t P be the p r o p e r t y of b e i n g a prime r i n g . I t i s e a s i l y v e r i f i e d t h a t i and the homomorphism between . F(W:P) and F(i(W):P) . induce a continuous c l o s e d mapping f:F(T:B) - F(W:B) such t h a t f f - S ^ © S'2) = f ( S 1 © Sg' ) = S 1 © S 2, f ( S 1 ' ) = S 1 ? and f ( S 2 ' ) = Sg. The s e t '{'S^, S-^Sg} i s a c l o s e d but not'open s e t i n ,F(W:B), and t h i s i s the image under the mapping f of { S 1 ' , S 1' © Sg} which i s bo t h ' closed, and open i n F(T: 3). . We c o n s i d e r b r i e f l y the case where a r i n g K: i s a f i n i t e subdirect.sum of;prime r i n g s . Then, as we have seen, • • > . • • ' • ' .• n there e x i s t s an i r r e d u n d a n t s u b d i r e c t embedding, i : K -* © K.., . V.. • • i = l 1 where each i s a priyie r i n g . I t f o l l o w s from Levy. ( 19 ) , t h a t i f a r i n g K i s an ir r e d u n d a n t s u b d i r e c t sum- of r i n g s fK : a e A } , : then the r i n g s K are'determined* up to i s o -L a a . . e morphism, by the r i n g K. We c o n s i d e r the opposite q u e s t i o n : 167. i f i : K - © K. i s an i r r e d u n d a n t s u b d i r e c t embedding, to • j = l J ' . , ;,• what extent i s . K determined by K^, Kg,...*K n? A p a r t i a l answer i s found i n the f o l l o w i n g r e s u l t s ; • / " . i i n • Lemma 5-5-8 L e t L - K - S = & -K., where i and i ' " 3=1 3 are r i n g monomorphisms sVeh t h a t i and^ .. i i ' - are s u b d i r e c t embeddings, and. each K. i s a. prime r i n g . Then K = i ' ( L ) 3 ., • i f and only i f , f o r every prime i d e a l Q' of S,;Q' n i ( K ) = Q' n l i ' ( L ) . Proof: C l e a r l y , i f K = i ' ( L ) , the c o n d i t i o n i s s a t i s f i e d . To show t h a t the c o n d i t i o n i s s u f f i c i e n t , i t s u f f i c e s t o show t h a t i ( K ) = i i ' ( L ) , s i n c e i and i i ' are mono-morphisms. Le t p. denote the p r o j e c t i o n of S onto K.. \ 3 3 Then any element i ( k ) of i ( K ) may be w r i t t e n i ( k ) = . n ' - ' . „ ' £ p . i ( k ) . Since i i ' i s a s u b d i r e c t embedding, there i s 3=1 3 an element w i n L such t h a t p ^ i i ' ( w ) = p . ^ i ( k ) , and so h n i ( k - i'(w)) i s i n £ K.. Now Q' = £ K. i s a prime ^ 3=2 J j = 2 J I d e a l of S, and we have i ( k 6 - i ' ( w ) ) e Q' fl i ( K ) = Q' n i i ' (L) T h e r e f o r e there i s an element' - z i n L such t h a t i ( k ' - 'i'(w)) = " i i ' - ( z ) . I t f o l l o w s t h a t k = i'(w) '+ i ' ( z ) = i'(.w+z)e i ' ( L ) . Since k i s a r b i t r a r y i n K, we have K = i ' ( L ) . Q.E.D. 1 6 8 . n Theorem 5.5.9 Let S = © K., where each K. i s prime, j - = l J and l e t i : K -» S and i 7 : K -• S 7 be s u b d i r e c t embeddings. Then i ( K ) = i 7'(K 7.) i f and only i f the f o l l o w i n g are s a t i s f i e d : ( 1 ) For each prime i d e a l . Q' of Sy i ( K ) n Q'- = i ' ( K ' ) ' r, Q'. ' (2) The s u b r i n g i ( K ) (1 i 7 ( K 7 ) , under the n a t u r a l embedding,into S, i s s u b d i r e c t l y embedded i n t o S. Proof: C l e a r l y these two c o n d i t i o n s f o l l o w i f i ( K ) = i 7 ( K 7 ) . Conversely suppose these c o n d i t i o n s .hold. Then, i f we l e t T = i(K).n i ' ( K ' ) , and i f we l e t f : T'-i(K) j and g: i ( K ) S be the obvious embeddings, we can apply the pr e v i o u s lemma and o b t a i n i(K) = f ( T ) = T = i ( K ) n- i ' ( K 7 ) . S i m i l a r l y ,we f i n d ' i ( K ) D i 7 ( K 7 ) = i ' ( K 7 ) . QoE.D. . . . . . ' 169. BIBLIOGRAPHY 1. A n d r u n a k i e v i c , V . A . , R a d i c a l s -of A s s o c i a t i v e R i n g s , Mat. S b . , 44 (1958) , 179-212, t r a n s l a t e d i n A . M . S . T r a n s l a t i o n s , Ser . 2 , 52 (1966) , 95-128. 2. A n d r u n a k i e v i c , V . A . , and R j a k u h i n , J u . M . , Modules and R a d i c a l s , ,Sov. M a t h . , 5 (1964) , 728-732. 3. A n d r u n a k i e v i c , V . A . , and R j a k u h i n , J u . M . , - S p e c i a l Modules • and S p e c i a l R a d i c a l s , ( R u s s i a n ) , D o k l . Akad. Nauk. SSSR, 147(1966), 1274-1277. 4. Bergman, G . , A R ing P r i m i t i v e on the R i g h t but Not on the L e f t , j 5. B l a i r , R . L . , I d e a l L a t t i c e s and 'the S t ruc tu r e of R i n g s , Trans . A . M . S . , 75 (1953) , 136-153. 6. D i v i n s k y , N . , Rings and R a d i c a l s , Toronto: U n i v e r s i t y of Toronto P r e s s , 1965. > 7- F a i t h , C . , Lec tu res on I n j e c t i v e Modules and Quotient R i n g s , . B e r l i n , H e i d e l b e r g , New York: Spr inge r V e r l a g , 1967. 8. G o l d i e , A . , Rings w i t h Maximum C o n d i t i o n , Mul t ig raphed Lec tu re Notes , New Haven: Ya l e U n i v e r s i t y . P r e s s , 1961. 9- G o l d i e , A . , T o r s i o n - f r e e Modules and R i n g s , J . A l g e b r a , 1(1964 ) , 26b-2b7. 10. H e i n i c k e , A , A Note on Lower R a d i c a l Cons t ruc t i ons f o r A s s o c i a t i v e R i n g s , Cdn . .Math . B u l l . , l l ( 1 9 b t f ) , 23-30 . 11. H e n t z e l , I . , A Note on Modules and R a d i c a l s , t o appear i n P roc . A.MTST ^ ~ : ' 12. Hoffman, A . E . and L e a v i t t ' , W . G . , P r o p e r t i e s I n h e r i t e d by  the Lower R a d i c a l , n t o appear. 170. 1 3 . Jacobson, N. , S t ruc tu r e of R i n g s , , A . M . S . C o l l . P u b l . 3 7 , (Rev. ed. ) 1954. ; ". 14. Johnson, P . E . and Levy , L . , Regular Elements i n Semiprime R i n g s , P roc . A . M . S . , 19(1968),. 951-963 . : ~~ 15 . K e l l e y , J . L . , General Topology, P r i n c e t o n : D. Van Nostrand C o . , I n d . , 1955 . ~ 1 6 . Koh, K. and Mewborn, A . C . , A Class of Prime R i n g s , Cdn. Math B u l l . , 9 ( 1 9 6 6 ) , 6 3 - 7 2 . ; : : ~.. 17 . Koh, K . ' a n d Mewborn, A . C . , The Weak R a d i c a l of a R i n g , P r o c . A . M . S . , 1 8 ( 1 9 6 7 ) , 5 5 4 - 5 5 9 . ) •  •".. 18 . Kurosh , A . G . , R a d i c a l s of Rings and A l g e b r a s , ! Mat. S b . , 3 3 ( 1 9 5 3 ) , 13-26. ; : ' ~ . 1 9 . - Levy , L . , Unique Subd i rec t Sums of Prime Rings, . Trans . " A . M . S . , 105 (1953 ) , 64 -76 . 2 0 . M i c h l e r , G.,, R a d i c a l e und S o c k e l , Math. A n n . , 1 6 7 ( 1 9 6 6 ) , 1-48. . . ^ 2 1 . M i c h l e r , G . , R a d i c a l s and S t r u c t u r e Spaces, A l g e b r a 4 ( 1 9 6 6 ) , 1 9 9 - 2 1 9 . 22 . Rjabuhi r i , J u . M . , O v e r n i l p o t e n t and S p e c i a l R a d i c a l s , • S tud ies i n A l g . and Math. A n a l . (Russ. ) , I zda t . " K a r t a .. Moldovenjaske" , K i s h i n e v , ( 1 9 6 5 ) , 6 5 - 7 2 . 2 3 . Sas i ada , and Cohn, P . M . , An Example of a Simple' R a d i c a l R i n g , J . A l g . 5 ( 1 9 6 7 ) , 3 7 3 - 3 7 7 . 24. S u l i n s k i , A . , Anderson, C , and D i v i n s k y , N . , Lower R a d i c a l . P r o p e r t i e s f o r A s s o c i a t i v e and A l t e r n a t i v e R i n g s , J . Lond. Math. S o c , 41(1966) , 417-424. ' 2 5 . S m a l l , . L . , On Some Questions i n Noethes ian R i n g s , B u l l . A . M . S . , 7 2 ( 1 9 6 6 ) , 8 5 3 - 8 5 7 . 

Cite

Citation Scheme:

        

Citations by CSL (citeproc-js)

Usage Statistics

Share

Embed

Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                        
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            src="{[{embed.src}]}"
                            data-item="{[{embed.item}]}"
                            data-collection="{[{embed.collection}]}"
                            data-metadata="{[{embed.showMetadata}]}"
                            data-width="{[{embed.width}]}"
                            async >
                            </script>
                            </div>
                        
                    
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.831.1-0080525/manifest

Comment

Related Items