2_talebi.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 5, No. 2, 2012, 108-115 ISSN 1307-5543 – www.ejpam.com A Generalization of ⊕-Supplemented Modules Tayyebeh Amouzegar1, Yahya Talebi 2,∗ 1 Department of Mathematics, Quchan Institute of Engineering and Technology, Quchan, Iran 2 Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babol- sar, Iran Abstract. Let M and X be R-modules. We define the X -⊕-supplemented modules via the classB(M , X ) as a generalization of⊕-supplemented modules. We show that any finite direct sum of X -⊕-supplemented modules is X -⊕-supplemented. It is given a number of necessary and sufficient conditions for every direct summand of an X -⊕-supplemented module to be X -⊕-supplemented. 2010 Mathematics Subject Classifications: 16D90, 16D99 Key Words and Phrases: X -⊕-supplemented module, Completely X -⊕-supplemented module, Hollow module 1. Introduction Throughout this paper R will denote an arbitrary associative ring with identity and M a unitary R-module. A submodule N of M is called small in M (notation N ≪ M) if ∀L � M , L + N 6= M . A non-zero module M is called hollow if every proper submodule is small in M . Let K and N be submodules of M . K is called a supplement of N in M if M = K + N and K is minimal with respect to this property, or equivalently, M = K + N and K ∩ N ≪ K . A submodule K of M is called a supplement in M provided there exists a submodule N of M such that K is a supplement of N in M . Following [9], a module M is called supplemented if every submodule of M has a supplement in M . According to [6], a module M is called ⊕-supplemented if every submodule of M has a supplement that is a direct summand of M . A module M is called completely ⊕-supplemented if every direct summand of M is ⊕-supplemented [see 4]. Let M and X be R-modules. In [5], Keskin Tütüncü and Harmancı defined the family B(M , X ) = {A≤ M | ∃Y ≤ X ,∃ f ∈ Hom(M , X/Y ), Ker f /A≪ M/A} and used this class to de- fine B(M , X )-projective modules as a generalization of projective modules. In this paper we define X -⊕-supplemented modules and completely X -⊕-supplemented modules via the class ∗Corresponding author. Email addresses: t.amoozegar�yahoo. om (T. Amouzegar), talebi�umz.a .ir (Y. Talebi) http://www.ejpam.com 108 c© 2012 EJPAM All rights reserved. T. Amouzegar, Y. Talebi / Eur. J. Pure Appl. Math, 5 (2012), 108-115 109 B(M , X ) as generalizations of ⊕-supplemented modules and completely ⊕-supplemented modules respectively. Let A and P be submodules of M with P ∈ B(M , X ). Following [7], P is called an X - supplement of A in M if it is minimal with the property M = A+ P. Equivalently, if M = A+ P and A ∩ P ≪ P. A module M is called X -supplemented if every submodule N of M with N ∈ B(M , X ) has an X -supplement in M . We say that a module M is X -⊕-supplemented if every submodule N of M with N ∈ B(M , X ), has an X -supplement that is a direct summand of M . We prove some results on these classes of modules. In Section 2, we recall some notions and results that they are used in this paper. In Section 3, we give a characterization of X -⊕- supplemented modules. It is shown that any finite direct sum of X -⊕-supplemented modules is X -⊕-supplemented. We give a number of necessary and sufficient conditions for every direct summand of an X -⊕-supplemented module to be X -⊕-supplemented. We show that the direct sum of any finite family Mi of relatively B-projective modules is X -⊕-supplemented if and only if every Mi is X -⊕-supplemented. In Section 4, we prove the equivalence of two conditions for a module with finite Goldie dimension: One saying that every direct summand N of M with N ∈ B(M , X ) is a finite direct sum of X -hollow modules, and the other stating that M is a completely X -⊕-supplemented module. 2. Preliminaries Let M be a module and N ≤ M . N is called a coclosed submodule in M if whenever N/K ≪ M/K then N = K . Let M be a module and B ≤ A ≤ M . If B is coclosed in M and A/B ≪ M/B, then B is called an co-closure of A in M . A non-zero module M is called local if the sum of all proper submodules of M is also a proper submodule of M . Every local module is hollow and hollow modules are ⊕-supplemented. A submodule K of M is called essential in M (notation K ≤e M) if K∩A 6= 0 for any nonzero submodule A of M . Recall that a module M is said to have the summand sum property (SSP) if the sum of two direct summands is again a direct summand. A module M is said to have the (finite) internal exchange property if for every (finite) index set I , whenever M = ⊕i∈IAi for modules Ai, then for every direct summand K of M there exist submodules Bi of Ai such that M = K⊕ (⊕i∈I Bi). The notation N ≤⊕ M denotes that N is a direct summand of M . N à M means that N is a fully invariant submodule of M (i.e., ∀φ ∈ EndR(M),φ(N) ⊆ N). Lemma 1. Let M, N and X be R-modules. Then the following hold: (1) If A∈B(M , X ) and B ≤ A with A/B≪ M/B, then B ∈B(M , X ). (2) Let h : M → N be an epimorphism and A∈B(M , X ) with Ker h≤ A. Then h(A) ∈ B(N , X ). Conversely, if h(A) ∈B(N , X ) and Ker h≤ A, then A∈B(M , X ). (3) Let B ≤ A≤ M. Then A∈ B(M , X ) if and only if A/B ∈B(M/B, X ). (4) Let h : N → M be an epimorphism and A∈B(M , X ). Then h−1(A) ∈B(N , X ). T. Amouzegar, Y. Talebi / Eur. J. Pure Appl. Math, 5 (2012), 108-115 110 Proof. See [5, Lemma 2.2]. Lemma 2. Let M and X be R-modules. Then the following hold: (1) Let M = A+ B. If B ∈ B(M , X ), then A∩ B ∈ B(M , X ). (2) Let M = ⊕ i∈I Mi . If Ni ∈B(Mi , X ), for every i ∈ I . Then ⊕i∈I Ni ∈B(M , X ). (3) Let M = M1 ⊕M2. If A∈B(M , X ), then A+Mi ∈ B(M , X ) for i = 1,2. Proof. (1) Let M = A+ B and B ∈ B(M , X ). There exist Y ≤ X and f : M → X/Y such that Ker f /B≪ M/B. Consider the isomorphism α : M/B → A/(A∩ B). Then α(Ker f /B) = Ker f /(A∩ B). Hence Ker f /(A∩ B)≪ M/(A∩ B). Therefore A∩ B ∈B(M , X ). (2) Since Ni ∈B(Mi , X ), there exist a submodule Y of X and a homomorphism fi : Mi → X/Y such that Ker fi/Ni ≪ Mi/Ni. Put f = ⊕i∈I fi . Then f : M → X/Y such that Ker f /⊕i∈I Ni ≪ M/⊕i∈I Ni. Thus ⊕i∈I Ni ∈ B(M , X ). (3) By Lemma 1 and [5, Lemma 3.5]. 3. X -⊕-Supplemented Modules Let X and M be R-modules. We recall that a module M is X -⊕-supplemented if every submodule N of M with N ∈ B(M , X ), has an X -supplement that is a direct summand of M . Clearly X -hollow modules are X -⊕-supplemented. It is obvious that X -⊕-supplemented modules are X -supplemented. Proposition 1. Let M be a module such that every submodule A of M with A ∈ B(M , X ) has a co-closure in M. Then the following statements are equivalent: (1) M is X -⊕-supplemented. (2) Any coclosed submodule H of M with H ∈ B(M , X ), has an X -supplement that is a direct summand of M. (3) For any submodule N of M with N ∈B(M , X ), there exists a direct summand K of M with K ∈B(M , X ) such that M = N + K and N ∩ K ≪ M. (4) For any coclosed submodule H of M with H ∈ B(M , X ), there exists a direct summand K of M with K ∈B(M , X ) such that M = H + K and H ∩ K ≪ M. Proof. (1)⇔ (3), (2)⇔ (4), (1)⇒ (2) and (3)⇒ (4) are clear. (4) ⇒ (1) Let A ∈ B(M , X ). By assumption, there exists a coclosed submodule B of M such that B ≤ A and A/B≪ M/B. By Lemma 1, B ∈ B(M , X ). Therefore there exists a direct summand K of M with K ∈ B(M , X ) such that M = B + K and B ∩ K ≪ M . Hence K is an X -supplement of B in M . Note that M = A+ K . Assume that K ′ < K and M = A+ K ′. Then M 6= B+ K ′ and so M 6= A+ K ′ since A/B≪ M/B. Thus K is an X -supplement of A in M . T. Amouzegar, Y. Talebi / Eur. J. Pure Appl. Math, 5 (2012), 108-115 111 Theorem 1. Any finite direct sum of X -⊕-supplemented modules is X -⊕-supplemented. Proof. Let M = M1 ⊕ M2 where M1 and M2 are X -⊕-supplemented modules. Let N be any submodule of M with N ∈ B(M , X ). We have N + M2 = M2 ⊕ [(N + M2) ∩ M1]. Since N ∈B(M , X ), N +M2 ∈B(M , X ) by Lemma 2. From [7, Lemma 3.1], (N+M2)∩M1 ∈B(M1, X ). Since M1 is X -⊕-supplemented, there exists a direct summand K1 of M1 with K1 ∈B(M1, X ) such that [(N +M2)∩M1]+K1 = M1 and (N +M2)∩K1≪ K1. By Lemma 2 and [7, Lemma 3.1], (N+K1)∩M2 ∈ B(M2, X ). Thus there exists a direct summand K2 of M2 with K2 ∈B(M2, X ) such that [(N + K1)∩M2] + K2 = M2 and (N + K1)∩ K2≪ K2. Let K = K1 ⊕ K2, then K is a direct summand of M and K ∈ B(M , X ) (Lemma 2). Moreover, M1 ≤ N + M2 + K1 and M2 ≤ N + K1 + K2. Hence M = N + K1 + K2 = N + K . Since N ∩ (K1+K2)≤ (N +K1)∩K2+(N+K2)∩K1, N ∩ (K1+K2)≤ (N +K1)∩K2+(N+M2)∩K1. As (N +M2)∩K1≪ K1 and (N + K1)∩ K2≪ K2, (N ∩K)≪ K . Thus M is X -⊕-supplemented. Corollary 1. Any finite direct sum of X -hollow modules is X -⊕-supplemented. Lemma 3. Let M = N⊕N ′ be a module. Assume that A is a submodule of N and K a submodule of M. If K ∩ (A⊕ N ′)≪ K, then A∩ (K + N ′)≪ N ∩ (K + N ′). Proof. Let π be the projection N ⊕ N ′ → N . Since K ∩ π−1(A) = K ∩ (A⊕ N ′) ≪ K , π(K∩π−1(A)) = π(K)∩A≪ π(K). But π(K) = N∩(K+N ′). Hence A∩(K+N ′)≪ N∩(K+N ′). Following [5], an R-module N is called B(M , X )-projective if for any submodule A of M with A ∈ B(M , X ), any homomorphism φ : N → M/A can be lifted to a homomorphism ψ : N → M . Two R-modules M1 and M2 are called relatively B-projective if M1 is B(M2, X )- projective and M2 is B(M1, X )-projective. Theorem 2. Let M = ⊕n i=1 Mi be a finite direct sum of relativelyB-projective modules Mi and let M have the summand sum property. Then the module M is X -⊕-supplemented if and only if Mi is X -⊕-supplemented for all 1≤ i ≤ n. Proof. The sufficiency is proved in Theorem 1. Conversely, we only prove M1 is X -⊕- supplemented. Let A ∈ B(M1, X ). By Lemma 1, A⊕ M2 ∈ B(M , X ). Since M is X -⊕- supplemented, there exists B ∈ B(M , X ) such that M = (A⊕M2) + B, (A⊕M2)∩ B≪ B and B is a direct summand of M . By Lemma 2, M2 + B ∈B(M , X ). Clearly M = M1 +M2 + B. By [5, Proposition 2.5], there exists T ≤ M2 + B such that M = M1 ⊕ T . Thus B +M2 = (M1 ∩ (B+M2))⊕ T . Now M1 = A+ ((B+M2)∩M1) and since (A⊕M2)∩ B≪ B, by Lemma 3, A∩ (M1 ∩ (B + M2))≪ M1 ∩ (B + M2). As M has the summand sum property, B + M2 is a direct summand of M . Thus (B + M2) ∩ M1 ≤ ⊕ M and so (B + M2) ∩ M1 is a direct summand of M1. By [7, Lemma 3.1 (1)], (B + M2) ∩ M1 ∈ B(M1, X ). Hence M1 is X -⊕-supplemented. Proposition 2. Let M and N be R-modules and h : M → N be an epimorphism such that Ker hà M. If M is X -⊕-supplemented, then N is X -⊕-supplemented. T. Amouzegar, Y. Talebi / Eur. J. Pure Appl. Math, 5 (2012), 108-115 112 Proof. Let A ∈ B(N , X ). By Lemma 1, h−1(A) ∈ B(M , X ). Since M is X -⊕-supplemented, there exist submodules H and H ′ of M such that M = H ⊕H ′, M = h−1(A)+H and h−1(A)∩H ≪ H. Now N = A+ h(H) and since h−1(A)∩H ≪ H, h(h−1(A) ∩ H) = A∩ h(H) ≪ h(H). Moreover, since Ker h à M , N = h(H)⊕ h(H ′). There- fore h(H) is an X -supplement of A in N and it is a direct summand of N . Hence N is X -⊕- supplemented. Corollary 2. Let M be an R-module and N be a fully invariant submodule of M. If M is X -⊕- supplemented, then M/N is X -⊕-supplemented. Proof. By Proposition 2. Recall that a module M is a duo module, if every submodule of M is a fully invariant submodule of M . Corollary 3. Let M be an X -⊕-supplemented duo module, then every direct summand of M is X -⊕-supplemented. Proof. By Corollary 2. Definition 1. A module M is said to have the (finite) strong internal exchange property if for every (finite) index set I, whenever M = K+(⊕i∈IAi) for a direct summand K of M and modules Ai, then M = K ⊕ (⊕i∈I Bi) for submodules Bi of Ai. It is clear that if a module M has the (finite) strong internal exchange property, then M has the (finite) internal exchange property. Theorem 3. Let M be an X -⊕-supplemented module with the finite strong internal exchange property. Then any direct summand of M is X -⊕-supplemented. Proof. Let N be a direct summand of M . Thus M = N ⊕ N ′ for some submodule N ′of M . Let A ∈ B(N , X ). By Lemma 1, A⊕ N ′ ∈ B(M , X ). Since M is X -⊕-supplemented, there exists a direct summand K of M with K ∈ B(M , X ) such that M = K + (A⊕ N ′) and (A⊕N ′)∩K ≪ K . Since M has the finite strong internal exchange property, M = K ⊕N1⊕N ′1 such that N1 ⊆ A and N ′1 ⊆ N ′. By modularity, N = N1⊕ (N ∩ (K ⊕ N ′1)). By Lemma 2 and [7, Lemma 3.1], N ∩ (K ⊕ N ′1) ∈ B(N , X ). As M = A+ (K ⊕ N ′1), N = A+ (N ∩ (K ′ ⊕ N ′1). Since (A⊕N ′)∩K ≪ K , by Lemma 3, A∩(K⊕N ′)≪ N ∩(K⊕N ′). Thus A∩(K⊕N ′1)≪ N ∩(K⊕N ′). Since N ∩ (K ⊕ N ′1)≤ ⊕ M , A∩ (K ⊕ N ′1)≪ N ∩ (K ⊕ N ′1). Hence N is X -⊕-supplemented. If in setB(M , X ), we take X = M , thenB(M , X ) coincides with the set of all submodules of M . Therefore we obtain the following corollary: Corollary 4. Let M be a ⊕-supplemented module with the finite strong internal exchange prop- erty. Then any direct summand of M is ⊕-supplemented. T. Amouzegar, Y. Talebi / Eur. J. Pure Appl. Math, 5 (2012), 108-115 113 4. Completely X -⊕-Supplemented Modules Let X and M be R-modules. We call a module M completely X -⊕-supplemented if every direct summand N of M with N ∈B(M , X ) is X -⊕-supplemented. Recall that a module M has B(M , X )-(D3) condition if for all A ∈ B(M , X ) and direct summand B of M , if A is a direct summand of M and M = A+ B then A ∩ B is a direct summand of M [5]. Proposition 3. Let M be an X -⊕-supplemented module with B(M , X )-(D3). Then M is com- pletely X -⊕-supplemented. Proof. Let N be a direct summand of M and A a submodule of N such that N ∈ B(M , X ) and A ∈ B(N , X ). We show that A has an X -supplement in N that is a direct summand of N . We have M = N ⊕ N ′ for some submodule N ′ of M . Let π : M → N be the projection along N ′. Since A∈B(N , X ), by Lemma 1(4), A⊕ N ′ = π−1(A) ∈B(M , X ). Since M = A+ N + N ′, A= (A⊕ N ′)∩ N ∈ B(M , X ) (Lemma 2). Since M is X -⊕-supplemented, there exists a direct summand B of M with B ∈B(M , X ) such that M = A+B and A∩B≪ B. Then N = A+(N∩B). Again by Lemma 2, N ∩ B ∈ B(M , X ). Furthermore N ∩ B is a direct summand of M because M has B(M , X )-(D3). Then A∩ (N ∩ B) = A∩ B is small in N ∩ B and by [7, Lemma 3.1], N ∩ B ∈B(N , X ). Let X and M be R-modules. We say N ∈ B(M , X ) is semisimple relative to the class B(M , X ) if, for every submodule K of N with K ∈ B(N , X ), there exists a submodule K ′ of N with K ′ ∈ B(N , X ) such that N = K ⊕ K ′. It is clear that every semisimple module relative to the classB(M , X ) is X -⊕-supplemented. Lemma 4. Let M be an X -supplemented module and let N be a submodule of M such that N ∩ Rad(M) = 0 and N ∈B(M , X ). Then N is semisimple relative to the classB(M , X ). Proof. We have to prove that M/Rad(M) contains no non-zero small submodule K/Rad(M) with K/Rad(M) ∈B(M/Rad(M), X ). Let K/Rad(M)≪ M/Rad(M) and K/Rad(M) ∈ B(M/Rad(M), X ). From Lemma 1, K ∈ B(M , X ). By hypothesis, there ex- ists a submodule B of M with B ∈ B(M , X ) such that M = K + B and K ∩ B ≪ B. As K/Rad(M) ≪ M/Rad(M), Rad(M) = K . Thus every submodule K/Rad(M) of M/Rad(M) with K/Rad(M) ∈B(M/Rad(M), X ) is a direct summand of M/Rad(M). Hence M/Rad(M) is semisimple relative to the class B(M/Rad(M), X ). Hence N is semisimple relative to the classB(M , X ). Proposition 4. Let M be an X -supplemented module and suppose that for every submodule N of M such that N ∩ Rad(M) = 0 we have N ∈ B(M , X ). Then M = M1 ⊕ M2, where M1 is a semisimple module relative to the classB(M , X ) and Rad(M2) essential in M2. Proof. Let M1 be a complement of Rad(M) in M , hence Rad(M)⊕ M1 is essential in M . Since M is X -supplemented, there exists a submodule M2 of M such that M = M1 + M2, M1 ∩ M2 ≪ M2 and M2 ∈ B(M , X ). Then M1 ∩ M2 is a submodule of both T. Amouzegar, Y. Talebi / Eur. J. Pure Appl. Math, 5 (2012), 108-115 114 Rad(M) and M1. It follows that M = M1⊕M2, Rad(M) = Rad(M2) is essential in M2, and by Lemma 4, M1 is semisimple relative to the classB(M , X ). A module M is said to be finite Goldie-dimensional provided M contains no infinite inde- pendent families of nonzero submodules. Theorem 4. Consider the following conditions for a projective module M: (i) M is a direct sum of X -⊕-supplemented modules and Rad(M) has finite Goldie dimension. (ii) M = M1 ⊕M2 such that M1 is semisimple relative to the class B(M , X ) and M2 has finite Goldie dimension and M2 is a (finite) direct sum of local modules. If for every submodule N of a direct summand Mi of M such that N ∩ Rad(Mi) = 0 we have N ∈B(Mi , X ), then (i)⇒ (ii) holds and if for every small submodule N of M1 we have N ∈B(M1, X ), then (ii)⇒ (i) holds. Proof. (i) ⇒ (ii) Let M = ⊕ i∈I Mi and Mi is X -⊕-supplemented for every i ∈ I . Since Rad(M) = ⊕i∈IRad(Mi), then there is a finite subset J of I such that Rad(Mi) = 0 for all i ∈ I \ J . Therefore Mi is semisimple relative to B(M , X ) for all i ∈ I \ J . Hence there is a submodule M1 semisimple relative to B(M , X ) such that M = M1 ⊕ (⊕ j∈J M j). By Propo- sition 4, without loss of generality, we may assume Rad(M j) is essential in M j( j ∈ J). Then M j( j ∈ J) has finite Goldie dimension by [3, Proposition 3.20]. Next we prove that each M j , for j ∈ J , is local or a finite direct sum of local modules. Set H = M j for any j ∈ J . First, note that Rad(H) 6= H because H is projective [1, Proposition 17.14]. Assume that H has Goldie dimension 1, and take some x ∈ H \Rad(H). Since H is X -⊕-supplemented, there is a submodule K of H with K ∈ B(H, X ) such that H = xR+ K , xR∩ K ≪ K and H = K ⊕ K1 for some submodule K1 of M . Then K = 0 or K1 = 0. If K1 = 0, then xR⊆ Rad(H) which is a con- tradiction. Hence K = 0 and H = xR. It follows that H is local. Let n> 1 be a positive integer and assume that each M j having Goldie dimension k (1≤ k < n) is local or a finite direct sum of local submodules. Let j ∈ J and H = M j and assume H has Goldie dimension n. Suppose that H is not local. Let x ∈ H \ Rad(H) such that H 6= xR. Since H is X -⊕-supplemented, there exist submodules K , K1 of H with K ∈ B(H, X ) such that H = xR+ K = K ⊕ K1 and xR∩ K ≪ K . It is clear that K1 6= 0. Also K 6= 0. Since projective modules satisfy (D3), and so they satisfy B(M , X )-(D3). By Proposition 3, we obtain that any direct summand of M is X -⊕-supplemented. Thus K and K1 are X -⊕-supplemented. By induction, K and K1 are local or finite direct sum of local submodules. This completes the proof of (i)⇒ (ii). (ii)⇒ (i) It is clear. Lemma 5. Let M be an indecomposable module. Then M is X -hollow if and only if M is com- pletely X -⊕-supplemented. Proof. Let M be completely X -⊕-supplemented. If N ∈ B(M , X ) is a proper submodule of M , then there exists an X -supplement A of M such that A is direct summand of M . By hypothesis we have A = M . Thus N = N ∩ M = N ∩ A ≪ M . Therefore M is X -hollow. Conversely, if M is X -hollow and N ∈ B(M , X ) then N ≪ M . Since M ∈ B(M , X ), M is an X -supplement of N in M . REFERENCES 115 Proposition 5. Let M = U ⊕ V such that U and V have local endomorphism rings. Then M is completely X -⊕-supplemented if and only if U and V are X -hollow modules. Proof. The necessity is clear from Lemma 5. Conversely, let K ∈ B(M , X ) be a direct summand of M . If K = M then by Corollary 1, K is X -⊕-supplemented. Assume K 6= M . Then either K ∼= U or K ∼= V [1, Corollary 12.7]. In either case K is X -⊕-supplemented. Thus M is completely X -⊕-supplemented. Theorem 5. Let M be a non-zero module with finite Goldie dimension. Then the following statements are equivalent: (i) Every direct summand N of M with N ∈ B(M , X ) is a finite direct sum of X -hollow modules. (ii) M is a completely X -⊕-supplemented module. Proof. (i)⇒ (ii) It is clear by Corollary 1. (ii)⇒ (i) Let N be a direct summand of M with N ∈ B(M , X ). 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