DENSITY THEOREMS AND APPLICATIONS by JOZSEF HORVATH B.Sc, T e l Aviv U n i v e r s i t y , 1976 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE i n the DEPARTMENT OF MATHEMATICS We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA June, 1977 o J o z s e f Horvath^ J u n e > 1 9 ? ? In p r e s e n t i n g t h i s t h e s i s in p a r t i a l an advanced degree at further of the requirements f o r the U n i v e r s i t y of B r i t i s h Columbia, I agree the L i b r a r y s h a l l make it I fulfilment freely available for that reference and study. agree t h a t p e r m i s s i o n for e x t e n s i v e copying o f t h i s thesis f o r s c h o l a r l y purposes may be granted by the Head o f my Department o r by h i s r e p r e s e n t a t i v e s . of this thesis for It i s understood that copying o r p u b l i c a t i o n financial gain s h a l l not written permission. Department The of Mathematics U n i v e r s i t y o f B r i t i s h Columbia 2075 Wesbrook P l a c e Vancouver, Canada V6T 1WS June 24, 1977 be allowed without my ii ABSTRACT Supervisor: P r o f . C T . Anderson One way o f g e t t i n g s t r u c t u r e theorems i n r i n g t h e o r y i s t o f i x a general c l a s s £ o f modules, and t o prove Schur's Lemma and t h e D e n s i t y Theorem f o r £ • F o r example, t h e G o l d i e Theorem f o r prime r i n g s f o l l o w s from Schur's Lemma and t h e D e n s i t y Theorem f o r t h e c l a s s of r a t i o n a l l y u n i f o r m , homogeneous modules i n a s i m i l a r way as t h e Wedderburn-Artin Theorem f o l l o w s from Schur's Lemma and t h e D e n s i t y Theorem f o r t h e c l a s s o f i r r e d u c i b l e modules. iii TABLE OF CONTENTS page INTRODUCTION 1 CHAPTER 1: GENERAL CLASSES OF MODULES CHAPTER 2: ANALOGUES OF SCHUR'S LEMMA AND THE CHAPTER 3: REFERENCES 3 DENSITY THEOREM 17 THE GOLDIE THEOREM 37 47 iv ACKNOWLEDGEMENT I would l i k e to thank my supervisor, Prof. Tim Anderson f o r the invaluable help, care and encouragement he gave me i n t h i s t h e s i s and during my whole year a t the U n i v e r s i t y of B r i t i s h Columbia. INTRODUCTION In t h i s paper ring means associative ring which does not contain a unity element and i s not necessarily left commutative. necessarily Module means module. For every ring A let ^ denote, for the moment, the class of a l l i r r e d u c i b l e A-modules. The proof of the c l a s s i c a l Wedderburn-Artin Theorem has the following main steps: (1) If V e then A = Hom^(V,V) is a division ring. (Schur's Lemma) (2) If V £ ^ elements and A = Hom^(V„V), v - , . . .,,v e V 1' n exists an element and any elements a e A This means that the r i n g A is, 1 J , . . . ,w n there A^ =w n of l e f t m u l t i p l i c a t i o n s by elements in a certain sense, dense i n the r i n g l i n e a r transformations e V such that av, = w, , . . . ,av 1 I n of then for any A - l i n e a r l y independent on the vector space Hom^(V,V) of a l l V . (Jacobson Density Theorem) (3) If has a member which i s f a i t h f u l and f i n i t e dimensional (over its centralizer), then A i s isomorphic to M^(D) where D is a d i v i s i o n r i n g and • n >_ 1 . (4) If A i s a simple, A r t i n i a n r i n g , then £ i s f a i t h f u l and f i n i t e dimensional (over i t s contains a member which centralizer). Here, i n order to get the desired structure theorem, we concentrate on a certain fixed c l a s s , of modules, and prove Schur's Lemma and - 2 - the Density Theorem for of those rings member. A £ . for which These enable us to describe the structure ^ has a f a i t h f u l , f i n i t e dimensional Once we know t h i s , the structure of a ring A of the class that we are interested i n i s e a s i l y obtained by showing that for such A, J v 2.^ does have a f a i t h f u l , f i n i t e dimensional member. part of the structure theory i s , Theorem for the class \ therefore, Schur's Lemma and the Density of modules. The aim of this thesis i s to emphasize the usefulness on a fixed class \ The essential of concentrating 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 r a t i o n a l l y 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 r a d i c a l . For example, the general class of irreducible modules defines the Jacobson r a d i c a l . class \ Then we show that the of r a t i o n a l l y uniform, homogeneous modules i s a general c l a s s . In Chapter 2 we give the analogues of Schur's Lemma and the Jacobson Density Theorem for this class of those rings A which have i n \, and use them to describe the structure \^ a f a i t h f u l member satisfying a c e r t a i n f i n i t e dimensionality condition. ' In Chapter 3 we deduce from this the Goldie Theorem. H i s t o r i c a l l y , 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 r i n g s . - 3 - CHAPTER 1 GENERAL CLASSES OF MODULES In o r d e r to study the structure c o n c e n t r a t e on a c e r t a i n f i x e d c l a s s i s the i t i s very u s e f u l of modules. The c l a s s of i r r e d u c i b l e modules which g i v e s the t h e o r y , and in particular of modules, as we In a l l the w i l l be classical Jacobson the Wedderburn-Artin Theorem. s h a l l see, g i v e s the to example structure Another G o l d i e Theorem f o r prime class rings. paper, a homorphism which i s a one-to-one mapping c a l l e d a monomorphism, and mapping w i l l be 1.1. of r i n g s c a l l e d an a homomorphism which i s an onto epimorphism. Radicals Definition: A class g of r i n g s (i) f o r every (ii) f o r every r i n g A t h e r e e x i s t s an ideal l e g , and J i s an A (This largest (iii) if The class Definition: Let if . g(A) = 0 Definition: g A class A e M and and epimorphism and g-ideal f o r every r i n g Example: an A e g is called a radical i f A, epimorphism i d e a l of of A B we I and i s denoted by have in J e g g(A) B e g . A such then that J C I . .) g(A/g(A)) = 0 . of a l l n i l r i n g s be A —> a radical. is a radical. A ring A is called g-semisimple c M nonzero i d e a l I —> of r i n g s I B . of i s c a l l e d r e g u l a r i f f o r every r i n g A t h e r e e x i s t s a nonzero r i n g B e M - 4 - 1.2. Theorem of Kurosh M Let be a regular class of rings. Let be the class of a l l rings M . which cannot be homomorphically mapped onto a nonzero member of Then (i) U.. i s a r a d i c a l . M (ii) For every ring (iii) If 3 A e M, ^(A) 0 • = i s a r a d i c a l such that 3(A) = 0 U^-semisimple ring i s every A e M, for every then g-semisimple. For the proof see Divinsky [1]. 1.3. General classes For an A-module V (0:V) = {a e A|aV = 0} . we denote by Of course, c a l l e d n o n t r i v i a l i f AV ± 0, if (0:V) i.e. (0:V) the annihilator i s an i d e a l of (0:V) i A . of A . V, i . e . V is V 'is c a l l e d f a i t h f u l (0:V) = 0 . Suppose that to every ring class \ A there i s assigned a (possibly of n o n t r i v i a l A-modules. - Such an assignment empty) i s called a class of modules. Definition: A class \ of modules i s said to be a general class provided: (i) If f : A —*• B i s an epimorphism and V e £ then as an A-module i n the obvious way, belongs to (ii) If then (iii) If f: A —• B V, C\ i s an epimorphism, V e jj^ considered as a B-module, i s i n (0:V) = 0 then Y_ ^ 0 and ^ V, considered . ker f C (0:V) £g . f or every nonzero i d e a l I of A . - 5- (iv) If for every nonzero i d e a l I of A =f 0 Y then f\ (0:V) = 0 . V Definition: Let ^-primitive i f Definition: £ ^ be a general class of modules. Let (A.}, . be r i n g s . A AeA T © A, f-c X > - A U A condition: B/I, to B There exist ideals is A called A isomorphic to a subring i s c a l l e d a subdirect sum of the projections This is equivalent to the following 1^ and A. B, of the canonical are onto. i s isomorphic to A A ring £ ffl A , i f the r e s t r i c t i o n s {A^} A ring contains a f a i t h f u l module. of the complete d i r e c t sum rings ^A in . A B such that for every X ^ 1 = 0 . A The following theorem shows that every general class of modules determines a r a d i c a l . 1.4. Let Theorem of Andrunakievic and Rjabuhin be a general class of modules. J (i) a (ii) For every ring (iii) Every Proof: Let (*) if V A, M a(A) = f\ Then (0:V) . £A be the class of ^-primitive r i n g s . V E [A then is nontrivial, V £ ^/(O'V) M = 0} . a-semisimple r i n g i s a subdirect sum of ^-primitive r i n g s . canonical projection that a = is a radical. V £ (As Denote a n C 0 j A/(0:V) £ M . (0:V) ^ A, A F i r s t we observe that so > A/(0:V) . A/(0:V) ± 0 . By property ( i i ) ^ '*"t ^ S c -'- e a r -'-y f a i t h f u l over i s a regular c l a s s . Let Consider the A e M A/(0:V) .) and l e t I of a general class We show now be a nonzero i d e a l - 6 - of A . As A i s ^-primitive, (0:V) = 0 . V £ of a g e n e r a l c l a s s , nonzero member of I/(0:V) . Let ± 0 . J M, Let V e £ . By (*), l/(0:V) and the c a n o n i c a l p r o j e c t i o n maps Therefore M (iii) ^A I i sa onto i s a r e g u l a r c l a s s of r i n g s . U^j be the r a d i c a l determined by show: So by p r o p e r t y M by the Theorem of Kurosh. We U„, = a . We prove t h i s by showing: A £ U^j i f and o n l y i f Suppose A £ . epimorphism A £ a . Then t h e r e e x i s t s a nonzero member A —>- B . As ( i ) of a g e n e r a l c l a s s , t h e r e e x i s t s a module of M . A £ U^j . B e M, V e there e x i s t s . V e ^ . So = a, to show t h a t f o r every r i n g A, so A a(A) c£ C\ a (A) <£ (0:V) . (0:V) . Therefore g a(A) = C\ ^ "|" ^° and by (*), A/(0:V) e M . _ nonzero rxng M . M and an By p r o p e r t y A £ a . A/(0:V) . Then Thus Now we want (0:V) . £A V e Y. such t h a t i s a nonzero i d e a l of v A/(0:V), i s a regular c l a s s , there e x i s t s a j . and an epxmorphism II Bern Thus we have As of i s a nonzero member i s a radical. Then t h e r e e x i s t s Q Suppose onto a V£ Suppose V e £ By (*), A/(0:V) The c a n o n i c a l p r o j e c t i o n maps We have proved A £ a . B a (A) + (0:V) (O'V) ' epimorphisms s n(A 0 ( l A ; . o(A) q(A)n(0:V) s o(A) + (0;V) (0:V) and the c o m p o s i t i o n i s an epimorphism from nonzero member of M . a o(A) onto T h i s i s a c o n t r a d i c t i o n because B which i s a a(A) e a = - 7 - a (A) C Therefore C\ V The r i n g A/a(A) I i a, A/a(A), (0:V) . ^A is a-semisimple, \^ which means so f o r every nonzero i d e a l ^ 0 • By p r o p e r t y I of ( i v ) of a g e n e r a l c l a s s we know: Q ° ^A/a(A) ( : V ) A/a(A) - 0 . Ve a e / O (0:V). VeY Let . Let V e j. A A , / a . ( A By p r o p e r t y ( i ) o f a g e n e r a l ) A c l a s s , i f we c o n s i d e r a e (0:V) . V as an A-module then By the d e f i n i t i o n of 0 = av = av Thus a e (0 )^/a(A) • : V A V v w a E f~\ (0:V)., / t v = 0 . r A/a(A) A/a(A) y a s a n v e V . y member of Thus and t h e r e f o r e as an A-module, f o r every s V e I^/a(A)' a e a(A) . This W e ^ a v e : proves: V £ Z r\ a(A) = V £ Finally, (0:V) . ^A let A be a-semisimple. Then a(A) = (0:V) = 0 . L e t V {I,}, . A AeA be the s e t of a l l i d e a l s of some module i n ^ . Then c l e a r l y , a s u b d i r e c t sum of the r i n g s ^-primitive. 1.5. which a r e the a n n i h i l a t o r of H l ^ =0, A {A/l^} . By (*), and t h e r e f o r e the r i n g s A is A/I^ are Q.E.D. Example For every• r i n g J A ^A A let £ i s a general c l a s s . A be the c l a s s o f i r r e d u c i b l e A-modules. (See H e i n i c k e [3].) The r a d i c a l determined Then by - 8 - £ i s called the Jacobson radical. The above class of modules gives the classical Jacobson, theory. We w i l l work with another general class, the class of rationally uniform, homogeneous modules. 1.6. Essential extensions Definition: Let W extension of W If First, the definitions and elementary properties. V be a submodule of V . V is called an essential i f for every nonzero submodule is an essential extension of W, U of V, we also say that WD U ^ 0 . 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 t r i v i a l . 1.7. The singular submodule A left ideal I submodule of A If V in a ring i s called essential i f i t i s an essential when we consider i s an A-module and It is clear that A v eV A as a module over i t s e l f . we denote: (0:v) = {a e A|av = 0} . (0:v) i s 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 - left i d e a l of Proof: so by we If v, w v+w implies -v e Z(V) left i d e a l of then . I <fc (0:av), . It i s clear and If we = 0 i s an . essential Definition: 1.8. show: are essential (0:v) A as 0 e Z(V) x e I av have left Z(V) e Z(V) left ideals, ( 0 : w ) C (0: v+w) , and that Let I v e such t h a t xav 0 ^ xa ^ 0 e Ia . . A, y e I i s c a l l e d the and be (0:v) such t h a t av singular C o n s i d e r the As 0 i= y e I C\ (0:av) . i d e a l of . Z(V) a nonzero lA(0:av) = 1 ^ 0 . then Thus t h e r e e x i s t s So we (0:w) that IC(0:av) I t i s nonzero because . (0:v), V . . take (0:v) f\ Ia ^ 0 i s a submodule of i s e s s e n t i a l , and v e Z(V), A Z(V) (0:w) e Z(V) a e A, yav Then e Z(V) Let . . (0:v) A 1.6 have Ia A} is If left essential, ya ^ 0 and T h i s shows t h a t e Z(V) . ideal (0:av) Q.E.D. submodule of V . Rational extensions Definition: Let W e x t e n s i o n of W a £ A integer and an be a submodule of i f f o r every n veV, such . O^v' V i s s a i d to be eV a rational t h e r e e x i s t s an element that: av + nv av' V e W + nv' ± 0 . Proposition Let W be (i) V (ii) If and a submodule of V . The i s a r a t i o n a l e x t e n s i o n of T i s a submodule of f(W) = 0 then V f = 0 . following W c o n d i t i o n s are equivalent: . which c o n t a i n s W, f e Hom^(T,V) - 10 - Proof: Let V W ^ T C be a rational extension of W, such that and an integer f(v) ^ 0 . As n f e Hom (T,V), A f =f 0 . Then there exists f (W) = 0 . We want to show: f = 0 . Suppose v e T V, W C V is rational, there exists a eA such that av + nv e W > af (v) + nf (v) ? 0 . But then f = 0, 0 = f(av + nv) = af(v) + nf(v), a contradiction. and (i) implies ( i i ) . To prove that ( i i ) implies (i) let v e V, that for every a eA (*) and every integer av + nv e W Consider the module f (w + av + nv) = av' + nv' By (*) f implies T = W + Av +2v f:T by Therefore 0 ^ v' e V av' + nv' = 0 . . Clearly, W T <=• V . Define > V for weW, f(W) = 0 . Thus, by (ii) , f = 0/ rational extension of W, Suppose n: a e A, i s well defined, and i t is clear that f(v) = f(0 + Ov + lv) = v' ^ 0, be given. n e . f e Hom^(T,V) and and in particular f(v) = 0 . But a contradiction. Therefore and ( i i ) implies (i) . Q.E.D. V is a The next proposition shows the connection between essential and rational extensions. 1.9. (i) Proposition If WCV i s a rational extension, then i t is an essential extension. - 11 - (ii) If W C V i s an e s s e n t i a l e x t e n s i o n and i s an r a t i o n a l Proof: ( i ) Let WC of V . of W, then WcrV extension. V be r a t i o n a l , and l e t Take a nonzero element there e x i s t s Z(V) = 0, a e A u e U . As and i n t e g e r U V n be a nonzero submodule i s a r a t i o n a l extension such t h a t au + nu e W au + nu 4- 0 . Clearly, So V nueW-Ou. O^au + i s an e s s e n t i a l e x t e n s i o n of e x t e n s i o n , and Z(V) = 0 . ( i i ) Let W C V W . Let v e V , 0 ^ v' e V . be an e s s e n t i a l Denote I={aeA|aveW} . Then I i sa left be a nonzero l e f t Jv = 0, then because W i d e a l of J C I and A . Then Jv jv ^ 0 essential left and i d e a l of jv e W . A . As i s not an e s s e n t i a l l e f t Take an element i e I such t h a t Then If V . 0 ^ j Z(V) = 0 v' f 0, and t h e r e f o r e We have: 1.10. Uniform, r a t i o n a l l y u n i f o r m and homogeneous modules V i s an I(0:v') . iv' ^ 0 . Q.E.D. i s s a i d t o be u n i f o r m i f i t i s an e s s e n t i a l e x t e n s i o n of each of i t s nonzero submodules. V I i v e W, V ( i ) An A-mqdule j eJ we know t h a t Thus, Definition: Jv/)Wjf 0 So t h e r e e x i s t s So Let J V . If J v ^ O , then and i v ' =f 0 . W . i s essential. e i A J . i d e a l i n A, i s a r a t i o n a l e x t e n s i o n of I i s a submodule of I / O j= J ^ 0 . i s an e s s e n t i a l submodule of such t h a t (0:v') i d e a l i n A . We show t h a t ( i i ) An A-module i s s a i d t o be r a t i o n a l l y u n i f o r m i f i t i s a r a t i o n a l e x t e n s i o n of each - 12 - of i t s nonzero submodules. It i s clear that a module V i s uniform i f and only i f the intersection of any two nonzero submodules of V i s nonzero. Every irreducible module i s rationally uniform, and by 1.9(i) every rationally uniform module i s uniform. By 1.9(ii) a uniform module which has zero singular submodule, i s rationally uniform. Definition: An A-module V submodule W of V i s called homogeneous i f for every nonzero there exists a monomorphism f: V Clearly, every irreducible module i s homogeneous. see that 2 *• W . It i s also easy to i s a homogeneous module over i t s e l f . Now we are ready to f i x 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. Proof: Then £ i s a general class. The f i r s t 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 O be a ring such that ideal of A . We want to show: V e T. to V such that 1^0.) (0:V) = 0, J ^ 0 • There exists an A-module IV ^ 0 . (Otherwise Consider V and let I be a nonzero I C f\ (0:V) =0, as an I-module. We show: i s a nontrivial I-module. V e £ a contradiction . As Let N = {v e v|lv = 0} . Clearly, an A-submodule of V . Suppose N ^ 0 . Then, as V A-module, there exists an A-monomorphism > N . For every f: V IV ^ 0, N is i s a homogeneous i eI - and v e V we have iv = 0 , we get 1 3 - f(iv) = if(v) = 0 because f(v) e N . As ker f = 0 , for every i e I and v e V, a contradiction to IV ^ 0 . N= 0. Therefore: To prove that V i s rationally uniform as an I-module, let W be a nonzero I-submodule of V, T an I-submodule of V containing f e Hom^CTjV) such that W, and f(W) = 0 . By 1 . 8 i t is enough to show that f = 0 . It i s easy to see that IW and IT are A-submodules of V, 0 ± IWC IT<= V . (IW ^ 0 because N = {v e V| Iv = 0 } = 0 .) As V a rationally uniform A-module, restriction f| : IT >V IW c: V i s an A-homomorphism (because f(a ^ i t ) = K. K. f(E(ai)t) = £(ai )f(t ) = k k k a I\f(tk) = a f(I±ktk)), (IW) = f(IW) = 0 . Therefore f| t e T . Then for every so f(t) E N = 0 . Therefore To show that f =0 , I T = 0, i.e. f(IT) = 0 . and V is rationally uniform as an IW is a nonzero A-submodule of V . As i s homogeneous as an A-module, there exists an A-monomorphism Clearly, f is also an I-monomorphism homogeneous as an I-module, and we have f: V — > W . Therefore Ve £ To prove property (iv) of a general class for ring such that for every nonzero ideal show K If(t) = 0 , V is a homogeneous I-module, let W be a nonzero I-submodule of V . Then again, V f| and i e I: i f ( t ) = f ( i t ) = 0 . Thus Let I-module. is i s a rational extension. The J- J. k and f\ (0:V) = 0 . Denote f: V —> IW. V is . This proves ( i i i ) . suppose that A is a I of A, £ ^ 0 . We want to K = f\ (0:V), i s a nonzero ideal of A, and therefore and suppose K £ 0 . Then Y f 0 . Take a K-module K W £ £ . As KW ^ 0 , K Denote: there exists U = KWQ . Clearly an A-module structure Ax U w^ e W such that KWQ ^ 0 . U is an abelian subgroup of W . We define >U on U in the following way: - 14 - For a e A, kw^ e K W Q = U To show that * Suppose the contrary. and Then (Ak)w^ is homogeneous over K, ker f = 0, to KW 4 0 . Therefore there exists a K-monomorphism x e K, we get xw = 0 w eW for every (Ak)w = 0, that for k e K, u e U We show now that U e \^ . M = {w e w|Kw =0} f: W * * turns w e W, a contradiction is well-defined. U It i s into an A-module, and . Then M is a K-submodule of W . If M + 0, is a homogeneous K-module, there exists a K-monomorphism f(w) e M . As U because f(w) e (Ak)w^ . we have: k*u = ku . y M . For every M = 0 . As we have: x e K, and Q easy to check that the operation then as W W, = [x(ak)]wQ = [(xa)k]w^ = (xa)[kw^] = xa«0 = 0 . x[(ak)Wp)] As Let (Ak)w^ = 0 . i s a nonzero K-submodule of f(xw).= xf(w) = x[(ak)wQ] for some a e A f(xw) = implies a eA . y (Ak)wQ . For every f: W kw^ = 0 be fixed and suppose kw^ = 0 . We show k e K and as W a * kw^ = (ak)w^ . is well-defined, we have to prove: (ak)wg = 0, for k e K Let define k e K, ker f = 0, U = K W Q ± 0, w e W, f(kw) = kf(w) = 0 because we get a contradiction to KW ^ 0 . Thus this implies: A*U => K*U = KU 4 0, and so i s a nontrivial A-module. To show that U is rationally uniform over A-submodule of U, submodule of W, ueU, O^u'eU. therefore, as W there exists an element k e K A, let S be a nonzero Clearly, S i s a nonzero K- i s a rationally uniform K-module, and an integer ku + nu e S ku' + nu' 4 0 . n such that - 15 - Thus we have k £ A and i n t e g e r n with k*u + nu £ S i k*u' + nu' So U i s a r a t i o n a l l y u n i f o r m A-module. geneous A-module, l e t S 0 . i s a nonzero there e x i s t s f | : U S be a nonzero K-submodule of a K-monomorphism > S . For every W . To prove t h a t A-submodule of As W and i s a homo- U . Then a g a i n , i s a homogeneous K-module, f : W —>• S . a £ A U C o n s i d e r the r e s t r i c t i o n u = kw^ £ U we have: flyCaau) = f ( a * k w ) = f ( ( a k ) w ) = (ak) f ( w ) = ( a k ) * f ( w ) = a * ( k * f ( w ) ) = Q Q Q a*(kf(w )) = a*f(kw ) = a*f(u) = a * f ^ ( u ) Q Q monomorphism, and t h i s shows t h a t proved: But U £ \ . K*U = KU, Therefore 1.12. so K = A As K = f\ KU = 0, U . Q Therefore f| i s an A- i s a homogeneous A-module. (0:V), we have i n contradiction (0:V) = 0 . Q KC to (0:U), thus M = {w e w|Kw K*U = 0 . = 0} = 0 . The weak r a d i c a l 1.11 d e f i n e s a r a d i c a l . denoted by W . radical), 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 , W C J and t h e r e f o r e f o r every r i n g two r a d i c a l s are d i f f e r e n t . A = {—|m,n (where A, J rationally i s the Jacobson W(A)C ( i . e . ^-primitive of 1.11), but J a c o b s o n - r a d i c a l Let of J(A) . But the In f a c t , the f o l l o w i n g example g i v e s a r i n g which i s weakly p r i m i t i v e Example: £ T h i s 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 u n i f o r m and homogeneous, we have class have Q.E.D. By the Theorem of A n d r u n a k i e v i c and R j a b u h i n , the g e n e r a l c l a s s A We integers, where £ i s the g e n e r a l (i.e. A eJ). m even, n odd} . I t i s easy to - 16 - see that A is a subring of the rational numbers. module over i t s e l f , and denote i t by member o f ^ . Clearly, and therefore .A A submodules of ^A . As and so ^A V ^ V ^ is a nontrivial module. 2 ^ i s commutative, ' 0 Let f: ^A — V by T h u S A A l s V^, U N I F O R M » monomorphism. Therefore ^ A N D for every x e ^A . f .A e /. . .A A A - A L there are no zero divisors in A . Thus A S A Z ^A ^ A ^A, v^ E V = °' let V be and define is clearly an A- is a faithful module because A is a weakly primitive ring. An easy computation shows that every element in A and therefore be nonzero are ideals in A, ^A . Take a nonzero element f (x) = X V Q 0 ^ v e ^A, V., , V„ 1 2 is rationally uniform. To show the homogeneity of a nonzero submodule of For because there are no zero divisors in A, & V V as a is a faithful A has zero singular submodule. A A A ^A (0:v) = {a e A|av = 0} = 0 A . Then Consider A i s a Jacobson-radical ring. i s left-quasi-regular, - 17 - CHAPTER 2 ANALOGUES OF SCHUR'S LEMMA AND THE DENSITY THEOREM From now on £ w i l l 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: U, A module V is said to be injective i f for every module submodule W of U and homomorphism an extension f e Hom^(W,V), there exists f e Hom^(U,V) of f . Theorem Let (i) (ii) V be an A-module. Then: There exists a maximal essential extension of V, i.e. an essential extension VC M is essential, then M =E. If M^ and M 2 such that i f Vcz M C E and Vcr E are maximal essential extensions of V, then there is an isomorphism between M^ and M 2 which fixes every element of V . i Therefore we can speak about the maximal essential extension of V, which w i l l be denoted by V . - 18 - (Iii) For V i s an i n j e c t i v e module. T1 H Moreover, / module c o n t a i n i n g V, VCUCV,,, U = V then i.e.i f u U V H i s a minimal i n j e c t i v e i s i n j e c t i v e and . the proof of t h i s well-known r e s u l t see e.g. F a i t h [ 2 ] . Definition: 2.2. V i s c a l l e d the i n j e c t i v e h u l l of H The q u a s i - i n j e c t i v e V . hull The f o l l o w i n g n o t i o n g i v e s a common g e n e r a l i z a t i o n 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 . Definition: module A module W of extension V V i s said to be q u a s i - i n j e c t i v e and homomorphism f e Hom (V,V) of A f e Hom (W,V) A i f f o r every sub- there e x i s t s an f . Theorem Let V be an A-module, and l e t T = Horn. (V ,V ) , A H H TT TT and l e t V. = TV = {Ja.v.la. e V, v . e V, Q ' I ' are finite} . Then: (i) Ver V ^ , (ii) V_ x V Q C T and i s a quasi-inj ective i s the s m a l l e s t q u a s i - i n j e c t i v e i.e. i f For be the i n j e c t i v e h u l l of VC QcrV^ and Q V . Denote: and the sums module. module between i s quasi-inj ective, V then and V , H V^ Q . the proof see F a i t h [ 2 ] . Definition: 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 f o r q u a s i - i n j e c t i v e modules. Definition: A ring 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 2.3. Let x e A such that a = axa . Theorem (Schur's Lemma for quasi-injective modules) V be a quasi-injective module over A, and let A = Hom^(V,V) . Then: (i) J(A) = {a e A|ker a (ii) A/J(A) Proof: is an essential submodule of is Von Neumann-regular." Denote N = {a e A|ker a i s essential in V} . Then left ideal in A . For let a, 3 e N . Then in V, so by 1.6 and ker a A ker 3 ker a 0 ker because a e N V} . implies and so -a e N . If a e N is a ker a, ker 3 are essential is essential, thus ker(a+3), N ker(a+3) is essential a + 3 £ N . Clearly and 3 £ A, then 0 e N, ga £ N ker a c ker 3a . From the Jacobson theory we know that the because Jacobson radical of a ring contains every one-sided ideal which is Jacobsonradical. Thus, to show N C J(A), i t suffices to prove that Jacobson-radical, or that every element of Let in a E N . Clearly, V, >- V and so Y + a = ya, 1 - a: V be i t s inverse. extended to a homomorphism v e V, N Jacobson-radical, and > (l-a)V As V ker a is essential is an isomorphism. i s quasi-inj ective, 3 £ A . We have: 3(l-ot) = 1 . Take and as has a left quasi-inverse. ker a O ker(l-a) = 0 . As ker(l-a) = 0 . So f: (l-a)V N N is 3(l-a)v = v f Let can be for every y = -a3 • It is easy to check that is a left ideal of A, y £ N . So N is 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, U D (W + ker a) = U 4 0 . If then 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, As W f\ ker a = 0, a|^:.W f: aW 0 ^ u = (-w) + (w+u) e U C\ (W + ker a) .) > aW i s an isomorphism. Let > 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 i s an essential submodule of V, so i s ker(a - aya), which means that a - aya e N . This proves (*). As N<= J(A), (*) implies that It remained to show: y e A such that Thus J(A)<= A/J(A) is a Von Neumann-regular ring. N . Let a e J ( A ) . By (*), there exists a - aya e N . J ( A ) i s an ideal in A, so ay e J ( A ) . 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 i s a left ideal i n A, we have -aya = g(a - aya) e N . But also J(A)<= N . a - aya e N, and so a e N . Thus 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 i s called regular i f i t is neither a l e f t , nor a right zero divisor in A, i.e. x e A, ax = 0 21 - implies x = 0, and also Definition: x e A, xa = 0 implies x = 0 . Let B be a ring with unity, A a subring of B . (This of course, does not mean that A A contains the unity element of B .) 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. Let Theorem (Schur's Lemma for rationally uniform, homogeneous modules) A be a ring and V £ Y. . Denote: V Q = the quasi-injective hull of V . A = Hom A (V ,v ).. n = Hom A (V,V) Q Q . Then: (i) V = AV (where A V = {Ya.v.la. £ A , v. £ V } ) . Q u l i ' l (ii) A is a division ring. (iii) 9, is a right order in A . (iv) V Q is a rationally uniform module. l 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^,V U A H H and V is the injective hull of V . Let a £ T and v £ V_. . Then n v has the form v = Y 3.v., with 3. £ T, v. e V . Thus, .,11 i i ' i=l v. x ) - 22 - n n = [ a3.v. e TV = VQ av = a(. I g.v.) 1 1 i i i=l i=l . Therefore a\ T E A for every a e Y . Let v E V . As we know, v Q n v = Y a.v. where a. e T and v. e V . Thus, Q has the form . s , i=l v = n n y a.v. = Y a.L 1 1 1 1 ' v. e AV . So V <=• AV . But clearly 0 AVcV , so V = AV . (ii) First we prove the following: Q If a e A and ker a =f 0 , then (*) a e A and ker a ^ 0 . As V Let Then V is an essential extension of V u IT (by 2 . 1 ) , so i s V W aV = 0 . . Thus, V n ker a ^ 0 . Let W = {v £ v|av E V} . Q is a submodule of V, and clearly: is rationally uniform, therefore V V A ker a . The mapping a| : W w >V i s zero on V f) ker a, thus by V Cl aV = 0 , and as V^ i s an essential extension of V, we must have aV = 0 . V Q is quasi-injective, by Theorem 2 . 3 we know: J(A) = {a e A|ker a i s essential in regular. Let where for W^ V . is a rational extension of Proposition 1 . 8 , aW = 0 . This implies As 0 ^ V D ker a V Q K and A/J(A) J(A) = 0 . We show now that a £ J(A) . Let v E V„ . By ( i ) , v has the form Q 3 E A, v ± i v = n 7 8v, . , l i=l E V . As a £ J(A) i = l,...,n . Thus is Von Neumann- ker^gj and B £ A, we have i I a$ e J(A), ± is an essential submodule of V^, and 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 , in particular ker(a$ ) ± / 0 1=1 1=1 - 23 - and we have J ( A ) = 0 . So A is a Von Neumann-regular ring. A i s a division ring, let 0 ^ a e A . As A i s Von Neumann- that regular, there exists Suppose y e A such that ker a ^ 0 . Then by (*), V a = aya . So a(l - ya) = 0 . ker a cz V Q . As V essential submodule of V Q , so i s ker a . But then a contradiction. Therefore 1 - ya = 0 . Thus (iii) ker a = 0 . As i s an a e J ( A ) = 0, a(l - ya) = 0, this implies ya = 1, which proves that A i s a division ring, Q can be considered a subring of A . First we show that a e Q .' By the quasi-injectivity of V Q , a can be extended to a Let homomorphism in A . We show that this extension is unique. f,g e A , f | = g| . Then v ring, v Suppose and as A i s a division 0 4 Vcr ker(f-g), f - g = 0 . For every a to V Q . Then of To show a e J2, denote by a the (unique) extension a • > a i s a mapping 0, >• A . Using the uniqueness of the extension, i t i s easy to check that this mapping i s a monomorphism of rings. We want to prove that Therefore Q Q can be considered a subring of A . i s a right order in A . As A is a division ring, the f i r s t requirement in the definition of a right order i s t r i v i a l l y satisfied. Let A be a nonzero element of A . Then XV 4 0 (because A i s invertible), therefore, as V Q i s an essential extension of V, V DXV 4 0 . Thus W = {v e v|Av e V} the homogeneity of V, \ | f e Q . For every w i s a nonzero submodule of V . By there exists a monomorphism v e V we have: f: V >W . Clearly, (X| f)v = A| (f(v)) = A(f(v)), and w (Af)v = X(f:(v)) = A(f(v)). So (A| f) and implies f Af are equal on V, and as A i s a division ring, this (A| f) = Af w . Clearly, i s invertible, and we have: f 4 0 because A = (A[ f)(f) \ f i s one-to-one. Thus the required representation - 24 - of X . This proves that (iv) To prove that 0, i s a right order in A . i s itself rationally uniform, let 0 ^ WC T C submodules, and let f e such that HOIII (T,VQ) a fW = 0 . As inj ective, there exists an extension f e A of f . As A i s quasii s a division ring and 0 ^ WCker f, we get f = 0, and so f = 0 . Thus rationally uniform module. is a Q.E.D. The following density theorem for quasi-injective modules i s a generalization of the Jacobson Density Theorem. Definition: v^,...,v Let A be a ring and V eV a A-module. The elements are said to be linearly independent over A i f none of them i s a A-linear combination of the others, i.e. i f v. i. > Av. for every i = l,...,n . It i s clear that when a vector space over A i s a division ring and V A, the above definition i s equivalent to the usual one. 2.5. Let Theorem (Density theorem for quasi-injective modules) V be a quasi-injective A-module which satisfies: (*) Denote: v e V, Av = 0 A = implies v = 0. Hom (V,V) . A Then for every A-linearly independent elements exists an element a eA i; such that av, 4 0, 1 Proof: v ...,v e V there 1 n i av„ = ... = av = 0 . ^ n The proof i s by induction on n . For n = 1 the result i s just the given condition (*). - 25 - We prove the result for n = 2 . Let a EA Suppose that there exists no v ^ e V such that be independent over av^ ^ 0, av =0, 2 A . i.e. we have: (**) a e A, av = 0 implies 2 Consider the submodule Av av^ = 0 . of V and define f(av ) /= av^ . Because of (**) f is well-defined, and i t is clearly 2 2 an A-homomorphism. As a|. AV V f: Av > • V 2 is quasi-injective, there exists = f . Then for every a e A by a e A with we have: 2 ot(av ) = f(av ) = av^ and a(av ) = aa(v ) = a(av > . 2 2 2 2 2 So a(v^ - av ) = av^ - a(av ) = 0 for every a e A, i.e. A(v^ - cw ) = By (*), v^ - av =0, i 2 ^ contradiction to the A-independence of 1' 2 " This establishes the result for n = 2 . Now suppose i t is true 2 2 v = 2 av n 2 V V for and we w i l l prove i t for n + 1 . Let v_,v ,...,v e V be 0 1 n independent over A (n _> 2) . Suppose that there exists no a e A such that n, 1 avu f 0, n (***) Consider .A av.. i = ... = avn =0, a e A, i.e. we have: av. = ... = av =0 1 n implies av_ = 0 . 0 as a module over i t s e l f , and define the maps by: f(a) = av^ , g (a) = av 1 1 , g (a) = av , n n a e A a eA a e A . f ,g , ...,g •:• A — - 26 - A V f,g^,...,g copies of V, are clearly A-homomorphisms. Consider the direct sum of n and for i = 1,. .. ,n let e_^: V >- V & ... © V embedding of V on the i-th component of V <9 ... © V . Let n g = I i ±' - S ( ) (§i( )»•••>§ (a)) for every a e A . i=l e g i , e a = a n v e Consider ... a v be the From (***) we know that ker g C? ker f . Therefore there exists a homomorphism >V h: gA such that hg = f . For , i = l,...,n Clearly, K let K. = f\ ker g., and let K = K, + ... + K . x ./. j 1 n i s a submodule of A . For i = l,...,n we have: e.g.KCZgK . (For let a e K . Then xx a = a. + ... + a where 1 n a. e K., j = l,...,n, and so e.g.(a) = e.(g.(a,+...+a )) 3 2 xx xx 1 •n e.(g (a )+...+g (a.)+...+g (a^)) = e.tg.U.)) = ( O ^ . g ^ a ^ a i 1 1 gK ± V i consider the homomorphisms • • gK ^ i s quasi-inj ective, the mapping be extended to a homomorphism we have: V. (h| ) (e. | e A . For every a^g^(a) = he^g_^(a), ): g.K — V a eK , - . 1 1 « - X=l | l l In particular, for every a get: l 1=1 a eK . n f(a) = J o^g^(a) = a^g^(a), i=l This means: i = l,...,n 1 l for every a eK n = f\ ker g.: j=2 3 J and using the definition of f and g^, we av = a^av^) = a ^ v ^ ) , Q and can and therefore: n n n £ a.g.(a) = £ he.g.(a) = h( £ e.g.(a)) = hg(a) = f(a), X=l 0) = i g(a^) e gK .) For fixed As 6 a ( v Q a i i^ v = °» f o r every n a e f\ ker g j=2 - 28 - a e A, av„ = •.. = av =0 £ n implies a(v„ - a_v.) = 0 . (J 1 1 V Q - a^v^,V2,...,v^ by the induction hypothesis, the elements linearly dependent over elements o' l'***' n v v V contradiction. Therefore, are A . But from this i t follows easily that the "*" l i a r e a so n e a r l y dependent over A , a Q.E.D. The Jacobson Density Theorem follows quickly from the above theorem by observing that an irreducible module the condition (*) v^,...,v e V n (because V is quasi-injective, and satisfies {v e v|Av =0} independent over A, and is a submodule of w^,...,w e V V). (Let arbitrary. By the above theorem there exist a.,,...,a e A such that a.v. ^ 0 for I n xi i = l,...,n, and a.v. = 0 for i ^ j . As V is irreducible, Aa.v. = i 3 i i for i = l,...,n . Therefore there exist b. e A such that b.a.v. = w., v x = l,...,n . Take 1 a = 1 1 1 x i = l,...,n n£ b.a. . Then clearly: av. = w., • i J J i i 3=1 The following density theorem w i l l give, as a corollary, the density J J theorem for our class of modules, 2.6. Let £ . Theorem (Density theorem) V be an A-module, E an extension of V, following properties are satisfied: (i) V 4- 0, (ii) E (*) Denote: V is homogeneous. is quasi-injective, uniform, and: v e E, Av = 0 implies A = Hom (E,E) A 9, = Hom (V,V) . A v = 0 . and suppose that the Then f o r any A - l i n e a r l y independent elements elements w,,.. 1 mapping . ,w A e Q e V n such t h e r e e x i s t s an element w, ,...,w I n e V A_^ = {a e A|av. = 0 satisfies are left As E (*), be g i v e n . f o r every by A, Denote j ^ i} . A/v^ f 0 2.5 ideals i n ^ e n a e A Aw a n d a n ^ and a nonzero n L e t the A - l i n e a r l y independent elements elements v that av., = Aw, ,. .., av = 1 1 n Proof: i ' * " ' ' v v,,..,,v I n for As i = l,...,n: E f o r every e E - and the i s quasi-inj ective i . Clearly, A^, A^ A, v,,...,A v a r e nonzero submodules of 11 n n i s u n i f o r m , and each of the submodules V, A.v,,...,A v is 11 n n n j=l J so 3 nonzero, we have V f) (D e x i s t s a monomorphism A.v.) ^ 0 . As 0 ^ A e ft . a. e A. i i Take such that i Aw. a = a^ + ... 4n ( / . i 3=1 av.l Thus av. = Aw., i > V Now 2.7. For i A (0 A.v.) 3 f i x e d , as . Then f o r every 7 Aw. l e A.v., l l i = l,...,n a.v. = a.v. = j l i i J there exists i=l,...,n. we have Aw. l Q.E.D. i we a r e a b l e to s t a t e the d e n s i t y theorem f o r the c l a s s J Theorem ( D e n s i t y theorem f o r r a t i o n a l l y u n i f o r m , homogeneous modules) Let . J = a.v. . l l n a.)v. = j l . i j=l for i s homogeneous, t h e r e n A: V j=l Clearly, V A be a r i n g and V e J. . Denote: and E - 30 - V Q = the quasi-injective hull of A = V, Hom (V ,V ), A Q Q fi = Hom (V,V) . A Then for any A-linearly independent vectors v.,...,v e V _ and any 1 n u. w,,...,w e V there exists an element a e A and a nonzero 1 n vectors mapping X e 9. such that: av, = l Remark: AW., i ,.. . ,av = n n AW 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) V Q is quasi-injective, uniform, and satisfies: (*). As V eJ , A v e VQ, Av = 0 implies v = 0 . we know that V i s nonzero and homogeneous. V Q i s , of course, quasi-injective, and by Theorem 2.4(iv) V Q is rationally uniform, and therefore uniform by 1.10. The only thing to be checked is that Let V Q satisfies (*). W = {v e VQ|Av = 0} . Then Suppose W i s clearly a submodule of V Q . W 4 0 . Then, as V Q 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 because and a e A and v e V we have: f(v) e W . As ker f = 0, we get av = 0 for every' a e A v e V , a contradiction because Therefore f (av) = af (v) = 0 V is a nontrivial A-module. W = 0, i.e. V Q 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 A we are able to describe the structure of those rings 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 = Hom (V ,V ) . A Suppose that Then A Proof: Q Q V^ is f i n i t e dimensional over A. is a left order in the simple, Artinian ring We know that class Hom^(Vp,Vp) . A is a division ring by Schur's Lemma for the (Theorem 2.4(H)) . Hom^(Vq,VQ) i s , of course, simple and Artinian, being the ring of linear transformations of a f i n i t e dimensional vector space. Therefore A an element As V is a faithful A-module, can be considered a subring of a eA is also faithful. HOIII^(VQ,VQ), to be the linear transformation by a (v) = av for v e V« . We want to prove that L v. in HOIII^(VQ,VQ) . Denote by n the dimension of a :V A n by considering >V defined is a left order 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 f i n i t e linear combination of elements v.,,...,v. e V . Clearly the vectors il lm. I generate v,,,...,v, ,...,v ;,...,v e V 11 lm, nl nm I n 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 contrary, and let theorem for £ 0 ^ v e Z ( V Q ) • Let i V Q is zero. be fixed. (Theorem 2.7), there exists a. e A Suppose the By the Density 0 f X. e Horn. (V,V) and X such that a.v = A.v. x x x to a mapping . As V~ Q is a submodule of Li I. = (0:a.v) , x x for V^, u. thus I.v. CZ ker a., x x x for I.v. = 0, A, av. = 0 x v for is a basis of A i = l,...,n . n 0 ± a e C\ 1=1 V~, is faithful. V Q as I is essential, and 1 I. . Then 1 aV_ = 0 . But then a = 0 Q This contradiction proves: is an essential left ideal of (over the ring is homogeneous over IWC I,,...,I I n HOIII (VQ,VQ) a Z(VQ) = 0 . A, . then I Consider V I-modules. We w i l l show that they satisfy the conditions of Theorem 2.6 Then I. i=l this implies contains an element which is invertible in V n Q Next we show that i f and 6 i = l,...,n . n because A . We have: is a division ring, and therefore n C) I.' f 0 . Take i=l and therefore i = l,...,n . for every 1 v, I rt i = l,...,n . As are essential left ideals of in particular a.v,...,a v e Z(V ), 1 n Q i = l,...,n, are essential left ideals in 0 ^ a. e A, we get x x x As /A. can be extended i = l,...,n . 0 = I.a.v = I.a.v. = a.I.v. x x x x x x x x Thus A. x a^ e A . We have: a.v = a.v. x x x ZCV^) X is quasi-injective, , W, 0 + w e W . As IW I, I). let W We know that To show that be a nonzero I-submodule of is an A-submodule of Z ( V Q ) =0, V ^ 0 . w £ Z(V^), V, and so (0:w) IW 4- 0 . V . (For let is not essential in - 33 - A . Therefore, so- IW ^ 0 .) V A-monomorphism from V into f: V > IW . Clearly; W . Thus V is a homogeneous I-module. and f e an A-submodule of HOIII^WJVQ) . Furthermore, . Then Let W be IWC W and IW i s f | ^ e Hom^ClWjV^)- (Because m > i , w, e IW we have: f (a > i , w, ) = f(Y(ai, )w, ) = - kk kk k k k=l a e A, ' L ^(aiDfCw^) = a L L i ^ f (w^) = af (£ i j ^ ^ ) •) As N Y Q as an A-module, there exists g f is also an I-monomorphism is quasi-injective as an I-module. an I-submodule of Clearly, lw 5^ 0, and is a homogeneous A-module, therefore there exists an We show now that for I <£• (0:w) . Thus as I is essential, g e Hom.(V A • Let i e Hom^.(V^,V^) e N x I, w e quasi-inj ective such that ,V ) U. 1 S g| lw = fI . iW 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 i s 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 To prove that submodules of is a quasi-injective I-module. i s a uniform I-module, let W^,W 2 . . Then IW-^, IW 2 be nonzero I- 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 IVl^CZ W, 2 we have W^ C\ W 2 f . IW^ D IW ^ 0 . As IW^CT W^ and 2 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 as Iv = 0, then v = 6 . But this i s clear, because Iv = 0 I i s essential, that v e Z(V^) = 0 . Thus the I-modules V<n satisfy the requirements of Theorem 2.6. We show now that implies, - 34 Hom (V ,V ) = A . I Q Clearly, f e Q AC Hom.j. ( V Q , V Q ) . HOHI (VQ,VQ) I i s an A-submodule of IVQ a e A such that ot [ = f| v e V Q we have: I(f(v) - a(v)) = 0, Y : IV • Q VQ f(v) - a(v) e Z ( V Q ) = 0 . Thus HOII1J.(VQ,VQ) = A . Therefore v^,...,v HOIILJ.(VQ,VQ) . By Theorem 2.6 there exists an element A e Hom^(V,V) i e I = f(iv) - a(iv) and therefore, as essential, nonzero mapping I Q i(f(v) - a(v)) = if(v) - ia(v) v e VQ, So for every f| . For every Q and and VQ, V Q is quasi-injective as an A-module, there is an A-homomorphism. As exists a mapping . To show the opposite inclusion, let = 0 . I is f = a e A . This shows: are linearly independent over n e e l and a such that ev, = Av,,...,ev = Av 1 1 n n V Q is quasi-injective as an I-module, As y e Hom.j.(VQ,VQ) = A . Thus we have mapping The vectors e, can be extended to a e e l and 0 4 y eA ev, = Y V , , . . . , e v = Y V 1 1 n n such that and as A Y V , , . . . , Y V 1 dim.V^ = n, A Q n are, clearly, linearly independent over Y V , , . . . , Y V 1' n i s a basis of V^ Q over A . Therefore considered as a linear transformation on the vector space invertible. So I A, V Q , is contains an element which i s invertible in Hom (VQ,V ) . A Q Now we can prove that every element cf> = a H> with a,b e A . Let A_^ = {a e A| _. = 0 av Clearly, and A,,...,A 1 n Av = 0, then c(> e for every <j>' e (because a HOIII (VQ,VQ) a j 4 x}, are left ideals of v = 0 HOIII (VQ,VQ) i s of the form . Denote for i = l,...,n . A . We know that i f v e V_ u. Z(VQ) = 0)• Therefore by the Density - 3 5 - theorem for quasi-inj ective modules (Theorem 2 . 5 ) , A f 0 v for every i = l,...,n . Thus as A,v,,...,A v are nonzero submodules of V „ , and l i n n Q is uniform (since i t is rationally uniform by 2 . 4 ( i v ) ) , VQ A,v,.....A v l i n n X. = (x e l are essential submodules of V_ . Let Q A|XI})(V.) ' e A.v.}, i i for i = l,...,n . Then X, ,. . . ,X i I n are essential left ideals of A . of A . If J(f(v.) = 0 , Jcf>(v^) 4 0 , then then (For let J JC x ± and so be a nonzero left ideal = J 4 0 . If JH X 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 Q l '' l l 0 4 j<j)(v.) e A.v. . We have 0 5^ j e J C\ X. . Thus n essential left ideal of A .) Take X = f\ X. . Then X i=l that 1 i e J J X. such i s an 1 i s an 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., 1 x i for i = l,...,n . So there exist elements b. e A., 1 ad)(v.)=b.v., 1 Take for i = l,...,n, such that 1 i = l,...,n. 1 1 b = b^ + ... + b • Then for every i = l,...,n: = ( nY b.)v. = y b.v. = b.v. 1=1 J=l bv_, = ( \ \>.~)v. = \ b v. = b.v_^ = a<}>(v^) . T. 4 Thus: a K v ^ = bv x ad)(v ) = bv n n So the linear transformations a<j> and b act equally on the basis - 36 - v, ,. . . ,v 1 n HOIII^CVQJVQ) of V „ . Therefore Q a<j> = b, and as a i s invertible in , we have: <j> = a H> , a,beA. Finally, we have to prove that every element of A which is regular in A, i s invertible in HOIII^CVQJVQ) . 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), Then gf = 0 implies g = 0. f i s 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 i s zero on fW .. So gf = 0 and g 4 0, a contradiction to the given property of f . Therefore dimensionality of W, Now let f is onto, and because of the f i n i t e - f is also one-to-one, and the lemma is proved.) a e A be regular in A . Let g e H O I I I ga = 0 . We know that g has the form a (VQ,VQ) g = b "*"c with and suppose 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 As V Q i s finite-dimensional over in HOHI (VQ,VQ) a . Thus, A A, by the Lemma is a left order in g = b "*"c = 0 . a i s invertible HOIII (VQ,VQ) a . 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 structure of those rings V A for which £ A contains a faithful member satisfying the finite-dimensionality condition is the quasi-inj ective hull of V, and A = VQ the dim^Vg < °° (where HOIII (VQ,VQ) a .) Thus, to get the structure of prime, Goldie rings, a l l we have to show i s the following: If V A is a prime, Goldie ring, then such that 3.1. ^ contains a faithful module dim^V^ < °° . Prime rings The next proposition i s elementary. Proposition: The following conditions on a ring A are equivalent: (i) If I,J. are ideals i n A and IJ = 0, then 1=0 (ii) If I,J are left ideals in A and IJ = 0, then or J = 0 . 1=0 or J = 0. (iii) 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 Definition: A nonzero ring A x = 0 or y = 0 . is said to be prime i f 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 annihilator, S be a nonempty subset of a ring £(S), of £(S) = {a e A|aS It is clear that single element Definition: is defined to be is a left ideal in we denote A left ideal I £(S) = £(x) of a ring A A . Let consists of the is said to be a left S of A such that I J and (i) I f) (ii) For every left ideal I'D J = J Definition: be left ideals in I is said to be 0 I' of A such that I ' ^ I, we have 4 o . A left ideal I of a ring there exists a left ideal J of Definition: A . J if: a complement of A S . Definition: (i) (If .) annihilator i f there exists a nonempty subset I = l(S) The left = 0} . £(S) x, S A . A ring A A A is called a complement i f such that I is a complement of is said to be Goldie i f : satisfies the ascending chain condition on left annihilators (i.e. there exists no strictly increasing, infinite sequence of left annihilators in A), J . - 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..,I ,...,I 1 / n 00 left ideals of A such that the sum x. e l . x x Proof: J I is direct. n=l is direct means that for every n > l : x, + ... + x = C — 1 n OO (The sum of nonzero 0 Y I \ n n=l J implies x, = ... = x = 0 . ) 1 n (i) implies ( i i ) : Suppose the contrary, i.e. there exists an infinite sequence I.,I ,...,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 0 + I + ...) = 0 . ± 2 Again by Zorn's Lemma there exists a left ideal T^ of A, which is maximal with respect to the properties: 'T 0 ( I + I +...) = 0 2 3 (The set of a l l left ideals and T of A which satisfy T o TQ + 1^ is nonempty because We continue by induction. TQJT^,...>T _ N 1 T f) ( I + I + ...) = 2 3 T^ + 1^ belongs to i t . ) Suppose we have already found left ideals 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 l - ll 1 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 + I • n n-l n n (The set of a l l l e f t ideals . T of A which satisfy Tfl(I , + 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 n •-. + In+2 , o •••) • Then x e (Tn-l.. + In) A (In+1 + where t t -, e T i.el., n-l n-l' j j x = tn -,l+ in = in+1 ,.+... + m>n+l. - , = - i + 1 , . + . . . + 1 e T . CS (I + I , , + . . . ) = 0 . n-l n n+1 m n-l n n+1 So t , = 0 , n-l oo and all - i + i ... + ...+ i =0, n n+1 m j . Thus and as Y I . n n=l is direct, u i . = 0 for i J x = t , + i =0.) n-l n J 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 T n ^ n+l T n (I + ^+2 ...) =0 'r>T . + I . n-l n It is clear that T - «= T . Also, n-l n as + I 4 0, n T n is a complement of I , , + I , „ + . . . , and that n+1 n+2 T T D I Cl T . , A ( I + I , + ...)=0, therefore, n-l n n-l n n+1 1 T , f ) I ^ I • Thus n-l' n n I n<£T ^ n-l But I C T ,+ICTT, n n-l n n and so: T . T . Thus we have a s t r i c t l y increasing, infinite n-l 4 n sequence of complements, (i). T ^ T. *-? ... ^ T . . . , in contradiction to o 4 1 4 4 n4 - 41 - (ii) implies ( i ) : Suppose the contrary, i.e. there exists in A a L 9 1 , 9 ... 9 l 9 ... of L f I T T nT complements. For every n > 1 there exists a left ideal J of A — n such that I i s a complement of . We have for every n > 1: strictly increasing, infinite sequence I . D J n n = 0 I D J 4 0 n+1 n Denote: (because I <T I ..) . n f n+1 K = I f] J , for n > 1 . Then n n+1' ' n — K ,K ...,K ,.. are 1 2 n 1 oo nonzero left ideals of A . We show that the sum Let k. + ... + k =0, 1 n where k. e K., i i OJ Y K n=l is direct. i = l,...,n . Then 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 n n k e l A J =0, n n n oo = 0 . So the sum 3.4. Let \ K n=l Similarly, J k , = 0,..., n-1 is direct, in contradiction to ( i i ) . Q.E.D. Proposition A be a prime ring which satisfies the ascending chain condition on left annihilators. Then, considering singular submodule of A Proof: Then A k =0. n X A as a module over i t s e l f , the i s zero. Suppose the contrary, i.e. Z(A) 4 0 . Let X = {£(x)|0 4 x e Z(A)}. i s a nonempty set of left annihilators in A, and therefore, as satisfies the ascending chain condition on left annihilators, a maximal element, £(x ), where 0 4 x Q Then L Q e Z(A). Let L = X contains {ye A|yZ(A) = 0}. i s 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 i t s e l f ) , therefore, as A is a prime ring and Z(A) ^ 0 , we have L = 0. - 42 - Therefore x^ i L, such that XgZ CZ £ ( X Q ) if . contradiction.) x Q A X Q = 0 a n d AXQ^ £ ( Z a e £(x z), ^ £(XQZ) in X . Thus: Z(A) = 0 z e Z(A) . It is clear that ^ u because A, x^ = 0, a i s an essential left ideal of a ^ £(x ), Q Q 0 4 X Q Z e Z(A), £(x^z) e X, Q g ) ) 4 0 . Thus there exists Q £(x ) x by the primeness of As z e Z(A), £(z) ax z = 0 . We have As £ ( is a nonzero left ideal in A . A X Q 0 then Therefore x^ Z(A) 4 0 . Thus there exists 4 0 . We show now: £(XQZ) A X Q = and so a e A such that and so A . ax^ 4 0, £(XQ) ^ " ^ ( X Q Z ) . a contradiction to the maximality of 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. Let Theorem A be a prime, Goldie ring. Then £ contains a faithful module V such that and dim^V^ < <» . (V^ d enotes th6 cjiicLsi injective - hull of V, A = Hom (V ,V ) .) Proof: A Q Q 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 D W^ = 0 . As left ideals V ,W 2 2 ^ I ' ^ I i n A such that is not a uniform A-module, there exist nonzero of A such that V^C. V , W<= V± and V^AWj = 0 2 Continuing by induction we obtain two infinite sequences of nonzero left ideals, V,,V ...,V 1 2 n V r> V ,. , n n+1 OJ and W,,W„,...,W 1 2 n V 3 W „ and V D W =0 n n+1 n n with the properties: for every n > 1 . — - 43 - We show that the sum > W xs dxrect. i n n=l where w. E W., i = l,...,n . Then w. i x 1 i -w. e W „ C v , . . . , - w e W <ZZ V . C V , . 2 2 1 n n n-1 1 w = (-W2) + ... + (~w) e 0 W = 0, n Suppose w„ + . . . + w 1 n =0, = (-w) + ... + (-w ), and z n Thus 0 w^ = 0 . Similarly, 00 w„ = 0,...,w = 0, and the sum 2 n > W -i n n=l i s direct. By Proposition 3.3 this is a contradiction to the fact that chain condition on complements. ideal satisfies the ascending A contains a nonzero left V which i s a uniform A-module. We w i l l prove that First, as A ring, Therefore A AV 4 0 and V e V so V V i s faithful, and dim^/^ < » . are nonzero left ideals of A, i s a nontrivial A-module. By Proposition 3.4, Z(A) = 0, therefore and Clearly, A is a prime Z ( V ) ^ Z(A) . Z(V) = 0 . Thus V is ) - 44 - rationally uniform. To prove that V i s 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 VWQ 4 0 . Define f: V — > W by f(v) = vw^ that It is clear that f i s an A-homomorphism. V ker f 4 0 . Then is rationally uniform, i s 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 So such for every v e V . Suppose ker f is a nonzero submodule of V, and as V WQ e W fV = Vw^ =0, a contradiction. f i s an A-monomorphism, and this proves the homogeneity of V . Thus: V e £ . A For the annihilator V 4 0 and A (0:V) of V we clearly have i s a prime ring, this implies faithful module. (0:V)V = 0, and as (0:V) = 0 . So V is a Denote: < VQ = the quasi-injective hull of V, A = Hom (V ,V ) . A Q Q It remained to prove that dim^Vp < °o . Suppose the contrary, i.e. that VQ is infinite dimensional over sequence v^,v ,•••jV^,... 2 of A-linearly independent vectors in Let I ± I I 2 n = (0: ) Vl = (0: ) f) (0:v ) V;L 2 = (0: ) D ... f) (0:v ) 1 n Vl A . Then there exists an infinite . - 45 - Consider the decreasing sequence of left ideals 1 I n We show that for every A n+1 n >^ 1 of such that: j c i n n Let n but Hi J n be fixed. ,. = o . n+1 Take a nonzero vector are linearly independent over for the class \ by the Density theorem a EA and av , = Aw . n+1 0 ^ av ,, e V . As n+1 a 4: 0, w e V, n A By the quasi-injectivity of Clearly and as v,,...,v ,v 1 n n+± A e Hom (V,V) such that av, = 0,...,av =0, 1 n We claim that A, w e V . As (Theorem 2.7), there exists an element a nonzero mapping as there exists a nonzero left ideal and as A V^, A:V A >V and w e V, av ., = Aw e n+1 can be extended to a mapping i s a division ring, a a eA is invertible. Thus, 0 4 w e V, av ,, = Aw = aw 4 0 . n+1 We know that Z(V) = 0, therefore an essential left ideal of of A av ,, £ Z(V), and (0:av ,,) i s not n+± n+i A . So there exists a nonzero left ideal K (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 such that .-.) f\ K = 0, k t (0:av , , ) , and therefore n+± n+1 0 4 ka e Ka = J . So n av, = 0,...,av =0, 1 n Thus J CZ I . n n J n is a nonzero left ideal of we have J v, = Kav, = 0 , . . . , J v n l I n n kav ,, i 0 . Thus n+1 A . As = Kav = 0 . n - 46 - x E J f l I , . Then n n+1 Let x e Ka D (0 :v ,,) . n+1 n k e K, and kav x = ka k E (0:av ,,) f) K = 0, n+1 = 0 . Therefore n+1 So J fi I , = 0 . n n+1 We have an Infinite sequence J ,J ,...,J »... for some and x = ka = 0 . Thus 2 of A such that: n of nonzero left ideals J <Z I , J = 0, for every n > 1 . We n n n n+1 — oo show that the sum ) J i s direct. Suppose i , + ... + i =0, where , n l n ' n=l 3 e J for i = l,...,n . Then = (-j ) + ... + (~j )» M L ± -i r r ± 0 l £ J C I„,.. . , - i 2. 2 ' 2 n 0 = ( J 2 E J j J ~J2} + ••• + ("V J O n e l J I cz I„ . n 2 / ^ I 2 °' = So h = 0 ' 00 Similarly, j 2 = 0,...,j =0, and the sum £ J n=l Proposition 3.3 this contradicts the fact that chain condition on complements. A By satisfies the ascending V Q is finite-dimensional over Q.E.D. . 3.6. Let Therefore A is direct. Corollary A (Theorem of Goldie) 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 faithful, and satisfies hull, of V, and A = V e which is dim^^ < °° (where V Q i s the quasi-inj ective HOIII (VQ,VQ) a .order in the simple, Artinian ring .) By Theorem HOHI^(VQ,VQ) 2.8, . A is a left 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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Density theorems and applications Horvath, Jozsef 1977
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Title | Density theorems and applications |
Creator |
Horvath, Jozsef |
Date Issued | 1977 |
Description | One way of getting structure theorems in ring theory is to fix a general class ∑ 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 class of rationally uniform, homogeneous modules in a similar way as the Wedderburn-Artin Theorem follows from Schur's Lemma and the Density Theorem for the class of irreducible modules. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2010-02-16 |
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 | 10.14288/1.0080122 |
URI | http://hdl.handle.net/2429/20307 |
Degree |
Master of Science - MSc |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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