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S − CS Modules and Rings International Journal of Algebra, Vol. 7, 2013, no. 2, 49 - 62 HIKARI Ltd, www.m-hikari.com S − CS Modules and Rings Nasr A. Zeyada King Faisal University, Faculty of Science Department of Mathematics, Saudi Arabia Cairo University, Faculty of Science Department of Mathematics, Egypt & King Abdulaziz University, Faculty of Science Department of Mathematics, Saudi Arabia [email protected] Noˆomen Jarboui King Faisal University, Faculty of Science Department of Mathematics, Saudi Arabia [email protected] Copyright c 2013 Nasr A. Zeyada and Noˆomen Jarboui. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. We investigate the concept of s − CS modules. Basic properties of these modules are given. We prove that any right s − CS, right CF is artinian and we answer the FGF conjecture positively in case of R is s − CS. Some properties of s−injective rings are given. Keywords: S − CS, S−injective, FGF conjecture 1. Introduction A module M is said to satisfy C1−condition or called CS−module if ev- ery submodule of M is essential in a direct summand of M. N. Zeyada [9] introduced s − CS modules. A right R−module M is called right s − CS if every singular submodule of M is essential in a summand of M. Any non- singular module is an s − CS module. Note further that any Goldie torsion 50 N. A. Zeyada and N. Jarboui s − CS module is a CS−module. A module M is called a min − CS mod- ule if every simple submodule of M is essential in a direct summand of M. N. Zeyada [9] introduced s−injective and strongly s−injective modules, for any two modules M and N, M is s − N−injective if any R−homomorphism f : K −→ M, where K is a singular submodule of N, extends to N. R is called right (self-) s−injective, if the right R−module RR is s−injective. M is strongly s−injective if M is s − N−injective for any module N. In this paper we investigate the notion of s − CS Modules and Rings. A ring R is called a right CF ring if every cyclic right R−module can be embedded in a free mod- ule. It is not known that every right CF ring is right artinian. G´omez Pardo and Guil Asensio proved that any right CS, right CF ring is right artinian. We extend their result to the case of s − CS. Also we give some properties of s−injective rings. Throughout this paper all rings are associative with identity, and all mod- ules are unitary R−modules. For a right R−module MR, we denoted by J(M), soc(M), Z(M) and Z2 (M) the Jacobson radical, the socle, the singular sub- module and the second singular submodule of M, respectively. Sr, Sl, Zr, Zl, r Z2 and J are used to indicate the right socle, the left socle, the right singular ideal, the left singular ideal, the right second singular ideal, and the Jacobson radical of R, respectively. For a submodule N of M, the notations N ⊆ess M, N ⊆max M and N ⊆⊕ M mean, respectively, that N is essential, maximal, and direct summand. Mod−R indicates the category of right R−modules. We refer to [1], [2] and [6] for all the undefined notions in this paper. 2. S − CS modules Definition 1. [9] A right R−module M is called s − CS if every singular submodule of M is essential in a summand of M. For example every nonsingular module is s − CS. Lemma 1. For a right R−module M, the following statements are equivalent: 1. The second singular submodule Z2 (M ) is CS and a summand of M. 2. M is s − CS. Proof. (1) =⇒ (2). If the second singular submodule Z2 (M)ofM is CS and a summand of M, then every singular submodule of M is essential in a summand of Z2 (M) and a summand of M. (2) =⇒ (1). Let M be s − CS and K is a submodule of Z2 (M). Then Z (K) ⊆ess L where L is a summand of M and L ⊆ess K + L. But L is closed, ess so K ⊆ L. Since Z2 (M) ⊆ L + Z2 (M) and Z2 (M) is closed in M,so L ⊆ Z2 (M) and Z2 (M)isCS. In particular, Z2 (M) is the only closure of Z (M). Thus Z2 (M) is a summand of M. Proposition 1. The following statements are equivalent: S − CS modules and rings 51 1. Every right R−module is s − CS. 2. Every Goldie torsion right R−module is CS. 3. For every right R−module M, M = Z2 (M)⊕K where Z2 (M) is CS and K is nonsingular. In this case we have: r (a) for every right ideal I, I = A⊕B where A ⊆ Z2 and B is nonsingular (b) for every essential right ideal I,ifa/∈ I there exist right ideals K and L such that a ∈ K, K ∩ L = I and K + L = R (c) every singular right R−module is CS. Dinh Van huynh, S. K. Jain and S. R. L´opez-Permouth [4] proved that if R is simple such that every cyclic singular right R−module is CS, then R is right noetherian. Corollary 1. If R is simple such that every cyclic right R−module is s−CS, then R is right noetherian. Definition 2. A right R−module M is called s − N−injective if every ho- momorphism f : K −→ M, where K is a singular submodule of N, extends to N. M is called s−injective if M is s − R−injective. M is called strongly s−injective, if M is s − N−injective for all right R−modules N. Lemma 2. Let M1 and M2 be right R−modules and let M = M1 ⊕ M2. Then M1 is s−M2−injective if and only if for every singular submodule N of M such that N ∩ M1 =0there exists a submodule M of M such that M = M1 ⊕ M , ∼ N ⊆ M and M = M2. Proof. Suppose that M1 is s − M2−injective. Let N be a singular submodule of M such that N ∩ M1 = 0. Consider the diagram α 0 −→ N −→ M2 exact β ↓ M1 where β = π1|N and α = π2|N and πi : M −→ Mi (i =1, 2) are the canonical projections. Since M1 is s − M2−injective so there exists an f : M2 −→ M1 such that fβ = α. The set M = {f (m)+m : m ∈ M2} is a submodule of ∼ M such that M = M1 ⊕ M , N ⊆ M and M = M2. Conversely, consider the homomorphism f : K −→ M1 where K is a singular submodule of M2.If N = {k − f (k):k ∈ K}, then N is a singular submodule of M and N∩M1 =0 and there exists a submodule M of M such that M = M1 ⊕ M and N ⊆ M . Let π denote the projection π : M1 ⊕ M −→ M1. For all k ∈ K we have 0=π (k − f (k)) = π (k) − π (f (k)) = π (k) − f (k). Hence π|M2 extends f. Lemma 3. Let M1 and M2 be right R−modules and let M = M1 ⊕ M2 where M1 and M2 are both s − CS and M1 is s − M2−injective. Then M is s − CS. 52 N. A. Zeyada and N. Jarboui Proof. Let N be a singular submodule of M. Two cases may occur: Case 1. N ∩ M1 =0. In this case, by Lemma 2 there exists a direct summand M of M such ∼ that and M = M2, N ⊆ M and M = M1 ⊕ M . Then M is CS, and so N ⊆ess K ⊆ M for some submodule K of M . Case 2. N ∩ M1 =0. Let N be a submodule of N such that N = N ∩ M1 ⊕ N . Since M1 is ess ⊕ s − CS, N ∩ M1 ⊆ K1 ⊆ M1 = K1 ⊕ K2 for submodules K1 and K2 of M1. ⊕ Since N ∩M1 = 0, as in Case 1 there exists C1 ⊆ M such that C1 isomorphic to M2, N ⊆ C1, M = C1 ⊕ M1 and C1 = C2 ⊕ C3 with N ⊆ C2 for some ⊕ submodules C2, C3 of C1. Hence K1 ⊕ C2 ⊆ M.ThusM is s − CS. Following [7], we say that M is a UC−module if each of its submodule has a unique closure in M. Lemma 4. [7] For a module M, the following statements are equivalent: 1. M is a UC−module. 2. For any closed submodule K of M and any submodule N of M, K ∩ N is closed in M. 3. There does not exists an R−module X with a proper essential submodule Y such that the module (X/Y ) ⊕ X embeds in M. Lemma 5. Let M be a UC−module. If M is s − CS, then every direct sum- mand of M is also s − CS. Proof. Let K be a direct summand of M and N be a singular submodule of K. Since M is s − CS, there exists a direct summand M1 of M such that ess ess c N ⊆ M1. Let L denote the closure of N in K,soN ⊆ L ⊆ K. Then c ess c ess c L ⊆ M.ThusN ⊆ M1 ⊆ M and also N ⊆ L ⊆ M and so, since M is UC, we have L = M1. Hence the closure L of N in K is a direct summand of K showing that K is s − CS, as required.
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