Journal of Sciences, Islamic Republic of Iran 18(4): 355-360 (2007) http://jsciences.ut.ac.ir University of Tehran, ISSN 1016-1104

Essentially Retractable Modules

M.R. Vedadi*

Department of Mathematical Sciences, Isfahan University of Technology, Isfahan 84156-83111, Islamic Republic of Iran

Abstract

We call a module M R essentially retractable if HomR (MN,0) ≠ for all essential submodules N of M. For a right FBN ring R, it is shown that: (i) A non-

zero module M R is retractable (in the sense that HomR (MN,0) ≠ for all non-

zero N ≤ M R ) if and only if certain factor modules of M are essentially retractable nonsingular modules over R modulo their annihilators. (ii) A non-zero

module M R is essentially retractable if and only if there exists a prime ideal

P ∈ Ass()M R such that HomR (MN,0) ≠ . Over semiprime right nonsingular rings, a nonsingular essentially retractable module is precisely a module with non- zero dual. Moreover, over certain rings R, including right FBN rings, it is shown that a nonsingular module M with enough uniforms is essentially retractable if and

only if there exist uniform retractable R-modules {U i }i∈I and R-homomorphisms α β M ⎯⎯→ ⊕ i∈I U i ⎯⎯→ M with βα ≠ 0 .

Keywords: Dual module; Essentially retractable; Homo-related

1. Introduction rings are prime [1], Baer or quasi-Baer [9], semiprime or nonsingular or finite uniform dimensional [4]; see Throughout all rings have non-zero identity elements also [3] and [10] where the terms “quotient-like” and and all modules are unital right R-modules. Any “slightly compressible” were respectively used for terminology not defined here may be found in [2,8]. S. retractable. We define essential retractability for a M. Khuri [5] defined the concept of a retractable module M by requiring that Hom for module: An R-module M is retractable if R R ()MN,0≠ all NM≤ where the notation NM≤ means N HomR ()MN,0≠ for all non-zero submodules N of M. eR eR The endomorphism ring of a retractable module which is an essential submodule of M R . Clearly non-zero also satisfies another condition, often and most notably retractable implies essentially retractable, but the nonsingularity, has been the subject of study in several inverse implication is not true in general. According to articles [6,7,11]. More recently retractable modules have Proposition 2.1 for an arbitrary module to be retractable, appeared in the study of modules whose endomorphism it suffices that all of its factor modules be essentially

2000 Mathematics Subject Classification: 16D10, 16S50. * Corresponding author, Tel.: +98(311)3913607, Fax: +98(311)3912602, E-mail: [email protected]

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retractable. This simple observation is our primary Lemma 2.2. The following statements are equivalent motivation to characterize essentially retractable for a non-zero R-module M. modules over certain rings. Characterization of (a) M R is essentially retractable. retractable modules over right FBN rings is already (b) There exists a non-zero f ∈ End( M ) such found by Smith [10]. In our endeavor, we first observe R that any nonsingular essentially retractable module has a that Im (f ) is an essentially retractable R-module. non-zero dual. This yields the fact that for a nonsingular Proof. (a) ⇒ (b) This is clear. module over a prime ring, the conditions retractable, (b) ⇒ (a). Suppose that (b) holds. Thus there exists a essentially retractable and having non-zero dual are all non-zero homomorphism hf: () M→ K∩ f( M) . It equivalent. We are then able to prove that over a right FBN ring R retractability of a module M is equivalent to follows that HomR (MK,0) ≠ . □ essential retractability of factor modules of the form The following result is a consequence of Lemma 2.2 and gives some information about essentially retractable NZN/ () where NMMP= / with PAssM∈ ( R ) modules over a right self-injective right nonsingular (Theorem 2.8), and that M is essentially retractable if R ring. Then, in Theorem 2.5, we shall characterize and only if HomR ()MRP,/≠ 0 for some essentially retractable modules over rings that belong to a larger class. PAssM∈ ()R . We note that each of the above factor Propositin 2.3. If R is a right self-injective right modules is nonsingular over R modulo its annihilator. nonsingular ring, then any R-module M is either Thus over a right FBN ring the study of retractable singular or essentially retractable. modules reduces to that of nonsingular essentially Proof. Suppose that M is a non-zero R-module. If retractable modules. Next we consider a ring R with the Z MM≤ then M is a singular module because property that its non-zero right ideals contain non-zero ( ) eR R retractable right ideals. The class of such rings is rather R is a right nonsingular ring. Therefore assume that large as it contains all commutative rings, local rings there is a non-zero mM∈ such that with T-nilpotent Jacobson radical, right FBN rings and ZM( ) ∩ mR= 0 . On the other hand, because R is right semiprime rings. For a nonsingular R-module M with self-injective, it is easy to verify that any nonsingular enough uniforms, we prove that M is essentially cyclic right R-module is injective. Consequently, mR is retractable if and only if M is home-related to a direct an injective retractable right R-module. Thus M has a sum of cyclic nonsingular uniform retractable right non-zero retractable direct summand. By Lemma 2.2, it ideals (Theorem 2.14). The module M R is said to be follows that M R is essentially retractable. □ homo-related to L R if there are α : M → L and We remark that a factor, or even a direct summand of β : LM→ such that βα ≠ 0 . a retractable module, need not be retractable. For example Z is not retractable in Mod-Z, but p ∞ Z ⊕ Z is clearly retractable where p is a prime 2. Essentially Retractable Modules p ∞ over Certain Rings number. However the class of essentially retractable Proposition 2.1. Let M be any non-zero module over modules is easily seen to be closed under direct sums, as an arbitrary ring R. If any non-zero factor of M is is the case for the class of retractable modules [10, R Proposition 1.4]. essentially retractable then M R is retractable. Lemma 2.4. Let M R be retractable with a semiprime Proof. Let 0 ≠≤NMR . Because EN() is an endomorphism ring S and let W be any direct sum of injective R-module the inclusion map NEN→ ( ) is copies of M R , then any non-zero submodule of W R is extended to a non-zero R-module homomorphism retractable. Proof. It is easy to verify that W R is also retractable. f : MEN→ (). Since NEN≤ e (), ()NfM∩ ( ) Since S is a semiprime ring, so is any column finite is an essential submodule of f ()M . Thus matrix ring over S. Hence TEndW= R () is a HomR fM, N∩ fM is non-zero by the ()() () semiprime ring. Now suppose that 0 ≠≤≤K NWR , I essentially retractable condition on f ()M . It follows = HomR (WN, ) , and J = HomR ()WK, . Clearly that HomR ()MN,0≠ , proving that MR is retractable. □ J ⊆ I are right ideals in T. Since W R is retractable,

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J ≠ 0 . Thus 0 ≠⊂J 2 JI by semiprimness of T. Hence (a) ⇒ (b). Let f : MR→ be non-zero. For any there exists f ∈ J such that fI ≠ 0 . It follows that non-zero mM∈ , there is a right ideal I of R such that f |:N NK→ is non-zero, Proving that N R is mI≅ I , and consequently HomR ()fM(),0 I≠ . Thus retractable. □ HomR (MmR,0) ≠ , proving that M is retractable. The following Theorem shows that nonsingular R That (b) ⇒ (c) is clear, while (c) ⇒ (a) follows by essentially retractable modules over a right nonsingular semiprime ring are exactly modules with non-zero Theorem 2.5. □ duals. For an R-module M, the dual module Let M be a non-zero R-module. Following [2] an associated prime ideal of M means a prime ideal P of R Hom is denoted by M*. R ()M ,R such that, for some non-zero submodule N of M, Theorem 2.5. Let M be a nonsingular R-module. If M PannL= ( ) for every non-zero submodule L of N. is essentially retractable then M* is non-zero. The R The set of associated prime ideals of M will be converse holds if R is a semiprime right nonsingular R ring. denoted by Ass( M R ). In [10], it is shown that if R is a Proof. Let m be any non-zero element of M. Since right FBN ring then a non-zero right R-module M is r.ann (m ) is not an essential right ideal of R, there exists retractable if and only if HomR ()MRP,/≠ 0 for a right ideal I in R such that mI≅ I . Thus any non-zero every associated prime ideal P of M. Using this and submodule of M contains a submodule isomorphic to a Theorem 2.5, we show in the following result that if R is non-zero right ideal of R. Let Α be a maximal a right FBN ring, then the retractability of a module is independent family of non-zero submodules each of equivalent to essential retractability of certain factor which is isomorphic to a non-zero right ideal. If modules.

N =⊕Α, then N is an essential submodule of M R . Theorem 2.8. Let M be non-zero R-module. If R is a right FBN ring, then M is retractable if and only if for Thus, by our assumption, HomR ()MN,0≠ . It follows R any PAssM∈ ( ) the RP/ -module NZ/ ( N) that HomR ()MR,0≠ . R RP/ Conversely, suppose that ZR()= 0 and let N be as is essentially retractable as a right R-module, where r NMMP= / . above. If there is a non-zero homomorphism Proof. Let PAssM∈ ( ) , NMMP= / , H =ZN( ) f : MR→ , the Ker f is not an essential submodule of R the singular submodule of N as an RP / -module. If M R because ZRr ()= 0 . Thus fN()≠ 0 . Hence M is retractable then Hom is non-zero. 2 R R ()M ,/RP ⎡⎤fN ≠ 0 by the semiprime condition on R. It ⎣⎦() It follows that there exists a non-zero RP / - follows that nf() M ≠ 0 for some nN∈ . Now homomorphism f :NRP→ / . Since RP / is a right hgofM=→: N is non-zero, where g : fM()→ N nonsingular ring, fH( ) = 0 and so f induces a non- is defined by g ()xnx= . Since hM() can be zero RP/ -homomorphism f :/NH→ RP /. Con- embedded in a free right R-module, it is retractable by sequently, NH/ is a nonsingular right RP / -module with non-zero dual. Thus NH/ is an essentially Lemma 2.4. Consequently, M R is essentially retractable right RP / -module by Theorem 2.5. Now retractable by Lemma 2.2. □ clearly, NH/ is essentially retractable. Conversely, Corollary 2.6. Over a commutative semiprime ring, a ( ) R nonsingular module is essentially retractable if and only NH let ( / ) R be essentially retractable. Then by if its dual module is non-zero. Theorem 2.5, it is easily seen that Hom NRP,/≠ 0 Proof. By Theorem 2.5. □ R ()

Corollary 2.7. Let M be a nonsingular module over a for any PAssM∈ ( R ) and so HomR ()MRP,/≠ 0. prime ring R. Then the following statements are Hence M is retractable. □ equivalent. R We note that each of the above N/Z(N) is nonsingular (a) HomR ()MR,0≠ . over R modulo its annihilator. Thus over right FBN ring

(b) M R is retractable. the study of retractable modules reduces to that of nonsingular essentially retractable modules. Later, we (c) M R is essentially retractable. shall investigate nonsingular essentially retractable Proof. We first note that if I and J are right ideals in R modules with enough uniforms over more general rings with IJ ≠ 0 , then HomR ()JI,0≠ . including right FBN rings. But now, we are going to

357 Vol. 18 No. 4 Autumn 2007 Vedadi J. Sci. I. R. Iran

give a characterization of essentially retractable Lamma 2.2. □ modules over a right FBN ring. The right R-module M Theorem 2.5 states, over certain rings, essentially is called compressible if for each non-zero submodule N retractable modules are precisely modules with non-zero of M, there exists a monomorphism θ : M → N . dual. Now, in view of Lemma 2.9, we consider rings R Lemma 2.9. Let R be a right FBN ring. Then every in which every non-zero cyclic right ideal contains a non-zero cyclic right R-module N contains a uniform non-zero retractable right ideal. For example if R is compressible submodule. either commutative (or even right duo) or a local ring Proof. Let U be a uniform submodule of N and let with T-nilpotent Jacobson radical then R has this PAssU∈ (). Thus there exists a non-zero submodule property [note that if x ∈ RI\ for some T-nilpotent

right ideal I and xII⊆/ then there is aI1 ∈ such that V in U such that PannV= R (). Now by [2, Corollary xaI1 ∉ . Again if xaI1 ⊆/ I then there exists aI2 ∈ 8.3], Z (V ) is zero. It follows that there exists right RP/ such that xaa∉ I. If this process does not stop, we ideal Y of RP/ such that Y can be embedded in V . 12 R /P get a sequence aa,,... in I with xaa... ∉ I for all Since all non-zero right ideals in RP/ are retractable 12 1 k k. But I is T-nilpotent, hence there exists n such that and nonsingular, Y is compressible as an RP / -module. 0...,= aa which is in contradiction with Clearly Y is also compressible. □ 1 n R xaa... ∉ I. It follows that there exists rR∈ such Theorem 2.10. Let M be a non-zero module over a 1 n that xrI∉ but xrI⊆ I . Consequently, right FBN ring R. Then M R is essentially retractable if HomR (RI/,( xRI+ ) / I) ≠ 0. This shows that and only HomR ()MRP,/≠ 0 for some PAssM∈ ( R ) . (RI/ ) is retractable]. Other examples of such rings Proof. (⇒) . Let M R be essentially retractable. By R Lemma 2.9 and Zorn’s Lemma, there exists an essential include right FBN rings as well as semiprime rings; see submodule N in M such that NV=⊕ where Lemmas 2.9 and 2.4. R iI∈ i In order to investigate essentially retractable modules each ViIi ()∈ is a uniform compressible submodule over such rings, we introduce a condition for a pair of of M R . Thus by our assumption, there is a non-zero modules M and L, that is somewhat stronger then the * condition M R ≠ 0 when LR= : f : MV→ i for some iI∈ . Let PAssfM∈ ()( ) . Let M and L be R-modules. We say that M is homo- Then there exists a non-zero clyclic submodule X of related to L if there exist R-homomorphism α : M → L f M such that PannX= . Now by [2, Corollary () R () and β : LM→ such that βα ≠ 0 . 8.3], X is a nonsingular right RP / -module so that Proposition 2.11. Let M be a non-zero R-module with there exists a submodule Y of X R /P such that Y is I = annR ( M ) . isomorphic to a right ideal of RP / . Because f (M ) is (a) M is homo-related to RI / if and only if a compressible right R-module, it can be embedded in Y. HomR (MRI,/) ≠ 0. It follows that HomR ()MRP,/≠ 0. (b) If R is a semiprime ring then M is homo-related ()⇐ . Assume that f :MRP→ / is a non-zero to R if and only if HomR (MR,0) ≠ . homomorphism for some PAssM∈ (). There exists Proof. We only prove(a) as (b) is proved by a similar R method. Suppose that a non-zero homomorphism a non-zero submodule N of M such that R α :/M → RI exists and let α (M ) = KI/ . Thus PannX= for every non-zero submodule X of N. R () MK ≠ 0 , so that there exists an element mM∈ such By Lemma 2.9, suppose that X is some cyclic uniform that mK ≠ 0 . Define β :RI / → M by compressible submodule of N. Again X is a torsionfree β (rI+=) mr, which satisfies βα ≠ 0 . The converse right RP/ -module, hence a non-zero right ideal Y of RP/ can be embedded in X. On the other hand, since is obvious. □ The following two Lemmas are needed. RP/ is a prime ring Yf( M ) ≠ 0 . Consequently, Lemma 2.12. Let M = ⊕ iI∈ M i be a direct sum of HomR ()fMY(),0≠ . It follows that HomR ()MX,0≠ . uniform submodules M i (iI∈ ) and let N be any non- Now X R is compressible and so is any of its zero submodule of M. Then there exists a subset I ′ of I submodules, hence M R is essentially retractable by such that the canonical projection π : M →⊕iI∈ ′ M i is

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one to one on N and π ()N is an essential submodule be embedded in Vi. By our assumption, we may suppose that Ui is cyclic and retractable. Now ⊕ iI∈ U i is of ⊕ iI∈ ′ M i. isomorphic to an essential submodule of M R and so by Proof. If NNMii=≠∩ 0 for all iI∈ then ⊕N i (a), there exists a non-zero homomorphism is an essential submodule of M and so NM≤ , in eR f ::MUL→⊕ = . It follows that M is homo- which case there remains nothing to prove. Suppose that iI∈ i related to L. NM∩ = 0 for some jI∈ . By Zorn’s Lemma there j (b) ⇒ (c). This is clear. exists a maximal subset I ′′ of I such that (c) ⇒ (a). Generally, if W is a non-zero nonsingular

NM∩ ()⊕=iI∈ ′′ i 0 . Note that I ′′ is a proper subset right R-module and L is a direct sum of uniform of I and hence III′′′= \ is a nonempty set. Let retractable right R-modules and there exist homomorphisms f :WL→ and g : LW→ such that π : M →⊕iI∈ ′ M i denote the canonical projection. f ≠ 0 and g is one to one, then W is essentially Then π |NiIi : NM→⊕ ∈ ′ is a monomorphism R retractable. This is because, by Lemmas 2.12 and 2.13, because ker (π |0NiIi) =⊕NM∩ ( ∈ ′′ ) =. Let kI∈ ′ . f (W ) is non-zero retractable and then W R is By the choice of I ′′ , NM∩{ kiIi⊕⊕()∈ ′′ M} ≠0 essentially retractable by Proposition 2.2. and hence π NM∩ ≠ 0 . It follows that π N is () k ( ) Now suppose that f :ML→=⊕ : iI∈ U i and an essential submodule of ⊕ M . □ iI∈ ′ i g : LM→ such that gof ≠ 0 where each UiIi ( ∈ )

Lemma 2.13. Let M R be retractable and is a uniform retractable module.

0 ≠≤NMR . If HomR (MNN/,) = 0 then N is Case 1. Assume that for each iI∈ , gU( i ) ≠ 0 . retractable. Furthermore, if M R is nonsingular then any Because M R is nonsingular, g is one to one on essential submodule of M R is retractable. UiIi ( ∈ ) . It follows that there exists a (I ) Proof. Let 0 ≠≤K N R . There is 0 ≠∈fS such that monomorphism hL::→= W M . Also 0≠ foπ : f ()MK⊆ . If fN()= 0 , then the rule mn+ WL→ where π :WM→ is any epimorphism. Thus,

→+mKerf yields a non-zero homomorphism by the first part, W R and so M R is essentially M /kerImNMf→≅ f which is in contradiction retractable. Case 2. Assume that there exists iI∈ such that with our assumption HomR ()MNN/,= 0. Thus gU( i ) = 0 . By Zorn’s Lemma, choose a maximal fN( ) ≠ 0 , hence f |N is a non-zero endomorphism of subset I ′ of I such that gU()⊕=′ 0 and set N with image in K. The last statement is now clear. □ iI∈ i ′ ′ Recall that a right R-module M is said to have LU= ⊕ jJ∈ j where J = II\ . Again, as in case 1, enough uniforms if every non-zero submodule of M L ′ can be embedded in M (J ) and because gof ≠ 0 , contains a uniform submodule. (J ) ′ Theorem 2.14. Let M be a nonsingular module over a there exists a non-zero homomorphism hM: → L. ring R whose non-zero right ideals contain non-zero Thus M R is essentially retractable. □ Corollary 2.15. Let R be a finite uniform dimensional retractable right ideals. If M R has enough uniforms, then the following statements are equivalent. commutative ring. Then a nonsingular R-module is (a) M is essentially retractable. essentially retractable if and only if it is homo-related to R a direct sum of uniform retractable modules. (b) M is homo-related to a direct sum ⊕ U of R iI∈ i Proof. If the cyclic R-module R/I is nonsingular, then it cyclic nonsingular uniform retractable right ideals Ui of is easy to verify that I is an essentially closed right ideal R. of R. Hence the uniform dimension of R is bigger than (c) M R is homo-related to a direct sum of uniform the uniform dimension of RI/ . This shows that any ( ) R retractable right R-modules. nonsingular R-module has enough uniforms. The result Proof. (a) ⇒ (b). By our assumption and Zorn’s is now clear by Theorem 2.14. □ Lemma there exists an essential submodule N in M R such that NV=⊕ where each ViI()∈ is a iI∈ i i Acknowledgement uniform submodule. Since each Vi is a nonsingular right R-module, there is a