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14
UDC 512.55 Вестник СПбГУ. Математика. Механика. Астрономия. 2021. Т. 8 (66). Вып. 3
MSC 16D10, 16D70
On a question concerning D4-modules
S. Das
Department of Mathematics, KPR Institute of Engineering and Technology, Coimbatore-641407, India
For citation: Das S. On a question concerning D4-modules. Vestnik of Saint Petersburg University. Mathematics. Mechanics. Astronomy, 2021, vol. 8(66), issue 3, pp. 467-474. https://doi.org/10.21638/spbu01.2021.308
An R-module M is called a D4-module if 'whenever M1 and M2 are direct summands of M with Mi + M2 = M and Mi = M2, then Mi П M2 is a direct summand of M'. Let M = ®ie/Mi be a direct sum of submodules Mi with Hom(Mi,Mj) = 0 for distinct i, j £ I. We show that M is a D4-module if and only if for each i G I the module Mi is a D4-module. This settles an open question concerning direct sums of D4-modules. Our approach is independent of the solution obtained by D'Este, Keskin Tiitiincu and Tribak recently.
Keywords : SIP-modules, D4-modules.
1. Introduction. By a ring we mean an associative ring with an identity element; modules are unitary.
A module M is said to be a SIP-module (SSP-module) if the intersection (respectively, the sum) of two direct summands of M is a direct summand of M. Kaplansky observed that over a commutative principal ideal domain every free module is a SIP-module (see [1, Exercise 51(a), p.49].) SIP-modules and SSP-modules have been extensively studied (see, for example, [2-4] and [5]).
For 1 < i < 4, a module M is called a Di-module if it satisfies the condition Di noted below.
D1. For every submodule A of M, there is a decomposition M = Mi ф M2 such that Mi < A and A П M2 is small in M2.
D2. If A < M such that M/A is isomorphic to a direct summand of M, then A is a direct summand of M.
D3. If M1 and M2 are direct summands of M with M1 + M2 = M, then M1 П M2 is a direct summand of M.
D4. If M1 and M2 are direct summands of M with M1 + M2 = M and M1 = M2, then M1 П M2 is a direct summand of M.
(For a detailed background of these notions, we refer to [6, Chapter 4] and to [7].) A module M is also called a lifting module if it satisfies condition D1 (see [8] for detailed information regarding these modules). We recall the characterization "the ring R is semiperfect if and only if R is lifting as a right (or left) R-module" (see [9, Theorem 1.2.13]). Now let R be a commutative domain with zero Jacobson radical which is not a field, and hence is not semiperfect. Then, by the above results, RR is a projective module which is not a Dl-module. We have, however, projective quasi-projective
© St. Petersburg State University, 2021
D2-module D3-module D4-module (see [6, Proposition 4.38 and Lemma 4.6]). Note that for all proper subgroups N of the (indecomposable) Priifer p-group M = Zp~, the group M/N is isomorphic to M. Hence it is D3 (as a Z-module) but not D2. In fact, there are rings over which every cyclic module is D3 but not all cyclic modules are D2 (see [10, Example 6.4]).
There is no known example of a module which is D4 but not D3 [11] (see also [12, p. 2]).
Let A and B be right R-modules. A homomorphism f G HomR(A, B) is said to be (von Neumann) regular (briefly, regular) if for some homomorphism g G HomR(B,A), we have the relation f = fgf. It is well-known that a homomorphism f G HomR(A, B) is regular if and only if Kerf) is a direct summand in A and Imf) is a direct summand in B.
Recall that a module M is called a Rickart module if the kernel of any endomor-phism f G EndR(M) is a direct summand in M. It follows from [13, Proposition 2.16] that every Rickart module is a SIP-module. A module M is called a dual Rickart module if the image of any endomorphism f G EndR(M) is a direct summand in M. It follows from [14, Proposition 2.11] that every dual Rickart module is a SSP-module.
2. Results. We begin with the recall of some results from [15].
Lemma 1 [15, Lemma 2.1]. Let M be a right R-module, f,g G EndR(M) be regular homomorphisms, and let
M = Kerf) © A = Im(f) © B,M = Ker(g) © A' = Im(g) © B'.
Then the following assertions hold:
(a) Im(fg) = f (A n (Im(g) + Kerf)));
(b) Ker(fg) = (g\A' )-1 (Im(g) n Kerf)) + Ker(g).
Lemma 2 [15, Lemma 2.2]. Let M be a right R-module, n be the projection onto the first direct summand with respect to the decomposition M = A1 © A2, and let n' be the projection onto the first direct summand with respect to the decomposition M = B1 © B2. Then the following assertions hold:
(a) Im(n'n) = (Ai + B2) n Bi;
(b) Ker(n'n) = (A1 n B2) + A2.
Proposition 1 [15, Theorem 2.3]. For a right R-module M, the following conditions are equivalent.
1. M is a SSP-module.
2. For any two regular homomorphisms f,g G EndR(M), the module Im(fg) is a direct summand of the module M.
Proposition 2 [15, Theorem 2.4]. For a right R-module M, the following conditions are equivalent.
1. M is a SIP-module.
2. For any two regular homomorphisms f,g G EndR(M), the module Ker(fg) is a direct summand of the module M.
Next we note examples of finite abelian groups which are not D4.
Example. Consider M = Z/2Z © Z/4Z as a Z-module. Then A = (1,3)Z and B = (0, 3)Z are isomorphic direct summands of M. However, A n B is not a direct summand of M. In fact, for any prime p, consider M = Z/pmZ © Z/pnZ with n > m as a Z-module, then M is not a D4-module, since there is an epimorphism Z/pnZ —> Z/pmZ whose kernel is not a direct summand of Z/pnZ.
The following theorem is an analogue of [15, Theorem 3.3].
Theorem 1. For a right R-module M, consider the following statements.
1. M is a DS-module.
2. For any two regular endomorphisms f,g G EndR(M), if Im(fg) is a direct summand of the module M, then the module Ker(fg) is a direct summand of the module M.
3. For any two regular endomorphisms f, g G EndR(M) satisfying the following:
(i) Im(fg) is a direct summand of the module M,
(ii) Kerf) = Im(g),
then the module Ker(fg) is a direct summand of the module M.
4. M is a D4-module.
5. For any two regular endomorphisms f, g G EndR(M) satisfying the following:
(1) Im(fg) is a direct summand of the module M,
(ii) N+Ker(f) = Im(g) for any direct summand N of M such that N n Kerf) = 0, then the module Ker(fg) is a direct summand of the module M.
Then (1) ^ (2) ^ (3) ^ (4) ^ (5).
Proof. (1) ^ (2) follows from [15, Theorem 3.3].
(2) ^ (3) is clear.
(3) ^ (4). Let M = A © A' = B © B', where A + B = M and A = B. Consider the natural projections n1 : A © A' —> A and n2 : B © B' —> B'. Then by Lemma 2(a), Im(n2n1) = B' is a direct summand of M. Therefore by assumption and Lemma 2(b), Ker(n2n1) = (A n B) © A' is a direct summand of M. This shows that A n B is a direct summand of M, as required.
(4) ^ (5). Let
M = Kerf) © A = Imf) © B = Ker(g) © A' = Im(g) © B'.
By Lemma 1(a), since f |A is an isomorphism (Im(g) + Kerf)) n A is a direct summand of M. Therefore, A = N © (Im(g) + Kerf)) n A, for some N < A. Since (N + Kerf)) + Im(g) = M, N + Kerf) = Im(g) and M is a D4-module, we have (N + Kerf ))n
Im(g) = (Ker(f ) n Im(g)) is a direct summand of M. Since g\A' : A' —>• Im(g) is an isomorphism, we have (g\A')-1(Im(g) n Ker(f)) is a direct summand of M. Hence by Lemma 1(b), Ker(fg) is a direct summand of M. □
Recall that a module M is called a C3-module if A and B are direct summands in M with A n B = 0, then A © B is a direct summand in M.
Following Ding et al. [16, Theorem 2.2(5)], a module M is called a C4-module if A and B are isomorphic direct summands in M with A n B = 0, then A © B is a direct summand in M. Clearly C3-modules are C4-modules. However, there are examples of C4-modules which are not C3.
The following theorem is an analogue of [15, Theorem 3.1].
Theorem 2. For a right R-module M, consider the following statements.
1. M is CS-module.
2. For any two regular endomorphisms f, g G EndR(M), if Ker(fg) is a direct summand of the module M, then the module Im(fg) is a direct summand of the module M.
3. For any two regular endomorphisms f, g G EndR(M) satisfying the following:
(i) Ker(fg) is a direct summand of the module M,
(ii) Ker(f) = Im(g),
then the module Im(fg) is a direct summand of the module M.
4. M is a C4-module.
5. For any two regular endomorphisms f, g G EndR(M) satisfying the following:
(1) Ker(fg) is a direct summand of the module M, (ii) N = Im(g) for any direct summand N of Ker(f ),
then the module Im(fg) is a direct summand of the module M. Then (1) ^ (2) ^ (3) ^ (4) ^ (5).
Proof. (1) ^ (2) follows from [15, Theorem 3.1].
(2) ^ (3) is clear.
(3) ^ (4). Let M = A © A' = B © B', where A n B = 0 and A = B. Consider the natural projections n1 : A © A' —> A and : B © B' —> B'. Then by Lemma 2(b), Ker(n2n1) = A' is a direct summand of M. Therefore by assumption and Lemma 2(a), Im(n2n1) = (A + B) n B' is a direct summand of M. Since A + B = B © (A + B) n B', A + B is a direct summand of M, as required.
(4) ^ (5). Let
M = Ker(f ) © A = Im(f ) © B = Ker(g) © A' = Im(g) © B'.
By Lemma 1(b), (g|A')-1(Im(g) n Kerf ) is a direct summand of A'. Since g|A' : A' —>• Im(g) is an isomorphism and Im(g) is a direct summand of the module M, we have that /m(g) n Ker(f) is a direct summand of the module M. Therefore, Ker(f) = N © (Im(g) n Ker(f )), for some N < M. Since N n Im(g) =0, N = Im(g) and M is a
C4-module, we have N © Im(g) is a direct summand of M. Since Ker(f ) < Im(g) © N, we have that
Im(g) © N = Ker(f ) © (Im(g) + N) n A = Ker(f ) © (Im(g) + Ker(f )) l~l A.
Therefore, (Im(g) + Ker(f)) n A is a direct summand of M. Hence by Lemma 1(a), Im(fg) is a direct summand of M. □
We can now prove the following result which has already appeared in [17, Proposition 5.7 and Corollary 2.9]. The proof has been outlined by us for the sake of completeness.
Proposition 3. For a right R-module M, the following conditions are equivalent.
1. M is a Démodule and a SSP-module.
2. M is a C3-module and a SIP-module.
3. M is a C4-module and a SIP-module.
4. M is a D3-module and a SSP-module.
5. M is an SSP-module and a SIP-module.
Proof. (1) (2). Let M be a SSP-module. It is clear that M is a C3-module. To see that M is a SIP-module, we shall use Proposition 2. Let f,g G EndR(M) be two regular endomorphisms such that
M = Ker(f ) © A = Im(f ) © B = Ker(g) © A' = Im(g) © B'.
We need to show that Ker(fg) is a direct summand of M. By Lemma 1(b), enough to show that Im(g) n Ker(f ) is a direct sumand of M. To this end we shall follow the proof of [3, Proposition 1.4]. Let n1 : Im(g) © B —> Im(g) and n2 : Ker(f ) © A —> Ker(f ) be the natural projections. Define 6 = ((ni — 1) o n2)|/m(s) : Im(g) —> B'. Then by [2, Proposition 1.4], I m (6) is a direct summand of B'. Hence M being a D4-module (use [7, Theorem 2.2]), we have Ker(6) = (Im(g) n Ker(f )) © (Im(g) n A) is a direct summand of Im(g). Thus Im(g) n Ker(f ) is a direct sumand of M, as desired.
(2) (3) is clear.
(3) (4). Let M be a SIP-module. It is clear that M is a D3-module. To see that M is a SSP-module, we shall use Proposition 1. Let f,g G EndR(M) be two regular endomorphisms such that
M = Ker(f ) © A = Im(f ) © B = Ker(g) © A' = Im(g) © B'.
We need to show that Im(fg) is a direct summand of M. By Lemma 1(a), enough to show that Im(g) + Ker(f ) is a direct sumand of M. To this end we shall follow the proof of [5, Theorem 8]. Let n1 : Ker( f ) © A —> Ker(f ) and n2 : Im(g) © B' —> B' be the natural projections. Define $ = (n2 o n1)|1m(s) : Im(g) —> B'. Then by [3, Proposition 1.4], Ker(4>) is a direct summand of B'. Hence M being a C4-module (use [16, Theorem 2.2]), we have Im($) = [Im(g) + Ker(f )] n [Im(g) + A] n B' is a direct summand of Im(g). So we can write M = Im($) © X for some X < M. Hence B' = Im($) © (B' n X). Then we have M = [Im(g) + Ker(f )] © [(Im(g) + A) n (B' n X)], as required.
(4) (5) follows from Proposition 2 and Theorem 1.
(5) (1) is clear. □
The following result extends [15, Lemma 4.2(2)].
Proposition 4. Let M be a dual Rickart module. If M is a Démodule, then the product of any two regular elements in the ring EndR(M) is a regular element.
Proof. It follows from the hypothesis and Proposition 3 that M is a SSP-module and a SIP-module. Hence the result follows from [15, Theorem 2.7]. □
The following theorem was proved in [17].
Theorem 3 [17, Theorem 5.6]. Let M = ®ie/Mi be a direct sum of submodules Mi. If N = ®ie/(N n Mi) for every submodule N of M, then M is a Démodule if and only if for each i G I, Mi is a Démodule.
In [17], immediately after Theorem 3 the following question was asked.
Question (see [17, Question, p. 4494]). It is known that if N = ®ie/(N n Mi) for every submodule N of M, then Hom(Mi, Mj) = 0 for every i = j in I, so it is natural to ask if [17, Theorem 5.6] (that is the theorem above) remains true if one assumes that Hom(Mi, Mj) = 0 for every i = j in I.
In the next proposition we show that Question above has a positive answer.
Proposition 5. Let M = ®ieNMi be a direct sum of submodules Mi in which Hom(Mi, Mj) = 0 for every i = j. Then the following assertions hold:
(i) if M is a Démodule, then for each i G I, Mi is a Démodule,
(ii) if each Mi is a Démodule, then M is a Démodule.
Proof. (i). Since a direct summand of a D4-module is a D4-module (see [7, Proposition 2.11]), for every i G N, Mi is a D4-module if M is a D4-module.
(ii). By hypothesis and [18, the paragraph before Corollary 16.5], we have
/EndR(Mi) 0 0 •• •• \
0 EndR (M2) 0 •• •
0 0 •• EndR (Mn) •
V • ... 7 NxN
EndR (M )
Take two regular elements f, g in EndR(M) such that Im(fg) is a direct summand of M and Ker(f) = Im(g). Then f = (fi)ieN and g = (gi)ieN for some regular elements / and gj in EndR(Mj) such that Im(fjgj) is a direct summand of Mj and [X + Ker(fj)] = Im(gj) for any direct summand Xj of Mj such that Xj fl Ker(fj) = 0 for all i G N. But then each Mj is a D4-module. Therefore by Theorem 1, Ker(fjgj) is a direct summand of Mj for all i G N. Hence Ker(fg) is a direct summand of M, as required. □
Remark. Let |pj}jeN be an infinite set of prime numbers and let p be a prime different from any of them. Then we have the following examples of D4-modules:
(i) M = ® (®jeN Z/pjZ) as a Z-module, where Zp~ is the Priifer p-group;
(ii) M = Q ® (®j£N Z/pjZ) as a Z-module.
The author is grateful to the referee for a detailed list of suggestions and comments that helped improve the article significantly. Also, the author would like to thank Professor Yasser Ibrahim for some fruitful conversations and Professor M.B.Rege for his encouragement and help in presentation.
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Received: September 13, 2020 Revised: March 14, 2021 Accepted: March 19, 2021
Author's information:
Soumitra Das — Assistant Professor; soumitrad330@gmail.com
К вопросу о Д4-модулях
С. Дас
Инженерно-технологический институт КПР, Коимбатур, 641407, Индия
Для цитирования: Das S. On a question concerning D4-modules // Вестник Санкт-Петербургского университета. Математика. Механика. Астрономия. 2021. Т. 8(66). Вып. 3. С. 467-474. https://doi.org/10.21638/spbu01.2021.308
R-модуль M называется В4-модулем, если всякий раз, когда M1 и M2 являются прямыми слагаемыми M с M1 + M2 = M и M1 = M2, то M1\M2 является прямым слагаемым M. Пусть M = фieI Mi — прямая сумма подмодулей Mi с Horn(Mi; Mj) = 0 для различных i,j е I. Показано, что M является ,04-модулем тогда и только тогда, когда для каждого i е I модуль Mi является ,04-модулем. Это решает открытый вопрос о прямых суммах ,04-модулей. Наш подход не зависит от решения, полученного недавно Д'Эсте, Кескином Тютюнджу и Трибаком. Ключевые слова: SIP-модули, ,04-модули.
Статья поступила в редакцию 13 сентября 2020 г.;
после доработки 14 марта 2021 г.; рекомендована в печать 19 марта 2021 г.
Контактная информация: Сумитра Дас — soumitrad330@gmail.com
