A note on noncosingular lifting modules

Let $R$ be a right perfect ring. Let $M$ be a noncosingular lifting module which does not have any relatively projective component. Then $M$ has finite hollow dimension.

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Datum:2012
Hauptverfasser: Amouzegar, Kalati T., Keskin, Tütüncü D., Амузегар, Калаті Т., Кескін, Тютюнсю Д.
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Sprache:Englisch
Veröffentlicht: Institute of Mathematics, NAS of Ukraine 2012
Online Zugang:https://umj.imath.kiev.ua/index.php/umj/article/view/2682
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Ukrains’kyi Matematychnyi Zhurnal
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author Amouzegar, Kalati T.
Keskin, Tütüncü D.
Амузегар, Калаті Т.
Кескін, Тютюнсю Д.
author_facet Amouzegar, Kalati T.
Keskin, Tütüncü D.
Амузегар, Калаті Т.
Кескін, Тютюнсю Д.
author_sort Amouzegar, Kalati T.
baseUrl_str https://umj.imath.kiev.ua/index.php/umj/oai
collection OJS
datestamp_date 2020-03-18T19:32:37Z
description Let $R$ be a right perfect ring. Let $M$ be a noncosingular lifting module which does not have any relatively projective component. Then $M$ has finite hollow dimension.
first_indexed 2026-03-24T02:28:15Z
format Article
fulltext К О Р О Т К I П О В I Д О М Л Е Н Н Я UDC 512.5 T. Amouzegar Kalati (Quchan Inst. Engineering and Technology, Iran), D. Keskin Tütüncü (Hacettepe Univ., Ankara, Turkey) A NOTE ON NONCOSINGULAR LIFTING MODULES ПРО НЕКОСИНГУЛЯРНI МОДУЛI IЗ ВЛАСТИВIСТЮ ПIДНЯТТЯ Let R be a right perfect ring. Let M be a noncosingular lifting module which does not have any relatively projective component. Then M has finite hollow dimension. Нехай R — праве досконале кiльце, а M — некосингулярний модуль iз властивiстю пiдняття, що не має жодної вiдносно проективної компоненти. Тодi M має скiнченну дуальну розмiрнiсть Голдi. 1. Introduction. Throughout this paper all rings are associative with identity and modules are unitary right modules. A module M is said to have finite hollow dimension if there exists an epimorphism from M to a finite direct sum of n hollow factor modules with small kernel. A module M is called lifting if for every A ≤ M, there exists a direct summand B of M such that B ⊆ A and A/B �M/B. A module M is amply supplemented and every coclosed submodule of M is a direct summand of M if and only if M is lifting by [1] (22.3(d)). In [5], Talebi and Vanaja defined Z(M) as follows: Z(M) = Re (M,S) = ⋂{ Ker(g) | g ∈ Hom (M,L), L ∈ S } , where S denotes the class of all small modules. They called M a cosingular (noncosingular) module if Z(M) = 0 (Z(M) =M ). In this note, as we state in the abstract, we prove the following main theorem: Let R be a right perfect ring. Let M be a noncosingular lifting module which does not have any relatively projective component. Then M has finite hollow dimension. For all undefined notions we refer to [1]. 2. Results. An R-module M is called dual Rickart if, for any element φ ∈ S = End(M), Imφ = eM, where e2 = e ∈ S. Lemma 2.1. Let M = ⊕i∈NMi be a dual Rickart module and let (fi : Mi →Mi+1)N be a se- quence of homomorphisms. Then for any finitely many elements a1, a2, . . . , an ∈M1, there exist some r ∈ N and a homomorphism h : Mr+1 → Mr such that fr−1fr−2 . . . f1(ak) = hfrfr−1 . . . f1(ak) for k = 1, 2, . . . , n. In particular, if M1 is finitely generated, then fr−1fr−2 . . . f1 = hfrfr−1 . . . f1. Proof. It is easy to see by [6] (43.3(3)). In [3], Keskin Tütüncü and Tribak introduced the concept of dual Baer modules. A module M is called a dual Baer module if for every right ideal I of S, ∑ φ∈I Imφ is a direct summand of M. It is clear that every dual Baer module is dual Rickart. Lemma 2.2. Let M = ⊕∞i=1Mi, where each Mi is local noncosingular. If, for each i, there is an epimorphism fi : Mi −→Mi+1, which is not an isomorphism, then M is not lifting. c© T. AMOUZEGAR KALATI, D. KESKİN TÜTÜNCÜ, 2012 1572 ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11 A NOTE ON NONCOSINGULAR LIFTING MODULES 1573 Proof. Let M = ⊕∞i=1Mi be a lifting module and (fi : Mi →Mi+1)N be a sequence of epimor- phisms, which are non-isomorphisms. By [3] (Theorem 2.14) and Lemma 2.1, there exist an r ∈ N and a homomorphism h : Mr+1 → Mr such that fr−1fr−2 . . . f1 = hfrfr−1 . . . f1. Since all fi are epimorphisms, we have hfr = 1Mr . Hence fr is an isomorphism, a contradiction. Lemma 2.2 is proved. Recall that a family of modules {Mi | i ∈ I} is called locally semi-T-nilpotent if, for any countable set of non-isomorphisms {fn : Min → Min+1} with all in distinct in I, and for any x ∈Mi1 , there exists k (depending on x) such that fk . . . f1(x) = 0 (see [4]). Corollary 2.1. Let M be a noncosingular lifting module such that M = ⊕∞i=1Mi, where each Mi is local for all i ∈ N. Then the family {Mi | i ∈ N} is locally semi-T-nilpotent. Proof. Consider any infinite sequence of non-isomorphisms fn Mi1 f1→Mi2 f2→ . . . Min fn→ . . . . It is obvious that fn is an epimorphism for all n ≥ 1. By Lemma 2.2, it is easy to see that the family {Mi | i ∈ N} is locally semi-T -nilpotent. Lemma 2.3. Let U and V be noncosingular hollow modules such that the module U ⊕ V is lifting. Then there exists an epimorphism from U to V or V is U -projective. Proof. Let M = U ⊕ V, M1 = U ⊕ 0 and M2 = 0 ⊕ V. Hence M = M1 ⊕M2. Suppose that there does not exist any epimorphism from U to V, i.e., from M1 to M2. We will show that V is U - projective. Let N be any nonzero proper submodule of M such that M = N+M1. Since M is lifting, there exists a direct summand K of M such that K ≤ N and N/K � M/K. Let M = K ⊕ K ′ for some submodule K ′ of M. Note that K and K ′ are hollow. Since M = K +M1, we have an epimorphism from M/K ′ to M2. If K ′ + M1 = M, then we have an epimorphism from M1 to M/K ′. So we have an epimorphism from M1 to M2, a contradiction. Thus K ′ +M1 6= M. Hence (K ′ +M1)/K ′ �M/K ′. Since every small module is cosingular, (K ′ +M1)/K ′ is cosingular. On the other hand, (K ′ +M1)/K ′ ∼= M1/(K ′ ∩M1) is noncosingular. Hence K ′ = K ′ +M1 and so M1 ≤ K ′. Thus M = K ⊕M1. By [6] (41.14), M2 is M1-projective, i.e., V is U -projective. Theorem 2.1. Let R be a right perfect ring. Let M be a noncosingular lifting module which does note have any relatively projective component. Then M has finite hollow dimension. Proof. By [3] (Theorem 2.14 and Corollary 2.6(ii)), there exists an index set I and hollow submodules Mi, i ∈ I, such that M = ⊕i∈IMi. Suppose that I is infinite. For all distinct i, j in I, Mi ⊕Mj is lifting and hence by Lemma 2.3, there exists an epimorphism from Mi to Mj or Mj is Mi-projective. By hypothesis, there exists an epimorphism from Mi to Mj . Now by Lemma 2.2, there exists an infinite subset J of I such that Mi ∼=Mj for all i, j ∈ J since ⊕i∈IMi is lifting. Let i ∈ J. Suppose that φ : Mi −→ Mi is a nonzero homomorphism. Since Mi is noncosingular and hollow, φ is an epimorphism. Suppose φ is not an isomorphism. Then for each i, j ∈ J, φ induces an epimorphism φij : Mi −→ Mj which is not an isomorphism, contradicting Lemma 2.2. Thus φ is an isomorphism. It follows that the ring End(Mi) of endomorphisms of Mi is a division ring, and by [2] (Lemma 1), Mi is Mi ∼= Mj-projective, a contradiction. Therefore, M has finite hollow dimension. ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11 1574 T. AMOUZEGAR KALATI, D. KESKİN TÜTÜNCÜ Corollary 2.2. Let R be a right perfect ring. Let M be a noncosingular lifting module which does not have any relatively projective component. Then M satisfies ACC equivalently, DCC on supplements. Proof. By Theorem 2.1 and [1] (20.34). Finally, we give the following: Proposition 2.1. Let R be a right perfect ring and let M = ∏∞ i=1 Mi, where each Mi is hollow noncosingular. If, for each i, there is an epimorphism fi : Mi+1 −→ Mi, which is not an isomorphism, then M is not lifting. Proof. Assume that g1 : P1 −→ M1 is a projective cover of M1. Since P1 is projective, there exists a homomorphism g2 : P1 −→M2 such that f1g2 = g1. Clearly, g2 is epic. Then for each i, we may define inductively, gi : P1 −→ Mi so that figi+1 = gi and all gi are epic. Note that P1 and all Mi are local and so cyclic. Now we have the strictly descending sequence since each fi is not monic for each i: P1 ⊃ Ker g1 ⊃ Ker g2 ⊃ . . . . Define the homomorphism χ : P1 −→ M by χ(y) = (gi(y))i∈I (y ∈ P1). Let Imχ = K. Then K is local and nonzero. Assume that K = xR for some nonzero element x ∈ K. We can suppose without loss of generality that x = (0, 0, . . . , 0, xn+2, xn+3, . . .) for some positive integer n. Then x ∈ N = ∏∞ n+2 Mi. So K ⊆ N. Note that K is coclosed in M by [5] (Lemma 2.3(2)). Now, let M = K ⊕ K ′ for some submodule K ′ of M and let y ∈ Ker gn. Consider t = = (0, 0, . . . , 0, gn+1(y), gn+2(y), . . .) ∈ M = ∏∞ i=1 Mi. Then t = t1 + t2 for some t1 ∈ K and t2 ∈ K ′. Then t2 = t − t1 ∈ K ∩ K ′ = 0. So t = t1 ∈ K ⊆ N. Thus gn+1(y) = 0 and so y ∈ Ker gn+1. It follows that Ker gn = Ker gn+1, a contradiction. Therefore K is not a direct summand of M and M is not a lifting module. Acknowledgments. This work has been done during a visit of the first author to the second author in the Department of Mathematics, Hacettepe University in 2011, and she wishes to thank the department for their kind hospitality. The first author also wishes to thank the Ministry of Science of Iran for the support. 1. Clark J., Lomp C., Vanaja N., Wisbauer R. Lifting modules // Frontiers in Math. – Birkhäuser Verlag, 2006. 2. Keskin D. Finite direct sums of (D1)-modules // Tr. J. Math. – 1998. – 22, № 1. – P. 85 – 91. 3. Keskin Tütüncü D., Tribak R. On dual Baer modules // Glasgow Math. J. – 2010. – 52. – P. 261 – 269. 4. Mohamed S. H., Müller B. J. Continuous and discrete modules // London Math. Soc. Lect. Notes. Ser. 147. – Cambridge: Cambridge Univ. Press, 1990. 5. Talebi Y., Vanaja N. The torsion theory cogenerated by M -small modules // Communs Algebra. – 2002. – 30, № 3. – P. 1449 – 1460. 6. Wisbauer R. Foundations of module and ring theory. – Reading: Gordon and Breach, 1991. Received 26.12.11 ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
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spelling umjimathkievua-article-26822020-03-18T19:32:37Z A note on noncosingular lifting modules Про некосингулярнi модулi iз властивiстю пiдняття Amouzegar, Kalati T. Keskin, Tütüncü D. Амузегар, Калаті Т. Кескін, Тютюнсю Д. Let $R$ be a right perfect ring. Let $M$ be a noncosingular lifting module which does not have any relatively projective component. Then $M$ has finite hollow dimension. Нехай $R$ — праве досконале кiльце, а $M$ — некосингулярний модуль iз властивiстю пiдняття, що не має жодної вiдносно проективної компоненти. Тодi $M$ має скiнченну дуальну розмiрнiсть Голдi. Institute of Mathematics, NAS of Ukraine 2012-11-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2682 Ukrains’kyi Matematychnyi Zhurnal; Vol. 64 No. 11 (2012); 1572-1574 Український математичний журнал; Том 64 № 11 (2012); 1572-1574 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2682/2122 https://umj.imath.kiev.ua/index.php/umj/article/view/2682/2123 Copyright (c) 2012 Amouzegar Kalati T.; Keskin Tütüncü D.
spellingShingle Amouzegar, Kalati T.
Keskin, Tütüncü D.
Амузегар, Калаті Т.
Кескін, Тютюнсю Д.
A note on noncosingular lifting modules
title A note on noncosingular lifting modules
title_alt Про некосингулярнi модулi iз властивiстю пiдняття
title_full A note on noncosingular lifting modules
title_fullStr A note on noncosingular lifting modules
title_full_unstemmed A note on noncosingular lifting modules
title_short A note on noncosingular lifting modules
title_sort note on noncosingular lifting modules
url https://umj.imath.kiev.ua/index.php/umj/article/view/2682
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