ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.790751 J. Algebra Comb. Discrete Appl. 7(3) • 259–267 Received: 5 September 2019 Accepted: 18 May 2020 Journal of Algebra Combinatorics Discrete Structures and Applications Some results on relative dual Baer property Research Article Tayyebeh Amouzegar, Rachid Tribak Abstract: Let R be a ring. In this article, we introduce and study relative dual Baer property. We characterize R-modules M which are RR-dual Baer, where R is a commutative principal ideal domain. It is shown that over a right noetherian right hereditary ring R, an R-module M is N-dual Baer for all R-modules N if and only if M is an injective R-module. It is also shown that for R-modules M1, M2, . . ., Mn such that Mi is Mj-projective for all i > j ∈ {1, 2, . . . , n}, an R-module N is ⊕n i=1 Mi-dual Baer if and only if N is Mi-dual Baer for all i ∈ {1, 2, . . . , n}. We prove that an R-module M is dual Baer if and only if S = EndR(M) is a Baer ring and IM = rM (lS(IM)) for every right ideal I of S. 2010 MSC: 16D10, 16D80 Keywords: Baer rings, Dual Baer modules, Relative dual Baer property, Homomorphisms of modules 1. Introduction Throughout this paper, R will denote an associative ring with identity, and all modules are unitary right R-modules. Let M be an R-module. We will use the notation N � M to indicate that N is small in M (i.e., L + N 6= M for every proper submodule L of M). By E(M) and EndR(M), we denote the injective hull of M and the endomorphism ring of M, respectively. By Q, Z, and N we denote the set of rational numbers, integers and natural numbers, respectively. For a prime number p, Z(p∞) denotes the Prüfer p-group. The concept of Baer rings was first introduced in [6] by Kaplansky. Since then, many authors have studied this kind of rings (see, e.g., [2] and [3]). A ring R is called Baer if the right annihilator of any nonempty subset of R is generated by an idempotent. In 2004, Rizvi and Roman extended the notion of Baer rings to a module theoretic version [10]. According to [10], a module M is called a Baer module if for every left ideal I of EndR(M), ∩φ∈IKerφ is a direct summand of M. This notion was recently dualized by Keskin Tütüncü-Tribak in [14]. A module M is said to be dual Baer if for every right ideal Tayyebeh Amouzegar; Department of Mathematics, Quchan University of Advanced Technology, Quchan, Iran (email: t.amoozegar@yahoo.com). Rachid Tribak (Corresponding Author); Centre Régional des Métiers de l’Education et de la Formation (CRMEF- TTH)-Tanger, Avenue My Abdelaziz, Souani, B.P. 3117, Tangier, Morocco (email: tribak12@yahoo.com). 259 https://orcid.org/0000-0002-0600-5326 https://orcid.org/0000-0001-8400-4321 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 I of S = EndR(M), ∑ φ∈I Imφ is a direct summand of M. Equivalently, for every nonempty subset A of S, ∑ φ∈A Imφ is a direct summand of M (see [14, Theorem 2.1]). A module M is said to be Rickart if for any ϕ ∈ EndR(M), Kerϕ is a direct summand of M (see [7]). The notion of dual Rickart modules was studied recently in [8] by Lee-Rizvi-Roman. A module M is said to be dual Rickart if for every ϕ ∈ EndR(M), Imϕ is a direct summand of M. In [8], it was introduced the notion of relative dual Rickart property which was used in the study of direct sums of dual Rickart modules. Let N be an R-module. An R-module M is called N-dual Rickart if for every homomorphism ϕ : M → N, Imϕ is a direct summand of N (see [8]). Similarly, we introduce in this paper the concept of relative dual Baer property. A module M is called N-dual Baer if for every subset A of Hom R(M,N),∑ f∈A Imf is a direct summand of N. It is clear that if M is N-dual Baer, then M is N-dual Rickart. We determine the structure of modules M which are RR-dual Baer for a commutative principal ideal domain R (Proposition 2.7). Then we show that for an R-module M, RR is M-dual Baer if and only if M is a semisimple module (Proposition 2.9). It is shown that over a right noetherian right hereditary ring R, an R-module M is N-dual Baer for all R-modules N if and only if M is an injective R-module (Corollary 2.17). We prove that if {Mi}I is a family of R-modules, then for each j ∈ I, ⊕ i∈I Mi is Mj-dual Baer if and only if Mi is Mj-dual Baer for all i ∈ I (Corollary 2.24). It is also shown that for R-modules M1, M2, . . ., Mn such that Mi is Mj-projective for all i > j ∈{1, 2, . . . ,n}, an R-module N is ⊕n i=1 Mi-dual Baer if and only if N is Mi-dual Baer for all i ∈{1, 2, . . . ,n} (Theorem 2.25). We conclude this paper by showing that an R-module M is dual Baer if and only if S = EndR(M) is a Baer ring and IM = rM (lS(I)) for every right ideal I of S, where lS(I) = {ϕ ∈ S | ϕI = 0}, rM (lS(I)) = {m ∈ M | lS(I)m = 0} and IM = ∑ f∈I Imf (Theorem 2.31). 2. Main results Definition 2.1. Let N be an R-module. An R-module M is called N-dual Baer if, for every subset A of Hom R(M,N), ∑ f∈A Imf is a direct summand of N. Obviously, an R-module M is dual Baer if and only if M is M-dual Baer. Example 2.2. (1) Let N be a semisimple R-module. Then for every R-module M, M is N-dual Baer. (2) If M and N are R-modules such that HomR(M,N) = 0, then M is N-dual Baer. It follows that for any couple of different maximal ideals m1 and m2 of a commutative noetherian ring R, E(R/m1) is E(R/m2)-dual Baer (see [12, Proposition 4.21]). (3) Let p be a prime number. Note that Z/pZ and Z(p∞) are dual Baer Z-modules. On the other hand, it is clear that Z(p∞) is Z/pZ-dual Baer but Z/pZ is not Z(p∞)-dual Baer. Recall that a module M is said to have the strong summand sum property, denoted briefly by SSSP, if the sum of any family of direct summands of M is a direct summand of M. Following [8, Definition 2.14], a module M is called N-d-Rickart if, for every homomorphism ϕ : M → N, Imϕ is a direct summand of N. Proposition 2.3. Let M and N be two R-modules. If M is N-dual Baer, then M is N-d-Rickart. The converse holds when N has the SSSP. Proof. This follows from the definitions of “M is N-d-Rickart” and “M is N-dual Baer”. The next example shows that the assumption “N has the SSSP” is not superfluous in Proposition 2.3. Example 2.4. Let R be a von Neumann regular ring which is not semisimple (e.g., R = ∏∞ i=1 Z/2Z). By [8, Proposition 2.26], the R-module RR does not have the SSSP. On the other hand, RR is RR-d-Rickart, but it is not RR-dual Baer (see [14, Corollary 2.9] and [8, Remark 2.2]). 260 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 Proposition 2.5. Let N be an indecomposable R-module. Then the following conditions are equivalent for an R-module M. (i) M is N-dual Baer; (ii) M is N-d-Rickart; (iii) Every nonzero ϕ ∈ HomR(M,N) is an epimorphism. Proof. (i) ⇒ (ii) and (iii) ⇒ (i) are clear. (ii) ⇒ (iii) Let 0 6= ϕ ∈ HomR(M,N). By assumption, Imϕ is a direct summand of N. But N is indecomposable. Then Imϕ = N. This completes the proof. Proposition 2.6. Let M and N be modules such that HomR(M,N) 6= 0 (e.g., N is M-generated). Then the following conditions are equivalent: (i) M is N-dual Baer and N is indecomposable; (ii) Every nonzero homomorphism ϕ ∈ Hom R(M,N) is an epimorphism. Proof. (i) ⇒ (ii) This follows from Proposition 2.5. (ii) ⇒ (i) It is clear that M is N-dual Baer. Now let K be a nonzero direct summand of N. Let K′ be a submodule of N such that N = K ⊕ K′. Since HomR(M,N) 6= 0, there exists a nonzero homomorphism ϕ ∈ HomR(M,N). Let π′ : N → K′ be the projection map and let i′ : K′ → N be the inclusion map. Then i′π′ϕ ∈ HomR(M,N). Assume that i′π′ϕ 6= 0. By hypothesis, Imi′π′ϕ = N. So K′ = N. Thus K = 0, a contradiction. Therefore i′π′ϕ = 0. Hence K′ = 0 and K = N. It follows that N is indecomposable. The following result describes the structure of R-modules which are RR-dual Baer, where R is a commutative principal ideal domain which is not a field. Proposition 2.7. Let R be a commutative principal ideal domain which is not a field. Then the following conditions are equivalent for an R-module M: (i) M is RR-dual Baer; (ii) M is RR-d-Rickart; (iii) M has no nonzero cyclic torsion-free direct summands; (iv) HomR(M,RR) = 0. Proof. (i) ⇒ (ii) This is clear. (ii) ⇒ (iii) Assume that M has an element x such that xR is a direct summand of M and RR ∼= xR. Let π : M → xR be the projection map and let f : xR → RR be an isomorphism. Then fπ : M → RR is an epimorphism. Let α be a nonzero element of R which is not invertible. Consider the homomorphism g : RR → RR defined by g(r) = αr for all r ∈ R. Then gfπ ∈ HomR(M,RR) and Imgfπ = αR. It is clear that αR 6= 0 and αR 6= R. Thus αR is not a direct summand of R. So M is not RR-d-Rickart, a contradiction. (iii) ⇒ (iv) Assume that HomR(M,RR) 6= 0. So there exists a nonzero homomorphism f : M → RR. Thus Imf = aR for some nonzero a ∈ R since R is a principal ideal domain. Then M/Kerf ∼= aR ∼= RR is a projective R-module. It follows that Kerf is a direct summand of M. Let Y be a submodule of M such that M = Kerf ⊕Y . Therefore Y ∼= RR. This contradicts our assumption. Hence HomR(M,RR) = 0. (iv) ⇒ (i) This is immediate. Example 2.8. Consider a Z-module M = Q(I) ⊕ T, where T is a torsion Z-module and I is an index set. Suppose that M is not Z-dual Baer. By Proposition 2.7, there exists a cyclic submodule L of M such that L ∼= Z and L is a direct summand of M. Let N be a submodule of M such that M = L⊕N. 261 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 Since T is the torsion submodule of M, we have T ⊆ N. Hence T is a direct summand of N. Let K be a submodule of N such that N = K⊕T . Thus M = L⊕K⊕T . Therefore L⊕K ∼= Q(I). So L is injective, a contradiction. It follows that M is Z-dual Baer. On the other hand, note that if T ∼= Z(2∞) ⊕ Z/8Z, then M is not a dual Baer module (see [14, Corollary 3.5]. In Proposition 2.7, we studied when an R-module M is RR-dual Baer. Next, we investigate when RR is M-dual Baer for an R-module M. Proposition 2.9. The following conditions are equivalent for an R-module M: (i) The R-module RR is M-dual Baer; (ii) M is a semisimple module. Proof. (i) ⇒ (ii) Let x ∈ M. Consider the R-homomorphism ϕ : R → M defined by ϕ(r) = xr for all r ∈ R. Then Imϕ = xR. Since RR is M-dual Baer, it follows that for any submodule L of M, L = ∑ x∈L xR is a direct summand of M. Therefore M is semisimple. (ii) ⇒ (i) is obvious. Corollary 2.10. The following conditions are equivalent for a ring R: (i) The R-module RR is dual Baer; (ii) The R-module RR is E(R)-dual Baer; (iii) R is a semisimple ring. Proof. (i) ⇔ (iii) By [14, Corollary 2.9]. (ii) ⇔ (iii) This follows from Proposition 2.9. Remark 2.11. If K is a submodule of an R-module M such that K is M-dual Baer, then K is a direct summand of M. In particular, if the R-module M is E(M)-dual Baer, then M is an injective module. The next example shows that even if a module M is injective, the module M need not be M-dual Baer. Example 2.12. Let R be a self injective ring which is not semisimple (e.g., R = ∏∞ n=1 Z/2Z). Then E(RR) = RR. By [14, Corollary 2.9], the R-module RR is not RR-dual Baer. Next, we will be concerned with the modules M which are N-dual Baer for all modules N. We begin with the following proposition which provides a class of rings R whose semisimple modules are N-dual Baer for any R-module N. Proposition 2.13. Let R be a right noetherian right V-ring and let M be a semisimple R-module. Then M is N-dual Baer for every R-module N. Proof. Let N be an R-module. It is clear that for any ϕ ∈ HomR(M,N), Imϕ is semisimple. Let A be a subset of HomR(M,N). Then ∑ f∈A Imf is a semisimple submodule of N. Since R is a right noetherian right V-ring, ∑ f∈A Imf is injective by [4, Proposition 1]. Therefore ∑ f∈A Imf is a direct summand of N. So M is N-dual Baer. The next example shows that the condition “R is a right noetherian ring” in the hypothesis of Proposition 2.13 is not superfluous. Example 2.14. Let F be a field and let R = ∏ n∈N Fn such that Fn = F for all n ∈ N. Then R is a commutative V-ring which is not noetherian. Note that Soc(R) = ⊕n∈NFn is an essential proper ideal of R. In particular, Soc(R) is not a direct summand of R. So Soc(R) is not RR-dual Baer. 262 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 Following [13], a module M is called noncosingular if for every nonzero module N and every nonzero homomorphism f : M → N, Imf is not a small submodule of N. Proposition 2.15. Let M be a module. Assume that M is N-dual Baer for every R-module N. Then every factor module of M is injective. In particular, M is a noncosingular module. Proof. Let L be a submodule of M. Let π : M → M/L be the natural epimorphism and let µ : M/L → E(M/L) be the inclusion map. Then µπ ∈ HomR(M,E(M/L)) and Imµπ = M/L. Since M is E(M/L)-dual Baer, M/L is a direct summand of E(M/L). So M/L is injective. This completes the proof. Proposition 2.16. Let R be a right noetherian ring. Then the following conditions are equivalent for an R-module M: (i) M is N-dual Baer for all R-modules N; (ii) Every factor module of M is an injective R-module. Proof. (i) ⇒ (ii) By Proposition 2.15. (ii) ⇒ (i) Let N be an R-module. It is clear that Imϕ is injective for every ϕ ∈ HomR(M,N). Since the ring R is right noetherian, ∑ f∈A Imf is injective for every subset A of Hom R(M,N) by [1, Proposition 18.13]. Therefore ∑ f∈A Imf is a direct summand of N. This proves the proposition. Recall that a ring R is called right hereditary if each of its right ideals is projective. It is well known that a ring R is right hereditary if and only if every factor module of an injective right R-module is injective (see, for example [16, 39.16]). The next result is a direct consequence of Proposition 2.16. It determines the structure of R-modules M which are N-dual Baer for all R-modules N, where R is a right noetherian right hereditary ring. Corollary 2.17. Let R be a right noetherian right hereditary ring (e.g., R is a Dedekind domain). Then the following conditions are equivalent for an R-module M: (i) M is N-dual Baer for any R-module N; (ii) M is an injective R-module. Example 2.18. Let M be a Z-module. It is easily seen from Corollary 2.17 that M is N-dual Baer for any Z-module N if and only if M is a direct sum of Z-modules each isomorphic to the additive group of rational numbers Q or to Z(p∞) (for various primes p). Combining Corollary 2.17 and [8, Corollary 2.30], we obtain the following result. Corollary 2.19. The following conditions are equivalent for a ring R: (i) Every injective R-module is dual Baer; (ii) Every injective module is N-dual Baer for every R-module N; (iii) R is a right noetherian right hereditary ring. The next characterization extends [14, Corollary 2.5]. Theorem 2.20. Let M and N be two R-modules. Then M is N-dual Baer if and only if for any direct summand M′ of M and any submodule N′ of N, M′ is N′-dual Baer. Proof. Let M′ = eM for some e2 = e ∈ EndR(M) and let N′ be a submodule of N. Let {ϕi}I be a family of homomorphisms in Hom R(M′,N′). Since ϕie(M) = ϕi(M′) ⊆ N′ ⊆ N for every i ∈ I and M is N-dual Baer, ∑ i∈I ϕie(M) is a direct summand of N. Therefore ∑ i∈I ϕi(M ′) is a direct summand of N′. It follows that M′ is N′-dual Baer. The converse is obvious. 263 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 Corollary 2.21. The following conditions are equivalent for a module M: (i) M is a dual Baer module; (ii) For any direct summand K of M and any submodule N of M, K is N-dual Baer. From [14, Example 3.1 and Theorem 3.4], it follows that a direct sum of dual Baer modules is not dual Baer, in general. Next, we focus on when a direct sum of N-dual Baer modules is also N-dual Baer for some module N. Proposition 2.22. Let N be a module having the SSSP and let {Mi}I be a family of modules. Then⊕ i∈I Mi is N-dual Baer if and only if Mi is N-dual Baer for all i ∈ I. Proof. Suppose that ⊕ i∈I Mi is N-dual Baer. By Theorem 2.20, Mi is N-dual Baer for all i ∈ I. Conversely, assume that Mi is N-dual Baer for all i ∈ I. Let {ϕλ}Λ be a family of homomorphisms in Hom R( ⊕ i∈I Mi,N). For each i ∈ I, let µi : Mi → ⊕ i∈I Mi denote the inclusion map. Then for every i ∈ I and every λ ∈ Λ, ϕλµi ∈ HomR(Mi,N). Since Mi is N-dual Baer for every i ∈ I, it follows that Im(ϕλµi) is a direct summand of N for every (i,λ) ∈ I × Λ. Note that for each λ ∈ Λ, Imϕλ = ∑ i∈I Im(ϕλµi). As N has the SSSP, ∑ λ∈Λ Imϕλ = ∑ λ∈Λ ∑ i∈I Im(ϕλµi) is a direct summand of N. Therefore ⊕ i∈I Mi is N-dual Baer. The following result is taken from [14, Theorem 2.1]. Theorem 2.23. The following conditions are equivalent for a module M and S = EndR(M): (i) M is a dual Baer module; (ii) For every nonempty subset A of S, ∑ f∈A Imf = e(M) for some idempotent e ∈ S; (iii) M has the SSSP and for every ϕ : M → M, Imϕ is a direct summand of M. Corollary 2.24. Let {Mi}I be a family of modules and let j ∈ I. Then ⊕ i∈I Mi is Mj-dual Baer if and only if Mi is Mj-dual Baer for all i ∈ I. Proof. The necessity follows from Theorem 2.20. Conversely, by assumption, we have Mj is Mj-dual Baer. Then Mj is a dual Baer module. By Theorem 2.23, Mj has the SSSP. Applying Proposition 2.22,⊕ i∈I Mi is Mj-dual Baer. In the following result, we present conditions under which a module N is ⊕n i=1 Mi-dual Baer for some modules Mi (1 ≤ i ≤ n). Theorem 2.25. Let M1, . . . , Mn be R-modules, where n ∈ N. Assume that Mi is Mj-projective for all i > j ∈ {1, 2, . . . ,n}. Then for any R-module N, N is ⊕n i=1 Mi-dual Baer if and only if N is Mi-dual Baer for all i ∈{1, 2, . . . ,n}. Proof. The necessity follows from Theorem 2.20. Conversely, suppose that N is Mi-dual Baer for all i ∈{1, 2, . . . ,n}. We will show that N is ⊕n i=1 Mi-dual Baer. By induction on n and taking into account [9, Proposition 4.33], it is sufficient to prove this for the case n = 2. Assume that N is Mi-dual Baer for i = 1, 2 and M2 is M1-projective. Let {φλ}Λ be a family of homomorphisms in Hom R(N,M1 ⊕M2). Let π2 : M1 ⊕ M2 → M2 be the projection of M1 ⊕ M2 on M2 along M1. We want to prove that∑ λ∈Λ Imφλ is a direct summand of M1 ⊕ M2. Since N is M2-dual Baer, ∑ λ∈Λ π2φλ(N) is a direct summand of M2. So ∑ λ∈Λ π2φλ(N) is M1-projective by [9, Proposition 4.32]. As M1 + (∑ λ∈Λ Imφλ ) = M1 ⊕ (∑ λ∈Λ π2φλ(N) ) is a direct summand of M1 ⊕M2, there exists a submodule L ≤ ∑ λ∈Λ Imφλ such that M1 + (∑ λ∈Λ Imφλ ) = M1⊕L by [9, Lemma 4.47]. Thus ∑ λ∈Λ Imφλ = ( M1 ∩ (∑ λ∈Λ Imφλ )) ⊕L by modularity. It is easily seen that ∑ λ∈Λ π2φλ(N) is a direct summand of M2. Let K2 be a submodule of M2 such that M2 = K2⊕ (∑ λ∈Λ π2φλ(N) ) . Therefore M1⊕M2 = M1⊕L⊕K2. Let π1 : M1⊕(L⊕K) → M1 264 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 be the projection of M1⊕M2 on M1 along L⊕K. Then π1φλ ∈ Hom R(N,M1) for every λ ∈ Λ. Moreover, we have ∑ λ∈Λ π1φλ(N) = π1 (∑ λ∈Λ Imφλ ) = ((∑ λ∈Λ Imφλ ) + (L⊕K) ) ∩M1. But ∑ λ∈Λ Imφλ = ( M1 ∩ (∑ λ∈Λ Imφλ )) ⊕L. Then, ∑ λ∈Λ π1φλ(N) = (( M1 ∩ (∑ λ∈Λ Imφλ )) ⊕L⊕K ) ∩M1 = M1 ∩ (∑ λ∈Λ Imφλ ) . Since N is M1-dual Baer, ∑ λ∈Λ π1φλ(N) = M1 ∩ (∑ λ∈Λ Imφλ ) is a direct summand of M1. It follows that ( M1 ∩ (∑ λ∈Λ Imφλ )) ⊕L is a direct summand of M1⊕L⊕K2. So ∑ λ∈Λ Imφλ is a direct summand of M1 ⊕M2. Consequently, N is M1 ⊕M2-dual Baer. This completes the proof. Corollary 2.26. Let M1, . . . , Mn be R-modules, where n ∈ N. Assume that Mi is Mj-projective for all i > j ∈ {1, 2, . . . ,n}. Then M = ⊕n i=1 Mi is a dual Baer module if and only if Mi is Mj-dual Baer for all i,j ∈{1, 2, . . . ,n}. Proof. The necessity follows from Theorem 2.20. Conversely, suppose that Mi is Mj-dual Baer for all i,j ∈ {1, 2, . . . ,n}. By Corollary 2.24, M is Mj-dual Baer for all j ∈ {1, 2, . . . ,n}. Since Mi is Mj-projective for all i > j ∈{1, 2, . . . ,n}, M is ⊕n i=1 Mi-dual Baer by Theorem 2.25. Thus M is a dual Baer module. Note that the sufficiency in Corollary 2.26 can be proved by using [14, Theorem 3.10]. Following [8, Definition 5.7], a module M is called N-D2 (or relatively D2 to N) if for any submodule M′ of M, M/M′ is isomorphic to a direct summand of N implies that M′ is a direct summand of M. Proposition 2.27. Let M1, . . . , Mn be R-modules, where n ∈ N. Assume that Mi is Mj-D2 for all i,j ∈ {1, 2, . . . ,n}. Then ⊕n i=1 Mi is a dual Baer module if and only if Mi is Mj-dual Baer for all i,j ∈{1, 2, . . . ,n} and M has the SSSP. Proof. (⇒) By [8, Theorem 5.11], Mi is Mj-d-Rickart for all i,j ∈ {1, 2, . . . ,n}. Note that Mi has the SSSP for every i ∈{1, 2, . . . ,n} (see Theorem 2.23). Applying Proposition 2.3, it follows that Mi is Mj-dual Baer for all i,j ∈{1, 2, . . . ,n}. (⇐) This follows easily from [8, Theorem 5.11], Proposition 2.3 and Theorem 2.23. Theorem 2.28. Let M = ⊕ i∈I Mi be the direct sum of fully invariant submodules Mi. Then M is a dual Baer module if and only if Mi is a dual Baer module for all i ∈ I. Proof. The necessity follows from [14, Corollary 2.5]. Conversely, let S = EndR(M) and let {ϕλ}Λ be a family of homomorphisms in S. For each i ∈ I, let πi : M → Mi be the projection map and let µi : Mi → M be the inclusion map. Note that for each λ ∈ Λ, ϕλ(M) = ∑ i∈I ϕλµi(Mi). Since each Mi (i ∈ I) is fully invariant in M, it follows that ϕλ(M) = ∑ i∈I πiϕλµi(Mi) for all λ ∈ Λ. For every i ∈ I and every λ ∈ Λ, let Ni,λ = πiϕλµi(Mi). Therefore, ∑ λ∈Λ ϕλ(M) = ∑ λ∈Λ ∑ i∈I πiϕλµi(Mi) = ∑ λ∈Λ (∑ i∈I Ni,λ ) = ⊕ i∈I (∑ λ∈Λ Ni,λ ) . Since each Mi (i ∈ I) is dual Baer, each Mi (i ∈ I) has the SSSP by Theorem 2.23. Thus ∑ λ∈Λ Ni,λ is a direct summand of Mi for every i ∈ I. So ∑ λ∈Λ ϕλ(M) is a direct summand of M. Consequently, M is a dual Baer module. 265 T. Amouzegar, R. Tribak / J. Algebra Comb. Discrete Appl. 7(3) (2020) 259–267 We conclude this paper by showing a new characterization of dual Baer modules. Let M be an R-module with S = EndR(M). Then for every nonempty subset A of S, we denote lS(A) = {ϕ ∈ S | ϕA = 0} and rM (A) = {m ∈ M | Am = 0}. We also denote lS(N) = {ϕ ∈ S | ϕ(N) = 0} for any submodule N of M. Recall that a ring R is called a Baer ring if for every nonempty subset I ⊆ R, there exists an idempotent e ∈ R such that lS(I) = Re. Proposition 2.29. ([5, Proposition 2.3]) For an R-module M, S = EndR(M) is a Baer ring if and only if rM (lS( ∑ ϕ∈A Imϕ)) is a direct summand of M for all nonempty subsets A of S. The next example shows that if M is a module such that S = EndR(M) is a Baer ring, then M is not a dual Baer module, in general. Example 2.30. Consider the Z-module M = Z. Then S = EndZ(M) ∼= Z. Clearly, Z is a Baer ring. On the other hand, it is easily seen that M is not a dual Baer module. Note that if M is an R-module with S = EndR(M), then for any nonempty subset A of S, lS(A) = lS(AM), where AM = ∑ f∈A Imf. The next result can be considered as an analogue of [8, Theorem 3.5]. Theorem 2.31. The following are equivalent for an R-module M and S = EndR(M): (i) M is a dual Baer module; (ii) S is a Baer ring and AM = rM (lS(AM)) for every nonempty subset A of S; (iii) S is a Baer ring and IM = rM (lS(IM)) for every right ideal I of S. Proof. (i) ⇒ (ii) From [15, Theorem 3.6], it follows that S is a Baer ring. Moreover, we have rM (lS(AM)) = rM (lS(A)) = rM (S(1 −e)) = e(M) = AM for all nonempty subsets A of S. (ii) ⇒ (iii) This is obvious. (iii) ⇒ (i) Let I be a right ideal of S. Since S is a Baer ring, rM (lS(IM)) is a direct summand of M by Proposition 2.29. But IM = rM (lS(IM)). Then IM is a direct summand of M. By Theorem 2.23, it follows that M is a dual Baer module. Combining Theorem 2.31 and [10, Theorem 4.1], we get the following result. Corollary 2.32. Let M be an R-module such that IM = rM (lS(IM)) for every right ideal I of S = EndR(M). If M is a Baer module, then M is a dual Baer module. References [1] F. W. Anderson, K. R. 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