UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

Rings with a polynomial identity Bridger, Lawrence Ernest 1970

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

Item Metadata

Download

Media
831-UBC_1970_A6_7 B75.pdf [ 6.28MB ]
Metadata
JSON: 831-1.0080479.json
JSON-LD: 831-1.0080479-ld.json
RDF/XML (Pretty): 831-1.0080479-rdf.xml
RDF/JSON: 831-1.0080479-rdf.json
Turtle: 831-1.0080479-turtle.txt
N-Triples: 831-1.0080479-rdf-ntriples.txt
Original Record: 831-1.0080479-source.json
Full Text
831-1.0080479-fulltext.txt
Citation
831-1.0080479.ris

Full Text

RINGS WITH A POLYNOMIAL IDENTITY by LAWRENCE ERNEST BRIDGER Bachelor of Mathematics, University of Waterloo Waterloo, Ontario 1968 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in the Department of MATHEMATICS. We accept this thesis as conforming to the required standard The University of British Columbia April 19 70 . In presenting th i s thes i s in pa r t i a l f u l f i lment o f the requirements fo r an advanced degree at the Un ivers i ty of B r i t i s h Columbia, I agree that the L ibrary sha l l make i t f r ee l y ava i l ab le for reference and study. I fu r ther agree tha permission for extensive copying of th i s thes i s for scho lar ly purposes may be granted by the Head of my Department or by his representat ives. It is understood that copying or pub l i ca t ion of th i s thes i s fo r f i nanc i a l gain sha l l not be allowed without my wr i t ten permission. Department of The Un ivers i ty of B r i t i s h Columbia Vancouver 8, Canada Supervisor: Dr. N.J. Divinsky ABSTRACT. Since Kaplansky's f i r s t paper on the subject of P.I. r i n g s appeared i n 1948, many f r u i t f u l r e s u l t s have a r i s e n from the study of such r i n g s . This t h e s i s attempts to present the most important of these r e s u l t s i n a u n i f i e d theory. Chapter I gives the b a s i c n o t a t i o n , d e f i n i t i o n s , a number of small lemmas together w i t h Kaplansky's i n c i s i v e r e s u l t on p r i m i t i v e P.I. r i n g s . We i n v e s t i g a t e a l s o the Kurosh problem f o r P.I. r i n g s , p r o v i d i n g f o r such r i n g s an a f f i r m a t i v e answer. A rat h e r n i c e u n i v e r s a l property f o r P.I. r i n g s which ensures that a l l P.I. r i n g s s a t i s f y some power of the standard i d e n t i t y i s proved. Chapter I I deals w i t h p a r t i c u l a r types of r i n g s such as r i n g s without zero d i v i s o r s and prime r i n g s and culminates i n a p a i r of p r e t t y r e s u l t s due to Posner and P r o c e s i . We show that prime P.I. r i n g s have a r a t h e r t i g h t s t r u c t u r e theory and i n f a c t the r e s t r i c t i o n s on the unde r l y i n g s et of c o e f f i c i e n t s can i n t h i s case be rela x e d to a very great extent. Chapter I I I i s e x c l u s i v e l y devoted to P.I. r i n g s w i t h i n v o l u -t i o n . Although such r i n g s are r a t h e r s p e c i a l i z e d much has been accomplished i n t h i s d i r e c t i o n i n recent years and many b e a u t i f u l theorems and proofs have been e s t a b l i s h e d , e s p e c i a l l y by Amitsur and Mar t i n d a l e . The source m a t e r i a l f o r chapter IV i s p r i m a r i l y P r o c e s i and i i i . A m i t s u r ' s work on J a c o b s o n r i n g s and H i l b e r t a l g e b r a s . A p p l i c a t i o n t o H i l b e r t ' s N u l l s t e l l e n s a t z and t o t he B u r n s i d e p r o b l e m a r e c o n s i d e r e d . F i n a l l y , c h a p t e r V c o n c e r n s i t s e l f c o m p l e t e l y w i t h g e n e r a l i z a t -i o n s o f t h e p r e c e d i n g f o u r c h a p t e r s . F o r t h e most p a r t t h e s e r e s u l t s do n o t g e n e r a l i z e e n t i r e l y , b u t by r e d u c i n g o u r demand on p o l y n o m i a l i d e n t i t i e s s l i g h t l y , many r e m a r k a b l y f i n e r e s u l t s have been p r o v e d . TABLE OF CONTENTS CHAPTER I §1: Results from General Ring Theory. §2: Notation, Definitions, General Results. §3: The Kurosh Problem. §4: The Identities of P.I. Rings. CHAPTER II §1: A Result of Levitzki and Amitsur. §2: P.I. Rings with No Nilpotent Ideals. §3: P.I. Rings with No Zero Divisors. §4: Prime P.I. Rings. §5: Prime P.I. Rings (Cont'd). § 6 : Two Pretty Results of Posner, Procesi. §7: Nil and Nilpotence in P.I. Rings. CHAPTER III \' §1: Introduction to Rings with Involution. §2: Preliminary Results. / §3: Amitsur's Theorem. § 4: Other Results. §5: The Kurosh Problem. CHAPTER IV § 1: A Paper of Amitsur and Procesi. §2: Extensions of Rings. §3: Geometric Interpretations. §4: The Burnside Problem. CHAPTER V §1: G.P.I. Rings. §2: Primitive G.P.I. Rings: Preliminaries. §3: Primitive G.P.I. Rings. §4: Prime G.P.I. Rings. §5: P.M. Rings. §6: J.P.M. Rings. §7: G.P.M. Rings. BIBLIOGRAPHY vi. ACKNOWLEDGEMENTS I wish to thank my advisor Dr. N.J. Divinsky for his suggestion of the topic and for his time and assistance during the writing of this ; thesis. I would also like to express my thanks to the National Research Council of Canada for their financial aid. \ / Chapter I Section 1: I n . t h i s s e c t i o n we merely l i s t a few r e s u l t s from general r i n g theory which we do not wish to s t a t e or prove i n the main body o f j the t h e s i s . For the most p a r t , these r e s u l t s can be found i n many of the usual books on r i n g theory (see, f o r example [16], [20], [26]). [22], [25], Theorem 1.1: Let A be a p r i m i t i v e r i n g . There e x i s t s a d i v i s i o n 1 ! ' r i n g D such that e i t h e r A i s isomorphic to , o r , given any i n t e g e r m > 0 , there e x i s t s a subring S of A which maps homomorphi m c a l l y onto D . ([ 2 2 ] , p. 43) m Theorem 1.2: Let D be d i v i s i o n r i n g w i t h centre Z and l e t K be a maximal s u b f i e l d of D ; then D ®^ K i s a dense r i n g of l i n e a r t ransformation on D considered as a v e c t o r space over K . ([22]., p. 95). P r o p o s i t i o n 1.3: I f A i s a n i l semi-simple r i n g then A[x] , the r i n g of polynomials i n the commutative indeterminate x over A , i s Jacobson semi-simple. ( [ 4 ] , p.4). Theorem 1.4: Suppose a r i n g A s a t i s f i e s the f o l l o w i n g four c o n d i t i o n s (11) every d i r e c t sum of l e f t i d e a l s contains only a f i n i t e number of non-zero terms. ( l r ) every d i r e c t sum of r i g h t i d e a l s contains only a f i n i t e number of non-zero terms. (2£) A has A.C.C. on l e f t a n n i h i l a t o r i d e a l s . (2r) A has A.C.C. on r i g h t a n n i h i l a t o r i d e a l s . Then A i s r i g h t quotient simple and l e f t quotient simple, i . e . A has a f u l l q u otient r i n g A c o n s i s t i n g of the set of a l l r x r matrices over some d i v i s i o n r i n g D ( r i s some p o s i t i v e i n t e g e r ) . A (not n e c e s s a r i l y commutative) i n t e g r a l domain K i s a r i g h t Ore domain i f K possesses a c l a s s i c a l r i g h t quotient f i e l d K Theorem 1.5: Let a r i n g A be r i g h t quotient simple w i t h A = D Then A . contains a r i g h t Ore domain K w i t h K = D , [18]. . A r i n g i s c a l l e d r i g h t Goldie i f i t s a t i s f i e s c o n d i t i o n s ', ( l r ) , (2r) of theorem 1.4. Then we have j Theorem 1.6: A r i n g A i s a semiprime Goldie, r i n g i f f i t i s a right,, order ( i . e . has as i t s r i g h t quotient r i n g ) i n a Jacobson semi-simple algebra Q w i t h D.C.C. A i s prime i f f Q i s simple. ([22]) By an e s s e n t i a l r i g h t i d e a l of a r i n g A , we s h a l l mean a r i g h t i d e a l I of . A such that I i n t e r s e c t s every non-zero r i g h t j i ' i d e a l of A i n a non-zero f a s h i o n . 1 P r o p o s i t i o n 1.7: I f A i s a semiprime r i g h t Goldie r i n g then; ___ - . (a) •, An e s s e n t i a l r i g h t i d e a l contains a r e g u l a r element. (b) I f c i s a r e g u l a r element, cA i s e s s e n t i a l . (c) I f cx = 0 i m p l i e s x = 0 then xc = 0 i m p l i e s x = 0 F i n a l l y we s t a t e a theorem which i s widely used i n chapterjilvj, namely: Theorem 1.8: A semiprime i d e a l i s the i n t e r s e c t i o n of the prime i d e a l s c o n t a i n i n g i t . S e c t i o n 2 This s e c t i o n w i l l i n troduce the n o t a t i o n , b a s i c d e f i n i t i o n s and a few elementary p r o p e r t i e s of r i n g s s a t i s f y i n g a polynomial i d e n t i t y Let A be an algebra over a f i e l d F and l e t {x^} be a f a m i l y ( p o s s i b l y i n f i n i t e ) of noncommuting indeterminates. Denote by I F[x ,...,x^] the f r e e algebra over the f i e l d F generated by x n,...,x . Let f ( x . , . . . , x ) e F[x n , . . . , x ] . I f , f o r every choice 1 m 1 m 1 m of elements a..,...,a from A we have f ( a 1 , . . . , a ) = 0 , we say ± m 1 m that A s a t i s f i e s a polynomial i d e n t i t y over F . I n s h o r t , A i s a P.I, algebra. More g e n e r a l l y , we might only r e q u i r e that A be an algebra over a commutative r i n g F . In some instances we may even be able to get along f i n e merely by assuming that A i s an ft-algebra f o r some domain of operators Q , w i t h the m u l t i p l i c a t i o n Q x A -> A s a t i s f y i n g i i some a d d i t i o n a l c o n d i t i o n s ( [ 3 1 ] , [11]). U s u a l l y , the instances i n > which we r e q u i r e only these weaker hypotheses w i l l be c l e a r but those which are not w i l l be noted. Now, of course, i f f = 0 every r i n g would s a t i s f y f so that by a P.I. r i n g , we s h a l l mean a . r i n g s a t i s f y i n g a non-zero polynomial i d e n t i t y . D e f i n i t i o n 2.2: A polynomial i d e n t i t y f ( x _ , . . . , x ) = 0 i s of degree 1 m — 2  _d i f there i s at l e a s t one non-zero monomial of f i n v o l v i n g the product of e x a c t l y d of the x_^  ( i n c l u d i n g m u l t i p l i c i t i e s ) and no non-zero monomial i n v o l v i n g more than d of the x. 2 For example., (x^x^x^ + x^x^x^) i s of degree 4. When we say that A i s a P.I. algebra of degree d we s h a l l mean that A s a t i s f i e s a polynomial i d e n t i t y of degree d but A s a t i s f i e s n o i d e n t i t y of degree l e s s than d P.I. algebras may s a t i s f y many complicated types of i d e n t i t i e s . I t would be convenient to know that they s a t i s f y a p a r t i c u l a r l y n i c e form of i d e n t i t y i n order that we may work more c o n c r e t e l y with. them. We.therefore make the f o l l o w i n g D e f i n i t i o n 2.3: A m u l t i l i n e a r polynomial i d e n t i t y i n m indeterminates i s an i d e n t i t y which has the form 1' . m a a ( l ) a(m) where the a e F and the sum runs over a l l permutations a of the m l e t t e r s 1,...,m . The importance of t h i s d e f i n i t i o n becomes evident i n the f o l l o w i n g p r o p o s i t i o n . P r o p o s i t i o n 2.4: I f A s a t i s f i e s a polynomial i d e n t i t y of degree d , then A s a t i s f i e s a m u l t i l i n e a r i d e n t i t y of degree l e s s than or equal to d . Proof: Let A s a t i s f y f ( x . , . . . , x ) = 0 . I f f i s not l i n e a r i n 1 m one of i t s indeterminates, say x^ , we w r i t e g (x , . . . , X . , X , . . ., x ) f (x.. , . . . , X . + X . , . . . j X ) — f (x , ...,x. , . . .,x ) - f(x.., . . . ,X. j ... }x ) . ± i ^ m ± I2 m where x. , x.. are indeterminates r e p l a c i n g x. . Now, g i s s a t i s f i e d by A s i n c e f i s s a t i s f i e d by A , and the degree of i n x. and x. i s one l e s s than the degree of f i n x. . Thus, a f t e r a f i n i t e X2 ' number of steps we can reduce to the case where A s a t i s f i e s an i d e n t i t y h ( y n,...,y ) = 0 which i s l i n e a r i n each of i t s indeterminates. At i n each step, we w i l l have introduced one new indeterminate. We note alsc that the degree of h w i l l be at most d Consider y . I f there are monomials of h not i n v o l v i n g n y we w r i t e n l ( y i ' * " ' , y n ) = h l ( y l ' ' ' " ' y n - l , y n ^ + h 2 ( y l ' ' ' ' ' y n * where, i n h^ , we have c o l l e c t e d a l l terms of h which do not i n v o l v e a y . Then every term of h does i n v o l v e a y . Set y = 0 Jn J 2 n . n 1 and assign ^±'"'''^n-± a r b i t r a r y elements i n A . Since y^ = 0 we see that h^ = 0 . But h i s a l s o zero s i n c e A s a t i s f i e s h Hence, A must a l s o s a t i s f y h^ , and h^ = 0 i s an i d e n t i t y having only (n -1 ) indeterminates. By c o n t i n u i n g t h i s process we reach a • p o i n t where each monomial i n v o l v e s the same number of indeterminates 6. and thus the i d e n t i t y i s i n m u l t i l i n e a r form. Its degree must s t i l l be less than or equal to d . Corollary : I f A i s a P.I', algebra of degree d , then A s a t i s f i e s a m u l t i l i n e a r i d e n t i t y of degree . d . ' The following propositions give conditions under which poly-nomial i d e n t i t i e s are preserved when we form new rings from a given family of rings. Proposition ,2.5: Let {A_^} be a family of algebras over a f i e l d F Then the subdirect sum A = I A. s a t i s f i e s a polynomial i d e n t i t y i f f i each A^  s a t i s f i e s the same polynomial i d e n t i t y . The proof i s c l e a r . Proposition 2.6: Let A be a P.I. algebra over F, K any commutative algebra over F . Then A ® K i s also a P.I. algebra over F r Proof: Since A i s a P.I. algebra i t must s a t i s f y a m u l t i l i n e a r i d e n t i t y , say f (x, x ) = E a x / n N . . . x / N . Let x = a, ® k,,. ; .,x J ' J 1' m a a a ( l ) a(m) 1 1 1 1 m = a ® k where a. ® k. E A ®„ K i m m j j F , Then, • f ( a 1 ® k 1 , . . . , a ® k ) = E a (a N ® k , . . , ) . . . (a , , ® k , N) ' K 1 1' ' m m a a a ( l ) a ( 1 ) a(m) a(m) = E a (a / l N . . . a , , ) ® (k .k , ,) a c r a ( l ) a(m)' a ( l ) cr(m) = E a (a a . , ) ® k since K commutative a a s (1) (m)' ' [§ a a ( a ( l ) " ' - a (m) ) ] ® k , j • . , ' i ! • = 0 ® k = 0- . n Since a t y p i c a l element of A <X>_ K has the form Z a. <2> kt F . n i x i = l i t i s clear that A <X> K s a t i s f i e s f = 0 and hence i s a P.I; algebra. r Proposition 2.6 merely states that any m u l t i l i n e a r i d e n t i t y s a t i s f i e d by A i s also s a t i s f i e d by A <8> K . However, i n the s p e c i a l r case i n which the f i e l d F i s i n f i n i t e we can prove even more with the help of a lemma of Amitsur [5]. Lemma 2.7: Suppose A i s an algebra over an i n f i n i t e f i e l d F , and A s a t i s f i e s f(x.,,...,x ) = 0 , f not n e c e s s a r i l y m u l t i l i n e a r . 1 m n Write f(x,,...,x ) = Z f (x,,...,x ) where f i s homogeneous of 1 m _ r 1 m r r=0 degree r i n x, . Then f (x,,...,x ) i s s a t i s f i e d by A f o r each 1 r i m r = 0 , . . . , n . Proof: Choose n d i f f e r e n t elements a,,...,a from F . Then we • I n n r have f (a.x., , ... ,x ) = Z a.f ( x l 5 . . . , x ) f o r each i = l , . . . , n l 1 m n l r 1 m . r=0 We can find- X. e F so that Z A. f ( a . x l 5 . . . , x ) = Af ( x l 5 . . . , x ) wher l r . l r l 1 m r 1 m x A = IT (a. - a, ) and hence A' e F . Since f i s s a t i s f i e d by A we . 1 K i<k have that Af i s s a t i s f i e d by A f o r each r and thus so i s r A ~ 1(Af ) = f r r I f F i s merely a commutative r i n g , then of course, A X may not e x i s t . But i n t h i s case, one of the conditions that we impose on A at the outset, i s that i f ya = 0 (for y e F, a e A ) then e i t h e r y = 0 or a = 0 . So that even though A may not e x i s t , the facts that Af = 0 and . A f 0 are s u f f i c i e n t to ensure that r 1 f = 0 . This completes the proof. , The promised extension of p r o p o s i t i o n 2.6 i s then P r o p o s i t i o n 2.8: Let -A be an algebra over an i n f i n i t e f i e l d F and l e t C be a commutative algebra over F . Then any polynomial i d e n t i t y s a t i s f i e d by A i s a l s o s a t i s f i e d by A cg^ C Proof: Suppose A s a t i s f i e s f ( x ^ , . . . , x m ) = 0 . W r i t i n g 'f(x^,...> x m) n = E f (x, ,...,x ) , where each f i s homogeneous of degree r i n n r 1 m r r=0 x, , we see by the lemma.that A s a t i s f i e s a l l f = 0 . Then, 1 „ r Z w r i t i n g f = E f (x-,...,x ) , where f i s homogeneous of degree r _ rs 1 m rs s>0 s i n x„ , we see that A s a t i s f i e s f = 0 f o r each r and f o r 2 rs each s . Continuing i n t h i s f a s h i o n we o b t a i n f = f ^ + + ••• + where each f. i s homogeneous i n each x. and A s a t i s f i e s a l l the 1 J ' f. . . l Now, because of the homogeneity of each f. i n each x. . ' i 1 we must have that A <X> C s a t i s f i e s f. f o r a l l i and therefore-F I A ® C s a t i s f i e s f = 0• . r Remark: I f C a l s o has an i d e n t i t y element then we can say that the i d e n t i t i e s s a t i s f i e d by A and A ® C c o i n c i d e . D e f i n i t i o n 2.9: ,The polynomial i d e n t i t y S (x.,...,x ) = S (x) = E ( - l ) S ^ n ° x .\...x , N m 1 m m a a(l) a(m) i s c a l l e d the standard i d e n t i t y of degree m According to the d e f i n i t i o n , i f two of the indeterminates are v the same'then the. i d e n t i t y vanishes. Because of t h i s and the f a c t that any element of an a l g e b r a . i s a l i n e a r combination of b a s i s elements we have the f o l l o w i n g r e s u l t ; P r o p o s i t i o n 2.10: I f A i s an algebra of dimension n over F then A s a t i s f i e s the standard i d e n t i t y of degree (n+1) This immediately gives us a p l e n t i f u l source of P.I. algebras. As a c o r o l l a r y we note that F^ , the set of a l l n x ,n matrices over 2 F , s a t i s f i e s the standard i d e n t i t y of degree n + 1 . We can a l s o show that F^ s a t i s f i e s no polynomial i d e n t i t y of degree l e s s than 2n For, i f i t d i d , then i t would s a t i s f y a m u l t i l i n e a r i d e n t i t y of degree m < 2n .. ' Denote t h i s i d e n t i t y by f ( x , , . . . , x ) = E a x ._....x / N = 0 . , a e F 1' ' m a a a ( l ) a(m) ' a We r e w r i t e t h i s i n the form X _, X • • • X — ZJ- QL X / i \ • • • X / \ 1 2 m i n a a ( l ) o-(m) a f l (we may have to r e l a b e l the x_^  and the ) . Now we make a s p e c i f i c choice f o r the x. from F . Choose x, = e_. , x„ = e ' , x„ = e.., l n 1 11 I LI j 12. x. = e„», ... where e.. .r e f e r s to the n x n matrix w i t h zeroes 4 23' 13 . . ; everywhere except a 1 i n the i j th p l a c e . We can make these choices since, m < 2n . A f t e r s u b s t i t u t i n g , the l e f t s i d e of (1) i s c l e a r l y I 1 j; . non-zero but the r i g h t s i d e i s zero. This c o n t r a d i c t i o n shows that * F^ cannot s a t i s f y a polynomial i d e n t i t y of degree l e s s than 2n A c t u a l l y , Amitsur [1] has proved even more, namely, that F^ i s a P.I. algebra of degree e x a c t l y 2n . This i s the content of Theorem 2.11: For each n , the complete matrix algebra s a t i s f i e s S„ (x) = 0 and moreover i f . A s a t i s f i e s a m u l t i l i n e a r i d e n t i t y of degree 2n -- n ° m <_ 2n ., then m = 2n and. the i d e n t i t y i s , but for a numerical 'factor,': ' the standard i d e n t i t y of degree 2n As a simple consequence of t h i s we have 2 Corollary: If A i s a simple algebra of degree n over. F then A s a t i s f i e s the- standard i d e n t i t y of degree 2n and no i d e n t i t y of degree less than 2n We now show that there i s no i d e n t i t y which holds f o r a l l the matrix algebras F for a l l n n Proposition 2.12: ; If d i s a p o s i t i v e integer and f =} 0 i s i n F[x,,...,x,l then there ex i s t s an integer n such that F does not I d n s a t i s f y f = 0 Proof: Suppose f i s of degree m . Let B be the i d e a l of F[x^,...,x^] generated by a l l monomials i n x^,...,x^ of degree l a r g e r F[x^,•.•,x^] than m . Then C = i s a f i n i t e dimensional algebra over B F . Using the regular representation of C we can regard C as a sub-algebra of F , where n = dim C . But f i M •»• f 4 0 i n C and thus n F 1 there must e x i s t matrices c, c, i n F such that f ( c c ) 4 0 I d n I d D e f i n i t i o n 2.13: An element a e A i s said to be algebraic over F i f i t s a t i s f i e s a monic polynomial i n F[x] , i . e . a polynomial of the form x 1 1 + a-,x11 "*•+...+«a , where each a-, e F > and a = 0 i f A has I n l n no i d e n t i t y element. D e f i n i t i o n 2.14: The algebra A i s a l g e b r a i c over F i f each element of A i s a l g e b r a i c over F . . .It i s a l g e b r a i c of bounded degree over F i f i t i s a l g e b r a i c , and i f the degrees of the minimal monic polynomials s a t i s f i e d by each element of A are bounded by some i n t e g e r m With these d e f i n i t i o n s we have another source of P.I. algebras, namely, a l g e b r a i c algebras of bounded degree. P r o p o s i t i o n 2.15: I f A i s an a l g e b r a i c algebra of bounded degree over F then A i s a P.I. algebra. Proof: Every element a e A s a t i s f i e s a polynomial of the form x 1 1 + a , x n x + ... + a where a. e F . Then we have a 1 1 + a,a n ^ + 1 n l . 1 + a = 0 . For any b e A , and w r i t i n g (uv - vu) = [u,v] we obt a i n n [a n,b] + a.,[a n \ b l + ... + a [ a, b ] = 0 1 n-1 Commute t h i s w i t h [a,b] now, to. o b t a i n [ [ a n , b ] , [a,b]] + a 1 [ [ a n _ 1 , b ] , [a,b]] + ... + [ [ a 2 , b ] , [ a , b ] ] a n _ 2 = 0 Commute t h i s w i t h [[a ,b], [a,b]] and so f o r t h u n t i l none of the c o e f f i c i e n t s remain. What i s l e f t then, i s a n o n - t r i v i a l i d e n t i t y i n v o l v i n g only higher commutators of the [ b ^ a ] . Hence, A i s a P.I. algebra. We end s e c t i o n 1 w i t h our f i r s t main theorem on P.I. r i n g s , a b e a u t i f u l r e s u l t due to Kaplansky [29], 12. Theorem 2.16: I f A i s a p r i m i t i v e algebra s a t i s f y i n g a polynomial i d e n t i t y f = 0 of degree d , then A i s a simple algebra f i n i t e 2 dimensional over i t s centre, of dimension at most [d/2] where [d/2] the l a r g e s t i n t e g e r l e s s than or equal to d/2 Proof: We may assume that, f i s m u l t i l i n e a r . The f a c t that A i s p r i m i t i v e means ( I , 1.1) th a t e i t h e r A - f o r some i n t e g e r n > 0 and some d i v i s i o n r i n g D , or e l s e f o r every i n t e g e r m > 0 , i s a homomorphic image of some subring of A . However, i n the l a t t e r case, we would have Z D ^ A , where Z i s the centre of D m m This would mean that Z s a t i s f i e s f = 0 f o r a l l i n t e g e r s m > 0 , I m . but t h i s c o n t r a d i c t s p r o p o s i t i o n 2.12. Therefore we must have A - f o r some i n t e g e r n > 0 I Let K be a maximal s u b f i e l d of D . Then by theorem 1.2 we know th a t D ® K i s a dense r i n g of l i n e a r transformations on D z considered as a ve c t o r space over K . Hence, D cS K = A ® K i s a n z z dense r i n g of K - l i n e a r transformations. By the same argument as used above we get A <g> K - K . Since A ® K s a t i s f i e s f = 0 we see that ° z n z K s a t i s f i e s f = 0 . But, from theorem 2.11, K s a t i s f i e s no i d e n t i t y n n of degree l e s s than 2n , and th e r e f o r e we have 2n <_ d , i . e . n-<_ [d/2] . From [A ® z K : K] = [A : Z] we get [A : Z] = n 2 <_ [d/2] 2"^.. The f a c t t h a t A . i s p r i m i t i v e and f i n i t e dimensional over Z means that I A i s simple and t h i s completes the proof. ' C o r o l l a r y : Every Jacobson semi-simple .P.I. algebra i s a s u b d i r e c t sum of simple algebras f i n i t e dimensional over t h e i r centres w i t h the dimensions over the centres bounded. i s S e c t i o n 3 - The Kurosh Problem An algebra A over a f i e l d F i s s a i d to be l o c a l l y f i n i t e i f the subalgebras over F generated by a f i n i t e number of elements of ! i A are always f i n i t e dimensional over F For algebras, the analogue of the Burnside Problem f o r groups asks whether an a l g e b r a i c algebra i s l o c a l l y f i n i t e . In general, of course, the answer i s no, but i n the presence of a polynomial i d e n t i t y we s h a l l show that the Kurosh Problem has an a f f i r m a t i v e answer. L o c a l f i n i t e n e s s i s a r a d i c a l property i n the sense of D i v i n s k y [16] and the f o l l o w i n g f a c t s can be e a s i l y v e r i f i e d ; (1) I f A i s an algebra over F and B i s an i d e a l of A such that both B and A/B are l o c a l l y f i n i t e , then A i s l o c a l l y f i n i t e . (2) I f B and C are l o c a l l y f i n i t e i d e a l s of A then B + C i s a l o c a l l y f i n i t e i d e a l of A ' (3) A contains a unique maximal l o c a l l y f i n i t e i d e a l L(A) , c a l l e d the l o c a l l y f i n i t e r a d i c a l of A . L(A) contains a l l the l o c a l l y f i n i t e i d e a l s of A (4) L(A/L(A)) = 0 . The f o l l o w i n g deeper r e s u l t a l s o holds f o r the l o c a l l y f i n i t e r a d i c a l ; (5) L(A) contains a l l the l o c a l l y f i n i t e one-sided i d e a l s of A lb s e t t l e the Kurosh problem f o r P.I. algebras we e s t a b l i s h two p r e l i m i n a r y lemmas; Lemma 3.1: Suppose A i s an a l g e b r a i c algebra over F w i t h no n i l p o t e n t elements. Then,, given a ,...,a i n A , there i s an idempotent e [ 14 i n A w i t h a.e = a. f o r a l l a. 1 1 1 Proof: We.establish a b i t more, and prove that i f the a^ a l l come from.an i d e a l I of A then e can be chosen from I C l e a r l y we can assume the a_^  are not zero. Also we can assume they are n o n i n v e r t i b l e s i n c e otherwise l e i and the lemma i s t r i v i a l by t a k i n g e = 1 The proof i s by i n d u c t i o n on n . Suppose n = 1 . Then a n = a i s non-zero and n o n i n v e r t i b l e . Since A i s a l g e b r a i c we have11 1 1 i j an + a n a 1 1 L + . . . + a, a*1 ^  = 0 , where a, j= 0 and n - k > 1 1 k k ~" Hence, ( a ^ + a a^ " ^  + ... + a. ) a n ^  = 0 and thus, 1 k [( a . + a^a + ... + a^)a] = 0 But a has no n i l p o t e n t elements so that a^+"'" + a n a ^ + ... + a. a = 0 where ct, 4= 0 1 k k ' l 2 Then f o r some polynomial p over F we have a = a p(a) . D e f i n i n g e = ap(a) , we see that e i s a n o n t r i v i a l idempotent w i t h ae = a , thus completing the proof f o r n = 1 Now, suppose we have an idempotent e" i n I such that a,e' = a, a ,e' = a . . I f a e' = a we are through, otherwise 1 1 n-1 n-1 n n . there must e x i s t an idempotent e" i n I w i t h (a e' - a ) e " = a &" - a c n n n n Rearranging gives us a n = a n ^ e ' + e " ~ e'?") a n ^ s e t t i n g e = e' + e" - e'e" gives us an idempotent .e w i t h the r e q u i r e d p r o p e r t i e s -15. Before e s t a b l i s h i n g the second lemma we note that i n a r i n g A w i t h no n i l p o t e n t elements, a l l idempotents are i n the centre Z(A) of A , f o r i f x i s a r b i t r a r y i n A , e an idempotent i n A , 2 2 then (ex - exe) = (xe - exe) = 0 and t h i s means that .ex = exe = xe as required.. Lemma 3.2: Let A be an a l g e b r a i c algebra over F , f i n i t e l y generate and s a t i s f y i n g a polynomial i d e n t i t y . I f A has no n i l p o t e n t elements then L(A) = 0 . Proof: I t i s w e l l known (see [16], p. 105) that the Jacobson r a d i c a l i of an a l g e b r a i c algebra i s n i l . Thus, si n c e A has no n i l p o t e n t elements A ' must be Jacobson semi-simple. Now l e t P be an i d e a l of A such that A/P i s p r i m i t i v e . Since A/P i s a l s o a P.I. r i n g we have, by Kaplansky's theorem, that i t i s a simple algebra f i n i t e dimensional over i t s centre C . Thus, A/P i s l o c a l l y f i n i t e over C . From the f a c t that any commutative a l g e b r a i c algebra i s l o c a l l y f i n i t e we see that C i s l o c a l l y f i n i t e over F . Hence A/P must be l o c a l l y f i n i t e over. F . But A/P i s f i n i t e l y generated over F so that A/P must be f i n i t e dimensional over F . A c c o r d i n g l y , l e t y , . . . b e a b a s i s of A/P over F Let 'x,,...,x generate A and l e t y. be the i n v e r s e image of y. 1 • n x. x Thus, we have m • x. = I a.. y. + u. , where a -. e F , u. e P . m y i y j = ^ B i j k y k + U i j ' W h e r £ ^ i j k e F U i j £ P 16 L e t t i n g P be the i d e a l of A generated by a l l the u. ° o x and u. , we have. P ^ P . Since the . x. generate A we know that i j o . i every element of A has the form Z v.y. + t , where Y. e F , t e P X X X o In p a r t i c u l a r , i f a e P we have a - t = Z Y . y . e P and hence x x I Y .y • xJ x Since the y. form a b a s i s we must have Y. = .0 , a l l i , i . e . x x a e P . Therefore, we have P = P <*•• • • 1o o By the previous lemma,' take an idempotent e e P w i t h u_^ e = u and u. .e = u. . f o r a l l u. and u. . . The u. ,u. . generate Plj J.nd thus pe = p f o r a l l p e P . Hence P = Ae and s i n c e e E Z(A) the P e i r c e decomposition A = Ae © A ( l - e ) c o n s i s t s of two s i d e d i d e a l s From A/P = A/Ae = A ( l - e ) and A/P f i n i t e dimensional over F , A ( l - e ) i s f i n i t e dimensional over F and t h e r e f o r e A ( l - e ) L(A) So L(A) =(= 0 and the proof i s complete. We are now i n - a p o s i t i o n to s e t t l e the Kurosh problem f o r P.I. algebras. Theorem 3 . 3 : I f A i s an a l g e b r a i c algebra over F s a t i s f y i n g a polynomial i d e n t i t y , then A i s l o c a l l y f i n i t e . 17. Proof: We assume that the i d e n t i t y s a t i s f i e d by' A i s m u l t i l i n e a r . of degree d and use i n d u c t i o n on d . I f d = 2 , A i s e i t h e r 3 commutative or A = 0 and the theorem i s true i n these cases. Consider- •-• d> 3 . We suppose A']= L(A) and reach a c o n t r a d i c t i o n . Since A =1= L(A) , by passing to A/L(A) we may assume that A i s a f i n i t e l y generated a l g e b r a i c algebra over F s a t i s f y i n g a m u l t i l i n e a r i d e n t i t y of degree d and f o r which L(A) = 0 .We show that such a s i t u a t i o n i s i m p o s s i b l e , and -this w i l l mean that A = L(A) , i . e . A l o c a l l y f i n i t e . • -A must have n i l p o t e n t elements, f o r , otherwise, by lemma 3.2, L(A) j= 0 . Let a 4= 0 b e . i n A such that a = 0 and l e t I \ T be the l e f t i d e a l generated by a , so that Ta = 0 V \ Now, A s a t i s f i e s f ( x , . . . , x ^ ) = x^qCx^, . . . ,x^) + h (x^» .'. .",x^) , where, i n h j X ^ always appears l a t e r than the f i r s t x^ i n any monomial. S e t t i n g x^ = a and x^ = t ^ a r b i t r a r y i n T f o r i = 2,...,d we have aq.(t 2, . . . , t d ) = 0 D e f i n i n g W = {x e T : ax = 0} and n o t i n g that TW = 0 we see that W •*3 T and T/W s a t i s f i e s q C x ^ j . - . j X ^ ) of degree (d-1) By i n d u c t i o n , T/W i s l o c a l l y f i n i t e . W i s a l s o l o c a l l y f i n i t e s i n c e i f \ W = 0 . Hence, T i s l o c a l l y f i n i t e , and therefore L(A) j= 0 s i n c e T ^ l ( A ) . But t h i s c o n t r a d i c t s L(A)' = 0 , and the l o c a l f i n i t e n e s s of A f o l l o w s . As a . c o r o l l a r y which;is of some independent i n t e r e s t we note that any a l g e b r a i c algebra of bounded degree over F i s l o c a l l y f i n i t e We f i n i s h o f f the question of l o c a l f i n i t e n e s s by j u s t n o t i n g 18. two extensions of theorem 3.3 which can be found i n Jacobson [26]. Theorem 3.4: Let A be an a l g e b r a i c algebra c o n t a i n i n g no n i l p o t e n t elements. I f , f o r every p r i m i t i v e i d e a l P of A , A/P:*: f s a'P'.I."' algebra, then A i s l o c a l l y f i n i t e . Theorem 3.5: Let A be an a l g e b r a i c algebra such that (1) A/P i s a P.I. algebra f o r every p r i m i t i v e i d e a l P (2) the r a d i c a l of every•hompmorphic image of A i s a P.I. algebra. Then A must be l o c a l l y f i n i t e . S e c t i o n 4: In t h i s s e c t i o n we examine the t o t a l i t y of polynomial i d e n t i t i e s s a t i s f i e d by a P.I. r i n g , and, i n p a r t i c u l a r , we f i n d c o n d i t i o n s which enable us to say whether two r i n g s s a t i s f y the same s e t of i d e n t i t i e s . These r e s u l t s are taken e x c l u s i v e l y from Amitsur [ 5 ] . Let F be a f i e l d . I f {x } denotes an i n f i n i t e set of a indeterminates, then by F[x ] we mean the f r e e algebra over F generated by the (noncommuting) indeterminates x a • D e f i n i t i o n 4.1: An (algebra) i d e a l Q of F f x a ] 1 S c a l l e d a T - i d e a l , i f f o r every homomorphism T : x^ t ( x a ) of ^ t x ^ ] onto i t s s u b r i n g s , T ^ we have Q ^ Q . Let A be a P.I. algebra over F . The set of i d e n t i t i e s ' (whose c o e f f i c i e n t s are i n F ) s a t i s f i e d by A i s a T - i d e a l of F[x ] . Denote t h i s i d e a l of i d e n t i t i e s ..of A by Q and c a l l the Ct A 19. quotient r i n g F[x ]/Q. the u n i v e r s a l r i n g of A Ct A. Lemma 4.2: Let P be an i d e a l of F[x. ] . The quotient F['x ]/P i s a a a P.I. algebra i f f P contains a non-zero T - i d e a l ; and i f t h i s c o n d i t i o n h o l d s , the i d e a l of i d e n t i t i e s of F[x 1/P i s the maximal T - i d e a l . • . a contained i n P T Proof: Define Q = { p ( x ) e P : p ( x ) e P f o r every homo, from a a F[x ] onto i t s subrings} . Q i s the maximal T - i d e a l of F[x j a a. contained i n P . Assume P contains a non-zero T - i d e a l , that i s , we assume Q f 0 . Take a non-zero element g(x^,x^, . . . from Q Let t n ( x ), t„(x ), t (x ) be an a r b i t r a r y set of n polynomials l a 2 a n a J from F[x ] . Define a homomorphism T from F[x ] onto one of i t s a a subrings by mapping x. t . ( x ) i f i = 1,2,...,n x x a x ± 0 i f i f 1,2, . . . ,n Since Q i s a T - i d e a l we have g(t 1(x o) , t 2 ( x a ) , .... , t n ( x a ) ) e Q ^ P This i m p l i e s that -F[x ]/P s a t i s f i e s g = 0 and hence i t i s a P.I. r i n g . The converse i s c l e a r . 20. Theorem 4.3: An algebra B over F s a t i s f i e s a l l i d e n t i t i e s of A i f f B i s a homomorphic image of the u n i v e r s a l r i n g of A . Proof: Suppose B s a t i s f i e s a l l i d e n t i t i e s of A , that i s , =s . Then the u n i v e r s a l j r i n g of A can be mapped homomorphically onto the u n i v e r s a l r i n g of B . I f the c a r d i n a l i t y of the s e t ^x^} ^ s l a r g e r . than-the c a r d i n a l number of B , then B i s a homomorphic image of F/Q and hence a homomorphic image of F/Q . 1 B A • \ Conversely, suppose B i s a homomorphic image of the u n i v e r s a l \ ' . \ r i n g of A . By the previous lemma, the u n i v e r s a l ring, of A s a t i s f i e s ( a l l the i d e n t i t i e s of A and thus, so does B Note: I f the s e t {x } i s given, then the theorem i s true only f o r ot those r i n g s whose c a r d i n a l i t y i s l e s s than or equal to the c a r d i n a l i t y of {x } . However, i f the r i n g s are given f i r s t , then we may choose! a the set {x } so that the. theorem w i l l be v a l i d . a j We now turn our a t t e n t i o n to the i d e a l s of i d e n t i f i e s of matrix r i n g s , i n p a r t i c u l a r , to F^ . For each i n t e g e r n l e t 1^ -denote the i d e a l of i d e n t i t i e s of F . Then I., — I . ^  I„ =s . . . s i n c e n 1 2 3 ? ,, contains an isomorphic copy of F f o r . a l l n > 0 n+1 n Lemma 4.4: The quotient F[x 1/1 i s a s u b d i r e c t sum of t o t a l matrix a n 1 algebras of order n over F Proof: Let T be a homomorphism of F[x ] onto F defined by a n x. r. where {r.} i s a s e t - o f generators of F . Let the k e r n a l x i . x to n of T be denoted by Q . Then F = T ' n Qn I 21. I f T ranges over a l l p o s s i b l e homomorphisms of t h i s type we get C\ Q T = I . Hence, i n the q u o t i e n t algebra F [ x ^ ] / I we get F [ xa] F [ x a ] / I n A Q^/I = 0 . Therefore — - — i s a s u b d i r e c t sum of — - — r = . — = T n \ Q T 7 l n F [ x J =L— = F Q T C o r o l l a r y : F[x ]/I i s isomorphic to a subalgebra of a t o t a l matrix r i n g of order n over a commutative algebra which does not c o n t a i n n i l p o t e n t elements. A complete d i r e c t sum of r i n g s isomorphic to F^ i s isomorphic , where R i s a complete d i r e c t sum of r i n g s isomorphic to F R has no n i l p o t e n t elements and the c o r o l l a r y f o l l o w s . Combining t h i s c o r o l l a r y w i t h the previous theorem we o b t a i n ; Theorem 4.5 : An algebra A s a t i s f i e s a l l i d e n t i t i e s of the t o t a l m a t r i x algebra F i f f i t i s isomorphic w i t h a s u b d i r e c t sum of algebras ° n ' isomorphic w i t h F^ ; an a l t e r n a t i v e necessary and s u f f i c i e n t c o n d i t i o n i s t hat A be isomorphic w i t h a subalgebra of a t o t a l matrix algebra of order n over a commutative algebra. . We showed before that every P.T. r i n g s a t i s f i e s a m u l t i l i n e a r polynomial i d e n t i t y . In t h i s same d i r e c t i o n we now show that every P.I. r i n g s a t i s f i e s an i d e n t i t y of the form ^2m^X^n = ^ s o m e p o s i t i v e i n t e g e r s m,n . To o b t a i n t h i s r e s u l t we need the f o l l o w i n g extension of lemma 2.7. Proof: to R n C l e a r l y 22. Lemma 4.6: Let Q be a non-zero T - i d e a l and l e t f ( x l S . . . , x ) e Q a W r i t i n g f ( x ,...,x ) = Z f . ( x ,...,x ) , where each f. i s homogeneous of degree i i n x^ , we have that e Q f o r a l l i = 0,1,...,i The proof i s e x a c t l y the same as lemma 2.7, and, i n p a r t i c u l a r , we s t i l l r e q u i r e the- f i e l d F to be i n f i n i t e . Lemma 4.7: I f Q i s a non-zero T - i d e a l and f ( x 1 , . . . , x ) i s a 1 m polynomial homogeneous i n x^ ', then f i s q u a s i - r e g u l a r modulo Q i f f f i s n i l p o t e n t modulo Q Proof: I f f i s n i l p o t e n t modulo Q then c l e a r l y f i s q u a s i - r e g u l a r modulo Q . Conversely, suppose f i s q u a s i - r e g u l a r modulo Q . Then we have 'I (f - g + fg) e Q f o r some polynomial g . Therefore x. e. g = f + f g (mod Q) E f + f 2 + f g (mod Q) E f + f 2 + ... + f n + 1 + f U + 1 g (mod Q) n+1 . def „ .1 _n+x , _ g - Z f - f g = . h e Q i = l W r i t i n g g = Z g. , where g. i s a polynomial homogeneous of degree j=0 J 3 i i n x n , we f i n d that 1 ; P n + 1 i P n+1 h = E g. - E f - E f g. e Q j=0 3 i=l j=0 J Now choose n > p . Then in the decomposition of h into components homogeneous in x^ , we must have f n occuring alone, and thus, by lemma 4 . 6 , f n e Q , i.e. f is nilpotent modulo Q . These two lemmas allow us to prove our main result; Theorem 4 . 8 : Every P.I. ring satisfies the identity ^ ^ x ) 1 1 = ^ ' some integers m > 0 , n > 0 Proof: Let A be a P.I. algebra, its ideal of identities in F[x ] . Then Q, is a non-zero T-ideal. Let J be the ideal. 1 a HA of F[x ] such that J/Q. is the Jacobson radical of F[x ]/Q. • Hence Ct A Ct A . |-F [ x a ] / Q A .. F [ x ] \ r - r - = -— is a semisimple P.I. algebra. Thus, i t is isomorphic \ J/QA \ to a subdirect sum of primitive P.I. algebras. But by Kaplansky's theorem a primitive P.I. algebra is a simple algebra finite dimensional over its centre, and by the corollary to theorem 2.11 these simple components satisfy a standard identity which can be considered as being of degree 2m , say, for some fixed integer m > 0 Thus, Proposition 2.2 assures us that F[x ]/J satisfies this a F L x J „ standard identity, i.e. we have S„ (x ) = 0 in :;— . But this zm a J .implies that S« (x ) e J/Q". and since S„ (x ) is then quasi-regular 2m a A 2 n a , • n 1 j modulo Q, i t must be nilpotent modulo Q. , i.e. S„ (x ) •= 0 XA A 2n a for some integer n > 0 . This is the .same as saying that ^2n^ Xa^ n E and thus, A satisfies S„ (x ) n = 0 2m a A^ 24. This theorem can be proved under more general conditions, j Amitsur did so in [5] assuming only that A has a ring of operators Q , with the properties; Q.) 0, is an infinite integral domain. (2) aA = 0 for a e 0, implies that a = 0 (3) a(ab) = (aa)b•= a(ab) for a l l a e Q, a,b e A With this more general result, we now note that we do not even need ft to be infinite. A l l we do is to work with A[t] , the ring of polynomials in a commutative indeterminate t over the ring A , and to see thatj 1 since A satisfies linear identities so does A[t] . Thus, A[t] is a P.I. algebra over the infinite integral domain ^[t] and hence A[t] satisfies S„ (x) n = 0 . Since A ^  A[t] , the generalization follows 2m. . . clearly. Chapter II In the latter part of chapter I we proved the beautiful result of Kaplansky for primitive P.I. rings. The present chapter J examines other types of P.I. rings. These include prime rings, rings ' with no nilpotent ideals, rings with no n i l ideals, and rings with no zero divisors. We shall prove a few results concerning the Levitzki and Baer radicals as well as a few theorems of the type n i l implies nilpotent and n i l implies locally nilpotent. • A very useful class of rings is the class of commutative rings. These rings satisfy the identity x-^x2 ~ X 2 X 1 = ^ > i.e« the standard identity of degree 2 . Many of the results of this chapter are generalizations of well-known commutative theorems. Unless express mention is made to the contrary we shall assume that A is a P.I. algebra over an integral domain fi . Thus, A is an fi-algebra satisfying the laws, i) aa = 0 => a = 0 or a = 0 , for a £• fi, a e A i i ) a(ab) = (aa)b = a(ab) , for a e fi, a,b e A . The coefficients of the polynomial identity satisfied by A are elements from fi . • Section 1. We begin with, Theorem 1.1: • Every n i l P.I. ring A is locally nilpotent. 26. Proof: Denoting the locally nilpotent (Levitzki) radical of A by A L(A) we see that is a n i l P.I. ring whose Levitzki radical is A zero. We shall prove the theorem by showing that ^(A) ~ ® A i A 1 Suppose YXKJ ' ® * Since jjj^ contains no locally nilpotent one-sided ideals we can reduce to the case in which A is a n i l P.I. ring^ with no locally nilpotent one-sided ideals. 2 • ' Let a e A with a =f 0 but a = 0 . If aA = 0 , then 2 £(A) , the left annihilator of A , is not zero and [jt(A)] = 0 contradicting assumptions on A . Therefore, we can assume further that aA =)= 0 . If f is a multilinear identity satisfied by A we write j f (x^, .. ., x^ j) f ^ > • • •»Xjj) x^ ^2 ^ 1^ * * • * » • s ' where x^ does not appear in the last position of any monomial in jj; f„ . If we substitute x.. = a , and x. = a. arbitrary from aA 2 1 i i for i f 1 , we get f2(a,a2 a^) = 0 and hence f-^(a2' • • • >a^}a = 0 Now, let K be the left annihilator of aA in the subring aA of A . The fact that f^(a2»•.•,a^)a = 0 for a l l a2 a^ in aA aA implies that — satisfies a polynomial identity of degree less then K or equal to (d-1) . By induction on the degree of the polynomial aA identity we see that — is locally nilpotent. But K is nilpotent and this means that aA is locally nilpotent. Thus, aA is a non-zero' locally nilpotent right ideal of A . This contradiction to A having no locally nilpotent one-sided ideals proves that JjJJ ~ ® , i.e. A = L(A) and the proof is complete. We now state two very similar results which will often be used in this chapter. The first is from Levitzki [31]. 4 i • 1  Proposition ,1.2: If A is a P.I. ring of degree d , then for each j nilpotent element a e A , we have a t d / 2 ] e NP(A) , where NP(A) is the sum of a l l the nilpotent ideals of A The proof of this proposition is very much like the proof of the following more useful result due to Amitsur [3]. Proposition 1.3: If A is a P.I. ring of degree d , then for each n i l subring T of A , we have T [ d / 2 ] ^ NP(A) Proof: First of a l l , we assume the stronger condition that T is nilpotent. Define the subrings A^ of A by; for i = 1 n A2. = T ^ A T 1 where n is an integer greater than [d/2] . From these definitions we can verify that (1) A. A. . •'• A. ATnA , where (i , i 0 , . . . , i ,) is a non-3-1 i 2 i d l i a identity permutation of (l,2,...,d) . .28. (2) A±A2 ... A d= (T n- 1A) d T [ d / 2 ] . . . ' ^ " ' .Write-the-polynomial identity satisfied by A in the following way: (3) 3 x xx 2 ... x d = I B.x ••• x.d , 3 M • I i f l f l 1 . By substituting x = a^ arbitrary from' A^  for i = 1,2,.... , d we see by (1) and (2) that (4) B(T n" 1A) dT [ d / 2 ] ^ ATnA . Now, because T is nilpotent there is a smallest integer n for which ATUA is a nilpotent ideal. If n > [d/2] then from (4) we deduce that g (ATn "*"A)d+ "^== ATnA which contradicts the choice of n . Thus, n <_ [d/2] and the proposition is proved for the case j when T is nilpotent. To finish the proof we consider T to be merely a n i l subring I T+NP(A) [ d / 2 ] of A . Then the factor ring NP(A) satisfies the identity x =0 by the previous proposition, and hence i t is locally nilpotent. But since NP(A) is locally nilpotent, the subring T+NP(A) of A must also be locally nilpotent. Hence, i f t n , . . .,t r , /0-, are arbitrary elements from T we 1 Ld/2J see that the subring < t 1»•••»t[d/2] > o f T generated by these elements is nilpotent. By the first part of the theorem for T nilpotent, we have ^ • • • • • • ' [ d ^ ] ^ 2 1 * N P ( A ) ' and thus for any choice of elements ti> • • • »t[(j/2] °^ T t^ i e P r o d u c t  t l t 2 •"' t[d/2] e N P ( A ) » ±-e- T [ d / 2 : i< NP(A) . This-completes the proof. . • ' 29 . S e c t i o n 2: I t i s w e l l known [36] t h a t a commuta t i ve r i n g w i t h no n i l p o t e n t i d e a l s i s i s o m o r p h i c t o a . s u b r i n g o f a c o m p l e t e d i r e c t sum o f f i e l d s . We show t h a t a P . I . r i n g w i t h no n i l p o t e n t i d e a l s i s i s o m o r p h i c t o a s u b r i n g o f a c o m p l e t e d i r e c t sum o f c e n t r a l s i m p l e a l g e b r a s o f f i n i t e o r d e r s o v e r t h e i r c e n t r e s . The f o l l o w i n g lemma i s t r i v i a l and can be f ound i n A m i t s u r ' [ 4 ] . ' Lemma 2 . 1 : I f A i s a P . I . r i n g , t h e n A [ x ] i s a l s o a P . I . r i n g and t he l i n e a r i d e n t i t i e s s a t i s f i e d by A a r e a l s o s a t i s f i e d by A [ x ] and c o n v e r s e l y . T h i s r e s u l t , t o g e t h e r w i t h ( I I , 1.3) a l l o w s us t o p r o v e ; Theorem 2 . 2 : L e t A be a P . I . r i n g . o f deg ree d and suppose t h a t A has no n o n - z e r o n i l p o t e n t i d e a l s . Then d i s e v e n , say d = 2m , and A i s a s u b r i n g o f a c o m p l e t e d i r e c t sum o f c e n t r a l s i m p l e a l g e b r a s 2 {A } s u ch t h a t m i s t h e uppe r bound o f t he o r d e r s o f t h e s e a l g e b r a s a o v e r t h e i r c e n t r e s . P r o o f : S i n c e A has no n o n - z e r o n i l p o t e n t i d e a l s t h e n i t a l s o has no n o n - z e r o n i l i d e a l s . F o r , i f T i s a n i l i d e a l o f A , t h e n by ( I I , 1 . 3 ) , T ^ 2 - ^ NP(A) = 0 . Hence , T i s n i l p o t e n t w h i c h means t ha t . • T = 0 . By ( I , 1.3). t he f a c t t h a t A has no n o n - z e r o n i l i d e a l s means tha t . A [ x ] i s J a c o b s o n s e m i - s i m p l e . From ( I I , 2 .1) we see t h a t t h e l i n e a r i d e n t i t i e s o f A and A [ x ] c o i n c i d e so t h a t A [ x ] i s a J a c o b s o n s e m i -s i m p l e P . I . r i n g o f deg ree d . By t h e c o r o l l a r y t o K a p l a n s k y ' s t heo r em, A [ x ] i s a s u b d i r e . c t sum o f c e n t r a l s i m p l e a l g e b r a s A^ such t h a t 30. 2 m is the least upper bound of the orders of these algebras over their centres. But the corollary to (I, 2.11) tells us that each satisfies S„ (x) = 0 . Hence, so does A[x] so that d < 2m . If now, d < 2m , 2m — i then each A^ satisfies a polynomial identity of degree d , contradicting 2 the corollary to (I, 2.11) and the fact that m is a least upper bound. ' • • v Thus, d = 2m and since A is a subring of A[x] the proof is complete. • i • i . • Corollary: A ring A with no non-zero nilpotent ideals is a P.I. ring i f f i t is isomorphic to a subring of a complete direct sum of central simple algebras of bounded order. As is often done in ring theory i t is quite useful to charac-terize the class of rings being studied by means of matrix rings. This is accomplished for P.I. rings in Theorem 2.3: Let A be a P.I. ring of degree d and suppose that; A has no nilpotent ideals. Then d is even, say d = 2m , and A is 2 isomorphic to a subring of a total matrix ring of order m over a commutative ring which does not contain nilpotent ideals. « Proof: We already know that A is a subring of a complete direct sum ! of a family of central simple algebras A^  . For each a , let F^  • I be a splitting field for A^  . Then A^  is isomorphic to a subring of the ring of a l l m x m matrices over F , that is, A is a a isomorphic to a subring of F . Consider the complete direct sum r " a m . • r ZF . This sum contains an isomorphic copy of A . But we also have £ F = (j F ) . Setting G = E F we observe that• G is a a am a am a a commutative ring with no nilpotent ideals since i t is a direct sum of: fields. 31. j Corollary: A ring A with no non-zero nilpotent ideals is a P.I. ring; i f f i t is isomorphic to a subring of a total matrix ring over a commutative ring. Corollary: A P.I. ring with no non-zero nilpotent ideals satisfies a standard identity. Now, i f A is a P.I. ring with no non-zero n i l ideals, the same results will hold, since A will have no nilpotent ideals either1. In fact (II, 1.3) assures us that for P.I. rings, the property of having no n i l ideals is precisely the same as having no nilpotent ideals. In effect, we have shown that P.I., tings with no n i l ideals can be embedded into matrix rings over commutative rings B . But can this be done for arbitrary P.I. rings? A negative answer is provided [25] by Herstein with the following example. i Let A be an infinite dimensional algebra over a field of characteristic zero and suppose A has a basis u^,...-u^,. ... such that u.u. = -u.u. for a l l i , i . This is merely an exterior algebra. Then A satisfies the identity [[x,y],z] = 0 . However, A does not satisfy any standard identity and hence i t cannot possibly (I, 2.11) be embedded in a B m Note: For any ring A , the Baer Lower Radical [16] 3(A) of A is characterized as the smallest ideal I of A with the property that A/I has no non-zero nilpotent ideals. Hence, we have proved that any! Baer semi-simple P.I. ring can be embedded into a matrix ring over a commutative ring. Section 3: We now proceed one step further and consider P.T. rings which have no divisors of zero. The result that we shall obtain will be a generalization of the fact that commutative rings without zero divisors possess fields of quotients which are commutative. We begin with Definition 3.1; A ring A is right (left) 0-regular i f for every pair of non-zero elements a,b e A , we have aA C\ bA f- 0 (Aa.0 Ab f 0) . Ore has shown that the 0-regular rings have division rings of quotients consisting of a l l elements of the form ab ^ , where a,b are elements in A Lemma 3.2; Let A be a P.I. ring with no zero divisors. Then A is both a right and left 0-regular ring. Proof: Let a,b be any two non-zero elements of A . We observe that i f A satisfies f(x^,...,xm) = 0 , then by substituting x^ = yz 1 for each i = l,...,m , we get an identity h(y,z) = 0 in just two indeterminates. Now let g(y»z) be a polynomial of minimal degree for which g(a,b) = 0 . Two cases can occur. Case 1: A has an identity element./ We write g(y,z) = a + yg1(y,z) + zg2(y,z) . i If g^(a,b) =0 , we have a contradiction to the definition of g unless g^(y,z) is itself zero. If g^(atb) = 0 , the same argument applies. But i f g,(y,z) = g9(y,z) = 0 then g(y,z) = 0 . This 33. contradiction means that at least one of g^(a,b) , g2(a,b) is not zero. Suppose that g^(a,b) f 0 . Then by substituting y = a , z = b in j g(y,z) we obtain a + agx(a,b) + bg2(a,b) = 0 . Multiplying on the right by b we have b(-a - g2b) = ag^ , where g i = g_^ (a,b) . Now, ag^ b =j= 0 since a =j= 0 , b =j= 0 ,g^=j=0 , and A has no zero divisors. Thus, ag^ b = b(-a - g2b) is a non-zero element in aA A bA and A is right 0-regular. Case 2: A does not have an identity element. We write g(y,z) = yy + 6z + yg^Cy.z) + zg 2(y,z) . If y = & = 0 we proceed as in case 1 with a = 0 . Therefore, suppose Y =f 0 . Multiplying g(a,b) on the right by a we obtain, »(-6a - g2a) = a(ya + g^a) , 1(1) where g^ = g^(a,b) . This element is an aA H bA and so, i f i t is not zero, we are finished. Therefore, suppose this element is zero. This will imply that 34. (2) ya + = 0. Then a(ya + g^) = 0 or . ya.a + ag^a = 0 or (ya + ag^)a = 0 or ya + ag1 = 0 (3) From (2) and (3) we conclude that ag^ = g^a . Define h(y,z) = yg^(y,z) - g1(y,z)y . Three possibilities can arise: (i) g^(y,z) has non-zero terms involving both y and z (or just z ). (ii) g^(y,z) involves only y's (i i i ) g1(y,z) is zero. From ag^ = g^a , i t is clear that h(a,b) = 0 . If the first possibility holds then h(y,z) f 0 since the indeterminates y,z do not commute. Also, by the choice of g(y,z) we see that the degrees of h(y,z) and g(y,z) are the same. But h(y,z) has no terms of the form yy,Sz , so that we may proceed by case 1 with a = 0 If the second possibility holds then from ya + g^a = 0 we have b(ya + g,a) = 0 , or (yb + g ; Lb)a = 0 Thus, yb + gjb - 0 , (4) and as before we can deduce also that ' 35. yb + b g 1 - 0 . (5) From (4) and (5) we have bg^ ^ = g^ b . Define k(y,z) = zg1(y,z) - g1(y,z)z Clearly, k(a,b) =0 and since g^(y,z) involves only y's we see that k(y,z) j= 0 , and also that k(y,z) has no terms of the form yy,&z As before we may proceed as' in case 1 with a = 0 If the third possibility holds, i.e. g^(y,z) =0 , then (1) reduces to b(-6a - g2a) = a(ya) Since y =(= 0 , a =j= 0 , then ya f 0 , and hence a(ya) =j= 0 . Thus, we have our required non-zero element in aA H bA , and A is right 0-regular in the second case. The proof of left 0-regularity is similar. We now know that a P.I. ring without zero divisors possesses a division ring of quotients. This ring of quotients is characterized by; ; Theorem 3.3: Let A be a P.I. ring with no zero divisors. Then the division ring of quotients D of A is a central division algebra of finite order over its centre and satisfies the same linear identities as A . Proof: The centre Z of D is an ft-subalgebra of D . By (I, 2.6) \ we know that A <2>n Z satisfies the same linear identities as! A Define a mapping p : A <2> Z -*• D by p(Ea. & z.) = Ea.z. Since Z commutes with A we see that p is an ft-homomorphism. Actually, p is onto. For, every element of D has the form ab. ^  , a,b E A , and also, since A, is algebraic over Z [6] we have ! ii' 36. Y, n-1 + y » where Y- e Z Hence pCZab1® y - ) = Eab1^. = ab -1 I.e. p is onto. Thus, D satisfies the same identities as does A ® Z , and we have already seen that A ®^ Z satisfies the same linear identities as does A . Since A«£ D , the linear identities satisfied by A and D must be the same. By Kaplansky's theorem D is a central simple algebra of finite order over its centre Z , and the theorem is proved. I The converse is also true so that we have Corollary: A is a P.I. ring without zero divisors i f f A is a subring of a central division algebra of finite order. Section 4: In this section we shall deal with prime P.I. rings. The main theorem on these rings is due to Posner [37] but his original proof is valid only for prime algebras. Instead we give a proof recently published by Amitsur [11] and shortened by Goldie [19]. This proof \\ assumes very few restrictions on ft and in particular, ft need not be a field or even an integral domain. j and ft its centroid. We let E(A) denote the set of endomorphisms of the additive abelian group A + and B(A) the set of a l l those elements of E(A) induced by right or left multiplication by elements of A Then, since ft is defined as the set of a l l those elements of E(A) which commute with each element of B(A) we have We assume throughout this section that A is a prime ring, a(ab) = aa.b = a.ab , 37. where a e fi , a,b e A . We suppose that the identity satisfied by A has coefficients in fi , The case in which A is an algebra over a field F (or integral domain) is then covered by our assumptions since, of course, we have F-^ fi . As usual we can also assume that the identity satisfied by A is multilinear, say 1 d i i - , i •, 1 d In order to rule out trivial identities we require that for at least one i , a.A f 0 . i Lemma 4.1: The prime P.I. ring A has a right and left quotient ring A which is the set of a l l r x r matrices over some division ring D Proof: We show that A satisfies conditions ( U l ) , (2£) , (lr) and (2r) of Goldie's theorem (I, 1.4). For condition (ll) we shall show that the length of a direct sum of non-zero left ideals cannot exceed (d-1) Accordingly, let I2 ® **" ® "'"d' ^ e a ^ r e c t s u m of left ideals. We distinguish two cases. Case 1: Suppose that in addition to being ring ideals, the I_. are also fi-ideals, i.e. they are closed under multiplication by elements of fi We let a_. be an arbitrary element of I_. , for j = 1,2,...,d Then, we have Consider the sum S^ of a l l monomials of (1) ending i n a^ . Since 1^ i s a l e f t i d e a l and since I © . . . © I i s a d i r e c t sum we see that S n = 0 d n 1 | i Also, by the primeness of A , we can drop a^ . Thus, we are l e f t wi^h something of the form, 2 a. a. ... a. = 0 , ; (2) i 1 x 2 X d . !; where i i s a permutation of 2,...,d . We now consider the sum of a l l monomials of (2) which end i n a '.' As before we get S_ = 0 ; and we are then able to drop the a^ to get something of the form . Z a. a. ... a. = 0 , (3)-i 1 x 3 x d where i i s a permutation of 3,...,d . Continuing i n t h i s fashion we obtain 1^ = 0 and t h i s gives us (12,) . Case 2: Suppose that at l e a s t one 1^  i s not an fi-ideal. Then c e r t a i n l y , fil^,...,fil^ are closed under m u l t i p l i c a t i o n by elements of fi . From case 1 we obtain fil =0 . Now fi =(=0 since fi i s non-t r i v i a l so that I , = 0 as i n case 1. a Hence, A s a t i s f i e s condition (12,) and s i m i l a r l y ( l r ) . For condition (22), suppose I. < I„ < ... < I , , where I. i s the l e f t a n n i h i l a t o r of a r i g h t i d e a l K_. . Of course, K =J= K^._1 f o r j = 2,...,d . Let k be the smallest p o s i t i v e integer such that there e x i s t s 39. a set {6^ } of elements in ft , not a l l zero, with the property that E 3 a a ... a = 0 , TT IT TT^ ^2 '"k where TT is a permutation of 1,2,... ,k and a^ e I., arbitrary. Since A satisfies the polynomial identity E a. x. x. ...x. = 0 we see that i 1 2 1d k exists and k < d . Now, E g a a ...a K. , = 0 . TT TT TT. TT_ TT, K~ l 1 2 k Thus, . Z 1 S 1 a 1 a 1 . . . a 1 = 0 » TT TT TT2 TT ^ where we are now summing over a l l those permutations IT''" for which TT^ Ck) = k . This can be rewritten h P 1 a 1 a 1 \ \ - l - 0 TT TT TU IT, , 1 k-1 I^K^_^ is a non-zero two-sided ideal of A and so by the primeness of A we have Z l ^ l a l * " a l = 0 ' TT TT TT-, TT, , 1 k-1 which contradicts the minimality of k Hence, = ^ > a n c* condition (2&) and similarly condition ( 2 r ) holds in A 40. Thus, A has a two-sided ring of quotients consisting of the set of a l l r x r matrices over some division ring D and the proof, is complete. ' ' Lemma 4.2: Let c be a regular element of the prime P.I. ring A . Then cA(Ac) contains a non-zero ideal of A . k i • Proof: Consider the set of right ideals {c A : k > 0} . f is a \\ non-trivial identity of each c A since i f ct.A f 0 . then a . c A f 0 k £ ' Also, since c A ^  c A for k <_ % , any polynomial identity satisfied by k I ! c A is satisfied by c A (when k <_ & ). k ! Now let c A be an ideal which satisfies a multilinear poly-nomial identity q(x^,...,xm) of.minimal degree. We write this as q(x',...,x ) = x q (x x + q,(xn>••• 1 m m i l m—i z i m where x never appears first in any term of q_ . We can also assume m 2 th at ^I'T 5 0 merely by renumbering the i f need be. Substitute 2k n n x. = c y. , i = 1,...,m-l I i • x = c y m m where the y.. are arbitrary from A . Then we have n k ( 2 k 2k \ , 2k 0 =,c y ^ c y i,...,c y ^ ) + c t , for some t e A . By the minimality of q we can choose ^\ i""»^ x l i-\ ^ n A such that Then, / 2k ZK. v i q 1(c y1,....,c y m_ 1) = e += 0 2k n k _,_ 2k 0 = c y e + c t m = c^(y e + c^t) m Hence, y^e + c t = 0 by the regularity of c . Now, y^ was arbitrary so that ' Ae ^ c^ A Thus, AeA ^  c A ^ c A , as required. We shall require one more lemma before we prove the main result. Suppose A is also a Jacobson semisimple ring. If {P^  : a e A is the class of primitive ideals of A we have that (HP : a e A^) = 0 a Let ft(f) be the set of coefficients of f We define a primitive ideal P to be trivial i f ft(f)A^ P , otherwise i t is non-trivial. a a The intersection of a l l the tri v i a l P is not zero since i t contains ft(f)A . Consequently, the intersection of a l l non-trivial P^  must be zero since A is a prime ring. We let {P : a e A} be the set of a l l . a non-trivial P or 42. Lemma 4.3: Let c be a regular element of A and cA «s- T =f= 0 , where T i s an i d e a l of A . Let A(T) = {a e A : T ^ P a > . Then, i f a e A(T) , [c + P ] i s a unit i n the ring A/P and OL . QL (/IP^  : a e A(T)) = 0 Proof: A/P^ i s a primitive ring and f(x^,...,x^) i s n o n - t r i v i a l i d e n t i t y for A/P so that by Kaplansky's theorem we see that A/p J a • a T + P i s a central simple algebra. Now a , being an i d e a l i n A/P P " a must be a l l of the ring A/P^ • From T < cA we have c + P = t + P = c r + P a a a i.,e. c + Pa = (c + Pa) (r + Pa) and (c + P ) i s a unit i n A/P a a For the second part of the lemma we see that (HP : T^P ) 4= 0 r a a 1 since T f 0 . Thus by the primeness of A , (HP : T s t P ) = 0 or, 1 a *^  a equivalently, (HP^ : a e A(T)) = 0 . Theorem (Posner) 4.4: A ring A i s a prime ring s a t i s f y i n g a n o n - t r i v i a ^ polynomial i d e n t i t y of minimal degree d with coe f f i c i e n t s i n the centroid v of A i f f A has a l e f t and right quotient ri n g A which i s a central 1 2 simple algebra of dimension (-j^ d) over i t s centre C . Moreover, AC = A . Proof: Suppose A i s a prime P.I. ring s a t i s f y i n g a mu l t i l i n e a r i d e n t i t y f(x^,...,x^) of minimal degree. First, we let A be Jacobson semi-simple as well. Let {P : a e A} be the set of non-trivial primitive ideals as before. a Define A A = — , for a l l a , a P a S = II A = {(x) = (...x ...) : x £ A , a l l • . a a a a aeA A(x) = {a : x f 0} , a V = {(x) e S : <fiPa : a e A(x)) =j= 0} Now, i f (x) , (y) e -V , and (a) e S we see that A(x + y) <A.(x) U A(y) , A (ax) _< A(x) A(xa) < A(x) . Thus V ^  S . . Under that map ¥ : A ->- S defined by Y(x) = (x) = (... ,x'+ P^ ,...) , we see that A = A' where A' is a subring of S Claim: A' C\ V = 0 . i !I For, i f (x) E A" r\ V then, (x) £ A'-> X A = X + P A > A L I ' a , by definition. But (x) £ V (fiP : x f 0) f 0 -> (AP : x i P ) f 0 . 44. Hence by the primeness of A we have (AP : x e P ) = 0 also, and a a since x is in this we must have x = 0 so that A' f) V = 0 . From •' A = A' <_S and A" d V = 0 we see that .A A * V <_ S/V and so we can regard A as a subring of S/V ... j Let c be a regular element of A . By (II, 4.2) let 0 j= T <_ cA , where T ^  A . Define an element (s) in S by setting s = (c + P ) - 1 , when P } T , and s = 0 , when P > T a a a — Then the product c(s) = (1) - (v) , where v = 0 i f P ' i T , a a a -1-and v = 1 i f P > T a a a — By definition, (v) e V and hence c is a unit in S/V . Since this j is true for any regular element c of A we must have the quotient ring A of A (which exists by II, 4.1) isomorphic to a subring of S/V The identity f(x 1,...,x J) is satisfied by each A , hence by I d a S = jjA , hence by S/V . It is also non-trivial for S/V since i t : OL is non-trivial for A <_ S/V . Thus, A must also satisfy f non-trivially and f must be of minimal degree for A since i t is minimal for A The fact that A is primitive allows us to use Kaplansky's theorem so : 45. that A is a simple algebra with (A : C) = (-|d)2 Now, suppose A is a prime P.I. ring but not Jacobson semisimple. We pass to A[t] ,. where t is a commuting indeterminate. A[t] is prime, satisfies the same multilinear identities as A , and is also Jacobson semisimple since A has no nonzero n i l ideals (I, 1.3). The quotient ring of A[t] satisfies the same identities as A[t] and i t contains the quotient ring of A . 1 2 Finally, let m = (-^d) and <!;[_>•••'^  e Q . be a C-basis of Q . Since q^ e Q we get q i = a^c ^  , for a l l i , where e A , c. regular in A . Then, Q=Qc=a nC+ ... + a C = AC , as ^ 1 m required. Section 5: We now delve a bit further into the theory of prime P.I. rings by investigating when a prime ring with a one-sided P.I. ideal is itself a P.I. ring. It is known that i f a prime ring possesses a non-zero commutative one-sided ideal then the whole ring is commutative. The following example illustrates the fact that, this result does not j extend to arbitrary polynomial identities. | I Let F be a field and F the ring of a l l those infinite matrices over F with finite rank. Let M = (m..) be a nMtfix with mn1 + 0 and m.. = 0 for a l l . i , i f 1 . Define I = MF . I is 11 ' —- - i j 1 ., - oo 2 a right ideal satisfying the identity (x.^ - x^x^) - 0 , but .F m satisfies no identity at a l l by (I, 2.12). Hence, we must impose stricter conditions on the one-sided ideal. To investigate this problem we introduce several new concepts. To begin with, ; Definition 5.1: The right singular ideal S R ( A ) of a ring J<A is defined as S^(A) = {x c A : r(x) is an essential right ideal} Now, let L be a lattice with 0 and 1 . An element B E 1 is called an essential extension of an element C £ L i f C <_ B and for every D £ L such that D D B =f= 0 we also have D f\ C =j= 0 A ring A whose right singular ideal S^(A) = 0 has the property ([27]) that each element I e L(A) , the set of right ideals of A , has a unique maximal extension I s in L(A) . The mapping j s : I -»-T of L(A) to L(A) is a closure operation on L(A) satisfying I (1) 0S = 0 (2) ( i n J ) S = I s r > J S , for each I,J e L(A) . (3) (x ^"I)S = x , for each x'e.A and each I E L(A) , where x 4 .= (y e. A : xy e 1} . • s The set L (A) can be made into a lattice in the usual way1; s by defining the union of a set of elements of L (A) to be the least upper bound of that set. Definition 5.2: A is a right quotient ring of A i f A >_ A and for , I each element b e A , we have bA 0 A f 0 . If, in addition, _1 A = {ab : a,b z A , b regular} then A is a classical right quotient ring.. i Definition 5.3: A ring A is right quotient simple i f i t has a classical right quotient ring A with A = Dn , for some division ring D and some positive integer n . ' . The usefulness of the more general concept of right quotient ring (as opposed to classical) becomes evident in the following result \ due to Johnson [28]. . ' j I Proposition 5.4: If A is a prime ring whose right singular ideal j S^(A) = 0 then A has a unique maximal right quotient ring A where A is a prime regular ring. Before we establish our. main results we first state three preliminary lemmas which can be found in [15]. Lemma 5.5: If A is a prime P.I. ring, A is a P.I. ring. This was proved, but not stated explicitly, in the proof of Posner's theorem. Lemma 5.6: Let A be a prime ring with S^(A) = 0 and let a e Horn (A,A) , A considered as a right A-module. If there is an n essential right ideal I of A such that a(I) = 0 then a = 0 Lemma 5.7: If A is a prime P.I. ring and i f S^(A) =0 , then Horn (A,A) has a polynomial identity. Two short proofs of these lemmas are given in [15] with the proof of 5.7 depending on the fact that we can write A as U Hom(I,A) 1 r modulo some equivalence relation and where the union runs over a l l essential right ideals I of A . With these lemmas we now establish our three main results. Theorem 5.8: Let A be a prime ring possessing a non-zero right ideal I satisfying a polynomial identity. Then, i f 5,(1) =0 , A is a P.I. ring. Proof: I is a prime ring since £(I) = 0 . Considering T as a left I-module, the dual of lemma 5.7 states that Horn (1,1) is a P.I. ring. However, A is anti-isomorphic to a subring of the P.I. ring Hom^(I,I) under the mapping a -> r (where r denotes right multiplication by the element "a" ). Since polynomial identities are preserved under i anti-isomorphisms we have the fact- that A is a P.T. ring. i This theorem gives conditions sufficient to ensure that the property of being a P.T. ring is passed on from a right ideal to the whole ring. In the special case in which A is right quotient simple we do not need any additional conditions on I Theorem 5.9: Let A be a right quotient simple ring and I a non-zero right ideal of A . Then, i f I is a P.I. ring, A is a P.I. ring. The next theorem provides us with necessary and sufficient conditions on I Theorem 5.10: Let A be a prime ring having a non-zero right ideal satisfying a polynomial identity. A necessary and sufficient condition that A satisfy a polynomial identity is that S^(A) = 0 and A has at most a finite number of orthogonal idempotents. Proof; A is defined to be regular i f for every a e A there exists an x such that axa = a . We know from [30] that every non-nil right ideal of a regular ring contains a non-zero idempotent. Also from [30] we have the fact that i f A is a ring in which every non-nil right ideal contains a non-zero idempotent, then either A contains an infinite number of orthogonal idempotents or else i t has D.C.C on right ideals. To prove sufficiency in 5.10 we assume that S^(A) = 0 and A has at most a finite number of orthogonal idempotents. Since A i s ' regular, A must have D.C.C. on right ideals. Since A is prime and Sr(A) =0 , A must also be prime [28, Theorem 2.7]. Now, in [28], i t is shown that LS(A) = LS(A) under the mapping I + I O A , I e LS(A) . Since A has D.C.C, LS(A) has D.C.C. Hence, from [28; Theorem 4.2], A is a classical right quotient ring. Then (II, 5.9) completes the proof. ; Necessity of the condition is clear. i ii: I ! i i ' t Section 6; In this section we collect a pair of pretty results, one due to Posner, the other to Procesi. The'first can be regarded merely as a corollary of Posner's theorem (II, 4.1). Theorem 6.1; Suppose A is a P.I. algebra over its field F . If every element of A is a sum of nilpotent elements, A must be n i l . Proof: We show that A has no prime ideals. Suppose P is a prime ideal of A . Then A/P has no nilpotent ideals and hence P is an algebra ideal. Therefore A/P is a prime P.I. algebra with the property that every element of A/P is a sum of nilpotent elements. Thus, we may assume from the start that A is a prime P.I. algebra every element of which is a sum of nilpotent elements. By the theorem of Posner (II, 4.1), A possesses a classical quotient ring A = for some integer r and some division ring D finite dimensional over its centre Z . Now let K be a splitting field for D over Z so that 1 D ® K = K for some integer n > 0 . Since D = A = AC , where r z n 6 r C is the centre of , we see that every element of is also a sum of nilpotent elements. Hence, every element.of is s t i l l a sum of nilpotent elements, with the result that the trace of any element in is zero. But, merely by taking e ^ from we see that tr(e i : L) = 1 f 0 . This contradiction proves that A has no prime ideals. Therefore, the Baer Lower Radical g(A) = A . Since g(A) <_ N(A) where N(A) is the n i l radical, we have A = N(A) , i.e. A is n i l . The second result, from Procesi's thesis [39], is very similar to that just proved. Theorem 6.2: Suppose A is a P.I. algebra over its field F If every element of A is a sum of elements which are algebraic over F , then A must be algebraic over F 51. Proof: Suppose, i n s t e a d , that an element a £ A i s transcendental over F . Define S = { f ( a ) : f .e F[x] , f =f 0} . By Zorn's lemma we can . choose an i d e a l P of A maximal re the e x c l u s i o n of S . By a standard argument P i s a prime i d e a l . Thus, A/P i s a prime P.I. r i n g w i t h the element a = a + P not a l g e b r a i c . Hence, we may assume from the s t a r t that A i s a prime P.I. algebra w i t h an element a £ A transcendental over F We s h a l l reach a c o n t r a d i c t i o n by showing that i n t h i s reduced case A i s a l g e b r a i c over -F . Now, s i n c e A i s a prime P.I. algebra we know by ( I I , 4.1) that A has a r i n g of quotients A which i s a simple algebra f i n i t e dimensional over it's centre C > F . I f C i s shown to be a l g e b r a i c over 'F , then c l e a r l y A (and t h e r e f o r e A ) w i l l a l s o be a l g e b r a i c over F . To show that C i s a l g e b r a i c over F we l e t c e C and w r i t e c = de , d,e £ A , e r e g u l a r . Hence, ce = d w i t h d,e £ A , e j= 0 . Denoting the reduced tr a c e of an element q £ A by t r q we get from cd = e , that c t r d = £rb Claim: t h e r e . e x i s t elements d',e' £ A w i t h cd' = e' and £rd' f 0 AeA i s an i d e a l i n the prime r i n g A so. that AeA i s e s s e n t i a l Thus, AeA has a quotient r i n g A and A = C(AeA) from Posner's theorem. Therefore tx(AeA) f 0 so that A has elements whose t r a c e i s not zero. Choose e' so that e" £ AeA and t r e ' f 0 . Then e' = E x. ey. , where. x.,y. e A . S e t t i n g d' = E' x. dy. , we 1 1 1 -1 1 get cd' = e' and c = ( t r b ' ) ( t r c ' ) . • ^ Now l e t "a" be an a r b i t r a r y element of A . Then a = E a. V i 1 V w i t h the a. a l g e b r a i c over F ' . Hence t r a . i s a l g e b r a i c over F l — l & • • so that t r a = I t r a. i s a l s o a l g e b r a i c over F Section 7: Let A be a prime r i n g s a t i s f y i n g a polynomial i d e n t i t y . Then by Posner's theorem ( I I , 4.1), A i s a r i g h t and l e f t Goldie r i n g and thus any n i l subring of A i s n i l p o t e n t . This gives us, P r o p o s i t i o n 7.1: Let • A be a prime P.I. r i n g . Then any n i l subring of A i s n i l p o t e n t . This p r o p o s i t i o n allows us to give another proof f o r ( I I , 1.1) which s t a t e s that f o r P.I. rings' n i l i m p l i e s l o c a l l y n i l p o t e n t . Proof: Suppose B i s a f i n i t e l y generated subring of the n i l P..I. j r i n g A I f B i s not n i l p o t e n t we s h a l l reach a c o n t r a d i c t i o n . B j f i n i t e l y generated i m p l i e s that B n i s f i n i t e l y generated. By Zorn's lemma choose, an i d e a l P of A maximal re the e x c l u s i o n of B n ( f o r ; a l l n > 0 ). Then P i s a prime i d e a l by a standard argument, and A/P i s a .prime P.I. ring.. Since A/P i s n i l , p r o p o s i t i o n 7.1 assures us that i t i s n i l p o t e n t . But t h i s c o n t r a d i c t s A/P , being prime. Hence B must be n i l p o t e n t and the proof i s complete. Theorem 7.2: A n i l P.I. r i n g w i t h A.C.C. on l e f t a n n i h i l a t o r s i s n i l -potent. Proof: A.C.C. on l e f t a n n i h i l a t o r s i m p l i e s D.C.C. on r i g h t a n n i h i l a t o r s so that there i s an i n t e g e r n such that xA n = 0 (x e A) . Define ' j B = {x e A : xA n = 0} and l e t A = A/B . I f B = A , A i s n i l p o t e n t . Therefore, suppose B j= A Now fiB < B and hence we see that A i s a n i l r i n g i n which xA j= 0 , 1 f o r x j 0 . A l s o , A s a t i s f i e s the same i d e n t i t i e s as A . Thus, 53. i f A i s no.t n i l p o t e n t we may reduce to the case i n which xA = 0 only . i f x = 0 . \ To show th a t t h i s i s impossible we assume that A s a t i s f i e s } j the m u l t i l i n e a r i d e n t i t y f ( x i V = _Z„ % V D oeS, CF(k) Let E = {x e A : Ax' = 0} . By [22] p. 89 , ' E =f 0 . Let x = e e E From Ae = 0 we get i.e. 6 a J a a X a ( 2 ) E( E a x \ a a(2) X a ( k ) = ° > W = ° But ftr(E) < r(E) means that A/r(E) s a t i s f i e s q ( x 2 s • • • > O E S a x„ro N o a(2) a(k) k-1 Now, q i s of degree (n-1) and by i n d u c t i o n A/r(E) must be n i l p o t e n t , i . e . there e x i s t s m > 0 w i t h A m < r(E) or E.A m = 0 . But, as we showed i n i t i a l l y i n the proof, we can assume that £(A) = 0 , so that E = 0 which c o n t r a d i c t s E j= 0 . . Thus, A must be n i l p o t e n t and. the theorem i s proved. To define the Baer Lower R a d i c a l 3(A) of any r i n g A the usual procedure'is to define NP(A) as the sum of a l l the n i l p o t e n t i d e a l s of A and to o b t a i n an ascending chain of i d e a l s 5 4 . NP(A) < NP 2(A) < ... < NP m(A) < .. NP m(A) A w i t h the property that — = NP( —: ) i f m i s not a l i m i t NP m i ( A ) NP m i(A) m ' a o r d i n a l , and NP (A) = U NP (A) i f m i s a l i m i t o r d i n a l . Then 3(A) £<m _ i s defined to be the f i r s t i d e a l ; NP m(A) f o r which NP™(A) = NP™ (A) '. 2 For n i l P.I. r i n g s we see that 6(A) = NP (A) . This f o l l o w s from ( I I , 1.3) sin c e we must have 3(A) t ^ 2 ] <_ NP(A) , which means that 3(A) < NP 2(A) . But, by d e f i n i t i o n 3(A) > NP 2(A) so that 3(A) = NP 2(A) Chapter•III Section 1: An i n v o l u t i o n * of a r i n g A i s a one to one mapping of j A onto i t s e l f such that ! (1) (a + b)'' = a" + b * , (2) (ab) = b a , (3) a = a , f o r a l l a,b £ A ii- I j : j: . I I f , i n a d d i t i o n , A i s an algebra over a r i n g of operators fi , we a l s o r e q u i r e , k k (4) (aa) = aa , f o r a l l a £ fi , a e A A r i n g w i t h i n v o l u t i o n i s ' then simply a r i n g A w i t h such a mapping * . Two s p e c i a l subsets of A play an important r o l e . i n the theory of r i n g s w i t h i n v o l u t i o n . These are the symmetric elements of A and.the anti-symmetric elements of A . We denote and define these by, x} , and i -x} , r e s p e c t i v e l y . The ordinary Jacobson s t r u c t u r e theory can be modified somewhat to give.a r a t h e r n i c e theorem f o r - r i n g s w i t h i n v o l u t i o n . We s t a r t i n t h i s d i r e c t i o n w i t h a number of d e f i n i t i o n s ; S = {x £ A : x = k K = {x £ A : x = 5 6 . D e f i n i t i o n l .T; I f A i s a r i n g w i t h i n v o l u t i o n * then a r i g h t • A-module M i s s a i d to be * - f a i t h f u l i f f o r a l l a e A w i t h Ma = Ma = 0 we must have a = 0 . ! D e f i n i t i o n 1.2: A r i n g A w i t h i n v o l u t i o n * i s s a i d to be ^ - p r i m i t i v e i f there i s an i r r e d u c i b l e r i g h t A-module which i s ^ - f a i t h f u l . D e f i n i t i o n 1.3: An i d e a l U of A i s a * - i d e a l of A i f U i s an i d e a l of A and U = U Note that i f U i s a * - i d e a l then the f a c t o r r i n g A/U has an i n v o l u t i o n induced by * , defined by (a + U) = a + U . ; Since *. must take 0 to 0 , the n e c e s s i t y of having U = U i s ! evident. The i n v o l u t i o n i n A/U induced by * i s a l s o denoted by * j . D e f i n i t i o n 1.4: A " - i d e a l U of A i s c a l l e d a * - p r i m i t i v e . i d e a l of A i f the f a c t o r r i n g A/U i s ^ - p r i m i t i v e . I f M i s an i r r e d u c i b l e r i g h t A-module then we s e t R(M) = {a e A : Ma = Ma'' = 0} . I t i s c l e a r that R(M) i s a * - i d e a l of A . In a d d i t i o n , i f J(A) denotes the Jacobson r a d i c a l of the r i n g A , then by using the c h a r a c t e r i z a t i o n of J(A) i n terms of r i g h t or l e f t q u a s i - r e g u l a r i t y we see that J(A) i s a l s o a '--ideal of A I f we l e t I denote the set of i r r e d u c i b l e r i g h t A-modules then we can prove ([14]) that as long as I i s non-empty, J(A) = C\ •. R Mel To c h a r a c t e r i z e " t h e " - p r i m i t i v e i d e a l s of A we have the f o l l o w i n g . (M) 5 7 Theorem 1.5: D . i s a '"'-primitive i d e a l of A i f f U = R(M) f o r some M e l . Proof: Suppose U = R(M) . Since U i s a * - i d e a l a l l that we need to show i s that A = A/U i s a '"'-primitive r i n g . D e f i n i n g Ma = Ma f o r a l l a e A we see that M i s a w e l l - d e f i n e d i r r e d u c i b l e r i g h t A-module. To prove M i s ^ - f a i t h f u l we l e t Ma = Ma = 0 . Then by d e f i n i t i o n we have Ma = Ma = 0 , which i m p l i e s that a e R(M) = U .and thus a = 0 . Hence, M i s - ' - f a i t h f u l over A and U i s a -'-primitive i d e a l of A Conversely, suppose that U i s a ^'-primitive i d e a l of A Then, by d e f i n i t i o n , the f a c t o r r i n g A = A/U i s a '''-primitive r i n g and thus possesses an i r r e d u c i b l e r i g h t module M which i s * - f a i t h f u l . We make M i n t o an ( i r r e d u c i b l e ) r i g h t A-module by d e f i n i n g Ma = Ma Using t h i s d e f i n i t i o n i t i s c l e a r that U = R(M) so that the theorem i s proved. Now, from t h i s .theorem and the f a c t that J(A) = D R(M) M e l we have Theorem 1.6: I f A i s a (Jacobson) semi-simple r i n g w i t h i n v o l u t i o n then A i s a s u b d i r e c t sum of ^'-primitive r i n g s . The connection between p r i m i t i v e r i n g s and ^ - p r i m i t i v e r i n g s i s provided, i n Theorem 1.7: I f A i s a -'-primitive r i n g , then A i s e i t h e r a p r i m i t i v e r i n g or there i s a non-zero i d e a l . U of A such that DAD = 0 , A/U i s ( r i g h t ) p r i m i t i v e , and A/U i s ( l e f t ) p r i m i t i v e . 58. Proof: Let M be an i r r e d u c i b l e r i g h t A-module which i s ; : * - f a i t h f - u l . Suppose M i s not f a i t h f u l (otherwise A i s p r i m i t i v e ) . Define • U = {a e A: Ma = 0} . U =)= 0 sin c e M i s not f a i t h f u l , and^ thus M i s an i r r e d u c i b l e f a i t h f u l r i g h t A/U-module, i . e . A/U i s r i g h t p r i m i t i v e . Now, M = A/J , where J i s a maximal modular r i g h t i d e a l , k >'< Set M' = A/J , and J i s a maximal modular l e f t i d e a l . Claim: u" = {a e A: aM' =0} k k k For, i f aM" =0 , or aA <_ J , then Aa <_ J , i . e . a e U k k k k k or a e U . Now l e t a e U . Then a e U , Ma = 0 , A a ^ _ J , k aA <_ J , aM' = 0 i . e . a e {a e A : aM" = 0} k 1 j' . Hence M' i s a f a i t h f u l i r r e d u c i b l e l e f t (A/U )-module. I F i n a l l y , i f a e U f\ U* , then Ma = 0 and aM' = 0 . k k k k From aM' =0 we have aA _< J , Aa <_ J , Ma = 0 . Thus Ma = Ma =0 I and s i n c e M i s ^ - f a i t h f u l , we have a = 0 and the proof i s complete. Secti o n 2: In an attempt to g e n e r a l i z e theorems of P.I. r i n g s to P.I., r i n g s w i t h i n v o l u t i o n , many w r i t e r s have assumed only that the symmetric elements S s a t i s f y a polynomial i d e n t i t y . However, l a t e l y Amitsur [12] has shown that any r i n g w i t h i n v o l u t i o n whose symmetric elements s a t i s f y a polynomial i d e n t i t y must of n e c e s s i t y i t s e l f s a t i s f y a polynomial i d e n t i t y and i n f a c t , i t s a t i s f i e s a r a t h e r n i c e form of i d e n t i t y although the one drawback so f a r i s that no bound has been obtained f o r i t s degree. j We begin the proof of t h i s r e s u l t w i t h a study of the n i l ; subsets of a r i n g A w i t h i n v o l u t i o n * One problem that we must deal w i t h i s the f a c t that a homomor-phism which preserves the i n v o l u t i o n * i n A may w e l l have the properity that i t s image possesses more symmetric elements than those a r i s i n g from 5 9 . the symmetric elements of A So that although the symmetric elements of A may s a t i s f y a polynomial i d e n t i t y , a l l the symmetric elements of' some homomorphic image A/P may not s a t i s f y t h i s i d e n t i t y . However, we s h a l l p r i m a r i l y be i n t e r e s t e d i n symmetric elements of .the forms a + a and ab + ba f o r some a,b e A , and i t i s easy to check that these types of symmetric elements i n A/P always a r i s e from the same types of symmetric elements i n A . Thus, when we speak of S now, we s h a l l mean * * * S = {a + a , a b + b a : a,b e A} , and t h i s new S i s a subset of the o r i g i n a l s e t . S of symmetric elements of A . Thus, we are r e a l l y r e q u i r i n g much l e s s w i t h t h i s new d e f i n i t i o n f o r S . ' [ V Lemma 2.1: Let A be a r i n g w i t h i n v o l u t i o n * whose symmetric •• -elements S s a t i s f y a polynomial i d e n t i t y f [ x ] = f ( x ^ , . . . , x ^ ) = 0 Then i f P i s a two-sided i d e a l , i n A and U i s a subset of A " m d such that U = U and U <_ P f o r some i n t e g e r m , we must have U generates a n i l p o t e n t i d e a l i n A/P - . i The proof of t h i s lemma can be found i n Amitsur [12]. With t h i s r e s u l t we may now e s t a b l i s h Theorem 2.2: - I f A i s a r i n g w i t h i n v o l u t i o n * whose symmetric i ! i; • 1 elements S s a t i s f y a polynomial i d e n t i t y . f [ x ] = 0 , then the n i l ' ! 1 r a d i c a l N(A) of A i s equal to g(A)- , and 3(A) d < NP(A) Proof: Since the i d e a l N(A) has the property that A/N(A) has no 60. \ n i l p o t e n t i d e a l s and si n c e 3(A) i s the s m a l l e s t i d e a l of .A w i t h t h i s property, we must have 3(A) <_ N(A) To prove•the other d i r e c t i o n we f i r s t e s t a b l i s h that ' S ' A N ( A ) _< 3(A) . Suppose 0 f s e S / l N(A) , but s £ 3(A) . Since I s i s n i l p o t e n t we may assume s e N(A) , s i . g (A) but s ie 3(A) . ' For each x e A , sx + x s i s n i l ( s i n c e i t i s i n N(A) ) and symmetric. d+1 * d By the previous lemma 2.1, (sx + x s) e 3(A) . Then (:sx;) / , • * \ d n f » \ TT . s A + 3(A) 1 _ . . . . (sx + x s) sx e 3(A) . Hence, i f — g ( A ) — ~ ~ ' , i t i s a n i l r i n g sA + 3(A) ,j; of bounded index. I f „ ,,. i s a n i l p o t e n t i d e a l of • > 3(A) v 3(A) 1 d sA + 3(A) theorem 5 i n Amitsur. [9] t e l l s us that _ (sA) <_ N P ( — ^ ) . For each T , TsA i s a r i g h t n i l p o t e n t i d e a l modulo 3(A) and hence TsA + 3(A) TsA < g(A) f o r , otherwise, . . i s a non-zero n i l p o t e n t r i g h t — g (.A; i d e a l of g^A)" " Thus (sA)^+± <_ 3(A) . This means that s generates a n i l p o t e n t i d e a l modulo g(A) , c o n t r a d i c t i n g , t h e f a c t s that s £ 3(A) A and .,,, has no n i l p o t e n t i d e a l s . This c o n t r a d i c t i o n proves that 3 (A) S n N(A) < 3(A) The f i n a l step i s to show that N(A) <_ g (A) . Let x e N(A) Then, from N(A) = N(A) we have (x t + t x) e N(A) f o r any t e A Suppose x £ g(A) . From x t x = (x t + t x ) x = 0 modulo g(A) and also, from x = -x modulo g (A) we get ( x A ) 2 = -x. AxA •= 0 modulo . 3(A) , c o n t r a d i c t i n g the f a c t that TT4T ^ a s n o n i l p o t e n t i d e a l s . 3 (A; Hence, N(A) = 3(A) To e s t a b l i s h the second h a l f of the theorem we l e t u1,u„,...,u, e 3(A) and define U = {u. , . . . ,u ,,u. , . . . ,u,} , a f i n i t e \ 1 Z d. 1.. d l d 61. set w i t h U = U . Now, 8(A) i s l o c a l l y n i l p o t e n t and hence U™ = 0 f o r some i n t e g e r m . By the preceding lemma, <_ NP(A) . In p a r t i -c u l a r , the product u nu_...u, e NP(A) . Since the u. are a r b i t r a r y 1 2. d l from 3(A) , we have proved 3 ( A ) ^ <_ NP(A) as re q u i r e d . Turning now to.the p r i m i t i v e images•of A we s h a l l prove that these always possess a minimal l e f t i d e a l . F i r s t we r e q u i r e a more general r e s u l t . Suppose A i s ' a n a r b i t r a r y r i n g w i t h i n v o l u t i o n * and l e t ? be a p r i m i t i v e i d e a l of A such, that A/P i s an i r r e d u c i b l e r i n g of endomorphisms of a vect o r space V over a d i v i s i o n r i n g D . Then, Amitsur [12] has proved Lemma 2.3: One of the f o l l o w i n g holds f o r A ; ^ has a minimal l e f t i d e a l . (2) there i s a f i n i t e - d i m e n s i o n a l D-subspace W <_ V , such that the l e f t i d e a l L = (0 : W) = {a : aW = 0} s a t i s f i e s L < P . (3) f o r every f i n i t e - d i m e n s i o n a l U <_ V , and every v £ V , [S n (0 :• U)]v £ U + vD . Using t h i s lemma we examine our p a r t i c u l a r case i n Lemma 2.4: I f A i s a r i n g w i t h i n v o l u t i o n * whose symmetric elements s a t i s f y a polynomial i d e n t i t y , then every p r i m i t i v e image A of A has \^  • a minimal l e f t i d e a l . ^ Proof: A l l we need to do i s to dispose of cases (2) and (3) o c c u r r i n g ! n .62. i n t he p r e v i o u s lemma. Suppose (1) does n o t h o l d b u t (2) doe s . Then, s i n c e A has no m i n i m a l l e f t i d e a l , (V : D) , t he d i m e n s i o n o f V o v e r D , must be i n f i n i t e . S i n c e (W : D) i s o n l y f i n i t e i t i s p o s s i b l e t o choose w , w , , . . . , w , , w h i c h a r e D - i n d e p e n d e n t and n o t i n W . By t h e d e n s i t y o 1 d theo rem choose t ^ . . . , ^ e L = (0 : W) s u ch t h a t t/w = 0 , f o r i =r i b u t t .w. = w. . J J 3-1 - • . .. From L < P we have t . e P and thu s s . = t . + t . = t . — . X X X .1 - 1 modulo P , w i t h the r e s u l t t h a t 0 = f [ s 1 , . . . , s d ] w d f [ t l , . . . , t d ] w d atnt„ . . . t , w , ( f i s m u l t i l i n e a r , as u s u a l ) ± z d d ctw . , ct i 0 o % = 0 , a 4= 0 i m p l i e s w = 0 w h i c h c o n t r a d i c t s t he D - i n d e p e n d e n c e ^ ^ . o 1 o o f t he w. . Hence , (V : D) must be f i n i t e and A mus t have a I i . m i n i m a l l e f t i d e a l . ' Suppose ca se (3) h o l d s . We choose w e V , s d ' s d - l ' ' * " ' S l e ^ w i t h the p r o p e r t i e s t h a t i ) w , s , w , s , . s ,w, . . . ,s. s, s,w a r e D - i n d e p e n d e n t d d-1 d k k+1 d i i ) s . ( s . s . , , . . . s ,w) = 0 f o r j > i j i l + l d • — The method o f c h o i c e i s t he f o l l o w i n g ; F i r s t choose w f 0 f r o m V . S e t t i n g U = 0 i n (3) o f ( I I I , 2.3) g i v e s us Sw wD . Choose s^ e S so t h a t s d w £ wD w h i c h means t h a t s d w i s D - i n d e p e n d e n t o f w . H a v i n g chosen s d > . , . 9 s ^ s a t i s f y - n g 63 i ) , i i ) we set U = wD + s^wD + ... + s ^ i ^ - ^ • • • s d w D • Then by (3) i n ( I I I , 2.3), s i n c e [S n (0 .: U) ] ( s k s k + 1 ... s dw) i U + s k s k + 1 . .. s dw , we can choose s, ., e S <^  (0 : U) such that s, ,s, s, , . . . s ,w A k-1 k-1 k k+l d T •U + s k s k + ^ s ^ w • Chosen i n t h i s way, the s^ s a t i s f y the'required c o n d i t i o n s i ) and i i ) ' . Now, s u b s t i t u t i n g these s^ i n t o f [ x ] we get 0 = f [s.,,. .. ,s ,] w = a s . s . . .. s ,w , and thus 1 a ± Z a s 1 s 2 . . . . s^w = 0 , which c o n t r a d i c t s the D-independence of c o n d i t i o n i ) . Hence, (3) cannot h o l d . Thus, A = A/P has a minimal l e f t i d e a l . With these p r e l i m i n a r y lemmas we can now e s t a b l i s h our two main theorems. S e c t i o n 3: The f i r s t r e s u l t has a r a t h e r long proof which can be found i n Amitsur [12]. Much of the messy work has been i n c o r p o r a t e d i n t o t h i s f i r s t theorem and thus the proof of the second theorem becomes as short . i as i t i s b e a u t i f u l . i Theorem 3.1:, Let A be a r i n g w i t h an i n v o l u t i o n * whose symmetric elements S s a t i s f y f [ x , , . . . , x , ] = 0 . Then -TTTV s a t i s f i e s an ' I ' d 8(A) 64. A i d e n t i t y of degree <_ 2d ; and ^ p ^ ) s a t i s f i e s an i d e n t i t y of degree <_ 2 d 2 . " A A c t u a l l y , Amitsur proves a b i t more, namely, that ^ s a t i s f i e s the standard i d e n t i t y of degree 2d ', i . e . S ^ t x ] = 0 By ( I I I , 2.2) 6(A) d£NP(A) so that i n ^jjy we have (S„,[x]) d = 0 s a t i s f i e d . 2d Theorem 3.2: I f A i s a r i n g w i t h i n v o l u t i o n * , such that i t s set S of symmetric elements- s a t i s f i e s a polynomial i d e n t i t y of degree d , then A s a t i s f i e s an i d e n t i t y (S^^Cx])™ = 0 f o r some i n t e g e r m Proof: Consider the d i r e c t product A = .IIA , where a ranges over a 2d the s et A of a l l 2d-tuples ( a . , . . . , a 0 ,) of elements of A and 1 la it A where A = A f o r a l l a. . By s e t t i n g (9 (a) = [0(a)] f o r each ct a , we make A i n t o a r i n g w i t h i n v o l u t i o n . The symmetric elements of A are j u s t those elements whose components are symmetric elements of " A . Since the symmetric elements of each A a s a t i s f y f [ x ] = 0 then so do the symmetric elements of A. . Hence, by ( I I I , 3.1) we have S 2 d(£ l 5...,£ 2 d) e 8(A) , f o r any 6K E A . B y ( I I I , 2.2), 8(A) = N(A) and thus . S 2 d ^ l ' ' * ' '^2d^ ± S n i l P ° t e n t » s a y [ S ? d ( 6 ' 1 , • • • ,0-2d) ] m = 0 f o r s integer, m depending on the $. , and the $. can be a r b i t r a r y elements of A . Now choose 2d p a r t i c u l a r elements of A , namely, choose ~ th IL e A so that IL p i c k s out the i component of each a = (a , . . . , a 2 ) i . e . IK (a) = . Do .this f o r i = 1,2 2d Then, f o r these II. -we have S_ , [n . ] m = 0. , f o r a f i x e d m l 2d l 65. Hence S O J [ n . ] m (a) = 0 Zd 1 Therefore 1 S„ , [a_, . . ., a„ = 0 . Since t h i s i s true f o r a l l Zd 1 Zd ct = (a., , . .., a„ ,) we have proved that [S„ , (x. , . . . ,x„ ,) ] m = 0 holds i n 1 Zd Zd 1 Zd A and the proof i s complete. Amitsur f u r t h e r shows that the i n t e g e r m i s bounded by some number depending on f [ x ] and not on A I f t h i s were not so, then m. there would be r i n g s A_^  s a t i s f y i n g f [ x ] = 0 and ^ ^ [ x ] = 0 with-minimal m. and m. < m„ < ... . But then HA. would s a t i s f y f = 0 i , 1 2 . I J but none of the ^ ^ [ x ] 1 1 1 = 0 In a more recent paper [13], Amitsur has f u r t h e r g e n e r a l i z e d t h i s l a s t theorem w i t h the f o l l o w i n g r e s u l t ; Theorem 3.3: I f A s a t i s f i e s a polynomial i d e n t i t y of the form c^, .. .,x^_ ; x^, k k f ( x ,...,x ; x ,...,x ) = 0 of degree • d , then A s a t i s f i e s an i d e n t i t y S 2 d [ x ] m = 0 . I f A i s semi-prime then m = 1 This e x t e n d s ' ( I I I , 3.2), s i n c e i f the symmetric elements of i A s a t i s f y a polynomial i d e n t i t y f(t^,...,t^) = 0 , then A must a l s o s a t i s f y the i d e n t i t y f ( x n + x, x, + x.) = 0 which has the same form 1 1 d d as the i d e n t i t y given i n the theorem. Amitsur proves t h i s r e s u l t by r e f i n i n g the lemmas used to e s t a b l i s h ( I I I , 3.2) and thus many of the proofs are s i m i l a r to the proofs we have obtained already. One of the lemmas, however, has some independent i n t e r e s t . I t s t a t e s that i f A i s a p r i m i t i v e r i n g w i t h i n v o l u t i o n * s a t i s f y i n g a polynomial i d e n t i t y •k f [ x ; x ] = 0 of degree d , then A i s a complete r i n g of l i n e a r transformations of a ve c t o r space of dimension l e s s than or equal to d . H e r s t e i n [24] and M a r t i n d a l e [33] have proved s p e c i a l cases; of ( I I I , 3.2) and.,in these s p e c i a l cases a bound can be given to the degree of the i d e n t i t y which A must s a t i s f y . Herstein's Theorem 3.4: I f A i s a simple'ring with i n v o l u t i o n * of c h a r a c t e r i s t i c not 2, whose symmetric elements s a t i s f y a polynomial i d e n t i t y of degree d over the centroid of A , then. A i s of dimensi 2 : at most 4d over i t s centre, i . e . A s a t i s f i e s a polynomial i d e n t i t y of degree at most 4d over i t s centre. Martindale, using some of Herstein's. techniques, generalizes j Herstein's r e s u l t as follows; Theorem 3.5: If A i s an algebra with i n v o l u t i o n * containing no non-zero n i l p o t e n t i d e a l s , whose symmetric elements s a t i s f y a polynomial i d e n t i t y of degree d , then A s a t i s f i e s a standard i d e n t i t y of degree at most 4d .. Section 4: At the end of the l a s t s e c t i o n we stated a p a i r of r e s u l t s which were r e a l l y j u s t s p e c i a l cases of Amitsur's theorem ( I I I , 3.2). These theorems were better, however, i n the sense that we obtained an upper bound f o r the degree of the i d e n t i t y s a t i s f i e d by the whole r i n g . In t h i s section we continue i n this d i r e c t i o n by proving a few theorems i n which assumptions made previously on the e n t i r e r i n g are now made only on S . As a consequence, we get s p e c i f i c upper bounds f o r dimensions and for the degrees of polynomial i d e n t i t i e s . Lemma 4.1: Suppose A i s a p r i m i t i v e , r i n g with i n v o l u t i o n * whose symmetric elements S s a t i s f y a polynomial i d e n t i t y of degree d 67. Then A has at most d orthogonal non-nilpotent symmetric elements, Proof: Let s , ...,s ^ be (d+1) orthogonal non-nilpotent symmetric elements. We s h a l l reach a c o n t r a d i c t i o n . P r i m i t i v e rings are also prime rings so that given non-zero elements a,b E A , we have aAb f= 0 . More generally, i f a^,...,an are any non-zero elements of A we must have a,Aa_A ... a .. Aa f 0 1 2 n-1 n 1 Hence, we have 52 A 51 A"2" ••• "d""d+l 2 2 2 2 s,As„A ... s ,As • , r 0 , and thus, there e x i s t elements a^,...,a d from A with the property that s ^ a ^ s 2 ^ .. • s^ ad^d+l ^ ^' ^ * Now, for each i = l , . . . , d , s e t u. = s . a . s . , n + s . , n a . s . ' l x.x x+1 x+1 1 1 and l e t f(x, x,) = ctx..x_...x, + I B(a)x ... x be the polynomial 1 d 1 2 d c f l °1 ad • • " ":' i d e n t i t y s a t i s f i e d by S . Since u^ e S for a l l i , we have f ( u , u.) = ct u u u, + E 3(a)u ...u =0 1 d 1 I a j _ _ a n a a#l But, using the f a c t that the s. are orthogonal and non-nilpotent we x can deduce that i f u^u_.u^ =f= 0 then we must have e i t h e r j = i + 1 and k = i + 2 , or else j = i - 1 and k = i - 2 . Therefore, f(u,,...,u' ) = au,u„...u. + gu.u. . .. .u, = 0 1 a 1 2 d a a - l 1 Substituting i n the expressions f o r the u^ we observe that 68. [ a s 1 a 1 s ^ a . s ^ a 0 . . . S j a , s , _ L 1 + gs a.s^a* 2 * , _ n 1 1 2 2 3 3 d d d+1 d+1 d d d-1... s^a^s^J - 0 Mu l t i p l y i n g this • on the l e f t by s^ and on the r i g h t by s w e obtain by the orthogonality of the , that 2 2 2 2 as.a,s„...s ,a,s .,. =0 , from which 1 1 2 d d d+1 S l a i S 2 - - - S d a d S d + l = " ° ' W h ± c h contradicts (1), thus proving the theorem. We now prove an analogue of Kaplansky's theorem i n which we put conditions on the set S of symmetric elements instead of on A Theorem 4.2: Suppose A i s a p r i m i t i v e r i n g with i n v o l u t i o n * such that S s a t i s f i e s a polynomial i d e n t i t y of degree d and i s algebraic 2 over 0, . Then A i s a simple algebra of dimension 4d over i t s i centre Z Proof: Let f (x.,, . . . ,x ,) be the m u l t i l i n e a r i d e n t i t y s a t i s f i e d by 1 d A . Case 1: Z S We can f i n d a n o n - t r i v i a l element t e Z A. K merely by taking i ' * an element z e Z with z £ S , and s e t t i n g t = z - z . Accordingly, i f we choose t e Z C\ K and k e K we have . ( t k ) " = k t = (-k)(-t) = kt = tk , 69. so that tk e S . Hence, f o r a l l k^ £ K and a l l e S we have, f ( t k ,s 2,...,s ) = 0 , or t f (k.,, s„ , .. . ,s ,) = 0 , s i n c e t e Z 1 2 d By the p r i m i t i v i t y of A , f ( k , s 2 > . . . , s ^ ) = 0 , f o r a l l k^ £ K and a l l s^ £ S . Any a e A can be w r i t t e n a = s + k , where s e'S , k £ K . Then, f ( a , s 2 , . . . , s ) = f ( s + k,s 2,...,s d) f ( s , s 2 , . . .,s ) + f ( k , s ,...,s d) = 0 , f o r a l l a £ A and a l l s. £ S . Continuing i n t h i s f a s h i o n , we deduce that f (a.,, . .., a,) = 0 x 1 d i f o r a l l a d £ A . Kaplansky's theorem then completes the proof. i Case 2: Z <_ S . By ( I I I , 4.1), l e t E = {e ,...,e } be a maximal set of orthogonal symmetric idempotents. By [14] theorem 5, e ] / - e ] _ i s e i t h e r a quaternion algebra or a d i v i s i o n algebra. Then [14] theorem 6 says that e^Ae^ i s f i n i t e dimensional over i t s centre, and thus [A : Z] i s a l s o f i n i t e . Remark 3 of [14] t e l l s us that A = , where e i t h e r D = Z 2 or D i s a d i v i s i o n algebra over Z possessing a maximal sub-f i e l d M w i t h ' [M : Z] f i n i t e . In the former case, q <_ d i m p l i e s that 2 2 [A : Z] = 4q < . 4d . I n the l a t t e r case,. * induces an i n v o l u t i o n xn A © M = M. , f o r some j . Remark 3 of [14] again and ( I I I , 4.1) z j •• ' 2 2 [ a p p l i e d to M ] t e l l us that j <• 2d so that [A : Z] < j < 4d 70. and . the proof i s complete. j The f o l l o w i n g theorem i s a r e s u l t f o r ''^-primitive r i n g s which i s analogous to that j u s t proved f o r p r i m i t i v e - r i n g s . This w i l l permit us to e s t a b l i s h f o r semi-simple r i n g s w i t h i n v o l u t i o n a c o r o l l a r y s i m i l a r to the c o r o l l a r y of (I, 2.16). Theorem 4.3:, Suppose A i s a '''-primitive r i n g (and not a p r i m i t i v e , r i n g ) such that S s a t i s f i e s a- polynomial i d e n t i t y f of degree d , X and i s a l g e b r a i c over Q . Then A = U © U , where D^A and U d 2 i i s a simple algebra of dimension <_ — over i t s centre. ; Proof: By (III, 1.7). there i s an i d e a l U |= 0 w i t h U r> = 0. k — k and A , A/U ' p r i m i t i v e r i n g s . . S e t t i n g A = A/U' we see that U i k ' U = ^ — 7 i s a non-zero i d e a l of A/U . Let u,,...,u, e U « 1 a. Noting that u. = u. + u. we have that — — k k f ( u , ...,u ) = f ( u 1 + u 1,...,u d + u d) = 0 . Kaplansky's theorem then says that U must be a simple algebra of dimen d 2 - - * s i o n < — r over i t s centre. Hence, U = A and A = U © U , x^here •;. ~ 4 2 , * d U = A/U i s a simple algebra of dimension <_ over i t s centre. Our f i n a l r e s u l t , a c o r o l l a r y to 4.2 and 4.3 reads, C o r o l l a r y : Suppose A i s a semi-simple r i n g w i t h i n v o l u t i o n * such that S s a t i s f i e s a polynomial i d e n t i t y of degree d and i s • a l g e b r a i c over ft . Then A s a t i s f i e s a standard i d e n t i t y of degree < 4 d 2 + 1 . '. .. 71. .Section 5: For P.I. r i n g s we have answered the Kurosh problem i n the a f f i r m a t i v e . For r i n g s w i t h i n v o l u t i o n we can do a b i t b e t t e r . I f i n such a r i n g only the elements of S s a t i s f y a polynomial i d e n t i t y and are .algebraic over F , we can prove that the r i n g i s l o c a l l y f i n i t e , j In view of Amitsur's r e s u l t s we can a u t o m a t i c a l l y assume that the r i n g A i t s e l f s a t i s f i e s a polynomial i d e n t i t y . Therefore, the problem i s reduced to showing that a P.I. algebra A w i t h i n v o l u t i o n * , whose symmetric elements are a l g e b r a i c over F , i s l o c a l l y f i n i t e over F We note f i r s t that s i n c e S i s a l g e b r a i c over F then from 2 the f a c t that k e S f o r every k e K we must a l s o have K a l g e b r a i c over F . A l s o , assuming that A i s not of c h a r a c t e r i s t i c 2 we can w r i t e any element a e A as a = s + k f o r s e S and k e K Hence, every element of A i s a sum of elements a l g e b r a i c over F so [ that the l o c a l f i n i t e n e s s of A f o l l o w s t r i v i a l l y from P r o c e s i ' s theorem ( I I , 6.2). Chapter IV This chapter t r e a t s Jacobson (P.I.) r i n g s and H i l b e r t (P.I.) algebras, and t h e i r r e l a t i o n to the noncommutative form of the H i l b e r t N u l l s t e l l e n s a t z . I f A i s a P.I. r i n g we s h a l l have occasion to pass from A to some homomorphic image of A , say A/B . I f the r i n g of c o e f f i c i e n t s fi of A i s general enough, then ft may induce only zero operators on A/B so that A/B may not be a P.I. r i n g . However, i f fi i s a f i e l d or an i n t e g r a l domain we have shown ( I , 4.9) that every. • m I P.I. algebra s a t i s f i e s an i d e n t i t y of the form ^ ^ x ) = 0 and t h i s i d e n t i t y i s c e r t a i n l y i n h e r i t e d n o n - t r i v i a l l y by non-zero homomorphic images of A . Hence, i n order to g e n e r a l i z e t h i s to more a r b i t r a r y c o e f f i c i e n t r i n g s we s h a l l demand throughout t h i s chapter that A s a t i s f y a proper polynomial i d e n t i t y , where we define an i d e n t i t y f ( x ^ , . . . , x d ) = 0 to be proper i f f o r every element .0 f a e A , there i s at l e a s t one c o e f f i c i e n t a E fi from the i d e n t i t y f w i t h aa. =f= 0 . Amitsur has shown that i f A s a t i s f i e s such, an i d e n t i t y then A s a t i s f i e s an i d e n t i t y w i t h c o e f f i c i e n t s +1 and t h i s i s s u f f i -c i e n t to ensure t h a t homomorphic images of A are P.I. r i n g s . Of course, proper polynomial i d e n t i t i e s a u t o m a t i c a l l y r u l e out t r i v i a l i d e n t i t i e s l i k e 2x = 0 i n a r i n g of c h a r a c t e r i s t i c 2 . However, i f fi', i s a f i e l d then every non-zero i d e n t i t y i s proper. A l s o , i f the c o e f f i c i e n t s of f generate fi as an i d e a l ( fi • i s u s u a l l y assumed to be commutative) then f. • i s a proper i d e n t i t y . 73. S e c t i o n 1: D e f i n i t i o n 4.1: A r i n g A i s c a l l e d a Jacobson r i n g i f every prime'., i d e a l of A i s the i n t e r s e c t i o n of a f a m i l y of maximal i d e a l s of A • . D e f i n i t i o n 4.2: An algebra A over a f i e l d F i s a H i l b e r t algebra i f f o r every maximal i d e a l M of A , A/M i s a f i n i t e dimensional algebra over F j Let F be a f i e l d and as usual F [ x n , . . . , x ] =. F[x] the ± n free algebra over the f i e l d F generated by the n noncommutative indeterminates x ,...,x n . Let H be a f i e l d c o n t a i n i n g F and suppose Q <_ F[x] . I f denotes the set of a l l r x r matrices with, e n t r i e s from H we define Z(Q,H ) = {(A) = (A n,...,A ) : A. e H and r 1 n l r q(A ,...,An) = 0 f o r a l l q E Q} Now' Z(Q,H r) can be i d e n t i f i e d w i t h {f : i s a homomorphism from F[x] i n t o f o r which ¥(Q) = 0} , which i s the same as : F ^ X ^ — » : where <Q> i s the i d e a l of F[x] generated by Q} A polynomial f [ x ] e F[x] vanishes f o r a l l zeroes of Q i n H i f f [ ( A ) ] = f [ A A ] = 0 f o r a l l (A) i n Z(Q,H ) , or, e q u i v a l e n t l y , I n r Z(Q,H^) <_ Z(f,H^) . I f S i s a set of n-tuples of m a t r i c e s , i . e . S <_ we l e t A[S] = {f e F[x] : S _< Z(f,H )}.. Hence, AtZCQ.H^)] i s . the i d e a l of a l l polynomials which vanish at a l l the zeros of Q I f F i s the a l g e b r a i c c l o s u r e of F then A[Z(Q,H )] i s 74. a c t u a l l y determined by what we c a l l the a l g e b r a i c zeros of Q i n F This the context of Theorem 1.3: 'A[Z(Q,F )] < A[Z(Q,H )] and e q u a l i t y holds i f E <_ F . [10] We now prove a r e s u l t which i s equ i v a l e n t to the noncommutative form of the H i l b e r t N u l l s t e l l e n s a t z : Theorem 1.4: Let Q o F [ x ] and suppose that the quotient r i n g R = F[x]/Q s a t i s f i e s a polynomial i d e n t i t y of degree d . Then i f f i s a polynomial i n F[x] the f o l l o w i n g are e q u i v a l e n t : (1) f e P. f o r every prime i d e a l P >_ Q (2) 1 e J(R) , where 1 = f + Q £ R and J(R) i s the Jacobson r a d i c a l of R (3) f o r every p a i r of i n t e g e r s m,r such that m >_ y >_ r , and any.set of polynomial u ,...,u i n F[x] , the 1 m polynomial u. fu„f...u f E A [ Z ( Q , F )] i . e . vanishes f o r 1 2. m r . a l l a l g e b r a i c zeroes of Q Proof: (1) => (2): Let T be the i n t e r s e c t i o n of a l l the' prime i d e a l s P which c o n t a i n Q'" . Then T/Q i s the Baer lower r a d i c a l g(R) of R and s i n c e g(R) <_ NL(R) <_ J(R) we are done. (2) => ( 3 ) : L e t ¥ be a homomorphism of R i n t o F^ — — i i ^ w i t h ^(x.) = A. E F and l e t A. = (a.,) . Then the a., are algebrai'c i i r i j k j k \ over F and thus ¥(R) " i s a subalgebra over F of a matrix algebra over a f i n i t e extension of F . Hence, y(R) i s f i n i t e dimensional over F , and ' v ( J ( R ) ) i s a n i l r i n g ( n i l subalgebra.of F ) so that i t i s n i l p o t e n t of index < r . I f m > r we have ¥(unf^...u f) £ — — 1 1 m ¥ ( J ( R ) ) m = 0 . • • ' : 75 Now, every (A^) e Z(Q,¥^) determines a homomorphism <f> : F[x] -> F^ defined by. $ (x^) = A^ and c l e a r l y $(Q) = 0 so $ induces a homomorphism of R i n t o F which i m p l i e s that u,f...u f r 1 m vanishes at (A^) ' and the proof i s completed.' Note that we have not as yet used the f a c t that R s a t i s f i e s a polynomial i d e n t i t y . We do, however, i n the f o l l o w i n g ; (3) => (1): Let P be a prime i d e a l of F[x] c o n t a i n i n g Q . The n a t u r a l p r o j e c t i o n of F ^ -> F ^ i m p l i e s that F ^ F[x] a l s o s a t i s f i e s a polynomial i d e n t i t y of degree d . But — — i s n i l semi-simple so we can embed ( I I , 2.3) i t i n a subdirect- sum of matrix (a) (c t ) r i n g s £ K , s , where K , , i s the set of a l l r ( c t ) * r(a) matrices a r ( c t ) r(a) (a) w i t h e n t r i e s from the f i e l d K >_ F Note a l s o that r ( c t ) <_ [d/2] For each a we have the map y . z KY . K ^ ct Q P r ( y ) r(a) and -V gives a zero y (x.) = A of Q i n nS°t\ . Now, by (IV, 1 ct & ct l a ± r(-a> — (rv) A[Z(Q,F r)] <_ A [ Z ( Q , K ^ y ] . By our assumptions from ( 3 ) , u^.-.u^f e A[Z(Q,K <-" )0] • Hence. ¥ ( u n f . . . u f) = ( u , f . . . u f) [A ] = 0 . Thus r(,a/ a 1 m l m c t ^ F[x] u,f...u f e Ker Y f o r every a but s i n c e „ i s a s u b d i r e c t sum 1 m ct P we must have u,f...u f e P 1 m D e f i n i n g T =HP , where the i n t e r s e c t i o n runs over a l l prime i d e a l s of R c o n t a i n i n g Q then T/Q = $(R). so that by s e t t i n g m = [d/2] we see that f generates a n i l p o t e n t i d e a l of index < [d/2] 76 But T/Q = (T/Q) ^ a s n 0 n i l P o t e n t i d e a l s . Hence, f e T and we are through. C o r o l l a r y : (a) f i n i t e l y generated algebras over a f i e l d F which s a t i s f y polynomial i d e n t i t i e s are Jacobson r i n g s . (b) f i n i t e l y generated p r i m i t i v e algebras over F which s a t i s f y polynomial i d e n t i t i e s are simple algebras of f i n i t e dimension over F . (c) i f f [ x ] vanishes f o r a l l zeroes i n IP of an i d e a l Q , then f o r some i n t e g e r u and some q e Q > t h e , r e l a t i o n f V [ x ] - q[x] = 0 holds i d e n t i c a l l y i n F^ This l a s t p a r t i s e q u i v a l e n t to the H i l b e r t N u l l s t e l l e n s a t z ; (c') L e t S .= (E±., ) be n generic m a t r i c e s , i . e . the e n t r i e s E^, are j k to j k commutative indeterminates over F. I f F[x] vanishes f o r a l l zeroes of Q i n .F , then f y [ S ] belongs to the i d e a l i n F[S] generated by a l l q[5] , q e Q . Proof: I f A i s a f i n i t e l y generated algebra over F , say A = < a , ,...,a> , the mapping F[x,,...,x ] A defined by x. -y a. I n r ° 1 n i i i s onto A . Hence, every f i n i t e l y generated algebra over F i s isomor-F [ x l p h i c to a r i n g —iz-^L = R . Every p r i m i t i v e image of R i s . a p r i m i t i v e r i n g s a t i s f y i n g a polynomial i d e n t i t y and hence by Kaplansky's theorem i s a simple r i n g . Thus, the p r i m i t i v e i d e a l s of R are maximal so that' J(R) i s the i n t e r s e c t i o n of a l l the maximal i d e a l s of R . From the equivalence of (1) and (2) of the theorem we see that B(R/S) = J(R/S) .1 f o r every homomorphic image R/S of R . Hence, R i s a Jacobson r i n g . i F T x l For'(b) p a r t , suppose 0 =j= R = i s a p r i m i t i v e algebra over F s a t i s f y i n g a polynomial i d e n t i t y . Then, of course, R i s a 7 7 . 2 2 c e n t r a l simple algebra of dimension m < [d/2] over..its centre C L e t t i n g C be the a l g e b r a i c c l o s u r e of C we have that R < C ° . ° • • — m The mapping * : F [ x ] + ^ l - * . C m gives a zero 0 ( x . ) ) e Z(Q,C ) of Q . Hence, Z(Q,C ) 4= (j) and l m m 1 . then (IV, 1.3) assures us that' Z(Q,F ) T (j) , f o r otherwise we have m ' A[Z(Q,C )] = A[Z(Q,F ) = F[x] , m m a c o n t r a d i c t i o n s i n c e 1 e J(R) by (IV,. 1.4) i m p l i e s that R = 0 . \ So l e t (A.) e Z(Q,F ) . D e f i n i n g ¥': F[x] -> F by . .. i m m , • " F[x] — " ¥(x-.) = A. induces a homomorphism from R = — — i n t o F . R i i Q m simple means that t h i s map i s an isomorphism so that y'(R) ^  F^ , and s i n c e V ( R ) i s f i n i t e l y generated over F i t must be of f i n i t e dimension over F ! i For (c) p a r t , l e t be the i d e a l of a l l i d e n t i t i e s s a t i s f i e d by F . Then ~n^A 1 S a f i n i t e l y generated r i n g and s r -~ a t i s f i e s a l l S . I i d e n t i t i e s of F^ . I f now f [ x ] e A(Z,F r) then (IV, 1.4) i m p l i e s that f [ x ] e J(R) and by (a) f i s n i l p o t e n t modulo <Q,M> . fi For (c") p a r t , the map f : • F ^ - •+ F[5] defined by ^(x j L) = r i s an isomorphism s i n c e g[x] e M , i . e . vanishes i d e n t i c a l l y i n F^ i f f g[S] = 0 . Parts (a) and (b) seem to r e q u i r e that the algebra have an i d e n t i t y but t h i s i s not the case s i n c e : (1) I f R i s a f i n i t e l y generated algebra over F •, then s'c R , the algebra obtained by a d j o i n i n g the f i e l d F i s a l s o f i n i t e l y :'78. generated w i t h J(R) = J(R*) and g (R) = g (R " ) . I f R s a t i s f i e s the * I i d e n t i t y f t x ^ . . . , x ] =. 0 • , R s a t i s f i e s f [ y 1 , . . . , y n J = 0. , where ! y . = x . x . - x . x . wi th x. T x. ' • (2) A f i n i t e l y generated p r i m i t i v e algebra over F s a t i s f y i n g a polynomial i d e n t i t y i s c e n t r a l simple of f i n i t e dimension and hence can always be assumed to contain a unit.-The f o l l o w i n g simple c o r o l l a r y gives an e x p l i c i t r e s u l t which we s h a l l use i n the next s e c t i o n . C o r o l l a r y : (a) A f i n i t e l y generated semiprime P.I. algebra A over a f i e l d F i s Jacobson semisimple. (b) A f i n i t e l y generated simple P.I. algebra over a f i e l d F i s f i n i t e dimensional over F Sect i o n 2: Our main purpose i n t h i s s e c t i o n w i l l be to g e n e r a l i z e the f o l l o w i n g commutative theorem; I f A i s a Jacobson r i n g then: A[x] i s a Jacobson r i n g . i f M i s a maximal i d e a l of A[x] then A D M i s a maximal i d e a l of A and [A[x]/M. : A/A n M] < <=° . (3) i f A i s a l s o a H i l b e r t algebra then A[x] i s a H i l b e r t a l g e b r a . To accomplish t h i s we need some new machinery, and. i n p a r t i c u l a r Theorem 2.1: (1) (2) 79. a new concept of the extension of a r i n g . Throughout much of what fo l l o w s i n t h i s chapter we s h a l l assume that the r i n g s we are working w i t h are. P.I. r i n g s but i n some of the p r e l i m i n a r y r e s u l t s i t w i l l be c l e a r that no assumption of P.I. i s needed. Unless i n d i c a t e d otherwises A s h a l l have a u n i t element. Let A be a subring of a r i n g S . The c e n t r a l i z e r C g(A) of A i n S i s defined by; C (A) = {s e S : sa = as f o r a l l a £ A} s I f s_,...,s £ C (A) we l e t A(s.,..-.,s ) be the subring of S I n s I n & generated by A and the s^' s D e f i n i t i o n 2.2: A r i n g S c o n t a i n i n g a subring A i s an extension of A i f S' = AC s(A) . A l s o , S i s a f i n i t e l y generated extension of A i f S = A ( s s ) w i t h s. £ C (A) I n i s Lemma 2.3: Let S be an extension of A I f S i s prime, A i s als o prime. I f I -= A then IS -=> S and, i n f a c t IS = SI = IC (A) j . Proof: Suppose S, i s prime. L et a,b e A such that aAb = 0 Then aSb = aAC (A)b = aAbC (A) = 0 so that a = 0 or b = 0 and A s s must be prime. Secondly, IS = IAC (A) = AC (A)I = SI 80. Lemma 2.4: Let S be a prime r i n g and an extension of the r i n g A I f c i s , a r e g u l a r element of A , c i s r e g u l a r i n S Proof: A i s prime by the previous lemma so that ( I I , 4.2) assures us that cA contains a non-zero two-sided i d e a l U . Now cS = cAS >_ US . But US -=> S by (IV, 2.3) and hence s i n c e S i s prime US i s an e s s e n t i a l i d e a l of S . Consequently, cS i s a l s o e s s e n t i a l . By ( I , 1.7) cS contains a r e g u l a r element, say cs . j Then c i s c l e a r l y l e f t r e g u l a r and part (c) of ( I , 1.7). t e l l s us that c i s r i g h t r e g u l a r a l s o . So that i f S. i s a prime r i n g extension of A , A i s a l s o prime and every r e g u l a r element of A i s r e g u l a r i n S . We can then extend the i n j e c t i o n map i : A ->- S to a map i ' C : Q(A) -> Q(S) , (where Q(A) , Q(S) are the quotient r i n g s of A , S r e s p e c t i v e l y ) i n such a way t h a t the f o l l o w i n g diagram commutes. A Q ( A ) - i - Q(S) Q(A) i s simple by ( I , 1.6) so that i i s 1-1 . A l s o , i f Z i s the centre of . Q(S) we have Q(S) = ZS by ( I I , 4.4) ZAC (A) s Q ( A ) C Q ( s ) ( Q ( A ) ) and thus Q(S) i s an extension of .Q(A) 81. Lemma 2.5: I f S i s a f i n i t e l y generated prime extension of a Jacobson semisimple- r i n g A then S i s a l s o Jacobson semisimple. Proof: Let S = A [ s , ...,s ] . Since Q(A) <_ Q(S) i n a n a t u r a l way we can define T = Q(A) [ s ^ , . .., s^ .] • By d e f i n i t i o n T i s an extension of Q(A) and S <_ T <_ Q(S) so that T i s a prime r i n g . Q(A) i s simple ( I , 1.6) and a l s o prime so that Q(A) i s p r i m i t i v e . Since Q(A) has a polynomial i d e n t i t y then ( I I , 4.4) i m p l i e s that i t i s f i n i t e dimensional over a f i e l d F where F. i s the centre of Q(A) Hence, T i s a f i n i t e l y generated algebra over F and T i s Jacobson semisimple from a c o r o l l a r y of (IV, 1.4). Suppose now that J(S) j= 0 and l e t 0f= f e J(S) From S <_ T and J(T) = 0 there i s a maximal i d e a l M of T w i t h f £ M . Let T = T/M . Now Q(A) can be i d e n t i f i e d w i t h a s u b r i n g of T and under t h i s i d e n t i f i c a t i o n T = Q(A)[s^,...,s^] i s an exten-s i o n of Q(A) . A l s o , T i s simple, f i n i t e l y generated over F so that (IV, 1.4, c o r o l l a r y ) says that T i s of f i n i t e dimension over the centre F of Q(A) and hence T i s of f i n i t e length over Q(A). . Let C = C_(Q(A)) . T Claim: C i s a simple algebra over F and T = Q(A) «^,C . • For, i f Z i s the centre of T the s^ commute w i t h the elements of Q(A) and thus F <_ Z . Since Q(A) i s simple we must have Q(A)® Z a simple algebra w i t h centre Z and isomorphic to Q(A)Z F Hence, Q(A)Z. i s a c e n t r a l simple algebra over: Z and C__(Q(A)Z) = C T ' So C i s a c e n t r a l simple algebra over Z and T'= Q(A)Z®FC = (Q(A)®FZ)®ZC = Q(A)^C Now, s i n c e T i s of f i n i t e length over Q(A) we see that dim C i s f i n i t e , say m and thus we have a homomorphism y : C F^ * D e f i n i n g y = y ® IQ(^) W E N A V E y. : T = Q(A)^,C -> Q(A) = Q(A)ffl Note t h a t i f a e Q(A) then y(a) -Since f £ M , 0 =}= f e T , T i s a simple r i n g and so TfT =. T Let t.,u. e T such that Z u. f t . = 1 . Consider the n a t u r a l x 1 x x maps, A ^ S -> T -> T -> Q(A) m Let S be the image of - S i n T . Since 1 e j ( s ) , J(S) <{= 0 A l s o , Q(A)S" = T s i n c e Q(A) [¥,... = T . Therefore, l e t d,e be r e g u l a r elements i n A such that dt. e-S and u.e £ S . x x Let the image of s, Let c = de so that c = I dt. f u.e £ J(S) x l ~ l i n Q(A) be s. . Let b £ A such that - bs. e A f o r a l l i . m x x m -F i n a l l y , assume a l s o that b £ Ac <_ J(S) , (replace b by l.b i f necessary). I f s. s. ...s i s . a monomial i n the s. we have 1 2 r. ~ ~ ~ ~ ~ b s. ...s. = bs. bs. ...bs. £ A and ( I I , 4.3) i m p l i e s that there i . x i , x„ x m 1 r . 1^ 2 r i s a maximal i d e a l M i n A such that -b i s i n v e r t i b l e i n A/M 83 Claim: MS f S . For, i f MS = S we have 1 = E m^ S^ . where the S^ are k monomials i n the s. . Choosing v > max. deg. S we see that x tc b V •= E m, S, b V e MA = M which . Tc k m m c o n t r a d i c t s the i n v e r t i b i l i t y of b i n A/M . Thus, MS i s a proper i d e a l of S and s i n c e M i s a maximal i d e a l of A,M = MS O A . I n "S/MS , b i s i n v e r t i b l e , but • b~ e J(S"/MS") i s a c o n t r a d i c t i o n , proving the lemma. Our main.theorem i s : Theorem 2.6: I f S i s a f i n i t e l y generated extension of a Jacobson r i n g A then (1) S i s a Jacobson r i n g . (2) I f M i s a maximal i d e a l of S then A A M i s a maximal i d e a l i n A and S/M i s of f i n i t e length over A/A A M . (3) I f A i s a l s o a H i l b e r t algebra over a f i e l d F then S i s a H i l b e r t algebra over F Proof: (1) I t i s enough to show that i f P i s a prime i d e a l of S then S = S/P i s Jacobson semisimple. Let A be the image of A i n " S . The f a c t s that S i s a f i n i t e l y generated prime extension of A and A i s Jacobson semisimple along w i t h the previous lemma complete the proof. S e c t i o n 3: In t h i s s e c t i o n we study the g e o m e t r i c - i n t e r p r e t a t i o n of the r e s u l t s of §2 and i n p a r t i c u l a r t h e i r r e l a t i o n to the H i l b e r t N u l l s t e l l e n s a t z . We define: i s a prime i d e a l of A} , i s a maximal i d e a l of A} . The Z a r i s k i topology on Spec A i s defined as f o l l o w s : the closed sets of Spec A are sets of the form V(B) where B < A and V(B) = {P £ Spec A : B <_ P} . We a l s o set D(B) = comp V(B) = Spec A-V(B) . Max-Spec A i s the set of a l l those p o i n t s i n Spec A which are closed. We u s u a l l y give Max-Spec.A the induced topology. Now, i f y : A -> S i s a map of r i n g s , i n general we have no r e l a t i o n between Spec A and Spec S . However, i f S i s an extension of y(A) we have a continuous map y : Spec S Spec A defined by * -1 Y (P) = Y ( p) • This depends on the f a c t that i f S i s an extension of y(A) and P i s a prime i d e a l of S then y -^ "(P) i s a prime i d e a l of A . The c o n d i t i o n that A i s a Jacobson r i n g i s r e a l l y j u s t a c o n d i t i o n on Spec A , f o r , we have - -P r o p o s i t i o n 3.1: The f o l l o w i n g c o n d i t i o n s are eq u i v a l e n t : (a) A i s a Jacobson r i n g (b) For every closed set C <_ Spec A , C D Max-Spec A i s I dense i n C . Proof: (a) => (b) I t w i l l be s u f f i c i e n t to show that i f D(B) C\ C f= 0 then- Max-Spec AD C C\ D(B) =f="0 . Let. P e D(B) DC . Since B P there i s an element'' b e B w i t h b £ P . Since A/P i s Jacobson Spec A =' {P : P Max-Spec A = {M : M semis imp l e there i s a maximal i d e a l M of A w i t h P <_ M and b £ M. ,. Hence, Me Max-Spec A n C n D(A) f 0 . I (b) => (a) . Let P be a prime i d e a l of A . I f a £ P then P e D(a) and D(a) i s an open s e t . By assumption, Max-Spec AD V(P) i s dense i n V(P) . Now D(a) (~\ V(P) i s a nonempty open set of V(P) s i n c e i t contains P , and thus Max-Spec A r\ D(a) CS V(P) j= 0 . I f M e Max-Spec A f\ D(a) r\ V(P) we have M _> P and a £ M . Since a i s a r b i t r a r y A/P i s Jacobson semisimple. i D e f i n i t i o n 3.2: A semiprime i d e a l I of a r i n g A has degree ! . n(deg I = n) i f A/I s a t i s f i e s the i d e n t i t y S_ (x) = 0 and no i d e n t i t y Zn of lower degree. Since every prime i d e a l has a degree and s i n c e a semiprime i d e a l I i s the i n t e r s e c t i o n of a l l the prime i d e a l s which contain i t , say I = C\ P , then every semiprime i d e a l a l s o has a degree and a ' a deg I = Max{deg. P^} The geometric i n t e r p r e t a t i o n of our r e s u l t s occurs when A i s a f i e l d F and S = F[s ,...,s^] i s a f i n i t e l y generated al g e b r a . ; Let Q be a u n i v e r s a l domain c o n t a i n i n g F . Define fty(S) } • ] : are c a l l e d p o i n t s i We note that i f S L = {y : y maps S -> Q and Q K. K. K. and |s| = <J |s|, . The e l e m e n t s o f |s| , |s| o f deg ree k and the p o i n t s o f S , r e s p e c t i v e l y . 8 7 . ye | s | k J y ( S ) i s a p r i m e r i n g s i n c e ft^. i s an e x t e n s i o n o f y (S) A l s o , P = K e r v 1 S a p r i m e i d e a l o f deg ree . k Suppose t h a t P i s a p r i m e i d e a l o f deg ree k . T h e n , o f c o u r s e , S/P has a q u o t i e n t r i n g . Q w h i c h i s a c e n t r a l s i m p l e a l g e b r a 2 o f d i m e n s i o n k o ve r i t s c e n t r e Z . C h o o s i n g ft b i g enough . so t h a t we can embed Z i n ft we can c o n s t r u c t Q< ,^ ft. T h i s i s a f i n i t e 2 d i m e n s i o n a l a l g e b r a c e n t r a l s i m p l e o v e r ft o f d i m e n s i o n k , i . e . Q ® ^ = ftk • Now Q = ZS/P so i f Y : S -y S/P -y Q -> Q®zft = ftfc i s the n a t u r a l map we must ' have . ft^ = fty(S) . Hence , y i s a p o i n t O J T \ deg ree k and P = P . Of c o u r s e , T depends on t he embedding Z ->• ft Y • ! • and on the i s o m o r p h i s m Q® ft = ft, so y i s n o t u n i q u e . However , i f y" i s a n o t h e r p o i n t c o n s t r u c t e d f r om P we have y (S) = y'' (S) = S/P P r o p o s i t i o n 3 . 3 : (a) | S| = | S/J (S ) ] • (b) deg ree S/J (S ) = n => | s | n =}= 0 and | s | m = 0 - f o r m > n The p r o o f i s c l e a r f r o m the f a c t s t h a t (a) => (b) and (a) i s t r u e s i n c e J ( S ) i s n i l ( I V , 1.4 c o r o l l a r y ) . D e f i n i t i o n 3 .4 : . A p o i n t y i s a l g e b r a i c i f y(S) i s a l g e b r a i c o v e r .F Now d e f i n e E. : |s|.-> Spec S by E (y) = Ker, y . I f Y e |s| t h e n deg E (y ) = k and E i s o n t o . A l s o , y i s a l g e b r a i c i j f f E(y) e Max - Spec S . Hence, we can i n t e r p r e t (IV, 1.4, c o r o l l a r y ) i n the f o l l o w i n g way s i n c e 'NL(S) i s the i n t e r s e c t i o n of a l l prime i d e a l s of S H i l b e r t ' s N u l l s t e l l e n s a t z : An element a £ S vanishes on a l l a l g e b r a i c p o i n t s of S- ( i . e . y(a) = 0 f o r a l l a l g e b r a i c p o i n t s y £ |s| ) i f f a generates a n i l i d e a l . S e c t i o n 4: One of the c l a s s i c a l problems i n group theory i s the Burn-s i d e problem: Is a t o r s i o n group n e c e s s a r i l y l o c a l l y f i n i t e ? In general t h i s need not. be t r u e , however, i t i s q u i t e important to know under what a d d i t i o n a l c o n d i t i o n s do we get an a f f i r m a t i v e answer. Burnside h i m s e l f proved Theorem 4.1: I f G i s a group of matrices over a f i e l d then G i s l o c a l l y f i n i t e . P r o c e s i [38] has e s t a b l i s h e d another s u f f i c i e n t c o n d i t i o n , namely, that the t o r s i o n group be embedded i n a r i n g s a t i s f y i n g a proper polynomial i d e n t i t y . Our f i r s t proof of t h i s f a c t depends on the r i n g being an algebra over a f i e l d and s i n c e we s h a l l g e n e r a l i z e t h i s result., l a t e r we give only an o u t l i n e of the proof.. Theorem 4.2: Let G be a t o r s i o n group. I f G can be embedded i n the m u l t i p l i c a t i v e group of i n v e r t i b l e elements of a P.I. algebra A over a f i e l d , F , then G i s l o c a l l y f i n i t e . Proof: Let FG be the l i n e a r span of G i n A I t i s a subalgebra of A • . Let 0 f a £ FG. a r b i t r a r i l y . Then' a = E a'.g. 0 =(= a. £ F , i ' 89. . e G . Since G i s t o r s i o n , g. 1 = 1 f o r some i n t e g e r n. > 0 so tha t ct.e. s a t i s f i e s the equation i i -n. n. a. x - 1 = 0 I Hence, each a ^ g ^ x s a l g e b r a i c over F so that every a e A i s a sum of a l g e b r a i c elements. By ( I I , 6.2) FG i s an a l g e b r a i c algebra over F . Consequently, ( I , 3.3) assures us that every f i n i t e l y generated subalgebra of FG i s f i n i t e dimensional over F • : Let g , ...,g. e G . The subalgebra F(g , ...,g.) generated ic l ic by these g_^  is f i n i t e dimensional over F and the subgroup H generated by these g. i s embedded i n F(g , ...,g. ) .' Then Burnside's theorem -L J_ K. (IV, 4.1) completes the proof. We now seek to g e n e r a l i z e the preceding theorem by dropping the assumption that A i s an algebra. Two p r e l i m i n a r y lemmas help us toward the g o a l . The f i r s t i s s i m i l a r to ( I I , 2.3). Lemma 4.3: Let A be a semiprime r i n g s a t i s f y i n g a proper polynomial i d e n t i t y of degree m . Then: (k) (1) A can be embedded i n a f i n i t e d i r e c t sum © A > k . (k) where A i s a commutative semiprime r i n g p r e s e r v i n g (k) 1 and A^ i s the r i n g of a l l k x k matrices w i t h ,(k) e n t r i e s from A (2). I f A = B[x^,....x^] i s a f i n i t e l y generated algebra over a Noetherian r i n g B then A can be embedded i n a f i n i t e d i r e c t sum of r i n g s , inhere F ^ (k) . i s a f i e l d and i s the r i n g of a l l k x k matrices 90. w i t h e n t r i e s from . Proof: (1) Let Sfc = {P e Spec A : deg P = k} . Now, s k = $ f o r k > [m/2] . We have the f o l l o w i n g embedding: [m/2] A + n / n A/P k=l \ a k £ ^ k °k (a) and s i n c e A/P can be embedded i n a matrix r i n g 0, we have the embed-a, k k ding [m/2] , \ _ [m/2] , , N\ A - n / n 0^\ = e n 0 ( a ) k=l \ « keS k / k=l AaeS Jk S e t t i n g A ( k ) = n 0 ( a ) gives-us (1). aeS. k (k) (2) Considering again the c o n s t r u c t i o n of (1) we see that the A 's are algebras over A . Let y ^ D e t n e n a t u r a l map [m/2] , . p , s A->- ® A < K ) + h A^ h ) . The Y^(x^) f o r i = l ,2,...,n are matrices over A^^ . Let a . ( s , t = l , . . . , h ; i = 1, ...,n) be the e n t r i e s of y, (x.) and l e t s , t , i ' h i A ^ be the A subalgebra of A ^ generated by the elements a . s j t , I We have the embedding [m/2] } •A -»• © v-; . k=i x 9 Now A TOO i s a semiprime Noetherian r i n g and hence i t s quotient r i n g i s a d i r e c t sum of f i e l d s and t h i s completes the proof. Lemma '4.4: Let G be a t o r s i o n group embedded i n a P.I. r i n g A I f the a d d i t i v e group of A i s f i n i t e l y generated then. G • i s l o c a l l y • f i n i t e . Proof: The t o r s i o n p a r t T o f A i s an i d e a l , of A and i t i s f i n i t e . ' Since A/T i s a f i n i t e l y generated f r e e module over the i n t e g e r s Z i t can be n a t u r a l l y embedded i n A/T ® Q which i s a f i n i t e dimensional algebra over Q (where Q = r a t i o n a l s ) . Hence A/T < Q f o r some intege m m > 0 . I f G' i s the image of G i n A/T , G i s l o c a l l y f i n i t e . Now, l e t t i n g N be the k e r n a l of the mapping G -> G we see that N i s f i n i t e . Thus, G/N and N both being l o c a l l y f i n i t e imply that G i s a l s o l o c a l l y f i n i t e . r i n g A .. Then G i s l o c a l l y f i n i t e [38] . Proof: C l e a r l y , we can assume that G i s f i n i t e l y generated. , L e t {a,»-..»a } e Q be the c o e f f i c i e n t s of the i d e n t i t y s a t i s f i e d by A Let B'= <a^,...,a ,1> i . e . B i s the subring of Q generated by 1 together w i t h the . C l e a r l y , we can a l s o assume (w.l.o.g.) that A i s the algebra over B generated by a f i n i t e number of generators of G and t h e i r i n v e r s e s . Our main theorem: Theorem 4.5 ( P r o c e s i ) : Let G be a t o r s i o n group embedded i n a P.I. n 92 . Now, NL(A) i s a f i n i t e l y generated semiprime algebra over the Noetherian r i n g B Let G = G/N be the image of G i n NL(A) * i = l F^"^ . I f i s the p r o j e c t i o n of G on w e have embedding To show that N i s l o c a l l y f i n i t e l e t g ,...,g 1 s N . Then g. = 1 + a.- , X iC X X f o r e NL(A) . The subring of NL(A) . generated by these ( I I , 1.1). Hence, the a d d i t i v e group of the subring of A generated by Lihese g^ i s f i n i t e l y generated and (IV, 4.4) a p p l i e s so t h a t N i s l o c a l l y f i n i t e . To show G i s l o c a l l y f i n i t e we note that by (IV, 4.3) we can embed G i n a f i n i t e d i r e c t sum of r i n g s of matrices over f i e l d s , say G -> IDK . By Burnside's theorem (IV, 4.1) each H_^  i s f i n i t e so G i s l o c a l l y f i n i t e . Thus, as before, G i s a l s o l o c a l l y f i n i t e and the proof i s complete. *' Chapter V Until now we have been considering rings A which are ft-algebras over an integral domain ft . The polynomial identities which A satisfied were of the form Za.x. x. ...x. =0 , 1 x l X2 Xd where the a. e .ft , and the sum runs over a l l permutations i of the d letters l,2,...,d . Recently, Amitsur introduced a more general type of polynomial identity. The results obtained in this more general case are not as sharp as the classical results, however, the fact that there are a number of beautiful theorems more than justifies this more general approach. We shall show, for example, that a division ring satisfies this extended form of a polynomial identity i f f i t is finite v dimensional over its centre. Section 1: Let {x } be an infinite family of indeterminates, and ;— a suppose that A is any ring which is an ft-algebra over an integral domain of operators ft . The polynomial identities which we shall study in this section are of the form G[x] •= E a. II. a. JJ. ... a'. IT. a. 1 1 J l X2 J2 \ J k ^Tc+l 94. where the n . are monomials i n the indeterminates x . a n d the a. 3 . a i are elements from A . More p r e c i s e l y , G[x] i s a n o n t r i v i a l element from the f ree product of the r i ng A and the f ree a s soc i a t i ve r i ng Q[x ] . Such a polynomial i d e n t i t y i s c a l l e d a genera l ized polynomial i d e n t i t y or a G.P.I. The r i n g A i s sa id to s a t i s f y the G.P.I. G[x] i f f o r every sub s t i t u t i on x^ = a^ we always have G[a] = 0 . I f th i s i s the case, we c a l l A a G.P.I, r i n g . Example 1; Richardson and L i t t lewood have shown [8] that the quaternions s a t i s f y the i d e n t i t y ( x i ) 2 - ( i x ) 2 + ( x j ) 2 - ( j x ) 2 + (xk ) 2 - ( kx ) 2 = 0 , where i , j , k i s the quaternion ba s i s . Example 2: Suppose F i s a f i e l d . Then F r s a t i s f i e s the G.P.I. e l l X i e l l X 2 e i l " e l l X 2 e l l X l e l l 0 ' where, as usua l , e . . i s the n x n matrix over F with 1 i n the p o s i t i o n ( i , j ) and zeroes elsewhere. F^ a l so s a t i s f i e s the G.P.I. ( . E . e i j X l e j i ) x 2 " X 2 ( . Z . elfl?i±> = ° ' 1,3 i»J J s ince Z e . .x e . . i s i n the centre of F • • 13 1 J i 1  1 » J -95. 2 Example 3: Let D be a central simple algebra of dimension n over its centre C . Then' D®F = F , where F is a maximal subfield of c n D . Hence, D satisfies both G.P.I.'s given in example 2. Example 4: Every P..I. ring is a G.P.I, ring. Note that example 3 shows that the extension of Kaplansky's theorem (I, 2.16) from primitive P.I. rings to primitive G.P.I, rings is not valid. Nevertheless, in the next few sections, we shall obtain a rather nice characterization of primitive G.P.I, rings. First, we require four lemmas which are fundamental to a l l that we shall be doing in this chapter. Lemma 1.1: Let V , U be vector spaces over a field F and,JLet T ,. .. ,.T be a finite set of F-independent linear transformations of V into U . Then, i f U is a finite dimensional subspace of U , at least lonfe of the following is true, i / I' i ' (1) there exists a ve V with T.v, ...,T v linearly I T independent modulo U q , or .' T (2) there exists an S = £ a.T.fO of finite rank. i=l 1 1 . If (2) holds, then we can choose S so that, dim S V < dim U + Ct1) - 1 Proof: Let T =.{ Z y.T. : y. e ¥} , i.e. J is the subspace of linear x=l ! transformations.of Horn ( V , U ) generated by the T. . If (1) does not r x hold, then clearly, T has the following property ; (*) For each v e V there is a non-zero T e T such that •• . • . \ Tv e U i We shall prove the lemma by induction on Dim T . Suppose dim T = 1 . Then T = FT and by property (*) , • • I T.V < U which implies that (2) holds since U is fi n i t e dimensional, l — o o Thus, in this case, the lemma is proved. ; Suppose dim T > 1 '. Choose V q =j= 0 arbitrarily in V By (*) let 0 4= T e T such that T v e U . Define 1 o o o o V =. {v eV : T v e U } . o o o V ^ 0 since v e V • If V = V , then from T V < U we see that o 1 o o o o — o T is of f i n i t e rank and the proof is finished. Therefore, assume o r that V 4= V . Now choose v, e V with v, i V . Define o 1 1 T o T o = {T e T: Tv e U } o ± o Since 1 v, i D we see that T i T . Also, (*) implies that o 1 * o o T o r T o t ° • Now, T < T since T A J . Let S be a submodule of o o T o T such that T < S < T and with T ••- FT © S o It i s clear that dim S < dim J so that we can apply the induction hypothesis to 3 • Accordingly, we define "= U o + Tv^ . Then (U^F) < (U :F) +. (T:F) < <=° so that U, is a f i n i t e dimensional sub-1 — o N 1 space of U . By induction on S then either (1) or (2) holds for S and U, . If (2) holds, there exists 0 f S e S such that SV is finite dimensional, thus proving the lemma. Therefore, suppose (1) holds. We shall obtain a contradiction i in this case and so the proof of lemma 1.1 will be complete. Since (1). is satisfied there is a non-zero w e V such that the following property is valid, (**) Sw e U L for S e 5 => S = 0 However, by (*) there is 0 ={= T e T such that (***) Tw e U < IL o — 1 Since T £ T = FT © S , we let T = aT + S , for a e F , S e S o o o ' o Then (***) becomes (aT +.S )w E U < U ' o o o — 1 This implies that a f 0 by the choice of w and we may, of course, assume that a = 1 . Hence (***) reduces to (T + S )w e U < U. . o o o — 1 Consider the element (w+v.) . By .(*) » T'(w + v n) £ U , for 1 i o some T' f 0 . .Let T" = BT +'S, , for B e F , S. e S . Then, x o i i 1 98 (gT + S )(w + v ) - g (T + S )w o l 1 o o (S. - gS )w + (gT + S.)v • x o o i l as in U . Hence, o ' < si " e so ) w e uo + ( F To + s ) v i = uo + T v i = u i But, since (S 1 - gSQ) e 5 we have by (**) that (S^ - gSQ) = 0 Therefore, 0 ¥ T = gT + S '= g (T + S ) and g f 0 . Also, 1 o 1 o o 1 g'Vcw + v ) = (T + S )(w +'v.) e U 1 0 0 1 o Therefore, (T + S )v T = (T + S )(w + v.) - (T + S )w e . o o l o o 1 o o o By the definition of T we have T + S e T . But T < S and o . • o o • o o — S e S implies that T e S which contradicts FT A 5 = 0 . This o o is the required contradiction. / Now, assuming that (2) is satisfied, the bound on dim SV is obtained in the following way. Let a(p,x) be the minimum dimension of such an SV . We know x = (T:F) , and i f we set y = ( u 0 :^) we have already shown in the proof that a(u, 1) "'< y and a(y,x) < Max [y,a(y + T , T - 1) ] 99 Thus, dim SV = a(u,x) < u + x + (x-1) + ... + 2 - » + ( ? ) - 1 T+1 = dim U + ( . ) - 1 . O L We extend lemma 1.1 in Lemma 1.2: . Let W be a cofinite submodule of V , (i.e. dim V/W < » ] and let T < HomfV.U) and U be as in lemma 1.1. If for every w e W — F o there exists T f 0 in T such that Tw e U , then there exists o 0 f S e T • such that SV is finite dimensional and in fact, we can choose S so that T + 1 dim SV < dim U + T dim (V/W) + (l„ ) - 1 . — o I The proof is an easy consequence of the previous lemma (see [8]). Both of the preceding lemmas have been proved for vector spaces over a field. We wish now to establish these same lemmas for vector spaces over a division ring D . The proof of lemma 1.1 goes over almost exactly i f we replace F by D and we then have Lemma 1.3 : U and V are vector spaces over D . Let U < U _ . 0 _ and T <_Hom^ (V,U) be two finite dimensional D-spaces. If for every 0 =)= v e V there exists 0 ={= T e T such that Tv E U o , then there is 0 f S E T such that SV generates a finite dimensional D-subspace T+1 of U and its dimension is <_ (U^D) + ( ) - 1 , where T = (T:D) 100. Now, let V be an abelian group having fi as an arbitrary •; set of operators. If D is a division subring of Hom^ (V,V) , V can !: be made into a vector space over D . In lemma 1.3, i f V,U are abelian groups over fi we put U = V to obtain Lemma 1.4: Let D be a division subring of Hom^(V,V) and let T^,...,T^ be linearly left D-independent endomorphisms of V . If V < V is a finite dimensional D-space then either there is a v e V o — such that T,v,...,T v are D-independent modulo V , or some . I T O T+1 (E d^T^V generates a D-space of dimension <_ dim V Q + ( 2 ) ~ 1 Section 2: Throughout this section we shall assume that A is a 1 primitive ring. Hence, A has a faithful irreducible module V and j Hom^ (V,V) = D is a division ring. V can be made into a vector space over D in the obvious way by defining O ' V = 0(v) for 0 z Hom^ (V,V) and v e V . We also know that A can be considered as a dense ring of linear transformations of VD . In addition we shall let C be the centre of D and F a maximal subfield of D Letting Z denote the ring of integers we see that A < Horn (V,V) <_ Horn (V,V) . Now Hom_(V,V) can be made into an algebra over F by setting i) (p 1 + 0 2) • v = 01(v) + 0 2(v) i i ) ( 0 ^ ) * v = 0 1(0 2(v)) i i i ) (aO) ' v = a • 0(v) , where 0^,0^,0 e Homz(V,V) , v e V , a £ F .. Define to be the subalgebra over F -of Horn (V,V) generated by A . Thus, the elements i o r . of A_ look like Z a.a. , where a. e A , a. e F and, of course, the . F . 1 1 i l i sum is finite. Note also that A„ is a homomorphic image of A® F under the mapping Z a. <g> a .• Z a .a , l i ' . i i i i Lemma 2 .1 : The canonical operation of A^ , on V turns A^ , into a dense ring of linear transformations with centralizing field F Proof: Writing the operations of A and D on the left of V we have by definition: (Z a.ct. ) v = Z a. (a . v) i i i i = Z a i(a iv) , since A <_ Hom^ (V,V) where, of course, Z a-.a. is a typical element of A„ and v e V '. i i . F If ; (Z a.a.)v = 0 for a l l v e V , then Z a.a. = 0 since i t is the i i . . ' i i . zero map in Homz(V,V) V is a faithful irreducible module so that i f 0 =j= v e V then Av = V , and since A <_ A^ we see that A^v = V also. Thus, ;• I Aj, is a dense ring of linear transformations on V The centralizer of A^ is by definition the set of a l l elements of Horn (V,V) which commute with a l l elements of A . Clearly, the centralizer of A contains F . Conversely, suppose X e Horn (V,V) I1 . - LA > A_ and X commutes with A . Then X commutes, in particular, with —• r r A . From D = Horn (V,V) we have X e D . Now, i f a e F , a e A 102. we see that a(Aa) = A(aa) , since \a = aA = (aa)A , since aa e and commutes with A„ F Thus, a(Aa - aA) =0 j => a(Aa - aA)v = 0 , a e A , v e V . =>•' Aa = aA , i.e. A commutes with every element in F . Hence, A e F since F is maximal and the proof is complete. \ Lemma 2.2: (1) Let 0 f a,b e A such that axb = bxa. for a l l x e A Then b = Aa for some A e C . (2) Let {a^ } be a C-base of F (i.e. a basis of the vector space F over C ). Then the elements of A can be written r uniquely in the form / E a.a. , for a. e A i i i Proof: (1) Since a =|= 0 and V is an irreducible faithful A-module we have aV = V. and hence AaV =-V . Any w e V can be expressed as w = E t. ar. , for some t. e A , v. e V . Now define A : V V i i i l by A(E t.av.) = E t.bv. . It is easy to verify that A e Horn (V,V) = D I X X X ' , A Let d be arbitrary element of D and consider 103 dX (Z t.av.) = d Z t.bv. x i 1 1 = z dt.bv. • 1 i = X Z dt.av. i i Xd(Z t^av^ , so that Xd = dX , i.e. X e C and' Xa = b by definition. k (2) Suppose Z a.a. = 0 . To finish the proof a l l we need i=l 1 1 to do is to show that a^ = 0 . Let k be the smallest integer for which there exists a sum of the above type which is equal to zero and for which not a l l a. = 0 l k k a^x( Z a^a^) ~ ( £ • a i a i ^ x a ] : C = 0 , V x e A . . i=l i=l j k-1 , i Z (a^xa^ - a^xa^ou = 0 . ' i=l \ • ' Since this element is of lower length we conclude that / / a k X a i = a i x a k ' ^ X From (1) we have a. = i m * l \ > f o r * i E C Therefore Z aa. = Z. X. a, a. = a, Z X.a. = 0 1 I K l . K 1 1 104. Now, a, |» 0 since k is minimal. Also, E X .a . is an element of k 1 . 1 1 the field F and so must be zero. Since the a. form a C-base l we have X. = 0 , for a l l i . But this contradicts X, = 1 , thus i k finishing the proof of (2) . Before we obtain our main results we have one preliminary step left. This is the important Theorem 2.3: Let A be a dense ring of linear transformations of V Q and let F be a maximal subfield of D . I f A^ contains a linear transformation of finite rank over F then A contains a linear transformation of finite rank over D and (D:C) < 0 3 Proof: By lemma 2.1, A^ , is also a dense ring of linear transformations Let T e Ap be such that (TV:F) < « . If {c^} is a C-base of . . I F , then uniquely we have T = E a.a. , where a. e A' ... l i i i=l\ Choose T of finite rank with k minimal. For any x e A , . / ' (c^xT - Txa^V <_ (a^TV + TV . Now, (TV:F) < 0 0 implies (a^xTV^) < «> since a^xTV is an F-homomorphic image of TV under Tv a^xTv . io 5^; Thus, (\ x T ~ T x \ ) i s o f t h e s a m e t v P e a s T a n d i s clearly of lower length than T . From (\x^ ~ Txa^) = E (a^xa^ - a^xa^)^ , we conclude that a j c x a - j _ ~ a i x a k ' x e A . Lemma 2.2 then assures that a. = X^a^ , where \ e C . Therefore, k k T.= E. a.a. = E X.a. a. . ., 1 1 . . IK. i 1=1 1=1 k = a ^ E X.a.) 1=1 where a = E e F . So T = a^ a and since T =(= 0 , a^ =(= 0 and a f 0 . TV = a^ aV ~ a^y > since a X exists in F a V is a D-space (since a A commutes with D ) and thus \ oo > ( a ^ F ) = (ak.V:D)(D:F) so that (T:D) = (a^VrD) < 0 0 , and we are finished the first part of the theorem. The maximality of F together with a remark from Jacobson [21] p. 175 tells us that (D:C) = ( D : F ) 2 < <*> . I n f a c t , a bound can be d e t e r m i n e d f o r (D:C) k C o r o l l a r y : I f T = Z a . a . . and (TV:F ) = m t h e n (D:C) < 2 m = , k - l 2 1 = 1 4 m , A s p e c i a l c a se o f t he theo rem i s C o r o l l a r y : L e t D be a d i v i s i o n a l g e b r a w i t h c e n t r e C and F a m a x i m a l s u b f i e l d o f D . Then Da^F i s a p r i m i t i v e r i n g a c t i n g on D and i t c o n t a i n s f i n i t e r a n k e d t r a n s f o r m a t i o n i f f (D:C) •<• 0 0 . P r o o f : C o n s i d e r D as a v e c t o r s pace o v e r i t s e l f b y l e t t i n g D a c t . on i t s e l f f r o m the l e f t . I t s c e n t r a l i z e r i s i t s a n t i - i s o m o r p h i c r i n g ; D o f a l l r i g h t m u l t i p l i c a t i o n s . |: * A i: D® F can be i d e n t i t i e d w i t h D ® F , where F i s t he > c c ' A c e n t r a l i z e r o f F i n D . By lemma 2 . 2 , D® F = D„ and we a p p l y c F t he t h e o r e m . ' \ ! ; S e c t i o n 3: We c o n t i n u e t h e n o t a t i o n ' o f §2 and ou r a s s u m p t i o n t h a t A i s p r i m i t i v e . L e t 2 d eno te t he c e n t r e o f A . Then , o f c o u r s e , 2 <_ Hom A (V ,V ) = D , so t h a t 2 i s an i n t e g r a l domain c o n t a i n e d i n C . We assume (and as we s h a l l s ee l a t e r , w i t h o u t T o s s o f g e n e r a l i t y ) t h a t 2 = C . Thus , A can be c o n s i d e r e d as an a l g e b r a o v e r C . L e t t h e G . P . I , s a t i s f i e d by A be o f t h e f o r m , (1) G[x ] = Z a . n. a . II. . . . a . II. a . , X l J l r 2 H \ \ V l 107 where e A , and the IK are monomials in C[x] . By using the same trick as in (1,2.4) we can assume that G[x] is linear in each x^ Let {a^ } be a C-base of A . Then, any a E A can be written a = Z a.a. , where a. £ C . Substituting these expressions x x for the a. in (1), pulling out a l l the a. e C to the left we can m 1 assume that (1) has the form (2) G[x] = Z a.a. JJ. a. JJ. ... a. JJ . a. 1 \ 3 1 X2 J2 xk J k \+l where the a. are from the C-base {a.} . If xn addxtxon, A x x m has an identity element 1, we may write (3) G[x] = Z ct.a.-x. a. x. ... a. x. a. , 1 - 1 ! ] 1 X2 32 Xk k^ ^+1 merely by letting some a. =1 (note that k is different in (2) and (3)). If A does not have an identity element then we s t i l l write G[x] in the form (3) allowing some a. =1 . Thus, (3) is a formal way of writing G[x] and no possibility of confusion can arise since ! 1 always appears in the form 1.x,x.l , or a.l . We also must \' assume then that the C-base of A which we have used is . {a^ } with ; the property that a^ = 1 . Using this convention we can always write G[x] in the form (3). Our main result for primitive G.P.I, rings is Theorem 3.1: A.primitive ring A is a G.P.I, ring i f f i t is isomorphic with a dense-ring of linear transformations over a division ring D which is finite over its centre C , and- A contains a linear 108. transformation of finite rank. Proof: [<=] Let e be a primitive idempotent in A . Then eAe = D . Let (D:C) = h . Since h < °° , D must satisfy S^^(x) = 0 . Hence, A must satisfy the G.P.I. E + ex. ex. ex. ... ex. e = 0 — i i i I 1 2 3 n+1 [=>] Conversely, suppose that A is a primitive G.P.I, ring satisfying i • i G[x] = 0 . Now, A is a dense ring of linear transformations on V^i Assume that A does not contain a linear transformation of finite ! rank, or, that (D:C) = » . By (V, 2.1) let A_ act on V_ . Then! r D (V, 2.2) says that the elements of A can be written uniquely as r £ aj_a^ » where a_^  e A , and the e F are members of a C-base of F . The elements of A and F commute with each other and as a result any multilinear G.P.I, satisfied by A is also satisfied by AF . Let the G.P.I, satisfied by A be written (*) G[x] = E a.a. x. a. x. ... a x. a. , 1 x i J i x2 H Hi Jk xk+i i . where a^ e C , a^ are elements of a C-base of A , with possibly some of the a^ = 1 ( a^ = 1 formally i f A has no identity). Consider the finite set S of a l l the a^ which appear in (*) in ;: monomials with a non-zero coefficient. With no loss of generality we \ • ' may assume that S <= {a 1 = 1 , a 2 a} 109, Now, A < Horn (V,V) so that each a. e S can be considered — D x • i as a linear transformation'of V . \ D | Claim: the elements of S are independent over F For, i f {a.} is a C-base of F we have X. = E c. a. , j X . X . J 3 3 for each X^ e F . Then, supposing that E = 0 , for some X e. F we deduce that X^  = 0 , for a l l i , by writing E a.X. = E a.(E c. a.) xx . X . X . J i 3 3 = E (E a.c. )ct . = 0 3 3 3 Now, E a.c. e A implies by (V, 2.2) that the coefficients are zero, 1 1 1 j i.e.- E a.c. =0 . But . the a. are from a C-base of A so that - i x . i i -3 | c. = 0 , for a l l i , j . This means that X. = 0 , for a l l i , ' x j 1 ! and the elements of S are independent over F Before continuing the proof we require Lemma 3.2: Let a ,...,a be C-independent elements in the primitive I T ring A . If A does not contain a finite ranked transformation, theni for every integer h , there exists vi»'' , > v] : i e ^ such that the set {a/v^ .} are Th D-independent vectors in V T Proof: Define T = { E a.g. : 3 e'F} . , i.e.. T is the T-dimensional i=l i i x linear subspace of A_ generated by a.,...,a . Let V = 0 . F I T O Applying (V, 1.4) means that there is a v e V with a,v ,a.v ,...,a v r r - , & o 1 O 2 O T T O F-linearly independent. 110, Applying (V,. 1.4) again, this time with V = Tv , we obtain o o an element e V such that {a^v^} a r e linearly independent modulo V q . Thus, the set ^ a i v 0 , a i v l ^ 1 S linearly independent over F Continue in this fashion, next setting V = Tv + Tv., in (V, 1 . 4 ) . o o 1 Consequently we can find v^,...,v^ e V so that (a-LVj} a r e F-linearly independent. Returning to the proof of the theorem, suppose A does not j contain a finite.ranked transformation. We shall reach a contradiction. By (V, 2.3) Aj, also does not have a transformation of finite rank. j The G.P.I, satisfied by A can be. written ! G[x] = E a.a. x. a. x. ... a. x. a. 1 X - 1 J 1 X 2 ; , 2 ^ k ^ + l where a. e C , a. are elements of a G-base of A . A typical monomial of G[x] is a .'a. x.. ... a. x. a. . By renumbering the 1 X l J l \ Jk Xk+1 x. (if need be) we choose a nonzero monomial x II = a.a. x, a. x, ,a. ... a. x,a, 1 x i k ^ - l ^ " 1 \-2 x l 1 xo Now write (**) G[x] = E a a x, a. x, .a. ... a. x na. + ... v v Tc i.. Tc-1 x. 0 i . 1 x v k-1 k-2 1 o where the sum runs over v , the a.. ,...,a. are fixed from k-1 o the definition of II , and at least one a 4= 0 . The terms after the plus sign involve either a different permutation of the x^ or contain a different a. . Applying the previous lemma 3.2 to 111. to a^,...,a e ^ ( F replaces D ) we see that there exists v.,...,v. e V such that {a.v.} are F-independent in V . Now, 1 h i j A is a dense ring of linear transformations of V and hence by h r definition there exist k elements d. of A satisfying 3 r d.(a. v. .) = v. , for each j = l,...,k 3 i - j . ! J - l 3 d.(a, v ) = 0 , otherwise. 3 \ V Choosing h > k gives us our contradiction. Indeed, consider G[d1> .. •,dk]v(. Take any monomial of G[x] , substitute x. = d. and multiply on the right by V q . We shall get zero unless the monomial chosen appears,, among the first terms in (**) , i . e . a =a. > 1 £ t £ k - 1 , *t ""t and x. = x„ . I n this case we get a a v. . Hence, i t & v v k G[d]v = T. a a v. o v v k Since one of the f 0 and the ' a v vk are linearly independent over F we have G[d]vQ =)= 0 , which implies that G[d] |= 0 contradicting the fact that G[x] = 0 holds in A and in A^ , . This completes the proof. A modification [8] of this method i permits us to obtain a bound for (D:C) . s • r Corollary: A primitive ring A satisfying a G.P.I, of degree k which 112, involves x C-independent elements (including 1) is isomorphic with a i dense ring of linear transformations over a division ring D containing a finite ranked transformation and (D:C) < 4 X " 1 ( ( T t 1 ) - 1 + x[h)2 The case x = 1 in this corollary includes the classical concept of polynomial identity found in the first four chapters. We also note that in this corollary, x = 1 yields Levitzki's bound of A direct application of this theorem and its corollary j gives this beautiful result for division rings: j Theorem 3.3: A division ring. D is a G.P.I; ring i f f (D:C) is finite and then ,: (D:C) < 4 T _ 1(( T2 1) - 1 + T [ | ] ) 2 Proof: Set A = D , V = D and let A act on V from the left. The centralizer of A is then D / the ring of right multiplications, * Since D is anti-isomorphic to D , (D:C) = (D*:C) < 0 0 . In this section, we assumed that the centre Z of A was just C so that we could consider A to be an algebra over C In general, though, Z- < C . However, we can suppose that the G.P.I. + satisfied by A is multilinear and hence i t holds in both A^ , and 113. . This shows that there was no loss of generality by taking A as a C-algebra since we could just as easily have started from A Section 4: In this section we shall examine prime G.P.I, rings. Accordingly, we assume throughout that A is a prime ring unless otherwise stated. Let U = {U : 0 f U -° A} . Define T = {f : f is module-homomorphism from some U e ii to A , U,A considered as right A-modules} i: If f,g e T , say f : U -> A and g : V ->• A we say that f and i' g are equivalent i f f = g "on some non-zero ideal W such that W < D/1V and' W e U . This defines an equivalence .relation on i| T and we let Q be the set of a l l these equivalence classes. We ': make Q into a ring by defining, for arbitrary f,g e Q : (1) f + g to be the class determined by f + g acting on U n V . (2) fg to be the class determined by f(g) acting from the left on VU . If a e A is arbitrary we let a denote left multiplica-tion by the element "a" . The mapping a a^ is a ring homomorphism from A into Q . Also, i f a = 0 then all = 0 for some U e U . Since A is prime, a = 0 so that A is isomorphic to a subring of ; Q and we can consider A to be a subring of Q We now make three important observations concerning the ring Q • First, ;Q has the property that for each q e Q there is 0 |= U-aA such that qU <_ A . Second, Q is a prime ring, for, suppose qQp = 0 for . . i elements q,p =(=0 . Then there are nonzero ideals U,V of A such that qU <_ A and pV <_ A . Also we can find u e U,v e V such that qu ={= 0 =}= pv . Thus, quApv <_ qQpv = 0 , which contradicts the primeness of A . Hence, Q is a prime ring. Third, the centre C of Q is a field. It is clear that C is an integral domain. Choose 0 =}= c e C . There is 0 ^ U O A with ell <_ A . The mapping cu -> u. from U -=A to A induces an element d of Q . So dcu = u for a l l u e U , dc = 1 , and C is a field. Now define S = AC . S is a subring of Q containing A . We call C the extended centroid of A and S the integral closure of A. . Similar to (V, 2.2) we have: \ • • " Lemma 4.1: Let a,b e S such that axb - bxa for a l l x e A (and therefore for a l l x e S ). Then b/ = qa for some q e C Using this we prove a result which is of fundamental impor-tance to what will follow. , Theorem 4.2: Let a.,...,a be C-independent elements of S , and :— 1 m m let bn,...,b e S with b n =(= 0 . If B = { E a.sb.l s e S} is 1 m x 1' . , i x'. ' x=l finite dimensional over C then 115, (1) B.fO (2) S has a minimal right ideal eS (3) eSe is a finite dimensional division algebra over C Proof: We use induction on m .. For m = 1 , we have B = aSb f 0 by the primeness of Q . So, from assumption, there exists a C-basis k v.,..., v. of B . Thus, axb = Z A.(x)v. , for a l l x e S , and l k . . 1 1 i=l A(x) e C . Choose s e S so that bsa j= 0 and set d = bs . Then axd = axbs = I X.(x)(v.s) , for a l l x e S i=l 1 1 Hence, dim^CaSd) <_ k . Since da =}- 0 and S is prime, aSd >_ NP(aSd) , so that + aSd is a finite dimensional semi-simple NP(aSd) algebra and in particular has an identity element u . Now, NP(aSd) is nilpotent, so that aSd must contain a non-zero idempotent &" Then e"S(<_ aSd) is a finite dimensional prime algebra over C and thus e'Se' = D where D is a finite dimensional division algebra n / 6 over C . Choose e to be a (primitive) idempotent of e'Se' Then . eSe = e(e'Se')e = D , ; I !l: • . 1 I I! so that eSe is a finite dimensional division algebra over C (and this implies that eS is a minimal right ideal of S ). Hence, the case m= 1 is proved. us, • i Suppose now that m > 1. . . By hypothesis B is finite dimensional over C , so let m E a.xb. = E X .(x)v. ', i - l 1 1 j-1 3 1 for a l l x e S . Note that a r e independent, |= 0 , ^ V j ^ is a basis for B , and X (x) e C . If each b^ is a C-multiple of b n , say b. = y .b^ , y. c C for. a l l i then 1 i ' x 1 1 x axb. = E X . (x) v. , ':, 1 j = 1 3 3 ! m for a l l x e S and where a = a + E y.a.^0 . But we have already x=2 done this case. Thus we may assume (by reordering subscripts i f necessary) \ that b^ and b^ are . C-independent. From the relation ' . ' . • \ m k | E a.xb. = E X.(x) v. , i i - l 1 1 j-1 3 1 • , I , • we have m . k E a.xb.tb_ = E X.(x)v.tbn , for a l l x,t e S i - i 1 1 1 j - i J ? 1 Also, m Z a.(xbnt)b. = E X.(xb,t)v. , . i x 1 x . , j 1 i i=l J=l. J J Subtracting these last two equations yields m k E a.x(h.tb. - b,tb.) = E {X . (x)v.tb. - X . (xb t)v. } , . ~ 1 l l l i . ' 3 3 1 3 1 3 1=2 i=l for a l l x,t e S . Since b^jb^ are not C-dependent then by (V, 4.1) there is a t e S such that b.t b. - b nt b 0 4= 0 . Set o 2 o 1 1 o 2 ' b: = b.t b, - b.t b. , 1 l o 1 1 o i w. = v.t b, , and 3 J o 1 y^(x) = -X ( x b ^ ) Then, m \ Z a.xb. = E .{A.(x)w. + y.(x)v.} , wxth b„ 4= 0 • r , ! 1 - - i J 3 3 3 2' 1=2 j=l The proof is completed by induction. 1 I Theorem 4.3; Let A be a prime ring with central closure S = AC Then S is a G.P.I, ring over C i f f S contains a minimal right ideal eS (hence S is primitive) and eSe is a finite dimensional division algebra over C \ Proof: [<=] dim (eSe) = I < 0 0 implies that eSe satisfies the c standard identity S (x) =Z +x.x. ...x. , l 1 2 n where n > I ., S then satisfies the G.P.I. Z + ex. ex. e ... ex. e = 0 . — xn 1- 1 i 1 2 n 118. i [=>] Conversely suppose S satisfies a G.P.I. We can assume that this ' G.P.I, is of the form f(x n,...,x ) = Z 3.a. x. a. ... a. x. a. 1 n 1 1 i . i . i ^ n i o J l .1 n-1 Jn n where 3^ e C , the set of {a^ } is C-independent, f is multilinear of degree n in x^,...,x , and the degree of f (being the maximal degree of its monomials) is minimal. We rewrite this identity as m f(x 1,...,x n) = Z a ix 1f i(x 2,...,x n) + g(x1,...,xn) = 0 , i=l where the f^ are nonzero G.P.I.'s homogeneous multilinear of degree (n-1) , and g is the sum of a l l those monomials of f whose first I indeterminate is not x^ . Likewise, we can write S ( x l x n } = ,\ g i X l b i + 1 P i X i q i ' i=l where g_^  is of degree (n-1) , and P^Q^ a r e G.P.I.'s of positive degree. Note also that the b. are C-independent since they are from I a C-basis of S . The expression for f then becomes j ; m K. f(x.,...,x ) = Z a.xnf. + Z g.x.b. + Z p.xq. = 0 1 n . , l 1 l . . & i 1 l i i i=l i=l Now let s,,s„,...,s ,t be arbitrary elements of S 12 n Substituting x^ = s^ in f and multiplying on the right by tb^ gives us m k Z a i.s 1f.tb 1 + Z g is 1b.tb 1 + L p s q t b = 0 . i=l i=l Similarly, by substituting x, = s, , i {= 1 and x^ = s^-^t 'into f , we obtain m E a.s.b,tf. + Z g.s.b.tb. +Z p.s.b.tq. =0 . n x 1 1 x . . B i 1 1 x rx 1 1 x^ x=l x=l Both these equations are valid for a l l s^, . . . ,S n»t e S . Subtracting, we get m (*) Z a s (f tb - b tf ) + Z g is 1(b.tb 1 - b tb ) + Zp is 1(q.tb 1 - b tq ) i=l i=l which, of course, is true for a l l s..,...,s,t£S. I n Suppose first that f,tb n = b..tf, , for a l l s,,...,s ,t e S 1 1 1 1 1 n • By (V, 4.1) i± ( s 2 S n ^ = x( s 2»* * *> s n^ bi » for a l l s„,...,s e S and where X(s ,„..,s ) e C . Now, by.the z n z n minimality of n :,(r_,...,r ) 4= 0 , for/ at least one choice of 1 I n 1 r„, ...,r e S . Define z n h(x2) = f 1(x 2,r 3,...,r^) h(x_) =f= 0 since h(r ?) 4^  0 .We may write 120 where the are C-independent in S and the d^ =}= 0 are from S Hence h(s) = y(s)b^ , for a l l s e S , where j u (s) = X(s,r 3 > ...,rn) e C . Thus, {h(s) : se S} = {Z 'c±sd_, : s e 3} i=l has dimension 1 over C . Then by (V, 4.2), dim^CeSe) < ^nd the proof is finished i f ^^.^l = k^tf^ • Therefore, suppose that f,t b. 4» b nt f. , for s r 1 o 1 1 1 o 1 ome r„,...,r , t e S . Set Z n f. = f.t b, - b,t f. , 1 i o 1 1 o i V. = b.t b, -• b, t b. , and x x o 1 1 o x q = q.t b., - b., t q. l nx o 1 1 ox Substituting these in (*) we find that . m K Z a.x.f: + Z g.x.br + Z p.x.q: = 0 , . , x 1 x . _ *?i 1 i r i l ^ i i=l x=2 . / where, of course, f'(r_,...,r ) 4= 0 . This identity is not trivial 1 Z n ' , m since, i f i t were, we would have Z a.x,f7 triv i a l and by choosing . . . i l l x=l ^ x^ = r_^  , i = 2,...,n, we would contradict (V, 4.2). We also note that in transforming the original G.P.I, into the form in which we now have i t has not altered the order of the x. x although some monomials may have vanished. We repeat the above procedure at most k times with the 121. result that the original G.P.I, is transformed into nontrivial G.P.I, of the form m E a i X l f i ^ x 2 ' " ' ' ' X n ^ + S ^ X 1 X n ^ = ° ' i=l where x^ never appears as the last indeterminate in any monomial . .  of g and the order of the x^ is unchanged. ' i Let x, x , r < n be the set of a l l those indeterminates 1' r ' — . \ which appear first in some monomial of the original G.P.I. By the \ preceding process applied to each x^ , 1 <_i ^_ r we can transform the original G.P.I, into one of the form Z a.x,f. + £ b.x„g + ... + E d.xh. = 0 , i 1 x x 2°± x r x where {a_^ } , {h^ } , ... , {d^ } are C-independnet subsets bf S , f.,g.,...,h. are nonzero G.P.I.'s of degree (n-1) in which none of x^,...,x ever appear as the last indeterminate in any monomial. -Since some indeterminate must appear last in each monomial we see ! I!-that r < n . B y the minimality of n , • i f,(r ,r„,...,r ) 4 0 , for some r„,...,r e S 1 2 3 n ' 2 n Define f _• (x«»• . •> x .) • f. (x0,..., x . j r ) , vX z n-l x z n—1 n g ^ ( x 1 , x 3 , . . . , x n _ 1 ) g i ( x 1 , x 3 , . . . . , x n _ 1 , r n ) , h i(x 1,...,x r_ rx r,... >x n_ : L) = h iU 1,...,x r_ 1,x r,...,x n_ 1 >r n) Claim: Z a.x.f: + Z b.x„g: + ... + Z d.x h; = 0 (**) i l l 1 2 1 l r l is a nontrivial identity of degree (n-1) . This would contradict the minimality of n , thus proving the theorem. Suppose (**)' is tr i v i a l . Then Z a.x.-f'T is triv i a l . i l l By letting c = f'(r„,...,r ) , we have E a.sc. = 0 , for a l l . i • i 2 n i i s e S , the {a.} are C-independent, c. = f'(r 0,...,r ) 4= 0 , thus , l 1 l l n 1 \ contradicting (V, 4.2). Hence, (**) is non-trivial and the proof of the theorem is finished. Much of the beauty of this theorem springs from the fact that no less than three of the important theorems which we proved in the first three chapters are corollaries of (V, 4.3). Let A be a primitive ring, V a faithful irreducible A-module. Then D = Horn (V,V) is a division ring. Let F be the centre of D , and set T = AF . T is a subring of Hom^ (V,V) jand i f C denotes the extended centroid of T we must have F <_ C that C <_ F follows from ([35], theorem 12). Thus, C =''"F The fact 'and the" central closure of T is T itself. Now, we are ready to prove Corollary (Kaplansky : cf I, 2.16): A primitive ring A satisfying a polynomial identity over its centroid Z is a finite dimensional central simple algebra. Proof: The identity satisfied by A is f(x. ,s...,x ) = X.X....X + Z a.x. x. ...x. =0 , 1 n 12 n . j " " -i i * l 1 X l X2 \ where a^ e Z . Note that f is satisfied by T as well. Suppose diiri^ V 2l n . Then Dn is a homomorphic image of a subring of T , 123. and as such, i t must satisfy f also. But this is impossible since we can choose- x^ = e£"^  , x^ = > = , ... . Hence, . • • / • dim^ V <. n < °° so that (V, 4.3) assures us that D is finite dimensional over its centre F Corollary (Amitsur : cf. V, 3.1): Let A be a primitive ring such that AF <_ A , where F is the centre of the associated division ring D . Then A satisfies a G.P.I, over F i f f A contains a minimal right ideal and D is finite dimensional over F The proof of this fact follows directly from (V, 4.3) by observing that those primitive rings studied by Amitsur are essentially just those which are F-algebras (i.e. AF <_ A ) . Corollary (Posner, cf. II, 4.4): Let A be a prime ring satisfying a P.I. over its centroid Z . Then A can be embedded as either ai; • j !! :-left or right order in its central closure S = AC , and S ! is a finite dimensional central simple algebra over C \ • Proof: A satisfies f(x n x ) - Z a.x. x. ... x. =0 , a. e Z I n . x xn x„ x x x 1 2 n We may assume that S = AC satisfies f too. Note that f is non- j trivial over C since different monomials have the x. in a different x order. From (V, 4.3), S is primitive so that by Kaplansky's theorem, (I, 2.16) S is a finite dimensional central simple algebra. We now show that A is an order in S i Claim: Any nonzero ideal U of A contains a regular element. Since S = we let e^,e^,...be the usual orthogonal idempotents in S . Write e. ="£ a. c. , where a. £ A , c. e C X X . X . x. • X . 3 j J 3 3 Because there are only a finite number of c ' s we can find a nonzero 1 j ideal W of A such that W <_ U and c^ W <_ A . , for a l l i , j 3 3 3 Hence e.W e. < U , and thus 0 4= e.W «=a A , and since A is prime1 x x — 1 x . r 3 3 e.W e. 4= 0 . We may then choose a nonzero u. e e.W e. < U , for x x 1 x x x — i = 1,...,k . Set u = u^ + ... + u^ . In S , u has rank k and therefore i t must be regular in A j For any finite set c.,c_,...,c e.C we find a',..., a ,b I L I m ... I m | e A , b regular, so that c^ = a^ b x , for a l l i . This implies, of course, that every, element in S = AC is of the form ,?ab -1 a,b e A , b regular. In order to find the . a_^  and the b , we first choose a nonzero ideal U of A such that c.U < A ,| for al'l. x — . 1 11 • • ' i i . Now, U contains a regular element b , so that we have j c.b = a. e A , for i = 1,... ,m x x In S this means c. = a.b x , i = l,...,m , and the proof of Posner's x x theorem is complete. Section 5: In 1955 Drazin [17] generalized polynomial identities in ! another way by introducing the notion of a pivotal monomial. This was expanded upon by Amitsur a couple of years later but since then ' 125. very l i t t l e has been achieved. If A is a ring over a set of operators fi , then a monomial of degree d in the indeterminates x^,...,x is any one of the t^ distinct formal expressions n(x) = x. x. ...x. , where the i , \ X2 Xd k take independently-the values l , . . . , t . Let II (x) = x. x. ...x. be a fixed monomial. Define • \ l X2 Xd \ = {c = x. x_. ...x^  : either i) q > d , or ii ) q <_ d and j ,=j= i ^ for some h} h h j q S = {a = x. x. ...x. : q > d but a 4= n} J l J2 Jq As usual, for arbitrary a^,...,a^ from A , we let 11(a) = a.a. ...a. , P (a) = {a(a) : a' e P } , and S (a) = (a(a) i . x 1 IT TT . TT ! 1 A d i a e S } • . TT Definition 5.1: A monomial LI(x) = x. x. ...x. in the indeterminates —;—; .... .- - - xi x2 xd x^ , ...,xt is called a right pivotal monomial for A over fi; i f for, every choice of a ^ , a we have ! Il(a)A <_ £ a(a)A , aeP i.e. n(a)A is in the right fi-ideal generated by P (a) , for a l l TT choices of the a. i Similarly, II(x) is a left pivotal monomial for. A i f AIT (a) <_ Z Aa(a) and a two-sided pivotal monomial for A i f aeP 126, AIT (a) A <_ Z Aj(a)A . When we just speak of pivotal monomials we shall aeP mean right pivotal monomials unless otherwise indicated. II (x) will be called a strongly right pivotal monomial for A i f Jl(a)A_< Z a(a)A . Similar definitions hold for strongly aeS TT left pivotal monomials and strongly two-sided pivotal monomials. Finally, a ring A which possesses a (strongly) right pivotal monomial will be called a (strongly) right P.M. ring. To show that the concept of a P.M. ring is, indeed a'generalization of a P.I. ring we prove Proposition 5.2: If A is a P.I. algebra over,the integral domain Q , then A has a strongly pivotal monomial. Proof: A satisfies a multilinear identity of the form Za.x.x. ...x. =0 , i 1 \ l X2 Xd where e Q. , and the sum runs over a l l permutations (i) of the d letters l,...,d . By renumbering the x^  if need be, we write this y polynomial as / ax,x,...x, + Z a.x. x. ...x. = 0 , 12 d . f l x x x x 2 x d where a =j= 0 . Define n(x) = x^x 2 >. .x^  . Since no permutation (i) of the sum is the identity permutation we see that each x. x. ...x. of the second term is in S . Clearly then x x x 2 x d I j Il(a)A is in the algebra ideal generaged by ^(a) , for a l l \ 127. choices of elements from A , and thus, A is a strongly P.M. ring. Other examples of P.M. rings are provided in; Proposition 5.3: The following are strongly P.M. rings: (1) rings with D.C.C. on right (left) ideals. (2) algebraic algebras of bounded degree.. The second statement is clear since algebraic algebras of bounded degree are P.I. rings [I, 2.15]. But, in each of these cases, we can prove that the rings in question, in fact satisfy a pivotal monomial in only one indeterminate. It is clear that a pivotal monomial in one indeterminate is a strongly pivotal monomial. By ([17], p. 354) for any a e A there is an integer N . . (independent of a ) and an element b (dependent on a ) such that 1 N N N a = a b , so that A satisfies the pivotal monomial x . Also ([30], p. 74) tells us that since N is independent of a , we N can choose b so that i t commutes with a . Thus, x is both a • . right and left pivotal monomial for A Suppose now that A is an algebraic algebra of bounded degree over ft . Each a e A satisfies a mohic polynomial of degree at most I in ft[x] so that / . m m-1 n. a + ct na + . . . + ct a = 0 , m-1 n -where--a-^ --e ft" ,~~m <_ I , n _< m ^xid ct^  =f= 0 -. Multiplying by a' we have . " ! i , Jl+l m+Jl-n. a na = g(a a ) , l-n 128 so that clearly, x is a (strongly) pivotal monomial for A and the proof is complete. By using Kaplansky's device of replacing x^ by u'N? for each i , we can write any pivotal monomial as one which involves only two indeterminates. More generally, Drazin [17] has shown that i f A has a pivotal monomial, then i t has a pivotal monomial which is linear in each indeterminate. Likewise a strongly P.M. ring has a strongly pivotal monomial which is linear in each indeterminate; Hence, i f a ring A has a pivotal monomial n of degree d in the indeterminates x^,...,xt then by renumbering the x^ and possibly adding some, new x 's we can always assume that II = x^x-.-.x, ± z d Equivalently, A is a P.M. ring of degree d means that for any a.,,...,a,,b e A the product. a.,a„....a,b is in the right i d l / d ideal generated by the d(d-l) products a.•a„...a._ a. (i,j = l,...,d ; i z 1 l j but i 4= j ) together with the d products a ...a,a. ( j = l,...,d) . . Kaplansky's theorem ( I , 2.16) tells us that a primitive • P.I. algebra is finite dimensional over its centre. This same result cannot be proved for primitive P.M. rings since this would imply that every division ring is finite dimensional over its centre. However, / we do prove the following analogue of ( I , 2.16): Theorem 5.4: If A is a primitive ring then the following three conditions are equivalent: (1) A has a pivotal monomial (2) A = . for some division ring D , integer :;q > 0< .• • 1 _ (3)_. A has a pivotal monomial x m in a single indeterminate x ,, (and hence i t is a strongly P.M. ring). | I! Proof; (1) => (2). Suppose A has a pivotal monomial n of degree d , say II =x^x ...x . Since A is primitive, i t is a dense ring of linear transformations of a vector space V over a division ring D We show that dirn^ V <_ d . Accordingly, let vQ,v^,...,vd be D-linearly independent elements of V . Since A is dense then by definition, there exist a^,a^t...,a^ e A such that v a. = v, , and v. a., = 6 , for a l l i f 0 , o i l l 1 1 • v^2 = v 2 , and v ^ a 2 ~ ® > ^ o r a l x i f 1 > v. .a, = v, , and v.a, = G , for a l l i f d - 1 d-1 a a i d ' Therefore, vll(a) = v a.a„.. .a, = v, , and o o 1 2 d d v a(a) = v a. a. ...a. = 0 , for a l l a e P O X1 1 2 . l d TT Since II (a)A is in the right ideal generated by a l l the a (a) from P^  (a) we see that v II (a)A = 0 = v,A . o a From v,A - 0 and the faithfulness of the A-module V d • ' we have A = 0 or v, = 0 , both contradictions. Thus, diiri V < d d u — and A = , for some q <_ d (2) => (3). Suppose A = D , for some division ring D q 2 q Now, Dq is a right vector space over D of dimension q so that each element of satisfies a nontrivial polynomial equation of degree 130, 2 q over D . Thus, i f d e D then there is an e e D with q q 2 2 d q - d q + 1 e , 2 i.e. x q is a pivotal monomial for D . Clearly, (3) => (1) and the theorem is proved. Corollary: If P is a right primitive ideal in the right P.M. ring A , the quotient ring A/P = , with q <_ d . Another theorem, proved by Amitsur [7], is a direct generalization of the important (II, 1.3)'. • Theorem 5.5:' If T is a n i l subring of a strongly P.M. ring A of degree d , then T is locally nilpotent and T d < NP(A) Proof: Let n be any integer greater than d . Define A_= T n~ 1AT 1 , for i = l,...,d , A. = ATnA , for i > d . x Then A.A. < ATnA , for i > i i J - -Take a^^ e A^^ for a l l i . Then 131 a. a. ...a. e ATnA , for p > d , or for p = d Xl 12 Xp and (±lt±2,...,±d) =f= (1,2,...,d) .. Since A is a strongly P.M. ring we have Aa na 0 a„A < E Aj(a)A < AT A , 12 d — _ — 0£S the last inequality having just been proved. If we now let the a^ range over a l l the elements of A^  the corresponding products a^a^.-.a^ will range over a l l the elements of (T n _ 1AT)(T n- 2AT 2) ... (T n _ dAT d) (T n- 1A) dT d Thus, (*) • (AT n _ 1A) d + 1< A(T n _ 1A) dA< ATnA If T is assumed to be not just n i l but also nilpotent we k let k be the smallest integer for which the ideal AT A is nilpotent. k-1 If k > d , then (*) implies that (AT A) is also nilpotent thereby contradicting the definition of k . Hence, k <_ d so that T d <_ NP(A) Suppose, however, that T is just n i l . Let t £ |T Then <t> , the subring generated by t , is nilpotent so that <t>d <_ NP(A) and hence t d e NP(A) . This means that the,. ri.ng, , .. i T + NP(A) — ^ p " ^ — is_ a n i l ring of .bounded index and thus is locally nilpotent. i u Since NP(A) is locally nilpotent, T must also be locally nilpotent1. 132. Hence, i f t-,...,t, are arbitrary elements from T , the ring 1 d <t^,...,td> is nilpotent and from the first part of the proof <t1,...,td>d <_ NP(A) , which implies, of course, that t n t _ . . . t , e NP(A) for a l l choices of 1 z d the t^ from T , i.e. T d <_ NP(A) , as required. This immediately gives us Proposition 5.6: If A is a strongly P.M. ring of degree d then A (1) the jail subrings of ^P(A) are nilpotent rings of degree <_ d . In particular, (2) [NL.(A) ] d <_ NP(A) . Hence NL(A) is also the Baer 2 lower radical of A and NL(A) = NP (A) . (3) If T is any nilpotent subset of A then T d generates a nilpotent ideal of A . Since n i l idealsv and nilpotent ideals coincide in rings with D. CC. on right ideals, NP(A) is nilpotent in such a ring and hence i from (V, 5.5) we get a result which was first proved by Levitzki: Corollary: The n i l subrings of a ring A with D.C.C. on r i are nilpotent. ght ideals: Section 6: From the.corollary to (V, 5.4) we know that P.M. rings of degree d possess the property: (M^ )' for every right primitive ideal P of A we have A/P = D, , where D is some division ring n and h ^  d . The' converse of this, namely, that a ring A having 133. property (M,) has a r i gh t p i v o t a l monomial i s not v a l i d . We now seek I d ! to charac ter ize r ings with property ) , and i n order to do so, we must extend s t i l l f a r ther the idea of a p i v o t a l monomial. Let U be a r i ght i d e a l i n a r ing A . An element a e A • ' X i s r i gh t quas i - regu lar modulo U i f there i s a b e A such that a + b - ab c U . A s usua l , a r i gh t i d e a l V of A i s r i gh t q u a s i - \ regu lar modulo U i f f each element of V i s r . q . r . modulo U . j Define j : i J(U) = {a e A : V x e A , ax i s r . q . r . modulo U} . I j In th i s no ta t ion , J(0) i s the Jacobson r a d i c a l of A . Define C(U) = C\ T , where the i n t e r s e c t i o n runs over a l l the modular maximal r i gh t i dea l s of A which contain U . Then, with the usual proof , we can e s t ab l i s h P ropos i t i on 6.1: C(U) = J(U) , fo r every r i gh t i d e a l U < A S im i l a r d e f i n i t i o n s and re su l t s ho ld fo r l e f t quas i -regular- i ty modulo l e f t i dea l s and f o r q u a s i - r e g u l a r i t y modulo i d e a l s . ! A monomial II(x) w i l l be c a l l e d a r i gh t J - p i v o t a l monomial or J.P.M. of a r i n g A i f f o r every sub s t i t u t i on x^ = a^ we have i- i n(a)A <_ J ( Z c(a)A) , oeP • TT or equ i va len t l y , n(a)b i s r . q . r . modulo Za(a)A- f o r a l l b e A A r i ng with a r i g h t J.P.M. of degree d w i l l be c a l l e d a r i g h t J.P.M. r i ng of degree d . The promised c h a r a c t e r i z a t i o n of r i n g s possessing property (M ) i s then Thereom 6.2: A r i n g A possesses property i f f A i s a r i g h t J.P.M. r i n g of degree d ; and then x i s a r i g h t J.P.M. of A Proof: Suppose n(x) = x i x 2 " ' ' x d ^ s a r i S n t J-F'M. f o r A . Let P be a r i g h t p r i m i t i v e i d e a l of A . Now, A = A/P i s a dense r i n g I of l i n e a r transformations of the r i g h t v e c t o r space V over a d i v i s i o n r i n g D , where V i s a f a i t h f u l , i r r e d u c i b l e A-module and | I D = Hom_ (V,V) . We show that h = (V:D) < d and t h i s w i l l imply that A ~ .. A/P = • A c c o r d i n g l y , assume . (V:D) > d , and l e t V Q , v ^ , . . . , v d be D-independent elements of V . By the d e n s i t y of A we can f i n d ^ a., — ,a, e A such that ' i . l a , I i i v a, = v„ , but v.a n = 0 , f o r a l l i {= 1 , ' ! o 1 2 i 1 1 i = v 3 » b u t x v i a 2 = ^ ' ^ o r a 1 1 1 ^ 2 » . v , • a, = v, , but v. a, = 0 , f o r a l l i {= d - 1 d-1 d d ' i d / ' Then, V Q I I ( a ) A = v^A j= 0 by the i r r e d u c i b i l i t y of V But v cr(a)A = 0 , f o r a l l a e P . Now l e t V = {a : v a = 0} o TT o o V i s a maximal modular r i g h t i d e a l i n A and thus s i n c e o v n(a)A 4= 0 we see that LT(a)A k V •. A l s o , a(a)A < V . From o ., o — o 135, the preceding proposition J(Ea(a)A < V and since n(a)A £ V we j o o i have as w e l l that II(a)A ^ _ J(Ea(a)A) , which contradicts the d e f i n i t i o n of a J.P.M. for . A .. Conversely, suppose A i s a r i n g s a t i s f y i n g property •(M^)' • We show that x d i s a r i g h t J.P.M. of A . Consider the rings D, ,h < d . I f e E D, , the chain of r i g h t ideals i ii h — h ° I i i 2 d d+1 eD. > e D, > .. . > e D, > e D, > h — h — — h — h — cannot have more than d d i f f e r e n t i d e a l s since the length of the composition seri e s of r i g h t i d e a l s of i s h <_ d . Therefore, we d+1 d^ . d d+1 . , d . ' have e D. = e D. , i . e . e e e D, and thus x i s a r i g h t h h h p i v o t a l monomial f o r , a l l h <_ d We show now that x d i s a J.P.M. f o r A . Let a e A a r b i t r a r y , and l e t V q be a maximal modular r i g h t i d e a l containing J(Ea(a)A) . I f P i s the r i g h t p r i m i t i v e i d e a l contained i n V then A/P = » h =s d' , by property (M^) so that a dA <_ (Ea(a)A + P) <_ Ea(a)A + V Q o Since t h i s i s true f o r a l l V Q 2l J(Zcr(a)A) then by ( V , 6.1) we see that a dA < J(Ea(a)A) , as required. Corollary: I f A s a t i s f i e s (M^) then the r i g h t J.P.M. of a l l matrix rings D^, h <_ d i s also a right J.P.M. of A 136. C o r o l l a r y ; I f A s a t i s f i e s (M^) t h e n a d e JCa^^A) ' - f o r e v e r y " a e A . We f i n i s h t h i s s e c t i o n by s how ing t h a t t he p r o p e r t y o f j| b e i n g a J . P . M . r i n g i s i n h e r i t e d by r i g h t i d e a l s . F i r s t , we need | Lemma 6 . 3 : I f A i s a J . P . M . r i n g o f deg ree d , t h e n f o r e v e r y d cL~f~2 a e A t h e r i g h t i d e a l a A i s r i g h t q u a s i - r e g u l a r modulo a A P r o o f : From the p r e c e d i n g p r o o f we know t h a t f o r e v e r y m a t r i x r i n g D, ,h < d , we have eD, = ed+"4), f o r e e D, . Thus , we have h — h h h d d~H2 ! e D, = e D, . A l s o f r om t h e p r o o f o f t h e l a s t t heo rem we deduce t h a t h h d dH~2 a A < J ( a A) f o r a e A . P r o p o s i t i o n 6.1 c o m p l e t e s the p r o o f . | Theorem 6 . 4 : I f U i s a r i g h t i d e a l i n t he J . P . M . r i n g A o f I deg ree d t h e n U i s a l s o a J . P . M . r i n g o f deg ree d i ' \ P r o o f : a^A i s r . q . r . modu lo a d + 2 A . A l s o , ah) <_ a^A and 1 a d + 2 A a d + " T J i m p l i e s t h a t a^U i s r . q . r . modulo a^+X\J . I f 1 s i s t he r i g h t q u a s i - i n v e r s e o f a^b modu lo ad+"Hj ( f o r any b e U I ) / / we have d. , cl d+1 d + l T a D + s - a DS = a t £ a U , d+1 d^ J_' <L i . e . s = a t - a b + a DS E U \ . Thus , f o r e v e r y b £ U , t he r i g h t q u a s i i n v e r s e o f a^b modu lo a d +"4j i s i n U . Hence , x d i s a r i g h t J . P . M . o f U 137, Section 7: . Just as he generalized polynomial identites Amitsur has also introduced the idea, of generalized pivotal monomials. This short section gives the only useful result known for generalized pivotal m'6n(|-' I ; mials or G.P.M.'s. ; Theorem 7.1: A necessary and sufficient condition that a primitive ring possess a minimal right ideal is that i t possess a (generalized) right pivotal monomial. Proof: [=>] Let A be a primitive ring and suppose that eA is a minimal right ideal. Then eAe is a division ring and therefore, eaebe e eaebeA for a l l a,b e A . Hence, II(x) = ex^ex^e is a G.P.M. for A . / [<=] Conversely, suppose the primitive ring A has a generalized right pivotal monomial. We can assume that this is of the form n(x) = a. x'a. x, .a. ... a. x na. l , d l , n d-1 i , „ . ; i n 1 l d d-1 d-2 1 o where the {a.} are from a C-base of . A . The primitivity of A means that A is a dense ring of linear transformations of a right vector space over a division ring D . Claim: {a. }•= {a. =l,a ,...,a } are D-independent. i £ 1 2 T Suppose £ a.d. = 0 , d. e D . Then for a C-base i i - l {a } of D we have A d i = Z c i a i * c i £ C j j 138. Therefore, 0 = E a.d. i x X E a.(Z c. a . ) i . x. j = Z (Z a.c. ) a . . . i i . 3 J 1 3 = 0 Thus, Z a.c. =0 from (V, 2.2) . . x x . 1 3 => C. = 0 since the a.'s are C-independent. x. x 3 So the {a. } are D-independent. By (V, 3.2) either i) A contains a finite ranked transformation, or i i ) for every integer h , VD contains a set of vectors v ,v, .....v. such that {a. v.} are D-linearly o 1 h x j independent in V Suppose case (ii) holds, e A such that Choose h > d . By density find , e h - l e i ( a i V o v , e (a v ) = 0 otherwise, 1 1 A U v 2 , e 2 ( a x v y ) = 0 otherwise, Then, e h ( V n v h - l ) = V h ' e h ( a A V y ) = ° o t h e r w i s e -h-1 : n(e)v = (a. e,a. e, ,a. ... a e.a.. )v ° xd d V i d - x xd-2 H 1 \ 0 = a. v f 0 since {a.v.} are xd ° • 1 2 139. D - l i n e a r l y independent. But c(e)v = 0 , f o r a l l o e P . This i s a contradiction since o ^ 11(e) <_ Za(e)A . Hence, case ( i ) must hold, A contains a l i n e a r transformation of f i n i t e rank, and thus a minimal r i g h t i d e a l . j BIBLIOGRAPHY 140 \ \ [1] Amitsur and Levitzki, Minimal Identities for Algebras, Proc. of the A.M.S., Vol. 1, #4, p. 449-463, 1950. [2] Amitsur and Levitzki, Remarks on Minimal Identities for^Algebras, Proc. of the A.M.S., Vol. 2, #2, p. 320-327, 1951. [3] Amitsur, Nil P.I. Rings, Proc. of the A.M.S., Vol. 2, #4, p. 538-340, 1951. [4] Amitsur, An Embedding of P.I. Rings, Proc. of the A.M.S.,.Vol. 3, #1, p. 3-9, 1952. [5] Amitsur, The Identities of P.I. Rings, Proc. of the A.M.S., Vol. 4, #1, p. 27-34, 1953. | [6] Amitsur, On Rings with Identities, Journal of the London Math. ; Soc, Vol. 30, p. '464-470, 1955. [7] Amitsur, Rings with a Pivotal Monomial, Proc. of the A.M.S., , Vol. 9, #4, p. 635-642, 1958. [8] Amitsur, Generalized Polynomial Identities and Pivotal Monomials,, Trans, of the A.M.S., Vol. 114, p. 210-226, 1965. [9] Amitsur, Nil Semi-Groups of Rings with a Polynomial Identity, Nagoya Math. J., Vol. 27, p. 103-111, 1966. 141. [10] Amitsur and Procesi, Jacobson Rings and Hilbert Algebras with Polynomial Identities, Ann. Mat. Pura. Appl., (4), Vol. 71, p. 61-72, 1966. [11] Amitsur, Prime Rings having Polynomial Identities with Arbitrary Coefficients, Proc. of the London Math. Soc, (3), Vol. 17, p. 470-486, 1967. • ! [12] Amitsur, Rings with Involution, Israel J. of Math., Vol. 6, p. 99-106, 1968. [13] "Amitsur, Identities in Rings with Involution, Israel J. of Math. i l!: Vol. .7, #r, p. 63-68, 1969. | [14] Baxter and Martindale, Rings with Involution and Polynomial Identities, Canad. J. Math., Vol. 20, p. 465-473, 1968. [15] Belluce and Jain, Prime Rings with a One-Sided Ideal Satisfying a Polynomial Identity, Pacific J. Math., Vol. 24, #3, p. 421-424, 1968. [16] Divinsky, Rings and Radicals, University of Toronto Press, 1965. 142. [17] Drazin, A Generalization of Polynomial Identities in Rings, Proc. of the A.M.S., Vol. 8, p. 352-361, 1957. I [18] Faith and Utumi, On Noetherian Prime Rings, Trans, of-the A.M.S.,' Vol. 114, p. 53-60, 1965. [19] Goldie, A Note on Prime Rings with Polynomial Identities, Journal of the London Math. Soc, (2), Vol. 1, #4, 1969. [20] Herstein, Theory of Rings, Math. Lecture Notes, University of Chicago, 1961. [21] Herstein and Small, Nil Rings Satisfying Certain Chain Conditions, Canad. J. Math., Vol. 16, p. 771-776, 1964. I [22] Herstein, Topics in Ring Theory, Chicago Math. Lecture Notes, j 1965. [23] Herstein and Small, Nil Rings Satisfying Certain Chain Conditions:: An Addendum, Canad/J. Math., Vol. 18, p. 369-375, 1967. i [24] Herstein, Special Simple Rings with Involution, Journal of Algebra, Vol. 6, p. 369-375, 1967. j ; i [25] Herstein, Non-Commutative Rings, Carus Math. Monographs of the • M.A.A., 1968. 1.43. [26] Jacobson, Structure of Rings, A.M.S. Colloquim Publications, :_ ...Vol. 37, 1964. -| ! j; . [27] Johnson, Structure Theory of Faithful Rings, II, Restricted Rings, Trans, of the A.M.S., Vol. 84, p. 523-544, 1957. [28] Johnson, Quotient Rings of Rings with Zero Singular Ideal, Pacific J. Math., Vol. 11, p. 1385-1392, 1961. [29] Kaplansky, Rings with a Polynomial Identity, Bull, of the A.M.S., Vol. 54, #6, p. 575-580, 1948. j [30] Kaplansky, Topological Representations of Algebras, II,Trans. j . of the A.M.S., Vol. 68, p. 62-75, 1950. [31] Levitzki, A Theorem on Polynomial Identities, Proc. of the A.M.S., '' ' • \ Vol. 1, p. 334-341, 1950. \ I [32] Martindale, Primitive Algebras with Involution, Pacific J. Math.,j Vol. 11, p. 1431-1441, 1961. j [33] Martindale, Rings with Involution and Polynomial Identities, Journal of Algebra, Vol. 11, #2, p. 186-194, 1969. 1 [34] Martindale,. Prime Rings Satisfying a Generalized Polynomial Identity, . Journal of Algebra, Vol. 12, #4, p. 576-587, 1969. 144. [35] Martindale, Lie Isomorphisms of Prime Rings, Trans, of the A.M.S, Vol. 142, p. 437-456, 1969. [36] McCoy, Rings, and Ideals, Cams Math. Monographs of the M.A.A., 1948. [37] Posner, Prime Rings Satisfying a Polynomial Identity, Proc. of the A.M.S., Vol. 11, p. 180-183, 1960. [38] Procesi, Ph.D. Thesis, University of Chicago, 1967. [39] Procesi and Small, Endomorphism Rings of Modules over P.I. Algebras, Math. Z., Vol. 106, p. 178-180, 1968. ! [40] Martindale, Proc. of the A.M.S., Vol. 124, //3, p. 508-511, 1970.1 / 

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.831.1-0080479/manifest

Comment

Related Items