In this paper, we give some results on (1 − C2)−modules and 1−continuous modules

Le Van An,

Ngo Sy Tung,

Nguyen Thi Hai Anh,

Okayama University Scientific Achievement Repository

Math. J. Okayama Univ. 59 (2017), 141–147ON THE (1− C2) CONDITIONLe Van An, Nguyen Thi Hai Anh and Ngo Sy TungAbstract. In this paper, we give some results on (1 − C2)−modulesand 1−continuous modules.1. IntroductionAll rings are associated with identity, and all modules are unital rightmodules. By MR, (RM) we indicate that M is a right (left) module over aring R. The Jacobson radical, the uniform dimension and the endomorphismring of M are denoted by J(M), u−dim(M) and End(M), respectively. Fora module M (over a ring R), we consider the following conditions:(1−C1) Every uniform submodule of M is essential in a direct summandof M .(C1) Every submodule of M is essential in a direct summand of M .(C2) Every submodule isomorphic to a direct summand of M is itself adirect summand of M .(C3) For any direct summands A,B of M with A ∩B = 0, A⊕B is alsoa direct summand of M .A module M is defined to be a (1 − C1)−module if it satisfies thecondition(1 − C1). If M satisfies (C1), then M is said to be a CS−module(or an extending module). M is defined to be a continuous module if it sat-isfies the conditions (C1) and (C2) . If M satisfies (C1) and (C3), then M issaid to be a quasi −continuous module. We call a module M a (C2)−moduleif it satisfies the condition (C2). We have the following implications:Injective ⇒ quasi −injective ⇒ continuous ⇒ quasi −continuous ⇒ CS⇒ (1− C1),and (C2) ⇒ (C3).For a set A and a module M , M (A) denotes the direct sum of | A | copiesof M . A module M is called a (countably)∑−quasi − injective if M (A)(resp. M (N)) is a quasi − injective −module for every set A (note that Ndenotes the set of all natural numbers). Similarly, a module M is called a(countably)∑−(1 − C1) if M (A) (resp. M (N)) is a (1 − C1)−module forevery set A.Mathematics Subject Classification. Primary 16D50; Secondary 16P20.Key words and phrases. injective module, continuous module, uniform module, UCmodule, distributive module.141142 L. AN, N. ANH AND N. TUNGIn Section 2, we give several properties on the (1 − C2)−modules,(strongly) 1−continuous modules, and discuss the question of when a1−continuous module is continuous ((1− C2)− module is (C2)−module)?2. (1− C2) conditionIn this section, we consider the following condition for a module M .(1−C2) Every uniform submodule isomorphic to a direct summand of Mis itself a direct summand of M .A module M is defined to be a (1−C2)−module if it satisfies the condition(1 − C2). If M satisfies (1 − C1) and (1 − C2) conditions, then M is saidto be a 1−continuouis module. M is defined to be a strongly 1−continuousmodule if it satisfies the conditions (C1) and (1−C2) . A ring R is called aright (left) 1−continuous ring if RR (resp. RR) is a 1−continuous module.We have the following implications:Continuous ⇒ strongly 1−continuous ⇒ 1−continuous,and (C2) ⇒ (1− C2).Remark 2.1. By [4, Corollary 7.8], let M be a right R−module withfinite uniform dimension, M is a (1− C1)− module if and only if M is CS.Therefore, M has finite uniform dimension thenM is a 1−continuous moduleif and only if M is strongly 1−continuous. In general, if M satisfies thecondition (1−C2), M may not satisfy the condition (C2). By the definitions(1−C2)−module, 1−continuous module and strongly 1−continuous module,we have:Lemma 2.2. Let M be a right R−module and N is a direct summand ofM . If M is a (1−C2)−module (1−continuous, strongly 1−continuous) thenN is also (1− C2)−module (resp. 1−continuous, strongly 1−continuous).Theorem 2.3. Let U = ⊕ni=1Ui where each Ui is a uniform module, thenthe following conditions are equivalent:(i) U is a (C2)−module;(ii) U is a (1− C2)−module and U satisfies the condition (C3).Proof. (i) =⇒ (ii). It is obvious(ii) =⇒ (i). We show that U is a (C2)−module, i.e., for two submodulesX,Y of U , with X ∼= Y and Y is a direct summand of U , X is also a directsummand of U . Note that Y is a closed submodule of M , there is a subsetF of {1, .., n} such that Y ⊕ (⊕i∈FUi) is an essential submodule of U . ButY , ⊕i∈FUi are direct summands of U and U satisfies the condition (C3), weimply Y ⊕ (⊕i∈FUi) = U . If F = {1, .., n} then X = Y = 0, as desired.If F 6= {1, .., n} and set J = {1, .., n}\F , then U = Y ⊕ (⊕i∈FUi) =(⊕i∈JUi)⊕ (⊕i∈FUi). Hence, X ∼= Y ∼= U/⊕i∈F Ui ∼= ⊕i∈JUi = Z. Supposethat J = {1, .., k} with 1 ≤ k ≤ n, i.e., Z = U1 ⊕ ..⊕ Uk. Let ϕ : Z −→ X,and set Xi = ϕ(Ui) then Xi ∼= Ui for any i = 1, .., k. We imply X = ϕ(Z) =ON THE (1− C2) CONDITION 143ϕ(U1 ⊕ .. ⊕ Uk) = ϕ(U1) ⊕ .. ⊕ ϕ(Uk) = X1 ⊕ .. ⊕ Xk. By Xi is a uniformsubmodule of U , Xi ∼= Ui with Ui is a direct summand of U and U is a(1−C2)−module, Xi is also a direct summand of U for any i = 1, .., k. ButU satisfies the condition (C3), X = X1 ⊕ ..⊕Xk is a direct summand of U .Hence U is a (C2)−module, proving (i). Theorem 2.4. Let U = ⊕ni=1Ui where each Ui is a uniform module, thenthe following conditions are equivalent:(i) U is a continuous module;(ii) U is a 1−continuous module.Proof. (i) =⇒ (ii). It is obvious.(ii) =⇒ (i). We show that S = End(Ui) is a local ring for any i = 1, .., n.We first prove a claim that Ui does not embed in a proper submodule of Ui.Let f : Ui −→ Ui be a monomorphism with f(Ui) is a proper submodule ofUi. Set f(Ui) = V , then V 6= 0, proper submodule of Ui and V ∼= Ui. Byhypothesis, Ui is a (1 − C2)−module, and hence V is a direct summand ofUi, i.e., Ui is not uniform module, a contradiction. Therefore, Ui does notembed in a proper submodule of Ui.Let g ∈ S and suppose that g is not an isomorphism. It suffices to showthat 1− g is an isomorphism. Note that, g is not a monomorphism. Then,since Keg(g) is a nonzero submodule, it is essential in the uniform moduleUi. We always have Keg(g)∩Keg(1−g) = 0, it follows that Ker(1−g) = 0,i.e. 1−g is a monomorphism. But Ui does not embed in a proper submoduleof Ui, 1− g must be onto, and so 1− g is an isomorphism, as required.Let Uij = Ui ⊕ Uj with i, j ∈ {1, .., n} and i 6= j. We show that Uijsatisfies the condition (C3), i.e., for two direct summands S1, S2 of Uij withS1 ∩ S2 = 0, S1 ⊕ S2 is also a direct summand of Uij . Note that, sinceu− dim(Uij) = 2, the following cases are trivial:1) Either one of the S′i has uniform dimension 2, consequently the otherSi is zero, or2) One of the S′i is zeroHence we consider the case that both S1, S2 are uniform. We prove thatUi does not embed in a proper submodule of Uj . Let h : Ui −→ Uj be amonomorphism with h(Ui) is a proper submodule of Uj . Set h(Ui) = L,then L 6= 0, proper submodule of Uj and L ∼= Ui. By hypothesis, U is a (1−C2)−module and Uij is a direct summand of U , Uij is also (1−C2)−module.Note that L is a uniform submodule of Uij and L ∼= Ui with Ui is a directsummanmd of Uij , L is also direct summand of Uij . Set Uij = L⊕ L′, thenby modularity we get Uj = L ⊕ L′′ with L′′ = Uj ∩ L′. Note that L′′ isalso proper submodule of Uj and L′′ 6= 0, hence Uj is not uniform module,a contradiction. Therefore Ui does not embed in a proper submodule of Uj .144 L. AN, N. ANH AND N. TUNGSimilary, Uj does not embed in a proper submodule of Ui. Note that, Ui(and Uj) does not embed in a proper submodule of Ui (resp. Uj).Note that, End(Ui) and End(Uj) are local rings, by Azumaya’s Lemma([1, 12.6, 12.7]), we have Uij = S2⊕K = S2⊕Ui or S2⊕K = S2⊕Uj . Sincei and j can interchange with each other, we need only consider one of thetwo possibilities. Let us consider the case Uij = S2⊕K = S2⊕Ui = Ui⊕Uj .Then it follows S2 ∼= Uj . Write Uij = S1⊕H = S1⊕Ui or S1⊕H = S1⊕Uj .If Uij = S1 ⊕H = S1 ⊕Ui, then by modularity we get S1 ⊕ S2 = S1 ⊕Wwhere W = (S1⊕S2)∩Ui. From here we get W ∼= S2, this means Ui containsa copy of S2 ∼= Uj . By Uj does not embed in a proper submodule of Ui, wemust have W = Ui, and hence S1 ⊕ S2 = Ui ⊕ Uj = Uij .If Uij = S1⊕H = S1⊕Uj , then by modularity we get S1⊕S2 = S1⊕W ′where W ′ = (S1 ⊕ S2) ∩ Uj . From here we get W ′ ∼= S2, this means Ujcontains a copy of S2 ∼= Uj . By Uj does not embed in a proper submoduleof Uj , we must have W′ = Uj , and hence S1 ⊕ S2 = Uij .Thus Uij satisfies (C3). Note that, Uij is a direct summand of U and Uis a CS −module (by U has finite dimension and U is a (1 − C1)−module,thus U is CS −module), Uij is also CS −module, and hence Uij is a quasi−continuous module for any i, j ∈ {1, .., n} and i 6= j.Now, by [6, Corollary 11], thus U is a quasi −continuous module. ByTheorem 2.3, U is a continuous module, proving (i). Corollary 2.5. Let U = ⊕ni=1Ui where each Ui is a uniform module, thenthe following conditions are equivalent:(i) U is a∑−quasi −injective module;(ii) U is a 1−continuous module, countably ∑−(1− C1)−module.Proof. (i) =⇒ (ii). It is obvious.(ii) =⇒ (i). By Theorem 2.4, U is a continuous module. By [7, Proposition2.5], U is a∑−quasi −injective module, proving (i). A right R−module M is called distributive if for any submodule A,B,Cof M then A∩ (B+C) = A∩B+A∩C. We say that, M is a UC −moduleif each of its submodule has a unique closure in M .Theorem 2.6. Let U = ⊕ni=1Ui where each Ui is a uniform module. Assumethat U is a distributive module, then the following conditions are equivalent:(i) U is a (C2)−module;(ii) U is a (1− C2)−module.Proof. (i) =⇒ (ii). It is obvious.(ii) =⇒ (i). Similar proof of Theorem 2.4, Ui does not embed in a propersubmodule of Uj for any i, j ∈ {1, .., n} and S = End(Ui) is a uniformmodule for any i ∈ {1, .., n}. We first prove a claim that, if S1 and S2are direct summands of U with u − dim(S1) = 1, u − dim(S2) = n − 1ON THE (1− C2) CONDITION 145and S1 ∩ S2 = 0, then S1 ⊕ S2 = U . By Azumaya’s Lemma, we haveU = S2⊕K = S2⊕Ui. Suppose that i = 1, i.e., U = S2⊕U1 = (⊕ni=2Ui)⊕U1.Write U = S1⊕H = S1⊕ (⊕i∈IUi) with I being a subset of {1, .., n} andcard(I) = n− 1. There are cases:Case 1. If 1 6∈ I, U = S1 ⊕ (U2 ⊕ .. ⊕ Un) = U1 ⊕ (U2 ⊕ .. ⊕ Un). Thenit follows from S1 ∼= U1. By modularity we get S1 ⊕ S2 = S2 ⊕ V whereV = (S1 ⊕ S2) ∩ U1. From here we get V ∼= S1, this means U1 contains acopy of S1 ∼= U1. By U1 does not embed in a proper submodule of U1, wemust have V = U1, and hence S1 ⊕ S2 = S2 ⊕ U1 = U .Case 2. If 1 ∈ I, there exist k 6= 1 such that k = {1, .., n}\I, U =S1 ⊕ (⊕i∈IUi) = Uk ⊕ (⊕i∈IUi). Then it follows S1 ∼= Uk. By modularity weget S1⊕S2 = S2⊕V ′ where V ′ = (S1⊕S2)∩U1. From here we get V ′ ∼= S1,this means U1 contains a copy of S1 ∼= Uk. By Uk does not embed in aproper submodule of U1, we must have V′ = U1, and hence S1⊕ S2 = U , asrequired.We aim show next that U satisfies the condition (C3), i.e., for two directsummands of X1, X2 of U with X1∩X2 = 0, X1⊕X2 is also direct summandof U . By Azumaya’s Lemma, we have U = X1 ⊕ K = X1 ⊕ (⊕i∈JUi) =(⊕i∈FUi) ⊕ (⊕i∈JUi) (where F = {1, .., n}\J) and U = X2 ⊕ L = X2 ⊕(⊕j∈DUj) = (⊕j∈EUj) ⊕ (⊕j∈DUj) (where E = {1, .., n}\D). We implyX1 ∼= ⊕i∈FUi and X2 ∼= ⊕j∈EUj . Suppose that E = {1, .., t} and let ϕ :⊕tj=1Uj −→ X2 be an isomorphism and set Yj = ϕ(Uj), we have Yj ∼= Ujand X2 = ⊕tj=1Yj . By hypothesis X2 is a direct summand of U , thus Yj isalso direct summand of U for any j ∈ {1, .., t}. We show that X1 ⊕ X2 =X1 ⊕ (Y1 ⊕ ..⊕ Yt) is a direct summand of U .We prove that X1⊕Y1 is a direct summand of U . By Azumaya’s Lemma,we have U = Y1 ⊕ W = Y1 ⊕ (⊕p∈PUp) = Uα ⊕ (⊕p∈PUp), with P is asubset of {1, .., n} such that card(P ) = n − 1 and α = {1, .., n}\P . Notethat, card(P ∩ J) ≥ card(J) − 1 = m. Suppose that {1, ..,m} ⊆ (P ∩ J),i.e., U = (X1 ⊕ (U1 ⊕ .. ⊕ Um)) ⊕ Uβ = Z ⊕ Uβ with β = J\{1, ..,m}and Z = X1 ⊕ (U1 ⊕ .. ⊕ Um). By U is a distributive module, we haveZ ∩Y1 = (X1⊕ (U1⊕ ..⊕Um))∩Y1 = (X1 ∩Y1)⊕ ((U1⊕ ..⊕Um)∩Y1) = 0.Note that, Z, Y1 are direct summands of U with u − dim(Z) = n − 1 andu− dim(Y1) = 1, U = Z ⊕ Y1 = (X1 ⊕ (U1 ⊕ ..⊕ Um))⊕ Y1 = (X1 ⊕ Y1)⊕(U1⊕ ..⊕Um). Therefore, X1⊕Y1 is a direct summand of U . By induction,we have X1⊕X2 = X1⊕ (Y1⊕ ..⊕Yt) = (X1⊕Y1⊕ ..⊕Yt−1)⊕Yt is a directsummand of U . Thus U satisfies the condition (C3).Finally, we show that U satisfies the condition (C2). By hypothesis (ii)and U satisfies (C3), thus U is a (1−C2)−module (see Theorem 2.3), proving(i). 146 L. AN, N. ANH AND N. TUNGTheorem 2.7. Let U1, .., Un be uniform local modules such that Ui does notembed in J(Uj) for any i, j = 1, .., n. If U = ⊕ni=1Ui is a UC distributivemodule then it is a continuous module.Proof. We first prove a claim that U is a CS module. Let A be a uniformclosed submodule of U . Let the Xi = A ∩ Ui for any i ∈ {1, .., n}. Supposethat Xi = 0 for every i ∈ {1, .., n}. By hypothesis, U is a distributivemodule, we have A = A∩ (U1⊕ ..⊕Un) = X1⊕ ..⊕Xn = 0, a contradiction.Therefore, there exists a Xt 6= 0, i.e., A ∩ Ut 6= 0. By property A and Utare closed uniform submodules of U , thus Xt is an essential submodule ofA and Xt is also essential submodule of Ut. Hence A and Ut are closure ofXt in U , U is an UC module we get A = Ut. This implies that A is a directsummand of U , i.e., U is a (1−C1)−module. By U has finite dimension, Uis CS module (see [4, Corollary 7.8]), as required.We aim to show next that S = End(Ul) is a local ring for any l ∈ {1, .., n}.Let f ∈ S and suppose that f is not an isomorphism. It suffices to showthat 1− f is an isomorphism.Suppose that, f is a monomorphism. Then f is not onto, and f : Ul −→J(Ul) is an embedding, a contradiction. Thus f is not a monomorphism.Then, since Ker(f) is a nonzero submodule, it is essential in the uniformlocal module Ul. Thus, since we always have Ker(f) ∩Ker(1 − f) = 0, itfollows that Ker(1 − f) = 0, i.e., 1 − f is a monomorphism. But, since Uldoes not embed in J(Ul), 1−f must be onto, and so 1−f is an isomorphism.Thus, S is a local ring.Now, we show that U is a (1−C2)−module, i.e., for two uniform submod-ules V,W of U , with V ∼= W and W is a direct summand of U , V is also adirect summand of U . By Azumaya’s Lemma, we have U = W ⊕ W ′ =W ⊕ (⊕j∈JUj) = Uk ⊕ (⊕j∈JUj) where J is a subset of {1, .., n} withcard(J) = n − 1 and k = {1, .., n}\J . Hence V ∼= W ∼= Uk. Let V ∗ bea closure of V in U . By U is a CS module, thus V ∗ is a direct summandof U . Similarly, there exists s ∈ {1, .., n} such that V ∗ = Us, this meansUs contains a copy of W ∼= Uk. If V is a proper submodule of Us, thenUk embed in J(Us), a contradiction. We must have V = Us, and henceV is a direct summand of U . Thus, U is a (1 − C2)−module, i.e., U is a1−continuous module (by U is a CS module).Finally, by Theorem 2.4 thus U is a continuous module. Corollary 2.8. Let U1, .., Un be uniform local modules such that Ui does notembed in J(Uj) for any i, j = 1, .., n. If U = ⊕ni=1Ui is a UC, distributivemodule then the following conditions are equivalent:(i) U is a∑−quasi −injective module;(ii) U is a countably∑−(1− C1)−module.Proof. (i) =⇒ (ii). It is obvious.ON THE (1− C2) CONDITION 147(ii) =⇒ (i). By Theorem 2.7, U is a continuous module. By [7, Proposition2.5], U is a∑−quasi −injective module, proving (i). AcknowledgementWe would like thank the referee for carefully reading this note and formany useful commentsReferences[1] F. W. Anderson and K. R. Fuller, Ring and Categories of Modules, Springer - Verlag,New York - Heidelberg - Berlin, 1974.[2] V. Camilo,Distributive modules, J. Algebra, 36 (1975) 16 - 25.[3] H. Q. Dinh and D. V. Huynh, Some results on self −injective rings and∑−CS rings,Comm. Algebra, 31 (2003), 6063 - 6077.[4] N. V. Dung, D. V. Huynh, P. F. Smith and R. Wisbauer, Extending Modules, Pitman,London, 1994.[5] C. Faith, Algebra II, Springer - Verlag, 1976.[6] A. Harmanci and P. F. Smith,Finite direct sum of CS −module, Houston J. Math.,19 (1993), 523 - 532.[7] D. V. Huynh and S. T. Rizvi, On countably sigma –CS rings, Algebra and its ap-plications, Narosa publishing house, New Delhi, Chennai, Mumbai, Kolkata, (2001),119 - 128.[8] D. V. Huynh, D. D. Tai and L. V. An, On the CS condition and rings with chainconditions, AMS. Contem. Math. Series, 480 (2009), 241 - 248.[9] M. A. Kamal, On the decomposition and direct sums of modules, Osaka J. Math., 32(1995), 125 - 133.[10] S. H. Mohamed and B. J. Mu¨ller, Continuous and Discrete Modules, London Math.Soc. Lecture Note Series 147, Cambridge Univ. Press, 1990.[11] P. F. Smith, Modules for which ever submodule has a unique closure, in Ring Theory(Editors, S. K. Jain, S. T. Rizvi, World Scientific, Singapore, 1993), 303 - 313.Le Van AnDepartment of Natural Education,Ha Tinh University, Ha Tinh, Vietname-mail address: an.levan@htu.edu.vn, levanan na@yahoo.comNguyen Thi Hai AnhDepartment of Natural Education,Ha Tinh University, Ha Tinh, Vietname-mail address: anh.nguyenthihai@htu.edu.vnNgo Sy TungDepartment of Mathematics,Vinh University, Nghe An, Vietname-mail address: ngositung@yahoo.com(Received June 20, 2013 )(Accepted June 18, 2015 )

On the (1 − C<sub>2</sub>) condition

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On the (1 − C<sub>2</sub>) condition

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