UBC Theses and Dissertations

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UBC Theses and Dissertations

Irreducible representations of algebras Goodaire, Edgar George 1972

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IRREDUCIBLE  REPRESENTATIONS  OF  ALGEBRAS  by  EDGAR B.Sc.  THESIS THE  GEORGE  GOODAIRE  (Hon.), U n i v e r s i t y o f T o r o n t o , 1969  SUBMITTED  IN  REQUIREMENTS  DOCTOR  PARTIAL  FOR  OF  THE  FULFILMENT DEGREE  OF  PHILOSOPHY  i n the "Department of  Mathematics  We a c c e p t t h i s t h e s i s as conforming r e q u i r e d standard  THE  UNIVERSITY  OF  BRITISH  December, 1972  t o the  COLUMBIA  In p r e s e n t i n g t h i s t h e s i s  in p a r t i a l  f u l f i l m e n t o f the  requirements f o r  an advanced degree at the U n i v e r s i t y of B r i t i s h C o l u m b i a , I agree the L i b r a r y  s h a l l make i t f r e e l y  a v a i l a b l e f o r r e f e r e n c e and  study.  I f u r t h e r agree t h a t p e r m i s s i o n f o r e x t e n s i v e c o p y i n g o f t h i s f o r s c h o l a r l y purposes may by h i s r e p r e s e n t a t i v e s .  be g r a n t e d by  g a i n s h a l l not  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 Columbia Vancouver 8, Canada  thesis  Department o r  I t i s u n d e r s t o o d that c o p y i n g o r  of t h i s thesis for financial written  the Head of my  that  publication  be a l l o w e d w i t h o u t  my  ii  ABSTRACT  An  element x o f an a s s o c i a t i v e a l g e b r a A i s c a l l e d  diagonable  p r o v i d e d A has a b a s i s o f c h a r a c t e r i s t i c v e c t o r s f o r the t r a n s f o r m a t i o n ad x: a  ax - xa o f A.  a diagonable  T h i s n o t i o n immediately g e n e r a l i z e s to t h a t o f  subspace L o f A.  The c e n t r a l i z e r A Q o f L p l a y s an important  r o l e i n the r e p r e s e n t a t i o n theory o f A, f o r t h e r e i s a one-to-one c o r r e s pondence between the "X-weighted" i r r e d u c i b l e modules o f A and o f A 0 .  I n C h a p t e r s Two and T h r e e , we f i r s t e x p l o r e v a r i o u s p r o p e r t i e s o f A and A Q , and then use the r e s u l t s o b t a i n e d diagonable  elements i n d i f f e r e n t a l g e b r a s .  ring-theoretic  t o c l a s s i f y the  We a l s o g i v e c o n d i t i o n s under  w h i c h a l l A-modules a r e w e i g h t e d .  The  Cartan  theory o f L i e and Jordan  Four by the o b s e r v a t i o n t h a t C a r t a n  algebras i s linked  subalgebras  o f simple  finite  L i e and J o r d a n  algebras  are diagonable  subspaces o f the r e s p e c t i v e u n i v e r s a l e n v e l o p i n g  i n Chapter dimensional  (over a l g e b r a i c a l l y c l o s e d f i e l d s o f c h a r a c t e r i s t i c 0)  F u r t h e r m o r e , i n the J o r d a n  algebras.  c a s e , the c e n t r a l i z e r o f a C a r t a n s u b a l g e b r a i s  the c e n t r a l i z e r o f one o f i t s elements and i s a d i r e c t sum o f complete  matrix  rings.  F i n a l l y , we a r e a b l e t o show t h a t the u n i v e r s a l e n v e l o p i n g o f any s i m p l e Jordan  algebra  a l g e b r a which c o n t a i n s an idempotent whose P e i r c e one-  space i s o n e - d i m e n s i o n a l ,  i s g e n e r a t e d by i t s i d e m p o t e n t s .  iii TABLE  OF  CONTENTS  Page  INTRODUCTION  1  PRELIMINARIES  3  CHAPTER ONE  REPRESENTATION  THEORY  6  1.1  The U n i v e r s a l E n v e l o p i n g A l g e b r a o f a L i e A l g e b r a  1.2  P r o p e r t i e s -of A l g e b r a s P o s s e s s i n g Dlagonable and t h e i r Modules  1.3  A Correspondence Between A- and A 0 -Modules  16  1.4  F u r t h e r R e s u l t s Concerning  21  CHAPTER TWO  2.1  RING-THEORETIC  Elements  and  CONNECTIONS  BETWEEN  A  o  AND  9  A  24  Chain Conditions  27 2.2  -Nil'potent 'Ideals - and  Semi-P-rimeness  32 2.3 CHAPTER THREE  Primitivity DIAGONABLE  ELEMENTS  AND  WEIGHTED  MODULES  3.1  Idempotents a r e D i a g o n a b l e  35  3.2  Prime A l g e b r a s  40  3.3  C e n t r a l Simple  CHAPTER FOUR  46  Algebras  APPLICATIONS  TO OF  THE A  UNIVERSAL  JORDAN  ENVELOPING  ALGEBRA  ALGEBRA  50 4.1  The U n i v e r s a l E n v e l o p i n g A l g e b r a o f a J o r d a n  4.2  A Q ( x ) as a C a r t a n  4.3  Simple J o r d a n  Algebra  55  BIBLIOGRAPHY  Subalgebra  Algebras  62 65  iv  ACKNOWLEDGEMENTS  I w i s h t o e x p r e s s my a p p r e c i a t i o n to my t h e s i s D r . C.T. A n d e r s o n , f o r h i s c o n t i n u a l encouragement preparation  of this  thesis.  supervisor,  and h e l p d u r i n g t h e  My s i n c e r e thanks a l s o go t o D r . F.W. Lemire  f o r i n t r o d u c i n g the i d e a o f a " d i a g o n a b l e element" around which t h e t h e s i s i s c e n t r e d , and f o r communicating first  t o me r e s u l t s which l e d t o much o f my  chapter.  The f i n a n c i a l s u p p o r t o f t h e U n i v e r s i t y o f B r i t i s h Columbia and the N a t i o n a l R e s e a r c h C o u n c i l o f Canada i s a l s o g r a t e f u l l y  acknowledged.  INTRODUCTION  The Lie  u n i v e r s a l enveloping  algebra of a f i n i t e  dimensional  simple  a l g e b r a over an a l g e b r a i c a l l y c l o s e d f i e l d o f c h a r a c t e r i s t i c 0 has  basis relative  a  to which the l i n e a r t r a n s f o r m a t i o n s ad h: a t-> ah - ha f o r  h i n a Cartan subalgebra Cartan subalgebra an a s s o c i a t i v e  are s i m u l t a n e o u s l y  i s an example of what we  diagonizable.  Thus any  c a l l a diagonable  such  stibspace of  algebra.  L e t L be a d i a g o n a b l e In Chapter One,  we  subspace o f an a s s o c i a t i v e a l g e b r a  A.  d e f i n e the concept o f a X-weighted A-module, where X  i s a l i n e a r f u n c t i o n a l on L , and  then g e n e r a l i z e a theorem of Lemire  ([14])  by p r o v i n g the e x i s t e n c e o f a one-to-one correspondence between the  irre-  d u c i b l e X-weighted modules o f A and A Q S  This  the c e n t r a l i z e r o f L i n A.  r e s u l t l e a d s us to b e l i e v e there should be the a l g e b r a s A and A Q  and  these we  some c l o s e c o n n e c t i o n s  i n v e s t i g a t e i n Chapter  Two.  S e v e r a l examples throughout the t h e s i s i n d i c a t e how diagonable it and  subspaces o c c u r  i s therefore necessary the d i a g o n a b l e  In o r d e r to o b t a i n many r e s u l t s ,  I f we  assume L has  the  algebra  o n l y a f i n i t e number of  the Jacobson r a d i c a l of A i s n i l p o t e n t whenever A Q  I f e v e r y A-module i s w e i g h t e d , A Q  S e v e r a l of the r e s u l t s i n Chapter Two A-module be w e i g h t e d .  frequently  to Impose some k i n d o f r e s t r i c t i o n s on  subspace.  d i s t i n c t r o o t s , then semi-simple.  i n algebras.  between  In one  i s semi-simple whenever A i s .  r e q u i r e that every  irreducible  o f the major r e s u l t s o f Chapter T h r e e , we  a b l e to prove that i f L i s a d i a g o n a b l e  subspace spanned by a s i n g l e  b r a i c element i n a prime a l g e b r a A, and one  is  are  alge-  i r r e d u c i b l e A-module i s w e i g h t e d ,  2  then every A-module i s w e i g h t e d .  In t h i s c h a p t e r , we  c h a r a c t e r i z e the d l a g o n a b l e elements algebra.  are a b l e a l s o to .  of a f i n i t e dimensional c e n t r a l  They t u r n out t o be e x a c t l y those elements x^hich are  as l i n e a r c o m b i n a t i o n s  o f p r i m i t i v e o r t h o g o n a l idempotents  simple  expressible  w i t h sum  the :  0  i d e n t i t y o f A. elements  An i n t e r e s t i n g consequence o f t h i s i s t h a t the d i a g o n a b l e  .  o f a m a t r i x r i n g are n o t h i n g but those m a t r i c e s s i m i l a r to d i a g o n a l  matrices. In  Chapter  F o u r , we  g i v e an i l l u s t r a t i o n o f the common C a r t a n  t h e o r y o f L i e and J o r d a n a l g e b r a s e s t a b l i s h e d i n F o s t e r ' s d o c t o r a l  ([3]).  tation  disser-  A Cartan subalgebra H of a f i n i t e dimensional simple Jordan  a l g e b r a over an a l g e b r a i c a l l y c l o s e d f i e l d o f c h a r a c t e r i s t i c 0 i s a d i a gonable  subspace o f i t s u n i v e r s a l e n v e l o p i n g a l g e b r a A.  i n the L i e c a s e , we  o b t a i n a one-to-one correspondence  i r r e d u c i b l e « m o d u l e s o f A and o f C ,  the c e n t r a l i z e r  Thus, e x a c t l y  as  between the X-weighted  i n A o f H.  This result  i s p a r t i c u l a r l y u s e f u l because C can be r e a l i z e d as the c e n t r a l i z e r o f some element i n II and  i s a d i r e c t sum  o f complete  matrix r i n g s over F. ' .  As a f i n a l a p p l i c a t i o n o f the t h e o r y o f d i a g o n a b l e e l e m e n t s , prove  we  t h a t the u n i v e r s a l e n v e l o p i n g a l g e b r a o f a s i m p l e J o r d a n a l g e b r a i s  generated as an a l g e b r a by i t s i d e m p o t e n t s , p r o v i d e d the J o r d a n a l g e b r a c o n t a i n s an Idempotent whose " P e i r c e one-space" i s o n e - d i m e n s i o n a l .  3  PRELIMINARIES  A l l f i e l d s i n this thesis are understood to have characteristic zero.  An algebra i s a vector space A over a f i e l d F together with a b i -  linear map A x A •> A denoted (a,b) H - ab such that a(b + c) = ab + ac ; and  (a + b)c = ac + be  a(ab) = (aa)b = a(ab)  for any a e F, and a, b, c e A. A right ideal of A i s a subspace I such that ua e I whenever u e I and a e A; a l e f t ideal i s defined i n an analogous way.  By an ideal of A, we simply mean a subspace which i s both a l e f t and  right i d e a l .  A i s a simple algebra i f i t has no non-zero proper ideals.  A i s nilpotent i f there i s a positive integer n such that every product of n elements of A i s 0. A right A-module i s an abelian group V together with a b i l i n e a r map V x A ->- V, denoted (v,a) H- va, such that v(a + b) = va + vb ;  (v + w)a = va + wa ;  v(ab) = (va)b for any a,b e A and v,w e V. We also assume v l = v for every v e V i f A has an identity 1. By an "A-module", we always mean "right A-module". A module V i s irreducible i f i t contains no non-zero proper submodules. Schur's Lemma states that i f V and W are irreducible A-modules, then any homomorphism from V to W i s either zero or an isomorphism. c  '  •  I f A i s an algebra over the f i e l d F and x e A, we define the  linear transformations R , L , and ad x of A by  R : x  a  ax  L : x  an  xa  ad x:  a •+ ax - xa  Note that ad x =  -  for a e A  .  The commutator (x,y) i s defined by (x,y) = xy - yx, the associator Xx,y,.z) by (x.,y,.z) = xy..z - x.yz.  For .any subsets X, 'Y, and Z of A, define  (X,Y) - {(x,y): x e X, y e Y} and (X,Y,Z) = {(x,y,z): x  X, y  e  e  Y, z  G  Z}.  The centre of A is {a e A: (a,A) = 0 } and we always denote this by Z(A). A i s commutative i f (A,A) = 0 and associative i f (A,A,A) = 0 . A Lie algebra is an algebra L such that £x,x] = 0 and [[x,y],z] + [[y,z],x] + [[z,x],y] = 0 for every x, y and z in L, where the product of elements x and y in L is denoted [x,y]«  An associative algebra  A over a f i e l d F determines a Lie algebra A^ by defining the product of elements x,y e A to be the commutator (x,y). 2  is a commutative algebra J such that (x ,y,x) = 0  A Jordan algebra for any x,y e J. If e  We always assume a Jordan algebra contains an identity 1 .  = e i s an idempotent in J, then there exists a direct sum decomposition  of the vector space J called the Peirce decomposition; namely, J = J  Q  + Ji^ + J  ;  x  J± = J (e) = {x e J: xe = ix}, i =  0,^,1  ±  e^ are pairwise orthogonal idempotents ( i G j = 0  More generally, i f e^  e  for i ^ j) with sum 1 , the Peirce decomposition has the form n J - @  j»  J  ;  J  ±i  = J ( e ) = {x E J: xe 1  i  = JjjCe^  0  JjgCe^)  =  ±  {x  = x}, i = 1,... ,n E J :  xe^^  = J$x =  xe  l<j  The  r e a d e r s h o u l d c o n s u l t Jacobson  [ll;  pages 118-120] f o r d e t a i l s .  F i n a l l y , we p o i n t out t h a t an element x i n an a l g e b r a A o v e r F i s c a l l e d a l g e b r a i c i f i t i s the s o l u t i o n a polynomial i n t with c o e f f i c i e n t s  t o some p o l y n o m i a l i n F [ t ] ( i ; e .  in F).  Among a l l such p o l y n o m i a l s ,  t h e r e i s one o f l e a s t degree c a l l e d the m i n i m a l p o l y n o m i a l o f x which d i v i d e s any o t h e r p o l y n o m i a l f o r which x i s a s o l u t i o n . " We assume t h a t the m i n i m a l p o l y n o m i a l i s raonic, i n which case i t i s u n i q u e .  We now make the c o n v e n t i o n t h a t h e n c e f o r t h , " a l g e b r a " w i l l mean_"associative  algebra with  identity".  always  6 CHAPTER  ONE  REPRESENTATION  1.1  THE  UNIVERSAL  ENVELOPING  THEORY  ALGEBRA  OF  A  LIE  L e t i i be a L i e a l g e b r a over a f i e l d F .  ALGEBRA  A representation of L  i s d e f i n e d t o be a l i n e a r map S:L •*• A, where A i s an a s s o c i a t i v e a l g e b r a over F , such t h a t f o r e v e r y a and b i n I ,  (1)  where S: u i + S„ for u e L u  S , = S S, - S,S a b a b b a  F o r example, R: x t->- R^ d e f i n e s a r e p r e s e n t a t i o n o f L i n the a s s o c i a t i v e a l g e b r a X g e n e r a t e d over F by {R^: x e L}. with  i d e n t i t y i s a u n i v e r s a l enveloping  An a s s o c i a t i v e a l g e b r a U(L)  algebra f o r L i f there i s a c a n o n i c a l  r e p r e s e n t a t i o n S*: L -*• U(L) such t h a t g i v e n any r e p r e s e n t a t i o n S: L -*• A  of L  i n an a s s o c i a t i v e a l g e b r a A, t h e r e e x i s t s a unique homomorphism Y: U(L) -> A such t h a t S = YoS*; i . e . which makes the f o l l o w i n g diagram commutative:  L  >  A  S  ( [ 9 ]) f o r a more comprehensive  We r e f e r the r e a d e r  to Chapter V o f Jacobson  t r e a t m e n t o f U(,L).  What i s i m p o r t a n t f o r us here i s :  THEOREM 1.1.1:  U(L) exists,  is unique  up to isomorphism,  and is  generated  by { S * x : x e £ } .  Suppose now t h a t L i s f i n i t e  dimensional  and s i m p l e , and t h a t F  7  is  algebraically closed.  a nilpotent implies  Then L p o s s e s s e s a C a r t a n s u b a l g e b r a //; i . e . ,  s u b a l g e b r a which i s s e l f - n o r m a l i z i n g  x e H.  There e x i s t s a s e t A o f l i n e a r f u n c t i o n a l s  H} i s non-zero f o r any r o o t a .  a:H -»- F c a l l e d  One i n t r o d u c e s an o r d e r i n A and i s  a b l e to d i s t i n g u i s h between p o s i t i v e and n e g a t i v e r o o t s . if  i t cannot be w r i t t e n  A+  the s e t o f s i m p l e and p o s i t i v e r o o t s r e s p e c t i v e l y .  basis  [x,#] £ H  - {x e L: x(R^ - a ( h ) l ) n = 0, f o r  r o o t s w i t h the p r o p e r t y t h a t the subspace all he  i n the sense t h a t  A root i s simple  as the sum o f two p o s i t i v e r o o t s .  Denote by A g  and  Then t h e r e e x i s t s a  B o f L c a l l e d a C a r t a n b a s i s , B = {e . f , h fi : 3 e A_, a e A+}, where — — — — —  —  —  —  0t  p  Ct  o  among the m u l t i p l i c a t i v e r e l a t i o n s f o r B, we have  [hg.hg,] = 0 ( 2 )  tVV  =  VB « e  [ f a , h 3 ] = Ba..efB  where the A  „ and B a,p  ,  a e A + , 3,3' e A s  „ are integers. a,p  "  By the P o i n c a r £ - B i r k h o f f - W i t t Theorem ( [ 9 e n v e l o p i n g a l g e b r a U(L) o f L has a l i n e a r b a s i s of  consisting  o f a l l elements  the form  (3)  TT  .f  n ( a ) a  s  IT ^e^s  aeA+  r  m  TT  eam(a)  aeA+  where n ( a ) , r ( B ) , m(a) a r e n o n - n e g a t i v e i n t e g e r s the  ; § 5 . 2 ] ) , the u n i v e r s a l  and the p r o d u c t r e s p e c t s  o r d e r i n A. For  a fixed 8  aeA+ because o f the i d e n t i t y  O  e A s , observe  aeA+  that  u  aeA+  8  (4)  (xy,z) = x ( y , z ) +  which h o l d s is  (x,z)y  i n any a s s o c i a t i v e a l g e b r a .  G e n e r a l l y , i t i s true that  (u,h 0 ) p o  an i n t e g r a l m u l t i p l e o f u f o r each b a s i s element u o f the form ( 3 ) , and  hence U(L) decomposes as a v e c t o r space i n t o a d i r e c t sum o f spaces o f the form {u e U(L): ( u , h R ) = nu} f o r n an i n t e g e r .  With t h i s m o t i v a t i o n  we  make the f o l l o w i n g d e f i n i t i o n .  1.1.2: Let A be an associative  DEFINITION  field  F.  algebra  Then an element x e A is diagonable  with unit element over a  if A  A Cx), a vector aeF  space direct  sum, where A a ( x ) = {a e A: (a,x) = a a } .  9  1.2  PROPERTIES OF ALGEBRAS POSSESSING DIAGONABLE ELEMENTS AND THEIR MODULES  We collect some important facts about diagonable elements i n the next two propositions.  PROPOSITION 1.2.1: Suppose A = ©  £  R. i s an algebra direct  sum and  lei  x =  \ x^ is a diagonable  element of A.  Then each x. is a diagonable .  iei and (R ) (x ) = ( x ) TV R^.  element in the algebra  A  i  ct  i  a  PROOF: Let u E R . Then (u,x) = (u,x.) = 1  u  e ( )* A  Q  x  a  £ u 0?«aeF a  a  e R. where u = J u aeF  &  Next,  ((u,x),x) = ((u,x.),x.) = (. I au ,x) = 7 a u e R.,. O&xeF " Oj*aeF 1  1  Continuing in this way, for every positive integer k, we have  ) o u e R.. O&xeF  Now u = 0 except for a in some finite set {ai,...,a } U {0}, no a  n  = 0.  Thus n with  v  i»-«'»  v  ^ %» i  e  n  system of n equations in n unknowns with coefficient  s3  matrix ( a ^ ) , i , j = l,...,n  a vandermonde matrix.  t  Since cq,...,a are n  distinct scalars, the system has a solution; i.e., each u e R^. Also, n * u = u - £ u ^ e R^ and so x^ i s diagonable in R^. Finally, a  Q  Q  ( R  i a )  ( x  i  )  =  V  x  ) 0  V  i s an immediate consequence of (u,x) * (u,x^) for u e R^.  10 PROPOSITION  1.2.2:  ( i ) If x is a diagonable  and B is a subspace in  the sense  that  of A invariant B decomposes  (tfot-e t/zat we do not require  (ii) then  t x is diagonable,  PROOF:  and, in fact,  element  and t c F is  = ab}.  non-zero,  i f AQ(x)  elements,,  = 0 except  for a E {a^: i E 1}  for a e { g j i j E J}, then : A y (x+y) £ 0 1/ I and J are f i n i t e sets,  ad(x+y) satisfies  the polynomial  ( I ) I f b E B, we can w r i t e b =  •*('Bvx>'S**B-'(byx-W')•  a "b OftxeF  Jf i£l  implies  the  linear  ( t - (c^+B-j)) e F [ t ] , J  \ b Q r e l a t i v e t o x , and s i n c e  *E B - f o r - e v e r y '  ( b , x ^ ) = (...(b,x),x),...,x),  A vandermonde m a t r i x  on B  B (x) = {b E B: (b,x)  .If x .and y are commuting diagonable  Y e { d i + $^: i e I , j e J } . transformation  A ,  and A Q ( x ) = A t a ( t x ) .  then x + y is diagonable = 0 except  Y B (x) with aeF x E BJ.  If x is a diagonable  (iii.)  and Ag(y)  that  in an algebra  ad x, then x is diagonable  under  as ©  element  i n t e g e r k > 0.  x repeated  k  Here we w r i t e  times.  argument i d e n t i c a l t o t h a t used i n t h e p r o o f o f 1.2.1  shows t h a t each b Q e B.  ( i i ) i s t r i v i a l , and ( i i i )  (x,y) = 0 i m p l i e s each A  f o l l o w s from ( i ) because  (x) i s i n v a r i a n t under ad y , and hence decomposes i  as ®  I ( A Q ( x ) ) g (y) r e l a t i v e t o y . 1 jeJ J  The  l a s t part of ( i i i )  and  Aa(x)  Clearly  (Aa,(x)) Q ( y ) C 3 Aa o (x+y). 1 1 J J  f o l l o w s from the o b s e r v a t i o n  that i f x i s diagonable  = 0 except f o r a i n a f i n i t e s e t { a ^ » • • • »ctn) , then t h e l i n e a r n t r a n s f o r m a t i o n ad x o f A i s a l g e b r a i c w i t h minimum p o l y n o m i a l TRt ~ a - » ) e F [ t ] . 1 i=l  Now l e t L be a l i n e a r subspace o f an a l g e b r a A over F which i s spanned by commuting d i a g o n a b l e  elements.  The p r e c e d i n g  proposition implies  11  t h a t e v e r y element o f L i s n e c e s s a r i l y  A map a: L  DEFINITION 1.2.3:  Aa(L)  such that  = {a e A: (a,x) = a ( x ) a , f o r every x e L}  is non-zero is called root space.  F  diagonable.  a root of L in A, and A 0 ( L ) is the  With L f i x e d , we w i l l w r i t e A a  " r o o t s o f A".  If V is a right  corresponding  i n s t e a d o f A a ( L ) and r e f e r t o  A-module which is also a vector  space over F ,  a map X:L -»• F is a weight of L in V if, = {v E V: v ( x - X ( x ) l ) n = 0, f o r every, x e L and n = n ( x ) > 0}  is non-zero.  weight space and V is said to be  is the corresponding  L-weighted or sometimes \-weighted if we wish to emphasize that X is a weight of L . We remind the r e a d e r  t h a t i f B i s any v e c t o r space over a f i e l d F ,  B*, the d u a l space o f B, i s d e f i n e d as the space o f a l l l i n e a r i.e.,  those maps ij>: B ->• F s a t i s f y i n g  ( i ) iKbj + > b  =  +  2  *<b2>  * n o t a t i o n i s standard  PROPOSITION 1.2.4:  and w i l l  a n d  f o r e v e r y b^,b2 E B , a e F .  ( i i ) \(i(ab^) = au>(bp  The  f u n c t i o n a l s on B;  be used f r e e l y i n the r e s t o f t h i s  thesis.  If X is any weight of L in an A-module V, and a is  a root of L, then both X and a are in L * . PROOF:  We show f i r s t t h a t f o r any u E L , X(u) i s the o n l y  r o o t o f u on V, .  F o r t h i s , l e t O^v E V..  Then  characteristic  12  v(u - a l )  s  « 0 » v ( u - X ( u ) l ) t i m p l i e s v ( ( u - X(u))  where we can expand u s i n g  l i n e a r now f o l l o w s  (a)  t + S  = 0,  theorem because u - X(u)1 and  the b i n o m i a l  Thus v ( ( a - X ( u ) ) l ) t + S  u - a l commute.  - (u - a l ) )  = 0 and X(u) = a .  That X i s  from  x , y e L i m p l i e s X(x) + X(y) i s a c h a r a c t e r i s t i c r o o t o f x + y on  arid  (b)  x e L and a e F i m p l i e s aX(x) i s a c h a r a c t e r i s t i c r o o t o f ax on  To  see ( a ) , we have p o s i t i v e i n t e g e r s n and m w i t h 0 = v ( x - x ( x ) l ) n  = v ( y - X ( y ) l ) m f o r any CMv  e V^.  S i n c e x - X ( x ) l and y - X ( y ) l commute,  v((x+y) - ( X ( x ) + X ( y ) ) l ) n + m = v ( ( x - X ( x ) l ) + (y - X ( y ) l ) ) n + m = 0. (b) f o l l o w s i n the same way. e x a c t l y as above, u s i n g  the l i n e a r i t y o f the map ad x .  A linear  DEFINITION 1.2.5:  commuting diagonable  F i n a l l y , t h a t a r o o t a i s a l s o i n L* i s proven  subspace L of an algebra  elements, is called  decomposes as a vector space relative i.e.,  A =©  I aeA  a diagonable  to the collection  A  over F , spanned by  subspace, if A A of roots of L  A . a  Suppose that L i s any two-dimensional subspace o f A w i t h b a s i s c o n s i s t i n g o f the commuting d i a g o n a b l e elements x and y . of proving  I n the c o u r s e  P r o p o s i t i o n 1.2.2, we saw  (1)  A  =  I a,8  (Aa(x))3(y),  ( A a ( x ) ) g ( y ) = A a ( x ) A Ag(y)  the sum taken over a l l c h a r a c t e r i s t i c r o o t s a o f ad x and c h a r a c t e r i s t i c r o o t s 8 o f ad y .  The l i n e a r i t y o f the r o o t s o f L i m p l i e s immediately  that  if  ( A a ( x ) ) g ( y ) $ 0, then t h i s space i s e x a c t l y A ^ ( L ) , where X i s the r o o t  of  L u n i q u e l y determined  by the c o n d i t i o n s X(x) = a , X(y) = B.  Thus (1)  is  a decomposition o f A r e l a t i v e t o L; i . e . , L i s a diagonable  subspace.  By i n d u c t i o n , we see t h a t any f i n i t e - d i m e n s i o n a l subspace L spanned by commuting d i a g o n a b l e elements  i s i n f a c t a diagonable subspace.  F o r example,  a C a r t a n s u b a l g e b r a o f a f i n i t e d i m e n s i o n a l s i m p l e L i e a l g e b r a over an a l g e b r a i c a l l y closed f i e l d  i s abelian  ([9 ; page 110]) and hence i t s image  under t h e c a n o n i c a l embedding i n the u n i v e r s a l e n v e l o p i n g a l g e b r a i s a diagonable  subspace.  Now f i x a d i a g o n a b l e subspace L o f A , and l e t  the d e c o m p o s i t i o n o f A r e l a t i v e  0  V^  +  A  to be zero if X + a is not a weight of L in V . V  An easy i n d u c t i o n r e v e a l s t h a t x a = a ( x - a ( x ) l )  fora l l a e A  Thus i f v ( x - X ( x ) l ) n = 0, we have  = v(x - X ( x ) l ) n a  and so v a  We come  A-modules.  k  and x e L .  be  -If--xj-.its-tm'b-mi&hted^A^&duZe* ••V^-Aac:--v\+0 for any weight \  where we define  PROOF:  j[ A aeA a  t o the c o l l e c t i o n A o f r o o t s o f L .  now t o an i n v e s t i g a t i o n o f L-weighted  •^EMMA-4^.-P6-:  A =©  (Ax (-HxH) ~ a k  n  k  =  v I  ~  v  =  va(x -(X(x) + a ( x ) ) l )  I ($(-X(x))n_ka(x - a(x)l)k k=0^' n  eV, , . X+a  COROLLARY 1.2.7:  Let V be an irreducible  decomposes as © £ V , relative XeA  A  h-weighted A-module.  Then V  to the set A of all weights of L in V .  A  3  14  ^ 0 for some X e A and so by the lemma,  PROOF: We simply note that ® £ V> XeA  i s a non-zero A-submodule of V.  Now the identity ( 4 ) i n 1 . 1 implies A A S  (2)  Q  g  A ^  where we again define & £ to be zero i f a + 8 i s not a root.  In particular,  a+  -A = {a e. ,A: ax = xa, for every x e-L} i s .a .subalgebra containing 1 and x; namely, the centralizer of L i n A. Also ( 2 ) implies that each A By 1 . 2 . 6 so i s any weight space  A -module. o  a  i s an  of an A-module, and we can  further show:  LEMMA  If V is an irreducible  1.2.8:  AQ-module PROOF:  for any weight  weighted  centre  is an  If W), i s a proper A-submodule of V\ , then W = .o ° o o A  A  If  1.2.9:  of A ,  K  is any maximal right  then au e K with  Q  irreducible  X.  is a proper A-submodule of V because of  LEMMA  A-module,  A = W, © T W\ A o r\j. L * o a A  A  1.2.6.  ideal  a e A implies Q  A  of A and u e Q  Z(A )., Q  the  a e K or u e K.  PROOF: If a i K, K + aA = A , and so k + ab = 1 for some k e K and b e A . Q  q  Q  Hence ku + abu = u with ku and abu (=aub) both i n K. So u e K.  Now suppose V i s an irreducible L-weighted A-module and non-zero weight space. homomorphism  A  Q  Let O^v e  . Then x: a n va i s an A ~module  -»• V , necessarily surjective by  where To i s the kernel of T .  is a  0  1.2.8,  and so  =  A  0  / T  Q  Since To i s a maximal right ideal of o A „ which &  !5 (x - X ( x ) l ) n , and  contains  x - X ( x ) l i s i n Z ( A 0 ) , we  c o n t a i n s x - X ( x ) l f o r e v e r y x e L by  1.2.9.  have t h a t T Q  But now,  actually  l e t t i n g T be  the  k e r n e l o f the A-module homomorphism A -> V d e f i n e d by a»-»- v a , s i m i l a r g i v e s V = A/T.  THEOREM  1.2.10:  Noting  t h a t T (\ A Q  = T Q , we  If V is an irreducible  have  reasoning  obtained:  L-weighted A-module and X is a  weight of L in V, then V = A / T , where x - X ( x ) 1 E T for every x E L .  We  c l o s e t h i s s e c t i o n w i t h an easy r e s u l t which  nevertheless  w i l l prove q u i t e u s e f u l .  PROPOSITION  1.2.11:  be a diagonable  Suppose  A = Q  £ R.  iei  subspace of A and write  is an algebra direct  A-module is L-weighted  3  Let  L  L = © £ L . . Then if every iei'  irreducible  sum.  every irreducible  1  R ^ -module is  L J -weighted, for any ± E I . n  PROOF:  Any  i r r e d u c i b l e R - -module V i s a l s o an i r r e d u c i b l e A-module under  the d e f i n i t i o n v a = v a ^  f o r v E V and a = £ a^ E A. Thus V i s L-weighted ° iEl and ^ 0 f o r some weight X . L e t x E L . Then w r i t i n g 1 = £ e^ and i£l x = / x, , v ( x - X ( x ) l ) n = 0 i m p l i e s v(x^ - X ( x ) e ^ ) n = 0 . S i n c e eniEl ° ° i s the i d e n t i t y i n R J , upon d e f i n i n g X J ( X J ) = X ( x ) , we see t h a t V i s x x o io o 1  L J1  o  -weighted w i t h weight X . . o  l  '  o  16  1.3  A  CORRESPONDENCE  BETWEEN  A- AND  A -MODULES o  L e t H be a C a r t a n s u b a l g e b r a o f a s i m p l e L i e a l g e b r a L over an a l g e b r a i c a l l y let  closed f i e l d  C be the c e n t r a l i z e r o f H i n the u n i v e r s a l  o f c h a r a c t e r i s t i c 0,  set of equivalence classes  a weight X e H* and X-weighted  and  V(I).  enveloping algebra  Lemire ([14]) has shown t h a t t h e r e i s a one-to-one the  finite-dimensional  correspondence between  of i r r e d u c i b l e representations of L possessing r e p r e s e n t a t i o n s o f C; i . e . , r e p r e s e n t a t i o n s  whose a s s o c i a t e d C-modules a r e X-weighted.  I n t h i s s e c t i o n , i t i s shown  t h a t t h i s r e s u l t f o l l o w s from the f a c t t h a t H i s a d i a g o n a b l e subspace o f  We  denote by W\ the c o l l e c t i o n o f a l l isomorphism c l a s s e s  o f i r r e d u c i b l e L-weighted A-modules V f o r which the  c o l l e c t i o n o f a l l isomorphism c l a s s e s  ^ 0, X e L * .  tvl o f non-zero  U(L).  [V]  W\° denotes  irreducible  A o -mcdules V f o r which V(x •- A ( x ) l ) = 0 f o r a l l x c L .  For c o n v e n i e n c e , we  w i l l w r i t e V when we  [vl.  strictly  The d i s c u s s i o n g i v e n by *V = V^.  LEMMA  p r i o r to  isomorphic  isomorphic  PROOF:  We  -*•  to W, then  irreducible  a map  -*•  for every  s i m p l y observe t h a t i f ^:W to W^  A-module and V is any  V is L-weighted  to V\ as AQ-modules  by r e s t r i c t i n g phism W^  1.2.10 e s t a b l i s h e s  $ i s w e l l - d e f i n e d because o f the next r e s u l t ;  1.3.1: If W is an L-weighted  A-module is  mean the isomorphism c l a s s  and irreducible,  and  W^  A e L*.  -*• V i s the A-module isomorphism, then  and the r i n g o f s c a l a r s  to A Q , we  because each module i s i r r e d u c i b l e and  o b t a i n an  iHwp  V^.  Isomor-.  17  A one-to-one correspondence between by showing that the map $ has an inverse.  LEMMA 1.3.2: distinct  and  First, a result in linear algebra:  Let V be any vector space over a field  elements of V*.  is a set of n distinct  w i l l be established  F and Tj,...,T  n  Then there is a v e V such that (T^v) ,... ,T (v)} n  scalars.  PROOF: Consider the finitely many linear functionals {T - T^: i 4 j} ±  of V*. Since T ...T are distinct, {kerCT^ - T..): i j} is a finite J  1}  n  set of proper subspaces of V. We are required to find a v e V such that n. v i Q ker(T - T,). But this follows from the fact that V cannot be expressed as a finite union of proper subspaces because F i s i n f i n i t e t (char F = 0). Indeed, i f V = \J V. , we may assume inductively that no v^ i s contained in U  V. and hence choose Vj £ Vi - \J V. and V2 I V,. j-2  3  .There must-be distinct-scalar-s  and a --such that -ct-j-Vj ••+ v and c^v^ + v£ 2  2  belong to the same V^. Then i ^ 1 because V2 i 0*1V1  + v ) - (a v 2  2  DEFINITION 1.3.3:  1  + v ) e V±. 2  and so for some i > 1,  But then (a^ - a ) v 2  An ideal I (right,  left,  1  E V i which i s impossibl  or two-sided) of A • © \ A  Q  Ct£A  is called homogeneous if I = ©  £ I/) A . Q  OIEA  PROPOSITION 1.3.4: (or left)  Any two-sided ideal of A is homogeneous. Any right  ideal containing  x - X(x)1 for every x E L, and some X e L* is  homogeneous. PROOF: Let I be a two-sided ideal of A and suppose a = £ a e I with a E A aeA Then for any x E L, (a,x) = ][ a(x)a E I. Also ((a,x),x) = \ <x(x) a a  2  a  (Mae A  0#KEA  is in I, and continuing i n this way, we get that for any integer k > 0,  18  I  •(1)  a<x)\ -  i  e  k  Now a a = 0 f o r a l l a except a e { c t j , . . . , a n } .  I  The  a r e i n L* by 1.2.4,  and so by 1.3.2, we c a n f i n d an x e L f o r which ia^(x),...,an(x)} set o f n d i s t i n c t s c a l a r s .  is a  Then l e t t i n g k r u n from 1 t o n , (1) i s a  system o f l i n e a r e q u a t i o n s over F whose m a t r i x o f c o e f f i c i e n t s i s a vandermonde m a t r i x w i t h non-zero  determinant.  Solving, i t i s clear  t h a t each a a . i s a l i n e a r combination o f - i j _ , . . . . ,.i n a = a - / a „ . i s then i n I t o o . For  and so i n I . Of c o u r s e  t h e second statement o f the p r o p o s i t i o n , i f I i s a r i g h t  o f A c o n t a i n i n g x - X ( x ) l f o r a l l x e L and some X e L * , and i f a =  ideal  ][ a aeA  e I,  we note t h a t (a,x) = (a,x - X ( x ) l ) i s i n I t o o , and hence the c o n c l u s i o n f o l l o w s j u s t as above.  LEMMA 1.3.5:  Any (proper) right  in 1 ^  contained  ideal  I  of A Q generates  \ A^ and hence is contained  a right  ideal  in a maximal right  of A  ideal I*  O^aeA  of A.  If I is maximal, I* (\ A = I , and if, in addition, Q  there is a X e L*  such that x - X(x)1 e I for every x e L , then I* Ss I (D - I A a and I * O^aeA  is  unique.  PROOF:  The r i g h t i d e a l IA o f A which I g e n e r a t e s i s c o n t a i n e d i n I @  because A^A^ c A ^ . involving  The e x i s t e n c e o f I * then f o l l o w s from a s t a n d a r d argument  an a p p l i c a t i o n  o f Zorn's Lemma.  Now I* (\ A  Q  £ A  Q  because 1 i I * ,  and s i n c e I * A A ^ g l and i s a r i g h t i d e a l o f A D , we must have I* (\ A if  I i s maximal.  Q  = I  N e x t , i f t h e r e i s a X e L* w i t h x - X ( x ) l e I f o r e v e r y  x e L , then (2)  \ AQ O^aeA  J < C I ©  I A 0/aeA  a  19  f o r any (proper)  r i g h t i d e a l J o f A which c o n t a i n s  I . To o b t a i n  ( 2 ) , we  o b s e r v e t h a t such a J i s homogeneous by 1.3.4 and so i s c o n t a i n e d i n (J AA )  I  +  Q  A .  O&teA  Thus the sum o f a l l p r o p e r r i g h t X  CD  A  I  A  J fV A  As I g J A A Q ^ A Q ,  = I by the m a x i m a l i t y o f I .  a  ideals  of A containing I i s contained i n  and so must a g a i n be p r o p e r .  Clearly  t h i s i s the unique  0/cte'A maximal r i g h t i d e a l I * .  ,Now .define a map ¥ :  •*•  as f o l l o w s :  .if V e  then  V - A Q / I where I i s a maximal r i g h t i d e a l o f A Q c o n t a i n i n g x - x(x)l f o r a l l x E L and some X e L * . (I + l ) ( x - X ( x ) l ) = 0. r i g h t i d e a l I * o f A. x E L.  Define  LEMMA 1.3.6:  ^-modules, Then A/11*  T h i s i s so because V(x - X ( x ) l ) = 0 i m p l i e s  By Lemma 1.3.5, I extends u n i q u e l y  A/I*  V = A/I*.  Suppose  because  e  A0/I^  and  A  - A / I 2 * cis A-modulest  Q  / I  a v e 2  where  by 1.3.5.  PROOF:  Suppose t h a t o ( I j + 1) = I Lift  (3)  u E A  2  irreducible  such  and 2  isomorphic  for every  x E L.  1^* and I 2 * are the right  + a Q , where a: A / I J  a t o a*: A/I^*  i s well-defined for i f a E  so t h e r e e x i s t s  f o r every  ¥ i s w e l l - d e f i n e d because o f :  A given  o*  ( I * + 1 ) ( x - X(x)1) = 0  and for some X E L*, X - X(x) 1 e I j (% I  given isomorphism.  t o a maximal  -»• A / I 2 * by a*:  AQ/I  ideals  of  i s the  2  1^* + a H - 1^* + a Q a .  b u t a Q a I I 2 * , then I 2 * + a Q a A = A, and  that  - Q au - l e I * S l 2 e 2  A  I  A  .  O^aeA Now because au £ 1^* £ I j @  £  A  Q  , we can w r i t e au = b  + J\> where b Q e l ^ * a  O^aeA and o(I1  so from (3) we see t h a t a Q b 0 - 1 E I 2 . + b0) = 0 = I  2  + aQbo.  This i s impossible  F i n a l l y , since I 2 * £ . I 2 ©  J  O^asA  AQ  because , a  Q  i I  2  *.  20 Hence o*  i s not z e r o and  We  are now  THEOREM 1.3.7:  PROOF:  We  able  t h e r e f o r e must be  a one-to-one  prove t h a t <I> and  t h a t $V =  = A0/I  f o r every x e L. Lemma 1.3.5.  ¥ are i n v e r s e maps.  I'C^V) i s then A/I* - V i s now  Theorem 1.2.10 upon o b s e r v i n g by u n i q u e n e s s .  Conversely,  r i g h t i d e a l of AQ L e t t i n g J * be see  correspondence  a  Q  given V e  V = AQ/J  = AQ/J  containing x - X(x)l  1.3.5, *(VV)  L*.  i s then ( A / J * ) A .  a: A  aQ  e J*}  = J * A AQ  = V as r e q u i r e d ; i . e . ,  J* and  so a(An)  £  w  (A/J*)^ i s i r r e d u c i b l e . = J by =  To  -*• ( A / J * ) , by a . »-*• J * + a  O  e AQ:  by  preceding  c o n t a i n i n g x - X ( x ) l f o r e v e r y x e L , f o r some X e  O  k e r n e l of a i s {aQ  recall  where J i s a maximal  . a „ ( x - X ( x ) l ) = (x - X ( x ) l ) a „ E J £  O  .  t h a t the i d e a l T d e f i n e d t h e r e i s the i d e a l ^ 1 *  a i s s u r j e c t i v e because i t i s non-zero and  (A/J*)A  G i v e n V e W-^, we  apparent from the d i s c u s s i o n  t h a t i d e a l of A g i v e n by  e A  and  where I* i s t h a t i d e a l of A g i v e n  t h a t t h i s i s Isomorphic to V, d e f i n e a map  For any  between  f o r some maximal r i g h t i d e a l I of A Q  That YC^V)  Schur's Lemma.  to prove the main r e s u l t o f t h i s s e c t i o n .  defines  $  an isomorphism by  V.  1.3.5.  Thus,  .  (A/J*),. A  The  21 1.4  FURTHER  RESULTS  CONCERNING  AND  ^ °  L e t V be an L-weighted i r r e d u c i b l e A-module w i t h A the s e t o f weights o f L i n V.  We  the a s s o c i a t e d f a m i l y A(V)  have seen how  of i r r e d u c i b l e A 0 ~modules determines v i a the map A-module V, up  to isomorphism ( 1 . 2 . 1 0 ) .  p o i n t i s the f o l l o w i n g : a f a m i l y F = {.V^ such t h a t F =  T h i s was  1^  (unique) maximal r i g h t i d e a l s 1^* such  ;  (A/T^)y  Then (1) c e r t a i n l y i m p l i e s  phism T : A -*• A/I^*  this  containing  and  e  L*.  1^*  of A  = A0/IM  Suppose now  ( A / I ^ * ) ^ J 0 and  g i v e n by x (a) = I * + b ^ a , where 1^*  (A/Ix*)y.  By  that  thus from  b^y  and hence 1 would b e l o n g to 1^*; (b^y - l ) a + a e 1^*  + b^  the i r r e d u c i b i l i t y o f A / l ^ * ,  t h e r e i s a y E A such t h a t b^y  and  - 1 e 1^*-  i s any  (1^* + b y ) A  C l e a r l y y i 1^*  but y l ^ * Is 1^,*  non= A/Ix*  because then  because i f a E  so ya E k e r x = I y A «  We  have e s t a b l i s h e d  the n e c e s s i t y p a r t o f the next theorem.  THEOREM 1.4.1: -A family  family  and  Theorem 1.2.10, I * i s the k e r n e l of the homomor-  and  =  at  that  e s t a b l i s h e d i n the p r o o f o f Theorem 1.3.7.  zero element o f  to ask  f o r every x E L r e s p e c t i v e l y , where X,u  (A/IA*)X - A0/IX  the d i s c u s s i o n p r e c e d i n g  b^ya  A natural question  are maximal r i g h t i d e a l s of A Q  e Jv^, A / l ^ * e Jif , and  A/Ix* = A / I J J * .  A}  Y the same i r r e d u c i b l e  g i v e n some subset A o f L * , the d u a l space o f L ,  know t h a t t h e r e e x i s t  (1)  e  A(V)?  x - X ( x ) l and x - u(x)1  w i t h A/1^*  ^:A  X e A) o f A 0 - m o d u l e s , does t h e r e e x i s t an A-module V  e  Suppose 1^ and  We  l  =  corresponding  (A0/Ix  E Wx°: X e A £ L*}  to an A-module V i f and only  of AQ-modules  i f there  exist  is  the  elements  22y  , in A, for each X and y in h, such  y»x  that  y,, > i I * but y.. ^I,* e l *.  v>  y  A  PROOF: We need only prove the sufficiency.  •» J  y  A A  If X,y e A and there i s a y i n A ,  y i I * , y l * ^ l * , then the map A/I * -*- A / I * defined by I j * + a H - I * + ya y  x  y  x  p  u  is well-defined and non-zero, hence an isomorphism by the irreducibility of the two modules.  Let V = A/I^*.  By (1) and Lemma 1.3.1, we see that V has  the desired property.  We now consider the extent to which the classes determined by the weight A .  and  The situation i s ideal for A -modules, some0  what less than ideal for A-modules as the following theorem shows.  THEOREM 1.4.2:  If W T\  (ii)  PROOF:  X  E  Q  then X is identically  ? <j>, then  equal  to y (\= ). v  for some a e A, y = X + a and -a e A.  ffc W °, then there exist maximal right ideals 1^  (i) If V  and I of A  W°  (i) I f  containing x - X(x)l and x - y (x) 1 respectively, for every  x E L, and such that V ^ A Q / ^ ~ / I • A  0  isomorphism and o ( I + 1) = I + a , a A  S u  y  Q  Qe  PP A . q  o s e  0  :  A 0  /I  **" o ^ p A  A  i  s  t  h  e  Then a ( I + (x - X(x))l) = 0 A  implies a (x - X(x)l) E 1^ and since a cannot belong to 1^, by Lemma 1.2.9 Q  Q  we must have x - X(x)l E I for every x E L. Thus I contains (x - X(x)l) - (x - y(x)l) = (y - X)(x)l  for every x L. e  Since I i s proper, y = X. (ii) If V E f/, n W , then there are maximal right ideals I, and I,, of A A y A M containing x - X(x)l and x - y(x)l respectively, for every x E L, and such that V * A / 1 , = A/I... We now notice that the decomposition of A relative  of V ;  to L i n d u c e s decompositions  V = ® I A +l /I  (2)  a  X  X  « 0  I A + I p /I M A  CIEA  aeA  M o r e o v e r , f o r a Q E A Q and x e L , a a ( x - X ( x ) l ) = (x - X ( x ) l ) a a + a ( x ) a Q so t h a t a a ( x - (X+a)(x)l) E 1^.  By Lemma 1.3.1, V  X  +  A  - A Q + I^/Ix and  the weights o f L i n V a r e c o n t a i n e d i n the s e t {X + a: a e A}. is  a w e i g h t , t h e r e i s an a i n A w i t h u = X + a .  an a  1  e A with X = y + a'.  Since u  By symmetry, t h e r e i s  C l e a r l y a' = -a.  We s i n g l e o u t the f o l l o w i n g two r e s u l t s , now o b v i o u s .  COROLLARY 1.4.3:  If V is an irreducible  L-weighted A-module, then the  weights of L in V are of the form X D + a, a e A , where X Q is any fixed weight.  In particular  only finitely  if A has only finitely  many non-zero roots,  V has  many weights.  COROLLARY 1.4.4:  If V is an irreducible  two weights of L in V differ  L-weighted A-module, then any  by a root of A.  In t h e s e q u e l , we w i l l f r e q u e n t l y have cause t o suppose t h a t A has o n l y f i n i t e l y many non-zero r o o t s .  DEFINITION diagonable,  1.4.5:  and A  L is a finitely  We t h e r e f o r e make the  diagonable  = 0 for all but a finite  subspace of A if it is number of a in L*.  24 CHAPTER  RING-THEORETIC  CHAIN  2.1  TWO  CONNECTIONS  BETWEEN  An  AND  A  CONDITIONS  If  R i s a ring  ( a s s o c i a t i v e ) , and M an R-module, then M has the  d e s c e n d i n g c h a i n c o n d i t i o n on submodules i f e v e r y d e c r e a s i n g Mist Mvjsi . . . o f submodules o f M t e r m i n a t e s ; Mfc  = Mk  f o r a l l t > k.  This i s equivalent  minimum c o n d i t i o n on submodules: has  sequence  i . e . , f o r some k > 0,  to s a y i n g  t h a t M has the  e v e r y c o l l e c t i o n o f submodules o f M  a s m a l l e s t member (with r e s p e c t t o i n c l u s i o n ) .  D e f i n i t i o n s o f the  a s c e n d i n g c h a i n c o n d i t i o n and maximum c o n d i t i o n a r e made i n the o b v i o u s way.  I f M = R considered  as a r i g h t R-module, submodules o f M a r e r i g h t  i d e a l s o f "R and i f " M has the d e s c e n d i n g  ( r e s p e c t i v e l y ascending) chain  c o n d i t i o n on i d e a l s , R i s s a i d t o be a r t i n i a n  (respectively noetherian).  We employ the same n o t a t i o n as t h a t o f Chapter 1. d i a g o n a b l e subspace o f an a l g e b r a A o v e r a f i e l d vector space,  .A =  PROPOSITION 2.1.1:  }> A a aeA .  relative  If A is artinian  k -module A Q has the descending Q  on its submodules. Conversely, condition  In particular,  (respectively  (respectively  noetherian),  ascending)  A Q is artinian  noetherian).  chain  then each condition  (respectively  (respectively  on h -submodules and L is a finitely  then A is artinian  A decomposes as a  to the s e t A o f r o o t s o f L i n A.  (respectively  if each A a has the descending  F.  L isa  diagonable  ascending) subspace  noetherian). chain of A,  25  PROOF:  Suppose A i s a r t i n i a n , and M^=2  o f A - submodules o f A a . of A, and  M  t*  =  M  f o r  k*  M *CZ;M  +  A  F o r each i ,  t h i s i s contained i n  a, $ E A ( § 1 . 2 ) .  Thus M^S.  Mt  e v e r  t > k.  Now  generates  +  Since M i * 2 M y  . . . i s a descending  2  I  chain  the r i g h t i d e a l M ^ * = M^A  Ag because A t t A ^ c A ^ ^  f o r any  * S • . . , there i s an i n t e g e r k > 0 f o r which E Mk,  i f t >. k and m  m e Mk*  = M *  and  F C  ][ Ag, so c o n s i d e r i n g the components o f m r e l a t i v e to L , m  E M .  and  On  the o r i g i n a l sequence o f A Q -modules t e r m i n a t e s a t M^.  •the o t h e r h a n d , i f L i s f i n i t e l y  d i a g o n a b l e and  c h a i n c o n d i t i o n on Ao-submodules, l e t I i = 2 *2 — of r i g h t i d e a l s  of A.  each A a ^  e  has a  the  descending  descending  N o t i c e t h a t each I j can be w r i t t e n as ®  £  chain (Ij)a,  CXEA  where ( ! j ) a i s the A c -module o f a-coraponents of elements i n I j . k £ SL i m p l i e s ( I ^ ^ S ? descending a e A. all  chains { ( I j ) a :  f o r a  each a .  Thus we  t e r m i n a t e d by  m > n , and  the n  o b t a i n f i n i t e l y many  j = 1,2,...} o f A D -submodules o f A a ,  Each c h a i n t e r m i n a t e s a f t e r a f i n i t e step, say.  so A i s a r t i n i a n .  The  Certainly  one  f o r each  number o f s t e p s , so they have  T h i s i m p l i e s t h a t "I  '= 'T  f o r any  p r o o f f o r the n o e t h e r i a n case  simply  mimics the above.  EXAMPLES 2.1.2:  In the converse  t h a t L be a f i n i t e l y  o f the p r e v i o u s p r o p o s i t i o n , the  d i a g o n a b l e subspace o f A i s e s s e n t i a l i a s  the  hypothesis following  examples show.  (i)  L e t A = R[y] be the r i n g of p o l y n o m i a l s  of r a t i o n a l f u n c t i o n s i n x over a f i e l d F . and yx'=  (x+l)y.  i n t e g e r k » 0,  I t i s not d i f f i c u l t  k k y r ( x ) = r(x+k)y .  i n y over the a l g e b r a R = F ( x )  Here x and y a r e  indeterminates  to show t h a t f o r any r ( x ) e R,  and  L = Fx i s a d i a g o n a b l e subspace of  A,  00  A , where A = RyJ for a l l , n n n=l i n t e g e r s n > 0. C l e a r l yJ A i s a f i e l d and each A -module A i s one dimeno o n 2 sional. However, A i s not a r t i n i a n (yA3y A z>. . . i s a ' d e c r e a s i n g i n f i n i t e  u  26  c h a i n o f r i g h t i d e a l s ) and L i s not f i n i t e l y  (ii)  Here we  l e t A = R [ y i , y 2 » « - « ] he  diagonable.  the a l g e b r a o f p o l y n o m i a l s i n commuting,  a l g e b r a i c a l l y independent i n d e t e r m i n a t e s y j , y 2 » . . . over the u s u a l r i n g R = F[x] of polynomials y^x =  (x + i ) y ^ .  n y  for  any  i  l  (commutative)  i n an i n d e t e r m i n a t e x over the f i e l d F .  Define  Then i t i s s t r a i g h t f o r w a r d to show t h a t  ">yi  1  f ( x ) e R.  n k  .  f ( x ) = f ( x + ni±i  -  k  +  . . . + " k i k ) y i ni...y± \ „ . " 1 k  M o r e o v e r , i t f o l l o w s t h a t L = Fx i s a d i a g o n a b l e  subspace  CO of  A, A = A Q  finitely  +  £ An, n=l  generated  (y^A £ of  =  n  has  + y^A  1  ... i s a s t r i c t l y  A) and L I s not ' f i n i t e l y  Q  r  f o r n >. 0.  is  = R i s a p r i n c i p a l i d e a l domain c h a i n c o n d i t i o n on A o ~submodules  However, A i s not  noetherian  i n c r e a s i n g sequence o f r i g h t  diagonable.  An  r  the a s c e n d i n g  f o r example, [2; § 1 1 . 1 4 - 1 1 . 1 6 ] ) . y^A  " 'y±  £ ZijS.=n j J j  as an A 0 -module and A  (hence n o e t h e r i a n ) , so (see,  and A  ideals  27  2.2  NILPOTENT  IDEALS  AND  SEMI-PRIMENESS  An i d e a l I ( r i g h t , l e f t , o r two-sided) o f a r i n g R i s n i l p o t e n t if  f o r some i n t e g e r n > 0, I n = 0, where l  n  o f sums o f monomials x ^ X 2 « . . x n , each x± e I . (0) i s the o n l y n i l p o t e n t i d e a l . i s the a n n i h i l a t o r o f V i n R. d e f i n e d t o be f \ { ( 0 : V ) :  denotes the i d e a l o f R c o n s i s t i n g R i s c a l l e d semi-prime i f  I f V i s an R-module, (0:V) = { x E R: Vx =  The Jacobson r a d i c a l o f R, J ( R ) , i s then  V an i r r e d u c i b l e R-module}.  I t i s a two-sided  i d e a l o f R c o n t a i n i n g e v e r y n i l p o t e n t l e f t , r i g h t and two-sided i d e a l o f R i s s a i d t o be semi-simple i f J(R) =  Now  d e f i n e a new  0.  a s s o c i a t i v e b i n a r y o p e r a t i o n on R by d e f i n i n g  xoy = x + y - xy f o r any x and y i n R. w i t h i d e n t i t y the 0 element o f R. inverse  R.  The p a i r (R,o) i s then a monoid  I f xoy = 0, we  say y i s a r i g h t  'of x , and x i s r i g h t q u a s i - r e g u l a r ; x i s a l e f t  o f y and y i s l e f t q u a s i - r e g u l a r .  quasi-  quasi-inverse  I f every element of a r i g h t i d e a l i s  r i g h t q u a s i - r e g u l a r , then every element x o f the i d e a l i s q u a s i - r e g u l a r in  the sense t h a t x i s both r i g h t and l e f t q u a s i - r e g u l a r w i t h unique q u a s i -  inverse.  Assuming t h a t R has an i d e n t i t y 1, t h i s i s e q u i v a l e n t  to s a y i n g  t h a t 1 - x i s i n v e r t i b l e i n R, because on w r i t i n g t h i s i n v e r s e i n the form 1 - y , the i d e n t i t y inverse of x.  (1 - x ) ( l - y) = 1 + x O y shows t h a t y i s the q u a s i -  I t t u r n s out t h a t the Jacobson r a d i c a l can be c h a r a c t e r i z e d  as a q u a s i - r e g u l a r i d e a l  (every element i n i t i s q u a s i - r e g u l a r )  contains every q u a s i - r e g u l a r r i g h t i d e a l . prove u s e f u l i n the s e q u e l .  which  This c h a r a c t e r i z a t i o n w i l l  A good d i s c u s s i o n o f a l l these i d e a s  can  be found i n [ 5 1 .  Now  throughout t h i s s e c t i o n , assume t h a t L i s a f i n i t e l y  a b l e subspace o f the a l g e b r a A, so t h a t A a  diagon-  = 0 except f o r a i n the s e t  0}  28 A ~ {0,.a^  ,... ,ak}.  A a Ag C A ^ ,  The r e a d e r i s reminded  where A a + g i s 0 i f a + 3 £ A.  t h a t f o r any a , 3 i n A,  We f i r s t prove the f o l l o w i n g  key lemma:  LEMMA 2.2.1:  Given elements  Xj^ _,x2 * . . - . » k + i °f x  e A, i ~ 1, ...jk+1, then there  m and n with  integers  1 < m < n •< k+1  E AQ.  so that  ^ . . . X J J  PROOF:  Let b 1 = g  If  exist  x.^ e A ^ and  with  A  p  b2 = ^  + 32» •••»  some bj, £ { a j , . . . j C i ^ } e i t h e r  b  k + l = 3 i + 3 2 + ••• + 3 k + l •  A i n which event  = 0 , o r b^ = 0 .  In e i t h e r case x ^ s A 0 and the a s s e r t i o n o f the lemma i s t r u e .  Other-  w i s e , ( b ^ , . . . h ^ i } i s a s e t o f k + 1 elements i n a s e t o f c a r d i n a l i t y k . So t h e r e e x i s t d i s t i n c t i n t e g e r s r and n w i t h b r = b n . we have 0 = b n - b r = $T+i + . . . + 3 n « f- V> ,1  V  ^  V  —m - * ---TI  w  Assuming r < n ,  L e t t i n g m = r + 1, i t f o l l o w s  o A  "o*  T h i s e n a b l e s us t o prove  2 LEMMA 2.2.2: I is  then  PROOF: be y  i  I is a right  Suppose  of A suoh that  ideal  (I  AQ)  = 0,  nilpotent.  We show t h a t j . ( k + 1 ) ( k + 2 ) = 0 .  F o r t h i s , l e t x ± , i = 1,..., (k+1) (k+2)  (k+1)(k+2) elements o f I , and d e f i n e y^ f o r i = l,...,k+2 by =  X  ( i - D (k+l)+l • •- x i ( k + l ) "  E a c n  y±e  x  »  a  n  ds  o b  y  t  n  e  previous  r e s u l t , we may w r i t e k+2  JY7  i  =  au  la3iu2a32'-'uk+la3k+lUk+2b  ~  where a and b a r e e i t h e r 1 or i n ( p o s s i b l y d i f f e r e n t ) r o o t spaces Ay, y */> 0, each U j E I A  A o  , i = l,...k+2, and each 3 i e I A A g i where f o r i = l , . . . , k + l ,  3-L e (cx^,..., a^} .  Now  a  u  g^ 2••  ,a  B k + i belongs  to Ag^...Ag  u s i n g 2.2.1, t h e r e e x i s t i n t e g e r s m and n w i t h 1 a  U  3m m+l*-- 3n a  b e l o n  S  s  t o  A V  1  Thus V ^ u ^  (k+1)(k+2)  IT j=l  and so a g a i n  m < n <, k+1  .. . a ^  such  (I ft A Q )  e  2  that  = 0  and  k+2  x  =  J  TTi=l  o.  =  y  1  (k+1)(k+2) Since I  c o n s i s t s o f sums o f p r o d u c t s o f (k+1)(k+2) elements  from I , the lemma i s e s t a b l i s h e d .  The main r e s u l t o f t h i s s e c t i o n now  THEOREM 2.2.3:  semi-prime,  If A is semi-prime,  then  contained  in  the nilpotent  \  A  r ivuur  . ... -  :  , _  ciiiiipuoc  right  ideals  Conversely  of A are  if A Q  exactly  is  those  A  A  zv  so is A Q .  .  0#XEA T»T»r\/-\T«,  follows.  J  xS  !  . 1  -.1— „  Semx—p i. xiue &uu  By an easy i n d u c t i o n argument, we  T  A  .  J  XS  a.  . . .•  1 -„  L i g i i L  • 1  t  .  t~  xufciij. O i *v  can assume n = 2.  Now  '  "f ^1  1  — \j.  1  WALU  N  IA i s a r i g h t  i d e a l of A c o n t a i n e d i n I 0  £ A^. A l s o IA A A Q = I has square 0, so O&xeA IA i s n i l p o t e n t by the p r e v i o u s lemma. Hence IA = 0 and 1 = 0 . Conversely, i f AQ i s semi-prime and I i s a n i l p o t e n t r i g h t i d e a l o f A, then I]_ = I + AI i s a two-sided  i d e a l of A which i s n i l p o t e n t because 1^  any p o s i t i v e i n t e g e r k.  1^  S.  1  AQ  I f I-^ = 0,  ( 1 ^ (\ ^ Q )  by homogeneity ( 1 . 3 . 4 ) .  I  t  k  I  F i n a l l y , we  + AI  = 0, so 1^ f\ A Q  k  = 0  I j so I i s c o n t a i n e d i n  O^OEA  too.  k  for and £  AQ  0/cteA  note  t h a t any  i d e a l contained i n  £  Aa  i s nilpotent  OAXEA  by 2.2.2  s i n c e I(\ A  COROLLARY 2.2.4:  ideals  PROOF:  Q  = 0.  If AQ  I of A contained  is semi-simple, in  \ k^.and 0&tEA  L e t T denote the sum  then hence  defined here.  J(A) is the sum  is  of all  right  nilpotent.  C e r t a i n l y T c J ( A ) because  J ( A ) c o n t a i n s a l l n i l p o t e n t i d e a l s o f A.  We prove t h a t the r e v e r s e  inclu-  s i o n h o l d s by e s t a b l i s h i n g  (1)  J(A)0AoCJ(Ao)  To see t h i s , we use the f a c t t h a t t h e Jacobson r a d i c a l o f a r i n g w i t h 1 can a l s o be c h a r a c t e r i z e d as the i n t e r s e c t i o n o f a l l maximal r i g h t ( [ 4 ; page 1 1 ] ) .  But every maximal r i g h t i d e a l o f A Q i s c o n t a i n e d  ideals in a  maximal r i g h t i d e a l o f A by 1.3.5..  Before  c o n t i n u i n g , we remark t h a t once a g a i n the u n d e r l y i n g  c o n s t r a i n t t h a t L be f i n i t e l y d i a g o n a b l e  i s e s s e n t i a l , f o r otherwise,  we o b t a i n a counter-example by t a k i n g A = F[x]<y>, the r i n g o f f o r m a l power s e r i e s i n y over the p o l y n o m i a l determinates,  and yx = ( x + l ) y .  r i n g F [ x ] , where x and y a r e i n -  A Q = F [ x ] i s s e m i - s i m p l e , b u t J(A) i s  n o t n i l p o t e n t ; i n d e e d , i t c o n t a i n s "the q u a s i - r e g u l a r n o n - n i l p o t e n t  ideal  CO  B  =  { 1 f-jOOy ": fj[(x) £ F [ x ] } . That B i s q u a s i - r e g u l a r can be seen i=l co ± by n o t i n g t h a t f o r a g i v e n element £ f 1 ( x ) y o f B, the e q u a t i o n i=l 1  CO  (1 -  CO  I f i ( x ) y 1 ) ( l + I gi(x)y1) i=l i=l  can be s o l v e d i n d u c t i v e l y f o r the g^(x)  =  1  e F [ x ] , i = 1,2,...  We c l o s e t h i s s e c t i o n w i t h  THEOREM 2.2.5:  If every  and hence AQ is semi-simple  PROOF:  The i n c l u s i o n  J ( A 0 ) £ J(A) 0 A Q .  A-module  is L-weighted,  then  J(AQ)  = A Q 0 1(A)  i f A is.  (1) i s always v a l i d , so we need prove o n l y  L e t V be an i r r e d u c i b l e A-module.  Then V = ©  that  I  XeA  31  relative  to the s e t A o f w e i g h t s o f L i n V . J vx  and v =  e V.  Then each v^aQ  Now  suppose v a Q = 0 w i t h a Q  = 0 because V ^ A  S E V^+Q  (1.2.6).  e A  Thus,  XEA  we  obtain  (2)  (Q:V)r\  AQ  =  (\  (0:VX)  Q  XeA where ( 0 : M ) O i n d i c a t e s  the a n n i h i l a t o r  i n A q o f an A 0 -module M .  Let I  t i v e l y J ) denote the c l a s s o f a l l i r r e d u c i b l e r i g h t A - ( r e s p e c t i v e l y modules. ( H VEI the  Then  AQ-)  ( 2 ) implies  (0:V» H  AD  last inclusion  Lemma 1 . 2 . 8 .  (resp  =  f\ ( ( 0 : V ) C\ A Q ) VEJ  following  - H f \ ( 0 : V x ) o 3 f\ (0:V)o V e I V E J XEA o  because each  Thus J ( A ) T \ A 0 2 J ( A 0 )  i s an i r r e d u c i b l e A 0 ~module by  as r e q u i r e d .  32  2.3  PRIMITIVITY  An  ideal P of a ring R i s primitive  i f i t i s the l a r g e s t (two-  s i d e d ) i d e a l o f R c o n t a i n e d i n a maximal r i g h t i d e a l o f R. are  e x a c t l y those which a r e a n n i h i l a t o r s  Primitive  o f i r r e d u c i b l e R-modules.  m i t i v e r i n g i s one i n which (0) i s a p r i m i t i v e  ideals  A pri-  i d e a l ; i . e . , there i s a  maximal r i g h t i d e a l c o n t a i n i n g no non-zero two-sided i d e a l , o r , a l t e r n a t i v e l y , t h e r e e x i s t s an i r r e d u c i b l e module V which i s f a i t h f u l i n the sense i t s annihilator  i s 0.  Suppose now t h a t the a l g e b r a A i s p r i m i t i v e w i t h f a i t h f u l d u c i b l e module V.  Taking a c l o s e r  l o o k a t (2) o f the p r e v i o u s  we see t h a t we have a c o l l e c t i o n o f i r r e d u c i b l e A 0 ~modules {V x : w i t h f\ (0:V X ) = 0 . XeA °  Now P x = ( 0 : V X ) Q i s a p r i m i t i v e  containing x - X(x)l  f o r every x e L as the d i s c u s s i o n  showed. X(x)  irre-  section, XeA}  i d e a l of AQ  a f t e r Lemma 1.2.9  M o r e o v e r , f o r X,u e A and X £ u , t h e r e i s an x e L f o r which  t y(x)  and so (x - u ( x ) l ) - (x - X ( x ) l ) = (X - : u ) ( x ) l i s i n P x + P y .  T h i s i m p l i e s P^ + P^ = A Q . by  that  Assuming L i s f i n i t e l y  d i a g o n a b l e , we know  1.4.3 t h a t A i s f i n i t e , and so we can a p p l y the Chinese Remainder Theorem  to o b t a i n A Q - ©  is a primitive  £ XeA  A 0 /P. .  The q u o t i e n t o f any r i n g by a p r i m i t i v e  r i n g , so here we have r e a l i z e d A Q as a d i r e c t sum o f p r i -  mitive algebras.  F i n a l l y we n o t i c e t h a t P r o p o s i t i o n s  e n a b l e us to extend o u r r e s u l t t o d i r e c t sums, thus  THEOREM 2.3.1:  which  is a direct  irreducible algebras.  ideal  Suppose  L is a finitely  sum of primitive  k-module  is weighted.  diagonable  algebras  1.2.1 and 1.2.11  giving  subspace  and. further  Then kQ is a direct  of an algebra  suppose sum of  that  every  primitive  A  33  As  a p a r t i a l c o n v e r s e , we  THEOREM 2.3.2:  a direct direct  sum sum  A itself  PROOF:  I, is a finitely  of primitive  q  so  diagonable  algebras,  of primitive  is  A  If  then  algebras.  a f i n i t e d i r e c t sum Let  I j be  non-zero two-sided i d e a l .  3  i d e a l o f A Q except ( 0 ) , because such an n © I T f t R. ( f o r 1 e A ) , a n d T f ( R . g L . j=l  also  be  that  of  A, and  kQ  is  embedded in A is  a  semi-prime,  embeddable.  i s necessarily  c o n t a i n i n g no  subspace  A/J(A) can  Assuming  j = l , . . . , n because i t c o n t a i n s 1. R.  obtain:  3  of p r i m i t i v e  algebras  a maximal r i g h t i d e a l  Then  ][ I .  can c o n t a i n  Rj, of no  j=l 3 i d e a l T would decompose as For  each j , l e t  3  °  J  i=l  1  i*j "Then T j i s a maximal r i g h t i d e a l o f "Ag, "and r i g h t i d e a l T^* P j = {a e A:  o f A such t h a t  Aa S= ^j*}  AQ  = T j by  i s c o n t a i n e d i n T^*  m i t i v e i d e a l because i t i s the  so c o n t a i n e d i n a maximal  annihilator  Lemma 1.3.5. and  Then  (since  1 e A)  is a  pri-  of the  i r r e d u c i b l e A-module  n n n n Notice that ( Q ft (\ (T.* ft A ) = (\ T. £ Ii, 3 j=l j=l j j=l J j=l J n and s o , because ( f\ P ) f l A Q i s a two-sided i d e a l , i t must be (0) by j= l 3 n what was s t a t e d above. Thus f\ P. £ A a by homogeneity and hence j=l J 0/aeA c o n t a i n e d In J(A) by 2.2.4. (Note t h a t a p r i m i t i v e r i n g i s s e m i - s i m p l e ) . A/T,*.  The  r e v e r s e i n c l u s i o n always h o l d s (see f o r example [5; page 40]) so we n n have t h a t J(A) = f | P . C l e a r l y A/J(A) can be embedded i n 0 £ A/P.. S i n c e 2.2.4 is  obvious.  j=l j i m p l i e s J(A)  is nilpotent,  the  j=l l a s t statement o f the  3  theorem  COROLLARY 2.3.3:  Under  then  primitive.  A/J(A) is  PROOF:  the conditions  of the theorem,  if AQ  is  primitive,  I f I i s a maximal r i g h t i d e a l o f A Q c o n t a i n i n g no non-zero two-  s i d e d i d e a l o f A D , and I * i s the maximal r i g h t i d e a l o f A g e n e r a t e d by I , then P H A Q = 0, where P i s the l a r g e s t  i d e a l of A contained i n I * .  see  t h i s , we s i m p l y n o t e t h a t P fi A Q £ I * f\ A Q = I .  the  theorem.  Since P i s a p r i m i t i v e  To  Thus P = J ( A ) as i n  i d e a l , the c o n c l u s i o n  follows.  35  CHAPTER  DIAGONABLE  3- 1  IDEMPOTENTS  In  ELEMENTS  ARE  THREE  AND  WEIGHTED  MODULES  DIAGONABLE  t h i s c h a p t e r we g i v e examples o f d i a g o n a b l e elements and  l o o k f o r c o n d i t i o n s on an a l g e b r a which a l l o w something the  nature of i t s diagonable elements.  to be s a i d  about  They o f t e n t u r n out t o be a l g e b r a i c .  The r e s u l t s o f Chapter Two w i l l be o f g r e a t h e l p i n g i v i n g  information  b o t h about a d i a g o n a b l e element x and i t s c e n t r a l i z e r because o f c o u r s e AQ(x) = A Q ( F x ) .  When employing r e s u l t s from Chapter Two, the d i a g o n a b l e  subspace L i s always understood t o be F x . if  i t i s Fx-weighted  A module i s s a i d t o be x-weighted  and we w i l l r e f e r t o "weights o f x" and " r o o t s o f x"  r a t h e r than weights and r o o t s o f F x .  Alsos  s i n c e any weight o r r o o t I s a  l i n e a r f u n c t i o n a l on L (1.2.4) i t i s c o m p l e t e l y determined by i t s v a l u e at  x.  We w i l l i d e n t i f y a r o o t a w i t h the s c a l a r ct(x) so t h a t the " r o o t s  of  x" w i l l a c t u a l l y be the r o o t s o f the minimum p o l y n o m i a l o f ad x i n t h e  event a l l but f i n i t e l y many r o o t spaces a r e z e r o .  will  be i d e n t i f i e d w i t h the s c a l a r s X(x).  Any for  S i m i l a r l y , the weights X  a l g e b r a c o n t a i n i n g idempotents abounds i n d i a g o n a b l e e l e m e n t s ,  we have:  THEOREM 3.1.1:  combination  An element  of orthogonal  i f x is algebraic Such an element  with  x =  n \ a i i=l  idempotents  minimal  is diagonable,  e  i  with  aj.,...jCin  with  e]_,...,e  polynomial  having  and A (x) =  T  E  F is a  linear  sum 1 i f and only  distinct e.Ae..  roots  in F .  36  n ][ ct.e as a b o v e . L e t S be any s u b s e t o f { l , . . . , n } 1 i=l w i t h the p r o p e r t y t h a t {ct^: i e S} i s the s e t o f d i s t i n c t c o e f f i c i e n t s o f n e]_,...,e a p p e a r i n g i n Y ct^e . We prove t h a t x has the m i n i m a l p o l y n o m i a l 1 i=l PROOF:  F i r s t assume x =  "[J ( t - a .)  (1)  e  F[t].  1  ieS  For t h i s i t i s s u f f i c i e n t to assume t h a t S - { l , . . . , n } f o r i f a j » . . . , a a r e n o t a l l d i s t i n c t , d e f i n e f o r each i e S, { f ^ : i E S) a r e o r t h o g o n a l d i s t i n c t set of s c a l a r s .  f^ =  £  idempotents w i t h sum 1, and n Now l e t f ( t ) = "P|~(t - a j .  ej .  Then x =  {a^:  i E S}  J ±^±> a  is a  Since  1-1  n x - ct.il = x - ct.j( £ e.) J i=l  =  n I (a i=l  ±  - ocJe. J  i t i s c l e a r t h a t f ( x ) =0. To see t h a t f i s a c t u a l l y the m i n i m a l n ± note t h a t i f g ( t ) = ^ a.t s F [ t ] , t h e n  polynomial,  1=1  g(x) =  I a±( I i=l j=l  J  J  ) =  n n £ ( j aict.i)e J j=l i=l  and so g ( x ) = 0 i f and o n l y i f c q , . . . , c t n are r o o t s o f g (because e j , . . . , e n are orthogonal  idempotents).  Since  ,ct n are d i s t i n c t , degree g = m i s  a t l e a s t as b i g as n = degree f . C o n v e r s e l y , i f x i s a l g e b r a i c w i t h m i n i m a l n p o l y n o m i a l f ( t ) = TT(t - a.) e F [ t ] where a,,...,a are d i s t i n c t s c a l a r s , 1=1  d e f i n e f o r each i = 1  n, h ^ ( t ) =  "j~]~(t - a , ) .  Then h ^ ( t ) , . . . , h n ( t )  a r e r e l a t i v e l y prime elements o f the E u c l i d e a n domain F [ t ] , so t h e r e e x i s t n ai ( t ) , . . . , a n ( t ) i n F f t ] w i t h £ a i ( t ) h i ( t ) = 1. ..Let e = a i ( x ) h i ( x ) , f o r i=l n i = l , . . . , n . Then e^,...,e are o r t h o g o n a l idempotents w i t h sum 1, x = £ xe i"=l 1 and x e ^ = (x - a ^ l ) e ^ + a^e^ = a^e^. F i n a l l y , t o see t h a t an element n x = J a l e i x S d i a g o n a b l e , we note t h a t i f a e A, i=l n n n a J 1=1 1 j=l 3 i,j=l i  37 and  (e^ae^.x) = (oij - c t ^ e ^ a e ^ .  The  I n the above p r o o f , we  theorem now  n o t i c e t h a t i f x i s an a l g e b r a i c  a b l e element whose minimal p o l y n o m i a l is  a f i n i t e l y dlagonable  Aa_g,  where a and  generalizes  of  PROOF:  s u b s p a c e , because the r o o t spaces are of the  algebraic  F generates  minimal  d i s t i n c t r o o t s i n F , then L =  6 are r o o t s o f the m i n i m a l p o l y n o m i a l .  Any  the form A a _ g  the  has  diagon-  This  Fx  form  result  considerably.  THEOREM 3.1.2:  a field  follows.  a finitely  diagonable  element  diagonable  where a and 3 are roots  polynomial  x in an algebra  subspace  of A.  (not necessarily  over  Root spaces in F) of  are  p(t)3  of x .  I f x i s i n the c e n t r e of A t h e r e i s n o t h i n g  some r o o t space A^(x)  A  i s non-zero f o r y r 0,  to p r o v e , so assume  Considering  f o r a moment  -A. -as an -algebra over -F, - the " a l g e b r a i c cl-os-ure -of F , •the- sub space A^C-x) is  i n v a r i a n t under the l i n e a r t r a n s f o r m a t i o n R x .  a l s o w i t h minimal p o l y n o m i a l and  0 i n A^(x)  ayX  = (x + y l ) y  a  (x -  of  i s algebraic,  Thus, since A^(x)p(Rx) = 0  m  e  i p H s a^f(x) = f(x+y)ay  n  ( 8 - y ) l ) a y = 0. polynomial  But  i r r e d u c i b l e polynomial  f o r any  f ( t ) e F [ t ] and  t h i s says t h a t a = 6 - y i s  o f L, , which i s a l s o p ( t ) .  r o o t s o f p ( t ) and we  see.  and  such t h a t ay(x - 6 1 ) ° = 0, f o r some i n t e g e r n > 0.  Over a f i e l d  now  Rx  p s p l i t s i n t o l i n e a r f a c t o r s o v e r F , t h e r e e x i s t s some 8 e F  a^  mal  p ( t ) s i n c e 1 e A.  Now  have the  a  so  r o o t o f the m i n i -  Hence y = 8 - o i s a d i f f e r e n c e  theorem.  of c h a r a c t e r i s t i c 0, i t i s well-known t h a t has  Now  distinct roots.  T h i s r e s u l t extends as  any we  LEMMA 3.1.3: If p(t) e F[t] is irreducible a - $ t F for any distinct  roots a  and of degree at least 2, then  and $ of p(t).  PROOF: Let K be the splitting field of p(t) over F and G the Galois group of K over F.  Then there exists a e G such that aa - 6 ([A; page 204]).  Suppose a - 8 e F.  Then x(a - aa) = a - ca for every T G G.  with T = a , we obtain a - i a  + aa = 2a.  - i  2 relation gives a + a a = 2aa.  Applying a to both sides of.this  3 Another application of a gives aa + a a =  2 "\ 2a a = 2(2aa - a), and so 2a + a a = 3aa. reveals that for any integer t > 0, ta + a identity of G for some k.  In particular,  An easy induction argument t+1  a - (t+l)a . Now a  k  is the  Hence (k-l)a + a = kaa; i.e., a = aa =  This contradiction implies the lemma.  With the aid of 3.1.2, we obtain immediately  THEOREM 3.1.4: Let x be an algebraic polynomial  is irreducible.  diagonable element whose minimal  Then x is in the centre of A.  We next give:  PROPOSITION 3.1.5: polynomial  Let x be an algebraic diagonable element with minimal n p(t) = ~\\(t - a.) e F[t]. Then every non-zero A-module is 1-1  x-weighted, every non-zero A -module is x-weighted, and weights of x are Q  in the set {aj,...,o }. all  (Note:  we do not assume here that a^,...,a are n  distinct.)  PROOF: Let V be an A-module different from zero, and let O^v e V. Then n either v(x - a j l ) = 0 or, since T"]~(x - a . l ) =0, there i s some integer k, k-1 i=l kil 1 < k < n with v "\J (x - a.l) $ 0 but (v 7T(X ~ ct l))(x - a l ) = 0, and i=l i-1 • 4  A  k  39  so V i s x - w e i g h t e d .  The p r o o f  t h a t A - m o d u l e s a r e x-weighted i s i d e n t i c a l . Q  Next l e t y be any weight o f x i n an A - ( o r A - ) module V. some v 5* 0 i n V w i t h v ( x - yl) = 0 . nomials t - y  a n  Thus t h e r e i s  Now i f y £ {aj, , . . • , a n ) ,  then the p o l y -  d p ( t ) i n F [ t ] a r e r e l a t i v e l y p r i m e , and so t h e r e a r e p o l y -  n o m i a l s a ( t ) and b ( t ) i n F [ t ] w i t h a ( t ) p ( t ) + b ( t ) ( t - y) = 1 . g i v e s b ( x ) ( x - yl) - 1 and hence v = v ( x - Y ! ) M X )  The  Setting t = x  = 0, which i s u n t r u e .  converse o f Theorem 3.1.2 i s g e n e r a l l y f a l s e ;  f o r example,  A i s t h e a l g e b r a g e n e r a t e d over F by two elements x and y w i t h x t r a n s -  if  2 cendental, y  = 0 , and yx = ( x + l ) y , then Fx i s a f i n i t e l y  A (the o n l y r o o t s o f x a r e 0 and 1) b u t o f course  of The  diagonable  subspace  x i s not a l g e b r a i c .  b e s t r e s u l t we c o u l d o b t a i n i n t h i s d i r e c t i o n now f o l l o w s :  THEOREM 3.1.6:  Suppose  L = Fx is a finitely  diagonable  subspace  •algebra- -A -which is ,a^direct-.sw^.--of.>primi-tix)e^.algebras,. that  every  Then x is  PROOF:  irreducible  A-module  and every  irreducible  Suppose AQ-module  further is  x-weighted.  algebraic.  By Theorem 2.3.1, A  Q  itself  i s a d i r e c t sum o f p r i m i t i v e  R^, i = l , . . . , n , o n l y f i n i t e l y many because 1 e A right ideal of A with a D ( x - X l ) be  of an  n  Q  , then A  Q  q  .  algebras  I f I i s any maximal  / I i s w e i g h t e d , so t h e r e i s some a D t I and X e F  e I , and t h i s i m p l i e s x - X l E I by 1.2.9.  Hence, l e t t i n g  a maximal r i g h t i d e a l o f R^ c o n t a i n i n g no non-zero i d e a l o f R^,  \ R. i s a maximal r i g h t i d e a l o f A and so jfi n n c o n t a i n s x - X. 1 f o r some X.» e F . W r i t i n g x = \ x., x . e R J , and 1 = £ e.,  i  = 1  n , f o r each i ,  ®  D  i=l e^  the i d e n t i t y o f R i , we have x i - X^e^ e  ponent-of x - X ^ l . in  the c e n t r e o f A  i=l  by l o o k i n g a t the i  t  n  com-  Now x^ - X^e^ i s i n t h e c e n t r e o f R^ because x - X l i s Q  and so ( x ^ - X ^ e ^ R ^ i s an i d e a l o f R^ c o n t a i n e d  T h e r e f o r e , x^ = X^e^ and s i n c e e^ a l g e b r a i c by Theorem 3.1.1.  e n are orthogonal  in J^.  idempotents, x i s  40  3.2  PRIME  ALGEBRAS  i f aRb ? 0 whenever a and b a r e non-zero  A r i n g R i s prime  elements o f R; o r , e q u i v a l e n t l y , i f the p r o d u c t o f non-zero r i g h t of R i s non-zero.  ideals  I n p a r t i c u l a r a prime r i n g i s semi-prime and i t i s a l s o  a f a c t t h a t any p r i m i t i v e  r i n g i s prime  ([5; page  951)-  In g e n e r a l one can not expect the i r r e d u c i b l e f a c t o r s  o f the  m i n i m a l p o l y n o m i a l o f a d i a g o n a b l e a l g e b r a i c element to be d i s t i n c t ; f o r example, l e t R by the q u o t i e n t o f the u s u a l p o l y n o m i a l r i n g F [ x ] by the p r i n c i p a l i d e a l g e n e r a t e d by x  3  - x  2  and l e t A = R [ y ] be the p o l y n o m i a l  r i n g o v e r R i n the i n d e t e r m i n a t e y , where yx = ( x + l ) y .  A  straight-forward  computation r e v e a l s t h a t A i s j u s t the f o u r - d i m e n s i o n a l a l g e b r a o v e r F w i t h 2 b a s i s 1, x , x , y and m u l t i p l i c a t i o n  l  X  X  y  1  l  X  x2  y  X  X  X  x2  0  x2  X  x2  0  y  y  y  0  2  X  2 2  y  table:  x i s a d i a g o n a b l e a l g e b r a i c element whose m i n i m a l p o l y n o m i a l p ( t ) = t  J  - t  has t as a r e p e a t e d f a c t o r .  I f A i s semi-prime, t h i s cannot happen, f o r we  PROPOSITION 3.2.1:  algebra  A over the field  irreducible PROOF:  If x is a diagonable  factors.  algebraic  element in a semi-prime  F, then the minimal polynomial '  have:  of x has  distinct .  •  L = Fx i s a f i n i t e l y d i a g o n a b l e subspace o f A and so by Theorem 2.2.3 r A (x) i s semi-prime. Suppose the m i n i m a l p o l y n o m i a l o f x i s p ( t ) = T T p . ( t )  41  where p j ( t ) , . . . , p Then i f any  ( t ) are the d i s t i n c t i r r e d u c i b l e f a c t o r s o f p ( t ) i n F [ t ] . r n^ > 1, u = JJp.(x) i s a non-zero n i l p o t e n t element i n the 1=1  centre of AQ(x)  and  hence generates a non-zero n i l p o t e n t i d e a l , o and t h i s i s  impossible.  There are o c c a s i o n s when a l l the i r r e d u c i b l e modules o f an  algebra  are w e i g h t e d ; f o r example, when x i s a d i a g o n a b l e a l g e b r a i c element whose minimal p o l y n o m i a l has In [15] case.  a l l i t s r o o t s i n the ground f i e l d  ( P r o p o s i t i o n 3.1.5).  Lemire g i v e s an example to show t h a t t h i s need not always be He  considers  the u n i v e r s a l e n v e l o p i n g  d i m e n s i o n a l simple L i e a l g e b r a A^ element i s the g e n e r a t o r there are i n f i n i t e l y  ([9  H of a C a r t a n  a l g e b r a £/(Aj) o f the  ; page 137]) subalgebra  the  three-  where the d i a g o n a b l e  o f A^.  many r o o t s p a c e s , f o r o t h e r w i s e ,  R e l a t i v e to  H,  £/(Aj) would c o n t a i n  n i l p o t e n t elements (because- (A a (H)) n cr^A^C-H)-) ,- and ' t h i s i s known -to be false  ( [ 9 ; page 1 6 6 ] ) .  which i s not H-weighted.  Lemire c o n s t r u c t s an i r r e d u c i b l e £ / ( A j ) - m o d u l e We  f u r t h e r pursue i n t h i s s e c t i o n the  inter-  r e l a t i o n s between an a l g e b r a A, a d i a g o n a b l e element x i n A, and p e r t y t h a t the i r r e d u c i b l e modules o f A should be  EXAMPLE 3.2.2:  We  a l g e b r a over a f i e l d w i t h the r e l a t i o n  [y,x] = y .  nomials over F i n the two  Let A  be  a l g e b r a o f the two-dimensional n o n - a b e l i a n  F (of c h a r a c t e r i s t i c 0 ) .  pro-  x-weighted.  g i v e here an example s i m i l a r to L e m i r e ' s .  the u n i v e r s a l e n v e l o p i n g  the  T h i s has  generators  Lie x and  A i s i n f a c t the r i n g o f non-commuting p o l y -  indeterminates  x i s a d i a g o n a b l e element i n A, and  x and y , where yx = ( x + l ) y .  t h e r e are i n f i n i t e l y  Now  many r o o t s , be-  00  cause A = ©  £ A n ( x ) , An(x) n=0  y  = F[x]yn.  S i n c e x i s not  invertible,  there  e x i s t s a maximal r i g h t i d e a l I of A which c o n t a i n s x, and V = A/I  i s x-weighted ( f o r (I + l ) x = 0 ) .  We  so the A-module  prove however t h a t not a l l  i r r e d u c i b l e A-modules are w e i g h t e d .  y + 1 i s not i n v e r t i b l e i n A and r i g h t i d e a l J which has  f(x)yk  (1)  To see  t h i s , we  use  the  i n a maximal  property  e J f o r k > 0 and  f ( x ) e F[x] i m p l i e s f =  i n d u c t i o n on n = degree f and -k.  and y + 1 are both i n J , then J c o n t a i n s are r e l a t i v e l y  so i s c o n t a i n e d  0.  For n = 0, i f ay  1 because these  two  polynomials  prime i n the E u c l i d e a n domain F [ y ] , hence c e r t a i n l y i n  Assuming the v a l i d i t y o f  (1) f o r p o l y n o m i a l s  f w i t h degree l e s s than  A. n,  k suppose f ( x ) e F[x] has  degree n and  f(x)y  and y + 1 are both i n J .  Then J must c o n t a i n  (cf~ \.-k  . . n \ _ c /•-.%.. k+1  S i n c e f ( x ) - f ( x + l ) has have f ( x ) = f ( x + i ) . n > 1 and  any  possibility  degree l e s s  fcf..\  c1,1 \ \ ..k+1  than the degree o f f , by i n d u c t i o n  we  T h i s i m p l i e s 5 + n i s a root of f f o r every i n t e g e r  r o o t £ o f f and  has  ..r/.-N.-k _  t h i s i s i m p o s s i b l e u n l e s s degree f = 0.  a l r e a d y been e l i m i n a t e d above and  so we  have o b t a i n e d  This (1).  I t f o l l o w s immediately t h a t J can c o n t a i n no  two-sided i d e a l o f A e x c e p t  (0) because any  1.3.4.  such i d e a l i s homogeneous by  s i m p l e , f a i t h f u l A-module ( i t s a n n i h i l a t o r so A i s a p r i m i t i v e  algebra.  A/J i s not weighted however, f o r we  can  and prove  n 1  ( I a.(x)y )(x - a l ) e J implies i=0  For n = 0, a D ( x ) ( x - a l ) e J i m p l i e s a Q ( x ) latter possibility  is a  i s a two-sided i d e a l o f J )  n (2)  T h e r e f o r e A/J  i s impossible.  Now  if J  £ ai(x)y i=0  i  = 0 or x = a l by contains  e J.  (1), and  the  43 n  i 1 ( I ai(x)y )(x - al) = £ y (x-il)(x - a l ) 1=0 i=0 then i t a l s o c o n t a i n s  n n  (y+D an(x)(x - « D -  I ? y a (x) (x - a D i=0 i  because y + 1 e J , and hence J c o n t a i n s t h e d i f f e r e n c e o f these  two element's;  namely,  "fy^-Ca^Cx-il) - ^Jan<x)Mx - a l ) = By i n d u c t i o n , we may assume t h a t a^(x) a^x)  = ^  Y(a.(x) -.^].a .(x+il))y (x i  n  = ^"j a n ( x + i l ) ; i . e . , t h a t  aQ(x+il), fori = 0,...,n-l.  Thus  I a i ( x ) y i - I (J) a 0 ( x + l l ) y i - J ( ^ ^ ( x ) i=0 i-OW ° i=0V ' is in J .  al).  =  (l+y)na0(x)  Hence we have ( 2 ) , and A/J i s n o t w e i g h t e d .  Thus we see t h a t even f o r a p r i m i t i v e a l g e b r a , one i s a b l e t o say l i t t l e  about which modules a r e w e i g h t e d .  THEOREM 3.2.3: diagonable  PROOF:  (3)  Let A be a prime algebra over the field  algebraic  some irreducible  We can however p r o v e :  element with minimum polynomial  k-module is x-weighted.  F and x e A be_ a-  p ( t ) e F [ t ] . Suppose  Then all k-modules are x-weighted.  The key s t e p i n the p r o o f i s to show t h a t p ( t ) has the form  p(t) =  TTq(t+a) azS  where q ( t ) e F [ t ] i s i r r e d u c i b l e and  A, the s e t o f r o o t s o f x .  By P r o p o s i t i o n 3.2.1, p ( t ) i s o f the form p ^ ( t ) . . . p s ( t ) where p i ( t ) , . . . , p g ( t ) are the d i s t i n c t monic i r r e d u c i b l e f a c t o r s o f p ( t ) i n F [ t ] . L e t A ^ j denote the subspace {a e A: p ^ ( x ) a = 0 = a p j ( x ) }  for 1 ^ i , j < s.  Then we have:  s (4)  A =  ®  I  A±1  i,j =l To prove t h i s , we note t h a t the l i n e a r t r a n s f o r m a t i o n  of A i s algebraic s Thus A = A. where i=l 1  w i t h minimal p o l v n o m i a l a l s o p ( t ) because 1 e A.  A^ = {a E A: p ^ ( x ) a = 0 } .  H e r s t e i n proves t h i s i n [4; page 256] f o r A  f i n i t e - d i m e n s i o n a l , b u t the p r o o f uses o n l y the f a c t t h a t the l i n e a r formation  i n question  i s algebraic.  Now each subspace Aj_ i s i n v a r i a n t  under R x which i s a l s o an a l g e b r a i c l i n e a r t r a n s f o r m a t i o n mal  polynomial p ( t ) .  trans-  Thus the r e s t r i c t i o n o f R x  of A with  mini-  t o A^ has m i n i m a l p o l y -  nomial d i v i d i n g p ( t ) and so each A^ decomposes i n t o a d i r e c t sum o f the subspaces A ^  f o r some i n t e g e r s j , 1 ^ j ^ s .  Now f o r each k, 1 ^ k < s , d e f i n e a k = "]"J~ p - ( x ) .  Then a, ^ 0  i/k and is .and  f o r any i , j E { l , . . . , s } , a^Aa^ 5* 0 (because A i s p r i m e ) . contained  i n A ^ j and so each o f the spaces A ^ j ^ 0 .  a„  Suppose 0 ^ a E A^  a " I » a a -relative...to x.. ,. ..Then - 0 = ap^(x.) = £ -a-p^Cx). OCEA  But a^Aa^  -For some a ,  CCEA  0 and because A a ( x ) A Q ( x ) S A a ( x ) and the sum  £ Aa(x) i s direct, CXEA  a a p ^ ( x ) => 0 . As we have seen b e f o r e , t h i s i m p l i e s p ^ ( x + a l ) a a a l s o have p ^ ( x ) a  = 0 ,  We  = 0 and h e n c e , j u s t as above, p j ( x ) a a «= 0 . The p o l y -  nomials p^(t+ct) and P j ( t ) cannot be r e l a t i v e l y p r i m e , and so because they c  are i r r e d u c i b l e and monic, P j ( )  =  Pi(t+a).  T h i s e s t a b l i s h e s ( 3 ) , where  q(t) = P i ( t ) . Some i r r e d u c i b l e A-module i s x-weighted by h y p o t h e s i s , and so by Theorem 1.2.10 (with L = Fx) there i s a s c a l a r X E F w i t h x - XI n o t invertible.  T h i s means t h a t the p o l y n o m i a l s p ( t ) and t - X a r e n o t r e l a -  t i v e l y prime and so t - X d i v i d e s p ( t ) .  I t f o l l o w s t h a t f o r some a E St  q(t+a) = t - X and hence q ( t ) = t - (X+ct) and p ( t ) has a l l i t s r o o t s i n F . The  theorem now f o l l o w s d i r e c t l y  from P r o p o s i t i o n 3.1.5.  L o o k i n g back a t t h e form o f the m i n i m a l p o l y n o m i a l p ( t ) i n (3) we see t h a t i n p a r t i c u l a r ,  i f x i s i n the c e n t r e o f A, S = A = {0}.  Together w i t h 3.1.4, we o b t a i n  COROLLARY 3.2.4:  algebra polynomial  A.  immediately  Let x be an algebraic  Then x is in the centre is  irreducible.  diagonable  element  of A i f and only  in a prime  i fits  minimal  46  3.3  CENTRAL  SIMPLE  ALGEBRAS  As might be expected, an element i n an algebra A over a f i e l d F which i s not diagonable, can become diagonable when A i s considered to be an algebra over the algebraic closure F of F.  For example, i n the algebra  of 2 x 2 matrices over the r e a l numbers, x = ^ cause A Q (x) = { a ^  ^  + b|j  i s not diagonable be-  j|: a,b r e a l } , while A Q (x) = 0 for a / 0 .  Over the complex numbers, however, the minimum polynomial of x has the d i s t i n c t roots ± i and so x i s diagonable by Theorem  More generally,  3.1.1.  i f x i s a diagonable element i n the complete r i n g A of matrices over any a l g e b r a i c a l l y closed f i e l d F, then since A i s semi-prime, the minimal p o l y nomial of x, whose roots of course l i e i n F, has d i s t i n c t roots by 3 . 2 . 1 and x i s s i m i l a r to a diagonal matrix as i s well-known.  This curious  observation that diagonability i n our sense i s equivalent to the usual meaning of the word i n a matrix r i n g does not depend on the algebraic closure of the underlying base f i e l d .  The proof of t h i s i s the p r i n c i p a l  aim of t h i s s e c t i o n .  LEMMA  3.3.1:  Then i f A is simple,  •potent.  PROOF:  Let A be any associative  with  so is the subring  1 and e e A an idem* A^(e) = {aeA: ae = ea = a}.  Relative to e, we have the Pierce decomposition of A,  A = A,Q0 + Aio + Aoi + A n ,  = A^(e).  Then  hence equals A.  I  = {aeA: ae = j a , ea = ia} .  Suppose A i s simple and I i s a non-zero i d e a l of A n  ( [8 ; page 4 8 ] ) .  AQJIAJQ  ring  +  AQJI  +  IAJQ  +  AQIIAIQ  i s a non-zero i d e a l of A and  Writing 1 as a sum of elements from I , A  0 1  l,I A  1  0  , and  and multiplying both sides of the equation so obtained by e f i r s t  on the l e f t and then on the r i g h t , we deduce that e e l .  But then, i f  47 a e  (e) , a = ae e I and so I = A j (e) .  An idempotent i n any r i n g  ( a s s o c i a t i v e or not) i s p r i m i t i v e i f  i t cannot be w r i t t e n as t h e sum o f two o r t h o g o n a l i d e m p o t e n t s .  An a l g e b r a  o v e r F i s s a i d t o be c e n t r a l o v e r F i f i t s c e n t r e i s F . We r e q u i r e t h e f o l l o w i n g r e s u l t , a p r o o f f o r w h i c h c a n be found i n s e c t i o n s 3.6 and 3.7 o f [12] .  LEMMA 3.3.2: be written  •  Let R be an artinian  as a sum of primitive  represented n 1 = ^ ej_ =  ring  with 1.  orthogonal  Then any idempotent can  idempotents.  If 1 can be  in two ways as a sum of primitive orthogonal idempotents, m \ f y then m = n and there is a unit v of R and a permutation  u of { 1 , . ..,n} such that v'^e^v = f ^ ^ .  -We use - t h i s *to " e s t a b l i s h  LEMMA 3.3.3:  Let A be a f i n i t e dimensional  central  a f i e l d F and suppose e e A is an idempotent.  simple algebra  Then A ^ ( e ) is also  over central  simple over F. PROOF:  By Lemma 3.3.1, we need o n l y show t h a t A^(e) i s c e n t r a l o v e r F .  U s i n g t h e p r e v i o u s lemma, i t i s easy t o show t h a t we c a n f i n d p a i r w i s e n o r t h o g o n a l p r i m i t i v e i d e m p o t e n t s e^ e n i n A such t h a t ^ = 1 and t i=l e = £ e, f o r some t < n . Now A = D , t h e r i n g o f m x m m a t r i c e s o v e r a i=l d i v i s i o n a l g e b r a D ( n e c e s s a r i l y c e n t r a l over F) by t h e Wedderburn Theorem ([6  ; page 9 8 ] ) .  Let {f^^: i , j "ij  = 1,...,m} be t h e u s u a l m a t r i x u n i t s i n A.  m Then 1 =  jj i i ' 'mm £ f j j and f ^ , . . . ^ ^ a r e o r t h o g o n a l p r i m i t i v e i d e m p o t e n t s .  By 3.3.2, m = n and t h e r e i s a u n i t v e A and a p e r m u t a t i o n  TT o f { l , . . . , n }  48  such  t h a t v *e.v = f i 7r(i)Tr(i)  Then one  can check e a s i l y  because D i s .  Suppose  THEOREM 3.3.4:  the  field  combination ring  PROOF:  t h a t B = A ^ ( f ) - Dfc which i s c e n t r a l over  F  is  A is a finite  dimensional  Then x e A is diagonable  F.  of orthogonal  of matrices  diagonal  t Y f ... ... . £j i r ( i ) T r ( i )  f =  S i n c e B = A ^ ( e ) , the lemma f o l l o w s .  Our main r e s u l t  over  L e t B = v *Ai (e)v and v l  over  idempotents.  F, then  any  central  simple  only  if x is a  if and  In particular,  diagonable  element  if A is is similar  algebra linear  the  complete  to a  matrix.  S i n c e A i s s e m i - p r i m e , the m i n i m a l p o l y n o m i a l of x has  the form  q^(t)...qn(t), Ftt]  where q j , . . . , q n are d i s t i n c t i r r e d u c i b l e p o l y n o m i a l s i n n n (by 3 . 2 . 1 ) ; We can w r i t e 1 = [ e^, and x = £ xe , where e-.,...,e ±=1 i=i  a r e p a i r w i s e o r t h o g o n a l idempotents which are p o l y n o m i a l s Herstein, q^(t), for  [4; page 2 5 6 ] .  i = l , . . . , n because f ( x ^ ) = f ( x ) e ^  i = l,...,n.  N o t i c e t h a t x^ c B^ and  v a r i a n t under ad x. this  M o r e o v e r , x^ = xe^ has  The  f o r any  i n x j u s t as i n  the minimum  polynomial  f ( t ) e F [ t ] . L e t B^ =  B^ i s a s u b a l g e b r a , so B^  r e s t r i c t i o n o f ad x to B^ i s ad x^ and by  A^e^  is in1.2.2(i)  r e s t r i c t i o n s a t i s f i e s a p o l y n o m i a l w i t h d i s t i n c t r o o t s i n F; i . e . , x^  i s diagonable  i n B^, which i s c e n t r a l s i m p l e over F by  3.3.3.  By  3.1.4,  x^ i s i n the c e n t r e o f B^; h e n c e , f o r some a . e F , x^ = a^e^. Therefore n x = Y a.e.. The l a s t statement o f the theorem now f o l l o w s from Theorem i=l 1 1 and l i n e a r a l g e b r a .  COROLLARY 3.3.5:  algebras  over  If A is a direct  a field  F, then  any  sum  of finite  diagonable  dimensional  element  central  in A is a  linear  3.1.1  simple  combination  of pairwise  centralizer  is  PROOF:  orthogonal  idempotents  with  sum 1, and i t s  semi-simple.  That a d i a g o n a b l e element x i s a l i n e a r c o m b i n a t i o n o f o r t h o g o n a l  idempotents  f o l l o w s d i r e c t l y from the Theorem t o g e t h e r w i t h P r o p o s i t i o n  x i s t h e r e f o r e a l g e b r a i c and so Fx i s a f i n i t e l y d i a g o n a b l e subspace o f P r o p o s i t i o n 3.1.5  i m p l i e s every A-module i s x-weighted  1.2 A.  and so A Q ( x ) i s  semi-simple by Theorem. 2.2.,5-.,  In  c o n c l u s i o n , we  a l i t y i n Theorem 3.3.4  remark t h a t the assumption o f  is crucial.  I f F(x) denotes  finite-dimension-  the r i n g o f r a t i o n a l  f u n c t i o n s over F i n an i n d e t e r m i n a t e x, and we a d j o i n an i n d e t e r m i n a t e y so t h a t yx = ( x + l ) y , then the r e s u l t i n g a l g e b r a o f non-commuting power s e r i i n y over F(x) i s c e n t r a l s i m p l e ( i n f a c t i t i s a d i v i s i o n a l g e b r a as Jacobson proves i n [8  ; pages 187-188]), x i s d i a g o n a b l e , but  not a l i n e a r combination o f  idempotents.  certainly  50  CHAPTER  APPLICATIONS  TO  THE A  4.1  THE  UNIVERSAL  FOUR  UNIVERSAL JORDAN  ENVELOPING  ENVELOPING  ALGEBRA  OF  ALGEBRA  ALGEBRA  OF  A  JORDAN  ALGEBRA  L e t . J be a J o r d a n a l g e b r a over a f i e l d F o f c h a r a c t e r i s t i c Then a r e p r e s e n t a t i o n o f J i s d e f i n e d t o be a l i n e a r map S: J  0.  A where  A i s an a s s o c i a t i v e a l g e b r a over F , such t h a t f o r e v e r y a , b,i and c i n J , we have  <l>  S  (2)  aSbc  +  S  bSca  +  S  cSab  =  sasbsc + scsbsa + s a c > b  Sj, S t  a  + ScaSb + S  ^  and  = sasbc + sbsca + scsab  where S: u •* S u f o r u E J . As i n the L i e c a s e , the map x -* R x o f J i n t o the a s s o c i a t i v e a l g e b r a g e n e r a t e d by r i g h t m u l t i p l i c a t i o n s i s an example o f a representation of J . u n i v e r s a l enveloping S*: J  An a s s o c i a t i v e a l g e b r a U(J) w i t h  identity i s a  algebra f o r J i f there i s a c a n o n i c a l r e p r e s e n t a t i o n  i/(J) such t h a t f o r any r e p r e s e n t a t i o n S: J  A o f J i n an a s s o c i a t i v e  a l g e b r a A, t h e r e e x i s t s a unique homomorphism Y: U(J) •* A such t h a t S = f o S * ; i . e . , which makes the f o l l o w i n g diagram commutative:  VU)  J  .—^A S  F o r any p o s i t i v e i n t e g e r n , l e t j " denote the l i n e a r ' s p a n o f a l l products  j(n)  =  o f h elements o f J . D e f i n e J ^ n ^ i n d u c t i v e l y by J  ^j(n-l)j2  some n .  f o rn >  ^  T h e n  ji g  c a l l e d  s o l v a b l e  l f  ^ = J , and  j(n> = o  for  I f J i s f i n i t e - d i m e n s i o n a l , i t c o n t a i n s a maximal s o l v a b l e i d e a l  called  the r a d i c a l  o f J . J i s then semi-simple  i f i t s r a d i c a l i s 0.  By  a theorem o f D i e u d o n n £ , any f i n i t e - d i m e n s i o n a l semi-simple J o r d a n a l g e b r a is  a d i r e c t sum o f simple Jordan a l g e b r a s .  We w i l l r e q u i r e s e v e r a l f a c t s about the u n i v e r s a l e n v e l o p i n g a l g e b r a K ( J ) o f a Jordan a l g e b r a J .  These can be found i n Jacobson  [7 ] ,  and a r e summarized i n  U(J) exists,  THEOREM 4.1.1:  up  to isomorphism.  addition  by { S *  is generated  If J is finite-dimensional  J is semi-simple,  then  so also  : aeJ}., and is  A  then  so is U(J);  e  with  F.  Then  is a linear roots  Let e be an idempotent  letting  a *+ a  combination  denote  in a Jordan  the canonical  of orthogonal  in the set {0,h,~h, l . » - l } .  If J = J  Q  a  of J in U(J),  and is  diagonable with  + Ji^ +  decomposition  + J  ^ AQ(e)  and  I ^ c A ^ e ) + A^Ce),  S e t t i n g a = b = c = e i n (2) g i v e s immediately  — Thus e i s a l g e b r a i c w i t h minimal p o l y n o m i a l d i v i d i n g a p o l y n o m i a l x^ith t h e d i s t i n c t r o o t s 0 , J j , and 1. b i n a t i o n o f o r t h o g o n a l idempotents are  J over  of J  to e , then  Jo  PROOF:  algebra  embedding  idempotents  J ^ = {a e J : ae = i a } for i = 0,%,l is the Peirce  relative  rather  result.  PROPOSITION 4.1.2:  field  if.in  is U(J).  We w i l l a l s o make e x t e n s i v e use o f t h e f o l l o w i n g technical  unique  i n {0,  1, -1} by 3.1.2.  A = i/(J)  t h a t 2 e 3 - 3e"2 + e" = 0 .  3 2 f ( t ) = 2t - 3t + t ,  Thus e i s a l i n e a r com-  and d i a g o n a b l e by Theorem 3.1.1. Next, i f u z J  Q  , put a = u ,  I t s roots  b = c = e  52 in  (1).  Then ue = eu so u e A Q ( e ) .  S i m i l a r l y i f u e J^  b = c = e i n (1) g i v e s ue + 2eu = eu + 2ue  + e2u  kui^ + ku_^ + u^ +  Now  any  = \u  such t h a t R„  ,  b a r e two  Then we  = u^ = u ^ j =  see  0 and  Now  write  that u e A i ^ e ) + A_i g (e) .  elements i n a J o r d a n a l g e b r a J .  t r a n s f o r m a t i o n Ra  ...R0  A Cartan  v,  =  ^ o f J by  0 f o r a l l a<, b.  subalgebra  c o n t a i n i n g 1 which has In  e Aa(e).  eu.  uR a  ^ =  (u,a,b) f o r  J i s a s s o c i a t o r n i l p o t e n t i f there i s a p o s i t i v e i n t e g e r k  u e J.  x e H.  with u a  and hence u D  suppose a and  d e f i n e the l i n e a r  u,  Finally,  + -su = ue +  + ku = 2eue + %u; i . e . , ( ( u , e ) , e ) = ku.  u - u Q + ui^ + u_jg + u^ + u_i  We  again u e A Q ( e ) .  and w i t h a = u , b = c = e i m p l i e s ue^ + e^u  Therefore, ue2  putting a =  (2) w i t h a = c = e and b = u i m p l i e s 2eue + hu -•  assuming u e J i , e q u a t i o n eu + ue  and  t  E J, i =  l,...,k.  o f J i s an a s s o c i a t o r n i l p o t e n t s u b a l g e b r a  the p r o p e r t y  t h a t i f ( x , H j H ) £ ; H f o r any  x e J  H  then  [10; page 6 0 1 ] , Jacobson shows t h a t i f J i s f i n i t e - d i m e n s i o n a l  over an a l g e b r a i c a l l y c l o s e d f i e l d o f c h a r a c t e r i s t i c 0, then J p o s s e s s e s t a Cartan subalgebra H = ][ J J J , where J = £ J . . i s the P e i r c e decomposit i o n of J r e l a t i v e  • • • i-1 i,j " to a s e t o f p a i r w i s e o r t h o g o n a l  which are a l s o p r i m i t i v e . t h a t J . . = F e . and  so H =  idempotents w i t h  sum  1  I f J i s s i m p l e , A l b e r t ( [ 1 ; page 561]) has shown t _ £ F e . . In t h i s c a s e , l e t I, = H, the image o f H  i=l i n [/(J) under the c a n o n i c a l embedding. by d i a g o n a b l e (1), i  By P r o p o s i t i o n 4.1.2, L i s spanned  elements which commute, because s e t t i n g a = e^, b = c = e^  j , gives  (e^,e^) = 0.  As was  1.2.5, L i s t h e r e f o r e a d l a g o n a b l e  p o i n t e d out  subspace o f  i n the d i s c u s s i o n a f t e r  U(J).  In h i s d o c t o r a l d i s s e r t a t i o n , F o s t e r showed t h a t the C a r t a n of L i e and  Jordan algebras  is essentially  s u r p r i s i n g to d i s c o v e r t h a t the C a r t a n algebras  in  the same ( [ 3 ] ) .  subalgebras  theory  It i s therefore  o f s i m p l e L i e and  (over a l g e b r a i c a l l y c l o s e d f i e l d s o f c h a r a c t e r i s t i c 0) share  Jordan the  not  53  common p r o p e r t y t h a t they are both d i a g o n a b l e subspaces o f the c o r r e s p o n d i n g universal enveloping algebras.  In p a r t i c u l a r , u s i n g Theorem 1.3.7, we  obtain  the analogue o f Lemire's r e s u l t ( [ 1 4 ] ) f o r J o r d a n a l g e b r a s .  For a fixed  THEOREM 4.1.3:  correspondence U(j)-modules C-modules,  between and  linear  X E H*, there  functional  the set of isomorphism  the set of isomorphism  where. C is the centralizer  classes  classes of H  one-to-one  of X-weighted  of X-weighted  g e n e r a l l y much " s m a l l e r 1 1 than A; moreover, we  irreducible  the a l g e b r a C i s  s h a l l see i n §4.2 t h a t C can  be c h a r a c t e r i z e d as the c e n t r a l i z e r o f a s i n g l e element o f  H.  c l o s e t h i s s e c t i o n w i t h the f o l l o w i n g example:  EXAMPLE 4.1.4:  L e t J be a s i m p l e J o r d a n a l g e b r a over an  c l o s e d f i e l d F o f c h a r a c t e r i s t i c 0 which  i s i f degree  where e^ and e.^ are o r t h o g o n a l p r i m i t i v e i d e m p o t e n t s . ( [ 1 ] ) , J ~ Fe^ + F e 2 + J j  2  where J ^  2  i s the subspace  algebraically  two; i . e . , 1 = e^ +  2  has a b a s i s u^, u 2  such t h a t u ^ u 2  = u2u^  { a c J : ae^ = a e 2 = ha),  = 0 and u^  2 = u2  =  1.  L e t C be the c e n t r a l i z e r i n [/(J) o f the embedding o f the C a r t a n s u b a l g e b r a Then C i s spanned over F by the f o l l o w i n g seven  H = Fe^ + F e 2 .  f j ° ej + e 2  - (u^ +  u22)  f2  =  - ^(Sj + e 2 ) + 2 e i e " 2  f3  = u2  - hi^i  f  4  =  2  *1  ~ 4 *1*2  f5  = 2e2  - 4exe2  u  -  v  = 8e2u^u2  Se^u,,  + e2) +  2exe2  e2  Then as i n A l b e r t  2 and J j  irreducible  in V ( J ) .  T h i s theorem i s p a r t i c u l a r l y u s e f u l because  We  is a  elements:  54  f p . . . , f t j  are pairwise orthogonal  idempotents which s a t i s f y  fu  = uf = f v = v f i = 0,  f5u  = u f 5 = 0;  ±  ±  i = 1,2,3  ±  f5v= vf  5  = v  Thus C = ( F f j <J)Ff 2 © F f 3 ) £ ) B , where B i s the s u b a l g e b r a * f ^ , f,., u and v , and ©  denotes an a l g e b r a d i r e c t sum.  r e l a t i o n s among -these, elements a r e g i v e n by the next  u  V  0  u  0  0  f  0  V  u  u  0  V  0  V  f  4  f  5  Thus, d e n o t i n g  f  4  f  4  f  5  5  The m u l t i p l i c a t i v e table:  0  "f4 0  o f C spanned by  -f  the square r o o t o f -1 by j , we have B = F g j ^ F g 2 ©  F g ^ © Fg^,  where  gj, " ^< 4 + J«) f  g 3 = jj(f5 + j v )  are p a i r w i s e o r t h o g o n a l seven c o p i e s o f F .  idempotents.  So we see t h a t C i s the d i r e c t sum o f  55  4.2  A fx) AS —o-5—  A  CARTAN  SUBALGEBRA  L e t L be an n - d i m e n s i o n a l L i e a l g e b r a w i t h b a s i s u^,...,u Then any a e L determines the s o - c a l l e d E n g e l s u b a l g e b r a n L D ( a ) = {x e L: x R a = 0 f o r some n } . I f a = £ £ . u . e L , l e t f ( t , a ) be i=l the c h a r a c t e r i s t i c p o l y n o m i a l o f R a . Then over a f i e l d  F.  f(t,a) = t  n  + PL C ^ , . . . , £ n ) t n ~ 1  where the c o e f f i c i e n t s P i » . . . , p  + ... + Pn^!...-.^)  are polynomials i n n - v a r i a b l e s over F .  S i n c e aR a = 0 f o r any a e L, p n = 0, and so t h e r e i s a w e l l - d e f i n e d s , 1 < s < n such t h a t p s t 0, b u t p r H 0 f o r s < r < n .  Jacobson i n  [13; page 60] c a l l s an element a e L r e g u l a r i f P s ( a ) 4 0 . that H i s a Cartan  subalgebra  i s a minimal Engel subalgebra  integer  He then shows  o f L i f and o n l y i f f o r some a eL, H = LQ(a) (with r e s p e c t  to i n c l u s i o n ) .  We now d i s t i n g u i s h a c e r t a i n c l a s s o f d i a g o n a b l e elements i n an a s s o c i a t i v e a l g e b r a and show t h a t they behave v e r y much l i k e  regular  elements o f a L i e a l g e b r a .  DEFINITION 4.2.1;  a field subspace  F is called  finitely  The  finitely  element diagonable  x in an associative  if Fx is a finitely  of A; i.e., if ad x i s an algebraic  whose minimal minimal  A diagonable  (always  polynomial with  diagonable  has distinct  respect elements  linear  roots.  to inclusion)  among centralizers  key r e s u l t " o f t h i s s e c t i o n now f o l l o w s .  A  diagonable  transformation  ;x is regular  in A.  algebra  of  if A D ( x ) is of all  over  THEOREM A.2.2:  If x i s a regular element in the algebra A =  £ A (x), aeF a  then A (x)g: A (y) /or every finitely 0  diagonable element y in A (x);  Q  in other words, every finitely  Q  diagonable element of A (x) is in the Q  centre of A (x). Q  PROOF:  Suppose y i s a finitely diagonable element in A (x). Q  t e F, define y = x + t(y - x).  Now y - x i s finitely diagonable by  t  Proposition 1.2.2.  For each  Also A^x) i s invariant under ad (y-x) for any a e F  and so by the same proposition y - x i s finitely diagonable on A (x). Q  Now assuming t f" 0, g i s a root of y - x on A (x) a  i f and only i f A (x) ft A (y-x) f" 0 0  p  i f and only i f A (x) (\ A (t(y-x)) / 0 a  tR  i f and only i f a + tg i s a root of y y  (by Proposition 1.2.2)  on A (x).  t  Q  i s finitely diagonable on A (x) and we now see that for t f* 0, the  t  a  minimal polynomial of the restriction of ad y^. to A (x) i s  (1)  f (X,t) = TT(X - <a+tg)) - X*01 + g tt  B^COx"1""1  the product taken over the roots g of y - x on A (x). a  i = l,...,m in fact m a  a  Here the g ^ ( t ) , a  depends only on a, not on t;  Q  i s just the number of roots of y - x on A (x).  a  g  are polynomials in t, and m  + ... + B ( t ) a  a  Now i f a f* 0,  j> 0, and so letting a range over the non-zero roots of x,  (0) = (-a)  we have f i n i t e l y many polynomials g ( t ) none of which i s identically 0. a  Since the characteristic of F i s 0, F i s infinite, so there i s an infinite subset D S F such that g ( t ) ^ 0 for any t e D and non-zero a. But for a t e D, A (y )«£ A (x) because i f a e A (y ) and we write a = £ a relative aeF a  m  0  t  Q  t  a  to x, then (a,y ) = 0 implies ( »y ) 0. But ad y a  t  =  a  t  fc  A (x) for a ^ 0 and so a = 0. Hence a = a e A (x). a  a  Q  of A (x), we have A (y ) =» A (x). Q  fc  Q  Q  i s non-singular on By the minimality  Therefore the minimal polynomial of  ad y on A (x) i s X for t e D; i.e., g^°(t) = 0 for infinitely many t, Q  57 i = l,...,m . 0  Therefore  the polynomials  ad y^ = ad y h a s t h e m i n i m a l p o l y n o m i a l  3^°(t) a r e i d e n t i c a l l y 0 and X on A ( x ) .  This says that  Q  A _ ( x ) C ^ A (y) as we wanted t o show.  We o b t a i n i m m e d i a t e l y t h e i m p o r t a n t  COROLLARY 4.2.3:  that  Let L be a finitely  any collection  has a minimal  PROOF:  of centralizers  member.  diagonable of finitely  Then the centralizer  L e t x E L be such t h a t  subspace  of A and assume  diagonable  elements  of L is Ag(x) for some x s L.  G  Q  t  = x + t ( y - x ) i s i n L , and e x a c t l y as  i n t h e p r o o f o f t h e theorem, we c a n show t h a t A ( y ) 3 A ( x ) . D  c e n t r a l i z e r o f L , w h i c h i s f\ A ( y ) must e q u a l yeL Q  COROLLARY 4.2.4:  is  spanned  If x is a regular  by finitely  diagonable  from A  ( x ) i s minimal i n the s e t { A ( y ) : y e L } .  A  Then f o r any y e L and t E F, y  corollaries:  element elements,  0  Thus t h e  A (x). Q  in an algebra  A and A ( x ) Q  then A ( x ) is D  commutative.  t _ £ Fe. denote t h e embedding o f a C a r t a n s u b a l g e b r a t i=l H = [ Fe, o f a f i n i t e dimensional simple Jordan algebra J i n i t s u n i v e r s a l i=l Now l e t L =  1  enveloping  a l g e b r a A, as i n §4.1. We saw i n t h a t s e c t i o n t h a t L i s a d i a -  g o n a b l e subspace o f A, and o f c o u r s e i t i s f i n i t e l y d i a g o n a b l e ,  because A i s  A p p l y i n g 4.2.3, we deduce t h a t t h e c e n t r a l i z e r C o f L t _ i s A (x) f o r some x = Y a.e. E L , a,,...,a £ F. A i s a l s o s e m i - s i m p l e , o ^ x l 1 t f i n i t e dimensional.  and so C i s semi-prime by Theorem 2.2.3. artinian ring i snilpotent,  But t h e J a c o b s o n r a d i c a l o f any  and hence C i s a c t u a l l y  semi-simple.  Wedderburn Theorem and t h e f a c t t h a t F i s a l g e b r a i c a l l y  By t h e  c l o s e d , we s e e t h a t  58  C i s a d i r e c t sum  of m a t r i x r i n g s over F ( ( 5 ; page 5 1 ] ) .  m a t r i x r i n g over F i s spanned by idempotents  e^j = ( e ^ +  e  e  £j) ~  i±  is  In p a r t i c u l a r , C = A 0 ( x )  a  any n x n  (hence by f i n i t e l y  the b a s i s o f m a t r i x u n i t s ie^y'  elements) f o r i t has  Now  diagonable  i » j " l,...,n} and  d i f f e r e n c e o f idempotents f o r i , j =  i s spanned by f i n i t e l y d i a g o n a b l e  l,...,n.  elements.  If  x c o u l d be chosen r e g u l a r , then C would be commutative by C o r o l l a r y 4.2.4, and a d i r e c t sum  of m a t r i x r i n g s over F can o n l y be commutative i f each  m a t r i x r i n g i s 1 x 1.  We  c o n j e c t u r e t h a t t h i s i s indeed  the c e n t r a l i z e r of a C a r t a n s u b a l g e b r a has it  the c a s e :  i n the u n i v e r s a l e n v e l o p i n g  the same a l g e b r a s t r u c t u r e as t h a t o f the C a r t a n s u b a l g e b r a ; i s a d i r e c t sum  of f i e l d s .  Because of the important t h e o r y of Jordan  that  T h i s was  indeed  algebra  namely,  the case i n the example 4.1.4.  r o l e the c e n t r a l i z e r p l a y s i n the r e p r e s e n t a t i o n  algebras  (Theorem 4 . 1 . 3 ) , the v a l i d i t y o f our  would s i m p l i f y t h i s theory c o n s i d e r a b l y .  The  conjecture  s t r u c t u r e theorem we  do have  i s summarized i n :  The  THEOREM 4.2.5:  dimensional  simple  Jordan  of characteristic of a single complete  centralizer  matrix  We  algebra'  0 in its  element  of  rings  finish  universal  the Cartan over  over  subalgebra  an algebraically  enveloping subalgebra,  algebra and  of a closed  finite field  is the  is a direct  F  centralizer sum  of  F.  t h i s s e c t i o n w i t h an attempt to c h a r a c t e r i z e r e g u l a r  elements i n an a l g e b r a . i n an a l g e b r a A.  of a Cartan  Suppose D i s a s e t of f i n i t e l y d i a g o n a b l e  Then f o r x and y i n D, A Q ( y ) £ L A Q ( x )  i s i n Z ( A Q ( y ) ) , the c e n t r e o f A Q ( y ) , and  so A Q ( x )  i f and o n l y i f x  i s m i n i m a l i n the s e t  { A Q ( y ) : y e D}  i f and o n l y i f f o r each y i n D, x e Z ( A Q ( y ) ) i m p l i e s  y e Z(AQ(x)).  We  use  elements  t h i s o b s e r v a t i o n to e s t a b l i s h  59 n PROPOSITION 4.2.6:  orthogonal  ][ ct^e.^ a p . . . , a n e F and e^, . . . , e n  in a prime algebra A, then if A c ( x ) is minimal over the set of  idempotents all  If x =  centralizers  of linear  e^,...,en are primitive  combinations  of orthogonal  idempotents, the  and fT(a.j_ - a^) / 0. n  PROOF:  = c^-  Suppose  w n e r e  £ &± e± 3 j » ' « « » B n are a r b i t r a r y i=l n distinct scalars. By Theorem 3.1.1, y i s d i a g o n a b l e and A Q ( y ) = £ e^Ae^, i=l G and so x e Z ( A D ( y ) ) . However, y i Z ( A Q ( x ) ) because 0 ? i A e 2 S . A q ( X ) b u t ( e ^ A e 2 » y ) ^ 0.  Then l e t y =  I t i s immediate t h a t e^,...,e n a r e p r i m i t i v e , because i f  n e^ = f ^ + f 2 w i t h f j , f 2 o r t h o g o n a l i d e m p o t e n t s , x = oc^f^ + o t j ^  +  I  a  i  e  i  and we j u s t showed t h a t A Q ( x ) c o u l d n o t be m i n i m a l . n LEMMA 4.2.7:  m  Suppose x - £ <*± ± e  i=l  hence if aj,...,a a r e distinct, n  PROOF:  I B.f. j - l ^ 3  are two sets of orthogonal  {fp...,fm}  are also  =  where {e,,...,e } and  idempotents  either  with sum 1 and  m > n o r m = .n and BA.,....,Bm  distinct. The f i r s t  statement  i s j u s t a statement  o f the f a c t t h a t the  c o e f f i c i e n t s o f x when e x p r e s s e d as a l i n e a r c o m b i n a t i o n idempotents  w i t h sum 1, a r e u n i q u e l y determined  mal p o l y n o m i a l o f x (Theorem 3.1.1).  of orthogonal  as the r o o t s o f the m i n i -  The r e s t i s o b v i o u s .  T h i s lemma e n a b l e s us t o prove  THEOREM 4.2.8:  over a field ap...,an  Let A be a finite  F.  distinct  with sum 1.  dimensional  Then x e A is regular scalars  central  if and only if  and e p . . . , , e n primitive  simple n x = £ i=l  orthogonal  algebra a  -i 4J e  1  idempotents  60  PROOF:  We  f i r s t note  combinations  t h a t the o n l y d i a g o n a b l e elements i n A a r e  o f o r t h o g o n a l idempotents by Theorem 3.3.4.  s i m p l e a l g e b r a w i t h 1 i s p r i m e , 4.2.6 direction.  C o n v e r s e l y , suppose x =  Then s i n c e a  g i v e s the r e s u l t i n the " o n l y i f " n a e w t b J i i ^ F  and e^  e fi p r i m i t i v e o r t h o g o n a l idempotents w i t h sum  b e f o r e P r o p o s i t i o n 4.2.6, i t i s s u f f i c i e n t (2)  linear  1.  distinct,  By the remarks  to e s t a b l i s h  I f y E A i s d i a g o n a b l e and x e Z ( A Q ( y ) ) , then y e Z ( A Q ( x ) ) .  Thus suppose y i s a d i a g o n a b l e element i n A. s i m p l e and  By 3.3.5, A Q ( y )  i s semi-  so the Wedderburn Theorem i m p l i e s t £ D^, D j , . . . , D m a t r i x r i n g s over d i v i s i o n a l g e b r a s A i=l which are f i n i t e - d i m e n s i o n a l over F .  (3)  AD(y)  L e t f ^ be  =@  t  the i d e n t i t y o f  for i = l , . . . , t .  gonal idempotents w i t h sum sum  1.  W r i t i n g each o f these  o f p r i m i t i v e o r t h o g o n a l i d e m p o t e n t s , we  t <. n .  The key  Then f p . . . , f t  are o r t h o -  ( I f necessary)  see from Lemma 3.3.2  as a  that  s t e p i n the p r o o f i s to show t  (4)  u d i a g o n a b l e i n Z(A  To see t h i s , we u± e Z ( D i )  note f i r s t  (y)) i m p l i e s u = J Kj^±* ° i=l  5i»'«'»£t  t h a t such a u can be w r i t t e n u =  for i = l , . . . , t .  e  F  *  t £ u^ w i t h i=l = & £ , & £ Z(Ai).  But Z(D ± ) = Z ( A i ) f ± and so u m S i n c e u i s d i a g o n a b l e , we know u = £ Z±&± where g^,...,g m a r e o r t h o g o n a l i=l idempotents w i t h sum 1 and 5 e F , and by r e - o r d e r i n g i f n e c e s s a r y , ±  ±  ±  ±  m  we  £^ £g distinct. s By Theorem 3.1.1, the m i n i m a l p o l y n o m i a l o f u i s p ( t ) = TRt ~ Since 1 t i-1 u.u =0 f o r i , j e {1 t} and i ^ j , p(u) = £ p(u.) and because (3) i s j i=l a d i r e c t sum, p ( u i ) = 0, i = l , . . . , t . But pCu.^) = p ( 6 i ) f i i m p l i e s p ( 6 i ) = 0. Now  may  suppose t h a t { £ ; ] _ , . . . , £ } =  over the f i e l d  F[6J],  , . .. , 5 } , S  s < m and  p ( t ) which i s o f degree s , has  the s +  1 roots  61  fi^t'Cj »• • • » ? s and so 6^ e establishes  ...because  £^  £  s  are d i s t i n c t .  (4) .  In p a r t i c u l a r , y e Z ( A Q ( y ) ) i m p l i e s t and i f x e Z(A ( y ) ) , x = £ Y 4 f j » Y 1 » " « » Y 1 - e i=l 1 1 d i s t i n c t elements o f F; t h e r e f o r e because t ,< and Yi»»««»Y I n  a  r  e  d i s t i n c t too.  y e Z(A0(x)), verifying  REMARK 4.2.9:  Hence A (x) = o  t y = ]> 8± i> s i » " « » 3 t e F> i=l n F . B u t x = £ a -( e .,» a i , . . . , a i=l 1 n , by t h e Lemma 4.2.7, t = n n Y f . A f . (3.1.1) and so i ± = 1 i f  (2) and g i v i n g the theorem.  I f A i s the r i n g o f n x n m a t r i c e s  over a f i e l d  F and  A^ denotes the a s s o c i a t e d L i e a l g e b r a , then i t i s known ([ 9 ; page 66]) t h a t x e A^ i s r e g u l a r i n t h e L i e sense i f and o n l y i f x has n d i s t i n c t characteristic  r o o t s i n the a l g e b r a i c c l o s u r e o f F .  t h a t i n t h i s c a s e , t h e two concepts o f r e g u l a r i t y  Theorem 4.2.8 shows  coincide i f F i s alge-  b r a i c a l l y c l o s e d ; a l t h o u g h the example a t the beginning that i f F i s not a l g e b r a i c a l l y in  This  o f §3.3 shows  c l o s e d , t h e r e may e x i s t r e g u l a r elements  the L i e sense which a r e n o t even  diagonable.  62  4.3  SIMPLE  JORDAN  We sal  saw  ALGEBRAS  i n the d i s c u s s i o n a f t e r C o r o l l a r y 4.2.4  univer-  e n v e l o p i n g a l g e b r a £/(J) o f a f i n i t e - d i m e n s i o n a l s i m p l e J o r d a n a l g e b r a  over an a l g e b r a i c a l l y c l o s e d This followed  f i e l d F i s spanned over F by  d i r e c t l y because J/(J) i s semi-simple and  In t h i s s e c t i o n we Jordan algebra by  t h a t the  = Fe.  perty  (see  We  remark t h a t any  (1)  reduced s i m p l e J o r d a n a l g e b r a has  mean a l i n e a r map  D(ab)  D2  = D(a)b  +  D:  A •*• A such t h a t f o r any  this  pro-  are d e r i v a t i o n s o f A and  form a L i e a l g e b r a  a and  A b in  aD(b)  a e F , then aD^  commutator (D^,D 2 ) are a l s o d e r i v a t i o n s and algebra  algebra  an idempotent e whose P e i r c e one-space  a d e r i v a t i o n o f a (not n e c e s s a r i l y a s s o c i a t i v e ) a l g e b r a  over a f i e l d F , we  and  simple  f o r example, Jacobson [11; page 2 0 2 ] ) .  By  I f D^  finite-dimensional.  f i e l d of c h a r a c t e r i s t i c 0 i s g e n e r a t e d as an  i t s idempotents, provided J contains  J^(e)  i t s idempotents.  prove that the u n i v e r s a l e n v e l o p i n g a l g e b r a o f any  over any  J  + D2  as w e l l as  so the d e r i v a t i o n s o f  under the commutator p r o d u c t .  the  any  If D i s a n i l -  D2 p o t e n t d e r i v a t i o n , then exp(D) = 1 + D +  of A  (see  +  . ... i s always an  §1.2 of[9]).  DEFINITION 4.3.1:  Diagonable  elements x and y in an algebra A over F are  said to be of the same type if there is an automorphism iji(x) = y,  automorphism  and in this case, we write  e q u i v a l e n c e r e l a t i o n on  x <\> y.  \\> of A suoh that  I t i s immediate t h a t *v* i s an  the c l a s s of d i a g o n a b l e elements i n A;  a s y m m e t r i c , r e f l e x i v e , and  transitive  relation.  i.e., i t is  A,  63  If x Q is a finitely  PROPOSITION.,4.3.2:  over  Fj then  the linear  of A containing  S spanned  element  in an algebra  by {x e A: x ^ x Q } is a Lie  V A ( x ) ; i.e., a subspace a  such  that  A  ideal  x e S and a e A  h  implies  (a,x) e S.  PROOF:  We note  and  space  diagonable  t h a t i f ^ i s an automorphism o f A, then i|>( A a (x 0 )) = A -Cij»(x 0 ))  thus i f x t x Q , the r o o t s o f x and x Q a r e the same.  I n p a r t i c u l a r x has  o n l y f i n i t e l y many r o o t s , and so i f 0 ^ a a e A a ( x ) and a f 0, then ad a Q A  i s a d e r i v a t i o n o f A which i s n i l p o t e n t because A ^ ( x ) ( a d a Q ) C f o r any r o o t $ o f x .  Thus e x p ( - i a d a „ ) i s an automorphism o f A which sends CL_—.  Ct  x to x + a a ; i . e . , x + a x  +  a  a =  a%  x  o  anc  *  a  a  =  ^  £ aa relative aeF  x+  a  «v» x .  S i n c e •'v. i s a t r a n s i t i v e  a ) - x e S f o r any a ^ 0.  t o x , then (a,x) =  relation,  I f a e A, and  Y a a Q e S. otfO  Thus S i s a L i e i d e a l .  1  We can now prove our main r e s u l t .  THEOREM 4.3.3:  Let J be a simple  contains  an idempotent  by its  idempotents.  PROOF:  g+kQ(x)  e such  that  Jordan  dlgebra  J ^ e ) = Fe.  over  a field  Then 0 ( J ) is  F  which  generated  We have seen i n 4 . 1 . 2 t h a t e i s an a l g e b r a i c d i a g o n a b l e element  i n U(J) w i t h r o o t s i n the s e t { 0 , 5 2 , - ^ , 1 , - 1 } .  By t h e p r e v i o u s p r o p o s i t i o n  the l i n e a r span S o f {x e t/(J): x % e} i s a L i e i d e a l o f £/(J) . S i s spanned by idempotents the same m i n i m a l idempotents.  Moreover,  since x ^ e c l e a r l y implies that x s a t i s f i e s  p o l y n o m i a l as e and hence i s a l s o a l i n e a r c o m b i n a t i o n o f  I t i s therefore s u f f i c i e n t  to prove S g e n e r a t e s £ / ( J ) .  For t h i s , d e f i n e S' = { a e J : a e i d e a l o f J i n t h e sense  that  S}.  Then S 1 i s an a s s o c i a t o r  64  (2)  (S'.J.J) + (J.S'.J) +  To e s t a b l i s h  t h i s , we  (3)  c s'.  f i r s t note t h a t by the i d e n t i t y  abc - abc + cba - cba + ca.b - b c a = 0  4 . 1 , we have  (2) o f  f o r any a,b,c e  J  I n t e r c h a n g i n g a and b i n (3) and then s u b t r a c t i n g from ( 2 ) , we o b t a i n  (4)  abc + cba - bac - cab + ca.b - cb.a = 0  and t h i s can be r e - w r i t t e n as  (5)  (a,c,b)  (c,(7,b))  =  Suppose a e S', b,c,e J.  Then we see from (5) t h a t  S i s a L i e i d e a l ; hence (S' , J, J) CII S '. that  (J,S',J)c^ S' and ( J , J , S ' ) £  *'S" - {a e S': aJ &. S'} ab.c  By s i m i l a r  ejT -2  and  o  e  Let J = J  Q  +  +  be the  Then we see t h a t e e S", f o r  S;  j T £ Ai^(e") + A u(e) c "2  But then  i s an i d e a l o f J because i f a e S" and b,c e J,  P e i r c e d e c o m p o s i t i o n o f J r e l a t i v e to e.  = 0  arguments we a l s o see  S and so we have ( 2 ) .  = a.be + (a,b,c) e S' i m p l i e s ab e S".  il7  (a,c,b) e S because  '2  ~ I  S by P r o p o s i t i o n s  4 . 1 . 2 and 4.3.4;  eJi <~ eFe £ F e s S.  Thus S" i s a non-zero i d e a l o f J and so e q u a l s J by s i m p l i c i t y . S" £  S' i m p l i e s S' = J ; , i . e . , J £ S .  S i n c e 7 g e n e r a t e s U(J),  But  so does S.  65  BIBLIOGRAPHY  1. A l b e r t , A.A., A Structure 48 ( 1 9 4 7 ) , 546-567.  Theory  for Jordan  Algebras,  Annals o f M a t h ,  2. C u r t i s , C.W., and R e i n e r , I . , " R e p r e s e n t a t i o n Theory o f F i n i t e Groups and A s s o c i a t i v e A l g e b r a s " , W i l e y ( I n t e r s c i e n c e ) , New York 1962. 3. F o s t e r , D.M., "A G e n e r a l C a r t a n T h e o r y " , Ph.D. T h e s i s , U n i v e r s i t y o f B r i t i s h C o l u m b i a , 1969. 4. H e r s t e i n , I.N., " T o p i c s i n A l g e b r a " , B l a i s d e l l , Toronto 1964; r e v i s e d e d i t i o n 1965. 5.  ' "Non-Commutative R i n g s " , Carus M a t h e m a t i c a l Monographs, M a t h e m a t i c a l A s s o c i a t i o n o f A m e r i c a , W i l e y , New York 1968.  6. J a c o b s o n , N., "Theory o f R i n g s " , M a t h e m a t i c a l Surveys I I , American M a t h e m a t i c a l S o c i e t y , New York 1943. 7*  General Representation Theory of Jordan Algebras, T r a n s a c t i o n s o f the American M a t h e m a t i c a l S o c i e t y , 70 (1951), 509-530.  8  " S t r u c t u r e o f R i n g s " , C o l l o q u i u m P u b l i c a t i o n s 37, American M a t h e m a t i c a l S o c i e t y , P r o v i d e n c e , R."I., 1*956; r e v i s e d e d i t i o n 1964  9  *  • _ _ _ _ _ _ _ _  "  L J e  A l g e b r a s " , Wiley  ( I n t e r s c i e n c e ) , New York 1962.  10.  Cartan Subalgebras of Jordan J o u r n a l 27 (1966), 591-609.  Algebras,  11.  "Jordan A l g e b r a s " , C o l l o q u i u m P u b l i c a t i o n s 39, American M a t h e m a t i c a l S o c i e t y , P r o v i d e n c e , R . I . , 1968.  Nagoya  Mathematics  12. Lambek, J . , " L e c t u r e s on Rings and Modules", E l a i s d e l l , T o r o n t o 1966. 13. L e m i r e , F.W., Irreducible Representations of a Simple Lie Algebra Admitting a One-Dimensional Weight Space, P r o c e e d i n g s o f the American M a t h e m a t i c a l S o c i e t y , 19 (1968), 1161-1164. 14.  "  Weight  Spaces  and Irreducible  Representations  Lie Algebras, P r o c e e d i n g s o f the American M a t h e m a t i c a l 22 (1969), 192-197. 15.  of  Simple  Society,  Existence of Weight Space Decompositions for Irreducible Representations of Simple Lie Algebras, Canadian Math B u l l e t i n , 14 ( 1 9 7 1 ) , 113-115.  16. McKrimmon, K., Jordan Algebras of Degree 1, B u l l e t i n o f the American M a t h e m a t i c a l S o c i e t y , 70 (1964), 702.  

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