SOME G E N E R A L I Z A T I O N S NILPOTENCE I N RING OF THEORY by R I C H A R D GREGORY B.Sc. A THESIS Stanford SUBMITTED University, in 1963 I N P A R T I A L F U L F I L M E N T . OF THE REQUIREMENTS DOCTOR BIGGS FOR THE DEGREE OF OF PHILOSOPHY the Department of Mathematics We accept required this thesis as conforming to standard THE U N I V E R S I T Y OF B R I T I S H July, 1968 COLUMBIA the In p r e s e n t i n g an this thesis in partial advanced degree a t the U n i v e r s i t y the Library I further for shall make i t f r e e l y agree that permission f u l f i l m e n t of the requirements f o r of B r i t i s h available Columbia, I agree that f o r r e f e r e n c e and S t u d y . f o r extensive copying of this thesis s c h o l a r l y p u r p o s e s may be g r a n t e d b y t h e Head o f my D e p a r t m e n t o r by h i s representatives. of this written thesis forfinancial gain permission. Department o f Mathematics The U n i v e r s i t y o f B r i t i s h V a n c o u v e r 8, Canada Date It i s understood Columbia 2 30th September, 1968 shall that copying or p u b l i c a t i o n n o t be a l l o w e d w i t h o u t my SUPERVISOR: P r o f e s s o r Nathan J. Divinsky ABSTRACT The the of development series paper of study and certain and central attempts r i n g s and of understanding series to to series define study are of has group greatly theory. particularly analogous what groups aided Normal important. concepts i n the interrelationships This theory exist between them. Baer ring has and theory Freidman equivalent s t u d i e d weakly possessing important results s e c t i o n s of of already a c c e s s i b l e subgroups. section k. phically The series. of these authors are properties In terms Not an that of the result that intersection not class theory given ideal and of Kegel the i n the more first equivalent are of meta* possess defined ideals meta* ideals many o f U*-ring of ideal i s not the the given This i s a meta always shown ideal a meta in ring. is a meta* other has. defined It i s also i s always homomor- are i s a U*-ring. of in nilpotent rings subrings f o r semigroups. every the equivalent Some o f theory series, of weakly chains of meta Also, of power n i l p o t e n t r i n g s i s not power n i l p o t e n t r i n g intersection ring However, i t does descending to the follows the central s e c t i o n 5 meta* every rary class ideals, paper. lower closed. ring central this possessing the upper Power n i l p o t e n t r i n g s , groups studied chain nllpotent rings, of groups three have contthat ideal. It since the ideal. Section kinds of m u l t i p l i c a t i v e studied that 6 i s concerned here are called a l l weakly products rings. suggests that class include last analogously actually shown to (hut with section t o group occur that that A result any r i n g s The every and power given may rings i s very The nilpotent which take rings large The rings rings on are prime U-rings rings. The b u t does n o t idempotent. types study which of which here. ring i s power nilpotent. certain and i t i s p r o v e d products are defined ring Finally, has a ring nilpotent. the conjecture that a r e power only place. nilpotent be p r i m e studies ring types. i n which i n the section a non-zero weakly o f some r i n g rings products i s n e a r l y completed does n o t p r o v e ) potent prime nilpotent products rings decomposition a l l U-rings of prime with types i tis type This a l l weakly similar suggests nil- TABLE OF CONTENTS Abstract i i Introduction v Not at i o n • 1 1. I-chains .2 . 2 . J-chains 11 3. U-rings 20 h. K-chains h2 5. U*-rings . . 51 6. Prime products 7. Ring rings types Conclusion B i b l i ography 56 66 , 81. 82 V INTRODUCTION The understanding advanced theory rings by t h e study has f o c u s e d and g r e a t description various done the h a v e b e e n made properties. s u c c e s s f u l , very rings. of rings that they There this here do n o t g i v e do g i v e little This reason provide Much of the inspiration nilpotence i n group t h e o r y . part of an explicit respect to efforts attempts have been t o explore of the r a d i c a l While of the nature some rings for the results i n f o r m a t i o n on r a d i c a l behind o f such rings, rings. further research i n the ring comes f r o m t h e s t u d y are obtained of the equivalent t h e "beginning Radical even more i n f o r m a t i o n . here indication with greatly- s e r i o u s work has been to believe that will results these properties. complete area parallel paper c o n t a i n most introduced at While some i n d i c a t i o n s i s every theory. to give are semi-simple commonly s t u d i e d r a d i c a l given of r a d i c a l a t t e n t i o n on t h e " r a d i c a l - f r e e " of r i n g s which on r a d i c a l classes and r e s u l t s attempts radical moderately o f t h e s t r u c t u r e o f r i n g s has been o f each theory of generalizations of It i s surprising how many i n t h e two t h e o r i e s . group theory appropriate concepts concept An i s given section i n this paper. 1. NOTATION The following symbols p a p e r t o mean e x c l u s i v e l y not defined and the following l a t e r on w h e n t h e y a p p e a r C i s any index set. N i s the set of n a t u r a l Z i s the set of a l l i n t e g e r s . TO i s the f i r s t the integer integers the [g.jh] non-finite p [x] . i s t h e i s t h e open largest (+) i s used t o denote . number. common m u l t i p l e between of the line g and h including subring of a r i n g " i f and <_ x . R g e n e r a t e d by the d i r e c t sum the subset A of of groups, r i n g s , e t c . only i f " . end the following o f an the subset S of R g e n e r a t e d by i s used t o denote the end ...} b. integer i s the 3 are text. i n t e r v a l o f t h e r e a l number i s the i d e a l of a r i n g means They h. <A> iff things. in this a n d q.. e n d p o i n t s a and g and are. u s e d i n the {1,2,3, ordinal i s the set of a l l integers both I(S) numbers (p,q.) i s t h e l e a s t set (a,b) with notations example of the p r o o f of a theorem a remark. or R. R. I-CHAINS 1. There its is i d e a l s ' to a ring some the enough make possible from the cases it ideals certain ideals might class of such to infer ideals. between some This useful to a ring or, a ring properties suggests generally, to the ideals of subrings is the set meta- were of in about related Meta i d e a l s of that know s o m e t h i n g more and a about ring. o r i g i n a l l y defined by (l) . An I-ehain 3 for every a begins with such I-chain of I =U i . a y< a y with R is a chain in R which ideal that exists S and reaches subrings is an ideal begins of a chain there of R is a meta ideal in R which S of a ring a finite The index number j of S of a ring an I-ohain A subring exists DEFINITION: which I 3 A subring exists DEFINITION: there = R, where J DEFINITION: there R is a chain a and i f a is a l i m i t ordinal, J natural of a ring I^ Cl i^ C. . . . C- of R if even be of relationship its which are a ring. DEFINITION: if of ideals subrings of it nature of the A promising Baer a close R after S is the an I-chain j stepsj a+ of R S. ideal begins Ij of R with S. smallest in R i.e. R = 3 . I.-chains structure to are p r i n c i p a l l y a s s o c i a t e d w i t h meta have been first independently, with the rings which studied below ideals. b y R. ' B a e r ( - 2 ) , ( 3 ) , (M Freidman i n which results mainly tools, used every t o analyse the ring Meta i d e a l s ( i ) . are interesting L a t e r and a p p a r e n t l y studied s u b r i n g i s a meta appear them i n connection ideal. i n themselves, Although they are i n t r o d u c e d t o a i d i n p r o v i n g more, c o m p l i c a t e d will appear later Theorem 1 . (Freidman) implies that RS i n t h e paper. S i s a chain ideal + S R C n n results of index n i n R S. PROOF: Let I — i s an p RS S C n L 2 ideal = ( I n. +,1 S ) S C — I .C 3 i n I — p + 1 ... d — and — n , =R n n + 1 S C I C. — ( I nS ) S " n - 1 J I be an I - c h a i n . f o r a l lp p i n (Z. — ... CL _ I _pS C_S . 2 Since [ 2 , n ] , '. S i m i l a r i l y•>, ' S R C S. n Theorem ring 2 . . (Baer) I f S i s a chain ideal R and i f l ( S ) i s t h e i d e a l [I.(S) C of index i n R generated n in a b y S, then S Q I(S). PROOF: It . .. i s easy I n + t o see t h a t l ( S ) C- RSR. - ^ = R b e an I - c h a i n . 3 For every L e t S <Z I integer g C. I C p i n [2,n],. (I P ST ) P = ( I SI ) S ( I S I ) C I P P P P ~ 3 2 SI P-1 2 n _. Hence P-1 i t c a n "be on seen that Theorem [I(S) ] C I SI 2 3. ( B a e r ) an ideal of R of index C S. 2 5 I f S i s a subring o f R such that I P o f R, a r i n g , C. S C. I , t h e n and I i s S i s a chain ideal m <_ p . PROOF: The f o l l o w i n g i s an I - c h a i n . . . C. S + I = I C beginning m E. with S. This Hence chain i n R: S C- S + I has p steps t h e index P _ 1 C. S + I and i s an o f S i n R i s some P " I-chain integer — P• - Theorem h. I f S i s a c h a i n then i s a chain any Q 2 S natural ideal ideal of index of index n i n a ring m <_ n i n R w h e r e R, k i s number. PROOF: S u p p o s e t h a t , t h e f o l l o w i n g i s an I - c h a i n . .. C — I , , = R. n+1 I-chain i n R. natural number. Theorem 5. a nilpotent k S C Then Hence — S I c 2 — has index A n i l p o t e n t chain ideal. i n R: S C. I C .. . C I = R i s a l s o an — n+1 m <_ n w h e r e k i s a n y H ideal i s always contained i n PROOF: Suppose t h a t S i s a chain 3 By theorem the ring fore 2, R. and that S =0. n [l.(S)] C. S w h e r e n Hence' [ l ( S ) ] ' i s the.index CIS 5 =0 and of S i n l(s) i s there- nilpotent itself. Chain from the theorem H i d e a l s have more general 5 and R e m a r k A. in ideal the considerahly meta different i d e a l s as a comparison between f o l l o w i n g remark i n d i c a t e s . (B.ae-r) A n i l p o t e n t m e t a i d e a l a nilpotent properties need not be contained ideal. EXAMPLE: Let vector V = space integers (+) isN V, 3n f be w n eR n defined otherwise ring • of R f n generated n (v.) I 3n and by: = 0. by S the i = 1,2,..., 2i+t. < 3n, there one dimensional v_^ o v e r the ring property Then , 2 =0. n There the of that f ( V . ) = 0 i f i > 3n i — f (v.) n l . J Hence S denote field a l l linear f ( V . ) C V. -. (±) V. _ l — l+l i+ 2 f o r a l l feR . Let n Let exists I n be the f in R ideal such that of exists Given geR an such n ,) = 2 i - l integer t that R f(v„. n [3n/2].. of = v. , , i f i i s even and i . < 3n; l+l ' •2 • f = 0 .and S = {0,f } i s a subn n n n for i s the the. v e c t o r R^ V with . . by Let on i f i < V. 1 2.. transformations each 1 generated modulo ... @ V. , w h e r e g.( v . Q > 1 such n ) = v that and v„. 2i gXv^) that = 0 i f g-f -f(v n It is if k 4 2i. g*f n the k ^ 21+1. 2 1 _ ) * f and that Since • g'f n = v 1 also, t r u e , that n d I and — n since the discrete 3 crete i + g«f f and and g - f . - f ( v ) t n *(v =0 k . ) = v g are and i f g«f I f.( V \ ) C V. R 3 n _ J 1 n direct direct 2 functions must c o n t a i n a l l e l e m e n t s n I . contains a l l elements f of n property R The 0, sum o f sum o f t h e © I the rings V n _ i+ f 1 n' . . . © 0 lf 0.'. defined so k 4 2 i - l . ( v, ) = 0 of the form R which V . have Hence Now l e t R be rings S R , l e t S be. t h e disn and l e t I be t h e discrete , n direct sum o f the rings I • where n ranges over, t h e natural 2 numbers. an ideal Then S i s of R which is crete direct ideal generated in is non-zero all by S in ideal. homomorphic is theory The following I. I of meta (a image that ring of from is =0,. since I . n l is the rings are of apply is to j S is disThe not contained and a i f are meta locally thereevery two-sl:ded p r o v e d .below t h a t unresolved particularly is > n. ring contains I the weakly n i l p o t e n t rings the is n weakly n i l p o t e n t ring It while =/ 0 i f a nilpotent 0). most ideals theorem and S and t h e r e f o r e of weakly n i l p o t e n t a fact the nilpotent, R is of weakly n i l p o t e n t It R and E a c h R^ i s different subrings of rings weakly nilpotent annihilator sums not sum o f t h e a nilpotent fore a subring all and that ideals. E problems nilpotent interesting in rings. since it 7. suggests why the study related to. t h e s t u d i e s Theorem 6. (Baer) o f meta ideals of generalized Every idempotent might he. c l o s e l y types of meta nilpotence. ideal i s an ideal. PROOF: Let S be following Let of i s an I - c h a i n : a be t h e l a r g e s t I and a . Since I therefore Similarly, and hence It number that of meta meta t o show ideals ideal intersection B. Suppose t h a t number of I that such a+1 n SI' a+1 ••• C I ^ that , SI = S(SI 2 n R. S i s an and S C I — = S I = the a+1 that i s a meta of a meta However, the i n t e r s e c t i o n ideal unlike o f an i n f i n i t e ideal. of a I t i s also of R i s i t s e l f the result ideal d I a+1 — a n -) a+1 C S I ' c S . n ,.S C S . T h i s s h o w s t h a t S i s an i d e a l a+1 — I = R a n d S i s a n i d e a l o f R. a o f R. Remark ordinal i t follows ideal. S C I ^ C I ^ C i s an i d e a l a i s easy a meta meta I a meta ideal be an i d e m p o t e n t — of I a— .. a+1 H finite true a meta f o r ideals, the number of meta of chain ideals ideals need not ideal. The intersection need n o t be a ideal. EXAMPLE: Let with S he t h e s e m i g r o u p the following defining g e n e r a t e d by t h e s e t {x^: relations: neN} 8. (1) S is (2) x (3) x commutative =0 2 ± = x n Let k 1 R be with the basis <_ p}> with f o r a l l neN. n n algebra S. L e t T be + x^R. T + T R: i s an the field ^ of i n t e g e r s s u b r i n g . <{x following P i s a chain P which The T + T R P T + T R over the T + T i s an P _ 1 R k p ... in R But y i s a sum of terms of the form x < n. i term for a l li in [l,k], x l f o r at l e a s t some n . l different of terms still of from which has T. i s even. an Due idealizer The also raises is a meta So hence i s not theorem a problem i d e a l o f R, contains S? Many if stronger version where k one n such appear i n the sum of the terms since i s not one i n the i d e a l i z e r T i s i t s own a meta i d e a l . has which does P number which y T R), natural a r b i t r a r y n a t u r e o f y, I which this odd Then yx^^T subscript. following It an subscripts e q u a l s y. t o the and L e t h be a l l the even and Hence m, ... x n m.. i R. i f yeR^T, -m-,. then and T = f~\ ( T + peN ideals. 2 . beginning C. T + TR C i d e a l of R H o w e v e r , I n t e r s e c t i o n .of c h a i n : p i s odd * I-chain C modulo 5 f a r reaching is still R necessarily open. have additional results of the theorem implications. were Namely, i f S a proper could true-. be ideal proved Theorem ideal 7• (Levic) S, t h e n I f a r i n g R has a proper, non-zero meta R i s not simple. PROOF: Let limit S C L C — d — .. C I „ = R be — fcs o r d i n a l , then I_ n an I - c h a i n . I f B i s not a i s an i d e a l o f R w h i c h i s proper and P -1 non-zero. any Therefore non-zero ring element of R generated ever x e l suppose xeR. B is a limit I f xR + Rx = 0,. t h e n b y x , a n n i h i l a t e s R on b o t h f o r some a < B a n d h e n c e S <Z I a is ordinal. 4- R. Select S, t h e s u b sides. How- Therefore S a a proper, non-zero i d e a l o f R. I f RxR = 0 a n d xR + Rx ^ f R. It follows that 0, 2 then R either ^ 0 a n d h e n c e Rx ^ R a n d x R Rx o r xR now on i t w i l l and a.,b.cR therefore i s a proper, non-zero be assumed t h a t RxR i d e a l o f R. ^ 0.. Hence from n L e t K =' {• / a. x b . = n e N i =l f o r every i i n [l,n]}.. I f K = R,. t h e n m x = ; c . x d . f o r some e l e m e n t s c . , d . e R . . l l l l i =l xeK and Let y L be n the smallest [l,m]. be Then a limit o r d i n a l number such I meta i s a proper ordinal. Now that x,c.,d.el.. f o r a l l i i n i i y i d e a l of R s i n c e y cannot i t can be shown that K d I . To Y ' for a l l n V a.xb.el i=l Y nJ a . x b . e l . a.,b.eR. I f a.,b.el for a l ll m [ l . n ] , then I'I I ' I Y . ^ i i Y ' i=l S u p p o s e t h a t a i s a n o r d i n a l n u m b e r >-y. a n d s u p p o s e t h a t a l l do t h i s i t i s sufficient t o show that 1 1 n 10 n { Y a. x b . i =l the elements i n the set all i i n [l,n]j. l i e i n I . y 1 Then neN i f neN n all = i in [l,n], n i =Yl a . x b . l i e s s i n c e i =Y l a . x b in I . - , 1 1 Y . - , 1 m T Y ( a. :c . ) x ( d .b . ) a n d s i n c e ' a l l t h e e l e m e n t s j=i 1 J J 1 ordinal numbers a.c. and 1 d.b. l i e i n I . . H e n c e b y t r a n s f i n i t e J i a limit for and a . , b . e l _, f o r i i a+l n ±ii at a n d a. , b . e l J i n d u c t i o n (the step i s obvious), the i d e a l K lies in I Y This not negates simple. the possibility that K = R and hence R i s H 11. 2.. Weakly nilpotent rings have been especially by equivalent d e f i n i t i o n of weakly a chain Kegel J-CHAINS (8), (9). condition. equivalent central over, weakly nilpotent of (see Kurosh ZA groups relationship is proved between A ring homomorphic different image from images rings ring nilpotent based on ring theory theory. More- theory rings an equivalent An important and meta ideals U-rings. R i s weakly R are the (l) for definition). 3 on of rings i n group are the gives contains nilpotent i f every a two-sided non-zero annihilator 0. From t h i s phic J-chains before, 8 below nilpotent series weakly i n section DEFINITION: Theorem Actually, of upper studied d e f i n i t i o n i t i s easy of weakly nilpotent t o see rings that a l l homomor- are weakly nilpotent rings. DEFINITION: of R The of J dJCl...CZJ . s 1 of largest ideal a l l the i f a. i s a J 0 P Z sisting a ring = J -, -,3 c I and J-chain of limit where ordinal, property 3 then of ideals ideal con- 1 annihilators the chain J ~ i s the P-r 1 two-sided R with R i s the that J a = [ j y<a — ' in R 3 J.^J . Y ' ^2 a+ + RJ' ^s the ^ CZ. J^ 3 0 12 .. DEFINITION: at J if J & The = J. g DEFINITION: consists J-chain a ring R terminates (or ends) . &+ 1 A ring of of R has a trivial J-chain if its J-chain 0. Theorem 8. (Kegel) J-chain terminates A ring at J Q = R i s weakly nilpotent i f f R's R. P PROOF: Suppose t h a t occurs i n R's homomorphic R i s weakly J-chain nilpotent. and t h a t i m a g e o f R, m u s t Suppose a l s o 4 R. contain Then that R/J^, a an a n n i h i l a t o r o f R / J D P different R/Jg to forms K/J D P relation R's from 0. However an i d e a l , where KR J-chain Suppose K*, the set of a l l a n n i h i l a t o r s of of R/Jg. K i s an i d e a l + RK CL. J which a o f R. shows does n o t t e r m i n a t e R's J-chain M o r e o v e r , K* is But K s a t i s f i e s that until terminates J„ ^ J ., . n i t reaches at J 0 = R. isomorphic the Therefore R. L e t R/K he P a n o n - z e r o homomorphic ordinal J number has x*, t h e image the property o f R and l e t y he t h e m a x i m a l CL K. Then t h e r e e x i s t s an x i n Y satisfy the relation: xR + Rx d J d K. — y — such t h a t , ., d- K w h i c h m u s t Y+l Hence image J o f x u n d e r t h e homomorphism that x*(R/K) + (R/K)x* = 0 . R -> R/K, . Hence R/K does 13. contain weakly a non-zero annihilator. This shows that R i s nilpotent. Theorem 9» ideals 5 A weakly nilpotent ring with ACC o n t w o - s i d e d i s nilpotent. PROOF: Consider R's J - c h a i n : 0 (Z J-, C •• • ^ J 1 this chain must terminate Since J for every natural 2 B y ACC = R f o r some n a t u r a l P 6 J E + E J <C J - , , i t follows that J — n n — n-1' p. n = R• Q P at J n number + = 0 1 TJ + 1 Theorem weakly 10.. number n a n d t h e r e f o r e A subring S of a weakly R = 0 nilpotent ring R i s nilpotent. PROOF: Let And If l e t S have t h e J - c h a i n xeJ that limit S D J S O J n . . . CL J = R. H: n ., C H _, . a+1 — a+1 ordinals 3 S /H R = S. Theorem 11. weakly J By t r a n s f i n i t e i s obvious) i.e. is OC-J^CJ^CL 0CH_,<SHC...C1H = H 1 2 . Y Y+l x R + Rx = 0 a n d h e n c e x S + S x = 0.. I t f o l l o w s then l 5 J: <Z H . S u p p o s e t h a t S D J C H . Then i f xeS D J 1 — 1 a — a a+1 + Rx (Z J a n d h e n c e x S + S x C. S C\ J C H . I t f o l l o w s — a — a — a that xR R have t h e J - c h a i n Hence A complete nilpotent. i t follows S i s weakly direct induction that S f~} J ( t h e step * C p — nilpotent. sum o f w e a k l y at H , p E nilpotent 0 rings ih. PROOF: Suppose (+J R = A yeC where Let R have t h e J - c h a i n Let A 0C J: have t h e J - c h a i n H J ^ C 0 C : i f J S = a (+) (H' ) y ye = max{3 nilpotent. a J = d ( H ) _ cC ... C Y 2 A A . ^ 0 implies r\ \ = r ••• C 1 Y since : Y £ G } , C (H ) Y Not e t h a t i s weakly Y Y Hence A each Y (+) • A = that * R and + i s weakly R 5 A discrete Corollary. weakly rings Y r\ = X . nilpotent. is (H ) nilpotent i s weakly direct sum o f w e a k l y and a s u b d i r e c t nilpotent sum o f w e a k l y rings nilpotent nilpotent. PROOF: This, f o l l o w s that such direct proved Theorem ring nilpotent a l l of the extensions weakly pleasant 10 a n d 1 1 a n d f r o m t h e f a c t sums, c a n b e r e p r e s e n t e d sum o f w e a k l y Of paper from theorems nilpotent properties. above, rings 12. ( K e g e l ) of a complete rings. ^ of nilpotence studied have t h e g r e a t e s t In addition the following as s u b r i n g s number o f t o the general results are also I f I i s a weakly R a n d R^ CI I , t h e n R i s w e a k l y i n this properties useful. nilpotent nilpotent. i d e a l of a 15 . PROOF: It I i s sufficient i s weakly JI (RJR)R either case JR ideal therefore has J* Suppose =0. As either i s true R i s weakly K is and image annihilator anni- J*(R/K) ? 0 event R/K has f o r every nilpotent. K e g e l ' s t e r m i n o l o g y (9) the f o l l o w i n g theorem non5 says i s a l e f t , conservative'" p r o p e r t y . being weakly property. annihilator. ( R / K ) j * ( R / K ) i s an this conservative case case J i s a a non-zero above, then and s i n c e nilpotent first Otherwise the annihilator being weakly 0.. ^ f ( l ) and J * a n n i h i l a t e s f ( l ) . 2 o f R, = i s nilpotent I n any image RJR o f R where R/K. i s easy t o see t h a t right image f 0 or J * a n n i h i l a t e s z e r o homomorphic In (R/K) that a non-zero t h e n R/K Again annihilates i n the second R has f has that In the i n the last annihilator. o f R/K. since (R/K)J*(R/K) a non-zero It I f I OK, Since J such It Suppose a homomorphic a non-zero i s an i d e a l (R/K)J* that he Hence o f I. u n d e r t h e h o m o m o r p h i s m which o f R. o f R; o f R; o f R. o f R. hilator.....' o f R/K or annihilator annihilator R -*• R/K a proper f(l) RJR ideal JR J 0 o r R J ± 0 o r JR + RJ = 0 . annihilator l e t f: a non-zero R J I + I J R = 0.. i s a non-zero non-zero t h e c a s e w h e n p = 2. Consider the i d e a l i s a non-zero Now i t has + R(RJR) C Then RJ nilpotent + I J = 0.. R: t o prove nilpotent i s also a 16. Theorem of 13. ( K e g e l ) E, t h e n I f L i s a weakly LR i s a w e a k l y nilpotent nilpotent left ideal ideal. PEOOF: Let LE have t h e J - c h a i n H: 0^ H ' d H d 1 Note t h a t that LR i t c a n be p r o v e d every i s an i d e a l i s not weakly homomorphism and note Then 4 LR. + l induction a s o f LR. Now suppose L e t f be t h e = R*. y = f ( L ) f ( E ) = f ( L E ) = LE/H b y f : E -»- E/H L e t L* = f ( - L ) . Hence L*E* Y has a t r i v i a l J - c h a i n a n d L* i s a w e a k l y n i l p o t e n t l e f t ideal o f R*. L e t L * h a v e t h e J - c h a i n J : 0 <Z J d J c. • • • <Z J = L* Since L*E* = H Y y by t r a n s f i n i t e o f R as w e l l nilpotent. of R given that easily ...dH 2 L*R* c a n n o t be n i l p o t e n t , L * J R * is D n 0 D 1 ^ P cannot be 0 f o r o t h e r - = L * J . R * =' 0 . L e t r\ b e t h e s m a l l e s t p o r d i n a l n u m b e r s u c h t h a t L * J R* ^ 0 . Then n c a n n o t be a n l i m i t o r d i n a l s i n c e i f L * J R* = 0 f o r a l l o r d i n a l n u m b e r s a. < n a a n d n i s a l i m i t o r d i n a l , t h e n L * J R* = 0 . N o t e t h a t L * J R* wise (L*R*) 2 C L*L*R* n is an i d e a l (L*R-*)L*J sided n R* d - L*J E* = 0. n-l n n have A left exchanging E on b o t h a trivial J-chain only sides _R* = 0 a n d n-1 L * J E* i s a n o n - z e r o the fact J-chain. the requirement J with a two- n contradicts may b e d e f i n e d modulo - Hence a n n i h i l a t o r of L*E* which L*E* must n o f L * R * a n d L * J R*(L*R-*) <Z- L * J that 3 i n t h e same w a y that as a J-chain J a n n i h i l a t e the ring the requirement that J a'+l .17. annihilate R on the left modulo J ( l e t J" If a ring R has non-trivial when R has right a non-trivial J-chain? both J-chain, The a non-trivial as a study of below w i l l verify. Whether written a direct sum below as i s an Remark C. of intriguing J-chain, i s no, left the ring and IR = also subring). rings as R have i n the and given a l l possible two 0 the does even J-chain a case non-trivial i n the examples i n the a example can given be example question. It is possible for a ring J-chain = RJ left answer be 0 a t o have n o n - z e r o R to have a trivial i d e a l s I and J such that 0.. EXAMPLE: Let integer on the 0,. are B be and of y main ring and a l l but of the diagonal and the ring c o n s i s t i n g of matrix with of every to the number copy Let R row with J be of the except that left of a l l elements zeros. a l l elements restrictions form the a l l of the R of the row the row diagonal 0.. Let mapping one-sided form (y,0). the ideal (0,'.z) w h e r e first entries identity except one-sided the main under the I be of with e n t r i e s are Let every the of of. t h e A R = A (+) B. i s a coxw s q u a r e m a t r i x with a l l coxoi s q u a r e m a t r i c e s a finite anti-automorphic consider R the e n t r i e s and the filled of A be of ideal where first R consisting z i s a coxco s q u a r e f i l l e d with zeros. 18. Then RI if (u,v) = uA (+) vB then of = 0.. = JR i s an (RJ)R satisfy the The ideals where I(n) n IR 0. and A i n the are IR nilpotent, a union A The chain: J a - 0 C {xeK: [) J Y<a Y DEFINITION: in R'if (u,v)R Similarly, only two-sided RJ are statement above ideals of the i n the Since i s the union of weakly J-chain. a special kind The 0,. and remark. first and of n 5 in A rows and therefore ideals. result nilpotent they the nilpotent following of weakly R matrices l(n) is nilpotent = ideals, union entries occur (u,v)(A(+)B) annihilator of non-zero i s the = i f R(u,v) of those J(R)-chain J, C ... of CL J K's an ideal = J « D l ,~ = a+l and since shows ideals is nilpotent. DEFINITION: J J-chain l(n) consists a trivial of R two-sided and example non-zero from i s the RJ Since columns. However A has weakly (0,0) So a trivial 0.. u m u s t be of A where a l l the weakly = element c o n d i t i o n s i n the ring first the 0. Moreover, Rein) = that then v must be R. the arbitrary 0,. = However, R has K C where + Rx CL J } . and — a An ideal a ring 3 J^ = {xeK: 3 is xE + Rx l p xR R i f a is a J limit ordinal, . J(R)-chain K of ends a ring at K. R i s niIpotently embedded = 0}, 19. T h e o r e m ih. (Kegel) If embedded itself, then in R is R is the union weakly of ideals nilpotently nilpotent. PROOF: Suppose in R. Let R = I ae c have I where the L each j(R)-chain (j is ) a : a nilpotently 0 C n embedded ),C l a ...C )„ = 1 f o r each a i n A . Then R has t h e J - c h a i n J : p a a 0 C J C ... C J = R w h e r e J ~ 3> I ) ( j ) „ f o r e v e r y o r d i n a l (J a number 6 and hence i f Y = maxtB a Theorem R 2 15. If R is a non-zero :ae0},- J must Y be weakly n i l p o t e n t R. E ring, then 4 R• PROOF: Suppose J: J 2 that R is 0 C. J ^ C. . . . d. J g = {xeR: weakly nilpotent = R. x R + Rx C J Suppose } ='{xeR: also and has that the 2 R J-chain = R. Then (xR + Rx) R + R(xR + Rx) = 0} 2 + x R x + R x = 0}.. A similar 3 2 2 3 J ^ = {xeR: xR + RxR + R xR + R x = 0} 2 That is, J g computation However, But is then = {xeR: shows xR that 2 s i n c e R = R, J ^ = J ^ and hence J ^ must e q u a l 3 2 R = 0, R i s n i l p o t e n t , and t h e r e f o r e R 4 R. a contradiction. R. This E 20 . 3. U-rings (3) of (5) (h) a U-ring radical, U-ring the have been is and is the to that potent. Freidman all result true true. for Some DEFINITION: A ring a meta ideal of R. Theorem (Freidman) a U-ring. as does not n i l radical, the Brown-McCoy follows Freidman the that has remaining radical every n i l characterized problem is to has been U-rings are weakly n i l - - a corollory the result that is weakly prove which results R is a U-ring (2) It However, general Freidman the U-rings. U^-ring U^-rings conditions. 16. U-rings, by that It Since (6) states Freidman obviously is radical. locally nilpotent locally nilpotent fortunately, of proves Jacobson locally nilpotent jectured not the Levitzky part extensively Freidman locally nilpotent. characterize is studied (2) equal radical-free every In (6). U-RINGS it this is nilpotent. corollory proved satisfy Every weakly below certain on U - r i n g s i f each con- are subring nilpotent Unand it that the additional given first. S of R is ring R is 21. PROOF: Let R have t h e J - c h a i n J: 0C J ,C J . = R. p L e t y he t h e m i n i m a l o r d i n a l 1 Let S he any p r o p e r number There such that J subring cL S. o f R. J 2 n C Y cannot Note t h a t ... C . be a l i m i t ordinal. Y x in J exists ^ S which s a t i s f i e s the r e l a t i o n : Y xS + Sx C J E + R J d J . CZ S . H e n c e S i s n o t i t s own Y Y Y-l izer i n R. ring T which in suppose R i s not a U-ring. i s not a meta R starting which a t T must ideal. 17. (Freidman) a U-ring. Also This Every every This Then subring any I-chain S of R i s a contradiction. homomorphic subring R has a sub- means t h a t e n d a t some p r o p e r i s i t s own i d e a l i z e r . Theorem is Now image of R i s a ideal- 2 o f R, a U-ring, U-ring. PROOF: Let R/K b e a h o m o m o r p h i c of R. L e t S* b e a n y s u b r i n g to S/K f o r some s u b r i n g R beginning an I-chain ... c r R/K. R/K with exists S C I Hence C Then 1^ C ... C S*: ideal K i s an S* i s There e x i s t s with S* i s a m e t a since E. ideal isomorphic an I - c h a i n i n The f o l l o w i n g i s S* C ^ / K C o f R/K. I ^ K C I t follows that S* i s a r b i t r a r y . T be a s u b r i n g an I - c h a i n 2 o f R where o f R/K. S o f R. i n R/K b e g i n n i n g i s a U-ring Let S: image o f R and S any s u b r i n g i n R beginning with S: S C2 I o f T. p C There I _ (Z . . • C. R. 22. The following . . . C. TH i s an I - c h a i n R = T. Hence i n T: S C S i s a meta Tf)I ideal <C T / 1 l 2 C 3 of T and T i s a U-ring. *S The which following i s not weakly Theorem not theorem narrows the search 18. weakly U-ring U-ring nilpotent. There e x i s t s a locally nilpotent i f f there with for a a trivial nilpotent U-ring exists which i s a locally nilpotent J-chain. PROOF: Suppose weakly R is a locally nilpotent. . . . C. J = J D 0 Let R have t h e J - c h a i n ,. Then R/J P + -L p therefore nilpotent U-ring Q which i s not J : 0 <C J ^ C i s a homomorphic image J g ^~ of R and P a locally nilpotent U-ring. However R/J D has a p trivial J-chain. The converse DEFINITION: Let i s clear. S - {x E : se(O.l) and s is a rational number}. s Define multiplication s + t < 1; The of with otherwise Z'assenhaus modulo basis S will S by ^. = 0. x x s Example integers in modulo p with be called the Let rule: S. x s p be p i s the basis x ^ - any x s prime algebra More a Zassenhaus +t over generally 3 Example. number. the any field algebra 23. The upper haus and class bound of on t h e Example i s a Baer locally not Lower a different class of is The following others may h a v e m u l t i p l i c a b l y indecomposable this weakly paper on p r i m e nilpotent indecomposable that this Theorem 19. are elements a ring n {x. : l x^ ielO, such that 4- 0 w h i l e X Q x. l = 0. i rings it is U-ring. sufficient Let Then S is {x^: i c N } are fore commute. S be the commutative powers It to show t h a t subring since of an w o u l d be subscripts with the multiplied together any R as result section all of elements, n. > 2 for l — a possible all ieN and U-ring R is subring by'{x : ± the not a ieN}. sequence elements i n the pleasant that the well. generated in two element property the of a In quite seems of is a l l n i l U-rings It PROOF: It example multiplicably = x . ., w h e r e i - l not this proved that a sequence Then R i s places by the n i l U-rings R has Zassen- theorem that good nilpotent elements. contain. for a very excludes suggests generated they may h o l d Suppose also products rings result It not locally on U - r i n g s w h i c h it. is Although a both a n d many of like it ring. bound rings n i l U-rings. a U-ring, Radical upper nilpotent sequence and there- i f the elements in when two elements were would be either 0 or an S had element 2k. with two [y subscript e q u a l t o t h e sum o f t h e s u b s c r i p t s multipliers. ( l / )] k I fx = *(!£_: k n '.n. . , i i=l = y can be renamed y ^ p / elements Every the as s such interval and y {n^: Lemma 1 9 A . Consequently n. l peN. Since any i n S are both powers o f ~ G ). that S, This found Exactly f o r elements which ^(g+.-k) = gives a number w h i c h rational i n S depends (which subscript i n a Zassenhaus y^eS i s a r a t i o n a l (0,1). y-subscripts integers the that • n. generating much l i k e ) ) f o r every may b e 0 i f s + t > ( n ^ - l / n very then k n k of t h e form y some x_^ i n t h e s e q u e n c e , . , i = l S structure / n J1 n two ( rr n . . , l i - l ; (x^)^ i s r e n a m e d y, of the Example. lies i n numbers s appear on t h e s e q u e n c e o f ieN}. Suppose y characteristic andy of y eS a n d s f o r i = 1 , 2.. < s < 1 a n d L: i s Then L^ d i v i d e s L^ . i PROOF: Note that L-. y 1 Since L y p c- sg = 0 implies l S = 0, L , t h e g r e a t e s t j that (L y 1 common S l )y, 2" 1 I S divisor S ; % = L y 2 of L 1 S- = 0 S and L , d. 25. must be a s o l u t i o n of the equation Xy S the L 2 smallest must positive be DEFINITION: integral and t h e r e f o r e A point in =0. Since d solution ofthis Lg does d i v i d e S will be an L i s 2 equation, . element of the form y . Is If the additive non-zero element characteristic i n the ring of every S i s 0,. d e f i n e o r a l l b u t one G = 0. w i s e l e t G* = m i n { c h a r a c t e r i s t i c ( y ) : e c h a r a c t e r i s t i c (y ) > 1}. Let y S be any element s s c h a r a c t e r i s t i c G*. Either y g s a n Other- d s with e Q (l) y S i s the only 0 . point i n S which has c h a r a c t e r i s t i c G* or (•2). there such t h a t In case exists open implies that te(a^,a ) 2 G. Then 2 hasc h a r a c t e r i s t i c ( y ) > 0*} a n d l e t y s s characteristic ( a , a ) . CL ( 0 , 1 ) interval ( l ) l e tG = min{characteristic characteristic has a maximum every ( y ) : y eS a n d s s be a p o i n t i n S which ± point y where s Tj must have a maximum characteristic open G b y lemma 19A. ( a ^ , a^) CL ( 0 , . l ) that y^ has c h a r a c t e r i s t i c Note also that i f G = .0, t h e n ( a ^ , a ) CL ( 0 , l ) s u c h 2 characteristic G. 0. that such t h a t I n case there te(a^,a ) 2 G*. < t. < s _L Hence (J there t e ( a ,a ) 2 exists implies ( 2 ) l e t G = G*. i s a maximum open implies y^_ h a s that interval 26.. DEFINITION: (a^^a^) is called DEFINITION: of G is called the form the •primary the primary characteristic interval of Sj of S. A formal additive relationship in S is an equation h £ L.y = 0 where s . = s . implies that i = j i=l i ' 0 3 l L.eZj and L .y 1s S * Is ^ 0 for every i in [l. -h].. s V Lemma 1 9 B . S i n which interval There every exists term no has formal additive subscripts relationships which lie i n the in primary ( a ^ , &• ) . PROOF: Let formal h be additive scripts in additive Let s m the (a ,a y.eS. s ). where s e(a s, } a n d h exists this number , a>,) for every s „ = minis, , I 1 a rational fact of terms there number exists . ••, + s . can be with m Since rewritten fewer than L y m as ( s + t } 'm ' ~ ' y,eS t such = 0, J L . y _ x=l ^ ~ i ' a formal h terms. additive This is a ( g + t v } ' = in s, } . h that - v i s. < u s u c h h < t that a h a s , when e v e r y t e r m has subh \ L.y = 0 is a formal X=l 1 Suppose = max { s _, , 1' Due t o positive relationship relationship any u > 0 t h e r e J fewest ( v [l,h]. Given that • t + s < a^ Si • 2 h I I L . 1, i =l i relationship contradiction. in ; y t = (a^,a>j) o 27. Lemma 1 9 C . S i n which There e x i s t s no f o r m a l a d d i t i v e any term has t h e form H a n d t < g/2 w h e r e Hy g i s the length where relationships G does i n not divide ofthe primary interval, ( a- , a ) . 1 2 PROOF: m S u p p o s e Hy + £ L . y = 0 i s a formal additive * J=l 3 w h e r e G d o e s n o t d i v i d e H a n d t. < g / 2 . S u p p o s e 3 ship s., < ... < s, . < t. < s. , < . . .. < s . 1 h h+1 m is a., + g / 2 < t +. u < a _ . an a d d i t i v e this the can he r e w r i t t e n primary interval A point DEFINITION: My^ = 0 for every If DEFINITION: y y i s 0 since is contradicts eS i s s > s t every term H y ^ as a f o r m a l a d d i t i v e which that J r e l a t i o n s h i p i n which ( a ^ , a^) h u t n o t e v e r y t e r m also T h e r e e x i s t s y. eS s u c h u m T h e n .(Hy. + J L . y • ) y = 0 0 = 1. 0 v that relation- S + lies i n ^ ^ 0. Consequently relationship i n lemma 1 9 B . an M-endpoint if M-y ^ 0 but s where M is an an M-endpoint integer. for some integer M and L s is the then smallest L is the Lemma 1 9 D . contains positive near Every points integer characteristic dense of subset s such that y t h e r e i s no p o i n t y g i n S. that y y is an L-endpoint 3 . o f an open interval i s n o t an M - e n d p o i n t s or such (b^jb^) CZ. ( 0 , l ) f o r a n y MeZ 28. PROOF: If the M-endpoints i n S are ordered according near characteristics t h e n no two M - e n d p o i n t s near characteristics a n d as t h e n e a r M-endpoints crease limit increase towards point 0.. (plus towards Since points i n S have subset of the interval limit points. at most Hence infinity, one l i m i t (h^^^) f o r some k e l l Case P eP(S) 0 - 19 w i l l l e t p o f t h e M-end- i n t h e dense (S) now b e i n S at a l l . finished. { p r i m e s p: = subset o f any M - e n d p o i n t s o f any p o i n t s p divides y^/j^^} Suppose P ( S ) i s an i n f i n i t e h T = { £ L.y , +• i=l i i and l e t £ set. Then c h o o s e v y + £ H y sS: j l w w = e Z , (A ,k- ) = 1, a n d ( p , k ) = 1 f o r a l l i i n [ l , h ] ; Q M.eZ, J H a n d such t h a t ( l ) : L > one many are not the y-subscripts k e B r only CZ ( 0 , l ) h a s i n f i n i t e l y S or are not the y - s u b s c r i p t s k point. have de- dense in E = ^-¥2./^^'' of the But every ( b jb-g) e i t h e r Let integers some o f t h e p o i n t s p r o o f of theorem t h e same the y-subscripts the y-subscripts of The have characteristics the positive infinity), to their w i a n d y;; i s a n M . - e n d p o i n t . s. j eZ, and e i t h e r , t w w i n [1,v] } f o r a l lj i n [l,m]; > g/2 o r G d i v i d e s H w f o r every 29. Note t h a t t h e s e t {£/k: k,£eN a n d p divides k} i s d e n s e i n b (0,g/2). From t h e p r o o f such te(0,g/2), that o f l e m m a 19D there t = Jl/k w h e r e p exists divides some y ^ e S k, a n d y i s not - O an M-endpoint no formal f o r m H y, w t J y^cS^T f o r any i n t e g e r additive r e l a t i o n s h i p s i n v o l v i n g elements exists of the where t < g/2 a n d G d o e s n o t d i v i d e H . w w w a n d t h e r e f o r e T ^ S. Note t h a t t h e product M-endpoint with any o t h e r ( Ly- ) = LH y, , u w t +u w where J for By lemma 1 9 C t h e r e M. every o f an e l e m e n t .in S i s 0 and t h a t either G divides w in [l,v]. Hence LH w or t w ( H y )• w t w +. u > g/2 • ; I f p divides neither k nor k , then o ^ Consequently, (L y • )(L y I... ) 1 ^2.^.1 2 2 1 p o does not. d i v i d e k k . 1 2 = L L y> n , •• 0 i = 1 , 2.. under then exists set k, £/k< t. >' J l / k s u c h there lies i n T i f L.y„ of S since and m u l t i p l i c a t i o n . divides Q , Hence T i s a s u b r i n g addition p \L exists y , Ly eT s u c h 4- 0. n o t d i v i d e - G, Since that 1/k J. / Consequently, not i n T since 19C this which is ( Ly £ / f c p element Q i t i s closed I f L y ^ ^ e S ^ T , '! , a n d g/2, L does that eT f o r P ( S ) i s an infinite + Jl/k < m i n { g / 2 , t } . ) = Ly divides cannot ^ ( k k ^ , (p » Q be e x p r e s s e d i s not 0 and i s Hk^+k) = 1 a n d b y lemma a s a sum o f t e r m s l i e i n T. H e n ' c e . . - L y ^ i s n o t i n t h e i d e a l i z e r element. there 1 ) (y i t s own i d e a l i z e r and (£,k) o f T and T i n S due t o t h e a r b i t r a r y n a t u r e of this 30, Case p^eP(S) the (2): Suppose such that sequence p P(S) i s a f i n i t e divides {n_^:ieN}. set. an i n f i n i t e Note t h a t Then choose number o f t e r m s i n every power o f p^ divides h some k s u c h t h a t y Let Q = { ,..eE. l/K-' Y L.y i=l (£^,k\) = 1 , a n d k_^ = p ^ Let q be a p r i m e {I M.eZ, J H Note w (0,g/2). exists for and y w a point any i n t e g e r . s. v + j=l j 0 ^ 1 1 i s an M . - e n d p o i n t j > g/2 w — 1 1 l e m m a 19C t h e r e H Y m t h e s e t {Ji/p^ : From elements + Y M.y L.y . . . H w = l •'•w y, w eQ: 1 eZ, and e i t h e r t that 1 1 q ^P(S) and l e t h i=l 1 1 f o r some n e N f o r a l l i i n [ l , h ] } . 1 1 such that Q* =' eS: L . e Z , &./k. 1 -> • , n the proof or G divides £,neN a n d (&,p^q.) te(0,g/2), and t = £/p^ , where n exists H w fora l l w in y that there i s n o t an M - e n d p o i n t (£,p^q) = 1. By no f o r m a l additive relationships of the form H y where w tw t . < g/2 w and G does H e n c e y,eQ ^ Q* t and t h e r e f o r e Q 4 Q*- [l.v = 1} i s d e n s e i n o f l e m m a 19D i t f o l l o w s y eS s u c h t h a t M, f o r a l lj i n [ l , m ] ; Now, involvin not d i v i d e note that i f 31. L y . , and - L l * , - ] , / -&-]_ L. L^y L y , are ^- 0. *• 2 ' 2 '.. :. . , . '.; ; i s an s t a t ement s f o u n d the form this H y W case elements Ti i n case where w . also, Q* i n Q* , t h e i r element i n Q*. ( 1 ) on' M . - e n d p o i n t s either t > g/2 w — i s a subring and = 1, then (q,£) there exists = 1, o f Q. and If L G. d o e s n o t a rational Since number t H eQ apply i n w ^ Q* and L, £/k such < that g/2, L ^ n that and min{t,g/2} there exists Consequently, (Ly £ < ( £/k + q/p^) a point J-^j-^ : )(y n ) 0 and l e m m a 19C which and does n o t = Ly this chain a idealizer is in Q w n \ ^P^ +qk)/kp (q,£p^ n " t i n the due h i c h k^. i s n to the x + qk) e x p r e s s e d as n o divides n = 1 and a sum of idealizer by terms of arbitrary Q* nature E A ring R i s a U^-ring i f every subring of R is ideal. It given ly^/^ be p^ number element. DEFINITION: define since cannot Hence i s i t s own of t h i s a element l i e i n Q*. Q* l i e i n Q* that s u c ] a 1 not f o r some n a t u r a l 0. t y Note of £/k divide > &/k the element s o r G. d i v i d e s y (£,k) product, appears that v e r y few the boundaries below U-[_-ring. shows t h a t of the not results class have been obtained which of U ^ - r i n g s . every weakly nilpotent The example ring is 32. Remark D. Not e v e r y U-ring is a U^-ring. EXAMPLE: Let {0)U{x n S be t h e s e m i g r o u p : n has l e s s t h a n smallest prime dividing multiplication * nm i n S be has l e s s t h a n dividing • nm primes Let R be t h e a l g e b r a S ^ {0}. {x ordered number ordering subring • .according over Any the integers I-chain of steps U\-ring. 1 • therefore Hence \_) neN modulo 2 w i t h an e v e n index i n the with Let way. basis I 2n 0 . Hence an prime i n t h e n x_^ d o e s n o t o c c u r after ordering Q has i f p i s the 2n+l-st i n any R i s not a J and n o t e t h a t f a c t o r s and t h e s m a l l e s t J prime o f R g e n e r a t e d by t h e s e t o f Let R have t h e J - c h a i n , i f n has k prime p-k o f n i s p. until x x =0.. n m i n the. u s u a l i n R beginning since of the primes, i n the I-chain x x = x i f n m nm otherwise to size : p i s a prime with the primes}. infinite J P by t h e r u l e : Let f a c t o r s where k i s t h e s m a l l e s t L e t Q be t h e . s u b r i n g elements the defined elements f a c t o r s where k i s t h e i s square-free; ^ the of be k prime n and n i s s q u a r e - f r e e } . k prime a n d nm c o n s i s t i n g of the set of x n prime occurs factor = R and R i s w e a k l y n i l p o t e n t and n R i s a U-ring. in 5 33. DEFINITION: a chain A ring of index R i s a U^-ring n. <_ M, for .DEFINITION: M i s an index ideal ring of the The has following with t h e example what n i l U^-rings Theorem 20... some integer bound index for subring of R i s M. a U^-ring i f every chain n\ <_ M. theorems f o r remark are i f every from Freidman F give (6) together a fairly good p i c t u r e of like. (Freidman) Every nilpotent ring i s a U^-ring. PROOF: Suppose S o f R. R P = 0.. By t h e o r e m Hence R has i n d e x It of a i s also given theorem rings, true that The p r o o f n. <_ p . 5 a homomorphic image of a U^-ring o f i n d e x b o u n d n <_ M of index M i s a U^-ring and t h a t of index statements i s t h e same f o r the corresponding statements f o r U-rings i n However, u n l i k e a direct U-ring. of index subring of these 17.' R e m a r k E. c. S C. R f o r e v e r y b o u n d p. subring of a U^-ring that P 3, S i s a c h a i n i d e a l index bound M i s a U ^ - r i n g b o u n d n. <_ M. a Then 0 = R the result' f o r weakly sum o f U - r i n g s (Freidman) A direct as nilpotent n e e d n o t be a U - r i n g . sum of U^-rings n e e d n o t be . 3h. EXAMPLE: Let first in R = Z/(2) + Z/(2). ring Z / ( 2 ) and l e t . f be t h e i d e n t i t y o f t h e s e c o n d t h e summand. which L e t e be t h e i d e n t i t y o f t h e Then e + f generates a subring proper i nR i s n o t a n i d e a l o f R. The proof following o f t h e main series theorem rin* 5 o f lemmas l e a d on U ^ - r i n g s directly which t ot h e are l o c a l l y n i l p o t ent. Lemma 21A.. ( F r e i d m a n ) and is has index either I f R i s a locally bound M and t h e .additive torsion free or every nilpotent group non-zero U -ring 2 structure element i nR has a d d i t i v e 2M+1 order p where p i s a p r i m e , then x = 0 for every xeR. PROOF: 2 M Let x be any e l e m e n t 2 i n R a n d o b s e r v e t h a t . <x>» <x >* IC. <x > f o l l o w s f r o m t h e o r e m 1 a n d t 2M+1 h e f a c t , t^h a t . a2 i subring of a U -ring is again a Ug-nng. Hence x = 2 , c.x w h e r e c. i s a n i=l 2 1 integer let by and x k = 2j + l. multiplying 2 ^ 4 0 while Otherwise both sides x 2 j +2 = 0.. l e t k = 2j.. Suppose x 2 j + 1 The e q u a t i o n s of t h e equation: X 4 0.. Then obtained 2M+1 _ .^ c i = l. ^2i 1 k —2n by x f o r n = 1,2,. M show s u c c e s s i v e l y that, c^ = 0 35. for n = 1,2,. form: x 2M+1. element M. = x 2M+1 y which The e q u a t i o n f i n a l l y y where yeR. can s a t i s f y I fx such 2M+1 o b t a i n e d has. t h e ,_ f 0, t h e n an e q u a t i o n i n a t h e only locally 2M+1 nilpotent desired ring i s 0 itself. I f R i s a locally index hound M such additive of =0 which i s the nilpotent U^-ring result. Lemma 2 1 B . ( F r e i d m a n ) with Hence x o r d e r p where R which that every p i s a prime i s g e n e r a t e d by e x a c t l y where t = H ( 2 M + l ) 3 M non-zero x i n R has and i f S i s a H elements, subring then = 0 - H + 1. PROOF: Let S be t h e r i n g generated by t h e elements x ,x ,. . . ,x H A c c o r d i n g t o lemma 2 1 A , ( x . ) ^ ^ ~ = 0 f o r e v e r y oM 1 2 . taken, from R. 2 + 1 i i n [1,H]. By t h e o r e m 2,. I.(<x >) i C <x > C I.( <x > ) . Hence M l(<x.,>) ( 3 S C l ( < x — + the 2 M + 1 ) = 0 f o ra l l i i n [l,H]. •>) + l ( < x >) + ... + l ( < x >) a n d t h a t 2 n J. ... + I.(<x >)]"k = 0 , xi statement Lemma 2 1 C . = 0 where (Freidman) I f R i s a locally that where p i s a prime and S i s any s u b r i n g H ( H elements, 2 M + 1 ) 3 M 1 2 given i n o f t h e lemma, bound M such = H(H that [ l . ( < x •>) + l ( < x >) t i s t h e number index exactly Due t o t h e f a c t s - H + 1 then nilpotent e v e r y x e R ^ {0}. h a s a d d i t i v e |S| -i)/(H-a). <_ p U where U^-ring order p o f R g e n e r a t e d by u+1 with 36. PROOF: Let x S be t h e s u b r i n g ,x , . . . , x . Then S* g e n e r a t e d b y x = 0 and c o n s e q u e n t l y t h e semigroup ,.. . ,x 1 H + H 2 + H u + 1 = H(H t - = H(H 1 t - 1 -l)/(H-l) elements. Let H ( 2 M + 1 ) 3 M _ H + 1 L positive i n t e g e r . <p a n d e a c h y ^ i s a n o n - z e r o semigroup S* a n d y . = y . i m p l i e s number o f s u c h that sums i s n o m o r e t h a n Lemma 2 I D . ( F r e i d m a n ) member i = j . of the However, t h e p . U I f R i s a locally nilpotent U^-ring e v e r y x e R M 0 } :has a d d i t i v e i n d e x bound M such t h a t p where no more t h a n -1)/(H-1). The r i n g S t h e r e f o r e u o f sums o f t h e f o r m T L . y . w h e r e e a c h L. i s a . ., l i l i =l consists with contains H + ... + H 3 o f R g e n e r a t e d by. t h e e l e m e n t s order p i s a prime, then R i s n i l p o t e n t . PROOF: •y Select by choice where, v + 1 = M ( M ^ M 2 M + 1 be. d e n o t e d b y x^,x^, subring a n y M+p M ^ 3 " M + 1 elements from t h e r i n g ^ -1) / ( M - l ) . x ; y-^yg, y M of R g e n e r a t e d by t h e elements Let. t h e s e x v p ' Let ,x , . . . , x elements s . a R be t h e Let = x_, x ^ . . . x,„. D e f i n e a. r e c u r s i v e l y b y a. = a. , y . f o r 0 • 1- 2 . M l l l-l" I all i i n [l,p ]. Then i t f o l l o w s from t h e o r e m 1 t h a t each a.eS a n d t h e r e f o r e t w o o f t h e p +1 e l e m e n t s 1 o 1 p be e q u a l . S u p p o s e a ^ = a., w h e r e i < j . Then a^ = a^z where 7 V 37. z i s a n e l e m e n t , o f R. Since a. m u s t b e 0 a n d h e n c e l in this prop'er subring choice Theorem a -, = 0. P v R r M+p o f R c a n be w r i t t e n nilpotent ring V =0 since i n t h e form every of a ^ 7 element by a of the elements 2 1 . •• * ( F r e i d m a n ) F ( R ) , and R is a locally the periodic If R is a locally part nilpotent U^-rin^ o f R , has. c h a r a c t e r i s t i c q_ > 0, then' R i s n i l p o t e n t . PROOF: First F(R) F ( R ) ,the periodic consider has. c h a r a c t e r i s t i c q_ > 0,. F ( R ) may direct sum of a f i n i t e number structure i s a primary show t h a t F(R) i s n i l p o t e n t each sum of these which structure. that p S p group Then S i s a U ^ - r i n g n _ 1 S ± 0.. ring. p(p S) = 0 and t h e r e f o r e is a nilpotent is easy is a nilpotent ring. t o see a f t e r ring. additive primes since group p. a finite, as a To that direct S u p p o s e S CL R i s a as i t s a d d i t i v e and t h e r e exists 21D p group n W n - 1 such S is. n-2 a nilpotent S/p whose Then b y lemma • ••• " The r i n g Since i t i s s u f f i c i e n t t o show i s nilpotent. has a p r i m a r y o f R. be r e p r e s e n t e d for distinct i s nilpotent rings = 0,. b u t p n of rings p group subrings of nilpotent ring part p •> S- i s ' _ a l s n i l p o t e n t Continuing a finite the factor t o argue number ring in this of steps that since (P^S/P 1 1 -^) fashion i t S itself •38. Now its consider additive R/F(R) w h i c h has group structure. exists w > 0 such that proves i n (13) t h a t x a torsion free According group t o lemma 21A = 0 f o r a l lx i n R/F(R). W a torsion free ring with the as there M. Nagata property w that R x =0 for a l l x is nilpotent The (6). an since i n the b o t h R/F(R) and f o l l o w i n g remark The fact that and of the A U,_,-ring r i n g has need nilpotent. Fen) example every subring i d e a l o f t h e r i n g makes R e m a r k F. r i n g must be are n i l p o t e n t . also appear i n the given i t a l l t h e more n o t be Hence H i n Freidman example is startling. nilpotent i f the p e r i o d i c part c h a r a c t e r i s t i c 0. EXAMPLE: Let ^ p elements and X p i s a prime : f o r each prime 0. .<_ n. < p } . P multiplication nx -mx = P is y J = M a k e A^ by the r u l e s : [(nmp)mod p ] x . P nx (+) p prime A L The {nx = subring : p and [(m+n)mod p ] x (A ) p+ . : P p P "'" = o, i.e. A P Then yeR implies that P n v / m x w h e r e p. i s a p r i m e . p . p . * i ^ 1=1 * i* i a l li i n [l,n]. = P n for + mx p Hence A p addition defining P Let R = a set of d i s t i n c t . number p d e f i n e a r i n g by P nilpotent. n u m b e r ) be and 0. < m . < p. p. ^ l ^ i p. i , m ' p * , g e n e r a t e d i n R by . i eN y, i s 39. really an.ideal of R since y x = [ ( p ,.m Pj J where j lies primes, every i n [l,n]; {p.^: i = l , 2 , . . . . , n } , subring U^-ring. prime DEFINITION: non-zero height if Since the set y x eS . of R and t h e r e f o r e R i s not nilpotent since (x ) P A ring is periodia element A ring of is of if R is the equation the if Hence R i sa 4 0 f o r every additive a power of is xzR a primary mz = x has order of number. the ring 3 additive a natural primary An element the of E element DEFINITION: =0.. p. non-zero DEFINITION: y x are a l l d i s t i n c t , o f R i s an i d e a l However number every otherwise )mod p ] x i f p = p P j ring order some 3 a solution of prime every number has infinite z for every p. me/I/. Lemma 2 2 A . infinite I f R i s a primary height ring i nthe additive a n d xeR i s an e l e m e n t o f group o f R, t h e n x R + Rx = 0.. PROOF: Suppose t h a t additive there group x e R i s an e l e m e n t o f R. Let p e x i s t s , a sequence k i n R,. t h e n = 0.. H o w e v e r follows that height order i n the' o f x. Then i n R, { x : n e N } , s u c h n x =• p x ^ = p x ^ = ... = p x ; f o r e v e r y element p y be t h e a d d i t i v e of elements 2 that m of infinite there i s a non-negative xy = ( p x ^ ) y k neN. I f y i s an i n t e g e r , k such = x^Cp^y) = x « 0 .= • 0 . xR = 0.,- F o r s i m i l a r reasons Rx = 0. I t that ko. DEFINITION: subring basic B which basic a The has subgroup of d e f i n i t i o n of It an 0 has i n R/B R's of a primary i t s additive additive ring group group. R is the structure (See the Kurosh (10): for is always subgroup). from i d e a l of the group for basic follows additive subring l e m m a 2 2.A t h a t ring, since structure. infinite R/B Hence height and a basic has subring a complete group for i t s every element other than therefore (R/B) = 0. It 2 follows that R d Theorem 22.. L e t R be be the p e r i o d i c B - part a locally o f R. nilpotent U^-ring Let F(R) = (+). F ieN p^-primary their ring usual and {p^: ordering. ieN} L e t A. i p sequence be i d e a l of F p a l l elements of i n f i n i t e is a i of primes in consisting p. I of l e t - F(R) where F i s the the and height in F I . i subring Then R i s w e a k l y P nilpotent i f f o r a l l ieN the basic o f F^ P sum of rings of f i n i t e I A. i is a dire 1 positive,c h a r a c t e r i s t i c PROOF: Let additive F be order the any set of a l l elements power o f p.. Then F(R) 1 F(R) i s weakly nilpotent i f each i n F(R) w h i c h = (+) F . ieN i have Henc p F i s weakly P i nilpotent. hi. Let A. = { X E F : x has i infinite height i n the additive group P of F p. (10) is }. a primary By finite that F p. /A. = F. 1 r i n g which L e t B. l follows is. group 1 f o r d e f i n i t i o n ) of the height. it The he group contains the basic from theorem 12 assumption, the i s weakly F no p. l that r i n g B^ nilpotent. F^ and ring since of the F^ i s weakly first term S i n c e ' , F. l i s weakly nilpotent I t follows Therefore p. ring F and 's J . - c h a i n . 2 — since F(R) i s weakly therefore nilpotent. CB. i i f B^ of rings of from theorem 21 i s weakly n i l i since By A^ is a theorem sub21, I R/F(R) i s n i l p o t e n t , and F. l infinite i s a d i r e c t sum nilpotent of F of Kurosh of, E. l P potent (see hence the elements subgroup positive characteristic. B. i s an. Ulm. f a c t o r R i s weakly nilpotent E k2. k. K-chains are the ring series i n group t h e o r y . theory equivalent this section seen that that For rings many J & 2 = of one The limit ^ ordinal, R e m a r k G. true in that the Some o f K a O V e for a a ring RZ> K, Z> K Z2 1 2 n v e v y 28 and the ring theorems in J-c'hains , i t can he the same properties for weakly n i l - below i s also K-chains these the true. have, u s e f u l relations are presented types. of •Y< a R = RK general ring R: a Since K on on are statement theorem another. J + The lower, c e n t r a l rings h a v e many o f do. to of comparing section J-chains of a+ By rings rings K'-chain ideals equivalent rings 7 below DEFINITION: of i n the corresponding to section chain groups. those classes relations in ZD nilpotent theory Power n i l p o t e n t power n i l p o t e n t weakly potent of with K-CHAINS ordinal R ... is the Z> K following = K D number 8 a Q 8 + 2 and 3 descending • , where if a is R = 0 a Y every natural whenever number n, a is finite. = R ^", i t i s n+ However, K a R 4 RK a case. EXAMPLE: Let R be the set where a i s of the form 2m/(2n+l) where m and i s any b of a l l 2x2 r a t i o n a l number. matrices Then R of and the n form are i s a ring with a 0 b 0 integers the usual K n3 • h3, matrix K: addition R D K n K 1 and m u l t i p l i c a t i o n . ... 3 K n 2 = K Q P P + Let R have t h e . . .• T h e n K Q 1 = R K-chain =. t h e s e t n • (A) neN of a l l 2x2 m a t r i c e s first row, second there. RK A simple non-zero column and any r a t i o n a l calculation shows that entry number K R = 0 occurs may i n the occur while s 0) DEFINITION: of the only = K . 0) at where A ring•R is power nilpotent if R's K-chain ends 0. The following power nilpotent T h e o r e m .23. every theorem provides definition rings. A ring non-zero an a l t e r n a t i v e , R i s power ideal I of nilpotent i f f I R + R I 4 ..I f o r R. PROOF: Suppose of R. I cjL Kg = 0,. I <$- K property. The follows Now that holds. ideal Hence K: and I i s a n o n - z e r o ideal R Z> K ZD • • • 3 f o r some s m a l l e s t a c a n n o t be IB + RI C — K Kg = 0.. ordinal a with a limit ordinal. . R + RK _, CL K . a-1 a-1 — a Hence It I 4 IR + R I . suppose t h a t a non-zero a number I d K . and t h e r e f o r e — a-1 ideal nilpotent Let R have t h e K - c h a i n Since this R i s power R's K-chain of R f o r which i f R i s n o t power I of R f o r which ends a t Kg the relation nilpotent, IR + RI = I . 4 .0.. KgR then Then + RKg there Kg i s = Kg i s some H kk. 2k. Theorem ideals A power nilpotent ring R with DCC on t w o - s i d e d i s nilpotent. PROOF: Consider E9K R's K - c h a i n : D L 2 n 1 B y DCC t h i s chain must K^ = K p Since K^ = R + 1 . H. ideals need not c o n t a i n P + 1 , nilpotent R P + 1 number of steps = 0 i f R i s power ring with any n i l p o t e n t 0. = p end a f t e r a f i n i t e Remark from A power 3... 3 L ACC o n at nilpotent. two-sided elements different 0. EXAMPLE: Let the R = 2Z, t h e r i n g K-chain integer even K, K R i s the product integers. Hence ACC on t w o - s i d e d potent elements Theorem power 25. =0. This o f a t most Then i s true a bounded However A subring other since every number ring and has R contains no e v e n i n t e g e r i f R has finite R i s a power n i l p o t e n t ideals. since o f even i n t e g e r s . ' even of no n o n - z e r o n i l than S o f a power n i l p o t e n t 0 i s nilpotent ring R i s nilpotent. PROOF: Let R 3 S be a s u b r i n g _Z> • • ' 3 K g = °* o f R and l e t R have t h e K - c h a i n Suppose t h a t SDH.,—)...;") H = H , _, . 1 — — y Y l + Then S has t h e K - c h a i n S r E. — Suppose t h a t . K: H: H C K . a — a h5. Then H H a +1 = H S + SH , w h i l e a a a+1 =^^ +l* B * y a ordinals r a n i s obvious) s i f n i K " t e = K R + RK . a : a+1 Therefore a induction (the step at limit H CZ K f o r e v e r y o r d i n a l number a. a — a Therefore H =0 a n d y• < 3. T h i s shows Y J Hence that H CZ- K 8 — 8 D S i s power Theorem is = 0. c 26. power A nilpotent. complete E direct sum o f power nilpotent rings nilpotent. PROOF: Suppose R = (+) A YEC where each A Y Let R have t h e K - c h a i n Let A A Y-,_ • A Y number ^ 0 implies 3 which the Y : Hence K = 0-subring y = n A discrete power n i l p o t e n t power 1 than ( + ) (K yeC r A Yo• 2 = P . direct and a s u b d i r e c t nilpotent. n (K _ yeC There e x i s t s ) , y a a l l of the o r d i n a l ) Y of the ring K © that P Corollary. is K i s greater Y nilpotent. O K ~> ... D K, = K 1 2 6 6 +1 A Z> ( K ) . ~Z) ( K ) ^ ...""}(K ) . Y Y 1 Y 2 - ^ y 3 R 3 2 ,yeC}. {8 K: have t h e K - c h a i n Y i s power Y (+) 0 = 0 yeC where an sum R i s power ordinal 0 . represents Y nilpotent. of power n i l p o t e n t sum Y numbers Y Hence since = 0 rings o f power n i l p o t e n t 5 is rings k6. PROOF: This such sum follows f r o m t h e o r e m s 25 sums c a n h e r e p r e s e n t e d o f power n i l p o t e n t Theorem ring 27. a n d 26 as s u b r i n g s P that o f a complete rings. direct - I f I , a power n i l p o t e n t R and R and t h e f a c t CL I f o r some n a t u r a l ring, number i s an i d e a l o f a p, then R i s power nilpotent. PROOF: Let I have t h e K - c h a i n H: I O H H 1 Let R have t h e K - c h a i n R r> K K: ZD K 1 T h e n K^ C 2p — Suppose t h a t or a limit that H 1 = I ,K 2 n i s a natural 2p 0 = R number o r d i n a l and t h a t K K Y , and R and t h a t v 2 a i s either 0. Then i f R^ m e a n s P . * <ZZ R K + K R a+2(n+l)p — a+2np a+2np P presents no p r o b l e m ) i t c a n be c o n c l u d e d 0 or a l i m i t limit and ordinal. Hence o r d i n a l and n i s a n a t u r a l this shows that ( t h e step at r i f y = a+n w h e r e R i s power n i l p o t e n t . then K limit that n and a l l cases number, t J ordinals numbers I CZ I H + H I — a+n a+n transfinite is induction P By u s i n g C H , f o r a l l natural a+2np — a+n p+ ± <Cl : — 2 p , , ., . a+n+1 K . D s 2 TD — s ) R K . R ^ . =n a+2np ; s = K Q p 2 p + 1 r =0. . ZD • • ' ~ZD K CZ H . a+2np — a+n i that ZD . . . O H 2 R does n o t a p p e a r , K „ . = * a+ 2 n p + 2 p follows H since 1 2 when a a i s 0 or a CZ H = 0 a+ 2 n p — y S n It i s also contain no n o n - z e r o idempotent K-chain. lies hence the proof. e lies K(R) - chain = of subrings number of the ring R = IR + RI K 3 a 3 e = e-e completes then 3 the IDK^ = K R + RK t 1 o f R's R and e induction of I is the following: D ordinal i n the ring o f R's K - c h a i n , t h e n If I is an ideal every of every term Transfinite a+ = K • , where 0 a i n K- ^- rings More g e n e r a l l y , R i s an e l e m e n t t h e a-th term DEFINITION: K idemp.ote.nts. of a ring a and ...DK power n i l p o t e n t F o r i f e i s an i d e m p o t e n t i nK every easy t o see t h a t and if a is a limit 3 for ordinal then 3 /O K y. • Y< a DEFINITION: embedded An ideal I of a ring in R if I's K(R)-chain Theorem 28. I f a ring power n i l p o t e n t R is power ends at power n i l p o t e n t l y 0. R has a homomorphic and i f t h e k e r n e l embedded nilpotently image T which i s I o f t h e homomorphism i s i n R, t h e n R i s power nilpotent. PROOF: Let potent Let Let f be t h e homomorphism from R onto T, a p o w e r n i l - ring. R h a v e t h e K - c h a i n K: T h a v e t h e K - c h a i n H: I D R 3 H K ^ H ± 2 ~Z> & 3 2 ... 3 . • • Z> K g = 0. = K g+ 1 . Since 1+8. f(R) f(K a CT, f ( K Q ) ^ H 0 )f(R)+ f(R)f(K induction (the step theoretic * numbers property a. r a Suppose. f(K- ) at l i m i t K' Y+n - a + J H a for a l l ordinal I t follows K that C I . Y i n the n-th term reaches 0 in R i s power nilpotent ) = By. t r a n s f i n i t e ~ K(R).-chain. i n I's K on t h e ' x - t h t e r m , ' Hence = i s due t o a s e t f ( K ) C a — =0. f(K Y i s contained i f I's K(R)-chain Hence ordinals of functions) Y Moreover Then ) <Z H T + TH C H _. — a a — a+1 f ( K ) cZ H Hence C E ^ Q =0. Y+ i f I i s power n i l p o t e n t l y embedded R. H The theorem Remark the T next e x a m p l e shows are necessary I. The rather than to obtain condition homomorphism be power merely that, t h e c o n d i t i o n s the general i n theorem 28 t h a t ring i n the result. the kernel n i l p o t e n t l y embedded a power n i l p o t e n t used I of i n the ring itself is R strictly necessary. EXAMPLE: Let S be t h e s e m i g r o u p A i s any non-empty Define A D B Let Then subset c o n s i s t i n g of the set o f N, t h e s e t o f n a t u r a l m u l t i p l i c a t i o n i n S be t h e r u l e : = 0 and e i t h e r A or B i s a f i n i t e R be t h e a l g e b r a over R has t h e K - c h a i n K: Z> K numbers}. = AL^B i f s e t ; otherwise the integers R Z> K A'B {0}.U{ACZN: mod G p with 3> . . . Z> K^. A-B basis = S. Actually 0. K n and = < {cA: c e Z , AeS, ' K to = < {cA: elements}>. However K 2 =0 to over at l e a s t and A has and hence K M o r e o v e r R/K^ elements} > an i n f i n i t e nilpotent co n+1 since i s nilpotent i s isomorphic AeS and A has i s the K-chain a finite = + as w e l l t o T, number number K^ ^ the subsemigroup of S c o n s i s t i n g {0}(J{ACN: K* ceZ, AeS, R i s n o t power nilpotent. algebra and A has K^. as power * the i n t e g r a l of the set of elements}. , t h e n K* = K /)T f o r every to' n n K* = K (~) T = 0 a n d R/K i s power n i l p o t e n t . to to to * Hence The next nilpotent e x a m p l e shows closed pot ent rings. Remark J. not a power be f o r R/K r i n g s , the very morphically. of that desireable does n o t h o l d A homomorphic nilpotent unlike image the case property of a power neN. 5 f o r weakly of being f o r the class I f homo- o f power n i l - nilpotent ring need ring. EXAMPLE: Let Let S be t h e f r e e R be t h e a l g e b r a basis S. Then semigroup g e n e r a t e d by t h e s e t ^ over the f i e l d R i s a power of integers nilpotent ring modulo since x neN}, : n 2 R with = 0. neN Define a function Zassenhaus all neN. F from the generators Example modulo Note that F ( X - ^ ) of R into 2, b y t h e r u l e : = 0. Since Q, the F ( x ) = y-j_/ n the function n F is ^ o r defined 50. on t h e g e n e r a t o r s o f R, it has r i n g homomorphic F(x K .x 1 F ( ) • • • • • v + x ^ i p F ( x £ ) 1 (= F c a n he properties. l/K. = 1 t o a l l o f R so Namely, £ yi/K 1 £ 1/SL.). j=l F(x K = • + ' that define • ... • x • ) = F ( x • • ) • . . . • q 1 q 0 i f•I i=l extended p K • • • ) + p • y 1 / £ i F i s a h o m o m o r p h i sm q since "both 3 Q and R a r e a l g e b r a s over Moreover, F(R) = Q since the. f i e l d of,i n t e g e r s t h e s e t ^y^fk.'' k E N ^ modulo 2. generates the 2 ring Q. However, root i n Q. Therefore Actually, Since easy be every Q every ring - Q since Q i s n o t power free ring element i n Q has nilpotent. i s a power i s a homomorphic t o see that, t h e c l a s s a subset every image of the r a d i c a l c l a s s of a free o f any p r o p e r square S nilpotent o f power n i l p o t e n t a ring. ring rings, i ti s cannot radical property. 51. . 5U*-rings definition This For are defined and U ^ - r i n g s means t h a t some r e s u l t s on U ^ - r i n g s example, the d i r e c t U*-rings they 3 for R = P every = a O y<a Z 2 > D 1 ordinal there results a ring D D g number over t o U*-rin n e e d n o t be a properties l a r g e r class of below provide rings only rings. of 2 The U*-rings. R i s a chain where a, and i> i s an a+1 i f a is a of subrings ideal of limit D ordinal 3 D . y DEFINITION: if of U-rings. A D-chain carry many p l e a s a n t a considerably their are also of. two. U * - r i n g s do n o t have, v e r y i n t r o d u c t i o n t o these DEFINITION: D sum seem t o be than the class R It, follows, from a l l U^-rings although of below.. that U*-ring. an U*-RINGS A subring exists S of a D-chain a ring in R i s a meta* R which ends at ideal D = D of S.. P The that definition chain of meta* Whether of a meta* i d e a l i d e a l s are meta* i d e a l s of a r i n g every equivalent meta ideal i d e a l s and t h a t R are meta* i s also to the stronger makes meta* i d e a l s of a meta* version i t clear R. ideal i s of theorem 7 ideals R 52. mentioned result every above w h i c h has n o t been (compared w i t h meta* Theorem ideal 29.. An decided. r e m a r k B) makes i s a meta The it.clear following that not ideal. intersection of meta* I d e a l s i s always a meta ideal. PROOF: L and M a r e meta* I d e a l s Suppose following E 3 E. 3 - 2 - D-chain 3 — L {M H E are D-chains 3 E, 3 ... - i n R: = y 3 L = M . E — i n R: o f R. Suppose t h e E 3 D D D , D ... 3 D =L , — d — 5 — — p Then t h e f o l l o w i n g i s a l s o a D Y E D D D ... D D. = L D l f l E . D ... — d — — P — d — M . H e n c e L f \ M i s a m e t a * i d e a l o f R. n D : Y E C } i s a s e t of meta* ideals and C i s a o f R, If subset Y of the o r d i n a l numbers, then passes M , M D through there M , f~^\ M yeC 1 sucession. (There intersections two meta* DEFINITION: is a meta* Theorem given A ring ideal 30.. may as t h e r e ideals i s a D.-chain i n R of Every be other i n that order which of Y subrings a r e i n t h e case of R between these of the i n t e r s e c t i o n above.) R i s a U*-ring H i f each subring S of R. subring of a U*-ring R is a U*-ring. R of 53. PROOF: Suppose t h a t R. R i s a U*-ring I f T i s any s u b r i n g Suppose t h e f o l l o w i n g Then t h e r e i s also o f S, t h e n i s a D-chain a D-chain S3D.nsD...3D riS d . — — T i s a meta* ideal i n R: - i n S which 3 R ends D 2 o f R. " ' ? a t T, of D g = T > namely: S P Theorem 31. S which has a p r o p e r , that S i s a snoring = T. o — and t h a t A ring Q i s contained R i s not a U*-ring non-zero subring i n no p r o p e r ideal i f f R has a .subring Q with the property o f S. PROOF: ring Suppose R i s not a U*-ring. Then R must have Q which i s not a meta* o f R. ideals of R which ideals, reason Q. contain 1^, i s an i d e a l there Define ordinal, exists Q. of R which a smallest = n u m b e r 8. I . Y Since Consider The i n t e r s e c t i o n contains ideal I ^ , 1^, e t c . i n a s i m i l a r l e tI ordinal ideal some s u b a l lthe of a l l these Q. F o r t h e same o f 1^, 1^, w h i c h fashion. Eventually, I R Q i s not a meta* contains I f a i s a limit = I . ^ f o r some ideal, I c 4 Q- P Hence I subring ideal P Q i s a subring with of I . of R i n which the property that Q i s a proper, Q i s contained i n no non-zero proper 5h. Suppose in a subring Then Q i s a non-zero Theorem I t follows 32. o f R and Q i s c o n t a i n e d S of R but Q i s contained Q i s not a meta* i d e a l U*-ring. subring from i n no i d e a l o f S and t h e r e f o r e t h e o r e m 30 t h a t A homomorphic image o f S. S i s not a R i s not a of a U*-ring i s a U*-ring. U*-ring. PROOF: Let Then for R/K b e a n a r b i t r a r y i f S* i s a n y s u b r i n g some s u b r i n g there exists homomorphic a D-chain Hence R. i n R which R ZD D^ ZD • • • ZD Dg = S i s s u c h is i n R/K w h i c h S* i s a m e t a * i d e a l of a U*-ring o f R/K, S* i s i s o m o r p h i c S of the ring a D-chain image ends Since. R i s a ends a t S. a D-chain. R t o S/K U*-ring Suppose that Then t h e f o l l o w i n g a t S*: R/K ZD D /K ZD ... O D / K 1 P n o f R/K. D I t follows that R/K U*-ring. i s a E R e m a r k K. A power nilpotent ring n e e d n o t be a U*-ring. EXAMPLE: The power ring R defined nilpotent ring. Example modulo a U*-ring set i n t h e example a f t e r However i t h a s Q, t h e Z a s s e n h a u s 2, as a h o m o m o r p h i c since the subring i s not contained remark J i s a image. The r i n g Q i s not o f Q g e n e r a t e d by t h e f o l l o w i n g i n any p r o p e r ideal o f Q: {x : n,qeN} q./2 Theorem 32 s h o w s that R c a n n o t be a U * - r i n g . n H 55. Remark K i s somewhat potent also ring an potent i s a U-ring. algebra semigroup seen t h a t is a A power n i l p o t e n t since the r i n g i s a U*-algebra. power n i l p o t e n t R e m a r k L. Since i t c a n be algebra surprising not every weakly n i l - R i n example K is e v e r y power n i l - However i t i s true that every U*-semigroup. semigroup is a U*-semigroup. PROOF: Suppose is R i s a power n i l p o t e n t a s u b s e m i g r o u p o f R. R ZD K of that n 1 ~D K_ Z? 2 ... P s u b s e m i g r o u p s i n R: Note that Let R have = 0. R ? S i f a is a limit Hence S i s a meta* semigroup and t h a t t h e K - c h a i n K: Then t h e f o l l o w i n g U ^ 3 S U K 2 o r d i n a l , then S U i d e a l o f R and S 3 i s a D-chain . . . ~ ? S U K g = K = (~^\ ( S l j K . . ) . Y <a ' R i s a U*-semigroup. 5 ; s • 56. 6. The excludes definition a l lrings P R I M E PRODUCTS given which Consequently, the class not a subset below have RINGS of prime of prime the. d e f i n i t i o n to r e a d "y i s a p r i m e i n R i f whenever or v i s a unit then unique to products ring factorization be p r i m e class domains, types. In particular, includes rings, elements x may elements An as DEFINITION: if the class element a product need An not element be written in R. as of be includes a l l of prime defined Although the and i n c l u d e s unique rings products and a l l weakly factoro f many rings nilpotent rings. i s a prime two definition as t h e r i n g s nilpotent yeR i s modified modified not include rings However, u , v c R a n d y =. u v , u paper. does i t i s very large but not a l l l o c a l l y be written as w e l l i n this a l lpower n i l p o t e n t DEFINITION: two a slightly products rings ization given below can be g i v e n w h i c h products rings o f prime element domains domains i s products rings. if of prime of prime rings a n y i d e m p o t e n t s o t h e r t h a n 0.. o f unique, f a c t o r i z a t i o n of the class i n R", products elements element in the i f y ring cannot R. (The distinct.) x in the a ring product R has a prime of a f i n i t e factorization number of prime •57. DEFINITION: in the ring in R but ...* y^ the If x = y^y^ R and neither s xeRj begins a ring, series the form of the that and. z^ y • are x elements y . ' z\z _zv factorization the of An ring trivial every of A series in DEFINITION: of the the y^ ...* y of an • of element factorization other x = factorization previous in x s the factorization. factorizations x = y ^* . . .• y primes series R) , then factorizations with i s a refinement the all of property DEFINITION: of z^ of x. A series has in of is a factorization (where s i s a unit DEFINITION: and zR i f y . =' j 2 i s a refinement element . . .'y i s obtained ends where i f a factorization y j' y 3 are n R. element xeR factorizations has (the of x ends property) after FF_ i f a finite every number steps. Remark M. primes. A non-zero However, idempotent may "be a f i n i t e an i d e m p o t e n t . c a n n o t have product of FF. EXAMPLE: Let where of S be t h e s e m i g r o u p a l lproducts integers modulo consisting o f two elements a r e s. L e t R be t h e a l g e b r a 2 which has S for a basis. over Then s and t the field t is a 58. prime in R since i t cannot elements i n R. Also, primes. However, be written s = t - t and s also equals as hence s-s a product of two s i s a product and hence the of series of n factorizations a finite number Note s that = t - t and is not of The structure note ring R defined of the that 33. class Every a prime products end after since ring is example second above factorization t 4 t-t . a ring in which every FF. are power n i l p o t e n t Theorem first theorems these i n the However, the products has following does n o t 5 .of t h e A prime element s = s steps. i n the a refinement non-zero for s o f the. f o r m s = t«t-t a l s o . DEFINITION: to of give of prime nearly, some i n d i c a t i o n products the of rings. same r e s u l t s the It is as were interesting obtained rings. subring S of a prime products o f S, a subring ring R is ring. PROOF: Let every x be series a non-zero of factorizations number of steps series of factorizations factorizations izations in S element since in R of x in R R i s a prime o f x. in S of x of x products i s also Consequently, is finite and S ends o f R. after ring. part every i s a prime of Then a finite But every a series series products of of factor- ring. S 59. 3^. Theorem is A complete, d i r e c t a prime products sum of prime products rings ring. PROOF; R = Suppose (+) A yeC each A where Y ring. i s a prime product: Y I f x i s any n o n - z e r o o f R and x = element £ x Y £ F where x a s u b s e t o f C, t h e n x consists in x v y v z v f refinement ment every series f o r a l l Y E F and F i s Y of f a c t o r i z a t i o n s o r in A of factorizations a 1 1 Y £ ' F t h e n x o f the. t r i v i a l of the t r i v i a l = I ( of x Y y- )( I : v factorization factorization . Y z v of factorizations o f x ends when the series of f a c t o r i z a t i o n s of FF. Hence a prime rings A discrete products ring i s a prime is i s , X ^ ( Y C F ) any one e n d s , x must sum of have H of prime products rings and a s u b d i r e c t products sum follows ring. f r o m t h e o r e m s 33 and 3^. is of prime products PROOF: This refineSince products ring. direct i f a o f x has t h i s . form. series R i s a prime ) That o f x and e v e r y the Corollary. i n R of o f t h e p r o d u c t s o f t h e sums o f c o r r e s p o n d i n g t e r m s the series = element, o f A i s a non-zero Y Y ^ 6o. Remark which N. A prime products ring h a s no p r i m e may "be a s u b r i n g of a ring elements. EXAMPLE: The since set ring Q has of r a t i o n a l an i d e n t i t y . of even i n t e g e r s integer products N shows ring may have The cannot be too. c l o s e l y ring, following 35. elements The R consisting of the subring four products ring a ring P C i s not a which i s a prime indicates related that to the ring o f R, I f o r some i n t e g e r every 5 R which a subring theorem i n which i s a prime. I f I i s a subring and i f R products by that ring. Theorem h a s no p r i m e i s a prime not. d i v i s i b l e Remark n u m b e r s , Q, such prime products subrings R itself. : i f I i s a prime p, t h e n R i s a products prime ring. PROOF: Due is t o the fact, t h a t a prime products ring a subring of a prime i t is sufficient products t o prove the ring case 2 when R =1. Suppose t h a t x i s not a prime an infinite d o e s n o t h a v e F F i n R. i n R and h e n c e xeR series = I. Let {s of r e f i n e m e n t s o f x i n R where = x. _. x •. • n ,n n ,1 n ,2 . i n I , ' { t : neN } w h e r e n x xeR An t n infinite yn ,l n,2 y series n ,n Then neN} n s be n of refinements of x = x, c a n be 61. constructed X *,m' S L y while e t 2 2 b y d e f i n i n g t h e y. l , l y = X X l , l a n d U 3 i+ h ' X refinement in = l T el 6 h e t y n i n terms 2 , l = 2 t : y X i of suitable 1 U,2 X + J 2,l 2,2 y = x £ R 2 i = s 1 a o f t . I n g e n e r a l , t h e y , - .'s c a n b e d e f i n e d 1 n+1,j t h e f o l l o w i n g s p e c i a l w a y ( w h e r e t h e g 's a r e . c h o s e n n n ; so t h a t y .-, v . . .*y .-, , , i s a r e f i n e m e n t n+l,l ii+l n+l J of y . 7y ): n , l. n,n J J ) let y„ . = x, ,»...*x. , y _ _ = x. . .'x, n+1,1 " *hn,r •• lm.,g *n+l,<-2 ~ l m , + 1 * * ' " U n , x l n A > n A 1 ... n-1 • , y , _ = x, V' itn+1 , n + l i;n ; Z'g'. + l for + , & 2 x, , n . w h e r e g . eN • kn , y g . . • I n i = 1L" & ± § 1 ~.:'L- ••! 1 h a l l i i n [ l , n ] a n d £ g. = ^ • i =l 1 = 1 I t has already n been shown 1 that they .'s c a n b e c h o s e n n+l,j Suppose t h a t hn + k S : X y n +l,2 they X h a s t h P e h , i = l , . . . , n +l, such that ± " X w a y w h e n n = 1. c a n b e s o c h o s e n when n = n . kn+h,i' • •' hn+h;kn+h numbers i nthis n r o y ^ n + l+,h +l*- * • ' U n + U , h + h ' X 1 1 n P e r + l j ' 2 y t y t h Then a t t h e r e e x i s = *l* +U , 1 ' * '^Un+Uh l n n+l,n+l X Un+U , | : i v -, w h e r e kn+k, ) h. .., l h . eN f o r a l l I m l L [ l , n +l ] 1 Suppose h Then b y l e t t i n g letting kn+k, X k-1 I i=l n + 2 h.+l kn+k, j = i T I i = l i i •v i » kn + k, I h. ' i =l 1 a n < l + l j f < > 1 k i k-1 X 1 l n h.+2 a n d y _ n+2,k+l " Un+U, y X 1 •n i =1 Hence h . > k f o r a t l e a s t one j i n [ l , n ] . . y _> k. n + 2 j = k k-1 £ h.+3 i = l 1 by l e t t i n g y , . = y , . , i f i > k+1, •° n+2 ,l ^ n+1,l-l J 17 0 n 1 I h. =• hn+h and i = l h i =l n +1 ;..?x, t 62 the y _ .'s a r e c h o s e n t h e g i v e n ° s p e c i a l way * n+2 , j is a refinement so t h a t t n+2 J _ . H e n c e , b y i n d u c t i o n on n , t can n+1 ' n c h o s e n f o r a l l neN a n d t h e r e f o r e x does n o t h a v e FF i n I be of t The. f o l l o w i n g t h e o r e m s prime products defined above. Theorem 36. prime rings with and remarks compare some of. t h e o t h e r classes I f R i s a weakly n i l p o t e n t r i n g , products the. c l a s s then of of rings R is a ring. PROOF: Suppose series not that y i s a n o n - z e r o element of R w h i c h has Q of refinements end after a finite of the t r i v i a l number factorization of steps. Suppose which that y e i t h e r x^ has FF; each In or otherwise does n o t have l e t y^ = x^. o f w h i c h does not have general, let y ^ n FF and l e t y = z n n G'='{neN: y = x } n . n one o f t h e two Suppose, t h a t * r following • , = x - l n i f x n and l sets way: L e t y^ = A sequence F F , c a n be ^7 ^ o r n defined i z^ i f x ^ elements, recursively. Let w i f neG, w ' n o = an i n f i n i t e = y and o number o f define (x_.,x. , . . . , x . l i ' ' 1 w n elements. i n the )y where ^n n l i . 1 < i . 2 < < i n i l z where e i t h e r x o r z . does n o t have n n n has FF; o t h e r w i s e l e t y = x . Let n n e t H ='{neN: y = z '}. T h e n a t l e a s t n n G and H has G does. FF. does = x,z,. o where a . < n and' { i l S i 2 , . .., i } = n i ( N ^ G ) P | [ 1,n] ; 63. otherwise w eliminate repetitions onto ,. n-1 G and define, u Then y the = w n = u v 1 1 o n n J-chain J: n and u Then w n-1 n = w z f o r every n n neG. J l e t F:N •+ G h e To an o r d e r p r e s e r v i n g m a p p i n g = w„, v and v = z ^ , >, f o r e v e r y n e N . F (n) n F (n) = u ..,v , f o r e v e r y n e N . Let R have n n+1 n+1 J J 0 CL J C- J , CZ . . . C n I d J = J 0 -, . 0 Then any P+l p . ordered y o J product of the form v v ../v., i J., s i n c e u (v. v *. . ,*v . ) • n n-1 1 1 n n n-1 1 4' 0 .'. S u p p o s e t h a t , v v , . ...v t J f o r e v e r y n£N. Then s i n c e n n-1 - 1 a ^ : n v _,,.(v. v *. . .*v- ) i J , (v v '. . .*v ) t J„ , _. . n +1 n n-1 1 a ' n n-1 1 • a+1 Hence n transfinite induction v v n n-1 ^ J„ f o r e v e r y I B The ...'v n case when H h a s (the step neN. Hence an i n f i n i t e number handled i n an a n a l o g o u s Theorem 37. products at l i m i t ordinals by 17 i s obvious) R i s not weakly of elements nilpotent. c a n be way. I f R i s a power ~ nilpotent ring, then R is a prime ring. PROOF:. Let R have t h e K-chain K: R D K O K 3 .. . 3 K g =•' 0 .. 2 If x i K^ trivial has FF. = R , then x i s a prime factorization. If y 4 K and t h e r e f o r e has Suppose t h a t , then either x i K only the implies, that y i s a prime or y x has Y + l factorizations of this wise form y lies of t h e form neither y^ y = 7-^2' or "'" e n Y e T are elements i n K R + RK = K y Y Y- which n + 1 is a 7 in K factorization , since other- contradiction. 6k. Hence y^ series after of of and y^ h a v e FF refinements by assumption of the number of refinements of the trivial number of a finite therefore factorization a finite after and steps. y It follows factorization steps. Hence every = y-^y^ mus '' sncl : that, every series y end every = y must element x i K n Y+ l has is FF. By obvious) transfinite every induction element x i K (the step , i.e. at l i m i t every ordinals non-zero element p in R, has Remark 0. FF. H A homomorphic not. c o n t a i n a n y image of a prime products ring need primes. EXAMPLE: The nilpotent ring and R defined i n the t h e r e f o r e a prime E x a m p l e m o d u l o 2, no primes of i t s square The nilpotent the R and Zassenhaus i s not Q, element root times i t s square products every i s . power The of R Zassenhaus and written as Q product ~ ring. image prime contains the root. i s a homomorphic Hence not be J example, f o r r e m a r k t h e r e f o r e a prime a U-ring. ring. image i n Q may R d e f i n e d i n the Example f o r remark products i s a homomorphic since every ring example J i s power The fact that o f R shows that products ring is a be nilpotent. U-ring. R e m a r k P. A prime products ring need not power .65. EXAMPLE: Let {x^: n X R be neN} = X with Un Un+l R 2 then X since and since the + X nilpotent in the. c o m m u t a t i v e following Un+2 lin+3 X every 2 . f o r of three these X Un+2 Un+3' X Remark Q. of relations basic types X T i n the element Un Un+l = X X n " X R n z = z ends set relations: i s o t p o set w e r for R steps s i n c e each of made p o s s i b l e b y are: X = x, x, l+n+2 Un + 3 X need not the Examples x be lies products ring h ki / ( II x • ), 1=1 o=l i , j i n <_ k property. X ring n i s true Un+2 4n+3' products e and of This substitutions h i s a prime in R this the generating substitutions a l s o has of A prime ' n s N by generating factorizations b a s i c type generating of a 1 1 where k = max{k. : i i n [ l , h ] } . the generated However, R i f z i s a non-zero series set element hence R = R every ring a = of , X n + " X 4n 4n+l' X U*-ring. EXAMPLE: The ring products {x 2 n : R ring. neN} of since S The + SR. R generated i s not i n the subring i s a proper x, - x , •, _ e SR 4n+2 .4,n-+3 ideal defined S of subring Hence by contained S. example f o r remark R generated of R. However X f o r a l l neN, x proper ideal is a the n = R x i . i . o_ i s not of R. prime set e SRC n It follows that i n any by P x ) n ) n + l(S), ' a + the U*-ring ~ 66. . 7. Meldrum (12) types. An and theory the paper ring analogous except of f o r the terminates A ring after recently completed definition of r i n g ring types the type nilpotent R has j steps study of group i s given below i s n e a r l y completed d e t e r m i n a t i o n of the types, f o r weakly DEFINITION: k has RING- .TYPES type and class in of this possible rings. (Q k) if its s i t s K-chain J-chain terminates after steps. Theorem 38. A has and then I f R i s the ring, type R has ring direct sum of the ( j , k ) w h i l e B has type ring rings type A and B (Jl,m), (max { j , I } ,-max{ k ,m} ) . ; PROOF: Let A have B have the J-chain K a = and J^ R's and J the K-chain (K „ ) {+) (K- ) f o r every A a^ B a : J-chain have J-chain is J J-chain the must length n have length h = max{k,m}. and the K-chain i s K, J ordinal a = m a x { j ,{,} K-chain K . = (J. K Then and i f R's ) (+) ( J _ ) A a ^ B a n u m b e r a. Hence and R's let K-chain and R's must 5 67. Theorem given shows r i n g types type. may Furthermore, possible rings r i n g types of various The ring 38 while a ring with i t makes may types given Suppose =0. n + 1 i t easy with different ring t o s e e how rings ofa l l from the examples of below. the class of possible rings. R i s a nilpotent Then of rings a quite be. c o n s t r u c t e d for nilpotent 39. R he t h e . d i r e c t sum next, t h e o r e m c h a r a c t e r i z e s types Theorem that R has r i n g type ring and R n 4 0 (n,n). PROOF: The K-chain 2 R 3 • • • 3 x_, , . . . ,x eR 1' n x 2 in length n since i t i s the. f o l l o w i n g : n 3 R f o r R has •...•x n J , , ... 2 J-chain J. such 3 that 0 x , and t h a t However, must there 1 'the n • . . . -x number n. x n 1 n other 4-0.. n , that hand, there I t follows x_*...'x does 3 n does n o t l i e i n J exists that not l i e , where n-1 n R has C J , , R ^ C J ... , a n d R d — 1 — 2 — end at J and h a v e l e n g t h n. n R n n n are n i l p o t e n t f r o m t h e o r e m 39 t h a t natural °- does n o t l i e i n J So R I s J - c h a i n Since R there rings are rings l of every of type index, the J n H i t follows (n,n) f o r every 68. R e m a r k R. For type which (6,0)) every ordinal are both 8 >. tn , t h e r e number weakly nilpotent and are rings power of nilpotent. EXAMPLE: Let 8 = y where + n y is a limit a non-negative integer. by {x the set S = Let : a i s an R be ordinal the ordinal number commutative number, but and n is ring not generated a limit a ordinal and a <_ g } , w i t h =0 for a l lx the relations: 2 (1) x a then (2) x 4 0 while x n : p Suppose t h a t eS a n o x^ + p eS where = 1 a < 6 ; i f 8 i s not a limit ordinal, 0.. for a l l i i n [l,m]. Then t h e product i X ~ • « • X ft = 0 i f a i s t h e 6 6 1 m { 1 6 and is 6 m kcN, } , i f a = n+k and 0 i f the i f k number any of the Let R have the if x et. zero . < m. element a product exceeds of elements f o r an and K and the ordinal the i n the element 0 numbers ordinal number of elements "finite" from part S of product). J-chain J. arbitrary i f z = of the n is a limit (Hence K-chain i n R, where of f a c t o r s subscripts stands smallest Then K i n S, =0 since i f z is a \ ( II x ), then i =l j = l i , j z has nononly a prime factorizations with h = max{ z i. K than w . 1 ,... ,% Now 8 where q_ }. l e t n+k fewer than h+1 factors where I t follows that z i R' "'" = K, a n d h hence be any n is a limit non-limit ordinal or prime ll+ ordinal zero and number smaller k natural is a 69. number. Then i t can be shown by induction on n+k that J n+k J is {x g e n e r a t e d by J -]_ ,, ; x , *x» where n+k n+k+1 6 following a n d 6 > n+k+1; n where 6 ..... 6 > n+k+m; ... 1' ' m ' since i f p * follows for J p > k+h, from then x t h i s that x a l l o r d i n a l numbers = }. n R and R e m a r k S. where n the ring There i s any R *x n+p a a ... set ; x elements: , x. •...•x. n+k+m o. o 1 m Not e t h a t x , , . • ... •x , » iJ n + w+6_, n+u+6, 1 h , ~ •...*x ,. i J . _. . n+u+o n+w+o^ n+k-i first occurs <_ 3 . H e n c e R's does, h a v e : t y p e i n R's J.-chain J.-chain integer at ends y J a at 5 rings and n+k It ($,u)). e x i s t power n i l p o t e n t non-negative of of t y p e is a n Y (n.,Y+n) limit ordinal number. EXAMPLE: Let R be only a only zeros finite diagonal. matrix the on the are matrices Let R have the the entries main Addition K diagonal and to zeros matrices n entries the K: ring on R _~) K of the 1 73 i f X Q +y D ... 3 a l l matrices a diagonals ) of are = with the the main usual (x. ) and. Y = ( y •„) ct , p • ot » p Y+n-1 and K XY = Y with and left multiplication in R t h e n X + Y = (x i s the (y+ ) integer multiplication: K-chain are a l l {y+n.) b y non-zero and and i n R, a of number o f addition shows t h a t ring + n in R - 0 = ( T • x Computation i n which a l l parallel y to the main c ) 70. diagonal there a n d j u s t t o t h e r i g h t .of t h e m a i n , d i a g o n a l . are exactly Y+n such diagonals, K J Let. R h a v e t h e J.-chain m < n, J i s the subring — m R i n which parallel subring in and a r e no l a s t integer; m J.-chain since parallel Hence diagonals The i n every of type Y n R i s power nilpotent r i n g types: o r (m, Y+n) 38 t h a t there ( a , 8 ) f o r any n o n - f i n i t e every where + ordinal A power matrix t o the main R and S and t h e o r e m non-finite ordinal n i n a non-negative integer number. nilpotent r i n g must (n,n) where where h a v e one n is a o r d i n a l n u m b e r _> n , a n d y i s a l i m i t of the non-negative n i s a non-negative integer, m i s ordinal number. ring. I f R has PROOF: Suppose t h a t finite (n,n) K-chain, R i s a power then nilpotent R i s nilpotent f o r some n o n - n e g a t i v e integer and t h e r e f o r e n. 0. in H a and 3 s i n c e y i s a limit following of i t . rings 8 has. t h e f o r m 40. on t h e l a s t and t o t h e r i g h t o f i t . diagonals from remarks nilpotent numbers Theorem n (n,y+n). follows power number are zeros n+1 and t o t h e r i g h t has t y p e ordinal and a l l the entries R there =0,, w h i l e K 4 Y+n Y+n-1 shows t h a t w h e n e v e r of R c o n s i s t i n g of a l l matrices J ^ o f R i s t h e e n d o f R's It are Computation to. t h e m a i n d i a g o n a l diagonal an J. Since has Suppose t h a t a type R's 71. K-chain Y has l e n g t h i s a limit Y+n where n i s a n o n - n e g a t i v e o r d i n a l number. _. CL. J Then K Y+n-1 — term K o f R's C J . Y — n which J-chain. Hence However, i f K Y — CZ J at l e a s t n. I t follows m i s an o r d i n a l n u m b e r It i s easy 38 a n d 39 potent rings nilpotent K that ., CZ J n-2 ' Y-1 K Y + n-1 — R has t y p e J-chain has (m,Y+n) w h e r e 5 Remarks the fact that R and S and from a d i r e c t sum i s a power n i l p o t e n t r i n g rings t o see J 4 J -. a n d R's • n n-1 that first > n. t o see from and from , , then Hence , the and 1 I t i s easy n-1 i s a contradiction. length J ^ 4 0. n integer of a l l the types given that theorems o f power n i l - there a r e power i n t h e .statement of t h e o r e m - kO . Remark T. There number 8. exist rings of type (0,8) f o r every ordinal EXAMPLE: Let define X a X A Q be t h e c l a s s _, = A w, a n d i f a i s a l i m i t a+l a ' = inf{6eA„: 0 subclass of a l l o r d i n a l numbers. & > X " f o r ! a l l y,<a} . Y ' of AQ w i t h L e t X^ ordinal,l e t Let A a be t h e smallest the properties: ( 1 ) • X aeA a,' (2) (3) for i f 6,neA ' a i f B CZ , then , and a and B i s a s e t , t h e n a l l neB}eA . a = 6+neA t h e inf{6£A : Q 6 > n 1, 72 Note t h a t a limit A- - j A -3 .. . I> A 3 A , 3 0-^1—' a a+1 n ... n o r d i n a l , then 9 , and t h a t i fa i s A A since X = inf{6eA,: a y Y a 0 a l l y < a} i s a n e l e m e n t i n A f o r a l l y. < a . Y < for Lemma T l . has Suppose t h a t the K-chain A , < <S y a Q i s a commutative r i n g Q Q ") K, 3 . . . ~> K = K„ . T h e n K K C K 1 —^ 8 8+1 a y — a+y a l l o r d i n a l p a i r s ( a , y ) "where a i s a l i m i t o r d i n a l . for K: and t h a t c n PROOF: Let a be any f i x e d l i m i t T h e n K K_, C K Q = K _, . a 1 — a a+1 K K a Y 1 ordinal. Suppose, t h a t ^ = K K Q d K Q = K . Y a+y a+y+ 1 + a L — n y < u, and u i s a l i m i t Suppose t h a t K K C K . a y — a+y I f K K C K' a y— a+y o r d i n a l , then by t r a n s f i n i t e Lemma T 2 . K K a y^y Y Y 1 t h e lemma i s t r u e . induction Suppose t h a t • Then for a l l — Hence < Q i s a commutative r i n g with a y = 1. J a + = K Y a+\i K-chain (a+1) PROOF: By lemma T l , ( K ^ ) a n X ^ = (X )u), i t f o l l o w s a+1 a ' n Let largest C •. ^ that OT ( , ) neN X Y he any n o n - z e r o o r d i n a l number such f o revery k a n ^ — o r d i n a l number. that X neN. Since OT K, . . * neN [X)n a = K, . X , a+1 L e t a ^ be the . < 6, a n d l e t 6., b e t h e n 73. largest the ordinal largest number ordinal i nA number such t h a t S < 6. L e t a , be a 1 — 2 s u c h t h a t 6.,+A . < 6, a n d l e t 6, 1 ~ a 2 2 be the largest ordinal number i n A such a In that 2 6.+ 6, < 6. 1 2 . - - general l e t a. be t h e l a r g e s t o r d i n a l number such t h a t J 6+...+S. +A < 6, a n d l e t 6. b e t h e l a r g e s t ordinal J n u m b e r i n A^ s u c h t h a t 6 +...+6.. <_ 6. E v e n t u a l l y , , f o r some J n a t u r a l n u m b e r n , 6, + ...+6 = 6 s i n c e ct.ct ,...,a i s a 1 n 1 2 n strictly decreasing set of ordinal DEFINITION: The representation number immediately given Lemma T 3 . is The l i m i t <5 = S^ + ... .+8 above form numbers. is i t s limit o f every non-zero of an ordinal form. ordinal number unique. PROOF: Given first term ordinal an o r d i n a l number i n 6's l i m i t number form, 6. i s u n i q u e l y 6, t h e o r d i n a l i s uniquely determined 6,,...,6. have been determined. 1 J - l n is composed o f u n i q u e l y determined number 6^, t h e determined. The once t h e o r d i n a l s H e n c e t h e sum 6 = S. + ...+S I n terms. ih. There exists an o r d i n a l number p s u c h that A > 3. P G be t h e s e t o f a l l o r d i n a l Let numbers less than A . Let R P be the ring generated by t h e s e t ( x ^ : 6eG> with the defining relations: (1) R i s (2) Let 6 have t h e l i m i t Then commutative. <5 m u s t m ordinals 6 = 6.+ ....+6 w h e r e 1 m 6 eA ^ A m a a , m m+1 i n the usual ordering of the be t h e n - t h o r d i n a l i n A satisfies form a where n i s a natural number. The g e n e r a t o r ( x > ) " " = xn + , 1 L e t R have t h e K - c h a i n Then x.eK. 6 A i f6 C G O A a a K. o t . 1 number. x m the relationship: Lemma Th. n m-1 L e t a be an ordinal . PROOF: If a = 1, t h e n number. x Since E K = K, . <5 CJ A^ also that where in (K A a a ) limit y G C G O A . 1 that f o r every 6 i s a limit Then ordinal neN, i t f o l l o w s x.eK f o r a l l SeG H A . 6 a a Suppose t h a t that Suppose i n A . be t h e y - t h o r d i n a l 0 must Ct~i~ 1 i s a limit + implies n+1 x . = (x„ ) 6 o +n , n a m e l y , 0+ n SeG/H A O t ordinal. (A- ) n . a f o r a l l neN. be t h e ( y + n ) - t h Let Then (x u ) n + 1 = x n a a By lemma T 2 , x . e K , n ( ordinal j h e n c e x_e 0 . I f a i s a a+l ordinal, then A = ^ a k G l O A Y ' for a l ly transfinite < a, induction A . 6EG/0 A Hence y<a y and t h e r e f o r e on a t h e lemma x e 6 r <0 y<a follows. K , implies that a A = K. y A . By a 75 Note t h a t {x„: o i f y e R , t h e n y c a n he e x p r e s s e d 6eG}, t h e generators P £ L.y. where j=l y = 3 J i n terms o f o f R, b y a n e q u a t i o n L. ' i s a n o n - z e r o integer of t h e form fora l l j in [l,p] 3 and where t h e y . a r e a l l d i s t i n c t elements o f R o f t h e form 3m y. = TT x . w h e r e 6. eG f o r a l l j i n L l , p J a n d a l l m i n 0 - , < 5 . j,m ° m=l j,m ' J 0 0 [l j s ]. I f K lies one £ i n [ l , p ] , i n R's K - c h a i n t h e n y^K^. and i fy ^ K f o r at l e a s t D For Y Kg £ implies that a sum o f distinct form t e r m s , e a c h o f w h i c h l i e s i n R ^ K^ a n d h a s t h e Jm L. II x . , m u s t e q u a l a sum o f d i s t i n c t t e r m s o f t h e m=l :j.,m same g e n e r a l no form in K . This additive r e l a t i o n s given Let for i s impossible since i n the definition there are of the r i n g L_, = R, a n d l e t L = K f o r a l l neN. I L e t 1 n n-1 a l l n o n - f i n i t e o r d i n a l numbers Lemma T5 . Suppose that L,. « i f9 < A . UJn +O a n : . A = K a Then L, . _ , = (L, ) U )(n+1) X^ w a a. L< = (L, . ) ( A )n A • a a Hence L R. L / , . •> L_ = {X ) n 0 a Tl + 1 1 a PROOF: If L, , (A li, , (A ; 0 = 1, L,, v L = L / , . % R ' (A ) n 1 [X ) n a a n v L = L , , \ , a n d n. < A . )n n (A )n+n a a ..a x R = L, , ,. )n+n (A jn+n+1 a v a = L / , x ,, . (A ) n + l a Suppos e t h a t T h e n L, . <, L , .. = L, . s L R (A ) n n+1 (A ) n n a a L e t 0 be a l i m i t o r d i n a l number < A , — a - 76. S u p p o s e t h a t L/, ( A n < 0. It will \ -L = L,, , for a l l ordinal ) n fj (A )n+n: a 1 be shown t h a t ( \ L,. n<e " ( A ' )n+e" a L If y Y E ( L ) A i ( L , ) "'"j A a n + 1 a ' then n + {x.: t h e n y E L y expressed y . = x • . . / x . x •...•x. J 'o 6 q>_, <p ° 1 n 1 s where x. 6. where x . , Y Then 6^' = 6^ x , •. . .«x , +1 • *s J in (L,- ) A a and ^ if L a \ , ) on n l e t t e r s ) = x , , • . . .»x , . + 1 * t n + 1 L also ( A )n 0 a \ L„ )n 0 ( A )n+e* • a = L/, ; a n a e n C = n ' + where equals x. 1 •. ..'VX- >..n 6 e 0 i s an o r d i n a l b i (L s i n c e ' R h a s no x L^ . Every J J a is a ) "'", symmetric and n+ divisors y. l x L a y j n )n+0 i ~L\ . . . • a:. r of zero. i s either an for a l l i in [l,p]. Hence 0 Hence element J a some . 'cx . * * 1 L ^ , o r y . h a s t h e same p r o p e r t i e s a s )n 0 ' l r v U a n d L/ (A H + of the generators, 6 s i n c e y. .in - .AA: ) n 0 DL,, — (A n is distinct, f o r a l l i i n [ l , n ] (where 1 a i n terms L U e L , . f o r a l l i i n [ 1 ,n ] a n d x , . •. . .'x , , i a 1 t t h e r e f o r e y.eL, eL, , (A )(n.+l) ^ a y. ] L/, n<e J ^^ eL/, \ (A ) n n<0 a j = / n r permutation y n E L , f o r a l l i i n [ l , n ] and A l • a Suppose t h a t y. X ) each n x .•...•x, i L , . +1 ^s •a ^ L a U (A: T L.y. where 6eG}, e q u a l s ...•x,, t numbers v a y., 3 y = •' f y. i=l 1 = L, (A a ^ L )n 0 ^ansfinite number < A . — a . By J l e m m a T l , L, (A inaction, L ( x . ) n a L 0 \ L. )n 0 a = L ( x . ) n + a Q 77. Lemma T 6 . and L, , (A a \ = )n [l[{x.: 6 S e A •'}.)•] : f o r a l l o r d i n a l a numbers n a a l l neN. PROOF: L, . .. = L = l({x.: (A ) 1 a) 6 l({x^: 6 e A •'}) f o r a l l o r d i n a l L,. v =• [ l ( { x . : IA' J n o a L,,. ^ = C A )\n+1) a w on <5eA • } ) . 1 x : 6 eA' ••})•] . = a n [L, ] A a n , L,. s = (A )n a n " r x = [l({x.: o numbers [L, ] .. A a r SeA })] a }) ] a 6eA T ordinals induction i s obvious), Hence R's u ^ . L/, v ,1 = ^e' Suppose lemma also Hence } ) ] = l({x.: ct 6 n on a , ( t h e s t e p at that T5, Hence by f o r a l l neN. L/ x = f~\ L/ . \ = D [I({x.: ( A ) o) - i (A ) n J 5 a neN • a neN Hence by t r a n s f i n i t e 0. <_ a . Then by n [I.({x : o 6eA Suppose t h a t L, induction . = A -: ,. . a+ 1 6eA' ) . a+ 1 n limit L , , , = l ( { x • <5eA }) f o r a l l o r d i n a l s \A )1 6 a a N K-chain. does n o t end u n t i l after K a. and P therefore K 4 K p Let S be t h e r i n g the defining and (3) relations l e tK number generated ( l ) and a n d l e t 6.eA . i n a i by t h e s e t { x ^ : 6eG} (2) given be t h e 3 - t h t e r m 3 natural _ . p + J- i n R's ^ A„.,, «i+l f o r the ring K-chain, with R above, l e t n be for a l l i i n [l,n] a where 78. a > a > ... 1 2 - n where x > a . n 0 I f x . x '. . .*x _ eK„ a n d i f x x • . . .»x , = x „ 6 6 o 3 ^ <J> 4> « J X 4, e A ^ A, , t h e n t U . 1 n l x 6 9 n Y + x + n Y 0 It + x U n l + a n d y =• 0 o r y e A = y+n, 2 the r e l a t i o n •...•x -x - . . : x • ( 1 n-1 1 t-1 where 2 + t •...'x. = o _, o 1 n X T + 4n 2 X + Y + ^n 3 ) + , the set of l i m i t h ° l d S ordinals. J- i s easy t o see t h a t elements i n the r i n g R of the form ( x . '...'x. ' X •...•x, ' X , x , ,, ) l i e i n K ,.. i n R's S 6n-l 9 * t - l 'Y ^ Y+^n+l 3+1 Q + n 1 K-chain. Hence i f S has t h e K - c h a i n a < 3. However H defined b y r e l a t i o n s (-3). — since ring S also since chain c P _, = H +1 P since of x . x due t o t h e a d d i t i v e has a t r i v i a l r 0 occurs H = K relations i n S J-chain. S i s a prime is finite. as a f a c t o r Also, i n a series every products decreasing every x ^ , 6eG, has of f a c t o r i z a t i o n s 5 Q Remark every i f Hence t h e r i n g S has t y p e ( 0 , 3 ) x ^ h a s F F due t o t h e f a c t t h a t of o r d i n a l numbers FF H, t h e n U. There are prime n a t u r a l number products rings of type (n,n-l) f o r n. EXAMPLE: Let {y (1) R ( k ) be t h e c o m m u t a t i v e : neN} y k x + 1 with =• 0 . the r e l a t i o n s : r i n g generated by t h e s e t n • 79. (2) y (3) i y = y s i f n > n k • ' ^Un/Jm-.+l nk-., to y y k-1 ...y steps. e + y , then Y n/n;--' Un y 'Y 1 2 +2 lin +3 k k y 1 2 . r e l a t i o n s of type next greater ring theorem than R(k) the remark ( 3 ) , R(k) k has P i s a prime ' • • *Y = Y Y k ' n 1 2 ' = R(k) R(k)'s k + 1 i t s K'chain ( k , k - l ) and that the products J-chain has . However ring types of information a l l commutative Theorem- U l . There are no cannot length i s a prime have k-1. Hence the ring i n the example ring. for length products R given with remarks determining the R, S, T, possible commutative and m rings i s any of type (m,n) o r d i n a l number U rin£ rings. i s a n a t u r a l number for 5 f o l l o w i n g theorem together complete n+1. since type gives where n shows t h a t same r e a s o n s The > s N. k — 1 ^R(k) and h e n c e R ( k ) ' s K - c h a i n has exactly L e t R ( k ) h a v e t h e J - c h a i n J : 0 d J , <L J ^ C . • • C J = J . 1 2 k k+1 k k—1 J - c h a i n ends at J since y ej , y e J , ... , y ej iC 1 1 X 2 X ri This for a l l ,...,n k Due The for 2 x 80 . PROOF: n Let R be a commutative steps, then R suppose and xR J £ n n+2 2* + = n + 2 J Hence X "'~ = R n + There natural number and m ^ = 0 no rings i s any the C 2 and J-chain R's are xR n+ n + I f R's R have <_TJ -^5 Hence x R = J , and . n+1 42. Let T h e n xR 0.. Theorem . n + 2 ring. J ends J-chain J , n this has K-chain n+ implies (m,n) o r d i n a l number and , xR "'" CT_ J ^ ... l e s s than of t y p e after that n+2 X £ J n ^ ' + steps. where n • is a > 2n+2.. PROOF: Let R be any ring such that R "^ = R n+ ? S R xR n + 2 . TO — S Let =0. R have where R 0 the means s= o r t h a t R d o e s n o t a p p e a r on t h a t s i d e . o v s 2n+3-s I R xR =0.. But i n each case s= o If X E J . then 2n+3' 3 either be s >_ n+2 ,. rewrittcn this than Corollary. not o r 2n+3-s £ I s= o i t follows longer to 2 + >_ n + 2.. 2n+2-s R xR 2 s 0 that xeJ Hence t h e A equation . n+2 0 since R „ and 2n+2 . ^n+l = R D hence R's above . J-chain Every weakly decide whether or not may From can be. n o 2n+2 s t e p s . E nilpotent ring a power n i l p o t e n t r i n g . n i l p o t ent. n = f o r s = 0 , 1 , . . . , 2n + 3 Hence t h e every weakly has theory a ring type of r i n g nilpotent ring similar types can- i s power 81. PROOF: There n a r e no w e a k l y i s a natural n. Then number F o r suppose R 2 J — nilpotent n-1 , R that there negative 3 — rings J n-2 O J integer ... (n,n). Theorem rings ordinal 0 kO shows of the form numbers. of the type other ordinal than J , ... J = 2 n Hence R i s that there 42 a shows k i s a nonnumber. R. are (a,8) where Theorem ( a , k ) where a n d a i s an n o n - f i n i t e (n,B) where number J, 1 n+1 , and R =0. of a l l types non-finite a r e no of type and 8 i s any o r d i n a l and has t y p e 8 are both rings t h e R has t h e J - c h a i n : power n i l p o t e n t and nilpotent 5 81. 5 CONCLUSION This paper between the above. In U-ring and ring and A p a r t i c u l a r , every a prime every few paper. potent. Theorem may weakly probable from attempt nilpotent. that challenging further an not to that ring every defined is power necessarily a nilpotent a U*-ring, ideal. the results in information r i n g may every theorem be Also, every 29 i s a U-ring. hopefully will is a 22 U^-ring i t seem These be that theorem n i l makes the power n i l - n i l r i n g which ring. prove that U*-ring nilpotence provides products Finally, every conjectures research. not nilpotent that relationships nilpotent s u g g e s t e d by suggests a prime of Also, but ring types weakly 19 be ring. ring, are on every also weakly i d e a l i s a meta section that is products conjectures The resulted products meta* suggests U-ring some i m p o r t a n t different generalizations i s a prime not establishes are resolved by 82. BIBLIOGRAPHY (1) B a e r , R., M e t a I d e a l s , R e p o r t o f a C o n f e r e n c e o n L i n e a r A l g e b r a , E a t . A c a d . S c i . , W a s h i n g t o n 1951 33-52. ( 2 ) . F r e i d m a n , P. A., (Kazan) 2 (i960) Rings with Idealizer Condition I ,Izv. 213-222. (3) F r e i d m a n , P. A., R i n g s w i t h I d e a l i z e r C o n d i t i o n Zap. U r a l . Gos. U n i v . 2 3.-1 ( 1 9 5 9 ) 3 5 - U 8 . (h) r E f e i d m a n , P. A., R i n g s w i t h I d e a l i z e r C o n d i t i o n Z a p . U r a l . G o s . U n i v . 23.-3 (196~0).< k9-6l. I I , Ucen. I I I , Ucen. (5) F r e i d m a n , P. A., R i n g s w i t h R i g h t I d e a l i z e r C o n d i t i o n , U r a l . G o s . U n i v . Mat.- Zap.- U(l9.63) 3, 5 1 - 5 8 . (6) F r e i d m a n , P. A., Ideals of F i n i t e •(•7) Gluskov, On R i n g s , A l l o f Whose S u b r i n g s a r e M e t a I n d e x , M a t . S b . 65 ( 1 9 6 U ) 3 1 3 - 3 2 3 . V. M. , On t h e C e n t r a l S e r i e s M a t . S b . 31 ( 1 9 5 2 ) U9I-U96. of I n f i n i t e (8) K e g e l , O t t o H., On R i n g s t h a t J . A l g e b r a 1 (196U) 103-109. a r e t h e Sum o f Two (9) K e g e l , O t t o H., A R e m a r k o n M a x i m a l S u b r i n g s , Math. J . 11 (196U) 251-255. Rings, Michigan ( 1 0 ) . K u r o s h , A. G., T h e o r y o f G r o u p s , V o l s . I a n d I I , P u b l i s h i n g C o m p a n y , New Y o r k , N.Y. i 9 6 0 . . (11) Groups, Chelsea L e v i c , E. M., On S i m p l e a n d S t r i c t l y S i m p l e R i n g s , L a t v i j a s PSR Z i n a t n u A k a d . V e s t . T e h n , Z i n a t n u S e r . 1 9 6 5 n o . 6... 5 3 - 5 8 . (12) M e l d r u m , J . D. P., On t h e C e n t r a l S e r i e s J. Algebra 6 (1967) 281-28U. (13) N a g a t a , M a s a y o s h i , On t h e N i l p o t e n c e o f N i l J. Math. Soc. Japan U (1953) 296-301. (lk) Shevrin, L. H., On S e m i g r o u p s w h i c h o f a Group Algebras, are A t t a i n a b l e , M a t . S b . 6 1 ( 1 9 6 3 ) 25.3-25.6. (15) S p e r l i n g , M., Rings, M a t . S b . 17 ( 1 9 ^ 5 ) Every Subring 371-38U. of which i s an I d e a l ,
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Some generalizations of nilpotence in ring theory Biggs, Richard Gregory 1968
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Title | Some generalizations of nilpotence in ring theory |
Creator |
Biggs, Richard Gregory |
Publisher | University of British Columbia |
Date Issued | 1968 |
Description | The study of certain series of groups has greatly aided the development and understanding of group theory. Normal series and central series are particularly important. This paper attempts to define analogous concepts in the theory of rings and to study what interrelationships exist between them. Baer and Freidman have already studied chain ideals, the ring theory equivalent of accessible subgroups. Also, Kegel has studied weakly nilpotent rings, the ring theory equivalent of groups possessing upper central series. Some of the more important results of these authors are given in the first three sections of this paper. Power nilpotent rings, the ring theory equivalent of groups possessing lower central series, are defined in section 4. The class of power nilpotent rings is not homomorphically closed. However, it does possess many of the other properties that the class of weakly nilpotent rings has. In section 5 meta* ideal and U*-ring are defined in terms of descending chains of subrings of the given ring. Not every power nilpotent ring is a U*-ring. This is contrary to the result for semigroups. It is also shown that an intersection of meta* ideals is always a meta* ideal. It follows that not every meta* ideal is a meta ideal since the intersection of meta ideals is not always a meta ideal. Section 6 is concerned with rings in which only certain kinds of multiplicative decomposition take place. The rings studied here are called prime products rings and it is proved that all weakly nilpotent and power nilpotent rings are prime products rings. A result given in the section on U-rings suggests that all U-rings may be prime products rings. The class of prime products rings is very large but does not include any rings with a non-zero idempotent. The last section studies ring types which are defined analogously to group types. The study of which ring types actually occur is nearly completed here. Finally, it is shown that every weakly nilpotent ring has a ring type similar to that of some ring which is power nilpotent. This suggests (but does not prove) the conjecture that all weakly nilpotent rings are power nilpotent. |
Subject |
Rings (Algebra) |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-06-29 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080533 |
URI | http://hdl.handle.net/2429/35820 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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