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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Irreducible representations of algebras Goodaire, Edgar George 1972
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Title | Irreducible representations of algebras |
Creator |
Goodaire, Edgar George |
Publisher | University of British Columbia |
Date Issued | 1972 |
Description | An element x of an associative algebra A is called diagonable provided A has a basis of characteristic vectors for the transformation ad x: a → ax - xa of A. This notion immediately generalizes to that of a diagonable subspace L of A. The centralizer A[sub O] of L plays an important role in the representation theory of A, for there is a one-to-one correspondence between the "λ-weighted" irreducible modules of A and of A[sub O]. In Chapters Two and Three, we first explore various ring-theoretic properties of A and A[sub O], and then use the results obtained to classify the diagonable elements in different algebras. We also give conditions under which all A-modules are weighted. The Cartan theory of Lie and Jordan algebras is linked in Chapter Four by the observation that Cartan subalgebras of simple finite dimensional Lie and Jordan algebras (over algebraically closed fields of characteristic 0) are diagonable subspaces of the respective universal enveloping algebras. Furthermore, in the Jordan case, the centralizer of a Cartan subalgebra is the centralizer of one of its elements and is a direct sum of complete matrix rings. Finally, we are able to show that the universal enveloping algebra of any simple Jordan algebra which contains an idempotent whose Peirce one-space is one-dimensional, is generated by its idempotents. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-03-03 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080443 |
URI | http://hdl.handle.net/2429/31959 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
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