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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.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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