Simple-direct-injective modules

Journal of Algebra 420 (2014) 39–53

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Journal of Algebra www.elsevier.com/locate/jalgebra

Simple-direct-injective modules Victor Camillo a , Yasser Ibrahim b , Mohamed Yousif c,∗ , Yiqiang Zhou d a

Department of Mathematics, University of Iowa, Iowa City, IA 52242, USA Department of Mathematics, Faculty of Science, Cairo University, Giza, Egypt c Department of Mathematics, The Ohio State University, Lima, OH 45804, USA d Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL, A1C 5S7, Canada b

a r t i c l e

i n f o

Article history: Received 22 April 2014 Available online xxxx Communicated by Louis Rowen MSC: 16D50 16E50 Keywords: Artinian serial ring C2-module C3-module Injective module Regular ring Simple-direct-injective module V -ring

a b s t r a c t A module M over a ring is called simple-direct-injective if, whenever A and B are simple submodules of M with A ∼ =B and B ⊆⊕ M , we have A ⊆⊕ M . Various basic properties of these modules are proved, and some well-studied rings are characterized using simple-direct-injective modules. For instance, it is proved that a ring R is artinian serial with Jacobson radical square zero if and only if every simple-directinjective right R-module is a C3-module, and that a regular ring R is a right V -ring (i.e., every simple right R-module is injective) if and only if every cyclic right R-module is simpledirect-injective. The latter is a new answer to Fisher’s question of when regular rings are V -rings [8]. © 2014 Elsevier Inc. All rights reserved.

1. Introduction Recall that a module is a C2-module if every submodule isomorphic to a summand is a summand, and a module is a C3-module if the sum of any two summands with * Corresponding author. E-mail addresses: [email protected] (V. Camillo), yfi[email protected] (Y. Ibrahim), [email protected] (M. Yousif), [email protected] (Y. Zhou). http://dx.doi.org/10.1016/j.jalgebra.2014.07.033 0021-8693/© 2014 Elsevier Inc. All rights reserved.

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zero intersection is again a summand. These modules have been extensively studied in module theory and ring theory, usually with the accompanying CS condition (a module is CS if every submodule is essential in a summand); for instance, see [5,6,10,11]. In this paper, we study the “simple” versions of C2 and C3-modules and use them to analyze the structures of rings and modules. A module is called simple-direct-injective if every simple submodule isomorphic to a summand is itself a summand, or equivalently if the sum of any two simple summands with zero intersection is again a summand (see Proposition 2.1). Various basic properties of these modules are presented in Section 2. In Section 3, we address the natural question of when every simple-direct-injective module is C3. It is proved that every simple-directinjective right R-module is C3 if and only if every simple-direct-injective right R-module is quasi-injective if and only if R is an artinian serial ring with Jacobson radical square zero (see Theorem 3.4). A ring is called a right V -ring if every simple right R-module is injective. Some natural connections between V -rings and simple-direct-injective modules are established in Section 4. For instance, it is shown that a ring is a right V -ring if and only if every right R-module is simple-direct-injective if and only if every 2-generated right R-module is simple-direct-injective if and only if every direct sum of simple-directinjective modules is simple-direct-injective (see Proposition 4.1). In [8], Fisher asked the question: When are regular rings right V -rings? By [8, Corollary 15], if R has all primitive factor rings right V -rings, then R regular implies that R is a right V -ring. However, it may not be really easier to verify that all primitive factor rings of R are right V -rings than R is a right V -ring (e.g., let R be primitive). Here we provide a new answer to Fisher’s question by showing that a regular ring R is a right V -ring if and only if every cyclic right R-module is simple-direct-injective (see Theorem 4.4). This means that, to show all simple R-modules are injective, it suffices to show that, the sum of two simple summands of any cyclic R-module with zero intersection is again a summand. Throughout, rings are associative with unity and modules are unitary. For a module M , we denote by rad(M ), soc(M ), Z(M ) and E(M ) the Jacobson radical, the socle, the singular submodule, and the injective hull of M , respectively. We write N ⊆ M if N is a submodule of M , N ⊆ess M if N is an essential submodule of M , and N ⊆⊕ M if N is a direct summand of M . We also write M (I) for a direct sum of I-copies of M . For a ring R, we denote by J(R) the Jacobson radical of R, Mn (R) the n × n matrix ring over R, and Mod-R the category of right R-modules. The endomorphism ring of a module M over a ring R is denoted by EndR (M ). For two modules X, Y over a ring R, the set of R-homomorphisms from X to Y is denoted by HomR (X, Y ) or Hom(X, Y ). By a regular ring we mean a von Neumann regular ring. We refer to [3,5,9,12] for all the undefined notions in this paper. 2. Simple-direct-injective modules Recall that a module M is called direct-injective (or C2) if every submodule isomorphic to a summand of M is itself a summand, and a module M is called a C3-module if the

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sum of any two summands of M with zero intersection is again a summand. This paper concerns the “simple” versions of C2 and C3-modules. Proposition 2.1. The following are equivalent for a right R-module M : (1) For any simple submodules A, B of M with A ∼ = B ⊆⊕ M , A ⊆⊕ M . (2) For any simple summands A, B of M with A ∩ B = 0, A ⊕ B ⊆⊕ M . (3) If M = A1 ⊕ A2 with A1 simple and f : A1 → A2 an R-homomorphism, then Imf ⊆⊕ A2 . Proof. (1) ⇒ (2). Let A and B be simple summands of M with A ∩ B = 0. Write M = A ⊕T for a submodule T ⊆ M , and let π : A ⊕T −→ T be the canonical projection. Clearly A ⊕ B = A ⊕ π(B). Since π(B) ∼ = B ⊆⊕ M , it follows from the hypothesis that ⊕ ⊕ π(B) ⊆ M , and so π(B) ⊆ T . Thus M = A ⊕ T = A ⊕ π(B) ⊕ K = A ⊕ B ⊕ K, for a submodule K ⊆ M . Therefore A ⊕ B ⊆⊕ M . (2) ⇒ (3). Without loss of generality we may assume that f = 0. This means that f is an R-monomorphism. Let T =: {a + f (a) : a ∈ A1 } be the graph submodule of M . We claim that M = T ⊕ A2 . For, if x ∈ M , then x = a + b, where a ∈ A1 and b ∈ A2 . Now, x = a + f (a) − f (a) + b ∈ T + A2 , and so M = T + A2 . If x ∈ T ∩ A2 , then x = a + f (a), for some a ∈ A1 , and hence a = x − f (a) ∈ A1 ∩ A2 = 0. This shows that x = 0, M = T ⊕ A2 , and T ⊆⊕ M . Next, we show that A1 ∩ T = 0. For, if x ∈ A1 ∩ T , then x = a + f (a), for some a ∈ A1 , and consequently, x − a = f (a) ∈ A1 ∩ A2 = 0. Now, since f is a monomorphism, a = 0, and hence x = 0. Since T M/A2 A1 is simple, A1 ⊕T ⊆⊕ M . Finally, we show that A1 ⊕T = A1 ⊕Im(f ). For, if x ∈ Imf , then x = f (a) for some a ∈ A1 , and so x = −a + a + f (a) ∈ A1 + T , and hence A1 ⊕ T = A1 ⊕ Imf . Since A1 ⊕ T ⊆⊕ M , we infer that Imf ⊆⊕ M , and so Imf ⊆⊕ A2 , as required. σ ∼ A ⊆⊕ M . We need (3) ⇒ (1). Let A and B be simple submodules of M with B = to show that B ⊆⊕ M . If A ∩ B = 0, there is nothing to prove. Otherwise, assume that A ∩ B = 0, and write M = A ⊕ T for a submodule T ⊆ M . If π : A ⊕ T −→ T is the natural projection map, then clearly A ⊕ B = A ⊕ π(B) and B π(B) is simple. Now, since A is simple, M = A ⊕ T , and π|B ◦ σ −1 : A → T is a monomorphism with Im(π|B ◦ σ −1 ) = π(B), it follows from the hypothesis that π(B) ⊆⊕ T . If T = π(B) ⊕ K for a submodule K of T , then M = A ⊕ T = A ⊕ π(B) ⊕ K = A ⊕ B ⊕ K and B ⊆⊕ M as required. 2

Definition 2.2. A module M is called simple-direct-injective if M satisfies the equivalent conditions of Proposition 2.1. A ring R is called right simple-direct-injective if the module RR is simple-direct-injective. Example 2.3. (1) Every indecomposable module is simple-direct-injective, in particular ZZ is a simpledirect-injective module that is not direct-injective.

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(2) If R is a commutative ring, then every cyclic R-module is a C3-module, so a simpledirect-injective module. (3) Let R = Z8 and let NR = K ⊕ L, where K ∼ = R and L ∼ = 2R. As 2K is not a summand of KR , 2K is not a summand of N . But 2K ∼ = L and L ⊆⊕ N . So NR is not a C2-module. Hence by Lemma 3.2 below, M := N ⊕ N is not a C3-module. We now show that M is simple-direct-injective. It suffices to show that M has no simple direct summands. Assume that XR is a simple direct summand of MR . Write M = Y1 ⊕ Y2 ⊕ Y3 ⊕ Y4 , where Y1 ∼ = Y2 ∼ = K and Y3 ∼ = Y4 ∼ = L. Because the endomorphism rings of K and L are local, the decomposition M = Y1 ⊕ Y2 ⊕ Y3 ⊕ Y4 complements direct summands by the Krull–Schmit Theorem. Hence there exists  some k ∈ {1, 2, 3, 4} such that M = X ⊕ ( i=k Yi ). This implies that Yk ∼ = X is simple, a contradiction. Lemma 2.4. Let M be a simple-direct-injective module. Then: k (1) For any finite set {X1 , . . . , Xk } of simple summands of M , i=1 Xi ⊆⊕ M . (2) The sum of all simple summands of M is fully invariant in M . Proof. (1). We have M = X1 ⊕ N1 where N1 ⊆ M . Let p1 : M → N1 be the projection onto N1 along X1 . Then either p1 (X2 ) = 0 or p1 (X2 ) ∼ = X2 . Thus p1 (X2 ) is a summand of M and hence a summand of N1 . So M = X1 ⊕ p1 (X2 ) ⊕ N2 where N2 ⊆ N1 . As X1 + X2 = X1 ⊕ p1 (X2 ), M = (X1 + X2 ) ⊕ N2 . Let p2 : M → N2 be the projection onto N2 along X1 + X2 . Then either p2 (X3 ) = 0 or p2 (X3 ) ∼ = X3 . Thus, p2 (X3 ) is a summand of M and hence a summand of N2 . So M = (X1 + X2 ) ⊕ p2 (X3 ) ⊕ N3 where N3 ⊆ N2 . Since X1 + X2 + X3 = (X1 + X2 ) ⊕ p2 (X3 ), we have M = (X1 + X2 + X3 ) ⊕ N3 . A simple induction shows that M = (X1 + · · · + Xk ) ⊕ Nk for some Nk ⊆ M . (2). It is obvious. 2 Proposition 2.5. Let M be a finitely generated module. If M is simple-direct-injective, then for any semisimple submodules A, B with A ∼ = B ⊆⊕ M , we have A ⊆⊕ M . Proof. For any semisimple submodules A, B with A ∼ = B ⊆⊕ M , A and B are both k k finitely generated. So, A = i=1 Ai and B = i=1 Bi where, for each i, Ai and Bi are simple and Ai ∼ = Bi ⊆⊕ M . Hence each Ai is a summand of M . So, by Lemma 2.4, A is a summand of M . 2 Proposition 2.6. Let M be a module such that the sum of all simple summands is essential in a summand of M . Then M is simple-direct-injective if and only if M = M1 ⊕ M2 , where soc(M1 ) ∩rad(M1 ) = 0, soc(M1 ) is fully invariant in M , and soc(M2 ) ⊆ rad(M2 ). Proof. (⇒). Let F be the set of all simple summands of M . By hypothesis, there is  a decomposition M = M1 ⊕ M2 , where X∈F X is essential in M1 and M2 ⊆ M .

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 Thus, X∈F X = soc(M1 ) is fully invariant in M by Lemma 2.4. Every simple module  contained in X∈F X is isomorphic to a simple summand of M , so it is a summand  of M and hence a summand of M1 . Thus every simple module contained in X∈F X is not contained in rad(M1 ). Hence soc(M1 ) ∩ rad(M1 ) = 0. Let M2 = X + Y where X is simple and Y ⊆ M2 . Since X is not a summand of M2 , X ∩ Y = 0, so X ⊆ Y and hence M2 = Y . This shows that X is a small submodule of M2 . Hence soc(M2 ) ⊆ rad(M2 ). (⇐). Let A, B be simple submodules of M with A ∼ = B ⊆⊕ M . Then B is not contained in M2 because each simple module of M2 is small in M2 . This shows that p(B) = 0, where p : M → M1 is the projection to M1 along M2 . Since soc(M1 ) ∩ rad(M1 ) = 0, p(B) is not small in M1 , so p(B) is a summand of M1 . Since A ∼ =B ∼ = p(B) ⊆⊕ M1 , there exists an endomorphism f of M such that f (p(B)) = A. Since soc(M1 ) is fully invariant in M , A ⊆ soc(M1 ). As above, A is not small in M1 , so A is a summand of M1 and hence a summand of M . 2  A locally simple summand of a module M is a direct sum L := t∈Λ Xt of simple  submodules of M such that t∈F Xt is a summand of M for any finite subset F of Λ. Proposition 2.7. Suppose every locally simple summand of M is a summand. Then M is simple-direct-injective if and only if M = M1 ⊕ M2 , where M1 is a fully invariant semisimple submodule of M and soc(M2 ) ⊆ rad(M2 ).  Proof. (⇒). Let F be the set of all simple summands of M . Then M1 := X∈F X =   {X : X ∈ H} X∈H X (where H ⊆ F ), which is fully invariant in M . By Lemma 2.4, is a locally simple summand of M , so it is a summand by hypothesis. Write M = M1 ⊕M2 where M2 ⊆ M . As argued in the proof of Proposition 2.6, soc(M2 ) ⊆ rad(M2 ). (⇐). Argue as in the proof of “(⇐)” of Proposition 2.6. 2 If a module M satisfies ACC on summands, then every locally simple summand of M is indeed a direct sum of finitely many simple summands, so it is a summand of M . Thus, the next corollary follows from Proposition 2.7. Corollary 2.8. Let M be a module with ACC on summands. Then M is simple-directinjective if and only if M = M1 ⊕M2 , where M1 is a fully invariant semisimple submodule of M and soc(M2 ) ⊆ rad(M2 ). Corollary 2.9. Let M be a module with ACC on summands. If M is simple-directinjective, then for any semisimple submodules A, B with A ∼ = B ⊆⊕ M , we have ⊕ A ⊆ M. Proof. By Corollary 2.8, M = M1 ⊕ M2 , where M1 is a fully invariant semisimple submodule of M and soc(M2 ) ⊆ rad(M2 ). Assume A ∼ = B ⊆⊕ M , where A, B are semisimple submodules of M . Since soc(M2 ) ⊆ rad(M2 ), B ∩ M2 = 0. So there exists

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N ⊆ M1 such that A ∼ =B∼ = N . As N ⊆⊕ M1 and M1 is fully invariant in M , A ⊆ M1 . So A is a summand of M1 and hence a summand of M . 2 Note that a ring R is an I-finite ring if and only if RR has ACC on direct summands and that a right ideal I is fully invariant in RR if and only if I is a two-sided ideal. Thus Corollary 2.8 has the following interesting consequence. Corollary 2.10. Let R be an I-finite ring. Then R is right simple-direct injective if and only if R ∼ = R1 × R2 , where R1 is semisimple artinian and soc(R2 )2 = 0. Next we consider whether being a right simple-direct-injective ring is a Morita invariant property. Definition 2.11. Let M and N be right R-modules. We call M a min-N -injective module if for any simple submodule K of N , every R-homomorphism f : K → M extends to N . M is called min-quasi-injective if it is min-M -injective. M is called mininjective if it is min-R-injective. The ring R is called right mininjective if RR is mininjective. Every min-quasi-injective module is clearly simple-direct-injective (see the proof of [10, Prop. 2.1]). But if R = Z2 ∝ (Z2 ⊕ Z2 ), the trivial extension of Z2 by its bimodule Z2 ⊕ Z2 , then RR is simple-direct-injective but not min-quasi-injective. However, if soc(RR ) is projecive, then R is right mininjective if and only if R is simple-directinjective. The next lemma can be easily verified.  Lemma 2.12. Let N = α∈I Nα be a direct sum of right R-modules. Then a right R-module M is min-N -injective if and only if M is min-Nα -injective for all α ∈ I. Lemma 2.13. If M is a simple-direct-injective right R-module, then every simple summand of M is min-M -injective. Proof. Let A be a simple summand of M and f : K −→ A an R-homomorphism with K a simple submodule of M . We show that f extends to M . If f = 0, we are done. If f = 0, then A ∼ = K. Since M is simple-direct-injective, K ⊆⊕ M and f extends to M . Thus A is min-M -injective. 2 Theorem 2.14. If R is a right simple-direct-injective ring, then R(I) is a simple-directinjective right R-module for any index set I. Proof. Let A and B be simple submodules of R(I) with A ∼ = B ⊆⊕ R(I) . We need to ⊕ (I) show that A ⊆ R . Let f : R −→ A be an R-epimorphism. Since R(I) is projective, A is projective and so f splits. Therefore, A ∼ = D ⊆⊕ R. Since R is right simple-directinjective, D is min-R-injective by Lemma 2.13. Consequently A is min-R-injective, and

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so A is min-R(I) -injective by Lemma 2.12. Therefore the identity i : A −→ A extends to R(I) and A ⊆⊕ R(I) , as required. 2 Corollary 2.15. A ring R is right simple-direct-injective if and only if every projective right R-module is simple-direct-injective. Theorem 2.16. Being a right simple-direct-injective ring is a Morita invariant property. Proof. Suppose that R is a right simple-direct-injective ring. It suffices to show that, for any n ≥ 1 and any e2 = e ∈ R with ReR = R, Mn (R) and eRe are right simple-directinjective rings. If P = (Rn )R and S = EndR (P ), then HomR (P, −) : NR → HomR (S PR , NR ) defines a Morita equivalence between Mod-R and Mod-S with inverse equivalence − ⊗S P : MS → M ⊗ P . Since a Morita equivalence preserves the simple-direct-injective module condition, SS ∼ = Hom(P, P )S is simple-direct-injective if and only if PR is simpledirect-injective. We see that the latter holds by Theorem 2.14, so S and hence Mn (R) is right simple-direct-injective. If P = (eR)R and S = eRe, then HomR (P, −) : NR → HomR (S PR , NR ) defines a Morita equivalence between Mod-R and Mod-S with inverse equivalence − ⊗S P : MS → M ⊗P . Since a Morita equivalence preserves the simple-direct-injective module condition, R being right simple-direct-injective implies that the right S-module HomR (S PR , R) is simple-direct-injective. But, HomR (S PR , R) ∼ = (Re)S = [(1 − e)Re ⊕ eRe]S . So, as a direct summand of a simple-direct-injective module, SS is simple-direct-injective. 2 3. When are simple-direct-injective modules C3? We characterize the rings whose simple-direct-injective modules are C3-modules. Lemma 3.1. Any direct sum of injective modules is simple-direct-injective.  Proof. Let M = i∈I Ei where each Ei is injective and let A ∼ = B ⊆⊕ M where A and  B are simple submodules of M . Since B is simple, B ⊆⊕ ( i∈F Ei ) for a finite subset F ⊆ I. Thus B is injective, and consequently A is injective and A ⊆⊕ M . 2 Lemma 3.2. (See [1, Corollaries 2.4 and 2.6].) (1) If M = A1 ⊕ A2 is a C3-module and f : A1 → A2 is an R-monomorphism, then Imf ⊆⊕ A2 . (2) If M ⊕ M is a C3-module, then M is a C2-module. Lemma 3.3. If M is an indecomposable module that is not simple, then M ⊕ E(M ) is simple-direct-injective.

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Proof. Let N = M ⊕ E where E = E(M ). To show that N is simple-direct-injective, it suffices to show that N has no simple summands. Assume X is a simple summand of N . As EndR (X) is local, by [3, Lemma 26.4], N = X ⊕ M  ⊕ E  , where M  is a summand of M and E  is a summand of E. Write E = E  ⊕ E  . If M  = 0, then M  = M , so N = X ⊕ M ⊕ E  . Thus, E  ∼ = N/(M ⊕ E  ) ∼ = X is simple. As M is   essential in E, E ⊆ M . As E is injective, it is a summand of M . So M = E  is simple. This contradiction shows that M  = 0. Thus N = X ⊕ E  , and it follows that M ⊕ E  ∼ = N/E  ∼ = X is simple. This shows that M is simple, a contradiction. 2 A module is uniserial if the lattice of its submodules is totally ordered under inclusion. A ring R is called left uniserial if R R is a uniserial module. A ring R is called serial if both modules R R and RR are direct sums of uniserial modules. Theorem 3.4. The following are equivalent for a ring R: (1) Every simple-direct-injective right R-module is a C3-module. (2) Every simple-direct-injective right R-module is quasi-injective. (3) Every right R-module is a direct sum of a semisimple module and a family of injective uniserial modules of length 2. (4) Every right R-module is a direct sum of a semisimple module and an injective module. (5) R is an artinian serial ring with J(R)2 = 0. Proof. (2) ⇒ (1). It is clear. (1) ⇒ (3). We claim that R is right artinian. First we show that R is right semiartinian. Assume on the contrary that M is a right R-module with soc(M ) = 0. If 0 = N ⊆ M , then soc(N ⊕ M ) = 0 and N ⊕ M is simple-direct-injective. Thus, by the hypothesis, N ⊕ M is a C3-module, and the inclusion map i : N → M splits by Lemma 3.2. This shows that M is semisimple, a contradiction. Thus soc(M ) = 0 for every right R-module M . So R is right semiartinian. Next we show that R is right noetherian. It suffices to show that for any family {Ki : i ∈ I} of simple right R-modules, M :=  i∈I E(Ki ) is injective. By Lemma 3.1, M ⊕ E(M ) is a simple-direct-injective module. Now, by the hypothesis M ⊕E(M ) is a C3-module, and by Lemma 3.2, the inclusion map i : M → E(M ) splits; i.e. M = E(M ) is injective as required. So R is right noetherian, and hence R is right artinian. We next show that every indecomposable injective right R-module E has a unique composition series of length at most 2. Note that E has a simple socle X and E = E(X). If E = X, we are done. Suppose X ⊂ Y ⊆ E. It suffices to show that Y = E. Let M = Y ⊕ E. Then, by Lemma 3.3, M is a simple-direct-injective module and hence a C3-module by the hypothesis. Hence, Y = E by Lemma 3.2. We now show that every finitely generated indecomposable right R-module has a unique composition series of length at most 2. To see this, let M be a finitely generated indecomposable right R-module. Since R is right artinian, MR is artinian, and so it is

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finite dimensional. If M is simple, we are done. If M is not simple, then M ⊕ E(M ) is a simple-direct-injective module by Lemma 3.3, so it is a C3-module by the hypothesis. Hence M = E(M ) is injective by Lemma 3.2. Thus, M is an indecomposable injective module, and, as above, it has a unique composition series of length at most 2. Finally, consider an arbitrary right R-module M . Since R is right noetherian, M contains a maximal injective submodule N . Write M = N ⊕K, where K contains no nonzero injective submodules. The injective module N is a direct sum of indecomposable injective modules each of which has a unique composition series of length at most 2. Thus there is a decomposition N = E1 ⊕ E2 , where E1 is semisimple and E2 is a direct sum of injective uniserial modules of length 2. So, to finish the proof, it suffices to show that K is semisimple. Without loss of generality, we can assume that K is a cyclic module. Since R is right artinian, K is artinian, so it is a direct sum of indecomposable modules. Therefore, we can further assume that K is cyclic indecomposable. As above, K is a uniserial module of length at most 2. If K is of length 2, then K = E(K) because E(K) is a uniserial module of length at most 2. This contradicts the fact that K contains no nonzero injective submodules. Hence K is simple. (3) ⇒ (2). Let M be a simple-direct-injective module. Write M = M1 ⊕ M2 , where  M1 = α∈Λ Nα is a direct sum of injective uniserial modules of length 2 and M2 =  β∈Γ Kβ is a direct sum of simple modules. Note (3) implies that R is right artinian, so every direct sum of injective right R-modules is injective. Hence M1 is injective, and M2 is quasi-injective. To show that M is quasi-injective, by [10, Proposition 1.17] it suffices to show that M2 is M1 -injective. By [10, Proposition 1.5], M2 is M1 -injective if and only if M2 is Nα -injective for every α ∈ Λ. Since R is right artinian, M2 is Nα -injective if and only if Kβ is Nα -injective for every β ∈ Γ . Hence, it suffices to show that Kβ is Nα -injective for each α ∈ Λ and each β ∈ Γ . Assume f : X → Kβ is a nonzero R-homomorphism where X ⊆ Nα . We show that f extends to Nα by verifying X = Nα . If X ⊂ Nα , then X = soc(Nα ) is simple because Nα is uniserial of length 2. So X∼ = Kβ ⊆⊕ M . Since M is simple-direct-injective, it follows that X ⊆⊕ M . This shows that X is a summand of Nα , a contradiction. Hence Kβ is Nα -injective. (3) ⇒ (4). Since (3) implies that R is right noetherian, every direct sum of injective R-modules is injective. So (3) clearly implies (4). (4) ⇔ (5). This is by [6, 13.5, p. 124]. (4) +(5) ⇒ (3). Since R is right artinian, every injective right R-module is a direct sum of uniform injective modules. So, to show (3), it suffices to show that every uniform right R-module M is uniserial of length at most 2. Since R is right artinian, soc(M ) is simple. If M is simple, we are done. Assume that M is not simple, and let soc(M ) ⊂ X ⊆ M . By (4), X is simple or injective. But X is not simple. So X is injective, and is a summand of M . This shows that M = X. Hence M is uniserial of length 2. 2 Note that, by [6, 13.5, p. 124], a ring R is artinian serial with J(R)2 = 0 if and only if every right R-module is CS.

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Corollary 3.5. A ring R is semisimple artinian if and only if every simple-direct injective right R-module is injective. Proof. If every simple-direct injective right R-module is injective, then every simple right R-module is injective and so R is a right V -ring. But R is artinian by Theorem 3.4, so R is semisimple artinian. 2 There exists a local, left and right artinian left uniserial ring R with J(R)2 = 0, and with a simple-direct-injective left R-module that is not a C3-module. Example 3.6. (See [4, p. 70].) Given a field F and an isomorphism F → F¯ ⊆ F , a → a ¯, let R be the left F -space on basis {1, t} with multiplication given by t2 = 0 and ta = a ¯t for all a ∈ F . Then R is a local ring, R/J(R) ∼ = F , J(R)2 = 0, and the only left ideals are 0, J(R) = Rt = F t, and R. Hence R is left artinian left uniserial. Moreover, if 1 < dimF¯ (F ) < ∞, then R is also right artinian but neither left nor right self-injective; in this case, by Lemma 3.3, R⊕E(R) is a simple-direct-injective left R-module that is not a C3-module. More specifically, if p is a prime and F = Zp (x) is the field of rational forms over Zp , then the map w → wp is an isomorphism F → F¯ , where F¯ = {wp : w ∈ F } and {1, x, ..., xp−1 } is a basis of F over F¯ . Example  3.7. Let Kbe a field and R be the K-algebra consisting of all 3 × 3 matrices of α1 α2 α3 the form 0 α4 0 , where αi ∈ K. Clearly R is a left and right artinian left serial ring, 0

0 α5

and we claim that it is not right serial. For, if M = e11 R then M = e11 K + e12 K + e13 K is an indecomposable right R-module, and hence simple-direct-injective. Clearly e11 R is neither simple nor injective as a right R-module (since e12 K = e12 R and e13 K = e13 R are proper submodules of e11 R with e12 R ∩ e13 R = 0). Now, by Lemma 3.3, e11 R ⊕ E(e11 R) is a simple-direct-injective right R-module, and by Lemma 3.2, e11 R ⊕ E(e11 R) is not a right C3-module. Hence, by Theorem 3.4, R is not a right serial ring. 4. Simple-direct-injective modules and V -rings In this section, some interesting connections between right V -rings and simple-directinjective modules are presented. Proposition 4.1. The following conditions are equivalent for a ring R: (1) (2) (3) (4) (5)

R is a right V -ring. Every right R-module is simple-direct-injective. Every finitely cogenerated right R-module is simple-direct-injective. Direct sums of simple-direct-injective modules are simple-direct-injective. Every 2-generated right R-module is simple-direct-injective.

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Proof. (1) ⇒ (2) ⇒ (3), (1) ⇒ (4) and (1) ⇒ (5). They are clear. (3) ⇒ (1) and (4) ⇒ (1). If S is a simple right R-module, then by the hypothesis S ⊕ E(S) is simple-direct-injective. Now, by (3) of Proposition 2.1, it follows that S = E(S); and so S is injective and R is a right V -ring. (5) ⇒ (1). Let S = xR be a simple right R-module and 0 = y ∈ E(S). Then xR ⊆ess yR. By the hypothesis, xR ⊕ yR is simple-direct-injective. By (3) of Proposition 2.1, it follows that xR = yR and so S = E(S); and R is a right V -ring. 2 A right R-module M is called strongly soc-injective if for any right R-module N and any semisimple submodule K of N , every R-homomorphism f : K → M extends to N . Lemma 4.2. (See [2, Proposition 16].) A right R-module M is strongly soc-injective if and only if M = E ⊕ T , where E is injective and soc(T ) = 0. Proposition 4.3. The following conditions on a ring R are equivalent: (1) R is a right noetherian right V -ring. (2) Every simple-direct-injective right R-module is strongly soc-injective. Proof. (1) ⇒ (2). By [7], (1) implies that every semisimple right R-module is injective; consequently every right R-module is strongly soc-injective. (2) ⇒ (1). By Lemma 4.2, every simple right R-module is injective, so R is a right V -ring. Now, if {Ki : i ∈ I} is a countable family of simple right R-modules, then  by Proposition 3.1, M = i∈I E(Ki ) is simple-direct-injective and hence strongly socinjective. Since M has essential socle, it follows from Lemma 4.2 that M is injective. Hence R is right noetherian. 2 Theorem 4.4. A regular ring R is a right V -ring if and only if every cyclic right R-module is simple-direct-injective. Proof. (⇒). This is clear by Proposition 4.1. (⇐). Let SR be a simple right R-module, and E = E(S) the injective hull of S. Assume to the contrary, there exists a non-zero element x ∈ E such that x ∈ / S. Clearly S ⊆ess xR. Define the epimorphism f : R → xR by f (r) = xr for r ∈ R, and set X = ker(f ). Now the map f induces an isomorphism σ : xR ∼ = R/X. If T /X = σ(S), then T /X = (tR + X)/X for some nonzero t ∈ R. Since R is regular, there exists s ∈ R such that tst = t. If we set e = ts, then e2 = e and T /X = (tR + X)/X = (eR + X)/X. Inasmuch as S ⊆ess xR, we infer that T /X is a minimal essential right ideal of R/X. If M =: {r ∈ R : er ∈ X}, then R/M ∼ = T /X and M is a maximal right ideal of R. Now we claim that, for N =: M ∩ X, X/N ∼ = R/M . To see this, observe first since (eR + X)/X is a simple essential submodule of the right R-module R/X and ((1 − e)R + X)/X is a non-zero submodule of R/X, it follows that (eR + X)/X ⊆ ((1 − e)R + X)/X,

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and hence e + X = (1 − e)(−r) + X for some r ∈ R. Therefore, y := e + (1 − e)r ∈ X, and if we multiply on the left by e we get ey = e. Now, N = M ∩ X ⊆ X ⊂ T , and if y ∈ N , then y ∈ M , which implies that ey ∈ X, and so e ∈ X, a contradiction. Thus y ∈ / N , and it follows that X is not contained in M . Now X/N = X/(M ∩X) ∼ = (X +M )/M = R/M . Next, we show that (eR + N )/N ∼ = R/M . If g : R → (eR + N )/N is given by g(r) = er + N , where r ∈ R, then clearly g is a well-defined R-epimorphism. Since M is a maximal right ideal of R and M ⊆ ker g, we infer that M = ker g and (eR+N )/N ∼ = R/M as required. Next we show that ((1 − e)yR + N )/N ∼ = R/M . Observe first that if m ∈ M , then it follows, from the definition of M and the fact ey = e, that em = eym ∈ X, and hence ym ∈ M . Therefore ym ∈ M ∩ X = N , and so yM ⊆ N . Since eM ⊆ N and ey = e, it follows that eyM ⊆ N and consequently (1 − e)yM ⊆ yM + eyM ⊆ N . Now if we define h : R → ((1 − e)yR + N )/N by h(r) = (1 − e)yr + N , where r ∈ R, then clearly h is a well-defined R-epimorphism. Since (1 − e)yM ⊆ N , it follows that R/M ∼ = ((1 − e)yR + N )/N as required. Therefore, ((1 − e)yR + N )/N ∼ = (eR + N )/N ∼ = ∼ ∼ ∼ ∼ R/M = X/N = (eR + X)/X = T /X = S. As eM ⊆ N , eN ⊆ eM ⊆ N and N is invariant under left multiplication by e. Therefore R/N = (eR + N )/N ⊕ ((1 − e)R + N )/N . Since ((1 − e)yR + N )/N ∼ = (eR + N )/N ⊆⊕ R/N , it follows from the hypothesis that ((1 − e)yR + N )/N ⊆⊕ R/N , and consequently ((1 − e)yR + N )/N ⊆⊕ ((1 − e)R + N )/N . Thus, R/N = (eR + N )/N ⊕ ((1 − e)yR + N )/N ⊕ A/N , where A/N is a right R-submodule of R/N . Finally, we only need to show that R/N = (eR + N )/N ⊕ X/N . For if this is the case, then (R/N )R has uniform dimension 2. So A/N must be zero and R/N = (eR + N )/N ⊕ ((1 −e)yR+N )/N , and consequently R/X ∼ = (eR+N )/N is simple, a contradiction. First we have (eR + N )/N ∩ X/N = 0. To see this, let er + N = x + N ∈ (eR + N )/N ∩ X/N , where r ∈ R and x ∈ X. Then er −x ∈ N , and since N ⊂ X, it follows that er ∈ X. This means r ∈ M , and hence er ∈ N . Therefore (eR + N )/N ∩ X/N = 0. Since (eR + N )/N and X/N are simple submodules of R/N and since X/N ∼ = (eR + N )/N ⊆⊕ R/N , (eR + N )/N + X/N = (eR + N )/N ⊕ X/N ⊆⊕ R/N . Hence it is enough to show that (eR + N )/N ⊕ X/N ⊆ess R/N . Now, let (aR + N )/N be a non-zero submodule of R/N . If a ∈ X, then 0 + N = a + N ∈ X/N ⊂ (eR + N )/N ⊕ X/N . Otherwise, assume that a∈ / X. In this case (aR + X)/X is a non-zero submodule of R/X. Consequently, since (eR + X)/X is a simple essential submodule of R/X, it follows that (eR + X)/X ⊆ (aR + X)/X. Therefore, e + X = ar + X for some r ∈ R. Thus for some x0 ∈ X, ar = e + x0 and 0 + N = ar + N = (e + N ) + (x0 + N ) ∈ (eR + N )/N ⊕ X/N , as required. 2 We next use Theorem 4.4 to give some examples of right V -rings. Example 4.5. (See [8, Example 1].) Let V be an infinite dimensional vector space over a field K. Let E = EndK (V ), and let R be the subring of E generated by the socle of E and the center K of E (i.e., the scalar transformations 1V k for k ∈ K). Then V is a

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simple left R-module. If V ∗ = HomK (V, K) and V ∗∗ = HomK (V ∗ , K), then R V R V ∗∗ since VK is of infinite dimension. Since the left R-module R V ∗∗ is an essential extension of R V and hence indecomposable, the left R-module V ⊕ V ∗∗ is a finitely cogenerated module that is not simple-direct-injective. For, if V ⊕ V ∗∗ were simple-direct-injective, then by Proposition 2.1, the inclusion map i : V → V ∗∗ splits, a contradiction. Now, by Proposition 4.1, R is not a left V -ring. As R is von Neumann regular, K is a flat left R-module. Hence both V ∗ = HomK (V, K) and K are injective right R-modules. Since V ∗ and K are the only simple right R-modules, R is a right V -ring. The lemma below can be easily verified. Lemma 4.6. Let e2 = e ∈ R and let I be a right ideal of R. The following conditions are equivalent: (1) (R/I)R = (eR + I)/I ⊕ ((1 − e)R + I)/I. (2) eI ⊆ I. We should point out that (eR + I)/I being a direct summand of (R/I)R does not imply eI ⊆ I. But if K/I ⊆⊕ (R/I)R , then there exists g 2 = g ∈ R such that K = gR+I and gI ⊆ I. For a subset S of a ring R, let S denote the subring of R generated by S and 1R . A ring R is said to satisfy condition (∗) if for any idempotents e and f , eR + f R = gR where g 2 = g ∈ e, f . Clearly, any abelian ring (i.e. every idempotent is central) satisfies condition (∗), but a ring with (∗) need not be abelian (e.g., M2 (Z2 )). A ring is called an exchange ring if idempotents can be lifted modulo every right (equivalently left) ideal. Example 4.7. Let R be an exchange ring with condition (∗). Then every cyclic right R-module is simple-direct-injective. Proof. Let I be a right ideal of R. Suppose K/I is a simple summand of (R/I)R . Then (R/I)R = (K/I) ⊕(K  /I). Thus R = K +K  . Write 1 = x +x where x ∈ K and x ∈ K  . Thus, x − x2 = x x ∈ K ∩ K  = I. Since R is an exchange ring, idempotents can be lifted modulo I. Hence there exists e2 = e ∈ R such that x −e ∈ I, and so (1 −x) −(1 −e) ∈ I. It follows that K = eR+I and K  = (1−e)R+I. Since (R/I)R = (K/I)⊕(K  /I), we deduce that eI ⊆ I by Lemma 4.6. If L/I is another simple summand of (R/I)R , then as above L = f R + I where f 2 = f and f I ⊆ I. By the hypothesis, there exists g 2 = g ∈ e, f  such that eR + f R = gR. Thus, K/I + L/I = (eR + f R + I)/I = (gR + I)/I. Inasmuch as g ∈ e, f , eI ⊆ I and f I ⊆ I, we infer that gI ⊆ I. By Lemma 4.6, (gR + I)/I is a summand of (R/I)R . Hence, (R/I)R is simple-direct-injective. 2 Example 4.8. A regular ring with condition (∗) is a right V -ring.

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∞ Example 4.9. Let Q = i=1 Ri where Ri = Mn (D) with D a division ring for all i, and   let R =  i≥1 Ri , 1Q  be the subring of Q generated by i≥1 Ri and 1Q . Then R is a regular V-ring. Proof. We will use Theorem 4.4 to verify that R is a right V-ring. Let N be a right ideal of R and let (X/N )R ⊆⊕ (R/N )R and (Y /N )R ⊆⊕ (R/N )R with X ∩ Y = N . We will show that X/N ⊕ Y /N ⊆⊕ (R/N )R . As argued in the proof of Example 4.7, there exist idempotents e and f of R such that eR + N X = , N N

Y fR + N = , N N

eN ⊆ N, f N ⊆ N, (eR + N ) ∩ (f R + N ) = N.

Write   e = e1 , e2 , . . . , en , e , e , . . .   f = f1 , f2 , . . . , fn , f  , f  , . . . 1 = α + β, where α = (1, 1, . . . , 1, 0, 0, . . .) (here 1 appears n times) and β = (0, 0, . . . , 0, 1, 1, . . .) (here 0 appears n times). Then R = A × B where A = αR and B = βR. Case 1: e = f  = 1. Then eR = (e1 R1 × · · · × en Rn ) × B, and f R = (f1 R1 × · · · × fn Rn ) ×B. So B ⊆ X ∩Y ⊆ N , and hence N = (A ∩N ) ⊕B. Thus (R/N )R ∼ = A/(A ∩N ) is semisimple. Therefore, X/N ⊕ Y /N is a direct summand of (R/N )R . Case 2: e = f  = 0. Then eR = eA, and f R = f A. As A is a semisimple Artinian ring, (A/N A)A is semisimple, so (eA + f A + N A)/N A is a direct summand of (A/N A)A . Hence there exists g 2 = g ∈ A such that eA + f A + N A = gA + N A and g(N A) ⊆ N A. Hence eR + f R + N = gR + N and gN ⊆ N A ⊆ N . By Lemma 4.6, (gR + N )/N is a direct summand of (R/N )R . So X/N ⊕ Y /N is a direct summand of (R/N )R . Case 3: e = 0 and f  = 1 (the case where e = 1 and f  = 0 is similar). Then eR = eA, and f R = f A ⊕ B. As A is a semisimple Artinian ring, (A/N A)A is semisimple, so (eA + f A + N A)/N A is a direct summand of (A/N A)A . Hence there exists g 2 = g ∈ A such that eA + f A + N A = gA + N A and g(N A) ⊆ N A. Then h := (g, β) is an idempotent of R, and eR + f R + N = gA + B + N = hR + N . Moreover, hN ⊆ gN + βN = g(N A) + N B ⊆ N A + N B = N . By Lemma 4.6, (hR + N )/N is a direct summand of (R/N )R . So X/N ⊕ Y /N is a direct summand of (R/N )R . 2 Example 4.10. Let A be an infinite set and consider the column finite matrix ring S = CFMA (F ) over a division ring F . Let T be the set of all matrices in S with only a finite number of non-zero entries and view F as the subring of all scalar matrices of S. Clearly, R = T + F is a semiartinian regular ring with soc(R) = T . We claim that R is neither a right nor a left V -ring. Assume to the contrary that R is a right V -ring and let e be a primitive idempotent of R. If eRR is simple, then it is injective. Since eS is an injective hull of eR, we infer that eR = eS. This is a clear contradiction, because eR ⊆ T and eS

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contains matrices with infinite non-zero entries. If eR is not simple, then by Lemma 3.3, the right R-module eR ⊕ eS is simple-direct-injective. Now, by Proposition 2.1, the inclusion map i : eR → eS splits, a contradiction as before. Thus R is not a right V -ring. By considering R as a subring of the row finite matrix ring RFMA (F ) and applying the above argument on the left, we see that R is also not a left V -ring. We should point out that, every element of R belongs to a subring of R isomorphic to one of the form Mn (F ) × F , for some positive integer n. This implies that R is a unit regular ring that is neither a left nor a right V -ring. Acknowledgments Part of this work was carried out while the second author was visiting the Mathematics Department of The Ohio State University. He would like to express his gratitude to the members of the mathematics department at OSU for the financial support and the warm reception. The research of the third author was supported by the Mathematics Research Institute of the Ohio State University, and that of the fourth author by a Discovery Grant Number 194196 from NSERC of Canada. References [1] I. Amin, Y. Ibrahim, M.F. Yousif, C3-modules, Algebra Colloq. (2015), to appear. [2] I. Amin, M.F. Yousif, N. Zeyada, Soc-injective rings and modules, Comm. Algebra 33 (2005) 4229–4250. [3] F.W. Anderson, K.R. Fuller, Rings and Categories of Modules, Springer-Verlag, Berlin, New York, 1974. [4] J.-E. Björk, Rings satisfying certain chain conditions, J. Reine Angew. Math. 245 (1970) 63–73. [5] J. Clark, C. Lomp, N. Vanaja, R. Wisbauer, Lifting Modules, Birkhäuser Verlag, Basel, Boston, Berlin, 2006. [6] N.V. Dung, D.V. Huynh, P.F. Smith, R. Wisbauer, Extending Modules, Longman Scientific and Technical, 1994. [7] C. Faith, Algebra II: Ring Theory, Springer-Verlag, Berlin, New York, 1976. [8] J.W. Fisher, Von Neumann regular rings versus V-rings, in: Lect. Notes Pure Appl. Math., vol. 7, Dekker, New York, 1974, pp. 101–119. [9] F. Kasch, Modules and Rings, London Math. Soc. Monogr., vol. 17, Academic Press, New York, 1982. [10] S.H. Mohamed, B.J. Müller, Continuous and Discrete Modules, Cambridge Univ. Press, Cambridge, UK, 1990. [11] W.K. Nicholson, M.F. Yousif, Quasi-Frobenius Rings, Cambridge Tracts in Math., vol. 158, Cambridge Univ. Press, Cambridge, UK, 2003. [12] R. Wisbauer, Foundations of Module and Ring Theory, Gordon and Breach, Philadelphia, 1991.