On H-Supplemented Modules - World Scientific

Algebra Colloquium 18 (Spec 1) (2011) 915–924

Algebra Colloquium c 2011 AMSS CAS ° & SUZHOU UNIV

On H-Supplemented Modules Derya Keskin T¨ ut¨ unc¨ u Department of Mathematics, University of Hacettepe 06800 Beytepe, Ankara, Turkey E-mail: [email protected]

Mohammad Javad Nematollahi Department of Mathematics, Islamic Azad University Arsanjan Branch, P.O. Box 73761-168, Arsanjan, Iran E-mail: [email protected]

Yahya Talebi Department of Mathematics, University of Mazandaran, Babolsar, Iran E-mail: [email protected] Received 9 March 2009 Revised 17 June 2009 Communicated by Nanqing Ding Ln

Abstract. Let M = i=1 Mi be a finite direct sum of modules. We prove: (i) If Mi is radical Mj -projective for all j > i and each Mi is H-supplemented, then M is H-supplemented. (ii) If all the Mi are relatively projective and M is H-supplemented, then each Mi is H-supplemented. Let ρ be the preradical for a cohereditary torsion theory. Let M be a module such that ρ(M ) has a unique coclosure and every direct summand of ρ(M ) has a coclosure in M . Then M is H-supplemented if and only if there exists a decomposition M = M1 ⊕ M2 such that M2 ⊆ ρ(M ), ρ(M )/M2 ¿ M/M2 , and M1 , M2 are H-supplemented. 2000 Mathematics Subject Classification: 16D10, 16D99 Keywords: (H-)supplemented module, (F I-)lifting module, supplement bounded module

1 Introduction Throughout this paper, R is always an associative ring with unit and all modules are unitary right R-modules. A submodule N of a module M is called small in M (denoted by N ¿ M ) if for every proper submodule L of M , N + L 6= M . A module M is called hollow if every proper submodule of M is small in M . Let N and L be submodules of M . N is called a supplement of L in M if it is minimal with respect to the property M = N + L, equivalently, M = N + L and N ∩ L ¿ N . M is called supplemented (resp., weakly supplemented) if for each submodule A of M , there exists a submodule B of M such that M = A + B and A ∩ B ¿ B (resp., A ∩ B ¿ M ). M is called amply supplemented if for any two submodules A

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D. Keskin T¨ ut¨ unc¨ u, M.J. Nematollahi, Y. Talebi

and B of M with M = A + B, A contains a supplement of B in M . The module M is called ⊕-supplemented if every submodule of M has a supplement which is a direct summand. For A ⊆ B ⊆ M , A is called a coessential submodule of B in M (denoted by A ≤ce B in M ) if B/A ¿ M/A. A submodule A of M is called coclosed in M (denoted by A ≤cc M ) if A has no proper coessential submodule. A is called a coclosure of B in M if A ≤ce B in M and A ≤cc M . A module M is called a unique coclosure module (or U CC-module) if every submodule of M has a unique coclosure in M . Recall that M is lifting if for any submodule N of M , there exists a direct summand K of M such that K ≤ N and N/K ¿ M/K, equivalently, if M is amply supplemented and every supplement submodule of M is a direct summand of M (see [5, Lemma 4.8]). M is called H-supplemented if for any submodule A of M , there exists a direct summand D of M such that A + X = M if and only if D + X = M for all X ≤ M . Obviously, every lifting module is an H-supplemented module and every H-supplemented module is ⊕-supplemented, but in general the converse directions are not true (see [5, Lemma A.4]). A submodule A of a module M is called projection invariant in M if (A)f ≤ A for any idempotent f ∈ End(M ). If for any f ∈ End(M ), (A)f ≤ A, then A is called a fully invariant submodule of M . In Section 2, we give some new characterizations of H-supplemented modules and investigate projection invariant submodules of an H-supplemented module. In Section 3, we investigate radical projective modules. The direct sum of two Hsupplemented modules need not be H-supplemented. While the properties lifting, amply supplemented and supplemented are inherited by direct summands, it is unknown (and unlikely) that the same is true for the property H-supplemented. In Section 4, we investigate finite direct sums of H-supplemented modules and direct summands of an H-supplemented module. 2 H-Supplemented Modules Theorem 2.1. Let M be a module. The following are equivalent: (1) M is H-supplemented. (2) For each Y ≤ M , there exists a direct summand D of M such that (Y + D)/D ¿ M/D and (Y + D)/Y ¿ M/Y . (3) For each Y ≤ M , there exist X ≤ M and a direct summand D of M with Y ⊆ X and D ⊆ X such that X/Y ¿ M/Y and X/D ¿ M/D. (4) For each Y ≤ M , there exist a supplement L of Y and a supplement K of L such that (Y +K)/Y ¿ M/Y , (Y +K)/K ¿ M/K and every homomorphism f : M → M/(K ∩ L) can be lifted to a homomorphism f¯ : M → M . Proof. (1)⇒(2) It is clear. (2)⇒(3) Let Y ≤ M . Then there exists a direct summand D of M such that (Y + D)/Y ¿ M/Y and (Y + D)/D ¿ M/D. Now take X = Y + D. (3)⇒(1) Let Y be a submodule of M . Then there exist a submodule X of M and a direct summand D of M such that both Y and D are coessential submodules of X in M . It is easy to see that M = A + D if and only if M = A + Y for any A ≤ M . Thus, M is H-supplemented.

On H-Supplemented Modules

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(2)⇒(4) Suppose Y ≤ M . Then there exist D, D0 ≤ M such that M = D ⊕ D0 , (Y + D)/Y ¿ M/Y and (Y + D)/D ¿ M/D. It is easy to show that D0 is a supplement of Y and D is a supplement of D0 . So (4) follows by taking D = K and D0 = L. (4)⇒(2) Put S = L ∩ K. We have S ¿ K and also S ¿ L. Let g : M → M/L and f : M → M/S be the natural maps. Note that there exists an isomorphism t : M/L → K/S. By assumption, there exists h : M → M such that hf = gt. We have K/S = (K)f = (K)gt = (K)hf . Hence, K + Ker f = (K)h + Ker f , i.e., K + S = (K)h + S. Hence, K = (K)h as S ¿ K. Note that (M )h = K. Hence, K = (K)h = (M )h. Therefore, K + Ker h = M . As Ker h is contained in L and L is a supplement of K, Ker h = L. Now L = Ker gt = Ker hf implies Ker f = 0, i.e., S = 0. Thus, M = K ⊕ L. This completes the proof. ¤ Definition 2.2. A module M is called F I-lifting if for every fully invariant submodule A of M , there exists a direct summand N of M such that A/N ¿ M/N . Proposition 2.3. The following are equivalent for a module M : (1) M is F I-lifting. (2) Every fully invariant submodule of M has a supplement which is a direct summand. Proof. (1)⇒(2) It is clear. (2)⇒(1) Let A be fully invariant in M . Then M = M1 ⊕ M2 such that A + M2 = M and A ∩ M2 ¿ M2 . Since A is a fully invariant submodule of M , A = (A + M1 ) ∩ (A + M2 ) = A + M1 . Hence, M1 ≤ A. Thus, M is F I-lifting. ¤ Now we have the following theorem: Theorem 2.4. Let M be a module and consider the following conditions: (1) M is lifting. (2) M is H-supplemented. (3) M is ⊕-supplemented. (4) M is F I-lifting. Then (1)⇒(2)⇒(3)⇒(4). Definition 2.5. A module M is said to have the coessential intersection property or CEIP if for any A, B, C, D ≤ M , A ≤ce B in M and C ≤ce D in M imply that A ∩ C ≤ce B ∩ D in M . Proposition 2.6. Let M be an H-supplemented module with CEIP . Then M is a U CC and lifting module. Proof. Let A be a submodule of M . Then there exists a direct summand D of M such that A ≤ce A + D in M and D ≤ce A + D in M by Theorem 2.1. By CEIP , A ∩ D ≤ce A + D in M . But A ∩ D ⊆ D ⊆ A + D, hence A ∩ D ≤ce D in M . As D ≤cc M , A ∩ D = D, i.e., D ≤ A and D ≤ce A in M . Therefore, M is lifting and so is amply supplemented. Now by [1, Proposition 3.10], M is a U CC-module. ¤

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Definition 2.7. A module M is called supplement bounded if it is supplemented and every proper coclosed submodule of M is contained in a nontrivial fully invariant submodule of M . Proposition 2.8. Let M be a supplemented module. Then M is supplement bounded if and only if every proper coclosed submodule of M is coessential in a nontrivial fully invariant submodule of M . Proof. Assume that M is supplement bounded. Let K ≤cc M be proper. Let X be the intersection of fully invariant submodules of M containing K. Then X is a fully invariant submodule of M . Let L/K be the supplement of X/K in M/K. Then L+X = M and (L/K)∩(X/K) ¿ M/K. Suppose L 6= M . Since L is a supplement submodule of M (see [4, Lemma 1.4]), there exists a fully invariant submodule Y of M such that Y 6= M and L ⊆ Y . So L + X ⊆ Y 6= M , a contradiction. Therefore, L = M and hence K ≤ce X in M . The converse is trivial. ¤ Proposition 2.9. Let M be a module. (1) Let M be indecomposable. Then M is H-supplemented if and only if M is hollow. (2) Let M be a U CC-module. Then M is H-supplemented if and only if M is lifting. (3) Let M be supplement bounded and suppose that every submodule of M has a coclosure. Then M is H-supplemented if and only if M is F I-lifting. Proof. (1) It is clear. (2) Let M be an H-supplemented U CC-module. Let N ≤ M . Since M is Hsupplemented, there exists a direct summand D of M such that (N +D)/D ¿ M/D and (N + D)/N ¿ M/N by Theorem 2.1. Now D is the unique coclosure of N + D in M . By [1, Lemma 3.2], D ≤ N . Thus, M is lifting. (3) (⇐) Assume that M is F I-lifting. Let Y ≤ M and Y 6= M . Since Y has a coclosure, there exists a submodule K of M such that K ⊆ Y , Y /K ¿ M/K and K ≤cc M . Since M is supplement bounded, there exists a fully invariant submodule B of M with K ≤ B and B/K ¿ M/K by Proposition 2.8. Since M is F I-lifting, there exists a direct summand D of M such that D ≤ B and B/D ¿ M/D. Let M = Y +L for some L ≤ M . Then M/K = Y /K +(L+K)/K = (L+K)/K implies that M = L+K. Hence, M = L+B and so M/D = (L+D)/D+B/D = (L+D)/D. Thus, M = L + D. Conversely, assume M = L + D. Then M = L + B. Now M/K = (L + K)/K + B/K implies that M = L + K and hence M = L + Y . Thus, M is H-supplemented. (⇒) By Theorem 2.4. ¤ A module M is called a small module if M ¿ T for some module T . For any module M , let Z(M ) = ∩ {Ker g | g : M → N , N is a small module} (see [6]). Lemma 2.10. Let M be H-supplemented. If X is projection invariant in M and Z(X) = X, then X is a direct summand of M . Proof. Since M is H-supplemented, there exists a direct summand D of M such that (X + D)/D ¿ M/D and (X + D)/X ¿ M/X. Since X is projection invariant


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in M , (X + D)/X is a direct summand of M/X. Therefore, X = X + D and hence D ⊆ X. Consider the natural epimorphism g : X → X/D = (X + D)/D ¿ M/D. Then Z(X) = X ⊆ Ker g = D. Hence, X = D. ¤ Proposition 2.11. Let M be any module and let N ≤ M be such that for each decomposition M = M1 ⊕ M2 of M , we have N = (N ∩ M1 ) ⊕ (N ∩ M2 ). If M is Hsupplemented, then M/N is H-supplemented. If moreover N is a direct summand of M , then N is also H-supplemented. Proof. Take L/N ≤ M/N . Since M is H-supplemented, there exist a direct summand D of M and a submodule X of M such that X/L ¿ M/L and X/D ¿ M/D. Write M = D ⊕ D0 . By hypothesis, N = (D ∩ N ) ⊕ (D0 ∩ N ) = (D + N ) ∩ (D0 + N ). M/N X/N So (D + N )/N ⊕ (D0 + N )/N = M/N . Now we have X/N L/N ¿ L/N and (D+N )/N ¿ M/N (D+N )/N ,

and hence M/N is H-supplemented. Now let N be a direct summand of M and Y ≤ N . Then there exists a direct summand D of M such that D+A = M if and only if Y +A = M for all A ≤ M . Let M = D ⊕ D0 = N ⊕ N 0 for some submodules N 0 and D0 of M . On the other hand, N = (D ∩N )⊕(D0 ∩N ), M = D ⊕D0 = Y +D0 = N +D0 = (D ∩N )⊕(D0 ∩N )+D0 = (D ∩ N ) ⊕ D0 imply that D ⊆ N . Therefore, D is a direct summand of N . It is easy to see that N = D + K if and only if N = Y + K for all K ⊆ N . ¤ Corollary 2.12. Let M be H-supplemented. If N is a projection invariant submodule of M , then M/N is H-supplemented. Moreover, every projection invariant direct summand of M is H-supplemented. Theorem 2.13. Let M be an H-supplemented module. Assume that every submodule of M has a coclosure in M . If X is projection invariant in M and Z(X) = X, then X is lifting. Proof. By Lemma 2.10, X is a direct summand of M . By Corollary 2.12, X is H-supplemented. Note that M is amply supplemented by [4, Lemma 1.7]. Hence, X is amply supplemented since it is a direct summand of M . So by [1, Proposition 4.2], X is U CC. Thus, by Proposition 2.9, X is lifting. ¤ Corollary 2.14. Let M be an H-supplemented module with Z(M ) = M . Assume that every submodule of M has a coclosure in M . Then M is lifting. Proof. Take X = M in Theorem 2.13.

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Remark. Let M be H-supplemented with Z(M ) = M . Assume that every submodule of M has a coclosure in M . If X is projection invariant in M , then M/X is lifting. By Corollary 2.12, M/X is H-supplemented. By [1, Proposition 4.2], M/X is U CC. Hence, M/X is lifting by Proposition 2.9. 3 Radical Projectivity Definition 3.1. Let M and N be modules. Then N is called radical M -projective if for any K ≤ M and any homomorphism f : N → M/K, there exists a homo-

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morphism h : N → M such that Im(f − hπ) ¿ M/K, where π : M → M/K is the natural epimorphism. Radical projective modules are defined in [3]. Proposition 3.2. Let ρ be the preradical of a cohereditary torsion theory and M = M1 ⊕ M2 , where ρ(M )/M2 ¿ M/M2 . Then M2 is radical M1 -projective. Proof. Let K ≤ M1 and f : M2 → M1 /K be any homomorphism. Then (M2 )f = (ρ(M2 ))f ⊆ ρ(M1 /K) = (ρ(M1 ) + K)/K ¿ M1 /K. Now take h : M2 → M1 to be the zero homomorphism. ¤ Proposition 3.3. Let M and N be modules. Then N is radical M -projective if and only if there exists X ¿ M such that N is radical M/X-projective and for any homomorphism h : N → M/X, there exists a homomorphism h : N → M such that Im(h − hπ1 ) ¿ M/X, where π1 : M → M/X is the natural epimorphism. Proof. (⇒) Take X = 0. (⇐) Let B ≤ M and f : N → M/B be a homomorphism. Let π : M → M/B be the natural epimorphism. Take B1 = B + X. Now consider the epimorphisms πB : M/B → M/B1 , π1 : M → M/X and πX : M/X → M/B1 . Since N is radical M/X-projective, there exist a homomorphism f 1 : N → M/X and a submodule B2 of M with B1 ⊆ B2 such that Im(f πB − f 1 πX ) = B2 /B1 ¿ M/B1 . Now there exist a homomorphism f 2 : N → M and a submodule A of M with X ⊆ A such that Im(f 1 − f 2 π1 ) = A/X ¿ M/X. It is easy to see that (N )(f − f 2 π) ⊆ (B2 + A)/B ¿ M/B. ¤ Corollary 3.4. Let M be a weakly supplemented module with Rad(M ) ¿ M . Then N is radical M -projective if and only if for any homomorphism h : N → M/Rad(M ), there exists a homomorphism h : N → M such that hπ = h, where π : M → M/Rad(M ) is the natural epimorphism. Proof. This is clear by Proposition 3.3 and the fact that N is K-projective for all semisimple modules K. ¤ Theorem 3.5. Let M = M1 ⊕ M2 . Consider the following conditions: (1) M1 is radical M2 -projective. (2) For every K ≤ M such that K + M2 = M , there exists M3 ≤ M such that M = M2 ⊕ M3 and (K + M3 )/K ¿ M/K. Then (1)⇒(2); and if M is amply supplemented, then (2)⇒(1). Proof. (1)⇒(2) Let K ≤ M and M = K + M2 . Consider the epimorphism π : M2 → M/K given by m2 7→ m2 + K and the homomorphism h : M1 → M/K given by m1 7→ m1 + K. Since M1 is radical M2 -projective, there exist a homomorphism h : M1 → M2 and a submodule X of M with K ⊆ X such that Im(h − hπ) = X/K ¿ M/K. Let M3 = {a − (a)h | a ∈ M1 }. Clearly, M = M2 ⊕ M3 . Since K + M3 ⊆ X, (K + M3 )/K ⊆ X/K. Hence, (K + M3 )/K ¿ M/K. (2)⇒(1) Assume that M is amply supplemented. Let g : M1 → M2 /L be any homomorphism and π : M2 → M2 /L the natural epimorphism. Let H = {m1 + m2 | m1 ∈ M1 , m2 ∈ M2 , (m1 )g = −(m2 )π)}.


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Then L ⊆ H and M = H + M2 . There exists a submodule H ⊆ H such that M = H + M2 and H ∩ M2 ¿ H. By hypothesis, there exists a submodule H 0 of M such that M = H 0 ⊕ M2 and (H 0 + H)/H ¿ M/H. Let α : H 0 ⊕ M2 → M2 be the projection map. Consider α|M1 : M1 → M2 . It is easy to see that Im(g − α|M1 π) ⊆ ((H 0 + H) ∩ M2 )/L and ((H 0 + H) ∩ M2 )/L ¿ M2 /L. ¤ Corollary 3.6. Let M = M1 ⊕ M2 and Z(M1 ) = M1 . Then M1 is M2 -projective if and only if M1 is radical M2 -projective. Proof. If M1 is M2 -projective, then clearly M1 is radical M2 -projective. For the converse, let M = N + M2 . By Theorem 3.5, there exists N 0 ≤ M such that (N +N 0 )/N ¿ M/N and M = N 0 ⊕M2 . Thus, N 0 ' M1 . Now let α : M1 → N 0 be the isomorphism. Consider the epimorphism η : N 0 → (N + N 0 )/N given by n0 7→ n0 +N with Ker η = N ∩N 0 . Then αη : M1 → (N +N 0 )/N ¿ M/N . Therefore, M1 = Z(M1 ) ⊆ Ker αη ⊆ M1 implies that (M1 )αη = (N 0 )η = (N + N 0 )/N = 0 and hence N 0 ⊆ N . Now by [7, 41.14], M1 is M2 -projective. ¤ 4 Direct Sums of H-Supplemented Modules Example 4.1. Let R be a commutative local ring and let M be a finitely generated Ln R-module. Assume M ' i=1 R/Ii for ideals Ii . Since every Ii is fully invariant in R, every R/Ii is H-supplemented. By [5, Lemma A.4], M is H-supplemented if I1 ≤ I2 ≤ · · · ≤ In . Otherwise, M is not H-supplemented. As we see from the above example, a finite direct sum of H-supplemented modules need not be H-supplemented. Theorem 4.2. Let M = M1 ⊕ M2 . (1) If M1 is radical M2 -projective (or M2 is radical M1 -projective) and M1 , M2 are H-supplemented, then M is H-supplemented. (2) If M2 is M1 -projective and M is H-supplemented, then M1 is H-supplemented. Proof. (1) Let Y ≤ M . Case 1: M = Y + M2 . By Theorem 3.5, there exists M3 ≤ M such that M = M3 ⊕M2 and (Y +M3 )/Y ¿ M/Y . Since M/M3 ' M2 , M/M3 is H-supplemented. Now consider the submodule (Y + M3 )/M3 of M/M3 . By Theorem 2.1, there exist X/M3 X/M3 ≤ M/M3 and a direct summand D/M3 of M/M3 such that (Y +M ' 3 )/M3 X/M3 M M ¿ Y +M and D/M ' X D ¿ D . Clearly, D is a direct summand of M . On 3 3 the other hand, X/Y ¿ M/Y . Therefore, M is H-supplemented. Case 2: M 6= Y + M2 . Since M/Y is supplemented, there exists a submodule K/Y of M/Y such that M/Y = K/Y +(Y +M2 )/Y and (K ∩(Y +M2 ))/Y ¿ K/Y . Then M = K +M2 . By Theorem 3.5, there exists M4 ≤ M such that M = M2 ⊕M4 and (K+M4 )/K ¿ M/K. Now M/M2 and M/M4 are H-supplemented. Therefore, there exist submodules X1 /M2 of M/M2 and X2 /M4 of M/M4 and direct summands D1 /M2 of M/M2 and D2 /M4 of M/M4 such that X1 /(Y + M2 ) ¿ M/(Y + M2 ), X1 /D1 ¿ M/D1 , X2 /(K + M4 ) ¿ M/(K + M4 ) and X2 /D2 ¿ M/D2 . Clearly, D1 ∩ D2 is a direct summand of M and (X1 ∩ X2 )/(D1 ∩ D2 ) ¿ M/(D1 ∩ D2 ). It is easy to check that (X1 ∩ X2 )/Y ¿ M/Y . Thus, M is H-supplemented. X Y +M3

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(2) Let Y /M2 ≤ M/M2 . Since M is H-supplemented, by Theorem 2.1, there exist a submodule X of M and a direct summand D of M such that X/Y ¿ M/Y and X/D ¿ M/D. Now M = Y + M1 = X + M1 = D + M1 implies that there exists a submodule D0 of M such that D0 ⊆ D and M = D0 ⊕ M1 . We know M/M2 2 that Y /M2 ⊆ X/M2 and X/M Y /M2 ¿ Y /M2 since X/Y ¿ M/Y . Now consider the isomorphism α : M/M2 → M/D0 given by m1 + M2 7→ m1 + D0 . Since D/D0 is a direct summand of M/D0 , (D/D0 )α−1 = D2 /M2 is a direct summand of M/M2 . Clearly, (D/D0 )α−1 = D2 /M2 ⊆ (X/D0 )α−1 = X/M2 . It is not hard to see that X/D2 ¿ M/D2 . Therefore, M/M2 is H-supplemented. Hence, M1 is H-supplemented. ¤ n Lemma 4.3. Let A and Ln {Mi }i=1 be modules. If each Mi is radical A-projective for i = 1, 2, . . . n, then i=1 Mi is radical A-projective.

Proof. The proof is straightforward. ¤ Ln Corollary 4.4. Let M = i=1 Mi be a finite direct sum of modules. (1) If Mi is radical Mj -projective for all j > i and each Mi is H-supplemented, then M is H-supplemented. (2) If all the Mi are relatively projective and M is H-supplemented, then each Mi is H-supplemented. Proof. The proof is a consequence of Theorem 4.2 and Lemma 4.3.

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Corollary 4.5. Let M = M1 ⊕ M2 . (1) If M1 is projective and H-supplemented, then M is H-supplemented if and only if M2 is H-supplemented. (2) If M1 is H-supplemented and M2 is semisimple, then M is H-supplemented. Example 4.6. Let M1 be an H-supplemented module with a finite composition series 0 = X0 ≤ X1 ≤ · · · ≤ Xm = M1 . Let M2 = Xm /Xm−1 ⊕ · · · ⊕ X1 /X0 . Then M = M1 ⊕ M2 is H-supplemented by Corollary 4.5(2). Proposition 4.7. Let M be an H-supplemented module and N ≤ M . If for each idempotent e : M → M , there exists an idempotent f : M/N → M/N such that (N +(M )e)/N ¿ M/N T /N T /N where Im(f ) = T /N , then M/N is H-supplemented. Proof. Let Y /N ≤ M/N . Since M is H-supplemented, there exist an idempotent e : M → M and a submodule X of M such that X/(M )e ¿ M/(M )e and X/Y ¿ M/Y . By hypothesis, there exists an idempotent f : M/N → M/N with Im(f ) = T /N such that (N + (M )e)/T ¿ M/T . Now T /N is a direct summand of M/N M/N X/N M/N and T /N ⊆ X/N . Clearly, X/N ¤ T /N ¿ T /N and Y /N ¿ Y /N . Theorem 4.8. Let K be a projection invariant submodule of M . (1) If M is H-supplemented, then there exists a decomposition M = M1 ⊕ M2 such that M2 ⊆ K and K/M2 ¿ M/M2 . (2) If M is H-supplemented, K has a unique coclosure and every direct summand of K has a coclosure in M , then there exists a decomposition M = M1 ⊕ M2 such that M2 ⊆ K, K/M2 ¿ M/M2 and M1 , M2 are H-supplemented.


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(3) Let ρ be the preradical for a cohereditary torsion theory. Assume that ρ(M ) has a unique coclosure and every direct summand of ρ(M ) has a coclosure in M . Then M is H-supplemented if and only if there exists a decomposition M = M1 ⊕ M2 such that M2 ⊆ ρ(M ), ρ(M )/M2 ¿ M/M2 and M1 , M2 are H-supplemented. Proof. (1) Since M is H-supplemented, there exists a direct summand D of M such that (D + K)/K ¿ M/K and (D + K)/D ¿ M/D. Consider the projection map e : M → M with (M )e = D. Let (M )e = D = M2 and (M )(1 − e) = M1 . Now (K)e ⊆ K since K is projection invariant, and hence K = (K)e ⊕ (K)(1 − e), (K)e = K ∩ (M )e and (K)(1 − e) = K ∩ (M )(1 − e). M = (M )e ⊕ (M )(1 − e) implies that M = K + (M )(1 − e). Therefore, (M )e = (K)e = D ⊆ K. Now ((M )e + K)/(M )e = K/(M )e ¿ M/(M )e. (2) By (1), M = M1 ⊕ M2 , M2 ⊆ K and K/M2 ¿ M/M2 , where (M )e = M2 , (M )(1 − e) = M1 and e : M → M is idempotent. Let M = A ⊕ B be any decomposition of M . Then K = (K ∩ A) ⊕ (K ∩ B). By hypothesis, K ∩ A has a coclosure L in M and K ∩ B has a coclosure L0 in M . It is easy to see that L ⊕ L0 is a coclosure of K in M . Then L ⊕ L0 = M2 since M2 is the unique coclosure of K in M . Now (L ⊕ L0 ) ∩ A = L ⊕ (L0 ∩ A) and (L ⊕ L0 ) ∩ B = L0 ⊕ (L ∩ B). Hence, (M2 ∩ A) ⊕ (M2 ∩ B) = L ⊕ L0 = M2 . Thus, by Proposition 2.11, M/M2 and M2 are H-supplemented. (3) (⇒) It follows from (2). (⇐) By Proposition 3.2, M2 is radical M1 -projective. Then by Theorem 4.2, M is H-supplemented. ¤ P Let X = i∈I Xi be a direct sum of submodules Xi (i ∈ I) of a module M . P Then X is called a local summand of M if i∈F Xi is a direct summand of M for each finite subset F of I. A module M is said to have the summand intersection property (or SIP ) if the intersection of two direct summands of M is again a direct summand of M . Recall that M is called a (D3 )-module whenever M = A + B, where A and B are direct summands of M , then A ∩ B is a direct summand of M . Lemma 4.9. Let M be a ⊕-supplemented module with SIP . Then every local summand of M is a direct summand. P Proof. Let X = i∈I Xi be a local summand of M . Since M is ⊕-supplemented, there exists a direct summand K of M L such that M = K + X and K ∩ X ¿ K. For any finite subset F of I, Y = i∈F Xi is a direct summand of M , hence Y ∩ K is a direct summand of M since M has SIP . Thus, Y ∩ K = 0. Therefore, M = K ⊕ X. ¤ Theorem 4.10. Let M be a ⊕-supplemented module with SIP . Then M is a direct sum of hollow modules. Proof. By [5, Theorem 2.17] and Lemma 4.9, M is a direct sum of indecomposable modules. By [2, Proposition 2.3], every direct summand of M is ⊕-supplemented. Therefore, M is a direct sum of indecomposable ⊕-supplemented modules, which

924 are hollow.


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Corollary 4.11. Let M be an H-supplemented module with SIP . Then M is a direct sum of hollow modules. Proposition 4.12. Let M be an H-supplemented (D3 )-module. Then any direct summand of M is H-supplemented. Proof. Let M = D ⊕ D0 . We want to show that M/D is H-supplemented. Let Y /D ≤ M/D. By Theorem 2.1, there exist a submodule X of M and an idempotent homomorphism e : M → M with Y ⊆ X and (M )e ⊆ X such that X/Y ¿ M/Y and X/(M )e ¿ M/(M )e. Clearly, M = (M )e + D0 . Since M is (D3 ), (M )e ∩ D0 is a direct summand of D0 and hence [((M )e ∩ D0 ) ⊕ D]/D is a direct summand of M/D and clearly it is contained in X/D. Since X/(M )e ¿ M/(M )e, ((X ∩ D0 ) ⊕ D)/[((M )e ∩ D0 ) ⊕ D] = X/[((M )e ∩ D0 ) ⊕ D] ¿ M/[((M )e ∩ D0 ) ⊕ D]. Therefore, M/D is H-supplemented. ¤ Acknowledgements. This paper was prepared during the visit of the second author to the Department of Mathematics, University of Hacettepe. He would like to express his gratitude to the colleagues in Beytepe for their kind hospitality. The authors would like to express their gratitude to the referee for valuable suggestions.

References [1] L. Ganesan, N. Vanaja, Modules for which every submodule has a unique coclosure, Comm. Algebra 30 (5) (2002) 2355–2377. [2] A. Harmanci, D. Keskin, P.F. Smith, On ⊕-supplemented modules, Acta Math. Hungar. 83 (1–2) (1999) 161–169. [3] M.C. Izurdiaga, Supplement submodules and a generalization of projective modules, J. Algebra 277 (2004) 689–702. [4] D. Keskin, On lifting modules, Comm. Algebra 28 (2000) 3427–3440. [5] S.H. Mohamed, B.J. M¨ uller, Continuous and Discrete Modules, London Math. Soc. Lecture Note Ser. 147, Cambridge University Press, Cambridge, 1990. [6] Y. Talebi, N. Vanaja, The torsion theory cogenerated by M -small modules, Comm. Algebra 30 (3) (2002) 1449–1460. [7] R. Wisbauer, Foundations of Module and Ring Theory, Gordon and Breach, Reading, 1991.