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UBC Theses and Dissertations

Density theorems and applications Horvath, Jozsef 1977

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DENSITY THEOREMS AND APPLICATIONS by JOZSEF HORVATH B . S c , Tel Aviv University, 1976 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in the DEPARTMENT OF MATHEMATICS We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA June, 1977 o Jozsef Horvath^ J u n e > 1 9 ? ? In present ing th is thes is in p a r t i a l fu l f i lment of the requirements f o r an advanced degree at the Un ivers i ty of B r i t i s h Columbia, I agree that the L ibrary sha l l make it f ree ly ava i lab le for reference and study. I fur ther agree that permission for extensive copying of th is thes is for scho la r ly purposes may be granted by the Head of my Department or by h is representa t ives . It is understood that copying or p u b l i c a t i o n of th is thes is fo r f inanc ia l gain sha l l not be allowed without my writ ten permission. Department of Mathematics  The Univers i ty of B r i t i s h Columbia 2075 Wesbrook Place Vancouver, Canada V6T 1WS June 24, 1977 i i ABSTRACT Supervisor: Prof. C T . Anderson One way of getting structure theorems i n r i n g theory i s to f i x a general c l a s s £ of modules, and to prove Schur's Lemma and the Density Theorem for £ • For example, the Goldie Theorem for prime rings follows from Schur's Lemma and the Density Theorem for the clas s of r a t i o n a l l y uniform, homogeneous modules i n a s i m i l a r way as the Wedderburn-Artin Theorem follows from Schur's Lemma and the Density Theorem for the clas s of i r r e d u c i b l e modules. i i i TABLE OF CONTENTS page INTRODUCTION 1 CHAPTER 1: GENERAL CLASSES OF MODULES 3 CHAPTER 2: ANALOGUES OF SCHUR'S LEMMA AND THE DENSITY THEOREM 17 CHAPTER 3: THE GOLDIE THEOREM 37 REFERENCES 47 i v ACKNOWLEDGEMENT I would l i k e to thank my supervisor, Prof. Tim Anderson for the invaluable help, care and encouragement he gave me i n th i s thesis and during my whole year at the University of B r i t i s h Columbia. INTRODUCTION In this paper ring means associative ring which does not necessarily contain a unity element and is not necessarily commutative. Module means lef t module. For every ring A let ^ denote, for the moment, the class of a l l irreducible A-modules. The proof of the c lass ica l Wedderburn-Artin Theorem has the following main steps: (1) If V e then A = Hom^(V,V) is a divis ion r ing. (Schur's Lemma) (2) If V £ ^ and A = Hom^(V„V), then for any A-linearly independent elements v - , . . .,,v e V and any elements , . . . ,w e V there 1' n J 1 n exists an element a e A such that av, = w, , . . . ,av = w 1 I n n This means that the ring A^ of left multiplications by elements of A i s , in a certain sense, dense in the ring Hom^(V,V) of a l l l inear transformations on the vector space V . (Jacobson Density Theorem) (3) If has a member which is fa i thful and f in i te dimensional (over i t s central izer) , then A is isomorphic to M (^D) where D is a divis ion ring and • n >_ 1 . (4) If A is a simple, Artinian r ing, then £ contains a member which is fa i thful and f in i te dimensional (over i t s central izer) . Here, in order to get the desired structure theorem, we concentrate on a certain fixed class, of modules, and prove Schur's Lemma and - 2 -the Density Theorem for £ . These enable us to describe the structure of those rings A for which ^ has a fa i thful , f in i te dimensional member. Once we know this , the structure of a ring A of the class that we are interested in is easily obtained by showing that for such v J A, 2.^  does have a fa i thful , f in i te dimensional member. The essential part of the structure theory i s , therefore, Schur's Lemma and the Density Theorem for the class \ of modules. The aim of this thesis is to emphasize the usefulness of concentrating on a fixed class \ of modules for which we can prove an analogue of Schur's Lemma and of the Density Theorem. We show that i f we take \ to be the class of rationally uniform, homogeneous modules, then the approach described above gives Goldie's structure theorem for prime rings satisfying the ascending chain condition. In Chapter 1 we define the notion of a general class of modules, and prove the Theorem of Andrunakievic and Rjabuhin which shows how a general class of modules defines a radical . For example, the general class of irreducible modules defines the Jacobson radical . Then we show that the class \ of rationally uniform, homogeneous modules i s a general class. In Chapter 2 we give the analogues of Schur's Lemma and the Jacobson Density Theorem for this class \ , and use them to describe the structure of those rings A which have in \ ^ a fa i thful member satisfying a certain f in i te dimensionality condition. ' In Chapter 3 we deduce from this the Goldie Theorem. His tor ica l ly , the generalizations of the Jacobson Density Theorem that w i l l be given here, are due to Faith [2] and to Koh and Mewborn [4]. It was Heinicke [3] who pointed out that they imply the Goldie Theorem on prime rings. - 3 -CHAPTER 1 GENERAL CLASSES OF MODULES In order to study the structure of rings i t i s very u s e f u l to concentrate on a c e r t a i n f i x e d c l a s s of modules. The c l a s s i c a l example i s the c l a s s of i r r e d u c i b l e modules which gives the Jacobson structure theory, and i n p a r t i c u l a r the Wedderburn-Artin Theorem. Another class of modules, as we s h a l l see, gives the Goldie Theorem for prime r i n g s . In a l l the paper, a homorphism which i s a one-to-one mapping w i l l be c a l l e d a monomorphism, and a homomorphism which i s an onto mapping w i l l be c a l l e d an epimorphism. 1.1. Radicals D e f i n i t i o n : A c l a s s g of rings i s c a l l e d a r a d i c a l i f ( i ) f or every A e g and epimorphism A —> B we have B e g . ( i i ) f o r every r i n g A there e x i s t s an i d e a l I i n A such that l e g , and i f J i s an i d e a l of A and J e g then J C I . (This l a r g e s t g-ideal of A i s denoted by g(A) .) ( i i i ) f o r every r i n g A, g(A/g(A)) = 0 . Example: The c l a s s of a l l n i l rings i s a r a d i c a l . D e f i n i t i o n : Let g be a r a d i c a l . A r i n g A i s c a l l e d g-semisimple i f g(A) = 0 . c D e f i n i t i o n : A c l a s s M of rings i s c a l l e d regular i f for every r i n g A e M and nonzero i d e a l I of A there e x i s t s a nonzero r i n g B e M and an epimorphism I —> B . - 4 -1.2. Theorem of Kurosh Let M be a regular class of rings. Let be the class of a l l rings which cannot be homomorphically mapped onto a nonzero member of M . Then (i) U.. is a radical. M ( i i ) For every ring A e M, ^(A) = 0 • ( i i i ) If 3 is a radical such that 3(A) = 0 for every A e M, then every U^-semisimple ring is g-semisimple. For the proof see Divinsky [1]. 1.3. General classes For an A-module V we denote by (0:V) the annihilator of V, i.e. (0:V) = {a e A|aV = 0} . Of course, (0:V) is an ideal of A . V is called nontrivial i f AV ± 0, i.e. (0:V) i A . V 'is called faithful i f (0:V) = 0 . Suppose that to every ring A there is assigned a (possibly empty) class \ of nontrivial A-modules. - Such an assignment is called a class of modules. Definition: A class \ of modules is said to be a general class provided: (i) If f: A —*• B is an epimorphism and V e £ then V, considered as an A-module in the obvious way, belongs to ^ . ( i i ) If f: A —• B is an epimorphism, V e jj^ and ker f C (0:V) then V, considered as a B-module, is in £g . ( i i i ) If C\ (0:V) = 0 then Y_ ^ 0 f or every nonzero ideal I of A . - 5 -(iv) If for every nonzero ideal I of A Y =f 0 then f\ (0:V) = 0 . V ^ A Definit ion: Let £ be a general class of modules. A ring A i s called ^-primitive i f ^ contains a fa i thful module. Definit ion: Let (A.}, . be rings. A ring B, isomorphic to a subring A AeA of the complete direct sum £ ffl A , i s called a subdirect sum of the rings {A^} i f the restrict ions of the canonical projections T © A, >- A to B are onto. This is equivalent to the following f-c X U A condition: There exist ideals 1^  in B such that for every X B/I, i s isomorphic to A. and ^ 1 = 0 . A A . A A The following theorem shows that every general class of modules determines a radical . 1.4. Theorem of Andrunakievic and Rjabuhin Let J be a general class of modules. Denote a = = 0} . Then (i) a i s a radical . ( i i ) For every ring A, a(A) = f\ (0:V) . V £ £ A ( i i i ) Every a-semisimple ring is a subdirect sum of ^-primitive rings. Proof: Let M be the class of ^-primitive rings. F irs t we observe that (*) i f V E [ A then 0 j A/(0:V) £ M . (As V is nontr iv ia l , (0:V) ^ A, so A/(0:V) ± 0 . Consider the canonical projection A > A/(0:V) . By property ( i i ) of a general class V £ ^ / ( O ' V ) a n C ^ '*"t ^ S c-'-e a r-'-y fa i thful over A/(0:V) .) We show now that M i s a regular class. Let A e M and let I be a nonzero ideal - 6 -of A . As A i s ^ - p r i m i t i v e , (0:V) = 0 . So by property ( i i i ) V £ ^ A of a general c l a s s , J ± 0 . Let V e £ . By (*), l/(0:V) i s a nonzero member of M, and the canonical pr o j e c t i o n maps I onto I/(0:V) . Therefore M i s a regular class of rin g s . Let U^ j be the r a d i c a l determined by M by the Theorem of Kurosh. We show: U„, = a . We prove t h i s by showing: A £ U^ j i f and only i f A £ a . Suppose A £ . Then there e x i s t s a nonzero member B of M and an epimorphism A —>- B . As B e M, there e x i s t s V e £ . By property (i ) of a general c l a s s , V e . So A £ a . Suppose A £ a . Then there e x i s t s a module V e ^ . By (*), A/(0:V) i s a nonzero member of M . The canonical p r o j e c t i o n maps A onto A/(0:V) . Thus A £ U^ j . We have proved = a, so a i s a r a d i c a l . Now we want to show that f o r every r i n g A, a(A) = C\ (0:V) . V££A Suppose a(A) c£ C\ (0:V) . Then there e x i s t s V e Y. such that a (A) <£ (0:V) . Therefore g ^ Q " | " v ^ ° i s a nonzero i d e a l of A/(0:V), and by (*), A/(0:V) e M . As M i s a regular c l a s s , there e x i s t s a _ I I j . a (A) + (0:V) nonzero rxng B e r n and an epxmorphism (O'V) ' Thus we have epimorphisms n(As . o(A) s o(A) + (0;V)  0 ( l A ; q(A)n(0:V) (0:V) a and the composition i s an epimorphism from o(A) onto B which i s a nonzero member of M . This i s a contradiction because a(A) e a = - 7 -Therefore a (A) C C\ (0:V) . V ^ A The r i n g A/a(A) i s a-semisimple, so for every nonzero i d e a l I of A/a(A), I i a, which means \^ ^ 0 • By property (iv) of a general class we know: Q ( ° : V ) A / a ( A ) - 0 . V e^A/a(A) Let a e / O (0:V). . Let V e j. , . By property ( i ) of a general VeY A A A / a ( A ) c l a s s , i f we consider V as an A-module then V e and therefore a e (0:V) . By the d e f i n i t i o n of V as an A-module, 0 = av = av for every v e V . Thus a e (0 : V)^/a(A) • A s v w a s a n y member of I^/a(A)' W e ^ a v e : a E f~\ (0:V)., / t v = 0 . Thus a e a(A) . This proves: r yA/a(A) V £ ZA/a(A) a(A) = r\ (0:V) . V £ ^ A F i n a l l y , l e t A be a-semisimple. Then a(A) = (0:V) = 0 . Let V ^ A {I,}, . be the set of a l l i d e a l s of A which are the a n n i h i l a t o r of A AeA some module i n ^ . Then c l e a r l y , H l ^ =0, and therefore A i s A a subdirect sum of the rings {A/l^} . By (*), the rings A/I^ are ^- p r i m i t i v e . Q.E.D. 1.5. Example For every• r i n g A l e t £ A be the clas s of i r r e d u c i b l e A-modules. Then J i s a general c l a s s . (See Heinicke [3].) The r a d i c a l determined by - 8 -£ is called the Jacobson radical. The above class of modules gives the classical Jacobson, theory. We will work with another general class, the class of rationally uniform, homogeneous modules. First, the definitions and elementary properties. 1.6. Essential extensions Definition: Let W be a submodule of V . V is called an essential  extension of W if for every nonzero submodule U of V, W D U ^  0 . If V is an essential extension of W, we also say that W is an essential submodule of V . Proposition (i) Let V C V ' C V" be A-modules. Then V<=V" is an essential extension i f and" only i f V<^V' and V'^ V" are essential extensions. (ii) The intersection of a finite number of essential submodules is an essential submodule. The proof of this proposition is trivial. 1.7. The singular submodule A left ideal I in a ring A is called essential i f i t is an essential submodule of A when we consider A as a module over itself. If V is an A-module and v e V we denote: (0:v) = {a e A|av = 0} . It is clear that (0:v) is a left ideal in A . Proposition Let V be an A-module,, and let Z(V) = {v e V|(0:v) is an essential - 9 -l e f t i d e a l of A} . Then Z(V) i s a submodule of V . Proof: If v, w e Z(V) then (0:v), (0:w) are e s s e n t i a l l e f t i d e a l s , so by 1.6 (0:v) A (0:w) i s e s s e n t i a l , and as (0:v) A (0:w)C (0: v+w) , we have v+w e Z(V) . It i s c l e a r that 0 e Z(V) and that v e Z(V) implies -v e Z(V) . Let a e A, v e Z(V), and we show: av e Z(V) . Let I be a nonzero l e f t i d e a l of A . If I C ( 0 : a v ) then lA(0:av) = 1 ^ 0 . If I <fc (0:av), take x e I such that xav ^ 0 . Consider the l e f t i d e a l Ia . I t i s nonzero because 0 ^ xa e Ia . As (0:v) i s e s s e n t i a l , (0:v) f\ Ia ^ 0 . Thus there e x i s t s y e I such that ya ^ 0 and yav = 0 . So we have 0 i= y e I C\ (0:av) . This shows that (0:av) i s an e s s e n t i a l l e f t i d e a l of A, and av e Z(V) . Q.E.D. D e f i n i t i o n : Z(V) i s c a l l e d the singular submodule of V . 1.8. Rational extensions D e f i n i t i o n : Let W be a submodule of V . V i s said to be a r a t i o n a l  extension of W i f f o r every v e V , O ^ v ' e V there e x i s t s an element a £ A and an integer n such that: av + nv e W av' + nv' ± 0 . Proposition Let W be a submodule of V . The following conditions are equivalent: (i) V i s a r a t i o n a l extension of W . ( i i ) If T i s a submodule of V which contains W, f e Hom^(T,V) and f(W) = 0 then f = 0 . - 10 -Proof: Let V be a rational extension of W, W ^ T C V , f e HomA(T,V), f (W) = 0 . We want to show: f = 0 . Suppose f =f 0 . Then there exists v e T such that f(v) ^  0 . As W C V is rational, there exists a e A and an integer n such that av + nv e W > af (v) + nf (v) ? 0 . But then 0 = f(av + nv) = af(v) + nf(v), a contradiction. Therefore f = 0, and (i) implies ( i i ) . To prove that (ii) implies (i) let v e V, 0 ^  v' e V be given. Suppose that for every a e A and every integer n: (*) av + nv e W implies av' + nv' = 0 . Consider the module T = W + Av +2v . Clearly, W T <=• V . Define f : T > V by f (w + av + nv) = av' + nv' for weW, a e A, n e . By (*) f is well defined, and i t is clear that f e Hom^ (T,V) and f(W) = 0 . Thus, by (ii) , f = 0/ and in particular f(v) = 0 . But f(v) = f(0 + Ov + lv) = v' ^  0, a contradiction. Therefore V is a rational extension of W, and (ii) implies (i) . Q.E.D. The next proposition shows the connection between essential and rational extensions. 1.9. Proposition (i) If WCV is a rational extension, then it is an essential extension. - 11 -( i i ) If W C V i s an e s s e n t i a l extension and Z(V) = 0, then WcrV i s an r a t i o n a l extension. Proof: ( i ) Let WC V be r a t i o n a l , and l e t U be a nonzero submodule of V . Take a nonzero element u e U . As V i s a r a t i o n a l extension of W, there e x i s t s a e A and integer n such that au + nu e W au + nu 4- 0 . C l e a r l y , O ^ a u + n u e W - O u . So V i s an e s s e n t i a l extension of W . ( i i ) Let W C V be an e s s e n t i a l extension, and Z(V) = 0 . Let v e V , 0 ^  v' e V . Denote I = { a e A | a v e W } . Then I i s a l e f t i d e a l i n A . We show that I i s e s s e n t i a l . Let J be a nonzero l e f t i d e a l of A . Then Jv i s a submodule of V . If Jv = 0, then J C I and I / O j = J ^ 0 . If J v ^ O , then J v / ) W j f 0 because W i s an e s s e n t i a l submodule of V . So there e x i s t s j e J such that j v ^ 0 and j v e W . Then 0 ^ j e i A J . So I i s an es s e n t i a l l e f t i d e a l of A . As Z(V) = 0 and v' f 0, we know that (0:v') i s not an e s s e n t i a l l e f t i d e a l i n A, and therefore I ( 0 : v ' ) . Take an element i e I such that i v ' =f 0 . We have: i v e W, i v ' ^  0 . Thus, V i s a r a t i o n a l extension of W . Q.E.D. 1.10. Uniform, r a t i o n a l l y uniform and homogeneous modules D e f i n i t i o n : ( i ) An A-mqdule V i s said to be uniform i f i t i s an es s e n t i a l extension of each of i t s nonzero submodules. ( i i ) An A-module V i s said to be r a t i o n a l l y uniform i f i t i s a r a t i o n a l extension of each - 12 -of its nonzero submodules. It is clear that a module V is uniform if and only if the intersection of any two nonzero submodules of V is nonzero. Every irreducible module is rationally uniform, and by 1.9(i) every rationally uniform module is uniform. By 1.9(ii) a uniform module which has zero singular submodule, is rationally uniform. Definition: An A-module V is called homogeneous if for every nonzero submodule W of V there exists a monomorphism f: V *• W . Clearly, every irreducible module is homogeneous. It is also easy to see that 2 is a homogeneous module over itself. Now we are ready to fix the class of modules we want to work with. 1.11. Theorem For every ring A, let ^ be the class of a l l nontrivial, rationally uniform, homogeneous A-modules. Then £ is a general class. Proof: The first two properties in the definition of a general class of modules (see 1.3) are easily verified for £ . To prove property ( i i i ) , let A be a ring such that O (0:V) = 0, and let I be a nonzero ideal of A . We want to show: J ^ 0 • There exists an A-module V e T. such that IV ^  0 . (Otherwise I C f\ (0:V) =0, a contradiction to 1^0.) Consider V as an I-module. We show: V e £ . As IV ^  0, V is a nontrivial I-module. Let N = {v e v|lv = 0} . Clearly, N is an A-submodule of V . Suppose N ^  0 . Then, as V is a homogeneous A-module, there exists an A-monomorphism f: V > N . For every i e I - 1 3 -and v e V we have f(iv) = if(v) = 0 because f(v) e N . As ker f = 0 , we get iv = 0 , for every i e I and v e V, a contradiction to IV ^  0 . Therefore: N = 0 . To prove that V is rationally uniform as an I-module, let W be a nonzero I-submodule of V, T an I-submodule of V containing W, and f e Hom^ CTjV) such that f(W) = 0 . By 1 . 8 i t is enough to show that f = 0 . It is easy to see that IW and IT are A-submodules of V, and 0 ± IWC IT<= V . (IW ^ 0 because N = {v e V| Iv = 0 } = 0 .) As V is a rationally uniform A-module, IW c: V is a rational extension. The restriction f| : IT > V is an A-homomorphism (because f(a ^ i t ) = J- J. K. K. f(E(aik)tk) = £(aik)f(tk) = a I\f(tk) = a f(I±ktk)), and f| (IW) = f(IW) = 0 . Therefore f | I T = 0 , i.e. f(IT) = 0 . Let t e T . Then for every i e I: if(t) = f(it) = 0 . Thus If(t) = 0 , so f(t) E N = 0 . Therefore f = 0 , and V is rationally uniform as an I-module. To show that V is a homogeneous I-module, let W be a nonzero I-submodule of V . Then again, IW is a nonzero A-submodule of V . As V is homogeneous as an A-module, there exists an A-monomorphism f: V —> IW. Clearly, f is also an I-monomorphism f: V —> W . Therefore V is homogeneous as an I-module, and we have V e £ . This proves ( i i i ) . To prove property (iv) of a general class for suppose that A is a ring such that for every nonzero ideal I of A, £ ^  0 . We want to show f\ (0:V) = 0 . Denote K = f\ (0:V), and suppose K £ 0 . Then K is a nonzero ideal of A, and therefore Y f 0 . Take a K-module K W £ £ K . As KW ^ 0 , there exists w^  e W such that KWQ ^ 0 . Denote: U = KWQ . Clearly U is an abelian subgroup of W . We define an A-module structure A x U > U on U in the following way: - 14 -For a e A, kw^  e KWQ = U define a * kw^  = (ak)w^ . To show that * is well-defined, we have to prove: kw^  = 0 implies (ak)wg = 0, for k e K and a e A . Let k e K be fixed and suppose kw^  = 0 . We show (Ak)w^ = 0 . Suppose the contrary. Then (Ak)w^  is a nonzero K-submodule of W, and as W is homogeneous over K, there exists a K-monomorphism f: W y (Ak)wQ . For every x e K, w e W we have: f(xw).= xf(w) = x[(ak)wQ] for some a e A because f(w) e (Ak)w^ . f(xw) = x[(ak )Wp) ] = [x(ak)]wQ = [(xa)k]w^ = (xa)[kw^] = xa«0 = 0 . As ker f = 0, we get xw = 0 for every x e K, w e W, a contradiction to KW 4 0 . Therefore (Ak)wQ = 0, and * is well-defined. It is easy to check that the operation * turns U into an A-module, and that for k e K, u e U we have: k*u = ku . We show now that U e \^ . Let M = {w e w|Kw =0} . Then M is a K-submodule of W . If M + 0, then as W is a homogeneous K-module, there exists a K-monomorphism f: W y M . For every k e K, w e W, f(kw) = kf(w) = 0 because f(w) e M . As ker f = 0, we get a contradiction to KW ^ 0 . Thus M = 0 . As U = KWQ ± 0, this implies: A*U => K*U = KU 4 0, and so U is a nontrivial A-module. To show that U is rationally uniform over A, let S be a nonzero A-submodule of U, ueU, O ^ u ' e U . Clearly, S is a nonzero K-submodule of W, therefore, as W is a rationally uniform K-module, there exists an element k e K and an integer n such that ku + nu e S ku' + nu' 4 0 . - 15 -Thus we have k £ A and integer n with k*u + nu £ S k*u' + nu' i 0 . So U i s a r a t i o n a l l y uniform A-module. To prove that U i s a homo-geneous A-module, l e t S be a nonzero A-submodule of U . Then again, S i s a nonzero K-submodule of W . As W i s a homogeneous K-module, there e x i s t s a K-monomorphism f: W —>• S . Consider the r e s t r i c t i o n f | : U > S . For every a £ A and u = kw^ £ U we have: flyCaau) = f(a*kw Q) = f((ak)w Q) = (ak) f(w Q) = (ak)*f(w Q) = a*(k*f(w Q)) = a*(kf(w Q)) = a*f(kw Q) = a*f(u) = a * f ^ ( u ) . Therefore f| i s an A-monomorphism, and t h i s shows that U i s a homogeneous A-module. We have proved: U £ \ . As K = f\ (0:V), we have K C (0:U), thus K*U = 0 . But K*U = KU, so KU = 0, i n contradiction to M = {w e w|Kw = 0} = 0 . Therefore K = A (0:V) = 0 . Q.E.D. 1.12. The weak r a d i c a l By the Theorem of Andrunakievic and Rjabuhin, the general cl a s s £ of 1.11 defines a r a d i c a l . This r a d i c a l i s c a l l e d the weak r a d i c a l and i s denoted by W . As every i r r e d u c i b l e module i s n o n t r i v i a l , r a t i o n a l l y uniform and homogeneous, we have W C J (where J i s the Jacobson r a d i c a l ) , and therefore f or every r i n g A, W ( A ) C J(A) . But the two r a d i c a l s are d i f f e r e n t . In f a c t , the following example gives a r i n g A which i s weakly pr i m i t i v e ( i . e . ^ - p r i m i t i v e where £ i s the general cla s s of 1.11), but Jacobson-radical ( i . e . A e J ) . Example: Let A = {—|m,n integers, m even, n odd} . It i s easy to - 16 -see that A is a subring of the rational numbers. Consider A as a module over itself, and denote i t by A . Then A is a faithful A A member of ^ . Clearly, A^ is a nontrivial module. For 0 ^  v e A^, (0:v) = {a e A|av = 0} = 0 because there are no zero divisors in A, and therefore .A has zero singular submodule. Let V., , V„ be nonzero A & 1 2 submodules of A^ . As A is commutative, V^, are ideals in A, and so V ^ V ^ V V 2 ^ 0 ' T h u S A A l s U N I F O R M » A N D A S Z^A A^ = °' A^ is rationally uniform. To show the homogeneity of A^, let V be a nonzero submodule of A^ . Take a nonzero element v^ E V and define f: A^ — V by f (x) = X V Q for every x e ^ A . f is clearly an A-monomorphism. Therefore .A e /. . .A is a faithful module because ^ A LA - A there are no zero divisors in A . Thus A is a weakly primitive ring. An easy computation shows that every element in A is left-quasi-regular, and therefore A is a Jacobson-radical ring. - 17 -CHAPTER 2 ANALOGUES OF SCHUR'S LEMMA AND THE DENSITY THEOREM From now on £ will always denote the class of nontrivial, rationally uniform, homogeneous modules. We want to prove the analogues of Schur's Lemma and the Jacobson Density Theorem for and with the help of these to describe the structure of those rings A for which ^ contains a faithful module satisfying a certain finite-dimensionality condition. To be able to do this, we need the notion of the quasi-injective hull of a module. 2.1. The infective hull Definition: A module V is said to be injective if for every module U, submodule W of U and homomorphism f e Hom^ (W,V), there exists an extension f e Hom^ (U,V) of f . Theorem Let V be an A-module. Then: (i) There exists a maximal essential extension of V, i.e. an essential extension VC M such that if Vcz MCE and Vcr E is essential, then M = E . (ii) If M^  and M2 are maximal essential extensions of V, then there is an isomorphism between M^  and M2 which fixes every element of V . i Therefore we can speak about the maximal essential extension of V, which will be denoted by V . - 18 -( I i i ) VT1 i s an i n j e c t i v e module. Moreover, V i s a minimal i n j e c t i v e H / H module containing V, i . e . i f U i s i n j e c t i v e and V C U C V , , , then U = V u . For the proof of t h i s well-known r e s u l t see e.g. F a i t h [2]. D e f i n i t i o n : V i s c a l l e d the i n j e c t i v e h u l l of V . H 2.2. The q u a s i - i n j e c t i v e h u l l The following notion gives a common generalization of i r r e d u c i b i l i t y and i n j e c t i v i t y . D e f i n i t i o n : A module V i s said to be quasi-inj ective i f for every sub-module W of V and homomorphism f e HomA(W,V) there e x i s t s an extension f e HomA(V,V) of f . Theorem Let V be an A-module, and l e t be the i n j e c t i v e h u l l of V . Denote: T = Horn. (VTT,VTT) , and l e t V. = TV = {Ja.v.la. e V, v. e V, and the sums A H H Q ' I ' are f i n i t e } . Then: ( i ) Ver V Q C T V ^ , and i s a quasi-inj ective module. ( i i ) V_ i s the smallest q u a s i - i n j e c t i v e module between V and V , x H i . e . i f V C Q c r V ^ and Q i s quasi-inj e c t i v e , then V^ Q . For the proof see F a i t h [2]. D e f i n i t i o n : V^ i s c a l l e d the q u a s i - i n j e c t i v e h u l l of V . F i r s t we prove an analogue of Schur's Lemma for q u a s i - i n j e c t i v e modules. D e f i n i t i o n : A r i n g A i s c a l l e d Von Neumann-regular i f f o r every element - 19 -a e A there exists x e A such that a = axa . 2.3. Theorem (Schur's Lemma for quasi-injective modules) Let V be a quasi-injective module over A, and let A = Hom^ (V,V) . Then: (i) J(A) = {a e A|ker a is an essential submodule of V} . (ii) A/J(A) is Von Neumann-regular." Proof: Denote N = {a e A|ker a is essential in V} . Then N is a left ideal in A . For let a, 3 e N . Then ker a, ker 3 are essential in V, so by 1.6 ker a A ker 3 is essential, thus ker(a+3) is essential because ker a 0 ker ker(a+3), and so a + 3 £ N . Clearly 0 e N, and a e N implies -a e N . If a e N and 3 £ A, then ga £ N because ker a c ker 3a . From the Jacobson theory we know that the Jacobson radical of a ring contains every one-sided ideal which is Jacobson-radical. Thus, to show NC J(A), i t suffices to prove that N is Jacobson-radical, or that every element of N has a left quasi-inverse. Let a E N . Clearly, ker a O ker(l-a) = 0 . As ker a is essential in V, ker(l-a) = 0 . So 1 - a: V > (l-a)V is an isomorphism. Let f: (l-a)V >- V be its inverse. As V is quasi-inj ective, f can be extended to a homomorphism 3 £ A . We have: 3(l-a)v = v for every v e V, and so 3(l-ot) = 1 . Take y = -a3 • It is easy to check that Y + a = ya, and as N is a left ideal of A, y £ N . So N is Jacobson-radical, and NCJ(A). Now we show: (*) for every a e A there exists y e A such that a - aya e N . - 20 -Let a e A . Let W be a submodule of V maximal with respect to the property: W f\ ker a = 0 . (Such W exists by Zorn's Lemma.) Then V is an essential extension of W + ker a . (For let U be a nonzero submodule of V . If UCW, then U D (W + ker a) = U 4 0 . If U^W, then W + U ^  W, so (W+U) Cl ker a ? 0 . So there exist elements w e W, u E U such that 0 / w + u e ker a . u ^  0, otherwise 0 4 w e W f\ ker a = 0 . Thus, 0 ^  u = (-w) + (w+u) e U C\ (W + ker a) .) As W f\ ker a = 0, a|^ :.W > aW is an isomorphism. Let f: aW > W C V be the inverse of al . By the quasi-injectivity of w V, f can be extended to a homomorphism y e A . We have: yaw = w for every w e W . (a - aya)w = aw - ayaw = aw - aw = 0 for every w e W . Also: (a - aya)v = av - ayav = 0 for every v e ker a . Thus W + ker a CZ ker(a - aya) . As W + ker a is an essential submodule of V, so is ker(a - aya), which means that a - aya e N . This proves (*). As N<= J ( A ) , (*) implies that A/J(A) is a Von Neumann-regular ring. It remained to show: J(A)<= N . Let a e J(A) . By (*), there exists y e A such that a - aya e N . J(A) is an ideal in A, so ay e J(A) . Thus ay has a quasi-inverse g e J(A) : B + ay = gay . We have: g(a - aya) = ga - gaya = ga - (g + ay)a = ga - ga - aya = -aya . As a - aya e N and N is a left ideal in A, we have -aya = g(a - aya) e N . But also a - aya e N, and so a e N . Thus J(A)<= N . Q.E.D. With the help of this theorem, we are able to prove now the analogue of Schur's Lemma for our class £ . Definition: An element a of a ring A is called regular if i t is neither a left, nor a right zero divisor in A, i.e. x e A, ax = 0 21 -implies x = 0, and also x e A, xa = 0 implies x = 0 . Definition: Let B be a ring with unity, A a subring of B . (This of course, does not mean that A contains the unity element of B .) A is said to be a right order in B i f : (i) Every element of A which is regular in A is invertible in B . (ii) Every element b e B has the form b = a-^ a^ where a^a^ e A, a.^ regular. The definition of a left order is analogous. 2.4. Theorem (Schur's Lemma for rationally uniform, homogeneous modules) Let A be a ring and V £ Y. . Denote: V Q = the quasi-injective hull of V . A = HomA(VQ,vQ).. n = HomA(V,V) . Then: (i) V = AV (where AV = {Ya.v.la. £ A , v. £ V } ) . Q u l i ' l l (ii) A is a division ring. ( i i i ) 9, is a right order in A . (iv) V Q is a rationally uniform module. Remark: The homogeneity of V is needed only for the proof of ( i i i ) . Proof: (i) From Theorem 2.2 we know that V = TV where r = Horn.(V^,VU) v. A H H and V is the injective hull of V . Let a £ T and v £ V_. . Then n x v has the form v = Y 3.v., with 3. £ T, v. e V . Thus, . , 1 1 i i ' i=l - 2 2 -n n av = a( I g.v.) = [ a3.v. e TV = V . 1 1 . T i i Q i=l i=l Therefore a\ E A for every a e Y . Let v E V . As we know, v Q n Q has the form v = Y a.v. where a. e T and v. e V . Thus, s . , 1 1 1 1 ' i=l n n v = y a.v. = Y a.L v. e AV . So V 0 <=• AV . But clearly AVcV , so V Q = AV . (ii) First we prove the following: (*) If a e A and ker a =f 0 , then aV = 0 . Let a e A and ker a ^  0 . As Vu is an essential extension of V IT (by 2 . 1 ) , so is VQ . Thus, Vn ker a ^  0 . Let W = {v £ v|av E V} . Then W is a submodule of V, and clearly: 0 ^  V D ker a W ^  V . V is rationally uniform, therefore V is a rational extension of V A ker a . The mapping a|w: W > V is zero on V f) ker a, thus by Proposition 1 . 8 , aW = 0 . This implies V Cl aV = 0 , and as V^ is an essential extension of V, we must have aV = 0 . As VQ is quasi-injective, by Theorem 2 . 3 we know: J(A) = {a e A|ker a is essential in V Q K and A/J(A) is Von Neumann-regular. We show now that J(A) = 0 . n Let a £ J(A) . Let v E V„ . By (i), v has the form v = 7 8 v , Q . , l I i=l where 3 ± E A, v i E V . As a £ J(A) and B i £ A, we have a$± e J(A), for i = l,...,n . Thus k e r ^ g j is an essential submodule of V^ , and in particular ker(a$±) / 0 for i = l,...,n . By (*), ag.V = 0 for n n i = l,...,n . .Thus: av = a( Y g.v.) = Y ag.v. = 0 . Therefore a = 0 , 1 = 1 1 = 1 - 23 -and we have J ( A ) = 0 . So A is a Von Neumann-regular ring. To show that A is a division ring, let 0 ^  a e A . As A is Von Neumann-regular, there exists y e A such that a = aya . So a(l - ya) = 0 . Suppose ker a ^  0 . Then by (*), V ker a cz V Q . As V is an essential submodule of V Q , so is ker a . But then a e J ( A ) = 0, a contradiction. Therefore ker a = 0 . As a(l - ya) = 0, this implies 1 - ya = 0 . Thus ya = 1, which proves that A is a division ring, ( i i i ) First we show that Q can be considered a subring of A . Let a e Q .' By the quasi-injectivity of V Q , a can be extended to a homomorphism in A . We show that this extension is unique. Suppose f,g e A , f | v = g|v . Then 0 4 Vcr ker(f-g), and as A is a division ring, f - g = 0 . For every a e J2, denote by a the (unique) extension of a to V Q . Then a • > a is a mapping 0, >• A . Using the uniqueness of the extension, it is easy to check that this mapping is a monomorphism of rings. Therefore Q can be considered a subring of A . We want to prove that Q is a right order in A . As A is a division ring, the first requirement in the definition of a right order is trivially satisfied. Let A be a nonzero element of A . Then XV 4 0 (because A is invertible), therefore, as V Q is an essential extension of V, V DXV 4 0 . Thus W = {v e v|Av e V} is a nonzero submodule of V . By the homogeneity of V, there exists a monomorphism f: V > W . Clearly, \ | w f e Q . For every v e V we have: (X| f)v = A| w(f(v)) = A(f(v)), and (Af)v = X(f:(v)) = A(f(v)). So (A| f) and Af are equal on V, and as A is a division ring, this implies (A| f) = Af . Clearly, f 4 0 because f is one-to-one. Thus w f is invertible, and we have: A = (A[ f)(f) \ the required representation - 24 -of X . This proves that 0, is a right order in A . (iv) To prove that is itself rationally uniform, let 0 ^  WC T C submodules, and let f e HO I I I a ( T , VQ ) such that fW = 0 . As is quasi-inj ective, there exists an extension f e A of f . As A is a division ring and 0 ^  WCker f, we get f = 0, and so f = 0 . Thus is a rationally uniform module. Q.E.D. The following density theorem for quasi-injective modules is a generalization of the Jacobson Density Theorem. Definition: Let A be a ring and V a A-module. The elements v^,...,v e V are said to be linearly independent over A if none of them is a A-linear combination of the others, i.e. if v. i. > Av. for every i = l,...,n . It is clear that when A is a division ring and V a vector space over A, the above definition is equivalent to the usual one. 2.5. Theorem (Density theorem for quasi-injective modules) Let V be a quasi-injective A-module which satisfies: (*) v e V , Av = 0 implies v = 0 . Denote: A = HomA(V,V) . Then for every A-linearly independent elements v i ;...,v e V there 1 n exists an element a e A such that av, 4 0, av„ = ... = av = 0 . 1 ^ n i Proof: The proof is by induction on n . For n = 1 the result is just the given condition (*). - 25 -We prove the result for n = 2 . Let v ^ e V be independent over A . Suppose that there exists no a E A such that av^ ^ 0, av2 =0, i.e. we have: (**) a e A, av2 = 0 implies av^ = 0 . Consider the submodule Av2 of V and define f: Av2 >• V by f(av 2) /= av^ . Because of (**) f is well-defined, and i t is clearly an A-homomorphism. As V is quasi-injective, there exists a e A with a|. = f . Then for every a e A we have: A V 2 ot(av2) = f(av 2) = av^ and a(av2) = aa(v2) = a(av2> . So a(v^ - av2) = av^ - a(av2) = 0 for every a e A, i.e. A(v^ - cw2) = By (*), v^ - av 2 =0, v i = a v2 ^ n contradiction to the A-independence of V1'V2 " This establishes the result for n = 2 . Now suppose i t is true for n, and we will prove i t for n + 1 . Let v_,v1,...,v e V be 0 1 n independent over A (n _> 2) . Suppose that there exists no a e A such that av n f 0, av.. = ... = av =0, i.e. we have: u i n (***) a e A, av. = ... = av =0 implies av_ = 0 . 1 n 0 Consider .A as a module over itself, and define the maps f ,g , ...,g •:• A — by: f(a) = av^ , a e A g1(a) = av 1 , a e A g (a) = av , a e A . n n - 26 -A V f,g^,...,g are clearly A-homomorphisms. Consider the direct sum of n copies of V, and for i = 1,. .. ,n let e_^ : V >- V & ... © V be the embedding of V on the i-th component of V <9 ... © V . Let n g = I ei g±' i , e - S( a) = (§i(a)»•••>§n(a)) for every a e A . i=l v e . . . a v Consider From (***) we know that ker g C? ker f . Therefore there exists a homo-morphism h: gA > V such that hg = f . For , i = l,...,n let K. = f\ ker g., and let K = K, + ... + K . x ./. j 1 n Clearly, K is a submodule of A . For i = l,...,n we have: e.g.KCZgK . (For let a e K . Then a = a. + ... + a where xx 1 n a. e K., j = l,...,n, and so e.g.(a) = e.(g.(a,+...+a )) 3 2 xax x 6x 1 • n e.(gi(a1)+...+g1(a.)+...+gi(a^)) = e.tg.U.)) = ( O ^ . g ^ a ^ 0) = g(a^) e gK .) For fixed i consider the homomorphisms g ±K • • gK ^ V . As V is quasi-inj ective, the mapping (h| ) (e. | ): g.K — V can be extended to a homomorphism e A . For every a e K and i = l,...,n we have: a^g^(a) = he^g_^(a), and therefore: n n n £ a.g.(a) = £ he.g.(a) = h( £ e.g.(a)) = hg(a) = f(a), for every a e K , - . 1 1 « - | l l a l l 1 X=l X=l 1=1 n In particular, for every a e K = f\ ker g.: . j=2 3 n J f(a) = J o^g^(a) = a^g^(a), and using the definition of f and g^, we i=l n get: av Q = a^av^) = a^v^) , a ( v Q - a i v i ^ = °» f o r every a e f\ ker g j=2 This means: - 28 -a e A, av„ = •.. = av =0 implies a(v„ - a_v.) = 0 . Therefore, £ n (J 1 1 by the induction hypothesis, the elements V Q - a^v^,V2,...,v^ are linearly dependent over A . But from this i t follows easily that the elements vo' vl'***' Vn a r e a"*"so l i n e a r l y dependent over A , a contradiction. Q.E.D. The Jacobson Density Theorem follows quickly from the above theorem by observing that an irreducible module V is quasi-injective, and satisfies the condition (*) (because {v e v|Av =0} is a submodule of V). (Let v^,...,vn e V independent over A, and w^,...,w e V arbitrary. By the above theorem there exist a.,,...,a e A such that a.v. ^ 0 for I n x i i = l,...,n, and a.v. = 0 for i ^ j . As V is irreducible, Aa.v. = v i 3 i i for i = l,...,n . Therefore there exist b. e A such that b.a.v. = w., 1 1 1 1 x n x = l,...,n . Take a = £ b.a. . Then clearly: av. = w., i = l,...,n • i J J i i 3=1 J J The following density theorem will give, as a corollary, the density theorem for our class of modules, £ . 2.6. Theorem (Density theorem) Let V be an A-module, E an extension of V, and suppose that the following properties are satisfied: (i) V 4- 0, V is homogeneous. (ii) E is quasi-injective, uniform, and: (*) v e E, Av = 0 implies v = 0 . Denote: A = HomA(E,E) 9, = HomA(V,V) . Then for any A - l i n e a r l y independent elements v i ' * " ' ' v n e ^ a n d a n ^ elements w,,.. . ,w e V there e x i s t s an element a e A and a nonzero 1 n mapping A e Q such that av., = Aw, ,. .., av = Aw 1 1 n n Proof: Let the A - l i n e a r l y independent elements v,,..,,v e E - and the I n elements w, ,...,w e V be given. Denote for i = l , . . . , n : I n A_^  = {a e A|av. = 0 for every j ^ i} . As E i s quasi-inj ective and s a t i s f i e s (*), by 2.5 A / v ^ f 0 for every i . C l e a r l y , A^, A^ are l e f t i d e a l s i n A, so A, v,,...,A v are nonzero submodules of E 1 1 n n As E i s uniform, and each of the submodules V, A.v,,...,A v i s 1 1 n n n nonzero, we have V f) (D A.v.) ^ 0 . As V i s homogeneous, there e x i s t s a monomorphism j = l J 3 n A: V > V A ( 0 A.v.) . j = l 3 J C l e a r l y , 0 ^ A e ft . For i f i x e d , as Aw. e A.v., there e x i s t s l l l a. e A. such that Aw. = a.v. . i i i l l Take a = a^ + ... 4- . Then for every i = l , . . . , n we have n n av. ( / a.)v. = 7 a.v. = a.v. = Aw. l . i j l . i j l i i l 3=1 J j = l Thus av. = Aw., for i = l , . . . , n . Q.E.D. i i Now we are able to state the density theorem for the class J 2.7. Theorem (Density theorem for r a t i o n a l l y uniform, homogeneous modules) Let A be a r i n g and V e J. . Denote: - 30 -VQ = the quasi-injective hull of V , A = Hom A(V Q,V Q), fi = HomA(V,V) . Then for any A-linearly independent vectors v.,...,v e V_ and any 1 n u. vectors w,,...,w e V there exists an element a e A and a nonzero 1 n mapping X e 9. such that: av, = AW., ,.. . ,av = A W l i n n Remark: It is justified to use the term "vector" because A is a division ring by Theorem 2.4(ii). Proof: To be able to apply Theorem 2.6, we have to know: (i) V 4 0, V is homogeneous, (ii) VQ is quasi-injective, uniform, and satisfies: (*). v e V Q , Av = 0 implies v = 0 . As V e J A , we know that V is nonzero and homogeneous. VQ is, of course, quasi-injective, and by Theorem 2.4(iv) VQ is rationally uniform, and therefore uniform by 1.10. The only thing to be checked is that VQ satisfies (*). Let W = {v e VQ|Av = 0} . Then W is clearly a submodule of VQ . Suppose W 4 0 . Then, as VQ is an essential extension of V , V C\ W 4 0 . By the homogeneity of V there exists a monomorphism f: V — - > V D W . For every a e A and v e V we have: f (av) = af (v) = 0 because f(v) e W . As ker f = 0, we get av = 0 for every' a e A and v e V , a contradiction because V is a nontrivial A-module. Therefore W = 0, i.e. VQ satisfies (*). Thus we can apply Theorem 2.6 - 31 -and get the result. Q.E.D. Now, as we have in our hands Schur's Lemma and the Density Theorem for the class we are able to describe the structure of those rings A for which J A has a faithful member satisfying a certain condition of finite dimensionality. \ 2.8. Theorem Let A be a ring, V e £ A faithful. Denote: Vg = the quasi-injective hull of V, A = HomA(VQ,VQ) . Suppose that V^ is finite dimensional over A . Then A is a left order in the simple, Artinian ring Hom^ (Vp,Vp) . Proof: We know that A is a division ring by Schur's Lemma for the class (Theorem 2.4(H)) . Hom^ (Vq,VQ) is, of course, simple and Artinian, being the ring of linear transformations of a finite dimensional vector space. As V is a faithful A-module, is also faithful. Therefore A can be considered a subring of H O I I I ^ ( V Q , V Q ) , by considering an element a e A to be the linear transformation a : Vn > V defined by a (v) = av for v e V« . We want to prove that A is a left order L v. in HO I I I ^ ( VQ , VQ ) . Denote by n the dimension of over A . By Theorem 2.4(i) V. •= AV, so we can choose a basis v,,...,v of V. Q 1 n Q over A such that v,,...,v e V . (For let u,,...,u be a basis of I n I n V Q . As V Q = AV, each u^ is a finite linear combination of elements v.,,...,v. e V . Clearly the vectors v,,,...,v, ,...,v ;,...,v e V i l lm. 11 lm, nl nm I I n generate V^, thus we can choose a subset of them which is a basis of V Q . ) - 32 -Now we show that the singular submodule of V Q is zero. Suppose the contrary, and let 0 ^ v e Z ( V Q ) • Let i be fixed. By the Density theorem for £ (Theorem 2.7), there exists a. e A and 0 f X. e Horn. (V,V) X X /A. such that a.v = A.v. . As V~ is quasi-injective, A. can be extended x x x Q , x to a mapping a^ e A . We have: a.v = a.v. for i = l,...,n . x x x ZCV^) is a submodule of V^ , thus a.v,...,a v e Z(Vrt), and therefore Li u. 1 n Q I. = (0:a.v) , i = l,...,n, are essential left ideals in A . We have: x x 0 = I.a.v = I.a.v. = a.I.v. for i = l,...,n . x x x x x x x x Thus I.v. CZ ker a., i = l,...,n . As A is a division ring, and x x x 6 0 ^ a. e A, we get I.v. = 0, for every i = l,...,n . I,,...,I x x x I n n are essential left ideals of A, therefore I. is essential, and i=l 1 n n in particular C) I.' f 0 . Take 0 ± a e C\ I. . Then i=l 1 1=1 1 av. = 0 for i = l,...,n . x As v, v is a basis of V~, this implies aV_ = 0 . But then a = 0 I n Q Q because is faithful. This contradiction proves: Z ( V Q ) = 0 . Next we show that if I is an essential left ideal of A, then I contains an element which is invertible in H O I I I a ( V Q , V Q ) . Consider V and V Q as I-modules. We will show that they satisfy the conditions of Theorem 2.6 (over the ring I). We know that V ^ 0 . To show that V is homogeneous over I, let W be a nonzero I-submodule of V . Then IWC W, IW is an A-submodule of V, and IW 4- 0 . (For let 0 + w e W . As Z ( V Q ) =0, w £ Z(V^), so (0:w) is not essential in - 33 -A . Therefore, as I is essential, I <£• (0:w) . Thus lw 5^  0, and so- IW ^  0 .) V is a homogeneous A-module, therefore there exists an A-monomorphism f: V > IW . Clearly; f is also an I-monomorphism from V into W . Thus V is a homogeneous I-module. We show now that is quasi-injective as an I-module. Let W be an I-submodule of and f e HO I I I ^ W J VQ ) . Then IWC W and IW is an A-submodule of . Furthermore, f | ^ e Hom^ ClWjV^ )- (Because m for a e A, > i , w, e IW we have: f (a > i , w, ) = f(Y(ai, )w, ) = ' L- k k L k k L k k k=l ^(aiDfCw^) = a i^f (w^ ) = af (£ ij^^) •) As N Y Q 1 S quasi-inj ective as an A-module, there exists g e Hom.(V , V N ) such that g| = fI . A U. x lw iW Clearly, g e Hom^.(V^,V^) • Let i e I, w e W . Then: i(f(w) - g(w)) = if(w) - ig(w) = f(iw) - g(iw) = 0 . Thus I(f(w) - g(w)) = 0 . As I is an essential left ideal of A, this implies f(w) - g(w) e Z ( V ^ ) = 0, and so f(w) = g(w), for every weW, i.e. g|^=f . Therefore is a quasi-injective I-module. To prove that is a uniform I-module, let W^ ,W2 be nonzero I-submodules of . . Then IW-^, IW2 are nonzero A-submodules of . (Again, they are nonzero because I is essential and Z ( V Q ) = 0 .) By Theorem 2.4(iv) is rationally uniform as an A-module, so is a uniform A-module. Thus IW^  D IW2 ^  0 . As IW^ CT W^  and IVl^CZ W2, we have W^  C\ W2 f 0, and therefore is a uniform I-module. To be able to use Theorem 2.6, i t remained to check that i f v e and Iv = 0, then v = 6 . But this is clear, because Iv = 0 implies, as I is essential, that v e Z(V^) = 0 . Thus the I-modules V<n satisfy the requirements of Theorem 2.6. We show now that - 34 Hom I(V Q,V Q) = A . Clearly, AC Hom.j. (VQ,VQ) . To show the opposite inclusion, let f e H O H I I ( V Q , V Q ) . I V Q is an A-submodule of V Q , and f | I Y : I V Q • V Q is an A-homomorphism. As V Q is quasi-injective as an A-module, there exists a mapping a e A such that ot [ = f | . For every i e I Q Q and v e VQ we have: i(f(v) - a(v)) = if(v) - ia(v) = f(iv) - a(iv) = 0 . So for every v e V Q , I(f(v) - a(v)) = 0, and therefore, as I is essential, f(v) - a(v) e Z ( V Q ) = 0 . Thus f = a e A . This shows: HOII1J.(VQ,VQ) = A . Therefore v^,...,vn are linearly independent over HOIILJ.(VQ,VQ) . By Theorem 2.6 there exists an element e e l and a nonzero mapping A e Hom^(V,V) such that ev, = Av,,...,ev = Av 1 1 n n As V Q is quasi-injective as an I-module, A can be extended to a mapping y e Hom.j.(VQ,VQ) = A . Thus we have e e l and 0 4 y e A such that ev, = Y V,,...,ev = Y V 1 1 n n The vectors Y V , , . . . , Y V are, clearly, linearly independent over A, 1 n and as dim.V^ = n, Y V , , . . . , Y V is a basis of V^ over A . Therefore A Q 1' n Q e, considered as a linear transformation on the vector space V Q , is invertible. So I contains an element which is invertible in HomA(VQ,VQ) . Now we can prove that every element <j>' e HO I I I a(VQ,VQ) is of the form cf> = a H> with a,b e A . Let c(> e HO I I I a(VQ,VQ) . Denote A_^  = {a e A|av_. = 0 for every j 4 x}, for i = l,...,n . Clearly, A,,...,A are left ideals of A . We know that if v e V_ 1 n u. and Av = 0, then v = 0 (because Z ( V Q ) = 0)• Therefore by the Density - 3 5 -theorem for quasi-inj ective modules (Theorem 2 . 5 ) , A v f 0 for every i = l,...,n . Thus A,v,,...,A v are nonzero submodules of V„, and l i n n Q as V Q is uniform (since it is rationally uniform by 2 . 4 ( i v ) ) , A,v,.....A v are essential submodules of V_ . Let l i n n Q X. = (x e A | X I } ) ( V . ) e A.v.}, for i = l,...,n . Then X, ,. . . ,X l ' i i i I n are essential left ideals of A . (For let J be a nonzero left ideal of A . If J(f(v.) = 0 , then J C x± and so J H X = J 4 0 . If Jcf>(v^ ) 4 0 , then Jc})(v^ ) is a nonzero submodule of V^, and as A^v^ is essential in V^, Jt() (v.) A A.v. 4 0 . So there exists i e J such Q l ' ' l l J that 0 4 j<j)(v.) e A.v. . We have 0 5^  j e J C\ X. . Thus X. is an n 1 1 essential left ideal of A .) Take X = f\ X. . Then X is an i=l 1 essential left ideal of A . Therefore there exists an element a e X n which is invertible in Hom^(V„,V ) . As a e X = f\ X., we have: ^ i=l 1 ad>(v.) e A.v., for i = l,...,n . 1 x i So there exist elements b. e A., i = l,...,n, such that 1 1 ad)(v.)=b.v., for i = l,. . . , n . 1 1 1 Take b = b^ + ... + b • Then for every i = l,...,n: n bv_, = ( \ \>.~)v. = \ b4v. = b.v_^  = a<}>(v^ ) . T. = ( Y b.)v. = y b.v. = b.v. 1=1 J=l Thus: aKv^ = bvx ad)(v ) = bv n n So the linear transformations a<j> and b act equally on the basis - 36 -v, ,. . . ,v of V„ . Therefore a<j> = b, and as a is invertible in 1 n Q HO I I I^CVQ JVQ) , we have: <j> = a H> , a , b e A . Finally, we have to prove that every element of A which is regular in A, is invertible in HO I I I^CVQ JVQ) . This will follow from the following elementary property of finite-dimensional vector spaces: Lemma: Let W be a finite-dimensional vector space over a division ring D . Let f e HonipCWjW) such that: g e HonipCWjW), gf = 0 implies g = 0 . Then f is invertible. (Proof: Suppose f is not onto, i.e. fW is a proper subspace of W . Then there exists a nonzero linear transformation g on W which is zero on fW .. So gf = 0 and g 4 0, a contradiction to the given property of f . Therefore f is onto, and because of the finite-dimensionality of W, f is also one-to-one, and the lemma is proved.) Now let a e A be regular in A . Let g e HO I I I a(VQ,VQ) and suppose ga = 0 . We know that g has the form g = b "*"c with b,c £ A . So b "*"ca = 0, and multiplying by b from the left we get ca = 0 . As c e A and a is regular in A, c = 0, and therefore g = b "*"c = 0 . As V Q is finite-dimensional over A, by the Lemma a is invertible in HO H I a ( VQ , VQ ) . Thus, A is a left order in HO I I I a(VQ,VQ) . Q.E.D. - 37 -CHAPTER 3 THE GOLDIE THEOREM Our aim is to prove the theorem of Goldie which gives the structure of prime, Goldie rings. In the previous chapter we obtained, with the help of Schur's Lemma and the Density Theorem for the class the structure of those rings A for which £ A contains a faithful member V satisfying the finite-dimensionality condition dim^Vg < °° (where VQ is the quasi-inj ective hull of V, and A = HO I I I a ( VQ , VQ ) . ) Thus, to get the structure of prime, Goldie rings, a l l we have to show is the following: If A is a prime, Goldie ring, then ^ contains a faithful module V such that dim^V^ < °° . 3.1. Prime rings The next proposition is elementary. Proposition: The following conditions on a ring A are equivalent: (i) If I,J. are ideals in A and IJ = 0, then 1=0 or J = 0 . (ii) If I,J are left ideals in A and IJ = 0, then 1=0 or J = 0 . (ii i ) If I,J are right ideals in A and IJ = 0, then 1=0 or J = 0 . (iv) If x,y e A and xAy = 0, then x = 0 or y = 0 . Definition: A nonzero ring A is said to be prime if i t satisfies one - 38 -of the above equivalent conditions. It is clear that every simple ring is prime. But, for example, the ring of integers is prime but not simple. 3.2. Goldie rings Definition: Let S be a nonempty subset of a ring A . The left annihilator, £(S), of S is defined to be £(S) = {a e A|aS = 0} . It is clear that £(S) is a left ideal in A . (If S consists of the single element x, we denote £(S) = £(x) .) Definition: A left ideal I of a ring A is said to be a left  annihilator if there exists a nonempty subset S of A such that I = l(S) . Definition: Let I and J be left ideals in A . I is said to be a complement of J i f : (i) I f) J = 0 (ii) For every left ideal I' of A such that I ' ^ I, we have I'D J 4 o . Definition: A left ideal I of a ring A is called a complement if there exists a left ideal J of A such that I is a complement of J . Definition: A ring A is said to be Goldie i f : (i) A satisfies the ascending chain condition on left annihilators (i.e. there exists no strictly increasing, infinite sequence of left annihilators in A), - 39 -(ii) A satisfies the ascending chain condition on complements. Of course, every Noetherian ring is Goldie. 3.3. Proposition The following conditions on a ring A are equivalent: (i) A satisfies the ascending chain condition on complements. (ii) There exists no infinite sequence I..,I0,...,I of nonzero 1 / n 00 left ideals of A such that the sum J I is direct. n=l OO (The sum Y I is direct means that for every n > l : x, + ... + x = C \ n  J — 1 n n=l x. e l . implies x, = ... = x = 0 . ) x x 1 n Proof: (i) implies ( i i ) : Suppose the contrary, i.e. there exists an infinite sequence I.,I0,...,1 of nonzero left ideals of A such 1 z n 00 that the sum Y I is direct. By Zorn's Lemma there exists a left , n n=l ideal TQ of A, which is maximal with respect to the property: T Q f) (I± + I 2 + ...) = 0 . Again by Zorn's Lemma there exists a left ideal T^ of A, which is maximal with respect to the properties: ' T 0 (I 2 + I 3 +...) = 0 (The set of a l l left ideals T of A which satisfy T f) (I 2 + I 3 + ...) = and T o TQ + 1^ is nonempty because T^  + 1^ belongs to it.) We continue by induction. Suppose we have already found left ideals TQ JT^,...>T N_ 1 such that for every i = l,...,n - 1, T is maximal with - 40 -respect to the properties: T i n ( I i + l + I i + 2 + - " ) = ° T. =3 T. 1 +1. . 1 l - l l Then, by Zorn's Lemma, there exists a left ideal T^ of A, which is maximal with respect to the properties: T Ci (I + I + .'. .) = 0 n n+1 n+2 T 3 T n + I • n n-l n (The set of a l l left ideals . T of A which satisfy Tfl(I ,n + I ,_...) = 0 n+1 n+2 and T O T . + I is nonempty because i t contains T , + I . For let n-l n n-l n x e (T .. + I ) A (I •-. + I , o + •••) • Then x = t , + i = i ,.+... + n-l n n+1 n+2 n-l n n+1 where t -, e T i . e l . , m > n + l . n-l n-l' j j -t , = - i + 1 , . + . . . + 1 eT . CS (I + I , , + . . . ) = 0 . So t ,=0, n-l n n+1 m n-l n n+1 n-l oo and - i + i ... + ...+ i =0, and as Y I is direct, i . = 0 for n n+1 m u. n i n=l J a l l j . Thus x = t , + i =0.) J n-l n So we have an infinite sequence T-,T_,...,T ,.. of left ideals such that u i n for every n >_ 1 T^ is maximal with respect to the properties Tn ^ ( In+l + ^+2 + ...) =0 T 'r>T . + I . n n-l n It is clear that T is a complement of I , , + I ,„+..., and that n n+1 n+2 T - «= T . Also, T T D I Cl T . , A ( I + I , 1 + ...)=0, therefore, n-l n n-l n n-l n n+1 as I 4 0, T , f ) I ^  I • Thus I <£T But I C T , + I C T T , n n-l' n n n ^ n-l n n-l n n and so: T . T . Thus we have a strictly increasing, infinite n-l 4 n sequence of complements, T ^ T. *-? ... ^ T . . . , in contradiction to o 4 1 4 4 n 4 (i). - 41 -(ii) implies (i): Suppose the contrary, i.e. there exists in A a strictly increasing, infinite sequence L 9 1 , 9 ... 9 l 9 ... of L f I T T n T complements. For every n > 1 there exists a left ideal J of A — n such that I is a complement of . We have for every n > 1: I . D J = 0 n n I D J 4 0 (because I <T I ..) . n+1 n n f n+1 Denote: K = I f] J , for n > 1 . Then K1,KOJ...,K ,.. are n n+1' ' n — 1 2 n oo nonzero left ideals of A . We show that the sum Y K is direct. n=l Let k. + ... + k =0, where k. e K., i = l,...,n . Then 1 n i i k = (-k.,) + ... + (-k ,), and -k_ e I„,..., -k , e I . As n 1 n-1 1 I n-1 n L C ...C I we get k =• (-k,) + ... + (-k ,) e I . But also 2 n n 1 n-1 n k e J , thus k e l A J =0, k = 0 . Similarly, k , = 0,..., n n n n n n J n-1 oo = 0 . So the sum \ K is direct, in contradiction to ( i i ) . Q.E.D. n=l 3.4. Proposition Let A be a prime ring which satisfies the ascending chain condition on left annihilators. Then, considering A as a module over itself, the singular submodule of A is zero. Proof: Suppose the contrary, i.e. Z(A) 4 0 . Let X = {£(x)|0 4 x e Z(A)}. Then X is a nonempty set of left annihilators in A, and therefore, as A satisfies the ascending chain condition on left annihilators, X contains a maximal element, £(x Q), where 0 4 x Q e Z(A). Let L = {ye A|yZ(A) = 0}. Then L is a left ideal in A, and LZ(A) = 0 . Z(A) is also a left ideal in A (by 1.7 i t is a submodule of A as a module over itself), therefore, as A is a prime ring and Z(A) ^ 0, we have L = 0 . - 42 -Therefore x^ i L, and so x^ Z(A) 4 0 . Thus there exists z e Z(A) such that X g Z 4 0 . We show now: £ ( x g ) ^ £ ( X Q Z ) . It is clear that £ ( X Q ) CZ £(X Q Z) . A X Q is a nonzero left ideal in A . ^ u because if A X Q = 0 then x Q A X Q = 0 a n d by the primeness of A, x^ = 0, a contradiction.) As z e Z(A), £(z) is an essential left ideal of A . Therefore A X Q ^ £ ( Z ) 4 0 . Thus there exists a e A such that ax^ 4 0, axQz = 0 . We have a e £(xQz), a ^  £(x Q), and so £(XQ) ^ " ^ ( X Q Z ) . As 0 4 X Q Z e Z(A), £(x^z) e X, a contradiction to the maximality of £(xQ) in X . Thus: Z(A) = 0 . Q . E . D . Now we are ready to prove the theorem which was the aim of this chapter, and which gives, as an immediate corollary, the Goldie Theorem. 3.5. Theorem Let A be a prime, Goldie ring. Then £ contains a faithful module V such that dim^V^ < <» . (V^ d enotes t h 6 cjiicLsi -injective hull of V, and A = HomA(VQ,VQ) .) Proof: First we show that A contains a nonzero left ideal V which is uniform as an A-module. Suppose that i t does not. Then A is not uniform as an A-module. Therefore there exist nonzero left ideals ^ I ' ^ I in A such that D W^  = 0 . As is not a uniform A-module, there exist nonzero left ideals V2,W2 of A such that V^C. V , W2<= V± and V^AWj = 0 Continuing by induction we obtain two infinite sequences of nonzero left ideals, V,,VOJ...,V and W,,W„,...,W with the properties: 1 2 n 1 2 n V r> V ,. , V 3W „ and V D W =0 for every n > 1 . n n+1 n n+1 n n — - 43 -We show that the sum > W xs dxrect. Suppose w„ + . . . + w =0, i n 1 n n=l where w. E W., i = l,...,n . Then w. = (-w0) + ... + (-w ), and i x 1 z n i -w. eW„Cv,...,-w e W <ZZ V . C V , . Thus 2 2 1 n n n-1 1 w = (-W2) + ... + (~wn) e 0 W = 0, w^  = 0 . Similarly, 00 w„ = 0,...,w = 0, and the sum > W is direct. By Proposition 3.3 2 n -i n n=l this is a contradiction to the fact that A satisfies the ascending chain condition on complements. Therefore A contains a nonzero left ideal V which is a uniform A-module. We will prove that V e V is faithful, and dim^/^ < » . First, as A and V are nonzero left ideals of A, and A is a prime ring, AV 4 0 so V is a nontrivial A-module. Clearly, Z(V)^ Z(A) . By Proposition 3.4, Z(A) = 0, therefore Z(V) = 0 . Thus V is ) - 44 -rationally uniform. To prove that V is homogeneous, Let W be a nonzero submodule of V . As V and W are nonzero left ideals in A, and A is a prime ring, VW 4 0 . Thus there exists WQ e W such that V W Q 4 0 . Define f: V —> W by f(v) = vw^  for every v e V . It is clear that f is an A-homomorphism. Suppose ker f 4 0 . Then ker f is a nonzero submodule of V, and as V is rationally uniform, V is a rational extension of ker f . f e Horn.(V,V) and f|, ,. = 0, A 'ker f therefore, by 1.8, f = 0 . But then fV = Vw^  =0, a contradiction. So f is an A-monomorphism, and this proves the homogeneity of V . Thus: V e £ A . For the annihilator (0:V) of V we clearly have (0:V)V = 0, and as V 4 0 and A is a prime ring, this implies (0:V) = 0 . So V is a faithful module. Denote: < VQ = the quasi-injective hull of V, A = HomA(VQ,VQ) . It remained to prove that dim^ Vp < °o . Suppose the contrary, i.e. that V Q is infinite dimensional over A . Then there exists an infinite sequence v^,v 2,•••jV^,... of A-linearly independent vectors in . Let I± = (0: V l) I 2 = (0:V;L) f) (0:v2) I = (0: V l) D ... f) (0:v ) n 1 n - 45 -Consider the decreasing sequence of left ideals 1 I n n+1 We show that for every n >^  1 there exists a nonzero left ideal of A such that: j c i but J H i ,. = o . n n n n+1 Let n be fixed. Take a nonzero vector w e V . As v,,...,v ,v n 1 n n+± are linearly independent over A, and as w e V, by the Density theorem for the class \ (Theorem 2.7), there exists an element a E A and a nonzero mapping A e HomA(V,V) such that av, = 0,...,av =0, av , = Aw . 1 n n+1 We claim that 0 ^  av ,, e V . As A:V > V and w e V, av ., = Aw e n+1 n+1 By the quasi-injectivity of V^, A can be extended to a mapping a e A Clearly a 4: 0, and as A is a division ring, a is invertible. Thus, as 0 4 w e V, av ,, = Aw = aw 4 0 . n+1 We know that Z(V) = 0, therefore av ,, £ Z(V), and (0:av ,,) is not n+± n+i an essential left ideal of A . So there exists a nonzero left ideal K of A such that (0:av ,,) f\ K = 0 . Take: J = Ka . Clearly, J n+1 n n is a left ideal of A . As K f 0, there exists 0 4 k e K . As (0:av .-.) f\ K = 0, k t (0:av ,,), and therefore kav ,, i 0 . Thus n+1 n+± n+1 0 4 ka e Ka = J . So J is a nonzero left ideal of A . As n n av, = 0,...,av =0, we have J v, = Kav, = 0,...,J v = Kav = 0 . 1 n n l I n n n Thus J CZ I . n n - 46 -Let x E J f l I ,n . Then x e Ka D (0 :v ,,) . So x = ka for some n n+1 n+1 k e K, and kav = 0 . Therefore k E (0:av ,,) f) K = 0, and n+1 n+1 x = ka = 0 . Thus J fi I , = 0 . n n+1 We have an Infinite sequence J ,J2,...,Jn»... of nonzero left ideals of A such that: J <Z I , J M = 0, for every n > 1 . We n n n n+1 — oo show that the sum ) J is direct. Suppose i , + ... + i =0, where L, n r r J l Jn ' n=l 3± e J± for i = l,...,n . Then = (-j2) + ... + (~j )» - i 0 £ J 0 C I„,.. . , - i E J O I cz I„ . So 2. 2 ' 2 Jn n n 2 j l = ( ~ J 2 } + ••• + ("V e J l / ^ I 2 = ° ' h = 0 ' 00 Similarly, j 2 = 0,...,j =0, and the sum £ J is direct. By n=l Proposition 3.3 this contradicts the fact that A satisfies the ascending chain condition on complements. Therefore V Q is finite-dimensional over A . Q . E . D . 3.6. Corollary (Theorem of Goldie) Let A be a prime, Goldie ring. Then A is a left order in a simple, Artinian ring. Proof: By the previous theorem, there exist a module V e which is faithful, and satisfies dim^^ < °° (where V Q is the quasi-inj ective hull, of V , and A = H O I I I a (V Q , V Q ) .) By Theorem 2 . 8 , A is a left .order in the simple, Artinian ring H O H I ^ ( V Q , V Q ) . Q . E . D . - 47 -REFERENCES 1. Divinsky, N.J., Rings and Radicals, University of Toronto Press, Toronto, 1965. 2. Faith, C, Lectures on Injective Modules and Quotient Rings, Springer Verlag, Berlin-Heidelberg-New York, 1967. 3. Heinicke, A.G., Some Results in the Theory of Radicals of Associative Rings, Ph.D. thesis, The University of British Columbia, Vancouver, 1968. 4. Koh, K. and Mewborn, A.C, The Weak Radical of a Ring, Proc. A.M.S., 18(1967), pp. 554-559. 

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