UBC Theses and Dissertations

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UBC Theses and Dissertations

Some generalizations of nilpotence in ring theory Biggs, Richard Gregory 1968

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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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