EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 6, No. 2, 2013, 119-125 ISSN 1307-5543 – www.ejpam.com Some Remarks on Finitely Quasi-injective Modules Zhu Zhanmin Department of Mathematics, Jiaxing University, Jiaxing, Zhejiang Province, 314001, P.R.China Abstract. Let R be a ring. A right R-module M is called finitely quasi-injective if each R-homomorphism from a finitely generated submodule of M to M can be extended to an endomorphism of M . Some con- ditions under which finitely generated finitely quasi-injective modules are of finite Goldie Dimensions are given, and finitely generated finitely quasi-injective Kasch modules are studied. 2010 Mathematics Subject Classifications: 16D50, 16L30 Key Words and Phrases: finitely quasi-injective modules, Kasch modules, endomorphism rings, semiper- fect rings, semilocal rings 1. Introduction Throughout the paper, R is an associative ring with identity and all modules are unitary. If MR is a right R-module with S = End(MR), and A ⊆ S, X ⊆ M , B ⊆ R, then we denote the Jacobson radical of S by J(S), and we write lS(X ) = {s ∈ S | sx = 0,∀x ∈ X }, rM (A) = {m ∈ M | am = 0,∀a ∈ A}, lM (B) = {m ∈ M | mb = 0,∀b ∈ B}. Following [5], we write W (S) = {s ∈ S | Ker(s)⊆ess M}. At first let we recall some concepts. A module MR is called finitely quasi-injective (or FQ-injective for short) [7] if each R-homomorphism from a finitely generated submodule of M to M can be extended to an endomorphism of M ; a ring R is said to be right F-injective if RR is finitely quasi-injective. F-injective rings have been studied by many authors such as [2, 3, 8]. A module MR is called a C1 module if every submodule of M is essential in a direct summand of M , C1 modules are also called CS modules. A module MR is called a C2 module if every submodule of M that is isomorphic to a direct summand of M is itself a direct summand of M . A module MR is called a C3 module if, whenever N and K are submodules of M with N ⊆⊕ M , K ⊆⊕ M , and N ∩ K = 0, then N ⊕ K ⊆⊕ M . A module MR is called continuous if it is both C1 and C2. A module MR is called quasi-continuous if it is both C1 and C3. It is well-known that C2 modules are C3 modules, and so continuous modules are quasi-continuous. Email address: zhanmin_zhu@hotmail.com http://www.ejpam.com 119 c© 2013 EJPAM All rights reserved. Z. Zhanmin / Eur. J. Pure Appl. Math, 6 (2013), 119-125 120 A module MR is said to be Kasch [1] provided that every simple module in σ[M] embeds in M , where σ[M] is the category consisting of all M -subgenerated right R-modules. In this note we shall mainly study finitely generated finitely quasi-injective modules with finite Goldie Dimensions, and finitely generated finitely quasi-injective Kasch modules, respectively. 2. Main Results We begin with some Lemmas. Lemma 1 ([10, Theorem 1.2]). For a module MR with S = End(MR), the following statements are equivalent: (1) MR is FQ-injective; (2) (a) lS(A ⋂ B) = lS(A) + lS(B) for any finitely generated submodules A, B of M, and (b) lM rR(m) = Sm for any m ∈ M. where lM rR(m) consists of all elements z ∈ M such that mx = 0 implies zx = 0 for any x ∈ R. Lemma 2 ([10, Theorem 2.1, Theorem 2.2]). Let MR be a finitely generated FQ-injective mod- ule with S = End(MR). Then (1) lS(Kerα) = Sα for any α ∈ S. (2) W (S) = J(S). Lemma 3 ([10, Theorem 2.3]). Let MR be a finitely generated finite dimensional FQ-injective module with S = End(MR). Then S is semilocal. Lemma 4. Let MR be a finitely generated FQ-injective module. Then it is a C2 module. Proof. Write S = End(MR). Let N be a submodule of M and N ∼= eM for some e2 = e ∈ S. Then there exists some s ∈ S such that N = seM and Ker(se) = Ker(e). By Lemma 2(1), we have Sse = Se, and hence e = tse for some t ∈ S with t = et. Thus (set)2 = set and N = (se)M = (set)M . Therefore N is a direct summand of M . Lemma 5. Let MR be a quasi-continuous module with S = End(MR). Then idempotents of S/W (S) can be lifted. Proof. Let s2− s ∈W (S), then Ker(s2− s)Ã M . If x ∈ Ker(s2− s), then (1− s)x ∈ Ker(s), sx ∈ Ker(1− s), and hence x = (1− s)x + sx ∈ Ker(s)⊕ Ker(1− s). It shows that Ker(s2 − s) ⊆ Ker(s)⊕ Ker(1− s), and thus Ker(s)⊕ Ker(1− s) Ã M . Now let N1 and N2 be maximal essential extensions of Ker(s) and Ker(1− s) in M , respectively. Then it is clear that N1∩N2 = 0 and N1⊕N2 Ã M . Since M is a C1 module and N1 and N2 are closed submodules of M , N1 and N2 are direct summand of M . But M is a C3 module, N1⊕N2 is a direct summand of M , so that N1⊕N2 = M . This implies that there exists an e2 = e ∈ S such that N1 = (1−e)M and N2 = eM . Let y ∈ Ker(s), z ∈ Ker(1− s), then noting that y ∈ (1− e)M and z ∈ eM , we have (e − s)(y + z) = z − sz = (1− s)z = 0, so that Ker(s)⊕ Ker(1− s) ⊆ Ker(e − s). And hence e− s ∈W (S), that is, idempotents modulo W (S) lift. Z. Zhanmin / Eur. J. Pure Appl. Math, 6 (2013), 119-125 121 Corollary 1. Let MR be a finitely generated FQ-injective C1 module with S = End(MR). Then S is semiperfect if and only if S is semilocal. Proof. Since MR is a finitely generated FQ-injective module , by Lemma 4, it is a C2 module and hence a C3 module. Thus MR is quasi-continuous by the condition that MR is a C1 module. And so the result follows from Lemma 5 and Lemma 2(2). Recall that a ring R is called right MP-injective [11] if every monomorphism from a prin- cipal right ideal of R to R extends to an endomorphism of R; a ring R is called right MGP- injective [11] if, for any 0 6= a ∈ R, there exists a positive integer n such that an 6= 0 and any R-monomorphism from anR to R extends to an endomorphism of R; a ring R is said to be right AP-injective [6] if, for any a ∈ R, there exists a left ideal Xa such that l r(a) = Ra⊕ Xa; a ring R is called right AGP-injective if, for any 0 6= a ∈ R, there exists a positive integer n and a left ideal Xan such that an 6= 0 and l r(an) = Ran ⊕ Xan . Clearly, right MP-injective rings are right MGP-injective, and right AP-injective rings are right AGP-injective. If R is a right MP-injective rings, then R is right C2 by [11, Theorem 2.7] and J(R) = Z(RR) by [11, Theorem 3.4]. If R is a right AP-injective rings, then R is right C2 by [9, Corollary 3.4] and J(R) = Z(RR) by [6, Corollary 2.3]. So by Lemma 5, we have immediately the following corollary. Corollary 2. Let R be a right CS ring. If R is right MP-injective or right AP-injective, then R is semiperfect if and only if R is semilocal. Let M be a right R-module. A finite set A1, . . . , An of proper submodules of M is said to be coindependent if for each i, 1 ≤ i ≤ n, Ai +∩ j 6=iA j = M , and a family of submodules of M is said to be coindependent if each of its finite subfamily is coindependent. The module M is said to have finite dual Goldie dimension if every coindependent family of submodules of M is finite. Refer to [4] for details concerning the dual Goldie dimension. Lemma 6 ([4, Propositions 2.43]). A ring R is semilocal if and only if RR has finite dual Goldie dimension, if and only if RR has finite dual Goldie dimension. Theorem 1. Let MR be a finitely generated FQ-injective module with S = End(MR). Then the following conditions are equivalent: (1) S is semilocal. (2) MR is finite dimensional. Furthermore, if M is a C1 module, then these conditions are equivalent to: (3) S is semiperfect. Proof. (1) ⇒ (2). If MR is not finite dimensional, then there exists 0 6= x i ∈ M , i = 1,2, 3, · · · , such that ∑∞ i=1 x iR is a direct sum. Since MR is FQ-injective, by Lemma 1, for any positive integer m and any finite subset I ⊂ N \m, S = lS(0) = lS(xmR∩ ∑ i∈I x iR) = lS(xm) + lS( ∑ i∈I x iR) = lS(xm) +∩i∈I lS(x i). Z. Zhanmin / Eur. J. Pure Appl. Math, 6 (2013), 119-125 122 Thus lS(x i), i = 1, 2,3, · · · is an infinite coindependent family of submodules of SS. By Lemma 6, S is not semilocal, a contradiction. (2)⇒ (1). By Lemma 3. Furthermore, if M is a C1 module, then since it is a C2 module by Lemma 4 and W (S) = J(S) by Lemma 2(2), we have (1)⇔ (3) by Lemma 5. The equivalence of (1) and (2) in the next Corollary 3 appeared in [8, Corollary 4.5]. Corollary 3. Let R be a right F-injective ring. Then the following conditions are equivalent: (1) R is semilocal. (2) R is right finite dimensional. Furthermore, if R is a right CS ring, then these conditions are equivalent to: (3) R is semiperfect. Theorem 2. Let MR be a FQ-injective Kasch module with S = End(MR), then (1) rM lS(K) = K for every finitely generarated submodule KR of MR. (2) Sm is simple if and only if mR is simple. In particular, Soc(MR) = Soc(S M). (3) lM (J(R))ÃS M. Moreover, if MR is finitely generated, then (4) lS(T ) is a minimal left ideal of S for any maximal submodule T of M. (5) lS(Rad(M))ÃS S. Proof. (1). Always K ⊆ rM lS(K). If m ∈ rM lS(K) − K , let K ⊆ T ⊆max (mR + K). By the Kasch hypothesis, let σ : (mR+ K)/T → M be monic, and define γ : mR+ K → M by γ(x) = σ(x+T ). Since MR is FQ-injective, γ= s· for some s ∈ S, so sK = γ(K) = 0. This gives sm = 0 as m ∈ rM lS(K). But sm = σ(m+ T ) 6= 0 because m /∈ T , a contradiction. Therefore, rM lS(K) = K . (2). If mR is simple. Then if 0 6= sm ∈ Sm, define γ : mR→ smR by γ(x) = sx . Then γ is a right R-isomorphism, and hence γ−1 extends to an endomorphism of M . Thus, m= γ−1(sm) = α(sm) for some α ∈ S, and so Sm is simple. Conversely, If Sm is simple. By (1), rM lS(m) = mR for each m ∈ M , which implies that for any m ∈ M , every S-homomorphism from Sm to M is right multiplication by an element of R. Now for any 0 6= ma ∈ mR, the right multiplication ·a : Sm → Sma is a left S-isomorphism. So let θ : Sma → Sm be its inverse, then θ is a right multiplication by an element b of R. Thus, m = θ(ma) = mab ∈ (ma)R. Hence mR is simple. (3). Let 0 6= m ∈ M . Suppose that T is a maximal submodule of mR. By the Kasch hypothesis, let σ : mR/T → M be monic, and define f : mR→ M by f (x) = σ(x + T ). Since MR is FQ-injective, f = s· for some s ∈ S, and then sm= f (m) = σ(m+ T ) 6= 0. But Z. Zhanmin / Eur. J. Pure Appl. Math, 6 (2013), 119-125 123 smJ(R) = f (m)J(R) = σ(m+ T )J(R) = 0, so 0 6= sm ∈ Sm∩ lM (J(R)). Therefore, lM (J(R))Ã S M . (4). Let T be any maximal submodule of M. Since MR is Kasch, there exists a monomor- phism ϕ : M/T → M . Define α : M → M by x 7→ ϕ(x + T ). Then 0 6= α ∈ S,αT = ϕ(0) = 0, and so lS(T ) 6= 0. For any 0 6= s ∈ lS(T ), we have T ⊆ Ker(s) 6= M , and so Ker(s) = T by the maximality of T . It follows that lS(T ) = lS(ker(s)) = Ss by Lemma 2(1). Therefore, lS(T ) is a minimal left ideal of S. (5). If 0 6= a ∈ S, choose a maximal submodule T of the right R-module aM . Since M is Kasch, there exists a monomorphism f : aM/T → M . Define g : aM → M by g(x) = f (x+T ). Since M is FQ-injective and finitely generated, g = s· for some s ∈ S. Take y ∈ M such that a y /∈ T , then sa y = g(a y) = f (a y + T ) 6= 0, and hence sa 6= 0. If a(Rad(M)) * T , then a(Rad(M)) + T = aM . But a(Rad(M)) << aM because M is finitely generated, so T = aM , a contradiction. Thus a(Rad(M)) ⊆ T , and then (sa)(Rad(M)) = g(a(Rad(M))) = f (0) = 0, whence 0 6= sa ∈ Sa ∩ lS(Rad(M)). This shows that lS(Rad(M))ÃS S. Lemma 7. Let MR be a finitely generated Kasch module with S = end(MR). If S is left finite dimensional, then M/RadM is semisimple. Proof. Let T be any maximal submodule of M . Since MR is Kasch, there exists a monomor- phism ϕ : M/T → M . Define α : M → M by x 7→ ϕ(x + T ). Then 0 6= α ∈ S,αT = ϕ(0) = 0, and so lS(T ) 6= 0. Let Ω = {K | 0 6= K = lS(X ) for some X ⊆ M}, then lS(T ) is minimal in Ω for any maximal submodule T of M . In fact, if lS(T ) ⊇ lS(X ) 6= 0, where X ⊆ M , then T ⊆ rM lS(X ) 6= M . So T = rM lS(X ), and hence lS(T ) = lS(X ). Since S is left finite dimen- sional, there exist some minimal members I1, I2, · · · , In in Ω such that I =⊕n i=1 Ii is a maximal direct sum of minimal members in Ω. Now we establish the following claims: Claim 1. rM (Ii) is a maximal submodule of M for each i. Since M is finitely generated and Kasch, rM (Ii) ⊆ Ti = rM lS(Ti) for some maximal submodule Ti . Thus Ii ⊇ lS rM lS(Ti) = lS(Ti) 6= 0, and so Ii = lS(Ti) by the minimality of Ii in Ω. Now we choose 0 6= ai ∈ lS(Ti). Then Ti = rM (ai), and hence rM (Ii) = rM lS(Ti) = rM lS rM (ai) = rM (ai) = Ti . Claim 2. RadM = ∩n i=1rM (Ii). Clearly, RadM ⊆ ∩n i=1rM (Ii). If T is a maximal submodule of M , then lS(T ) ∩ I 6= 0. Taking some 0 6= b ∈ lS(T )∩ I , we have T = rM (b)⊇ ∩n i=1rM (Ii). This gives that ∩n i=1rM (Ii)⊆ RadM , and the claim follows. Finally, observing that each M/rM (Ii) is simple by Claim 1, and the mapping f : M/RadM →⊕n i=1M/rM (Ii); m+ RadM 7→ (m+ rM (I1), · · · , m+ rM (In)) is a monomorphism by Claim 2, we have that M/RadM is semisimple. Theorem 3. Let MR be a finitely generated and FQ-injective Kasch module with S = End(MR). Then the following conditions are equivalent: REFERENCES 124 (1) M/Rad(M) is semisimple. (2) S is left finitely cogenerated. (3) S is left finite dimensional. In this case, Soc(SS) = lS(Rad(M)), and G(SS) = c(SSoc(SS)) = c(M/Rad(M)). Proof. (1) ⇒ (2). It is trival in case M = 0. 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