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Irreducible representations of algebras Goodaire, Edgar George 1972

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IRREDUCIBLE REPRESENTATIONS OF ALGEBRAS by EDGAR GEORGE GOODAIRE B.Sc. (Hon.), U n i v e r s i t y of Toronto, 1969 THESIS SUBMITTED IN PARTIAL FULFILMENT T H E REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY i n the "Department of Mathematics We accept t h i s thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA December, 1972 In p resenting 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 Columbia, I agree that 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 reference and study. I f u r t h e r agree t h a t permission f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r e p r e s e n t a t i v e s . It i s understood that copying or p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. Department of Mathematics  The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada i i ABSTRACT An element x of an as s o c i a t i v e algebra A i s c a l l e d diagonable provided A has a basis of c h a r a c t e r i s t i c vectors f or the transformation ad x: a ax - xa of A. This notion immediately generalizes to that of a diagonable subspace L of A. The c e n t r a l i z e r A Q of L plays an important r o l e i n the representation theory of A, for there 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 of A and of A 0. In Chapters Two and Three, we f i r s t explore various r i n g - t h e o r e t i c properties of A and A Q, and then use the r e s u l t s obtained to c l a s s i f y the diagonable elements i n d i f f e r e n t algebras. We also give conditions under which a l l A-modules are weighted. The Cartan theory of L i e and Jordan algebras i s linked i n Chapter Four by the observation that Cartan subalgebras of simple f i n i t e dimensional L i e and Jordan algebras (over a l g e b r a i c a l l y closed f i e l d s of c h a r a c t e r i s t i c 0) are diagonable subspaces of the respective universal enveloping algebras. Furthermore, i n the Jordan case, the c e n t r a l i z e r of a Cartan subalgebra i s the c e n t r a l i z e r of one of i t s elements and i s a d i r e c t sum of complete matrix r i n g s . F i n a l l y , we are able to show that the u n i v e r s a l enveloping algebra of any simple Jordan algebra which contains an idempotent whose Peirce one-space i s one-dimensional, i s generated by i t s idempotents. TABLE OF CONTENTS INTRODUCTION PRELIMINARIES CHAPTER ONE REPRESENTATION THEORY 1.1 The Universal Enveloping Algebra of a L i e Algebra 1.2 Properties -of Algebras Possessing Dlagonable Elements and t h e i r Modules 1.3 A Correspondence Between A- and A0-Modules 1.4 Further Results Concerning and CHAPTER TWO RING-THEORETIC CONNECTIONS BETWEEN A AND A o 2.1 Chain Conditions 2.2 -Nil'potent 'Ideals - and Semi-P-rimeness 2.3 P r i m i t i v i t y CHAPTER THREE DIAGONABLE ELEMENTS AND WEIGHTED MODULES 3.1 Idempotents are Diagonable 3.2 Prime Algebras 3.3 Central Simple Algebras CHAPTER FOUR APPLICATIONS TO THE UNIVERSAL ENVELOPING ALGEBRA OF A JORDAN ALGEBRA 4.1 The Universal Enveloping Algebra of a Jordan Algebra 4.2 A Q(x) as a Cartan Subalgebra 4.3 Simple Jordan Algebras BIBLIOGRAPHY i i i Page 1 3 6 9 16 21 24 27 32 35 40 46 50 55 62 65 i v ACKNOWLEDGEMENTS I wish to express my appreciation to my thesis supervisor, Dr. C.T. Anderson, for h i s continual encouragement and help during the preparation of t h i s t h e s i s . My sincere thanks also go to Dr. F.W. Lemire for introducing the idea of a "diagonable element" around which the the s i s i s centred, and for communicating to me r e s u l t s which led to much of my f i r s t chapter. The f i n a n c i a l support of the Un i v e r s i t y of B r i t i s h Columbia and the National Research Council of Canada i s also g r a t e f u l l y acknowledged. INTRODUCTION The u n i v e r s a l enveloping algebra of a f i n i t e dimensional simple L i e algebra over an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c 0 has a basis r e l a t i v e to which the l i n e a r transformations ad h: a t-> ah - ha for h i n a Cartan subalgebra are simultaneously diagonizable. Thus any such Cartan subalgebra i s an example of what we c a l l a diagonable stibspace of an a s s o c i a t i v e algebra. Let L be a diagonable subspace of an a s s o c i a t i v e algebra A. In Chapter One, we define the concept of 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 generalize a theorem of Lemire ([14]) by proving the existence of a one-to-one correspondence between the i r r e -ducible X-weighted modules of A and A Q S the c e n t r a l i z e r of L i n A. This r e s u l t leads us to believe there should be some close connections between the algebras A and A Q and these we i n v e s t i g a t e i n Chapter Two. Several examples throughout the thesis i n d i c a t e how frequently diagonable subspaces occur i n algebras. In order to obtain many r e s u l t s , i t i s therefore necessary to Impose some kind of r e s t r i c t i o n s on the algebra and the diagonable subspace. I f we assume L has only a f i n i t e number of d i s t i n c t r o o t s , then 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 s semi-simple. I f every A-module i s weighted, A Q i s semi-simple whenever A i s . Several of the r e s u l t s i n Chapter Two require that every i r r e d u c i b l e A-module be weighted. In one of the major r e s u l t s of Chapter Three, we are able to prove that i f L i s a diagonable subspace spanned by a si n g l e a l g e -b r a i c element i n a prime algebra A, and one i r r e d u c i b l e A-module i s weighted, 2 then every A-module i s weighted. In t h i s chapter, we are able also to . characterize the dlagonable elements of a f i n i t e dimensional c e n t r a l simple algebra. They turn out to be exactly those elements x^hich are expressible as l i n e a r combinations of p r i m i t i v e orthogonal idempotents with sum the : 0 i d e n t i t y of A. An i n t e r e s t i n g consequence of t h i s i s that the diagonable . elements of a matrix r i n g are nothing but those matrices s i m i l a r to diagonal matrices. In Chapter Four, we give an i l l u s t r a t i o n of the common Cartan theory of L i e and Jordan algebras established i n Foster's doctoral d i s s e r -t a t i o n ([3]). A Cartan subalgebra H of a f i n i t e dimensional simple Jordan algebra over an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c 0 i s a d i a -gonable subspace of i t s u n i v e r s a l enveloping algebra A. Thus, exactly as i n the L i e case, we obtain a one-to-one correspondence between the X-weighted irr e d u c i b l e « m o d u l e s of A and o f C , the c e n t r a l i z e r i n A of H. This r e s u l t 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 of some element i n II and i s a d i r e c t sum of complete matrix rings over F. ' . As a f i n a l a p p l i c a t i o n of the theory of diagonable elements, we prove that the u n i v e r s a l enveloping algebra of a simple Jordan algebra i s generated as an algebra by i t s idempotents, provided the Jordan algebra contains an Idempotent whose "Peirce one-space" i s one-dimensional. 3 PRELIMINARIES A l l fields in 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 ; (a + b)c = ac + be and 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 is 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 ideal. A i s a simple algebra i f i t has no non-zero proper ideals. A i s nilpotent i f there is a positive integer n such that every product of n elements of A is 0. A right A-module is an abelian group V together with a bilinear 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 is 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 • ' If 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 : a ax x L : a n xa x ad x: a •+ ax - xa for a e A Note that ad x = - . The commutator (x,y) is 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 e X, y 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 is 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 field F determines a Lie algebra A^ by defining the product of elements x,y e A to be the commutator (x,y). 2 A Jordan algebra is a commutative algebra J such that (x ,y,x) = 0 for any x,y e J. We always assume a Jordan algebra contains an identity 1 . If e = e is 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 More generally, i f e^ e^ are pairwise orthogonal idempotents ( eiGj = 0 for i ^ j) with sum 1 , the Peirce decomposition has the form n J - @ j» J ; J±i = J 1(e i) = {x E J: xe ± = x}, i = 1,... ,n = J j j C e ^ 0 J j g C e ^ ) = { x E J : xe^^ = J$x = x e l<j The reader should consult Jacobson [ l l ; pages 118-120] for d e t a i l s . F i n a l l y , we point out that an element x i n an algebra A over F i s c a l l e d algebraic i f i t i s the s o l u t i o n to some polynomial i n F [ t ] ( i ; e . a polynomial i n t with c o e f f i c i e n t s i n F ) . Among a l l such polynomials, there i s one of l e a s t degree c a l l e d the minimal polynomial of x which divides any other polynomial f or which x i s a s o l u t i o n . " We assume that the minimal polynomial i s raonic, i n which case i t i s unique. We now make the convention that henceforth, "algebra" w i l l always mean_"associative algebra with i d e n t i t y " . 6 CHAPTER ONE REPRESENTATION THEORY 1.1 THE UNIVERSAL ENVELOPING ALGEBRA OF A LIE ALGEBRA Let ii be a L i e algebra over a f i e l d F. A representation of L i s defined to 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 algebra over F, such that f or every a and b i n I , (1) S , = S S, - S,S where S: ui+ S„ for u e L a b a b b a u For example, R: x t->- R^ defines a representation of L i n the a s s o c i a t i v e algebra X generated over F by {R^: x e L}. An a s s o c i a t i v e algebra U(L) with i d e n t i t y i s a u n i v e r s a l enveloping algebra for L i f there i s a canonical representation S*: L -*• U(L) such that given any representation S: L -*• A of L i n an a s s o c i a t i v e algebra A, there e x i s t s a unique homomorphism Y: U(L) -> A such that S = YoS*; i . e . which makes the following diagram commutative: L > A S We r e f e r the reader to Chapter V of Jacobson ( [ 9 ]) f or a more comprehensive treatment of U(,L). What i s important f or 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 that L i s f i n i t e dimensional and simple, and that F 7 i s a l g e b r a i c a l l y c l o s e d . Then L possesses a Cartan subalgebra //; i . e . , a n i l p o t e n t subalgebra which i s self - n o r m a l i z i n g i n the sense that [x,#] £ H implies x e H. There e x i s t s a set A of l i n e a r functionals a:H -»- F c a l l e d roots with the property that the subspace - {x e L: x(R^ - a ( h ) l ) n = 0, for a l l h e H} i s non-zero for any root a. One introduces an order i n A and i s able to d i s t i n g u i s h between p o s i t i v e and negative r o o t s . A root i s simple i f i t cannot be written as the sum of two p o s i t i v e r o o t s . Denote by Ag and A + the set of simple and p o s i t i v e roots r e s p e c t i v e l y . Then there e x i s t s a ba s i s B of L c a l l e d a Cartan b a s i s , B = {e . f , hfi: 3 e A_, a e A+}, where — — — — — — — — 0t Ct p 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 for B, we have [hg.hg,] = 0 ( 2 ) t V V = VBe« [ f a , h 3 ] = B a.. ef B , a e A +, 3,3' e As where the A „ and B „ are i n t e g e r s . a,p 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 ; § 5 . 2 ] ) , the u n i v e r s a l enveloping algebra U(L) of L has a l i n e a r basis c o n s i s t i n g of a l l elements of the form (3) TT . f a n ( a ) IT s r m TT e a m ( a ) aeA + ^e^ s aeA + where n ( a ) , r ( B ) , m(a) are non-negative integers and the product respects the order i n A. For a fixed 8 O e A s, observe that aeA + aeA + u aeA + because of the i d e n t i t y 8 (4) (xy,z) = x(y,z) + (x,z)y which holds i n any a s s o c i a t i v e algebra. Generally, i t i s true that (u,h 0 ) po i s an i n t e g r a l m u l t i p l e of u for each basis element u of the form ( 3 ) , and hence U(L) decomposes as a vector space i n t o a d i r e c t sum of spaces of the form {u e U(L): (u,h R ) = nu} for n an i n t e g e r . With t h i s motivation we make the following d e f i n i t i o n . DEFINITION 1.1.2: Let A be an associative algebra with unit element over a field F. Then an element x e A is diagonable if A A Cx), a vector aeF space direct sum, where A a(x) = {a e A: (a,x) = aa}. 9 1.2 PROPERTIES OF ALGEBRAS POSSESSING DIAGONABLE ELEMENTS AND THEIR MODULES We collect some important facts about diagonable elements in the next two propositions. PROPOSITION 1.2.1: Suppose A = © £ R. is an algebra direct sum and l e i x = \ x^ is a diagonable element of A. Then each x. is a diagonable . i e i element in the algebra and (R i) c t(x i) = A a(x) TV R^ . PROOF: Let u E R . Then (u,x) = (u,x.) = £ au e R. where u = J u & 1 0?«aeF a aeF u Q e A a ( x ) * Next, ((u,x),x) = ((u,x.),x.) = (. I au ,x) = 7 a u e R.,. 1 1 O&xeF " Oj*aeF Continuing in this way, for every positive integer k, we have ) o u e R.. O&xeF Now u a = 0 except for a in some finite set {ai,...,an} U {0}, no = 0. Thus n ^ with vi»-«'»vn e %» i s 3 system of n equations in n unknowns with coefficient matrix (a^), i , j = l,...,n t a vandermonde matrix. Since cq,...,an are distinct scalars, the system has a solution; i.e., each u a e R^ . Also, n * u Q = u - £ u Q^ e R^  and so x^ is diagonable in R^ . Finally, ( R i ) a ( x i ) = V x ) 0 V is an immediate consequence of (u,x) * (u,x^) for u e R^ . 10 PROPOSITION 1.2.2: ( i ) If x is a diagonable element in an algebra A , and B is a subspace of A invariant under ad x, then x is diagonable on B in the sense that B decomposes as © Y B (x) with B (x) = {b E B: (b,x) = ab}. aeF (tfot-e t/zat we do not require that x E BJ. ( i i ) If x is a diagonable element and t c F is non-zero, then tx is diagonable, and A Q(x) = A t a ( t x ) . ( i i i . ) .If x .and y are commuting diagonable elements,, then x + y is diagonable and, in fact, i f A Q(x) = 0 except for a E {a^: i E 1} and Ag(y) = 0 except for a e {gji j E J}, then : Ay(x+y) £ 0 implies Y e {di + $^: i e I, j e J}. 1/ I and J are f i n i t e sets, the linear transformation ad(x+y) satisfies the polynomial Jf (t - (c^+B-j)) e F [ t ] , i£l J PROOF: (I) I f b E B, we can write b = \ b Q r e l a t i v e to x, and since •*('Bvx>'S**B-'(byx-W')• a "b *E B-for-every integer k > 0. Here we write OftxeF ' ( b , x ^ ) = (...(b,x),x),...,x), x repeated k times. A vandermonde matrix argument i d e n t i c a l to that used i n the proof of 1.2.1 shows that each b Q e B. ( i i ) i s t r i v i a l , and ( i i i ) follows from ( i ) because (x,y) = 0 implies each A (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 . C l e a r l y (A a,(x)) Q ( y )C 3 A a o (x+y). j e J 1 J 1 J 1 J The l a s t part of ( i i i ) follows from the observation that i f x i s diagonable and A a(x) = 0 except for a i n a f i n i t e set {a^» • • • »ctn) , then the l i n e a r n transformation ad x of A i s algebraic with minimum polynomial TRt ~ a- » )e F [ t ] . i = l 1 Now l e t L be a l i n e a r subspace of an algebra A over F which i s spanned by commuting diagonable elements. The preceding p r o p o s i t i o n implies 11 that every element of L i s n e c e s s a r i l y diagonable. DEFINITION 1.2.3: A map a: L F such that A a(L) = {a e A: (a,x) = a ( x ) a , f o r every x e L} is non-zero is called a root of L in A, and A 0(L) is the corresponding root space. With L f i x e d , we w i l l write A a instead of A a(L) and r e f e r to "roots of A". If V is a right 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, for every, x e L and n = n(x) > 0} is non-zero. is the corresponding weight space and V is said to be L-weighted or sometimes \-weighted if we wish to emphasize that X is a weight of L. We remind the reader that i f B i s any vector space over a f i e l d F, B*, the dual space of B, i s defined as the space of a l l l i n e a r f u n c t i o n a l s on B; i . e . , those maps ij>: B ->• F s a t i s f y i n g ( i ) iKbj + b 2> = + *<b2> a n d ( i i ) \(i(ab^) = au>(bp for every b^,b2 E B , a e F. The * notation i s standard and w i l l be used f r e e l y i n the r e s t of t h i s t h e s i s . PROPOSITION 1.2.4: 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 that f or any u E L, X(u) i s the only c h a r a c t e r i s t i c root of u on V, . For t h i s , l e t O^v E V.. Then 12 v(u - a l )s « 0 » v(u - X ( u ) l ) t implies v ( ( u - X(u)) - (u where we can expand using the binomial theorem because u u - a l commute. Thus v ( ( a - X ( u ) ) l )t + S = 0 and X(u) = a. l i n e a r now follows from (a) x, y e L implies X(x) + X(y) i s a c h a r a c t e r i s t i c root of x + y on arid (b) x e L and a e F implies aX(x) i s a c h a r a c t e r i s t i c root of ax on To see ( a ) , we have p o s i t i v e integers n and m with 0 = v(x - x ( x ) l ) n = v(y - X ( y ) l )m for any CMv e V^. Since 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) follows i n the same way. F i n a l l y , that a root a i s also i n L* i s proven exactly as above, using the l i n e a r i t y of the map ad x. DEFINITION 1.2.5: A linear subspace L of an algebra A over F, spanned by commuting diagonable elements, is called a diagonable subspace, if A decomposes as a vector space relative to the collection A of roots of L i.e., A = © I Aa . aeA Suppose that L i s any two-dimensional subspace of A with basis c o n s i s t i n g of the commuting diagonable elements x and y. In the course of proving Proposition 1.2.2, we saw (1) A = I ( A a ( x ) ) 3 ( y ) , ( A a ( x ) ) g ( y ) = A a(x) A Ag(y) a,8 the sum taken over a l l c h a r a c t e r i s t i c roots a of ad x and c h a r a c t e r i s t i c roots 8 of ad y. The l i n e a r i t y of the roots of L implies immediately that - a l ) )t + S = 0, - X(u)1 and That X i s i f ( A a ( x ) ) g ( y ) $ 0, then t h i s space i s exactly A ^ ( L ) , where X i s the root of L uniquely determined by the conditions X(x) = a, X(y) = B. Thus (1) i s a decomposition of A r e l a t i v e to L; i . e . , L i s a diagonable subspace. By i n d u c t i o n , we see that any finite-dimensional subspace L spanned by commuting diagonable elements i s i n fa c t a diagonable subspace. For example, a Cartan subalgebra of a f i n i t e dimensional simple L i e algebra 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 the canonical embedding i n the u n i v e r s a l enveloping algebra i s a diagonable subspace. Now f i x a diagonable subspace L of A , and l e t A = © j[ A be aeA a the decomposition of A r e l a t i v e to the c o l l e c t i o n A of roots of L. We come now to an i n v e s t i g a t i o n of L-weighted A-modules. • ^ E M M A-4^.-P6-: -If--xj-.its-tm'b-mi&hted^A^&duZe* ••V^-Aac:--v\+0 for any weight \3 where we define V ^ + A to be zero if X + a is not a weight of L in V . k V PROOF: An easy induction reveals that x a = a(x - a ( x ) l ) f o r a l l a e A A and x e L. Thus i f v(x - X ( x ) l )n = 0, we have 0 = v(x - X ( x ) l ) n a = v I (Ax k(-HxH) n~ ka ~ v I ( $ ( - X ( x ) ) n _ k a ( x - a ( x ) l ) k k=0^' = va(x -(X(x) + a ( x ) ) l )n and so va eV, , . X+a COROLLARY 1.2.7: Let V be an irreducible h-weighted A-module. Then V decomposes as © £ V , relative to the set A of all weights of L in V . XeA A 1 4 PROOF: We simply note that ^ 0 for some X e A and so by the lemma, ® £ V> is a non-zero A-submodule of V. XeA Now the identity ( 4 ) in 1 . 1 implies ( 2 ) AQ Ag S A ^ where we again define & a +£ to be zero i f a + 8 is not a root. In particular, -A = {a e. ,A: ax = xa, for every x e-L} is .a .subalgebra containing 1 and x; namely, the centralizer of L in A. Also ( 2 ) implies that each Aa is an Ao-module. By 1 . 2 . 6 so is any weight space of an A-module, and we can further show: LEMMA 1 . 2 . 8 : If V is an irreducible weighted A-module, is an irreducible AQ-module for any weight X . PROOF: If W), is a proper A-submodule of V\ , then W = A = W , © T W\ A . Ao ° Ao Ao Ao r\j. L * Ao a is a proper A-submodule of V because of 1 . 2 . 6 . LEMMA 1 . 2 . 9 : If K is any maximal right ideal of AQ and u e Z ( A Q ) . , the centre of A Q , then au e K with a e AQ implies a e K or u e K. PROOF: If a i K, K + aAQ = A q , and so k + ab = 1 for some k e K and b e AQ. Hence ku + abu = u with ku and abu (=aub) both in K. So u e K. Now suppose V is an irreducible L-weighted A-module and is a non-zero weight space. Let O^ v e . Then x: a n va is an A0~module homomorphism A Q -»• V , necessarily surjective by 1 . 2 . 8 , and so = A 0 / T Q where T is the kernel of T . Since T is a maximal right ideal of A „ which o o & o !5 contains (x - X ( x ) l )n, and x - X(x)l i s i n Z ( A 0 ) , we have that T Q a c t u a l l y contains x - X ( x ) l for every x e L by 1 . 2 . 9 . But now, l e t t i n g T be the kernel of the A-module homomorphism A -> V defined by a»-»- va, s i m i l a r reasoning gives V = A/T. Noting that T (\ A Q = T Q , we have obtained: T H E O R E M 1 . 2 . 1 0 : If V is an irreducible 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 close t h i s section with an easy r e s u l t which nevertheless w i l l prove quite u s e f u l . P R O P O S I T I O N 1 . 2 . 1 1 : Suppose A = Q £ R . is an algebra direct sum. Let L i e i be a diagonable subspace of A and write L = © £ L . . Then if every i e i ' 1 irreducible A-module is L-weighted3 every irreducible R ^ -module is L J -weighted, for any ±n E I . P R O O F : Any i r r e d u c i b l e R - -module V i s also 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 va = va^ for v E V and a = £ a^ E A. Thus V i s L-weighted ° i E l and ^ 0 for some weight X . Let 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 implies v(x^ - X(x)e^ ) n = 0 . Since en-i E l 1 ° ° l o i s the i d e n t i t y i n R J , upon defi n i n g X J (X J ) = X ( x ) , we see that V i s xo i o xo L J -weighted with weight X . . 1o o ' 16 1.3 A CORRESPONDENCE BETWEEN A- AND A -MODULES o Let H be a Cartan subalgebra of a simple fi n i t e - d i m e n s i o n a l L i e algebra L over an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c 0, and l e t C be the c e n t r a l i z e r of H i n the un i v e r s a l enveloping algebra V(I). Lemire ([14]) has shown that there i s a one-to-one correspondence between the set of equivalence classes of i r r e d u c i b l e representations of L possessing a weight X e H* and X-weighted representations of C; i . e . , representations whose associated C-modules are X-weighted. In t h i s s e c t i o n , i t i s shown that t h i s r e s u l t follows from the fact that H i s a diagonable subspace of U(L). We denote by W\ the c o l l e c t i o n of a l l isomorphism classes [V] of i r r e d u c i b l e L-weighted A-modules V for which ^ 0, X e L*. W\° denotes the c o l l e c t i o n of a l l isomorphism classes tvl of non-zero i r r e d u c i b l e A o-mcdules V f o r which V(x •- A(x)l) = 0 for a l l x c L. For convenience, we w i l l write V when we s t r i c t l y mean the isomorphism cla s s [vl. The di s c u s s i o n p r i o r to 1.2.10 establishes a map -*• given by *V = V^. $ i s well-defined because of the next r e s u l t ; LEMMA 1.3.1: If W is an L-weighted irreducible A-module and V is any A-module isomorphic to W, then V is L-weighted and irreducible, and W^  is isomorphic to V\ as AQ-modules for every A e L*. PROOF: We simply observe that i f :^W -*• V i s the A-module isomorphism, then by r e s t r i c t i n g to W^  and the r i n g of s c a l a r s to A Q, we obtain an Isomor-. phism W^  -*• because each module i s i r r e d u c i b l e and iHwp V^. 17 A one-to-one correspondence between and will be established by showing that the map $ has an inverse. First, a result in linear algebra: LEMMA 1.3.2: Let V be any vector space over a field F and Tj,...,T n distinct elements of V*. Then there is a v e V such that (T^v) ,... ,Tn(v)} is a set of n distinct scalars. PROOF: Consider the finitely many linear functionals {T± - T^: i 4 j} of V*. Since T 1 }...T n are distinct, {kerCT^ - T..): i J j} is a finite set of proper subspaces of V. We are required to find a v e V such that n. But this follows from the fact that V cannot be v i Q ker(T - T,). expressed as a finite union of proper subspaces because F is infinite t (char F = 0). Indeed, i f V = \J V. , we may assume inductively that no v^ is contained in U V. and hence choose Vj £ Vi - \J V. and V2 I V,. 3 j-2 .There must-be distinct-scalar-s and a2--such that -ct-j-Vj ••+ v 2 and c^v^ + v£ belong to the same V^. Then i ^  1 because V2 i and so for some i > 1, 0*1V1 + v 2) - (a 2v 1 + v 2) e V±. But then (a^ - a 2)v 1 E V i which is impossibl DEFINITION 1.3.3: An ideal I (right, left, or two-sided) of A • © \ AQ is called homogeneous if I = © £ I/) AQ. Ct£A OIEA PROPOSITION 1.3.4: Any two-sided ideal of A is homogeneous. Any right (or left) 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 a E A aeA 2 Then for any x E L, (a,x) = ][ a(x)a a E I. Also ((a,x),x) = \ <x(x) a (Mae A 0# KEA is in I, and continuing in this way, we get that for any integer k > 0, 18 •(1) I a < x ) \ - i k e I Now a a = 0 for a l l a except a e {ctj,...,a n}. The are i n L* by 1.2.4, and so by 1.3.2, we can f i n d an x e L for which ia^(x),...,an(x)} i s a set of n d i s t i n c t s c a l a r s . Then l e t t i n g k run from 1 to n, (1) i s a system of l i n e a r equations over F whose matrix of c o e f f i c i e n t s i s a vandermonde matrix with non-zero determinant. Solving, i t i s c l e a r that each a a . i s a l i n e a r combination o f - i j _ , . . . . ,.i and so i n I . Of course n a = a - / a „ . i s then i n I too. For the second statement of the p r o p o s i t i o n , i f I i s a r i g h t i d e a l of A containing x - X ( x ) l for a l l x e L and some X e L*, and i f a = ][ a e I , aeA we note that (a,x) = (a,x - X(x)l) i s i n I too, and hence the conclusion follows j u s t as above. LEMMA 1.3.5: Any (proper) right ideal I of A Q generates a right ideal of A contained i n 1 ^ \ A^ and hence is contained in a maximal right ideal I* O^aeA of A. If I is maximal, I* (\ AQ = I, and if, in addition, 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 of A which I generates i s contained i n I @ \ A Q O^aeA because A^A^ c A ^ . The existence of I* then follows from a standard argument i n v o l v i n g an a p p l i c a t i o n of Zorn's Lemma. Now I* (\ AQ £ AQ because 1 i I*, and since I* A A ^ g l and i s a r i g h t i d e a l of A D, we must have I* (\ AQ = I i f I i s maximal. Next, i f there i s a X e L* with x - X(x ) l e I for every x e L, then (2) J < C I © I Aa 0/aeA 19 for any (proper) r i g h t i d e a l J of A which contains I . To obtain ( 2 ) , we observe that such a J i s homogeneous by 1.3.4 and so i s contained i n (J AAQ) + I A . As I g J A A Q ^ A Q, J fV A = I by the maximality of I . O&teA a Thus the sum of a l l proper r i g h t i d e a l s of A containing I i s contained i n XCD I A A and so must again be proper. C l e a r l y 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 follows: .if V e then V - A Q/I where I i s a maximal r i g h t i d e a l of A Q containing x - x(x)l for a l l x E L and some X e L*. This i s so because V(x - X(x)l) = 0 implies (I + l ) ( x - X(x)l) = 0. By Lemma 1.3.5, I extends uniquely to a maximal r i g h t i d e a l I* of A. A/I* e because ( I * + 1)(x - X(x)1) = 0 for every x E L . Define V = A/I*. ¥ i s well-defined because of: LEMMA 1.3.6: Suppose A 0/I^ and A Q / I 2 a v e irreducible and isomorphic ^-modules, and for some X E L*, X - X(x) 1 e I j (% I 2 for every x E L. Then A/11* - A / I 2 * cis A-modulest where 1^* and I 2 * are the right ideals of A given by 1.3.5. PROOF: Suppose that o ( I j + 1) = I 2 + a Q , where a: A /IJ A Q / I 2 i s the given isomorphism. L i f t a to a*: A/I^* -»• A/I 2* by a*: 1^* + a H - 1^* + a Q a . o* i s well-defined f o r i f a E but a Qa I I 2 * , then I 2 * + a QaA = A, and so there e x i s t s u E A such that (3) - Qau - l e I 2 * S l 2 e I A A . O^aeA Now because au £ 1^* £ I j @ £ A Q , we can write au = b + J\>a where b Q e l ^ * O^aeA and so from (3) we see that a Q b 0 - 1 E I 2 . This i s impossible because o ( I 1 + b 0) = 0 = I 2 + a Q b o . F i n a l l y , since I 2 * £ . I 2 © J A Q , aQ i I 2 * . O^asA 20 Hence o* i s not zero and therefore must be an isomorphism by Schur's Lemma. We are now able to prove the main r e s u l t of t h i s s e c t i o n . THEOREM 1.3.7: $ defines a one-to-one correspondence between and . PROOF: We prove that <I> and ¥ are inverse maps. Given V e W-^, we r e c a l l that $V = = A 0/I for some maximal r i g h t i d e a l I of AQ containing x - X ( x ) l for every x e L. I'C^V) i s then A/I* where I* i s that i d e a l of A given by Lemma 1.3.5. That YC^V) - V i s now apparent from the discussion preceding Theorem 1.2.10 upon observing that the i d e a l T defined there i s the i d e a l ^ 1 * by uniqueness. Conversely, given V e V = A Q / J where J i s a maximal r i g h t i d e a l of A Q containing x - X(x)l for every x e L, for some X e L*. L e t t i n g J * be that i d e a l of A given by 1.3.5, *(VV) i s then ( A / J * ) A . To see that t h i s i s Isomorphic to V, define a map a: A -*• (A/J*), by a. »-*• J * + a . For any a e A . a „ ( x - X(x)l) = (x - X ( x ) l ) a „ E J £ J* and so a(An) £ (A/J*),. Q O O O w A a i s s u r j e c t i v e because i t i s non-zero and (A/J*)^ i s i r r e d u c i b l e . The kernel of a i s {a Q e A Q: a Q e J*} = J * A A Q = J by 1.3.5. Thus, ( A / J * ) A = A Q/J = V as required; i . e . , = V. 21 1.4 FURTHER RESULTS CONCERNING AND ^ ° Let V be an L-weighted i r r e d u c i b l e A-module with A the set of weights of L i n V. We have seen how the associated family A(V) = l^:A e A} of i r r e d u c i b l e A0~modules determines v i a the map Y the same i r r e d u c i b l e A-module V, up to isomorphism (1.2.10). A natural question to ask at t h i s point i s the following: given some subset A of L*, the dual space of L, and a family F = {.V^  e X e A) of A 0-modules, does there e x i s t an A-module V such that F = A(V)? Suppose 1^ and 1^ are maximal r i g h t i d e a l s of A Q containing x - X ( x ) l and x - u(x)1 for every x E L r e s p e c t i v e l y , where X,u e L*. We know that there e x i s t (unique) maximal r i g h t i d e a l s 1^* and 1^ * of A with A/1^* e Jv^, A / l ^ * e Jif , and such that (1) ( A / I A * ) X - A 0 / I X ; ( A / T ^ ) y = A 0 / I M This was established i n the proof of Theorem 1.3.7. Suppose now that A/I x* = A / I J J * . Then (1) c e r t a i n l y implies (A/I^*)^ J 0 and thus from the d i s c u s s i o n preceding Theorem 1.2.10, I * i s the kernel of the homomor-phism T : A -*• A/I^* given by x (a) = I * + b^a, where 1^* + b^ i s any non-zero element of ( A / I x * ) y . By the i r r e d u c i b i l i t y of A / l ^ * , (1^* + b y)A = A/I x* and there i s a y E A such that b^y - 1 e 1^ *- C l e a r l y y i 1^* because then b^y and hence 1 would belong to 1^*; but y l ^ * Is 1^ ,* because i f a E b^ya = (b^y - l ) a + a e 1^* and so ya E ker x = I y A « We have established the necessity part of the next theorem. THEOREM 1.4.1: -A family ( A 0 / I x E Wx°: X e A £ L*} of AQ-modules is the family corresponding to an A-module V i f and only i f there exist elements 22-y , in A, for each X and y in h, such that y,, > i I * but y.. ^I,* e l *. y»x v>A y •J»A A y PROOF: We need only prove the sufficiency. If X,y e A and there is a y in A , y i I y*, y l x * ^ l y * , then the map A/I x* -*- A / I p * defined by I j * + a H - I u* + ya 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 and determined by the weight A . The situation is ideal for A0-modules, some-what less than ideal for A-modules as the following theorem shows. THEOREM 1.4.2: (i) I f W° then X is identically equal to y (\=v). (ii) If WXT\ ? <j>, then for some a e A, y = X + a and -a e A. PROOF: (i) If V E ffc W °, then there exist maximal right ideals 1^ and I of AQ containing x - X(x)l and x - y (x) 1 respectively, for every x E L, and such that V ^  A Q / ^ ~ A 0 / I y • S uPP o s e 0: A 0 / I A **" Ao^p i s t h e isomorphism and o(I A + 1) = I + a Q, a Q e A q . Then a(I A + (x - X(x))l) = 0 implies a Q(x - X(x)l) E 1^ and since a Q cannot belong to 1^, by Lemma 1.2.9 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 e L. Since I is 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 to L induces decompositions of V ; (2) V = ® I Aa+lX/IX « 0 I A A + I p/I M CIEA aeA Moreover, for 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 that a a(x - (X+a)(x)l) E 1^. By Lemma 1.3.1, V X + A - AQ+ I^/Ix and the weights of L i n V are contained i n the set {X + a: a e A}. Since u i s a weight, there i s an a i n A with u = X + a. By symmetry, there i s an a1 e A with X = y + a'. C l e a r l y a' = -a. We s i n g l e out the following two r e s u l t s , now obvious. COROLLARY 1.4.3: If V is an irreducible L-weighted A-module, then the weights of L in V are of the form XD + a, a e A , where XQ is any fixed weight. In particular if A has only finitely many non-zero roots, V has only finitely many weights. COROLLARY 1.4.4: If V is an irreducible L-weighted A-module, then any two weights of L in V differ by a root of A. In the sequel, we w i l l frequently have cause to suppose that A has only f i n i t e l y many non-zero r o o t s . We therefore make the DEFINITION 1.4.5: L is a finitely diagonable subspace of A if it is diagonable, and A = 0 for all but a finite number of a in L*. 24 CHAPTER TWO RING-THEORETIC CONNECTIONS BETWEEN A n AND A 2.1 CHAIN CONDITIONS If R i s a r i n g (associative), and M an R-module, then M has the descending chain condition on submodules i f every decreasing sequence Mist Mvjsi ... of submodules of M terminates; i . e . , f or some k > 0, Mfc = Mk for a l l t > k. This i s equivalent to saying that M has the minimum condition on submodules: every c o l l e c t i o n of submodules of M has a smallest member (with respect to i n c l u s i o n ) . D e f i n i t i o n s of the ascending chain condition and maximum condition are made i n the obvious way. I f M = R considered as a r i g h t R-module, submodules of M are r i g h t i d e a l s of "R and if"M has the descending (respec t i v e l y ascending) chain con d i t i o n on i d e a l s , R i s said to be a r t i n i a n ( r e s p e c t i v e l y noetherian). We employ the same notation as that of Chapter 1. L i s a diagonable subspace of an algebra A over a f i e l d F. A decomposes as a vector space, .A = }> A a r e l a t i v e to the set A of roots of L i n A. aeA . PROPOSITION 2.1.1: If A is artinian (respectively noetherian), then each kQ-module A Q has the descending (respectively ascending) chain condition on its submodules. In particular, A Q is artinian (respectively noetherian). Conversely, if each A a has the descending (respectively ascending) chain condition on h -submodules and L is a finitely diagonable subspace of A, then A is artinian (respectively noetherian). 25 PROOF: Suppose A i s a r t i n i a n , and M^=2 ... i s a descending chain of A - submodules of A a. For each i , generates the r i g h t i d e a l M ^ * = M^A of A, and t h i s i s contained i n + I Ag because A t tA^c A ^ ^ f o r any a, $ E A ( § 1 . 2 ) . Since M i * 2 M 2 * S •.., there i s an integer k > 0 for which Mt * = Mk* f o r e v e ry t > k. Now i f t >. k and m E Mk, m e Mk* = M F C * and M A * C Z ; M + ][ Ag, so considering the components of m r e l a t i v e to L , m E M . Thus M^S. Mt and the o r i g i n a l sequence of AQ-modules terminates at M^. On •the other hand, i f L i s f i n i t e l y diagonable and each A a has the descending chain con d i t i o n on Ao-submodules, l e t Ii=2 *2 — ^ e a descending chain of r i g h t i d e a l s of A. Notice that each I j can be written as ® £ ( I j ) a , CXEA where ( ! j ) a i s the Ac-module of a-coraponents of elements i n I j . C e r t a i n l y k £ SL implies ( I ^ ^ S ? a f o r each a. Thus we obtain f i n i t e l y many descending chains { ( I j ) a : j = 1,2,...} of AD-submodules of A a, one for each a e A. Each chain terminates a f t e r a f i n i t e number of steps, so they have a l l terminated by the n step, say. This implies that "I '= 'T for any m > n, and so A i s a r t i n i a n . The proof for the noetherian case simply mimics the above. EXAMPLES 2.1.2: In the converse of the previous p r o p o s i t i o n , the hypothesis that L be a f i n i t e l y diagonable subspace of A i s e s s e n t i a li a s the following examples show. (i ) Let A = R[y] be the r i n g of polynomials i n y over the algebra R = F(x) of r a t i o n a l functions i n x over a f i e l d F. Here x and y are indeterminates and yx'= (x+l)y. It i s not d i f f i c u l t to show that f o r any r(x) e R, and k k integer k » 0, y r(x) = r(x+k)y . L = Fx i s a diagonable subspace of A, 00 A , where A = Ry f o r a l l u , n n J n=l integers n > 0. C l e a r l y A i s a f i e l d and each A -module A i s one dimen-J o o n 2 s i o n a l . However, A i s not a r t i n i a n (yA3y A z>. . . i s a'decreasing i n f i n i t e 26 chain of r i g h t i d e a l s ) and L i s not f i n i t e l y diagonable. ( i i ) Here we l e t A = R [ y i , y 2 » « - « ] he the algebra of polynomials i n commuting, a l g e b r a i c a l l y independent indeterminates y j , y 2 » . . . over the usual (commutative) r i n g R = F[x] of polynomials i n an indeterminate x over the f i e l d F. Define y^x = (x + i ) y ^ . Then i t i s straightforward to show that ">yi n k f(x) = f(x + ni±i + ... + " k i k ) y i ni...y± \ nl yi . - „ . -1 k " 1 k for any f(x) e R. Moreover, i t follows that L = Fx i s a diagonable subspace CO of A, A = A Q + £ A n, and A = £ " 'y± r for n >. 0. A n i s n=l n ZijS.=n 1 r j J j f i n i t e l y generated as an A0-module and A Q = R i s a p r i n c i p a l i d e a l domain (hence noetherian), so has the ascending chain condition on Ao~submodules (see, f o r example, [2; §1 1 . 1 4 - 1 1 . 1 6 ] ) . However, A i s not noetherian (y^A £ y^A + y^A ... i s a s t r i c t l y increasing sequence of r i g h t i d e a l s of A) and L Is not ' f i n i t e l y diagonable. 27 2.2 NILPOTENT IDEALS AND SEMI-PRIMENESS An i d e a l I ( r i g h t , l e f t , or two-sided) of a r i n g R i s n i l p o t e n t i f f o r some integer n > 0, In = 0, where ln denotes the i d e a l of R c o n s i s t i n g of sums of monomials x^X2«..x n, each x± e I. R i s c a l l e d semi-prime i f (0) i s the only n i l p o t e n t i d e a l . I f V i s an R-module, (0:V) = { x E R: Vx = 0} i s the a n n i h i l a t o r of V i n R. The Jacobson r a d i c a l of R, J ( R ) , i s then defined to be f\{(0:V): V an i r r e d u c i b l e R-module}. It i s a two-sided i d e a l of R containing every 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 of R. R i s s a i d to be semi-simple i f J(R) = 0. Now define a new a s s o c i a t i v e binary operation on R by d e f i n i n g xoy = x + y - xy for any x and y i n R. The p a i r (R,o) i s then a monoid with i d e n t i t y the 0 element of R. I f xoy = 0, we say y i s a r i g h t quasi- inverse 'of x, and x i s r i g h t quasi-regular; x i s a l e f t quasi-inverse of y and y i s l e f t quasi-regular. I f every element of a r i g h t i d e a l i s r i g h t quasi-regular, then every element x of the i d e a l i s quasi-regular i n the sense that x i s both r i g h t and l e f t quasi-regular with unique quasi-i n v e r s e . Assuming that R has an i d e n t i t y 1, t h i s i s equivalent to saying that 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 inverse i n the form 1 - y, the i d e n t i t y (1 - x ) ( l - y) = 1 + x O y shows that y i s the quasi-inverse of x. I t turns out that the Jacobson r a d i c a l can be characterized as a quasi-regular i d e a l (every element i n i t i s quasi-regular) which contains every quasi-regular r i g h t i d e a l . This c h a r a c t e r i z a t i o n w i l l prove u s e f u l i n the sequel. A good discussion of a l l these ideas can be found i n [51. Now throughout t h i s s e c t i o n , assume that L i s a f i n i t e l y diagon-able subspace of the algebra A, so that A a = 0 except for a i n the set 28 A ~ {0,.a^ ,... ,a k}. The reader i s reminded that for any a, 3 i n A, A aAg C A ^ , where A a + g i s 0 i f a + 3 £ A. We f i r s t prove the following key lemma: LEMMA 2.2.1: Given elements Xj^ _,x2 *..-.» xk+i °f A with x.^  e A ^ and e A, i ~ 1, ...jk+1, then there exist integers m and n with 1 < m < n •< k+1 so that ^ . . . X J J E A Q. PROOF: Let b 1 = g p b 2 = ^ + 32» •••» bk+l = 3i + 3 2 + ••• + 3 k + l • I f some bj, £ {aj,... j C i ^ } e i t h e r A i n which event = 0, or b^ = 0. In e i t h e r case x ^ s A 0 and the as s e r t i o n of the lemma i s t r u e . Other-wise, (b^,.. . h ^ i } i s a set of k + 1 elements i n a set of c a r d i n a l i t y k. So there e x i s t d i s t i n c t integers r and n with b r = b n . Assuming r < n , we have 0 = b n - b r = $T+i + ... + 3 n« L e t t i n g m = r + 1, i t follows f- V> ,1 V ^ V o A —m - * ---TI w "o* This enables us to prove 2 LEMMA 2.2.2: Suppose I is a right ideal of A suoh that (I A Q) = 0, then I is nilpotent. PROOF: We show that j . (k + 1) (k + 2) = 0 . For t h i s , l e t x±, i = 1,..., (k+1) (k+2) be (k+1)(k+2) elements of I, and define y^ for i = l,...,k+2 by yi = X( i - D (k+l)+l • • - x i ( k + l ) " E a c n y± e x» a n d s o b y t n e previous r e s u l t , we may write k+2 JY7i = a u l a 3 i u 2 a 3 2 ' - ' u k + l a 3 k + l U k + 2 b ~ where a and b are ei t h e r 1 or i n (possibly d i f f e r e n t ) root spaces Ay, y */> 0, each U j E I A A o , i = l,...k+2, and each 3 i e I A Ag i where for i = l,...,k+l, 3-L e (cx^,..., a^} . Now ag^u2••, aBk+i belongs to Ag^...Ag and so again using 2.2.1, there e x i s t integers m and n with 1 m < n <, k+1 such that a3 m Um+l*-- a3n b e l o n S s t o 1 A V Thus V ^ u ^ .. . a ^ e (I ft A Q ) 2 = 0 and (k+1)(k+2) k+2 IT x = TT-y = o. j=l J i=l 1 (k+1)(k+2) Since I consists of sums of products of (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 of t h i s s e c t i o n now follows. THEOREM 2.2.3: If A is semi-prime, so is A Q. Conversely if A Q is semi-prime, then the nilpotent right ideals of A are exactly those contained in \ A . 0#XEA A T»T»r\/-\T«, . ... - , _ A J ! - . 1 — „ . 1 T J . .. .• 1 -„ • 1 t t~ . ' 1 "f 1^ N r ivuur : c i i i i i p u o c zv xS Semx—p i. xiue &uu A XS a. L i g i i L xufciij. O i *v WALU 1 — \j. By an easy induction argument, we can assume n = 2. Now IA i s a r i g h t i d e a l of A contained i n I 0 £ A^. Also 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 previous 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 of A, then I]_ = I + AI k k k 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^ I + AI for any p o s i t i v e integer k. If I-^ = 0, (1^ (\ ^ Q ) t = 0, so 1^ f\ A Q = 0 and 1^ S. 1 A Q by homogeneity (1.3.4). I I j so I i s contained i n £ AQ O^OEA 0/cteA too. F i n a l l y , we note that any i d e a l contained i n £ Aa i s n i l p o t e n t O AXEA by 2.2.2 since I(\ A Q = 0. COROLLARY 2.2.4: If AQ is semi-simple, then J(A) is the sum of all right ideals I of A contained in \ k^.and hence is nilpotent. 0&tEA PROOF: Let T denote the sum defined here. C e r t a i n l y T c J(A) because J(A) contains a l l n i l p o t e n t i d e a l s of A. We prove that the reverse i n c l u -sion holds by e s t a b l i s h i n g (1) J ( A ) 0 A o C J ( A o ) To see t h i s , we use the fa c t that the Jacobson r a d i c a l of a r i n g with 1 can also be characterized as the i n t e r s e c t i o n of a l l maximal r i g h t i d e a l s ([4; page 11]). But every maximal r i g h t i d e a l of A Q i s contained i n a maximal r i g h t i d e a l of A by 1.3.5.. Before continuing, we remark that once again the underlying co n s t r a i n t that L be f i n i t e l y diagonable i s e s s e n t i a l , for otherwise, we obtain a counter-example by taking A = F[x]<y>, the r i n g of formal power s e r i e s i n y over the polynomial r i n g F [ x ] , where x and y are i n -determinates, and yx = (x+l)y. A Q = F[x] i s semi-simple, but J(A) i s not n i l p o t e n t ; indeed, i t contains "the quasi-regular non-nilpotent i d e a l CO B = { 1 f-jOOy1": fj[(x) £ F[x]}. That B i s quasi-regular can be seen i = l co ± by noting that for a given element £ f 1 ( x ) y of B, the equation i = l CO CO (1 - I f i ( x ) y1) ( l + I g i ( x ) y1) = 1 i= l i = l can be solved i n d u c t i v e l y f o r the g^(x) e F[x] , i = 1,2,... We close t h i s s e c t i o n with THEOREM 2.2.5: If every A-module is L-weighted, then J(A Q) = A Q 0 1(A) and hence AQ is semi-simple i f A is. PROOF: The i n c l u s i o n (1) i s always v a l i d , so we need prove only that J ( A 0 ) £ J(A) 0 A Q. Let V be an i r r e d u c i b l e A-module. Then V = © I XeA 3 1 r e l a t i v e to the set A of weights of L i n V. Now suppose v a Q = 0 with a Q e A and v = J v x e V . Then each v^aQ = 0 because V^A S E V^+Q ( 1 . 2 . 6 ) . Thus, XEA we obtain ( 2 ) (Q:V)r\ A Q = (\ ( 0 : V X ) 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 of an A0-module M. Let I (resp t i v e l y J ) denote the c l a s s of a l l i r r e d u c i b l e r i g h t A- ( r e s p e c t i v e l y A Q - ) modules. Then ( 2 ) implies ( H ( 0 : V » H A D = f\ ((0:V) C\ A Q) - H f \ ( 0 : V x ) o 3 f \ (0:V) o V E I V E J V E J XEA V e I o the l a s t i n c l u s i o n following because each i s an i r r e d u c i b l e A0~module by Lemma 1.2.8. Thus J ( A ) T \ A 0 2 J ( A 0 ) as r e q u i r e d . 32 2.3 PRIMITIVITY An i d e a l P of a r i n g R i s p r i m i t i v e i f i t i s the l a r g e s t (two-sided) i d e a l of R contained i n a maximal r i g h t i d e a l of R. P r i m i t i v e i d e a l s are exactly those which are a n n i h i l a t o r s of i r r e d u c i b l e R-modules. A p r i - mitive r i n g i s one i n which (0) i s a p r i m i t i v e i d e a l ; i . e . , there i s a maximal r i g h t i d e a l containing 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 , there 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 that i t s a n n i h i l a t o r i s 0. Suppose now that the algebra A i s p r i m i t i v e with f a i t h f u l i r r e -ducible module V. Taking a closer look at (2) of the previous s e c t i o n , we see that we have a c o l l e c t i o n of i r r e d u c i b l e A0~modules {Vx: X e A } with f\ (0:V X) = 0. Now P x = ( 0 : V X ) Q i s a p r i m i t i v e i d e a l of A Q XeA ° containing x - X(x)l f o r every x e L as the discussion a f t e r Lemma 1.2.9 showed. Moreover, f o r X,u e A and X £ u, there i s an x e L for which X(x) 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 . This implies P^ + P^ = A Q. Assuming L i s f i n i t e l y diagonable, we know by 1.4.3 that A i s f i n i t e , and so we can apply the Chinese Remainder Theorem to obtain A Q - © £ A0/P. . The quotient of any r i n g by a p r i m i t i v e i d e a l XeA i s 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 of p r i -mitive algebras. F i n a l l y we notice that Propositions 1.2.1 and 1.2.11 enable us to extend our r e s u l t to d i r e c t sums, thus g i v i n g THEOREM 2.3.1: Suppose L is a finitely diagonable subspace of an algebra A which is a direct sum of primitive algebras and. further suppose that every irreducible k-module is weighted. Then kQ is a direct sum of primitive algebras. 33 As a p a r t i a l converse, we obtain: THEOREM 2.3.2: If I, is a finitely diagonable subspace of A, and kQ is a direct sum of primitive algebras, then A/J(A) can be embedded in a direct sum of primitive algebras. Assuming also that A is semi-prime, A itself is so embeddable. PROOF: A q i s n e c e s s a r i l y a f i n i t e d i r e c t sum of p r i m i t i v e algebras R j , j = l , . . . , n because i t contains 1. Let I j be a maximal r i g h t i d e a l of contain no R. containing no non-zero two-sided i d e a l . Then ][ I. can 3 j=l 3 i d e a l of A Q except (0), because such an i d e a l T would decompose as n © I Tft R. (for 1 e A ) , a n d T f ( R . g L . For each j , l e t j=l 3 ° 3 J i = l 1 i * j "Then Tj i s a maximal r i g h t i d e a l of "Ag, "and so contained i n a maximal r i g h t i d e a l T^* of A such that AQ = Tj by Lemma 1.3.5. Then Pj = {a e A: Aa S= ^j*} i s contained i n T^* (since 1 e A) and i s a p r i -m itive i d e a l because i t i s the a n n i h i l a t o r of the i r r e d u c i b l e A-module n n n n A/T,*. Notice that ( Q ft (\ (T.* ft A ) = (\ T. £ Ii, 3 j=l j=l j j=l J j=l J n and so, 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 stated above. Thus f\ P. £ A a by homogeneity and hence j=l J 0/aeA contained In J(A) by 2.2.4. (Note that a p r i m i t i v e r i n g i s semi-simple). The reverse i n c l u s i o n always holds (see for example [5; page 40]) so we n n have that J(A) = f | P . C l e a r l y A/J(A) can be embedded i n 0 £ A/P.. j=l j j=l 3 Since 2.2.4 implies J(A) i s n i l p o t e n t , the l a s t statement of the theorem i s obvious. COROLLARY 2.3.3: Under the conditions of the theorem, if AQ is primitive, then A/J(A) is primitive. PROOF: I f I i s a maximal r i g h t i d e a l of A Q containing no non-zero two-sided i d e a l of A D, and I* i s the maximal r i g h t i d e a l of A generated by I, then P H A Q = 0, where P i s the la r g e s t i d e a l of A contained i n I*. To see t h i s , we simply note that P fi A Q £ I* f\ A Q = I. Thus P = J(A) as i n the theorem. Since P i s a p r i m i t i v e i d e a l , the conclusion f o l l o w s . 35 CHAPTER THREE DIAGONABLE ELEMENTS AND WEIGHTED MODULES 3-1 IDEMPOTENTS ARE DIAGONABLE In t h i s chapter we give examples of diagonable elements and look f or conditions on an algebra which allow something to be said about the nature of i t s diagonable elements. They often turn out to be a l g e b r a i c . The r e s u l t s of Chapter Two w i l l be of great help i n gi v i n g information both about a diagonable element x and i t s c e n t r a l i z e r because of course A Q(x) = A Q ( F x ) . When employing r e s u l t s from Chapter Two, the diagonable subspace L i s always understood to be Fx. A module i s sai d to be x-weighted i f i t i s Fx-weighted and we w i l l r e f e r to "weights of x" and "roots of x" rather than weights and roots of Fx. A l s o s since any weight or root Is 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 completely determined by i t s value at x. We w i l l i d e n t i f y a root a with the sca l a r ct(x) so that the "roots of x" w i l l a c t u a l l y be the roots of the minimum polynomial of ad x i n the event a l l but f i n i t e l y many root spaces are zero. S i m i l a r l y , the weights X will be i d e n t i f i e d with the sc a l a r s X(x). Any algebra containing idempotents abounds i n diagonable elements, for we have: n THEOREM 3.1.1: An element x = \ a i e i with a j . , . . . j C i n E F is a linear i = l combination of orthogonal idempotents e]_,...,e with sum 1 i f and only i f x is algebraic with minimal polynomial having distinct roots in F. Such an element is diagonable, and A (x) = T e.Ae.. 36 n PROOF: F i r s t assume x = ][ ct.e as above. Let S be any subset of {l,...,n} i = l 1 w i t h the property that {ct^: i e S} i s the set of d i s t i n c t c o e f f i c i e n t s of n e]_,...,e appearing i n Y ct^e . We prove that x has the minimal polynomial i = l 1 (1) "[J ( t - a.) e F [ t ] . i e S 1 For t h i s i t i s s u f f i c i e n t to assume th a t S - {l,...,n} f o r i f a j » . . . , a are not 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 ^ = £ e j . Then x = J a±^±> { f ^ : i E S) are orthogonal idempotents w i t h sum 1, and {a^: i E S} i s a n d i s t i n c t set o f s c a l a r s . Now l e t f ( t ) = "P|~(t - a j . Since 1-1 n n x - ct.il = x - ct.j( £ e.) = I (a± - ocJe. - J i = l i = l J i t i s c l e a r that f ( x ) =0. To see that f i s a c t u a l l y the minimal p o l y n o m i a l , n ± note that i f g ( t ) = ^ a.t s F [ t ] , then 1=1 n n g(x) = I a±( I ) = £ ( j a ict. i)e i = l j = l J J j = l i = l J and so g(x) = 0 i f and only i f cq,...,ct n are roots of g (because e j , . . . , e n are orthogonal idempotents). Since ,ctn are d i s t i n c t , degree g = m i s at l e a s t as b i g as n = degree f . Conversely, i f x i s a l g e b r a i c w i t h minimal n polynomial 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 ) are r e l a t i v e l y prime elements of the Euclidean domain F [ t ] , so there 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 ei = a i ( x ) h i ( x ) , f o r i = l n i = l , . . . , n . Then e^,...,e are orthogonal idempotents w i t h sum 1, x = £ xe i"=l 1 and xe^ = (x - a ^ l ) e ^ + a^e^ = a^e^. F i n a l l y , to see that an element n x = J alei x S diagonable, we note that i f a e A, i = l n n n a 1=1 1 j = l 3 i , j = l J 37 and (e^ae^.x) = (oij - ct^e^ae^. The theorem now f o l l o w s . In the above proof, we notice that i f x i s an algebraic diagon-able element whose minimal polynomial has d i s t i n c t roots i n F, then L = Fx i s a f i n i t e l y dlagonable subspace, because the root spaces are of the form A a_g, where a and 6 are roots of the minimal polynomial. This r e s u l t generalizes considerably. THEOREM 3.1.2: Any algebraic diagonable element x in an algebra A over a field F generates a finitely diagonable subspace of A. Root spaces are of the form A a_ g where a and 3 are roots (not necessarily in F) of p(t)3 the minimal polynomial of x. PROOF: I f x i s i n the centre of A there i s nothing to prove, so assume some root space A^(x) i s non-zero for y r 0, Considering for a moment -A. -as an -algebra over -F, - the "algebraic cl-os-ure -of F, •the- sub space A^C-x) i s i n v a r i a n t under the l i n e a r transformation Rx. Now Rx i s a l g e b r a i c , a l s o with minimal polynomial p(t) since 1 e A. Thus, since A^(x)p(R x) = 0 and p s p l i t s i n t o l i n e a r factors over F, there e x i s t s some 8 e F and a^ 0 i n A^(x) such that ay(x - 6 1 )° = 0, for some integer n > 0. Now ayX = (x + y l )ay imp Hes a^f(x) = f(x+y)a y for any f ( t ) e F [ t ] and so (x - ( 8 - y ) l )na y = 0. But t h i s says that a = 6 - y i s a root of the m i n i -mal polynomial of L, , which i s also p ( t ) . Hence y = 8 - o i s a d i f f e r e n c e of roots of p(t) and we have the theorem. Over a f i e l d of c h a r a c t e r i s t i c 0, i t i s well-known that any i r r e d u c i b l e polynomial has d i s t i n c t r o o t s . This r e s u l t extends as we now see. LEMMA 3.1.3: If p(t) e F[t] is irreducible and of degree at least 2, then a - $ t F for any distinct roots a 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. In particular, with T = a - i , we obtain a - ia + aa = 2a. Applying a to both sides of.this 2 3 relation gives a + a a = 2aa. Another application of a gives aa + a a = 2 "\ 2a a = 2(2aa - a), and so 2a + a a = 3aa. An easy induction argument t+1 k reveals that for any integer t > 0, ta + a a - (t+l)a . Now a is the identity of G for some k. 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 diagonable element whose minimal polynomial is irreducible. Then x is in the centre of A. We next give: PROPOSITION 3.1.5: Let x be an algebraic diagonable element with minimal n polynomial p(t) = ~\\(t - a.) e F[t]. Then every non-zero A-module is 1-1 x-weighted, every non-zero AQ-module is x-weighted, and weights of x are in the set {aj,...,o }. (Note: we do not assume here that a^,...,an are all distinct.) PROOF: Let V be an A-module different from zero, and let O^ v e V. Then n either v(x - ajl) = 0 or, since T"]~(x - a . l ) =0, there is some integer k, k-1 i=l k i l 1 < k < n with v "\J (x - a.l) $ 0 but (v 7T(X ~ ct 4l))(x - a kl) = 0, and i=l i-1 • A 39 so V i s x-weighted. The proof that A Q-modules are x-weighted i s i d e n t i c a l . Next l e t y be any weight of x i n an A - (or A - ) module V. Thus there i s some v 5* 0 i n V with v(x - yl) = 0. Now i f y £ {aj, , .. • , a n ) , then the p o l y -nomials t - y a nd p(t) i n F[t] are r e l a t i v e l y prime, and so there are p o l y -nomials a(t) and b(t) i n F[t] with a ( t ) p ( t ) + b ( t ) ( t - y) = 1 . Setting t = x gives b(x)(x - yl) - 1 and hence v = v(x - Y ! ) M X ) = 0, which i s untrue. The converse of Theorem 3.1.2 i s generally f a l s e ; for example, i f A i s the algebra generated over F by two elements x and y with x trans-2 cendental, y = 0 , and yx = (x+l)y, then Fx i s a f i n i t e l y diagonable subspace of A (the only roots of x are 0 and 1) but of course x i s not a l g e b r a i c . The best r e s u l t we could obtain i n t h i s d i r e c t i o n now follows: THEOREM 3.1.6: Suppose L = Fx is a finitely diagonable subspace of an •algebra- -A -which is ,a^direct-.sw^.--of.>primi-tix)e^.algebras,. Suppose further that every irreducible A-module and every irreducible AQ-module is x-weighted. Then x is algebraic. PROOF: By Theorem 2.3.1, A Q i t s e l f i s a d i r e c t sum of p r i m i t i v e algebras R^, i = l , . . . , n , only f i n i t e l y many because 1 e A q . I f I i s any maximal r i g h t i d e a l of A Q , then A Q / I i s weighted, so there i s some a D t I and X e F with a D ( x - X l )n e I, and t h i s implies x - Xl E I by 1.2.9. Hence, l e t t i n g be a maximal r i g h t i d e a l of R^ containing no non-zero i d e a l of R^, i = 1 n, f o r each i , ® \ R. i s a maximal r i g h t i d e a l of A D and so j f i n n contains x - X. 1 for some X.» e F. Writing x = \ x., x. e R J , and 1 = £ e., i= l i = l e^ the i d e n t i t y of Ri, we have x i - X^e^ e by looking at the it n com-ponent-of x - X ^ l . Now x^ - X^e^ i s i n the centre of R^ because x - Xl i s i n the centre of A Q and so (x^ - X^e^R^ i s an i d e a l of R^ contained i n J ^ . Therefore, x^ = X^e^ and since e^ e n are orthogonal idempotents, x i s algebraic by Theorem 3.1.1. 40 3.2 PRIME ALGEBRAS A r i n g R i s prime i f aRb ? 0 whenever a and b are non-zero elements of R; o r , e q u i v a l e n t l y , i f the product of non-zero r i g h t i d e a l s of R i s non-zero. In 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 also a fact that any p r i m i t i v e r i n g i s prime ([5; page 951)-In general one can not expect the i r r e d u c i b l e factors of the minimal polynomial of a diagonable algebraic element to be d i s t i n c t ; f o r example, l e t R by the quotient of the usual polynomial r i n g F[x] by the 3 2 p r i n c i p a l i d e a l generated by x - x and l e t A = R[y] be the polynomial r i n g over R i n the indeterminate y, where yx = (x+l)y. A straight-forward computation reveals that A i s j u s t the four-dimensional algebra over F with 2 basis 1, x, x , y and m u l t i p l i c a t i o n table: l X X y 1 l X x 2 y X X 2 X x2 0 x2 X2 2 X x2 0 y y y y 0 x i s a diagonable algebraic element whose minimal polynomial p(t) = tJ - t has t as a repeated f a c t o r . I f A i s semi-prime, t h i s cannot happen, for we have: PROPOSITION 3.2.1: If x is a diagonable algebraic element in a semi-prime algebra A over the field F, then the minimal polynomial of x has distinct irreducible factors. ' . • PROOF: L = Fx i s a f i n i t e l y diagonable subspace of A and so by Theorem 2.2.3 r A (x) i s semi-prime. Suppose the minimal polynomial of x i s p(t) = TTp. (t) 41 where p j ( t ) , . . . , p (t) are the d i s t i n c t i r r e d u c i b l e factors of p(t) i n F [ t ] . r Then i f any 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 A Q(x) and hence generates a non-zero n i l p o t e n t i d e a l , oand t h i s i s impossible. There are occasions when a l l the i r r e d u c i b l e modules of an algebra are weighted; for example, when x i s a diagonable algebraic element whose minimal polynomial has a l l i t s roots i n the ground f i e l d (Proposition 3.1.5). In [15] Lemire gives an example to show that t h i s need not always be the case. He considers the u n i v e r s a l enveloping algebra £/(Aj) of the three-dimensional simple L i e algebra A^ ( [ 9 ; page 137]) where the diagonable element i s the generator H of a Cartan subalgebra of A^. R e l a t i v e to H, there are i n f i n i t e l y many root spaces, for otherwise, £/(Aj) would contain n i l p o t e n t elements (because- (Aa(H))ncr^A^C-H)-) ,- and 'this i s known -to be f a l s e ( [ 9 ; page 166]). Lemire constructs an i r r e d u c i b l e £/(Aj)-module which i s not H-weighted. We further pursue i n t h i s s e c t i o n the i n t e r -r e l a t i o n s between an algebra A, a diagonable element x i n A, and the pro-perty that the i r r e d u c i b l e modules of A should be x-weighted. EXAMPLE 3.2.2: We give here an example s i m i l a r to Lemire's. Let A be the u n i v e r s a l enveloping algebra of the two-dimensional non-abelian L i e algebra over a f i e l d F (of c h a r a c t e r i s t i c 0 ) . This has generators x and y with the r e l a t i o n [y,x] = y. A i s i n f a c t the r i n g of non-commuting p o l y -nomials over F i n the two indeterminates x and y, where yx = (x+l)y. Now x i s a diagonable element i n A, and there are i n f i n i t e l y many r o o t s , be-00 cause A = © £ A n ( x ) , A n(x) = F [ x ] yn. Since x i s not i n v e r t i b l e , there n=0 e x i s t s a maximal r i g h t i d e a l I of A which contains x, and so the A-module V = A/I i s x-weighted (for (I + l ) x = 0 ) . We prove however that not a l l i r r e d u c i b l e A-modules are weighted. y + 1 i s not i n v e r t i b l e i n A and so i s contained i n a maximal r i g h t i d e a l J which has the property (1) f ( x ) y k e J for k > 0 and f(x) e F[x] implies f = 0. To see t h i s , we use induction on n = degree f and -k. For n = 0, i f ay and y + 1 are both i n J , then J contains 1 because these two polynomials are r e l a t i v e l y prime i n the Euclidean domain F [ y ] , hence c e r t a i n l y i n A. Assuming the v a l i d i t y of (1) for polynomials f with degree l e s s than 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 contain (cf~ \.-k ..n\ _ c /•-.%.. k+1 ..r/.-N.-k _ fcf..\ c1,1 \ \ ..k+1 Since f(x) - f(x+l) has degree les s than the degree of f , by induction we have f(x) = f ( x + i ) . This implies 5 + n i s a root of f for every integer n > 1 and any root £ of f and t h i s i s impossible unless degree f = 0. This p o s s i b i l i t y has already been eliminated above and so we have obtained (1). It follows immediately that J can contain no two-sided i d e a l of A except (0) because any such i d e a l i s homogeneous by 1.3.4. Therefore A/J i s a simple, f a i t h f u l A-module ( i t s a n n i h i l a t o r i s a two-sided i d e a l of J) and so A i s a p r i m i t i v e algebra. A/J i s not weighted however, for we can prove n n (2) ( I a.(x)y 1)(x - a l ) e J implies £ a i ( x ) y i e J . i=0 i=0 For n = 0, a D ( x ) ( x - a l ) e J implies a Q(x) = 0 or x = a l by (1), and the l a t t e r p o s s i b i l i t y i s impossible. Now i f J contains 43 n i ( I a i ( x ) y 1 ) ( x - al ) = £ y ( x - i l ) ( x - a l ) 1=0 i=0 then i t also contains n ( y + Dna n ( x ) ( x - « D - I ? y ia (x) (x - a D i=0 because y + 1 e J , and hence J contains the di f f e r e n c e of these two element's; namely, "fy^-Ca^Cx-il) - ^Ja n<x)Mx - a l ) = Y(a.(x) -.^].an.(x+il))yi(x - al). By i n d u c t i o n , we may assume that a^(x) = "^j a n ( x + i l ) ; i . e . , that a ^ x ) = ^ a Q ( x + i l ) , f o r i = 0, . . . , n - l . Thus I a i ( x ) y i - I (J) a 0 ( x + l l ) y i - J ( ^ ^ ( x ) = (l + y ) n a 0 ( x ) i=0 i-OW ° i=0V ' i s i n J . Hence we have (2), and A/J i s not weighted. Thus we see that even for a p r i m i t i v e algebra, one i s able to say l i t t l e about which modules are weighted. We can however prove: THEOREM 3.2.3: Let A be a prime algebra over the field F and x e A be_ a-diagonable algebraic element with minimum polynomial p(t) e F [ t ] . Suppose some irreducible k-module is x-weighted. Then all k-modules are x-weighted. PROOF: The key step i n the proof i s to show that p(t) has the form (3) 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 set of roots of x. By P r o p o s i t i o n 3.2.1, p(t) i s of 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 tors of p(t) i n F [ t ] . Let A^j denote the subspace {a e A: p^(x)a = 0 = apj(x)} f o r 1 ^  i , j < s. Then we have: s (4) A = ® I A±1 i , j = l To prove t h i s , we note that the l i n e a r transformation of A i s algebra i c s with minimal polvnomial also p(t) because 1 e A. Thus A = A. where i=l 1 A^ = {a E A: p^(x)a = 0 } . Herstein proves t h i s i n [4; page 256] for A fi n i t e - d i m e n s i o n a l , but the proof uses only the fact that the l i n e a r t r a n s -formation i n question i s a l g e b r a i c . Now each subspace Aj_ i s i n v a r i a n t under Rx which i s also an algebraic l i n e a r transformation of A with m i n i -mal polynomial p ( t ) . Thus the r e s t r i c t i o n of Rx to A^ has minimal p o l y -nomial d i v i d i n g p(t) and so each A^ decomposes in t o a d i r e c t sum of the subspaces A ^ for some integers j , 1 ^  j ^ s. Now f o r each k, 1 ^  k < s, define a k = "]"J~ p - ( x ) . Then a, ^ 0 i/k and for any i , j E { l , . . . , s } , a^Aa^ 5* 0 (because A i s prime). But a^Aa^ i s contained i n A^j and so each of the spaces A^j ^  0 . Suppose 0 ^  a E A^ .and a " I » a a -relative...to x.. ,. ..Then - 0 = ap^(x.) = £ -a-p^Cx). -For some a, OCEA CCEA a„ 0 and because A a ( x ) A Q ( x ) S A a(x) and the sum £ A a(x) i s d i r e c t , CXEA a ap^(x) => 0 . As we have seen before, t h i s implies p^(x+al)a a = 0 , We also have p^(x)a = 0 and hence, 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 prime, and so because they are i r r e d u c i b l e and monic, P j (c) = P i ( t + a ) . This establishes (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 hypothesis, and so by Theorem 1.2.10 (with L = Fx) there i s a s c a l a r X E F with x - XI not i n v e r t i b l e . This means that the polynomials p(t) and t - X are not r e l a -t i v e l y prime and so t - X divides p ( t ) . It follows that 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 roots i n F. The theorem now follows d i r e c t l y from Pr o p o s i t i o n 3.1.5. Looking back at the form of the minimal polynomial p(t) i n (3) we see that i n p a r t i c u l a r , i f x i s i n the centre of A, S = A = {0}. Together with 3.1.4, we obtain immediately COROLLARY 3.2.4: Let x be an algebraic diagonable element in a prime algebra A. Then x is in the centre of A i f and only i f i t s minimal polynomial is irreducible. 46 3 . 3 CENTRAL SIMPLE ALGEBRAS As might be expected, an element in an algebra A over a f i e l d F which is not diagonable, can become diagonable when A i s considered to be an algebra over the algebraic closure F of F. For example, in the algebra of 2 x 2 matrices over the real numbers, x =^ is not diagonable be-cause AQ(x) = { a ^ ^ + b | j j|: a,b real}, while AQ(x) = 0 for a / 0 . Over the complex numbers, however, the minimum polynomial of x has the distinct roots ± i and so x is diagonable by Theorem 3 . 1 . 1 . More generally, i f x i s a diagonable element in the complete ring A of matrices over any algebraically closed f i e l d F, then since A is semi-prime, the minimal poly-nomial of x, whose roots of course l i e in F, has distinct roots by 3 . 2 . 1 and x is similar to a diagonal matrix as i s well-known. This curious observation that diagonability in our sense is equivalent to the usual meaning of the word in a matrix ring does not depend on the algebraic closure of the underlying base f i e l d . The proof of this i s the principal aim of this section. LEMMA 3 . 3 . 1 : Let A be any associative ring with 1 and e e A an idem* •potent. Then i f A is simple, so is the subring A^(e) = {aeA: ae = ea = a}. PROOF: Relative to e, we have the Pierce decomposition of A, A = A,Q0 + Aio + Aoi + A n , = {aeA: ae = j a , ea = ia} . ( [8 ; page 4 8 ] ) . Suppose A is simple and I is a non-zero ideal of A n = A^(e). Then I + A Q J I + I A J Q + A Q I I A I Q i s a non-zero ideal of A and hence equals A. Writing 1 as a sum of elements from I , A 0 1 l , I A 1 0 , and A Q J I A J Q 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 right, we deduce that e e l . But then, i f 47 a e (e) , a = ae e I and so I = Aj (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 the sum of two orthogonal idempotents. An al g e b r a over F i s s a i d to be c e n t r a l over F i f i t s centre i s F. We r e q u i r e the f o l l o w i n g r e s u l t , a proof f o r which can be found i n s e c t i o n s 3.6 and 3.7 of [12] . • LEMMA 3.3.2: Let R be an artinian ring with 1. Then any idempotent can be written as a sum of primitive orthogonal idempotents. If 1 can be represented in two ways as a sum of primitive orthogonal idempotents, n m 1 = ^ ej _ = \ 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 "es t a b l i s h LEMMA 3.3.3: Let A be a f i n i t e dimensional central simple algebra over a f i e l d F and suppose e e A is an idempotent. Then A^(e) is also central simple over F. PROOF: By Lemma 3.3.1, we need only show that A^(e) i s c e n t r a l over F. Using the previous lemma, i t i s easy to show that we can f i n d p a i r w i s e n orthogonal p r i m i t i v e idempotents e^ e n i n A such that ^ = 1 and t i = l e = £ e, f o r some t < n. Now A = D , the r i n g of m x m matrices over 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 the Wedderburn Theorem ( [ 6 ; page 9 8 ] ) . Let { f ^ ^ : i , j = 1,...,m} be the usual m a t r i x u n i t s i n A. j j i i ' 'mm " i j m Then 1 = £ f j j and f ^ , . . . ^ ^ are orthogonal p r i m i t i v e idempotents. By 3.3.2, m = n and there i s a u n i t v e A and a permutation TT of {l,...,n} 48 t such that v *e.v = f Let B = v *Ai (e)v and f = Y f ... ... . i 7 r ( i ) T r ( i ) l v £j i r ( i ) T r ( i ) Then one can check e a s i l y that B = A^(f) - Dfc which i s c e n t r a l over F because D i s . Since B = A^(e), the lemma fo l l o w s . Our main r e s u l t i s THEOREM 3.3.4: Suppose A is a finite dimensional central simple algebra over the field F. Then x e A is diagonable if and only if x is a linear combination of orthogonal idempotents. In particular, if A is the complete ring of matrices over F, then any diagonable element is similar to a diagonal matrix. PROOF: Since A i s semi-prime, the minimal polynomial of x has the form q ^ ( t ) . . . q n ( t ) , 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 polynomials i n n n F t t ] (by 3.2.1) ; We can write 1 = [ e^ , and x = £ xe , where e-.,...,e ±=1 i=i are pairwise orthogonal idempotents which are polynomials i n x j u s t as i n H e r s t e i n , [4; page 256]. Moreover, x^ = xe^ has the minimum polynomial q ^ ( t ) , i = l , . . . , n because f(x^) = f ( x ) e ^ for any f ( t ) e F [ t ] . Let B^ = A ^ e ^ f o r i = l , . . . , n . Notice that x^ c B^ and B^ i s a subalgebra, so B^ i s i n -va r i a n t under ad x. The r e s t r i c t i o n of ad x to B^ i s ad x^ and by 1.2.2(i) t h i s r e s t r i c t i o n s a t i s f i e s a polynomial with d i s t i n c t roots i n F; i . e . , x^ i s diagonable i n B^, which i s c e n t r a l simple over F by 3.3.3. By 3.1.4, x^ i s i n the centre of B^; hence, for some a. e F, x^ = a^e^. Therefore n x = Y a.e.. The l a s t statement of the theorem now follows from Theorem 3.1.1 i=l 1 1 and l i n e a r algebra. COROLLARY 3.3.5: If A is a direct sum of finite dimensional central simple algebras over a field F, then any diagonable element in A is a linear combination of pairwise orthogonal idempotents with sum 1, and i t s centralizer is semi-simple. PROOF: That a diagonable element x i s a l i n e a r combination of orthogonal idempotents follows d i r e c t l y from the Theorem together with Proposition 1.2 x i s therefore algebraic and so Fx i s a f i n i t e l y diagonable subspace of A. Proposition 3.1.5 implies every A-module i s x-weighted and so A Q(x) i s semi-simple by Theorem. 2.2.,5-., In c onclusion, we remark that the assumption of f i n i t e - d i m e n s i o n -a l i t y i n Theorem 3.3.4 i s c r u c i a l . I f F(x) denotes the r i n g of r a t i o n a l functions over F i n an indeterminate x, and we adjoin an indeterminate y so that yx = (x+l)y, then the r e s u l t i n g algebra of non-commuting power s e r i i n y over F(x) i s c e n t r a l simple ( i n fact i t i s a d i v i s i o n algebra as Jacobson proves i n [8 ; pages 187-188]), x i s diagonable, but c e r t a i n l y not a l i n e a r combination of idempotents. 50 CHAPTER FOUR APPLICATIONS TO THE UNIVERSAL ENVELOPING ALGEBRA OF A JORDAN ALGEBRA 4.1 THE UNIVERSAL ENVELOPING ALGEBRA OF A JORDAN ALGEBRA Let. J be a Jordan algebra over a f i e l d F of c h a r a c t e r i s t i c 0. Then a representation of J i s defined to be a l i n e a r map S: J A where A i s an a s s o c i a t i v e algebra over F, such that f or every a, b,i and c i n J , we have <l> SaSb c + SbSc a + ScSa b = S j , t S a + S c a S b + S ^ and (2) s as bs c + s cs bs a + s a c > b = s as b c + s bs c a + s cs a b where S: u •* S u for u E J . As i n the L i e case, the map x -* Rx of J i n t o the a s s o c i a t i v e algebra generated 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 of a representation of J . An a s s o c i a t i v e algebra U(J) with i d e n t i t y i s a un i v e r s a l enveloping algebra f or J i f there i s a canonical representation S*: J i/(J) such that f or any representation S: J A of J i n an a s s o c i a t i v e algebra A, there e x i s t s a unique homomorphism Y: U(J) •* A such that S = foS*; i . e . , which makes the following diagram commutative: VU) J .—^A S For any p o s i t i v e integer n, l e t j " denote the linear'span of a l l products of h elements of J . Define J ^n^ i n d u c t i v e l y by J ^ = J , and j(n) = ^ j ( n - l ) j 2 f o r n > ^ T h e n j i g c a l l e d s o l v a b l e l f j(n> = o for some n. I f J i s f i n i t e - d i m e n s i o n a l , i t contains a maximal solvable i d e a l c a l l e d the r a d i c a l of 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 of D i e u d o n n £ , any finite-dimensional semi-simple Jordan algebra i s a d i r e c t sum of simple Jordan algebras. We w i l l require several facts about the un i v e r s a l enveloping algebra K(J) of a Jordan algebra J . These can be found i n Jacobson [7 ] , and are summarized i n THEOREM 4.1.1: U(J) exists, is generated by { S * A : aeJ}., and is unique up to isomorphism. If J is finite-dimensional then so is U(J); if.in addition J is semi-simple, then so also is U(J). We w i l l also make extensive use of the following rather t e c h n i c a l r e s u l t . PROPOSITION 4.1.2: Let e be an idempotent in a Jordan algebra J over a field F. Then letting a *+ a denote the canonical embedding of J in U(J), e is a linear combination of orthogonal idempotents and is diagonable with roots in the set {0,h,~h, l . » - l } . If J = J Q + Ji^ + with J ^ = {a e J : ae = ia} for i = 0,%,l is the Peirce decomposition of J relative to e , then J o + J ^ A Q(e) and I ^ c A ^ e ) + A ^ C e ) , A = i/(J) PROOF: Se t t i n g a = b = c = e i n (2) gives immediately that 2e3 - 3e"2 + e" = 0. — 3 2 Thus e i s algebraic with minimal polynomial d i v i d i n g f ( t ) = 2t - 3t + t , a polynomial x^ith the d i s t i n c t roots 0, Jj, and 1. Thus e i s a l i n e a r com-bi n a t i o n of orthogonal idempotents and diagonable by Theorem 3.1.1. I t s roots are i n {0, 1, -1} by 3.1.2. Next, i f u z J Q , put a = u , b = c = e 52 i n (1). Then ue = eu so u e A Q ( e ) . S i m i l a r l y i f u e J^t putting a = u, b = c = e i n (1) gives ue + 2eu = eu + 2ue and again u e A Q ( e ) . F i n a l l y , assuming u e J i , equation (2) with a = c = e and b = u implies 2eue + hu -• eu + ue and with a = u, b = c = e implies ue^ + e^u + -su = ue + eu. Therefore, ue2 + e2u + ku = 2eue + %u; i . e . , ((u,e),e) = ku. Now write u - u Q + ui^ + u_jg + u^ + u_i with u a e A a ( e ) . Then we see that kui^ + ku_^ + u^ + = \u and hence u D = u^ = u^j = 0 and u e Ai^e) + A_ig(e) . Now suppose a and b are two elements i n a Jordan algebra J . We define the l i n e a r transformation Ra ^ of J by uRa ^ = (u,a,b) for any u e J . J i s associator nilpotent i f there i s a p o s i t i v e integer k such that R„ , ...R0 v, = 0 for a l l a<, b. E J , i = l , . . . , k . A Cartan subalgebra of J i s an associator n i l p o t e n t subalgebra H containing 1 which has the property that i f ( x , H j H ) £ ; H for any x e J then x e H. In [10; page 601], Jacobson shows that 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 closed f i e l d of c h a r a c t e r i s t i c 0, then J possesses t a Cartan subalgebra H = ][ J J J , where J = £ J . . i s the Peirce decomposi-• • • i-1 i , j " t i o n of J r e l a t i v e to a set of pairwise orthogonal idempotents with sum 1 which are also p r i m i t i v e . I f J i s simple, Albert ([1; page 561]) has shown t _ that J . . = Fe. and so H = £ Fe. . In t h i s case, l e t I, = H, the image of H i=l i n [/(J) under the canonical embedding. By Proposition 4.1.2, L i s spanned by diagonable elements which commute, because s e t t i n g a = e^, b = c = e^ i n (1), i j , gives (e^,e^) = 0. As was pointed out i n the d i s c u s s i o n a f t e r 1.2.5, L i s therefore a dlagonable subspace of U(J). In h i s doctoral d i s s e r t a t i o n , Foster showed that the Cartan theory of Lie and Jordan algebras i s e s s e n t i a l l y the same ( [ 3 ] ) . It i s therefore not s u r p r i s i n g to discover that the Cartan subalgebras of simple L i e and Jordan algebras (over a l g e b r a i c a l l y closed f i e l d s of c h a r a c t e r i s t i c 0) share the 53 common property that they are both diagonable subspaces of the corresponding u n i v e r s a l enveloping algebras. In p a r t i c u l a r , using Theorem 1.3.7, we obtain the analogue of Lemire's r e s u l t ([14]) for Jordan algebras. THEOREM 4.1.3: For a fixed linear functional X E H*, there is a one-to-one correspondence between the set of isomorphism classes of X-weighted irreducible U(j)-modules and the set of isomorphism classes of X-weighted irreducible C-modules, where. C is the centralizer of H in V ( J ) . This theorem i s p a r t i c u l a r l y useful because the algebra C i s generally much "smaller1 1 than A; moreover, we s h a l l see i n §4.2 that C can be characterized as the c e n t r a l i z e r of a single element of H. We close t h i s section with the following example: EXAMPLE 4.1.4: Let J be a simple Jordan algebra over an a l g e b r a i c a l l y closed f i e l d F of c h a r a c t e r i s t i c 0 which i s i f degree two; i . e . , 1 = e^ + e 2 where e^ and e.^ are orthogonal p r i m i t i v e idempotents. Then as i n Albert ( [ 1 ] ) , J ~ Fe^ + Fe 2 + J j 2 where J ^ 2 i s the subspace {acJ: ae^ = ae 2 = ha), 2 2 and J j 2 has a basis u^, u 2 such that u^u 2 = u 2u^ = 0 and u^ = u 2 = 1. Let C be the c e n t r a l i z e r i n [/(J) of the embedding of the Cartan subalgebra H = Fe^ + F e 2 . Then C i s spanned over F by the following seven elements: f j ° ej + e 2 - ( u ^ + u 2 2) f 2 = - ^(Sj + e 2) + 2 eie"2 f 3 = u 2 - hi^i + e 2) + 2e xe 2 f 4 = 2*1 ~ 4*1*2 f 5 = 2e 2 - 4e xe 2 u - S e ^ u , , v = 8e 2u^u 2 54 f p . . . , f t j are pairwise orthogonal idempotents which s a t i s f y f±u = uf± = f±v = v f i = 0, i = 1,2,3 f 5 u = u f 5 = 0; f 5 v = v f 5 = v Thus C = ( F f j <J)Ff2 © F f 3 ) £)B, where B i s the subalgebra of C spanned by *f ^ , f,., u and v, and © denotes an algebra d i r e c t sum. The m u l t i p l i c a t i v e r e l a t i o n s among -these, elements are given by the next table: f4 f5 u V f4 f4 0 u 0 f5 0 f5 0 V u u 0 "f4 0 V 0 V 0 - f Thus, denoting the square root of -1 by j , we have B = F g j ^ Fg 2 © F g ^ © Fg^, where gj, " ^ <f4 + J«) g 3 = j j ( f 5 + j v ) are pairwise orthogonal idempotents. So we see that C i s the d i r e c t sum of seven copies of F. 55 4.2 A fx) AS A CARTAN SUBALGEBRA —o-5— Let L be an n-dimensional L i e algebra with basis u^,...,u over a f i e l d F. Then any a e L determines the s o - c a l l e d Engel subalgebra n L D(a) = {x e L: xR a = 0 for 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 polynomial of Ra. Then f ( t , a ) = tn + P L C ^ , . . . , £ n ) tn~1 + ... + Pn^!...-.^) where the c o e f f i c i e n t s Pi»...,p are polynomials i n n-variables over F. Since aRa = 0 f o r any a e L, p n = 0, and so there i s a well-defined integer s, 1 < s < n such that p s t 0, but p r H 0 for s < r < n. Jacobson i n [13; page 60] c a l l s an element a e L regular i f P s(a) 4 0. He then shows that H i s a Cartan subalgebra of L i f and only i f for some a eL, H = LQ(a) i s a minimal Engel subalgebra (with respect 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 of diagonable elements i n an a s s o c i a t i v e algebra and show that they behave very much l i k e regular elements of a L i e algebra. DEFINITION 4.2.1; A diagonable element x in an associative algebra A over a field F is called finitely diagonable if Fx is a finitely diagonable subspace of A; i.e., if ad x i s an algebraic linear transformation of whose minimal polynomial has distinct roots. ;x is regular if A D(x) is minimal (always with respect to inclusion) among centralizers of all finitely diagonable elements in A. The 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 . THEOREM A.2.2: If x is a regular element in the algebra A = £ A (x), aeF a then A 0(x)g: AQ(y) /or every finitely diagonable element y in A Q(x); in other words, every finitely diagonable element of AQ(x) is in the centre of AQ(x). PROOF: Suppose y is a finitely diagonable element in A Q(x). For each t e F, define y t = x + t(y - x). Now y - x is finitely diagonable by Proposition 1.2.2. Also A^x) is invariant under ad (y-x) for any a e F and so by the same proposition y - x is finitely diagonable on A Q(x). Now assuming t f" 0, g is a root of y - x on Aa(x) i f and only i f A0(x) ft Ap(y-x) f" 0 i f and only i f Aa(x) (\ A t R(t(y-x)) / 0 (by Proposition 1.2.2) i f and only i f a + tg is a root of y t on A Q(x). y t is finitely diagonable on Aa(x) and we now see that for t f* 0, the minimal polynomial of the restriction of ad y^ . to A (x) is (1) ftt(X,t) = TT(X - <a+tg)) - X*01 + B^COx"1""1 + ... + B a ( t ) g a the product taken over the roots g of y - x on A a(x). Here the g^ a(t), i = l,...,m are polynomials in t, and mQ depends only on a, not on t; in fact ma is just the number of roots of y - x on A a(x). Now i f a f* 0, g (0) = (-a) j> 0, and so letting a range over the non-zero roots of x, a we have finitely many polynomials g a ( t ) none of which is identically 0. Since the characteristic of F is 0, F is infinite, so there is an infinite subset D S F such that g a ( t ) ^  0 for any t e D and non-zero a. But for ma t e D, A 0(y t)«£ AQ(x) because i f a e A (yt) and we write a = £ a a relative aeF to x, then (a,yt) = 0 implies ( a a»y t) = 0. But ad yfc is non-singular on Aa(x) for a ^  0 and so a a = 0. Hence a = a Q e A Q(x). By the minimality of A Q(x), we have A (yfc) =» A Q(x). Therefore the minimal polynomial of ad y on AQ(x) is X for t e D; i.e., g^°(t) = 0 for infinitely many t, 57 i = l,...,m 0. Therefore the polynomials 3^°(t) are i d e n t i c a l l y 0 and ad y^ = ad y has the minimal polynomial X on A Q ( x ) . This says that A _ ( x ) C ^ A (y) as we wanted to show. We o b t a i n immediately the important c o r o l l a r i e s : COROLLARY 4.2.3: Let L be a finitely diagonable subspace of A and assume that any collection of centralizers of finitely diagonable elements from A has a minimal member. Then the centralizer of L is Ag(x) for some x s L. PROOF: Let x E L be such that A G ( x ) i s minimal i n the set { A Q ( y ) : y e L}. Then f o r any y e L and t E F, y t = x + t ( y - x) i s i n L, and e x a c t l y as i n the proof of the theorem, we can show that A D ( y ) 3 A 0 ( x ) . Thus the c e n t r a l i z e r of L, which i s f\ A Q ( y ) must equal A Q ( x ) . yeL COROLLARY 4.2.4: If x is a regular element in an algebra A and A Q ( x ) is spanned by finitely diagonable elements, then A D ( x ) is commutative. t _ Now l e t L = £ Fe. denote the embedding of a Cartan subalgebra t i = l 1 H = [ Fe, of 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 enveloping algebra A, as i n §4.1. We saw i n that s e c t i o n that L i s a d i a -gonable subspace of A, and of course i t i s f i n i t e l y diagonable, because A i s f i n i t e d imensional. Applying 4.2.3, we deduce that the c e n t r a l i z e r C of 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 semi-simple, o ^ x l 1 t and so C i s semi-prime by Theorem 2.2.3. But the Jacobson r a d i c a l of any a r t i n i a n r i n g i s n i l p o t e n t , and hence C i s a c t u a l l y semi-simple. By the Wedderburn Theorem and the f a c t that F i s a l g e b r a i c a l l y c l o s e d , we see that 58 C i s a d i r e c t sum of matrix rings over F ((5; page 51]). Now any n x n matrix r i n g over F i s spanned by idempotents (hence by f i n i t e l y diagonable elements) for i t has the basis of matrix units ie^y' i » j " l,...,n} and e^j = ( e ^ + e£ j ) ~ e i ± i s a d i f f e r e n c e of idempotents for i , j = l , . . . , n . In p a r t i c u l a r , C = A 0(x) i s spanned by f i n i t e l y diagonable elements. If x could 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 matrix rings over F can only be commutative i f each matrix r i n g i s 1 x 1. We conjecture that t h i s i s indeed the case: that the c e n t r a l i z e r of a Cartan subalgebra i n the u n i v e r s a l enveloping algebra has the same algebra structure as that of the Cartan subalgebra; namely, i t i s a d i r e c t sum of f i e l d s . This was indeed the case i n the example 4.1.4. Because of the important r o l e the c e n t r a l i z e r plays i n the representation theory of Jordan algebras (Theorem 4.1.3), the v a l i d i t y of our conjecture would s i m p l i f y t h i s theory considerably. The structure theorem we do have i s summarized i n : THEOREM 4.2.5: The centralizer of a Cartan subalgebra of a finite dimensional simple Jordan algebra' over an algebraically closed field F of characteristic 0 in its universal enveloping algebra is the centralizer of a single element of the Cartan subalgebra, and is a direct sum of complete matrix rings over F. We f i n i s h t h i s section with an attempt to characterize regular elements i n an algebra. Suppose D i s a set of f i n i t e l y diagonable elements i n an algebra A. Then for x and y i n D, A Q(y) £L A Q(x) i f and only i f x i s i n Z ( A Q ( y ) ) , the centre of A Q ( y ) , and so A Q(x) i s minimal i n the set {A Q(y): y e D} i f and only i f for each y i n D, x e Z(A Q(y)) implies y e Z ( A Q ( x ) ) . We use t h i s observation to e s t a b l i s h 59 n PROPOSITION 4.2.6: If x = ][ ct^e.^ a p . . . , a n e F and e^, ...,e n orthogonal idempotents in a prime algebra A, then if A c(x) is minimal over the set of all centralizers of linear combinations of orthogonal idempotents, the e^,...,en are primitive and fT(a.j_ - a^) / 0. n PROOF: Suppose = c^- Then l e t y = £ &± e± w n e r e 3 j » ' « « » B n are a r b i t r a r y i = l n d i s t i n c t s c a l a r s . By Theorem 3.1.1, y i s diagonable and A Q(y) = £ e^Ae^, i= l and so x e Z ( A D ( y ) ) . However, y i Z(A Q(x)) because 0 ? GiAe2 S . A q ( X ) but (e^Ae2»y) ^ 0. I t i s immediate that e^,...,en are p r i m i t i v e , because i f n e^ = f^ + f2 with f j , f2 orthogonal idempotents, x = oc^f^ + o t j ^ + I a i e i and we j u s t showed that A Q(x) could not be minimal. n m LEMMA 4.2.7: Suppose x - £ <*±e± = I B.f. where {e,,...,e } and i = l j - l ^ 3 { f p . . . , f m } are two sets of orthogonal idempotents with sum 1 and hence if aj,...,an are distinct, either m > n or m = .n and BA.,....,Bm are also distinct. PROOF: The f i r s t statement i s j u s t a statement of the fact that the c o e f f i c i e n t s of x when expressed as a l i n e a r combination of orthogonal idempotents with sum 1, are uniquely determined as the roots of the m i n i -mal polynomial of x (Theorem 3.1.1). The re s t i s obvious. This lemma enables us to prove THEOREM 4.2.8: Let A be a finite dimensional central simple algebra n over a field F. Then x e A is regular if and only if x = £ a - i e 4 J i = l 1 a p . . . , a n distinct scalars and ep...,,e n primitive orthogonal idempotents with sum 1. 60 PROOF: We f i r s t note that the only diagonable elements i n A are l i n e a r combinations of orthogonal idempotents by Theorem 3.3.4. Then since a simple algebra with 1 i s prime, 4.2.6 gives the r e s u l t i n the "only i f " n d i r e c t i o n . Conversely, suppose x = J a i e i w ^ t b F d i s t i n c t , and e^ efi p r i m i t i v e orthogonal idempotents with sum 1. By the remarks before 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 to e s t a b l i s h (2) I f y E A i s diagonable and x e Z ( A Q ( y ) ) , then y e Z ( A Q ( x ) ) . Thus suppose y i s a diagonable element i n A. By 3.3.5, A Q(y) i s semi-simple and so the Wedderburn Theorem implies t (3) A D(y) = @ £ D^, D j , . . . , D t matrix rings over d i v i s i o n algebras A i= l which are f i n i t e - d i m e n s i o n a l over F. Let f ^ be the i d e n t i t y of for i = l , . . . , t . Then f p . . . , f t are o r t h o -gonal idempotents with sum 1. Writing each of these (If necessary) as a sum of p r i m i t i v e orthogonal idempotents, we see from Lemma 3.3.2 that t <. n. The key step i n the proof i s to show t (4) u diagonable i n Z(A (y)) implies u = J Kj^±* 5 i » ' « ' » £ t e F * ° i = l t To see t h i s , we note f i r s t that such a u can be w r i t t e n u = £ u^ with i = l u± e Z(D i) for i = l , . . . , t . But Z(D±) = Z ( A i ) f ± and so u± = &±£±, &± £ Z ( A i ) . m Since u i s diagonable, we know u = £ Z±&± where g^,...,gm are orthogonal i=l idempotents with sum 1 and 5 m e F , and by re-ordering i f necessary, we may suppose that {£;]_,...,£} = ,. .. , 5 S } , s < m and £^ £ g d i s t i n c t . s By Theorem 3.1.1, the minimal polynomial of u i s p(t) = TRt ~ Since t i-1 1 u.u = 0 for 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 implies p(6 i) = 0. Now over the f i e l d F [ 6 J ] , p(t) which i s of degree s, has the s + 1 roots 61 fi^t'Cj »• • • »? s and so 6^ e . . . b e c a u s e £^ £ s are d i s t i n c t . This establishes (4) . t In p a r t i c u l a r , y e Z(A Q(y)) implies y = ]> 8±fi> s i » " « » 3 t e F> t i = l n and i f x e Z(A ( y ) ) , x = £ Y 4 f j » Y 1»"«»Y 1- e F. B ut x = £ a-(e.,» a i , . . . , a i=l 1 1 i = l 1 d i s t i n c t elements of F; therefore because t ,< n, by the Lemma 4.2.7, t = n n and Yi»»««»Y a r e d i s t i n c t too. Hence A (x) = Y f.Af. (3.1.1) and so I n o ± = 1 i i y e Z ( A 0 ( x ) ) , v e r i f y i n g (2) and giv i n g the theorem. REMARK 4.2.9: I f A i s the ri n g of n x n matrices over a f i e l d F and A^ denotes the associated L i e algebra, then i t i s known ([ 9 ; page 66]) that x e A^ i s regular i n the L i e sense i f and only i f x has n d i s t i n c t c h a r a c t e r i s t i c roots i n the algebraic closure of F. Theorem 4.2.8 shows that i n t h i s case, the two concepts of r e g u l a r i t y coincide i f F i s a l g e -b r a i c a l l y closed; although the example at the beginning of §3.3 shows that i f F i s not a l g e b r a i c a l l y c l o s e d , there may e x i s t regular elements i n the L i e sense which are not even diagonable. 62 4.3 SIMPLE JORDAN ALGEBRAS We saw i n the discussion a f t e r Corollary 4.2.4 that the u n i v e r -s a l enveloping algebra £/(J) of a f i n i t e - d i m e n s i o n a l simple Jordan algebra J over an a l g e b r a i c a l l y closed f i e l d F i s spanned over F by i t s idempotents. This followed d i r e c t l y because J/(J) i s semi-simple and f i n i t e - d i m e n s i o n a l . In t h i s s e c t i o n we prove that the u n i v e r s a l enveloping algebra of any simple Jordan algebra over any f i e l d of c h a r a c t e r i s t i c 0 i s generated as an algebra by i t s idempotents, provided J contains an idempotent e whose Peirce one-space J^(e) = Fe. We remark that any reduced simple Jordan algebra has t h i s pro-perty (see for example, Jacobson [11; page 202]). By a d e r i v a t i o n of a (not n e c e s s a r i l y a s s o c i a t i v e ) algebra A over a f i e l d F, we mean a l i n e a r map D: A •*• A such that for any a and b i n A, (1) D(ab) = D(a)b + aD(b) I f D^ and D2 are derivations of A and a e F, then aD^ + D2 as w e l l as the commutator (D^,D2) are also derivations and so the derivations of any algebra form a L i e algebra under the commutator product. I f D i s a n i l -D2 potent d e r i v a t i o n , then exp(D) = 1 + D + + . ... i s always an automorphism of A (see §1.2 o f [ 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 \\> of A suoh that iji(x) = y, and in this case, we write x <\> y. I t i s immediate that *v* i s an equivalence r e l a t i o n on the c l a s s of diagonable elements i n A; i . e . , i t i s a symmetric, r e f l e x i v e , and t r a n s i t i v e r e l a t i o n . 63 PROPOSITION.,4.3.2: If x Q is a finitely diagonable element in an algebra A over Fj then the linear space S spanned by {x e A: x ^ xQ} is a Lie ideal of A containing V A (x); i.e., a subspace such that x e S and a e A a h implies (a,x) e S. PROOF: We note that i f ^  i s an automorphism of A, then i|>(Aa(x0)) = A -Ci j»(x 0)) and thus i f x t x Q, the roots of x and x Q are the same. In p a r t i c u l a r x has only 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 i s a d e r i v a t i o n of A which i s nilp o t e n t because A^(x)(ad a Q) C Ag + k Q ( x ) f o r any root $ of x. Thus exp(-iad a„) i s an automorphism of A which sends Ct CL_—. x to x + a a; i . e . , x + a «v» x. Since •'v. i s a t r a n s i t i v e r e l a t i o n , x + aa % xo a n c* aa = ^x + aa) - x e S for any a ^ 0. I f a e A, and a = £ a a r e l a t i v e to x, then (a,x) = Y a a Q e S. Thus S i s a L i e i d e a l . aeF otfO We can now prove our main r e s u l t . 1 THEOREM 4.3.3: Let J be a simple Jordan dlgebra over a field F which contains an idempotent e such that J ^ e ) = Fe. Then 0(J) is generated by its idempotents. PROOF: We have seen i n 4.1.2 that e i s an algebra i c diagonable element i n U(J) with roots i n the set {0,52 , -^,1 ,-1} . By the previous p r o p o s i t i o n the l i n e a r span S of {x e t/(J): x % e} i s a L i e i d e a l of £/(J) . Moreover, S i s spanned by idempotents since x ^ e c l e a r l y implies that x s a t i s f i e s the same minimal polynomial as e and hence i s also a l i n e a r combination of idempotents. I t i s therefore s u f f i c i e n t to prove S generates £/(J). For t h i s , define S' = { a e J : a e S}. Then S1 i s an associator i d e a l of J i n the sense that 64 (2) (S'.J.J) + (J.S'.J) + c s'. To e s t a b l i s h t h i s , we f i r s t note that by the i d e n t i t y (2) of 4 . 1 , we have (3) abc - abc + cba - cba + ca.b - bca = 0 f o r any a,b,c e J Interchanging a and b i n (3) and then subtracting from ( 2 ) , we obtain (4) abc + cba - bac - cab + ca.b - cb.a = 0 and t h i s can be re-written as (5) (a,c,b) = (c , (7,b)) Suppose a e S', b,c,e J. Then we see from (5) that (a,c,b) e S because S i s a L i e i d e a l ; hence (S' , J, J) CII S '. By s i m i l a r arguments we also see that (J,S',J)c^ S' and (J,J,S')£ S and so we have ( 2 ) . But then *'S" - {a e S': aJ &. S'} i s an i d e a l of J because i f a e S" and b,c e J, ab.c = a.be + (a,b,c) e S' implies ab e S". Let J = J Q + + be the Peirce decomposition of J r e l a t i v e to e. Then we see that e e S", for i l 7 = 0 e S; ejT o j T £ Ai^(e") + A u(e) c S by Propositions 4 .1.2 and 4.3.4; -2 "2 '2 ~ I and eJi <~ eFe £ Fes S. Thus S" i s a non-zero i d e a l of J and so equals J by s i m p l i c i t y . But S" £ S' implies S' = J ; , i . e . , J £ S . Since 7 generates U(J), so does S. 65 BIBLIOGRAPHY 1. A l b e r t , A.A., A Structure Theory for Jordan Algebras, Annals of Math, 48 (1947), 546-567. 2. C u r t i s , C.W., and Reiner, I . , "Representation Theory of F i n i t e Groups and A s s o c i a t i v e Algebras", Wiley ( I n t e r s c i e n c e ) , New York 1962. 3. Foster, D.M., "A General Cartan Theory", Ph.D. Thesis, U n i v e r s i t y of B r i t i s h Columbia, 1969. 4. He r s t e i n , I.N., "Topics i n Algebra", B l a i s d e l l , Toronto 1964; revised e d i t i o n 1965. 5. ' "Non-Commutative Rings", Carus Mathematical Monographs, Mathematical Asso c i a t i o n of America, Wiley, New York 1968. 6. Jacobson, N., "Theory of Rings", Mathematical Surveys I I , American Mathematical Society, New York 1943. 7* General Representation Theory of Jordan Algebras, Transactions of the American Mathematical Society, 70 (1951), 509-530. 8* "Structure of Rings", Colloquium Pub l i c a t i o n s 37, American Mathematical S o c i e t y , Providence, R."I., 1*956; revised e d i t i o n 1964 9 • _ _ _ _ _ _ _ _ "L J e Algebras", Wiley ( I n t e r s c i e n c e ) , New York 1962. 10. Cartan Subalgebras of Jordan Algebras, Nagoya Mathematics Journal 27 (1966), 591-609. 1 1 . "Jordan Algebras", Colloquium Publications 39, American Mathematical Society, Providence, R.I., 1968. 12. Lambek, J . , "Lectures on Rings and Modules", E l a i s d e l l , Toronto 1966. 13. Lemire, F.W., Irreducible Representations of a Simple Lie Algebra Admitting a One-Dimensional Weight Space, Proceedings of the American Mathematical Society, 19 (1968), 1161-1164. 1 4 . " Weight Spaces and Irreducible Representations of Simple Lie Algebras, Proceedings of the American Mathematical So c i e t y , 22 (1969), 192-197. 15. Existence of Weight Space Decompositions for Irreducible Representations of Simple Lie Algebras, Canadian Math B u l l e t i n , 14 (1971), 113-115. 16. McKrimmon, K., Jordan Algebras of Degree 1, B u l l e t i n of the American Mathematical So c i e t y , 70 (1964), 702. 

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