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Uniform almost relative injective modules
Journal of Algebra 478 (2017) 353–366
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Journal of Algebra www.elsevier.com/locate/jalgebra
Uniform almost relative injective modules Surjeet Singh House No. 424, Sector 35 A, Chandigarh 160036, India
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Article history: Received 4 March 2015 Available online 2 February 2017 Communicated by Louis Rowen MSC: primary 16D50 secondary 16E50 Keywords: Almost relative injectives Almost self-injectives CS modules Row-finite matrices Von Neumann regular rings
a b s t r a c t The concept of a module M being almost N -injective where N is some module, was introduced by Baba (1989). For a given module M , the class of modules N , for which M is almost N -injective, is not closed under direct sums. Baba gave a necessary and sufficient condition under which a uniform finite length module U is almost V -injective, where V is a finite direct sum of uniform, finite length modules, in terms of extending properties of simple submodules of V . Let U be a uniform module and V be a finite direct sum of indecomposable modules. Recently, Singh (2016), has determined some conditions under which U is almost V injective, which generalize Baba’s result. In the present paper some more results in this direction are proved. A module M is said to be completely almost self-injective, if for any two subfactors A, B of M , A is almost B-injective. A necessary and sufficient condition for a module M to be completely almost self-injective is given. Using this, it is proved that a Von Neumann ring R is completely almost right self-injective R if and only if soc(R is semi-simple and every minimal right R) ideal of R is injective. © 2017 Elsevier Inc. All rights reserved.
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jalgebra.2017.01.033 0021-8693/© 2017 Elsevier Inc. All rights reserved.
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Introduction Let MR , NR be two modules. As defined by Baba [2], M is said to be almost N -injective, if for any homomorphism f : A → M , A N , either f extends to a homomorphism g : N → M or there exist a decomposition N = N1 ⊕ N2 with N1 = 0 and a homomorphism h : M → N1 such that hf (x) = π(x) for any x ∈ A, where π : N → N1 is a projection with kernel N2 . This concept plays a significant role is studying extending modules. A module M that is almost M -injective, is called an almost self-injective module. For a module M , the class of those modules N for which M is almost N -injective, is not closed under direct sums. Let {Uk : 0 ≤ k ≤ n} be a finite family of uniform modules n
of finite lengths, and U = ⊕ Σ Uk . Baba [2] has given a characterization for U0 to be k=1
almost U -injective, in terms of the property of simple submodules of U being contained in uniform summands of U . Let M be a uniform module and V be a finite direct sum of indecomposable modules. In [7], conditions under which M is almost V -injective have been investigated, thereby Baba’s result has been generalized. In Section 1, this study has been further continued. Let M be an almost self-injective, uniform modules. Let T be the set of those maximal homomorphisms f : N → M , N < M which cannot be extended to endomorphisms of M . In Section 1, an algebraic structure on T is given. As defined in Section 2, a module M is called a completely almost self-injective module, if for any two subfactors A, B of M , A is almost B-injective. In Theorem 2.4, it is proved that a uniform module U is completely almost self-injective if and only if U is of length not more than 2 and is quasi-injective. In Theorem 2.11 a necessary and sufficient condition for a module M to be completely almost self-injective is proved. From which, it follows that a Von Neumann regular ring R is completely right almost self-injective if and only R if soc(R is semi-simple and every minimal right ideal of R is injective. R) Preliminaries. All rings considered here are with unity and all modules are unital right modules unless otherwise stated. Let M be a module. Then E(M ) and J(M ) denote the injective hull and radical, respectively, of M . The symbols N M , N < M , N ⊂e M denote that N is a submodule of M , N is a submodule different from M , N is an essential submodule of M respectively. Any submodule of a homomorphic image of M is called a subfactor of M . A module M whose ring of endomorphisms End(M ) is local, is called an LE module. A module M such that its complement submodules are summands of M , is called a CS module (or a module satisfying condition (C1 )). The terminology used here is available in standard text books like [1,4]. For details on Von Neumann regular rings see [5]. 1. Direct sums of uniform modules Definition 1.1. Let MR and NR be any two modules. Then M is said to be almost N -injective, if given any R-homomorphism f : A → M , A N either f extends to an R-homomorphism from N to M or there exist a decomposition N = N1 ⊕ N2 with
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N1 = 0, and an R-homomorphism h : M → N1 such that hf (x) = π(x) for any x ∈ A, where π : N → N1 is a projection with kernel N2 . 2 Let a module MR be almost NR -injective and f : L → M , L < N be maximal in the sense that there does not exist any homomorphism g : L1 → M , L < L1 N , which extends f . Now L ⊂e N . By the definition, we get N = N1 ⊕ N2 for some N1 = 0, ker f ⊆ N2 , in this situation, f is monic on L ∩ N1 . It follows that ker f is not essential in L. From which, we get the following. If f : K → M , K < N is any homomorphism such that ker f ⊂e K, then f can be extended from N to M . The definition also gives that if N is indecomposable, then any homomorphism f : K → M , K < N with ker f = 0, can be extended from N to M . Proposition 1.2 ([7, Proposition 1.2]). (i) A module MR is almost NR -injective if and only if for any R-homomorphism f : L → M , L < N which is maximal with respect to the property that it cannot be extended from N to M , there exist a decomposition N = N1 ⊕N2 with N1 = 0, and an R-homomorphism h : M → N1 such that hf (x) = π(x) for any x ∈ L, where π : N → N1 is a projection with kernel N2 . 2 Proposition 1.3. Let AR , BR be any two modules and f : L → B, L < A be an R-homomorphism that is maximal with respect to the property that it cannot be extended from A to B. Let C be a summand of A. If L ∩ C < C, then f1 = f | L ∩ C from L ∩ C to B is a maximal homomorphism that cannot be extended from C to B. If L ∩ C ⊆ ker f , then C ⊆ L. Proof. The first part is proved in [7, Proposition 1.3]. In the later case f1 = 0, so it extends to a zero map from C to B, which proves that C ⊆ L. 2 The following is well known. (See [6].) Proposition 1.4. Let MR , NR be any two modules such that M is almost N -injective. (i) Any summand K of M is almost N -injective. (ii) If W is a summand of N , then M is almost W -injective. (iii) If N = N1 ⊕ N2 and M is not N -injective, then M is either not N1 -injective or not N2 -injective. 2 Theorem 1.5 ([7, Corollary 1.7]). Let MR be any uniform module, and NR be any module. Then M is almost N -injective if and only if for any maximal homomorphism f : L → M , L < N which cannot be extended from N to M , N = N1 ⊕ ker f such that f (N1 ∩ L) = M. 2
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Lemma 1.6. Let AR , BR be any two modules, MR a uniform module which is almost A-injective and almost B-injective. Let N = A ⊕ B. Then M is almost N -injective if and only if for any uniform summands A1 of A and B1 of B, M is almost A1 ⊕ B1 -injective. If M is A-injective, then M is almost N -injective. Proof. If M is almost N -injective, as A1 ⊕ B1 is a summand of N , M is almost A1 ⊕ B1 -injective. Conversely, suppose the given condition holds. Let f : L → M , L < N be a maximal homomorphism which cannot be extended from N to M . Let f1 = f |(L ∩ A), f2 = f |(L ∩ B). Case 1. B ⊂ L. Then L = (L ∩ A) ⊕ B and f1 : L ∩ A → M is a maximal R-homomorphism that cannot be extended from A to M . By (1.5), A = A1 ⊕ ker f1 , where A1 is uniform and f (L ∩A1 ) = M . For each b ∈ B, there exists a unique a ∈ L ∩A1 such that f (b) = f (a). We get a homomorphism π : B → L ∩ A1 such that π(b) = a for any b ∈ B, a ∈ L ∩ A, whenever f (b) = f (a). Then B = {b − π(b) : b ∈ B} ⊆ ker f and N = A1 ⊕ (ker f1 ⊕ B ) = A1 ⊕ ker f . Case 2. L ∩ A < A, L ∩ B < B. Then A = A1 ⊕ ker f1 , B = B1 ⊕ ker f2 , f (L ∩ A1 ) = M = f (L ∩ B1 ). Let T = A1 ⊕ B1 . By the hypothesis, M is almost T -injective and L ∩ T < T . Now f |(L ∩ T ) is a maximal homomorphism that cannot be extended from T to M . Hence T = C ⊕ ker(f |(L ∩ T )). Which gives N = C ⊕ ker f . Hence M is almost N -injective. In case M is A-injective, the arguments in Case 1 give that M is almost N -injective. 2 Lemma 1.7. Let A, B, M be three uniform right R-modules such that M is almost A-injective and almost B-injective, but is neither A-injective nor B-injective. If M is almost A ⊕B-injective, then any complement submodule of A ⊕B is a summand of A ⊕B. Conversely, if M is neither A-injective nor B-injective, every complement submodule of A ⊕ B is a summand of A ⊕ B, then M is almost A ⊕ B-injective. Proof. Now there exist maximal homomorphisms f1 : L1 → M , L1 < A, f2 : L2 → M , L2 < B which cannot be extended from A to M , from B to M respectively. Both fi are monomorphisms. Let C be a complement submodule of T = A ⊕ B other than T . Then C is uniform, therefore, either A ∩ C = 0 or B ∩ C = 0. Suppose A ∩ C = 0. We get a maximal R-homomorphism f : L → M , such that L1 ⊕ C ⊆ L, f extends f1 and it is zero on C. By (1.5), ker f is a summand of T . However ker f is uniform and C ⊆ ker f . Hence C = ker f . Conversely, let the given condition hold. Let f : L → M , L < T be a maximal homomorphism that cannot be extended from T to M . Let L1 = L ∩ A, L2 = L ∩ B. If either L1 = A or L2 = B, as discussed in Case 1 of the proof of (1.6), we get T = C⊕ker f , f (C ∩ L) = M . Suppose L1 < A, L2 < B. By (1.5), f (L1 ) = M = f (L2 ), which gives an isomorphism π : L2 → L1 , π(x) = y for any x ∈ L2 , y ∈ L1 , whenever f (x) = f (y). Thus K =
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{x − π(x) : x ∈ L2 } ⊆ ker f , is uniform, therefore ker f is uniform and there exists a uniform summand C of T containing ker f . Then T = C ⊕C for some uniform submodule C . If ker f < C ∩ L, then kerL f cannot be uniform, which gives a contradiction. Hence ker f = C ∩ L. By (1.3), C ⊆ L. Thus T = C ⊕ ker f . Hence M is almost T -injective. 2 The following result generalizes [7, Theorem 1.11]. Theorem 1.8. Let MR be a uniform module and NR = N1 ⊕ N2 , where Ni are indecomposable. Then M is almost N -injective if and only if one of the following holds. (i) M is N1 -injective and N2 -injective. (ii) M is almost N1 -injective and almost N2 -injective. If M is neither N1 -injective nor N2 -injective, then N is a CS module. Proof. Let MR be a uniform module and almost N -injective. By (1.4), M is almost Ni -injective for i = 1, 2. If M is neither N1 -injective nor N2 -injective, by (1.7), every complement submodule of N is a summand, i.e. N is a CS module. Conversely, let the given conditions hold. Let f : L → M , L < N be a maximal homomorphism that cannot be extended from N to M . Let Li = L ∩ Ni and L1 < N1 . Then f (L1 ) = M and N1 is uniform. If L2 = N2 , as in Case 1 in (1.6), N = N1 ⊕ ker f . In case L2 < N2 , the result follows from (1.7). 2 Lemma 1.9. Let MR be any module. Let A, B be two submodules of M , which are complements of each other, and let π : A ⊕ B → A be projection with kernel B. If π extends to an endomorphism of M , then M = A ⊕ B. Proof. Suppose π extends to an endomorphism σ of M . Now B ⊆ ker σ. Let 0 = x ∈ ker σ. For some r, 0 = xr ∈ A ⊕ B. Then σ(xr) = 0 gives xr ∈ B. Hence B ⊂e ker σ, which gives B = ker σ. Suppose A < σ(M ). As A is closed, A is not essential in σ(M ). M Now σ induces an isomorphism σ : M B → σ(M ), as A ⊂e M = B , σ(A) ⊂e σ(M ), i.e. A ⊂e σ(M ), which is a contradiction. Thus A = σ(A) = σ(M ). Hence M = A ⊕ B. 2 Proposition 1.10. Let MR be an almost self-injective module. (i) Let B be a closed submodule of M such that no complement of B contains an infinite direct sum of summands of M . Then B is a summand of M . (ii) Any almost self-injective module of finite Goldie dimension is a CS-module and is a finite direct sum of uniform submodules. (iii) If A is a uniform submodule of M and B is a complement of A, then B is a summand of M . (iv) Let S be a minimal submodule of M such that S ⊆ J(M ). Then M = M1 ⊕ M2 such that M1 is uniform but not simple, and soc(M1 ) ⊕ M2 = S ⊕ M2 .
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Proof. Let A be a complement of B and π : A ⊕ B → A be a projection with kernel B. If π extends to an endomorphism of M , by (1.9), B is a summand of M . Suppose for no complement A of B, the projection π can be extended to an endomorphism of M . By (1.1), there exists a decomposition M = N1 ⊕ M2 such that N1 = 0 and B ⊆ M2 . If B ⊂e M2 , B = M2 and we finish. Otherwise we repeat the process with M2 . We get M2 = N2 ⊕ M3 with N2 = 0 and B ⊆ M3 . Continue the process. The process cannot be continued indefinitely, for otherwise an infinite direct sum of summands of M is contained in a complement of B. Hence B is a summand of M . This proves (i). Now (ii) and (iii) are immediate from (i). (iv) Let B be any complement of S, by (i), M = M1 ⊕M2 , where M1 = 0 and B = M2 . Then J(M ) = J(M1 ) ⊕ J(M2 ) and S ⊕ J(M2 ) ⊆ J(M ). Thus M1 cannot be simple. As S embeds in M1 , soc(M1 ) ∼ = S. It is obvious that soc(M1 ) ⊕ M2 = S ⊕ M2 . 2 Let MR be a uniform almost self-injective module. Let T be the set of those maximal homomorphisms f : L → M , L < M which cannot be extended to endomorphisms of M . We give an algebraic structure in T modulo the additive group of endomorphisms of M . Proposition 1.11. Let a module MR be almost NR -injective. Let f1 : L1 → M , L1 N be a maximal homomorphism in the sense that it has no proper extension f : G → M , G N . Let f2 : L2 → M , L2 N be another homomorphism. (i) Let μ : L1 ∩ L2 → M , μ(x) = f1 (x) − f2 (x) for x ∈ L1 ∩ L2 . If g : L → M, L M is a maximal extension of μ, then L ∩ L2 = L1 ∩ L2 . (ii) If f2 is also maximal, then L ∩ L1 = L1 ∩ L2 = L ∩ L2 . In particular, if L = N , then L1 = L2 . (iii) Let 0 = f : H → M , H ⊂e N be a homomorphism and f1 : L1 → M , L1 M be a maximal extension of f . Let f2 : L2 → M , L2 M be another maximal extension of f . Then L2 = L1 . (iv) Let M be almost self-injective and uniform. Let 0 = f : H → M , H M be a homomorphism and f1 : L1 → M , L1 M , and f2 : L2 → M , L2 M be any maximal extension of f . Then L1 = L2 . Let f : L → M , L < M be a homomorphism. If for some 0 < K L, f |K can be extended to an endomorphism of M , then f can also be extended to an endomorphism of M . Proof. Let K = L ∩ L2 . Then K ∩ L1 = L ∩ L1 ∩ L2 = L1 ∩ L2 . Now g(x) and f2 (x) are defined for any x ∈ K. We get β : K → M , β(x) = g(x) + f2 (x), x ∈ K. Clearly β|(K ∩ L1 ) = f1 |(K ∩ L1 ), so we get β1 : K + L1 → M , β1 (x + y) = β(x) + f1 (y) for x ∈ K, y ∈ L1 . As f is maximal, we get K ⊆ L1 . Hence L1 ∩ L = L1 ∩ L2 . This proves (i) and (ii). (iii) Let μ be as in (i). As H ⊆ ker μ, μ extends to a homomorphism g : N → M . By (ii), L1 = L2 . It follows that if f2 is also maximal, then L1 ∩ L = L1 ∩ L2 = L2 ∩ L, in particular, if we have L = N , then L1 = L2 . Now (iv) is immediate from (iii). 2
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Let MR be an almost self-injective uniform module, T be set of those maximal homomorphisms f : L → M , L < M , which cannot be extended to endomorphisms of M . We give an algebraic structure on T that makes it an abelian group modulo S = End(M ). The above result shows that given any homomorphism g : H → M , 0 < H < M , there exists a unique L M such that for any maximal extension h : K −→ M , K M of g, K = L. This L is called a maximal domain of g. It follows that, if g extends to an endomorphism of M , then any other maximal extension of g is an endomorphism of M . For any two f1 : L1 → M , f2 : L2 → M members of T ∪ S, define f1 ∼ f2 if for some non-zero K L1 ∩ L2 , λ : K → M , λ(x) = f1 (x) − f2 (x) can be extended to an endomorphism of M . It is an equivalence relation. It follows that if f1 ∼ f2 , then L1 = L2 . So, to each equivalence class [f ] is associated an L M such that any g ∈ [f ] has domain L. We have [0] = S. Let G be the set of these equivalence classes. Let f1 , f2 , f3 ∈ T ∪ S with L1 , L2 , L3 as their respective domains. Let μ be f1 + f2 defined on L1 ∩ L2 and g : L → M , L M be a maximal extension of μ. Set [f1 ] + [f2 ] = [g]. It follows from (1.11) that the addition is well defined. It is clear from the definition that if [f1 ] + [f2 ] = [h], then for some 0 = K L1 ∩ L2 ∩ dom(g), f1 + f2 − h defined on K can extended to an endomorphism of M . Now ([f1 ] + [f2 ]) + [f3 ] = [g] + [f3 ] = [k] for some k ∈ T ∪ S. There exist non-zero submodules K1 , K2 of M such that f1 + f2 − g is naturally defined on K1 and g + f3 − k is naturally defined on K2 and they are extendable to some endomorphisms of M . By (1.11)(iv), f1 + f2 + f3 − k defined on K = K1 ∩ K2 can be extended to an endomorphism of M . Similarly, if [f1 ] + ([f2 ] + [f3 ]) = [k1 ]. There exists a non-zero K M on which f1 + f2 + f3 − k1 is naturally defined and is extendable to an endomorphism of M . From which it follows that k − k1 naturally defined on K ∩ K can be extended to an endomorphism of M . Which proves that addition is associative. [0] is the additive identity in G. We get that G is an abelian group. 2 2. Completely almost self-injective modules Definition 2.1. A module MR is said to be completely almost self-injective, if for any two subfactors U , V of M , U is almost V -injective. 2 Lemma 2.2. Let MR be a completely almost self-injective module. (i) M N is semisimple for any N ⊂e M . (ii) Let SR be a simple module which is a subfactor of M . Then either S is M -injective or given any maximal homomorphism f : L → S, L < M which cannot be extended from M to S, M = M1 ⊕ ker f , for some M1 which is uniform but not simple, soc(M1 ) ∼ =S and L = soc(M1 ) ⊕ ker f . (iii) J(M ) ⊆ soc(M ) and soc(M ) is an intersection of maximal submodules. Proof. (i) Now every homomorphic image of M is essentially M -injective. Let N ⊂e L M . Let L be any submodule of M containing N . Then N is almost M N -injective. It
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L L M M follows from (1.1) that N is M N -injective. Therefore N is a summand of N . Hence N is semi-simple. (ii) Now S is almost M -injective. Let f : L → S, L ⊂e M be a maximal homomorphism. Suppose f cannot be extended from M to S. Let A = ker f . Then A is not essential in L. Hence L = S1 ⊕ ker f , where S1 ∼ = S. By (1.10)(iii), M = M1 ⊕ M2 such that ker f ⊂e M2 , M1 is uniform and soc(M1 ) ∼ = S1 . Clearly, ker f = L ∩ M2 . By (1.3), ker f = M2 . Which gives that M1 is not simple. Then L = S1 ⊕ M2 . (iii) For any essential submodule N of M , as M N is semi-simple, J(M ) ⊆ N . Which gives J(M ) ⊆ soc(M ). It also gives that any essential submodule of M is an intersection of maximal submodules. Therefore soc(M ) is also an intersection of maximal submodules. 2
Theorem 2.3. Let MR be a completely almost self-injective module. Let S be a simple module which is a subfactor of M . Then either S is M -injective or there exists a decomposition M = M1 ⊕ M2 such that M2 is a complement of S and M1 is uniserial and has length 2. If J(M ) = 0, then S is M -injective. Proof. Let S be not M -injective. By (2.2), M = M1 ⊕ M2 such that M1 is uniform 1 and M2 is a complement of S. Thus soc(M1 ) ∼ = S. By (2.2), M S1 is semi-simple, where M1 S1 = soc(M1 ). Suppose d( S1 ) > 2. Then we get two different uniserial submodules A, B of M1 such that d(A) = d(B) = 2. As A is almost self-injective, it must be quasi-injective. Similarly, B is quasi-injective. Suppose A and B are not isomorphic. However B is almost A-injective, there exists a homomorphism h : A → B such that h(x) = x for any x ∈ S1 . Then h : A → B is an isomorphism, which gives a contradiction. Hence A ∼ = B, which also gives A = B. Hence d(M1 ) = 2. This also gives J(M1 ) = 0 and therefore J(M ) = 0. 2 Theorem 2.4. Any completely almost self-injective, uniform module U is either simple or of length 2. Proof. Suppose soc(U ) = 0. Let 0 = x ∈ U . We have a maximal submodule T of xR. xR Then T = 0. As xR T is almost U -injective, we get an epimorphism f : U → T extending xR the natural map from xR to T . This gives that U has a maximal submodule K1 such that KU1 ∼ = xR T . Suppose U is not quasi-injective. By (1.5), there exists a monomorphism f : W → U , W < U such that f (W ) = U . Set W1 = f −1 (K1 ). Then W1 < W . As W1 H soc(W1 ) = 0, we get a 0 < V < W1 . By (2.2)(i), VU is semi-simple. We have W V = V ⊕ V , ∼ W where H V = W1 . Then f (H) K1 . Now K1 is almost H-injective. Let g = f | V : V → K1 . Suppose g extends to a homomorphism g : H → K1 . Now g = f | H, g − (f | H) is zero on V , so its image in U is semi-simple. Which is a contradiction. Thus g does not have an extension from H to K1 . By (1.5), there exists a monomorphism λ : W → K1 , W < H, which is an extension of g, such that λ(W ) = K1 . Once again λ = f | W , that gives that soc(U ) = 0. Hence U is quasi-injective. Suppose there exists a maximal submodule K2 of U different from
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K1 . As for U , K1 and K2 are also quasi-injective. We have a maximal homomorphism f : A → K2 , A K1 which extends the inclusion map on K1 ∩ K2 . Now f extends to an automorphism g of U . If A = K1 , we get f (K1 ) = K2 , K1 = K2 , a contradiction. Hence A < K1 . As K2 is almost K1 -injective, f (A) = K2 . Then g(K1 ) = U , which is also a contradiction. Hence U has only one maximal submodule K1 . Thus J(U ) = K1 . By (2.2)(iii), J(U ) ⊆ soc(U ) = 0. Which gives a contradiction. Hence soc(U ) = 0. Let S = soc(U ). Suppose U = S. Then S is not U -injective. Therefore, by (2.3), d(U ) = 2. 2 Remark. A related result is that for a ring R, finite direct sum of uniform modules is extending if and only if any uniform module over R is of length not more than 2 [3, 13.1]. Theorem 2.5. Let MR be any completely almost self-injective module. Then semi-simple and soc(M ) ⊂e M .
M soc(M )
is
Proof. Firstly, let M be finitely generated. By (2.2)(i), M E is a finite direct sum of simple modules for any E ⊂e M . Let H be a complement of soc(M ) in M . Then soc(H) = 0. Suppose H contains an infinite direct sum. Then H contains an infinite direct sum L = ⊕ Xi , which is essential in H. For each Xi , there exists proper submodule Yi of Xi i that is essential in Xi . Then L1 = ⊕ Yi ⊕ soc(M ) ⊂e H ⊕ soc(M ) ⊂e M . Thus LM1 is of i
finite length. However LL1 is not of finite length. This gives a contradiction. Hence H is of finite Goldie dimension. As H is almost self-injective, by (1.10)(ii), H = A1 ⊕A2 ⊕... ⊕At , where each Ai is uniform. By (2.4), each Ai has length 2. Thus soc(H) = 0, unless H = 0. M Hence soc(M ) ⊂e M and soc(M ) is semi-simple. Let M need not be finitely generated. Suppose soc(M ) is not essential in M . We can find a non-zero x ∈ M such that soc(M ) ∩ xR = 0. Thus soc(xR) = 0. As xR is also finitely generated completely almost self-injective, soc(xR) = 0, which is a contradiction. M Thus soc(M ) ⊂e M . By (2.2)(i), soc(M ) is semi-simple. 2 Lemma 2.6. Let MR be a completely almost self-injective module, and U , V be two nonsimple uniform subfactors of M . (i) d(U ) = d(V ) = 2. (ii) U is M -injective and isomorphic to a summand of M . (iii) If soc(U ) ∼ =V. = soc(V ), then U ∼ Proof. By (2.4), d(U ) = d(V ) = 2. Suppose U is not M -injective. Let f : L → U , L < M be a maximal homomorphism that cannot be extended from M to U . We have a decomposition M = M1 ⊕ M2 such that f is one-to-one on M1 ∩ L, f (M1 ∩ L) = U and ker f = M2 . As M1 is uniform, by (2.4), d(M1 ) = 2. Which gives that M1 = M1 ∩ L ⊆ L. Hence L = M , which is a contradiction. Hence U is M -injective. There exists a factor T of M containing U . As U is T -injective, it is a summand of T . Therefore U is a factor of M . We get an epimorphism g : M → U . Let K = ker f . Then
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K is not essential in M . We get a minimal submodule S of M such that S ∩ K = 0. Now K is a complement of S in M . By (1.10), M = K1 ⊕ K with K1 uniform. We get U∼ = K1 . This proves (i) and (ii). (iii) We have an isomorphism f : soc(U ) → soc(V ). By (1.5), if f does not extend from U to V , f (soc(U )) = V , which is a contradiction. Hence U ∼ =V. 2 M Lemma 2.7. Let MR be any module such that soc(M ) is semi-simple. Then: (i) Soc(M ) ⊂e M . (ii) Let N be a module which embeds in M or is a homomorphic image of M . Then N M N soc(N ) ⊂e N and soc(N ) is semi-simple. If soc(M ) is finitely generated, then soc(N ) is also finitely generated. M L (iii) If soc(M ) is finitely generated, then for any subfactor L of M , soc(L) is finitely generated and soc(L) ⊂e L. M Proof. (i) Let K be a complement of soc(M ). As K embeds in soc(M ) , K is semi-simple. Which gives soc(M ) ⊂e M . (ii) Let N be a submodule of M . As soc(M ) ⊂e M , soc(N ) = soc(M ) ∩ N ⊂e N . As N M N soc(N ) embeds in soc(M ) , soc(N ) is semi-simple. Thus soc(N ) ⊂e N . Let f : M → K be an epimorphism and B = ker f . Now soc(M ) = H ⊕ soc(B) for some semi-simple submodule H of M . Thus H+B ⊆ soc( M B B ). As soc(B) ⊂e B, H ∩B = 0. M K But H ⊕ B ⊂e M . Therefore H⊕B is semi-simple, which also gives that soc( soc(K) ) is M K semi-simple. If soc(M ) is finitely generated, it also follows that soc(K) is finitely generated. (iii) It follows from (ii). 2
Lemma 2.8. Let MR be any module such that it satisfies the following conditions. M (i) soc(M ) is semi-simple. (ii) Any complement of a minimal submodule of M is a summand of M . (iii) Any uniserial submodule of M of length 2 is quasi-injective and any two nonsimple uniserial submodules with isomorphic socles are isomorphic. Then for any minimal submodule S ⊆ J(M ) and a complement M2 of S, M = M1 ⊕ M2 for some uniserial submodule M1 of length 2 with S ⊂ M1 . Any non-simple uniform submodule of M is of length 2 and is a summand of M . Proof. Let S ⊆ J(M ) be a minimal submodule. Let M2 be a complement of S in M . By the hypothesis, M = N1 ⊕ M2 for some uniform submodule N1 . Let π1 : M → N1 be a projection with kernel M2 . Then for S1 = π1 (S) = soc(N1 ), S ⊕ M2 = S1 ⊕ M2 . Which 1 gives that N1 is not simple. By (2.7), N S1 is semi-simple. By (iii), any two non-simple uniserial submodules of N1 are isomorphic, and they are quasi-injective, it follows that N1 itself is uniserial and hence has length 2. Let T be a maximal direct sum in M of uniserial submodules of length 2. Now T = ⊕ Ai , where each Ai is uniserial and i∈Λ
has length 2. Let S = xR be a minimal submodule of M contained in J(M ). Suppose S ∩ T = 0. By (ii) M = M1 ⊕ M2 where M2 is a complement of S containing T . As
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M1 is of length 2 and M1 ∩ T = 0, we get a contradiction. Thus S ⊂ T . We get finitely many distinct elements say 1, 2, ..., k in Λ such that x = x1 + x2 + ... + xk for some non-zero xi ∈ Ai . For any i ≥ 2, we have monomorphism σi : soc(A1 ) → soc(Ai ) such that σi (x1 ) = xi . By (iii), σi can be extended to an isomorphism ηi : A1 → Ai . Now A1 = y1 R for some y1 ∈ A1 . Let y = y1 + y2 + ... + yk , where for each i ≥ 2, yi = ηi (y1 ). Then K = yR is uniserial and S ⊂ yR. The second part is immediate. 2 Another interesting property of a completely almost self-injective module is given by the following. Proposition 2.9. Let MR be a completely almost self-injective module. Let A, B be two subfactors of M such that J(A) = 0 and f : L → B, L ⊂e A is a monomorphism such that for any minimal submodule S of A, f (S) ⊆ J(B), then f can be extended to a homomorphism from A to B. Proof. By the hypothesis B is almost A-injective. By (2.4) and (2.6), any non-simple uniform submodule U of A has length 2 and is A-injective, it is a summand of A. However J(A) = 0. Therefore A has no non-simple uniform submodule. By (1.10), any complement of a minimal submodule of A is a summand of A. Condition (iii) in (2.8) is vacuously satisfied by A. It follows that A satisfies the hypothesis of (2.8). Similarly for B. Thus by (2.8), any minimal submodule of J(B) is contained in a uniserial submodule of length 2. Suppose f : L → B, L ⊂e A, cannot be extended from A to B. By definition, we have a decomposition A = A1 ⊕ A2 and a homomorphism h : B → A1 such that A1 = 0 and hf (x) = π1 (x) for x ∈ L, where π1 : A → A1 is a projection with kernel A2 . Let S be a minimal submodule of A contained in A1 . As f (S) ⊆ J(B), by (2.8), there exists a uniserial submodule U of B such that f (S) < U . Then U ∼ = h(U ) ⊆ A. Thus J(A) = 0, which is a contradiction. Hence f can be extended to a homomorphism from A to B. 2 We shall give a necessary and sufficient condition for a module to be completely almost self-injective Lemma 2.10. Let MR be a module satisfying the following conditions. M (i) soc(M ) is semi-simple. (ii) Any complement in M of a minimal submodule S of M is a summand of M . If a minimal submodule S ⊆ J(M ), then there exists a uniserial submodule U of M of length 2 containing S. (iii) Any uniserial subfactor of M of length 2 is quasi-injective. Any two non-simple uniserial subfactors U , V are isomorphic, whenever they have isomorphic socles. Then the following hold. (a) Any simple module SR is either M -injective or there exists a uniserial summand U of M of length 2 such that S ∼ = soc(U ). (b) Any uniserial subfactor of M of length 2 is M -injective. For any V < M such that M V is uniserial and of length 2, V is a summand of M .
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Proof. (a) It follows from the given conditions that any non-simple uniform submodule of M is uniserial and has length 2. Let S be any simple module. Suppose S is not M -injective. Then there exists a maximal homomorphism f : L → S, L < M , which cannot be extended from M to S. Let K = ker f . Then K is not essential in M for otherwise M K being semi-simple leads to a contradiction. We get a minimal submodule S1 ⊂ L such that S1 ∩ K = 0. By (ii) M = M1 ⊕ M2 such that M2 is a complement of S1 containing K. If S2 = soc(M1 ), L = S2 ⊕ ker f , S2 ⊆ M1 , ker f ⊆ M2 . As ker f ⊂e M2 , M2 = ker f . This gives S2 < M1 . We also get d(M1 ) = 2 and S2 = J(M1 ). Clearly, S∼ = S2 . (b) Let U be a uniserial subfactor of M of length 2. Suppose U is not M -injective. Let f : L → U , L < M be a maximal homomorphism that cannot be extended from M to U . Let K = ker f . As before there exists a minimal submodule S1 of L such that S1 ∩K = 0. By (2.8), we get M = M1 ⊕ M2 such that M1 is uniserial, S1 ⊂ M1 and K ⊆ M2 . Now K is a complement of S1 in L. Which gives K = L ∩ M2 . By (1.3), M2 = K. By (iii), f | S1 can be extended from M1 to U . By (1.3), M1 ⊆ L. Thus L = M , which is a contradiction. Hence U is M -injective. By (i), V is not essential in M . So, there exists a minimal submodule S such that S ∩ V = 0. Then V is a complement of S and by (ii), M = M1 ⊕ V for some uniserial submodule M1 containing S. 2 Theorem 2.11. Let MR be any module. Then M is completely almost self-injective if and only if it satisfies the following conditions: M (i) soc(M ) is semi-simple. (ii) If N is a subfactor of M , then any complement in N of a simple submodule S of N is a summand of N and if S ⊂ J(N ), then there exists a uniserial summand N1 of N of length 2 that contains S. (iii) Any two non-simple uniserial subfactors U , V of M are quasi-injective, and if they have isomorphic socles, they are isomorphic. (iv) If A, B are two subfactors of M such that J(A) = 0, then any monomorphism f : L → B, L ⊂e A such that f (soc(A)) ⊆ J(B) can be extended to a homomorphism from A to B. Proof. If M is completely almost self-injective, as seen above, M satisfies the given conditions. Conversely, let given conditions hold. By (i), soc(M ) ⊂e M . Let A be a subfactor of A M . By (2.7), soc(A) is semi-simple. By (2.10), any uniserial subfactor of M of length 2 is M -injective, so it is also A-injective. It follows that A satisfies conditions (i), (ii) and (iii). By (2.8), if a minimal submodule S of A is in J(A), then there exists a uniserial summand U ⊆ A such that S ⊂ U and d(U ) = 2. Let B be another subfactor of M and let there exist a maximal homomorphism f : L → B, L < A that cannot be extended from A to B. Let C = ker f . Then f induces L L f : C → B. Set L = C . Now f is a maximal homomorphism that cannot be extended A A from C to B, so A = C is not semi-simple. Suppose there exists a uniserial submodule
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of A of length 2. By (2.10)(b), V = U ⊕ C for some uniserial submodule U of V with d(U ) = 2. Then S = soc(U ) ⊆ L. We can find a complement A2 of S in A such that C ⊆ A2 and L ∩ A2 is a complement of S in L. Now A = U ⊕ A2 . Suppose f (S) ⊆ J(B). As f (S) ∩ f (A2 ∩ L) = 0, by (2.8), B = B1 ⊕ B2 , where B1 is uniserial, d(B1 ) = 2, f (S) ⊂ B1 and f (A2 ∩ L) ⊆ B2 . Now f |S can be extended from U onto B1 . By (1.3), U ⊆ L, so we take B1 = f (U ). We get h : B → U which is zero on B2 and is such that hf (x) = x for any x ∈ U . Suppose there exists a minimal submodule S ⊆ (J(A) ∩ L) such that f (S) J(B). Then L = S ⊕ L1 for some L1 < L containing C, and B = f (S) ⊕ D such that D contains f (L1 ). At the same time, we have A = U ⊕ L1 for some non-simple uniserial submodule U containing S. This gives a homomorphism h : B → U which is zero on C and it satisfies hf (x) = x for any x ∈ S. If S J(A) and f (S) J(B), then L = S ⊕ L2 and B = f (S) ⊕ B2 for some L2 < A containing C, and some B2 < B such that f (L2 ) ⊆ B2 . We get a homomorphism h : B → S, which is zero on B2 and is such that hf (x) = x for x ∈ S. Finally, we assume that J(A) = 0 and for any minimal submodule S of A, f (S) ⊆ A J(B). It follows from (2.10), that J( X ) = 0 for any submodule X of A. By (iv), f can be extended from A to B. This all proves that B is almost A-injective. Hence M is completely almost self-injective. 2 V C
M Corollary 2.12. Let MR be any module such that soc(M ) is finitely generated. Then M is completely almost self-injective if and only if it satisfies the following conditions: M (i) soc(M ) is semi-simple. (ii) If N is a subfactor of M , then any complement in N of a simple submodule S of N is a summand of N and if S ⊂ J(N ), then there exists a uniserial summand N1 of N of length 2 that contains S. (iii) Any two non-simple uniserial subfactors U , V of M are quasi-injective, and if they have isomorphic socles, then they are isomorphic.
Proof. Suppose the above three conditions are satisfied by M . As seen in the proof of the above theorem, any non-simple, uniform subfactor U of M is of length 2 and is M -injective. This property is carried over to any subfactor A of M . Now any minimal submodule S ⊆ J(M ) is contained in a uniserial submodule V of length 2, and V is a M summand of M . As soc(M ) is a finite direct sum of simple modules, we get M = M1 ⊕M2 , where M1 is such that J(M1 ) = 0 and M2 is a finite direct sum of uniserial submodules of length 2. Thus M is M2 -injective. This property carries over to any subfactor of M . Thus any subfactor A of M with J(A) = 0, is semi-simple. Using this one see that condition (iv) in (2.11) holds for M . This proves the result. 2 R Let R be a Von Neumann regular ring such that soc(R is semi-simple. Let S be a R) minimal right ideal of R. Then S is injective if and only if every complement of S is a summand of RR . It follows that if Von Neumann regular ring R is such that its every
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R minimal right ideal is injective and soc(R is semi-simple, then there does not exist any R) non-simple uniform R-module. We get the following.
Corollary 2.13. Let R be a Von Neumann regular ring. Then RR is completely almost R self-injective if and only if soc(R is semi-simple and every minimal right ideal of R is R) injective. 2 Since any simple module over a commutative Von Neumann regular ring is injective, any commutative Von Neumann regular ring R is completely almost self-injective if and R only if soc(R) is semi-simple. Example. Let F be any field and S = Π Fi , where each Fi = F and Λ is an infinite set. i∈Λ
Let R be the subring of S consisting of those (xi ) ∈ S, which have all except finitely many terms equal. Then soc(R) is the set of those (xi ) ∈ R which have finitely many terms R non-zero. One sees that soc(R) is simple. Hence R is completely almost self-injective. 2 Let MR be a finitely generated completely almost self-injective module. By (2.8), any minimal submodule S ⊆ J(M ) is contained in a uniserial summand of M of length 2. But n M Ai , where each Ai som(M ) is finitely generated and semi-simple, we get that M = ⊕ i=1
is either simple or a uniserial module of length 2, in the later case, an Ai is M -injective. Thus M is a serial module of finite length. Let R be a semi-perfect. Let R be a completely almost right self-injective ring. It follows that R is a right serial right artinian ring such that for any indecomposable idempotent e ∈ R, either eR is simple or is injective and has length 2. It follows that any non-simple indecomposable right R-module is injective and has length 2. Using this one can see, if R is such that for any indecomposable idempotent e ∈ R, either eR is simple, or is injective and has length 2, then the conditions in (2.12) are satisfied. Thus R is completely almost right self-injective. Acknowledgment The author is extremely thankful to the referee for his valuable suggestions. References [1] F.W. Anderson, K.R. Fuller, Rings and Categories of Modules, Grad. Texts in Math., vol. 13, Springer Verlag, 1974. [2] Y. Baba, Note on almost M -injectives, Osaka J. Math. 26 (1989) 687–698. [3] N.V. Dung, D.V. Huynh, P.F. Smith, R. Wisbauer, Extending Modules, Pitman Res. Notes Math. Ser., vol. 313, 1994. [4] C. Faith, Algebra II, in: Ring Theory, in: Grundlehren Math. Wiss., vol. 191, Springer Verlag, 1976. [5] K.R. Goodearl, Von Neumann Regular Rings, Monogr. Stud. Math., vol. 4, Pitman, 1979. [6] M. Harada, Almost relative projective modules and almost relative injective modules, Monograph. [7] S. Singh, Almost relative injective modules, Osaka J. Math. 53 (2016) 423–438.
Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Uniform almost relative injective modules Surjeet Singh House No. 424, Sector 35 A, Chandigarh 160036, India
a r t i c l e
i n f o
Article history: Received 4 March 2015 Available online 2 February 2017 Communicated by Louis Rowen MSC: primary 16D50 secondary 16E50 Keywords: Almost relative injectives Almost self-injectives CS modules Row-finite matrices Von Neumann regular rings
a b s t r a c t The concept of a module M being almost N -injective where N is some module, was introduced by Baba (1989). For a given module M , the class of modules N , for which M is almost N -injective, is not closed under direct sums. Baba gave a necessary and sufficient condition under which a uniform finite length module U is almost V -injective, where V is a finite direct sum of uniform, finite length modules, in terms of extending properties of simple submodules of V . Let U be a uniform module and V be a finite direct sum of indecomposable modules. Recently, Singh (2016), has determined some conditions under which U is almost V injective, which generalize Baba’s result. In the present paper some more results in this direction are proved. A module M is said to be completely almost self-injective, if for any two subfactors A, B of M , A is almost B-injective. A necessary and sufficient condition for a module M to be completely almost self-injective is given. Using this, it is proved that a Von Neumann ring R is completely almost right self-injective R if and only if soc(R is semi-simple and every minimal right R) ideal of R is injective. © 2017 Elsevier Inc. All rights reserved.
E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jalgebra.2017.01.033 0021-8693/© 2017 Elsevier Inc. All rights reserved.
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Introduction Let MR , NR be two modules. As defined by Baba [2], M is said to be almost N -injective, if for any homomorphism f : A → M , A N , either f extends to a homomorphism g : N → M or there exist a decomposition N = N1 ⊕ N2 with N1 = 0 and a homomorphism h : M → N1 such that hf (x) = π(x) for any x ∈ A, where π : N → N1 is a projection with kernel N2 . This concept plays a significant role is studying extending modules. A module M that is almost M -injective, is called an almost self-injective module. For a module M , the class of those modules N for which M is almost N -injective, is not closed under direct sums. Let {Uk : 0 ≤ k ≤ n} be a finite family of uniform modules n
of finite lengths, and U = ⊕ Σ Uk . Baba [2] has given a characterization for U0 to be k=1
almost U -injective, in terms of the property of simple submodules of U being contained in uniform summands of U . Let M be a uniform module and V be a finite direct sum of indecomposable modules. In [7], conditions under which M is almost V -injective have been investigated, thereby Baba’s result has been generalized. In Section 1, this study has been further continued. Let M be an almost self-injective, uniform modules. Let T be the set of those maximal homomorphisms f : N → M , N < M which cannot be extended to endomorphisms of M . In Section 1, an algebraic structure on T is given. As defined in Section 2, a module M is called a completely almost self-injective module, if for any two subfactors A, B of M , A is almost B-injective. In Theorem 2.4, it is proved that a uniform module U is completely almost self-injective if and only if U is of length not more than 2 and is quasi-injective. In Theorem 2.11 a necessary and sufficient condition for a module M to be completely almost self-injective is proved. From which, it follows that a Von Neumann regular ring R is completely right almost self-injective if and only R if soc(R is semi-simple and every minimal right ideal of R is injective. R) Preliminaries. All rings considered here are with unity and all modules are unital right modules unless otherwise stated. Let M be a module. Then E(M ) and J(M ) denote the injective hull and radical, respectively, of M . The symbols N M , N < M , N ⊂e M denote that N is a submodule of M , N is a submodule different from M , N is an essential submodule of M respectively. Any submodule of a homomorphic image of M is called a subfactor of M . A module M whose ring of endomorphisms End(M ) is local, is called an LE module. A module M such that its complement submodules are summands of M , is called a CS module (or a module satisfying condition (C1 )). The terminology used here is available in standard text books like [1,4]. For details on Von Neumann regular rings see [5]. 1. Direct sums of uniform modules Definition 1.1. Let MR and NR be any two modules. Then M is said to be almost N -injective, if given any R-homomorphism f : A → M , A N either f extends to an R-homomorphism from N to M or there exist a decomposition N = N1 ⊕ N2 with
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N1 = 0, and an R-homomorphism h : M → N1 such that hf (x) = π(x) for any x ∈ A, where π : N → N1 is a projection with kernel N2 . 2 Let a module MR be almost NR -injective and f : L → M , L < N be maximal in the sense that there does not exist any homomorphism g : L1 → M , L < L1 N , which extends f . Now L ⊂e N . By the definition, we get N = N1 ⊕ N2 for some N1 = 0, ker f ⊆ N2 , in this situation, f is monic on L ∩ N1 . It follows that ker f is not essential in L. From which, we get the following. If f : K → M , K < N is any homomorphism such that ker f ⊂e K, then f can be extended from N to M . The definition also gives that if N is indecomposable, then any homomorphism f : K → M , K < N with ker f = 0, can be extended from N to M . Proposition 1.2 ([7, Proposition 1.2]). (i) A module MR is almost NR -injective if and only if for any R-homomorphism f : L → M , L < N which is maximal with respect to the property that it cannot be extended from N to M , there exist a decomposition N = N1 ⊕N2 with N1 = 0, and an R-homomorphism h : M → N1 such that hf (x) = π(x) for any x ∈ L, where π : N → N1 is a projection with kernel N2 . 2 Proposition 1.3. Let AR , BR be any two modules and f : L → B, L < A be an R-homomorphism that is maximal with respect to the property that it cannot be extended from A to B. Let C be a summand of A. If L ∩ C < C, then f1 = f | L ∩ C from L ∩ C to B is a maximal homomorphism that cannot be extended from C to B. If L ∩ C ⊆ ker f , then C ⊆ L. Proof. The first part is proved in [7, Proposition 1.3]. In the later case f1 = 0, so it extends to a zero map from C to B, which proves that C ⊆ L. 2 The following is well known. (See [6].) Proposition 1.4. Let MR , NR be any two modules such that M is almost N -injective. (i) Any summand K of M is almost N -injective. (ii) If W is a summand of N , then M is almost W -injective. (iii) If N = N1 ⊕ N2 and M is not N -injective, then M is either not N1 -injective or not N2 -injective. 2 Theorem 1.5 ([7, Corollary 1.7]). Let MR be any uniform module, and NR be any module. Then M is almost N -injective if and only if for any maximal homomorphism f : L → M , L < N which cannot be extended from N to M , N = N1 ⊕ ker f such that f (N1 ∩ L) = M. 2
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Lemma 1.6. Let AR , BR be any two modules, MR a uniform module which is almost A-injective and almost B-injective. Let N = A ⊕ B. Then M is almost N -injective if and only if for any uniform summands A1 of A and B1 of B, M is almost A1 ⊕ B1 -injective. If M is A-injective, then M is almost N -injective. Proof. If M is almost N -injective, as A1 ⊕ B1 is a summand of N , M is almost A1 ⊕ B1 -injective. Conversely, suppose the given condition holds. Let f : L → M , L < N be a maximal homomorphism which cannot be extended from N to M . Let f1 = f |(L ∩ A), f2 = f |(L ∩ B). Case 1. B ⊂ L. Then L = (L ∩ A) ⊕ B and f1 : L ∩ A → M is a maximal R-homomorphism that cannot be extended from A to M . By (1.5), A = A1 ⊕ ker f1 , where A1 is uniform and f (L ∩A1 ) = M . For each b ∈ B, there exists a unique a ∈ L ∩A1 such that f (b) = f (a). We get a homomorphism π : B → L ∩ A1 such that π(b) = a for any b ∈ B, a ∈ L ∩ A, whenever f (b) = f (a). Then B = {b − π(b) : b ∈ B} ⊆ ker f and N = A1 ⊕ (ker f1 ⊕ B ) = A1 ⊕ ker f . Case 2. L ∩ A < A, L ∩ B < B. Then A = A1 ⊕ ker f1 , B = B1 ⊕ ker f2 , f (L ∩ A1 ) = M = f (L ∩ B1 ). Let T = A1 ⊕ B1 . By the hypothesis, M is almost T -injective and L ∩ T < T . Now f |(L ∩ T ) is a maximal homomorphism that cannot be extended from T to M . Hence T = C ⊕ ker(f |(L ∩ T )). Which gives N = C ⊕ ker f . Hence M is almost N -injective. In case M is A-injective, the arguments in Case 1 give that M is almost N -injective. 2 Lemma 1.7. Let A, B, M be three uniform right R-modules such that M is almost A-injective and almost B-injective, but is neither A-injective nor B-injective. If M is almost A ⊕B-injective, then any complement submodule of A ⊕B is a summand of A ⊕B. Conversely, if M is neither A-injective nor B-injective, every complement submodule of A ⊕ B is a summand of A ⊕ B, then M is almost A ⊕ B-injective. Proof. Now there exist maximal homomorphisms f1 : L1 → M , L1 < A, f2 : L2 → M , L2 < B which cannot be extended from A to M , from B to M respectively. Both fi are monomorphisms. Let C be a complement submodule of T = A ⊕ B other than T . Then C is uniform, therefore, either A ∩ C = 0 or B ∩ C = 0. Suppose A ∩ C = 0. We get a maximal R-homomorphism f : L → M , such that L1 ⊕ C ⊆ L, f extends f1 and it is zero on C. By (1.5), ker f is a summand of T . However ker f is uniform and C ⊆ ker f . Hence C = ker f . Conversely, let the given condition hold. Let f : L → M , L < T be a maximal homomorphism that cannot be extended from T to M . Let L1 = L ∩ A, L2 = L ∩ B. If either L1 = A or L2 = B, as discussed in Case 1 of the proof of (1.6), we get T = C⊕ker f , f (C ∩ L) = M . Suppose L1 < A, L2 < B. By (1.5), f (L1 ) = M = f (L2 ), which gives an isomorphism π : L2 → L1 , π(x) = y for any x ∈ L2 , y ∈ L1 , whenever f (x) = f (y). Thus K =
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{x − π(x) : x ∈ L2 } ⊆ ker f , is uniform, therefore ker f is uniform and there exists a uniform summand C of T containing ker f . Then T = C ⊕C for some uniform submodule C . If ker f < C ∩ L, then kerL f cannot be uniform, which gives a contradiction. Hence ker f = C ∩ L. By (1.3), C ⊆ L. Thus T = C ⊕ ker f . Hence M is almost T -injective. 2 The following result generalizes [7, Theorem 1.11]. Theorem 1.8. Let MR be a uniform module and NR = N1 ⊕ N2 , where Ni are indecomposable. Then M is almost N -injective if and only if one of the following holds. (i) M is N1 -injective and N2 -injective. (ii) M is almost N1 -injective and almost N2 -injective. If M is neither N1 -injective nor N2 -injective, then N is a CS module. Proof. Let MR be a uniform module and almost N -injective. By (1.4), M is almost Ni -injective for i = 1, 2. If M is neither N1 -injective nor N2 -injective, by (1.7), every complement submodule of N is a summand, i.e. N is a CS module. Conversely, let the given conditions hold. Let f : L → M , L < N be a maximal homomorphism that cannot be extended from N to M . Let Li = L ∩ Ni and L1 < N1 . Then f (L1 ) = M and N1 is uniform. If L2 = N2 , as in Case 1 in (1.6), N = N1 ⊕ ker f . In case L2 < N2 , the result follows from (1.7). 2 Lemma 1.9. Let MR be any module. Let A, B be two submodules of M , which are complements of each other, and let π : A ⊕ B → A be projection with kernel B. If π extends to an endomorphism of M , then M = A ⊕ B. Proof. Suppose π extends to an endomorphism σ of M . Now B ⊆ ker σ. Let 0 = x ∈ ker σ. For some r, 0 = xr ∈ A ⊕ B. Then σ(xr) = 0 gives xr ∈ B. Hence B ⊂e ker σ, which gives B = ker σ. Suppose A < σ(M ). As A is closed, A is not essential in σ(M ). M Now σ induces an isomorphism σ : M B → σ(M ), as A ⊂e M = B , σ(A) ⊂e σ(M ), i.e. A ⊂e σ(M ), which is a contradiction. Thus A = σ(A) = σ(M ). Hence M = A ⊕ B. 2 Proposition 1.10. Let MR be an almost self-injective module. (i) Let B be a closed submodule of M such that no complement of B contains an infinite direct sum of summands of M . Then B is a summand of M . (ii) Any almost self-injective module of finite Goldie dimension is a CS-module and is a finite direct sum of uniform submodules. (iii) If A is a uniform submodule of M and B is a complement of A, then B is a summand of M . (iv) Let S be a minimal submodule of M such that S ⊆ J(M ). Then M = M1 ⊕ M2 such that M1 is uniform but not simple, and soc(M1 ) ⊕ M2 = S ⊕ M2 .
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Proof. Let A be a complement of B and π : A ⊕ B → A be a projection with kernel B. If π extends to an endomorphism of M , by (1.9), B is a summand of M . Suppose for no complement A of B, the projection π can be extended to an endomorphism of M . By (1.1), there exists a decomposition M = N1 ⊕ M2 such that N1 = 0 and B ⊆ M2 . If B ⊂e M2 , B = M2 and we finish. Otherwise we repeat the process with M2 . We get M2 = N2 ⊕ M3 with N2 = 0 and B ⊆ M3 . Continue the process. The process cannot be continued indefinitely, for otherwise an infinite direct sum of summands of M is contained in a complement of B. Hence B is a summand of M . This proves (i). Now (ii) and (iii) are immediate from (i). (iv) Let B be any complement of S, by (i), M = M1 ⊕M2 , where M1 = 0 and B = M2 . Then J(M ) = J(M1 ) ⊕ J(M2 ) and S ⊕ J(M2 ) ⊆ J(M ). Thus M1 cannot be simple. As S embeds in M1 , soc(M1 ) ∼ = S. It is obvious that soc(M1 ) ⊕ M2 = S ⊕ M2 . 2 Let MR be a uniform almost self-injective module. Let T be the set of those maximal homomorphisms f : L → M , L < M which cannot be extended to endomorphisms of M . We give an algebraic structure in T modulo the additive group of endomorphisms of M . Proposition 1.11. Let a module MR be almost NR -injective. Let f1 : L1 → M , L1 N be a maximal homomorphism in the sense that it has no proper extension f : G → M , G N . Let f2 : L2 → M , L2 N be another homomorphism. (i) Let μ : L1 ∩ L2 → M , μ(x) = f1 (x) − f2 (x) for x ∈ L1 ∩ L2 . If g : L → M, L M is a maximal extension of μ, then L ∩ L2 = L1 ∩ L2 . (ii) If f2 is also maximal, then L ∩ L1 = L1 ∩ L2 = L ∩ L2 . In particular, if L = N , then L1 = L2 . (iii) Let 0 = f : H → M , H ⊂e N be a homomorphism and f1 : L1 → M , L1 M be a maximal extension of f . Let f2 : L2 → M , L2 M be another maximal extension of f . Then L2 = L1 . (iv) Let M be almost self-injective and uniform. Let 0 = f : H → M , H M be a homomorphism and f1 : L1 → M , L1 M , and f2 : L2 → M , L2 M be any maximal extension of f . Then L1 = L2 . Let f : L → M , L < M be a homomorphism. If for some 0 < K L, f |K can be extended to an endomorphism of M , then f can also be extended to an endomorphism of M . Proof. Let K = L ∩ L2 . Then K ∩ L1 = L ∩ L1 ∩ L2 = L1 ∩ L2 . Now g(x) and f2 (x) are defined for any x ∈ K. We get β : K → M , β(x) = g(x) + f2 (x), x ∈ K. Clearly β|(K ∩ L1 ) = f1 |(K ∩ L1 ), so we get β1 : K + L1 → M , β1 (x + y) = β(x) + f1 (y) for x ∈ K, y ∈ L1 . As f is maximal, we get K ⊆ L1 . Hence L1 ∩ L = L1 ∩ L2 . This proves (i) and (ii). (iii) Let μ be as in (i). As H ⊆ ker μ, μ extends to a homomorphism g : N → M . By (ii), L1 = L2 . It follows that if f2 is also maximal, then L1 ∩ L = L1 ∩ L2 = L2 ∩ L, in particular, if we have L = N , then L1 = L2 . Now (iv) is immediate from (iii). 2
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Let MR be an almost self-injective uniform module, T be set of those maximal homomorphisms f : L → M , L < M , which cannot be extended to endomorphisms of M . We give an algebraic structure on T that makes it an abelian group modulo S = End(M ). The above result shows that given any homomorphism g : H → M , 0 < H < M , there exists a unique L M such that for any maximal extension h : K −→ M , K M of g, K = L. This L is called a maximal domain of g. It follows that, if g extends to an endomorphism of M , then any other maximal extension of g is an endomorphism of M . For any two f1 : L1 → M , f2 : L2 → M members of T ∪ S, define f1 ∼ f2 if for some non-zero K L1 ∩ L2 , λ : K → M , λ(x) = f1 (x) − f2 (x) can be extended to an endomorphism of M . It is an equivalence relation. It follows that if f1 ∼ f2 , then L1 = L2 . So, to each equivalence class [f ] is associated an L M such that any g ∈ [f ] has domain L. We have [0] = S. Let G be the set of these equivalence classes. Let f1 , f2 , f3 ∈ T ∪ S with L1 , L2 , L3 as their respective domains. Let μ be f1 + f2 defined on L1 ∩ L2 and g : L → M , L M be a maximal extension of μ. Set [f1 ] + [f2 ] = [g]. It follows from (1.11) that the addition is well defined. It is clear from the definition that if [f1 ] + [f2 ] = [h], then for some 0 = K L1 ∩ L2 ∩ dom(g), f1 + f2 − h defined on K can extended to an endomorphism of M . Now ([f1 ] + [f2 ]) + [f3 ] = [g] + [f3 ] = [k] for some k ∈ T ∪ S. There exist non-zero submodules K1 , K2 of M such that f1 + f2 − g is naturally defined on K1 and g + f3 − k is naturally defined on K2 and they are extendable to some endomorphisms of M . By (1.11)(iv), f1 + f2 + f3 − k defined on K = K1 ∩ K2 can be extended to an endomorphism of M . Similarly, if [f1 ] + ([f2 ] + [f3 ]) = [k1 ]. There exists a non-zero K M on which f1 + f2 + f3 − k1 is naturally defined and is extendable to an endomorphism of M . From which it follows that k − k1 naturally defined on K ∩ K can be extended to an endomorphism of M . Which proves that addition is associative. [0] is the additive identity in G. We get that G is an abelian group. 2 2. Completely almost self-injective modules Definition 2.1. A module MR is said to be completely almost self-injective, if for any two subfactors U , V of M , U is almost V -injective. 2 Lemma 2.2. Let MR be a completely almost self-injective module. (i) M N is semisimple for any N ⊂e M . (ii) Let SR be a simple module which is a subfactor of M . Then either S is M -injective or given any maximal homomorphism f : L → S, L < M which cannot be extended from M to S, M = M1 ⊕ ker f , for some M1 which is uniform but not simple, soc(M1 ) ∼ =S and L = soc(M1 ) ⊕ ker f . (iii) J(M ) ⊆ soc(M ) and soc(M ) is an intersection of maximal submodules. Proof. (i) Now every homomorphic image of M is essentially M -injective. Let N ⊂e L M . Let L be any submodule of M containing N . Then N is almost M N -injective. It
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L L M M follows from (1.1) that N is M N -injective. Therefore N is a summand of N . Hence N is semi-simple. (ii) Now S is almost M -injective. Let f : L → S, L ⊂e M be a maximal homomorphism. Suppose f cannot be extended from M to S. Let A = ker f . Then A is not essential in L. Hence L = S1 ⊕ ker f , where S1 ∼ = S. By (1.10)(iii), M = M1 ⊕ M2 such that ker f ⊂e M2 , M1 is uniform and soc(M1 ) ∼ = S1 . Clearly, ker f = L ∩ M2 . By (1.3), ker f = M2 . Which gives that M1 is not simple. Then L = S1 ⊕ M2 . (iii) For any essential submodule N of M , as M N is semi-simple, J(M ) ⊆ N . Which gives J(M ) ⊆ soc(M ). It also gives that any essential submodule of M is an intersection of maximal submodules. Therefore soc(M ) is also an intersection of maximal submodules. 2
Theorem 2.3. Let MR be a completely almost self-injective module. Let S be a simple module which is a subfactor of M . Then either S is M -injective or there exists a decomposition M = M1 ⊕ M2 such that M2 is a complement of S and M1 is uniserial and has length 2. If J(M ) = 0, then S is M -injective. Proof. Let S be not M -injective. By (2.2), M = M1 ⊕ M2 such that M1 is uniform 1 and M2 is a complement of S. Thus soc(M1 ) ∼ = S. By (2.2), M S1 is semi-simple, where M1 S1 = soc(M1 ). Suppose d( S1 ) > 2. Then we get two different uniserial submodules A, B of M1 such that d(A) = d(B) = 2. As A is almost self-injective, it must be quasi-injective. Similarly, B is quasi-injective. Suppose A and B are not isomorphic. However B is almost A-injective, there exists a homomorphism h : A → B such that h(x) = x for any x ∈ S1 . Then h : A → B is an isomorphism, which gives a contradiction. Hence A ∼ = B, which also gives A = B. Hence d(M1 ) = 2. This also gives J(M1 ) = 0 and therefore J(M ) = 0. 2 Theorem 2.4. Any completely almost self-injective, uniform module U is either simple or of length 2. Proof. Suppose soc(U ) = 0. Let 0 = x ∈ U . We have a maximal submodule T of xR. xR Then T = 0. As xR T is almost U -injective, we get an epimorphism f : U → T extending xR the natural map from xR to T . This gives that U has a maximal submodule K1 such that KU1 ∼ = xR T . Suppose U is not quasi-injective. By (1.5), there exists a monomorphism f : W → U , W < U such that f (W ) = U . Set W1 = f −1 (K1 ). Then W1 < W . As W1 H soc(W1 ) = 0, we get a 0 < V < W1 . By (2.2)(i), VU is semi-simple. We have W V = V ⊕ V , ∼ W where H V = W1 . Then f (H) K1 . Now K1 is almost H-injective. Let g = f | V : V → K1 . Suppose g extends to a homomorphism g : H → K1 . Now g = f | H, g − (f | H) is zero on V , so its image in U is semi-simple. Which is a contradiction. Thus g does not have an extension from H to K1 . By (1.5), there exists a monomorphism λ : W → K1 , W < H, which is an extension of g, such that λ(W ) = K1 . Once again λ = f | W , that gives that soc(U ) = 0. Hence U is quasi-injective. Suppose there exists a maximal submodule K2 of U different from
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K1 . As for U , K1 and K2 are also quasi-injective. We have a maximal homomorphism f : A → K2 , A K1 which extends the inclusion map on K1 ∩ K2 . Now f extends to an automorphism g of U . If A = K1 , we get f (K1 ) = K2 , K1 = K2 , a contradiction. Hence A < K1 . As K2 is almost K1 -injective, f (A) = K2 . Then g(K1 ) = U , which is also a contradiction. Hence U has only one maximal submodule K1 . Thus J(U ) = K1 . By (2.2)(iii), J(U ) ⊆ soc(U ) = 0. Which gives a contradiction. Hence soc(U ) = 0. Let S = soc(U ). Suppose U = S. Then S is not U -injective. Therefore, by (2.3), d(U ) = 2. 2 Remark. A related result is that for a ring R, finite direct sum of uniform modules is extending if and only if any uniform module over R is of length not more than 2 [3, 13.1]. Theorem 2.5. Let MR be any completely almost self-injective module. Then semi-simple and soc(M ) ⊂e M .
M soc(M )
is
Proof. Firstly, let M be finitely generated. By (2.2)(i), M E is a finite direct sum of simple modules for any E ⊂e M . Let H be a complement of soc(M ) in M . Then soc(H) = 0. Suppose H contains an infinite direct sum. Then H contains an infinite direct sum L = ⊕ Xi , which is essential in H. For each Xi , there exists proper submodule Yi of Xi i that is essential in Xi . Then L1 = ⊕ Yi ⊕ soc(M ) ⊂e H ⊕ soc(M ) ⊂e M . Thus LM1 is of i
finite length. However LL1 is not of finite length. This gives a contradiction. Hence H is of finite Goldie dimension. As H is almost self-injective, by (1.10)(ii), H = A1 ⊕A2 ⊕... ⊕At , where each Ai is uniform. By (2.4), each Ai has length 2. Thus soc(H) = 0, unless H = 0. M Hence soc(M ) ⊂e M and soc(M ) is semi-simple. Let M need not be finitely generated. Suppose soc(M ) is not essential in M . We can find a non-zero x ∈ M such that soc(M ) ∩ xR = 0. Thus soc(xR) = 0. As xR is also finitely generated completely almost self-injective, soc(xR) = 0, which is a contradiction. M Thus soc(M ) ⊂e M . By (2.2)(i), soc(M ) is semi-simple. 2 Lemma 2.6. Let MR be a completely almost self-injective module, and U , V be two nonsimple uniform subfactors of M . (i) d(U ) = d(V ) = 2. (ii) U is M -injective and isomorphic to a summand of M . (iii) If soc(U ) ∼ =V. = soc(V ), then U ∼ Proof. By (2.4), d(U ) = d(V ) = 2. Suppose U is not M -injective. Let f : L → U , L < M be a maximal homomorphism that cannot be extended from M to U . We have a decomposition M = M1 ⊕ M2 such that f is one-to-one on M1 ∩ L, f (M1 ∩ L) = U and ker f = M2 . As M1 is uniform, by (2.4), d(M1 ) = 2. Which gives that M1 = M1 ∩ L ⊆ L. Hence L = M , which is a contradiction. Hence U is M -injective. There exists a factor T of M containing U . As U is T -injective, it is a summand of T . Therefore U is a factor of M . We get an epimorphism g : M → U . Let K = ker f . Then
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K is not essential in M . We get a minimal submodule S of M such that S ∩ K = 0. Now K is a complement of S in M . By (1.10), M = K1 ⊕ K with K1 uniform. We get U∼ = K1 . This proves (i) and (ii). (iii) We have an isomorphism f : soc(U ) → soc(V ). By (1.5), if f does not extend from U to V , f (soc(U )) = V , which is a contradiction. Hence U ∼ =V. 2 M Lemma 2.7. Let MR be any module such that soc(M ) is semi-simple. Then: (i) Soc(M ) ⊂e M . (ii) Let N be a module which embeds in M or is a homomorphic image of M . Then N M N soc(N ) ⊂e N and soc(N ) is semi-simple. If soc(M ) is finitely generated, then soc(N ) is also finitely generated. M L (iii) If soc(M ) is finitely generated, then for any subfactor L of M , soc(L) is finitely generated and soc(L) ⊂e L. M Proof. (i) Let K be a complement of soc(M ). As K embeds in soc(M ) , K is semi-simple. Which gives soc(M ) ⊂e M . (ii) Let N be a submodule of M . As soc(M ) ⊂e M , soc(N ) = soc(M ) ∩ N ⊂e N . As N M N soc(N ) embeds in soc(M ) , soc(N ) is semi-simple. Thus soc(N ) ⊂e N . Let f : M → K be an epimorphism and B = ker f . Now soc(M ) = H ⊕ soc(B) for some semi-simple submodule H of M . Thus H+B ⊆ soc( M B B ). As soc(B) ⊂e B, H ∩B = 0. M K But H ⊕ B ⊂e M . Therefore H⊕B is semi-simple, which also gives that soc( soc(K) ) is M K semi-simple. If soc(M ) is finitely generated, it also follows that soc(K) is finitely generated. (iii) It follows from (ii). 2
Lemma 2.8. Let MR be any module such that it satisfies the following conditions. M (i) soc(M ) is semi-simple. (ii) Any complement of a minimal submodule of M is a summand of M . (iii) Any uniserial submodule of M of length 2 is quasi-injective and any two nonsimple uniserial submodules with isomorphic socles are isomorphic. Then for any minimal submodule S ⊆ J(M ) and a complement M2 of S, M = M1 ⊕ M2 for some uniserial submodule M1 of length 2 with S ⊂ M1 . Any non-simple uniform submodule of M is of length 2 and is a summand of M . Proof. Let S ⊆ J(M ) be a minimal submodule. Let M2 be a complement of S in M . By the hypothesis, M = N1 ⊕ M2 for some uniform submodule N1 . Let π1 : M → N1 be a projection with kernel M2 . Then for S1 = π1 (S) = soc(N1 ), S ⊕ M2 = S1 ⊕ M2 . Which 1 gives that N1 is not simple. By (2.7), N S1 is semi-simple. By (iii), any two non-simple uniserial submodules of N1 are isomorphic, and they are quasi-injective, it follows that N1 itself is uniserial and hence has length 2. Let T be a maximal direct sum in M of uniserial submodules of length 2. Now T = ⊕ Ai , where each Ai is uniserial and i∈Λ
has length 2. Let S = xR be a minimal submodule of M contained in J(M ). Suppose S ∩ T = 0. By (ii) M = M1 ⊕ M2 where M2 is a complement of S containing T . As
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M1 is of length 2 and M1 ∩ T = 0, we get a contradiction. Thus S ⊂ T . We get finitely many distinct elements say 1, 2, ..., k in Λ such that x = x1 + x2 + ... + xk for some non-zero xi ∈ Ai . For any i ≥ 2, we have monomorphism σi : soc(A1 ) → soc(Ai ) such that σi (x1 ) = xi . By (iii), σi can be extended to an isomorphism ηi : A1 → Ai . Now A1 = y1 R for some y1 ∈ A1 . Let y = y1 + y2 + ... + yk , where for each i ≥ 2, yi = ηi (y1 ). Then K = yR is uniserial and S ⊂ yR. The second part is immediate. 2 Another interesting property of a completely almost self-injective module is given by the following. Proposition 2.9. Let MR be a completely almost self-injective module. Let A, B be two subfactors of M such that J(A) = 0 and f : L → B, L ⊂e A is a monomorphism such that for any minimal submodule S of A, f (S) ⊆ J(B), then f can be extended to a homomorphism from A to B. Proof. By the hypothesis B is almost A-injective. By (2.4) and (2.6), any non-simple uniform submodule U of A has length 2 and is A-injective, it is a summand of A. However J(A) = 0. Therefore A has no non-simple uniform submodule. By (1.10), any complement of a minimal submodule of A is a summand of A. Condition (iii) in (2.8) is vacuously satisfied by A. It follows that A satisfies the hypothesis of (2.8). Similarly for B. Thus by (2.8), any minimal submodule of J(B) is contained in a uniserial submodule of length 2. Suppose f : L → B, L ⊂e A, cannot be extended from A to B. By definition, we have a decomposition A = A1 ⊕ A2 and a homomorphism h : B → A1 such that A1 = 0 and hf (x) = π1 (x) for x ∈ L, where π1 : A → A1 is a projection with kernel A2 . Let S be a minimal submodule of A contained in A1 . As f (S) ⊆ J(B), by (2.8), there exists a uniserial submodule U of B such that f (S) < U . Then U ∼ = h(U ) ⊆ A. Thus J(A) = 0, which is a contradiction. Hence f can be extended to a homomorphism from A to B. 2 We shall give a necessary and sufficient condition for a module to be completely almost self-injective Lemma 2.10. Let MR be a module satisfying the following conditions. M (i) soc(M ) is semi-simple. (ii) Any complement in M of a minimal submodule S of M is a summand of M . If a minimal submodule S ⊆ J(M ), then there exists a uniserial submodule U of M of length 2 containing S. (iii) Any uniserial subfactor of M of length 2 is quasi-injective. Any two non-simple uniserial subfactors U , V are isomorphic, whenever they have isomorphic socles. Then the following hold. (a) Any simple module SR is either M -injective or there exists a uniserial summand U of M of length 2 such that S ∼ = soc(U ). (b) Any uniserial subfactor of M of length 2 is M -injective. For any V < M such that M V is uniserial and of length 2, V is a summand of M .
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Proof. (a) It follows from the given conditions that any non-simple uniform submodule of M is uniserial and has length 2. Let S be any simple module. Suppose S is not M -injective. Then there exists a maximal homomorphism f : L → S, L < M , which cannot be extended from M to S. Let K = ker f . Then K is not essential in M for otherwise M K being semi-simple leads to a contradiction. We get a minimal submodule S1 ⊂ L such that S1 ∩ K = 0. By (ii) M = M1 ⊕ M2 such that M2 is a complement of S1 containing K. If S2 = soc(M1 ), L = S2 ⊕ ker f , S2 ⊆ M1 , ker f ⊆ M2 . As ker f ⊂e M2 , M2 = ker f . This gives S2 < M1 . We also get d(M1 ) = 2 and S2 = J(M1 ). Clearly, S∼ = S2 . (b) Let U be a uniserial subfactor of M of length 2. Suppose U is not M -injective. Let f : L → U , L < M be a maximal homomorphism that cannot be extended from M to U . Let K = ker f . As before there exists a minimal submodule S1 of L such that S1 ∩K = 0. By (2.8), we get M = M1 ⊕ M2 such that M1 is uniserial, S1 ⊂ M1 and K ⊆ M2 . Now K is a complement of S1 in L. Which gives K = L ∩ M2 . By (1.3), M2 = K. By (iii), f | S1 can be extended from M1 to U . By (1.3), M1 ⊆ L. Thus L = M , which is a contradiction. Hence U is M -injective. By (i), V is not essential in M . So, there exists a minimal submodule S such that S ∩ V = 0. Then V is a complement of S and by (ii), M = M1 ⊕ V for some uniserial submodule M1 containing S. 2 Theorem 2.11. Let MR be any module. Then M is completely almost self-injective if and only if it satisfies the following conditions: M (i) soc(M ) is semi-simple. (ii) If N is a subfactor of M , then any complement in N of a simple submodule S of N is a summand of N and if S ⊂ J(N ), then there exists a uniserial summand N1 of N of length 2 that contains S. (iii) Any two non-simple uniserial subfactors U , V of M are quasi-injective, and if they have isomorphic socles, they are isomorphic. (iv) If A, B are two subfactors of M such that J(A) = 0, then any monomorphism f : L → B, L ⊂e A such that f (soc(A)) ⊆ J(B) can be extended to a homomorphism from A to B. Proof. If M is completely almost self-injective, as seen above, M satisfies the given conditions. Conversely, let given conditions hold. By (i), soc(M ) ⊂e M . Let A be a subfactor of A M . By (2.7), soc(A) is semi-simple. By (2.10), any uniserial subfactor of M of length 2 is M -injective, so it is also A-injective. It follows that A satisfies conditions (i), (ii) and (iii). By (2.8), if a minimal submodule S of A is in J(A), then there exists a uniserial summand U ⊆ A such that S ⊂ U and d(U ) = 2. Let B be another subfactor of M and let there exist a maximal homomorphism f : L → B, L < A that cannot be extended from A to B. Let C = ker f . Then f induces L L f : C → B. Set L = C . Now f is a maximal homomorphism that cannot be extended A A from C to B, so A = C is not semi-simple. Suppose there exists a uniserial submodule
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of A of length 2. By (2.10)(b), V = U ⊕ C for some uniserial submodule U of V with d(U ) = 2. Then S = soc(U ) ⊆ L. We can find a complement A2 of S in A such that C ⊆ A2 and L ∩ A2 is a complement of S in L. Now A = U ⊕ A2 . Suppose f (S) ⊆ J(B). As f (S) ∩ f (A2 ∩ L) = 0, by (2.8), B = B1 ⊕ B2 , where B1 is uniserial, d(B1 ) = 2, f (S) ⊂ B1 and f (A2 ∩ L) ⊆ B2 . Now f |S can be extended from U onto B1 . By (1.3), U ⊆ L, so we take B1 = f (U ). We get h : B → U which is zero on B2 and is such that hf (x) = x for any x ∈ U . Suppose there exists a minimal submodule S ⊆ (J(A) ∩ L) such that f (S) J(B). Then L = S ⊕ L1 for some L1 < L containing C, and B = f (S) ⊕ D such that D contains f (L1 ). At the same time, we have A = U ⊕ L1 for some non-simple uniserial submodule U containing S. This gives a homomorphism h : B → U which is zero on C and it satisfies hf (x) = x for any x ∈ S. If S J(A) and f (S) J(B), then L = S ⊕ L2 and B = f (S) ⊕ B2 for some L2 < A containing C, and some B2 < B such that f (L2 ) ⊆ B2 . We get a homomorphism h : B → S, which is zero on B2 and is such that hf (x) = x for x ∈ S. Finally, we assume that J(A) = 0 and for any minimal submodule S of A, f (S) ⊆ A J(B). It follows from (2.10), that J( X ) = 0 for any submodule X of A. By (iv), f can be extended from A to B. This all proves that B is almost A-injective. Hence M is completely almost self-injective. 2 V C
M Corollary 2.12. Let MR be any module such that soc(M ) is finitely generated. Then M is completely almost self-injective if and only if it satisfies the following conditions: M (i) soc(M ) is semi-simple. (ii) If N is a subfactor of M , then any complement in N of a simple submodule S of N is a summand of N and if S ⊂ J(N ), then there exists a uniserial summand N1 of N of length 2 that contains S. (iii) Any two non-simple uniserial subfactors U , V of M are quasi-injective, and if they have isomorphic socles, then they are isomorphic.
Proof. Suppose the above three conditions are satisfied by M . As seen in the proof of the above theorem, any non-simple, uniform subfactor U of M is of length 2 and is M -injective. This property is carried over to any subfactor A of M . Now any minimal submodule S ⊆ J(M ) is contained in a uniserial submodule V of length 2, and V is a M summand of M . As soc(M ) is a finite direct sum of simple modules, we get M = M1 ⊕M2 , where M1 is such that J(M1 ) = 0 and M2 is a finite direct sum of uniserial submodules of length 2. Thus M is M2 -injective. This property carries over to any subfactor of M . Thus any subfactor A of M with J(A) = 0, is semi-simple. Using this one see that condition (iv) in (2.11) holds for M . This proves the result. 2 R Let R be a Von Neumann regular ring such that soc(R is semi-simple. Let S be a R) minimal right ideal of R. Then S is injective if and only if every complement of S is a summand of RR . It follows that if Von Neumann regular ring R is such that its every
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R minimal right ideal is injective and soc(R is semi-simple, then there does not exist any R) non-simple uniform R-module. We get the following.
Corollary 2.13. Let R be a Von Neumann regular ring. Then RR is completely almost R self-injective if and only if soc(R is semi-simple and every minimal right ideal of R is R) injective. 2 Since any simple module over a commutative Von Neumann regular ring is injective, any commutative Von Neumann regular ring R is completely almost self-injective if and R only if soc(R) is semi-simple. Example. Let F be any field and S = Π Fi , where each Fi = F and Λ is an infinite set. i∈Λ
Let R be the subring of S consisting of those (xi ) ∈ S, which have all except finitely many terms equal. Then soc(R) is the set of those (xi ) ∈ R which have finitely many terms R non-zero. One sees that soc(R) is simple. Hence R is completely almost self-injective. 2 Let MR be a finitely generated completely almost self-injective module. By (2.8), any minimal submodule S ⊆ J(M ) is contained in a uniserial summand of M of length 2. But n M Ai , where each Ai som(M ) is finitely generated and semi-simple, we get that M = ⊕ i=1
is either simple or a uniserial module of length 2, in the later case, an Ai is M -injective. Thus M is a serial module of finite length. Let R be a semi-perfect. Let R be a completely almost right self-injective ring. It follows that R is a right serial right artinian ring such that for any indecomposable idempotent e ∈ R, either eR is simple or is injective and has length 2. It follows that any non-simple indecomposable right R-module is injective and has length 2. Using this one can see, if R is such that for any indecomposable idempotent e ∈ R, either eR is simple, or is injective and has length 2, then the conditions in (2.12) are satisfied. Thus R is completely almost right self-injective. Acknowledgment The author is extremely thankful to the referee for his valuable suggestions. References [1] F.W. Anderson, K.R. Fuller, Rings and Categories of Modules, Grad. Texts in Math., vol. 13, Springer Verlag, 1974. [2] Y. Baba, Note on almost M -injectives, Osaka J. Math. 26 (1989) 687–698. [3] N.V. Dung, D.V. Huynh, P.F. Smith, R. Wisbauer, Extending Modules, Pitman Res. Notes Math. Ser., vol. 313, 1994. [4] C. Faith, Algebra II, in: Ring Theory, in: Grundlehren Math. Wiss., vol. 191, Springer Verlag, 1976. [5] K.R. Goodearl, Von Neumann Regular Rings, Monogr. Stud. Math., vol. 4, Pitman, 1979. [6] M. Harada, Almost relative projective modules and almost relative injective modules, Monograph. [7] S. Singh, Almost relative injective modules, Osaka J. Math. 53 (2016) 423–438.