MODULES WHICH ARE SUBISOMORPHIC TO QUASI-INJECTIVE MODULES

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1 Al- Khaladi Iraqi Journal o Science, Vol.50, No., 009, PP Iraqi Journal o Science MODULES WHICH AE SUBISOMOPHIC TO QUASI-INJECTIVE MODULES Amer H. H. Al-Khaladi Department o Mathematics, College o Science, Diyala University. Diyala-Iraq. Abstract Let be a commutative ring with identity and let M be a unitary let module. We call the module M kerquasi-injective i or every monomorphism rom N into Q(M ), where N is a submodule o Q( and Q( is a quasi-injective hull o M g and or every homomorphism rom N into M, there exists a homomorphism h rom Q ( into M such that ker h ker g. It is clear that every quasi-injective module is kerquasi-injective, however the converse is alse. Also every ker-injective module is kerquasi-injective, however the converse is alse. In this paper we give some characterizations o kerquasi-injective modules, we also study some conditions under which a kerqausi-injective module becomes quasi-injective. For example, i a kerquasi-injective module is a initely generated, then it is a quasi-injective. We ought to mention that we were not able to give an example o a kerquasi-injective module which is not quasi-injective and kerinjective. الموديلات التي تكون جزي ية التقابل الى موديلات شبه اغمارية عامر حيدر حسين الخالدي قسم الرياضيات كلية العلوم جامعة ديالى. ديالى الع ارق. الخلاصة لتكن حلقة ابدالية وليكن M موديولا يساريا على. نقول ان الموديول M شبه اغماري النواة اذا كان لكل تشاكل متباين من N الى ) Q(M حيث N موديول جزي ي من الغلاف شبه الاغماري للموديول M الذي يرمز له بالرمز ) Q(M ولكل تشاكل g من N الى M يوجد تشاكل h من ) Q(M الى M يحقق. ker h ker g من الواضح ان كل موديول شبه اغماري هو موديول شبه اغماري النواة. لقد بينا ان العكس غير صحيح بصورة عامة كذلك كل موديول اغماري النواة هو شبه اغماري النواة ولقد بينا ان العكس غير صحيح ايضا. كذلك اعطينا بعض المكافي ات الى تعريفنا. واخي ار لابد من الاشا رة الى اننا لم نستطع اعطاء مثال على موديول شبه اغماري النواة لكن ليس شبه اغماري ولا اغماري النواة. وعليه نترك البرهنة على وجود (أو عدم وجود) مثال كهذا مسا لة مفتوحة. Introduction Two modules are subisomorphic i each has a monomorphism into the other []. Bumby, [] has shown that i two modules are subisomorphic then their quasi-injective hulls are isomorphic. The purpose o this paper is to 6

2 Al- Khaladi Iraqi Journal o Science, Vol.50, No., 009, PP initiate the study o modules which are subisomorphic to their quasi-injective hulls. We introduce the ollowing deinition: a module M is kerquasi-injective (KQI ) i given any monomorphism : N Q(, where N is any submodule o the quasi-injective hull Q(, and any homomorphism g : N M there exists a homomorphism h : Q ( M such that ker h ker g. In section, we show that KQI modules are precisely those modules which are subisomorphic to their quasiinjective hulls. In section o the paper we give various conditions under which a KQI module becomes quasi-injective, we would like point out that our results parallel the results in [] o ker-injecctive modules. Finally, we remark that in this paper stands or a commutative ring with and a module means a unitary let module.. Characterization o kerquasiinjective modules We start the section by the ollowing: Theorem.. Let M be a module, then the ollowing statements are equivalent: (i) M is KQI. (ii) M is subisomorphic to Q(. (iii) Given any monomorphism : N Q(, where N is any submodule o the quasi-injective hull Q( and any homomorphism g : N M, there exists a monomorphism k : M M and a homomorphism h : Q( M such that h kg. (iv) There exist two homomorphisms : M Q( and h : Q( M such that h is a monomorphism. Proo. ( i) ( ii). Consider the ollowing 0 M i Q( I M h where i is the inclusion homomorphism and I is the identity homomorphism. Since M is KQI, there exists a homomorphism h : Q( M such that ker hi ker I 0. This implies that hi is a monomorphism, we claim th at h is a monomorphism. In a ct, i h( x) 0 and x 0, there exists r such that 0 rx M, since M e Q(M ) ( M is an essential extension o Q ( ), h ( rx) rh( x) 0,but h ( rx) hi( rx) 0, hence rx 0 contradicti on. Thereore h is a momomorphism and hence M is subisomorphic to Q (. (ii) (iii).consider the ollowing 0 N Q( g t h O M I Q(M) M where N is a submodule o th e quasi-injective hull o Q (, is any monomorphis m and g is any homomorphism. Let i : M Q( be the inclusion homomorphism. Sin ce Q( is quasi-injective there exists a homomorphism t : Q( Q(M ) such that t ig [3]. From pa rt (ii) there exists a monomorphism s : Q( M. Let st h and si k. Hence h kg. ( iii) ( iv). Let N M and g identity homomorphism. ( iv) ( i). Let : N Q( be any monomorphism and g : N M be any homomorphism. Let i : M Q( be the inclusion homomorphism. Since Q( is quasi-injective there exists a homomorphism t : Q(M ) Q( such that t ig. From part (iv) there exists a homomorphism s such that si is a monomorphism. Let st h. We claim that ker h ker g. In act let x ker h then 0 ( h )( x) h( ( st ( ( s( ig)( x) ( si)( g(. But si is a monomorphism thereore g( x) 0 and hence x ker g. Corollary.. I M is a KQI module, then or every diagram o the orm 7

3 Al- Khaladi Iraqi Journal o Science, Vol.50, No., 009, PP N Q( g M M there exists a homomorphism h : Q( M such that ker h ker g. Proo. From the proo o theorem. part ( iv) ( i), i x ker g then g( x) 0 si( g( s( ig( ( st)( ( h( ( ( h )( x) and hence x ker h, that is ker g ker h. Thus ker h ker g. emarks and examples.3 () I Q( is subisomorphic to E ( ( E( is an injective hull o then E( Q (. Proo. By [], E ( Q(M ),butq(m ) E(, thereore E( Q(. () I M is subisomorphic to M and M is KQI, then M is also. Proo. There exists a monomorphism : M M and a monomorphism g : Q( M and hence g is also a monomorphism. By [] Q ( Q( ) M.Thus there exists a monomorphism rom Q ( M ) into M. (3) It is easy to see that i ker g ker h in the deinition o KQI, then the module M is not necessarily KQI (take M as module). (4) It is clear that every quasi-injective is KQI. But the converse is not true in general. For example let M E() as module. E ( is subisomorphic to M and by () Q( is subisomorphic to M. But M is not quasi-injective since is not quasi-injective. (5) It is clear that every ker-injec tive module M ( E ( is subisomorphic to [] is KQI, but the conve rse is not tr ue in general. For example take M as module, Q( ) that is is qu asi-injective and hence by (4) is KQI, but is not ker- injectiv e because i is ker-injeective module, then is subisomorphic to E( ) and by [],this is contradiction. Thus is KQI but not ker-injective. Proposition.4. I M is KQI then or every monomorphism : N M and or every homomorphism g : N M there exists a homomorphism h : M M such that ker h ker g. Proo. Consider the ollowing i 0 N M Q( g h t M Since M is KQI, there exists a homom orphism t : Q( Q ( such that ker ti ker g. Let ti h. Thus ker h ker ti ker g. Lemma.5. I is principle ideal domain then every nonzero homomorphism : I, where I is a nonzero ideal o is a monmorphism. Proo. Le t I (r) or some 0 r and let ( x) 0, x I, this implies that ( sr) 0, or some s and hence s ( r) 0. But is an integral domain, thereore either s 0 or ( r) 0. I ( r) 0, then is the zero homomorphism and this is impossible. Thus s 0 and hence x 0 that is is a monomorphism. The converse o the proposition. 4 is not true. For example let M as module. Consider the ollowing 0 (n) g h Now we suppose that n 0 and g 0. Since is principle ideal domain thereore by lemma.5, any homomorphism h : is a monomorphism also g is a monomorphism, thus h is also a monomorphism. Hence ker h ker g =0 But as module is not KQI. 8

4 Al- Khaladi Iraqi Journal o Science, Vol.50, No., 009, PP Quasi-injective modules and kerquasi-injective modules We start with the ollowing: Proposition.. I M is a KQI module, then there exists an ep imorphism h : M Q(. Proo. Since M is KQI, there exists a monomorphism : Q( M. This implies that Q(M ) (. Consider the ollowing i j 0 ( M Q( k t Q ( where i, j are inclusion homomorphisms and k is an isomorphism. Since Q ( is a quasiinjective module, there exists a homomorphism t : Q( Q( such that tij k. This implies t is an epimorphism. Let tj h and let x Q(M ), since k is an iso m orphism, there exists y ( such that k( y) x. Thus h( y) hi( y) k( y) x. That is h is an epimorphism. Corollary.. I M is KQI, then ann ( ann ( Q(). Proo. Clearly ann ( Q( ) ann (.Let r ann (, by proposition. there exists an epimorphism h : M Q (. But r ann (, thereore 0 = h( 0) h( r rh( rq(. This implies r ann ( Q( ), thus ann ( ann ( Q( ). Corollary.3. For every KQI module, there exists a proper submodule N o M such that M / N is quasi-injective. Moreover M / N Q(. Proo. By proposition. there exists an epimorphism h : M Q( and hence M / ker h Q(, put N ker h. A module is hopian i every onto endomorphism is an automorphism. Theorem.4. I M is a KQI module and Q ( is hopian, then M is quasi-injective. Proo. Suppose Q(M ) is hopian, b y the proo o proposition. there exists an epimorphismt : Q( Q( such that tji k. Since Q( is hopian, thereore t is an isomorphism and hence ji is an isomorphism, this implies j is an epimorphism and hence M Q(M ). Theorem.5. I M is a KQI module and M or Q ( is initely generated, then M is quasi-injective. Proo. Suppose Q ( is initely generated, then by [4] Q( is hopian and we use theorem.4 to have the requirement.now suppose M is initely generated. By proposition. there exists an epimorphisim h : M Q( and hence Q ( is initely generated and by [ 4] Q ( is hopian and by proposition.4 M is quasi-injective. A module is cohopian i every one to one endomorphism is an automorphism. Theorem.6. I M is KQI and M or Q ( is cohopian, then M is quasi-injective. Proo. Suppose M is cohopian and i : M Q( is the inclusion homomorphism and : Q( M is a monomorphism. Then i is also a monomorphism, this implies i is an isomorphism and hence is an epimorphism. Thus M is quasi-inje ctive. Wh ile i Q ( is cohopian, i is a monomorphism this implies i is an isomorphism. Thus i is an epimorphism, that is M is quasi-injective. Corollary.7. I M is KQI and Q ( is directly inite (that is a module not isomorphic to any proper direct summand o it sel [5, P.65]) then M is quasi-injective. Proo. I Q ( is directly i nite then by [6] Q ( is cohopian and by theorem.6 M is quasi-injective. Proposition.8. I is an integral domain and has a submodule which is KQI, then is a ield. Proo. By [7] is subisomorphic to every submodule o. Since has KQI 9

5 Al- Khaladi Iraqi Journal o Science, Vol.50, No., 009, PP submodule say I, then b y remarks and examples.3 () is KQI. By proposition. there exists an epimorphism : Q( ). Since is initely generated, then by theorem.5 is quasi-injective and hence is sel injective. Thus by [7] is a ield. Next we gi ve a proposition that shows that i every KQI on the ring is a quasi-injective, then that ring must be a semi-simple artinian and conversely. Proposition.9. is a semi-simple artinian i every KQI module is quasi-injective. Proo. ( ) By [8] every module is a quasiinjective. Conversely let M be any module, this implies M E( is a ker-injective module and hence by remarks and examples.3 (5) M E( is KQI.Thus M E( is quasi-injective. Since a direct summand o a quasi-injective is a quasi-injective [5] then M is a quasi-injective and by [8] is a semiinjective artinian. Acknowledgement. I am grateul to the reeree or valuable suggestion and comments. 8.. E. Johnson and E.T. Wong, 96, "Quasi-injective modules and irreducible rings" J. London Math.Soc.,, 36, eerences.. Bumby, 965, "Modules which are isomorphic to submodules o each other": Arch. der Math.,, 6, G. F. Birkenmeier, 978, "Modules which are subisomorphic to injective modules": Journal o pure and applied algebra,, 3, F. W. Anderson and K.. Fuller, 974, "ings and categories o modules": Springer-Verlag, Berlin,. 4. W.V. Vasconcelos, 969, "On initely generated lat modules": Trans. Amer. Math. Soc.,, 38, K.. Gooderal, 976, "ing theory, Non singular rings and modules": Marcel Dekker,. 6. G. F. Birkenmeier, 967, "On the cancelation o quasi-injective modules": Communication algebra,,4, Amer H. H.Al-Khaladi, 993" Ker- injective and L injective modules": M. Sc. thesis, College o Science, University o Baghdad,. 30