Definition 2.1
We say that a proper submodule N of an R-module M is a quasi 2-absorbing submodule if (N :R M) is a 2-absorbing ideal of R.
Example 2.2
By [12, 2.3], every 2-absorbing submodule is a quasi 2-absorbing submodule. But the converse is not true in general. For example, the submodules 〈1/p2 + Z〉 and 〈1/p + Z〉 of the Z-module Zp∞ are quasi 2-absorbing submodules which are not 2-absorbing submodules.
An R-module M is said to be a multiplication module if for every submodule N of M there exists an ideal I of R such that N = IM [4].
Proposition 2.3
Let M be a multiplication R-module. Then a submodule N of M is a 2-absorbing submodule of M if and only if it is a quasi 2-absorbing submodule of M.
ProofThis follows from [2, Theorem 3.9].
Proposition 2.4
Let M be an R-module and N1, N2be two submodules of M with (N1 :R M) and (N2 :R M) prime ideals of R. Then N1 ∩ N2is a quasi 2-absorbing submodule of M.
ProofSince (N1 ∩ N2 :R M) = (N1 :R M) ∩ (N2 :R M), the result follows from [3].
Let N be a submodule of an R-module M. The intersection of all prime submodules of M containing N is said to be the (prime) radical of N and denote by rad(N). In case N does not contained in any prime submodule, the radical of N is defined to be M [10].
An R-module M is said to be a Laskerian module if every proper submodule of M is a finite intersection of primary submodules of M [8]. We know that every Noetherian module is Laskerian.
Theorem 2.5
Let M be an R-module and N be a quasi 2-absorbing submodule of M. Then we have the following:
(a) (N :M I) is a quasi 2-absorbing submodules of M for all ideals I of R with I ⊈ (N :R M).
(b) If I is an ideal of R, then (N :R InM) = (N :R In+1M), for all n ≥ 2.
(c) If M is a finitely generated Laskerian R-module, then rad(N) is a quasi 2- absorbing submodule of M.
Proof(a) Let I be an ideal of R with I ⊈ (N :R M). Then ((N :M I) :R M) is a proper ideal of R. Now let a, b, c ∈ R and abcM ⊆(N :M I). Then abcIM ⊆N. Thus either acM ⊆N or cbIM ⊆N or abIM ⊆N. If cbIM ⊆N or abIM ⊆N, then we are done. If acM ⊆N, then acM ⊆(N :M I), as needed.
(b) It is enough to show that (N :R I2M) = (N :R I3M). It is clear that (N :R I2M) ⊆(N :R I3M). Since N is quasi 2-absorbing submodule, (N :R I3M)I3M ⊆N implies that (N :R I3M)I2M ⊆N or I2M ⊆N. If (N :R I3M)I2M ⊆N, then (N :R I3M) ⊆(N :R I2M). If I2M ⊆N, then (N :R I2M) = R = (N :R I3M).
(c) Let M be a finitely generated Laskerian R-module. Then by [9, Theorem 5]. Now the result follows from the fact that is a 2-absorbing ideal of R by [3, Theorem 2.1].
An R-module M is said to be a comultiplication module if for every submodule N of M there exists an ideal I of R such that N = (0 :M I), equivalently, for each submodule N of M, we have N = (0 :M AnnR(N)) [1].
Corollary 2.6
Let M be a comultiplication R-module such that the zero submodule of M is a quasi 2-absorbing submodule. Then every proper submodule of M is a quasi 2-absorbing submodule of M.
ProofThis follows from Theorem 2.5 (a).
Proposition 2.7
Let M be an R-module and {Ki}i∈I be a chain of quasi 2- absorbing submodules of M. Then ∩i∈IKi is a quasi 2-absorbing submodule of M.
ProofClearly, (∩i∈IKi :R M) ≠ R. Let a, b, c ∈ R and abc ∈ (∩i∈IKi :R M) = ∩i∈I (Ki :R M). Assume to the contrary that ab ∉ ∩i∈I (Ki :R M), bc ∉ ∩i∈I (Ki :R M), and ac ∉ ∩i∈I (Ki :R M). Then exist m, n, t ∈ I such that ab ∉ (Kn :R M), bc ∉ (Km :R M), and ac ∉ (Kt :R M). Since {Ki}i∈I is a chain, we can assume without loss of generality that Km ⊆Kn ⊆Kt. Then
As abc ∈ (Km :R M), we have either ab ∈ (Km :R M) or ac ∈ (Km :R M) or bc ∈ (Km :R M). In any case, we have a contradiction.
Definition 2.8
We say that a quasi 2-absorbing submodule N of an R-module M is a minimal quasi 2-absorbing submodule of a submodule K of M, if K ⊆N and there does not exist a quasi 2-absorbing submodule T of M such that K ⊂ T ⊂ N.
It should be noted that a minimal quasi 2-absorbing submodule of M means that a minimal quasi 2-absorbing submodule of the submodule 0 of M.
Lemma 2.9
Let M be an R-module. Then every quasi 2-absorbing submodule of M contains a minimal quasi 2-absorbing submodule of M.
ProofThis is proved easily by using Zorn’s Lemma and Proposition 2.7.
Theorem 2.10
Let M be a Noetherian R-module. Then M contains a finite number of minimal quasi 2-absorbing submodules.
ProofSuppose that the result is false. Let ∑ denote the collection of all proper submodules N of M such that the module M/N has an infinite number of minimal quasi 2-absorbing submodules. Since 0 ∈ ∑, we have ∑ ≠ ∅︀. Therefore ∑ has a maximal member T, since M is a Noetherian R-module. Clearly, T is not a quasi 2-absorbing submodule. Therefore, there exist a, b, c ∈ R such that abc(M/T) = 0 but ab(M/T) ≠ 0, ac(M/T) ≠ 0, and bc(M/T) ≠ 0. The maximality of T implies that M/(T + abM), M/(T + acM), and M/(T + bcM) have only finitely many minimal quasi 2-absorbing submodules. Suppose P/T is a minimal quasi 2-absorbing submodule of M/T. So abcM ⊆T ⊆ P, which implies that either abM ⊆P or acM ⊆P or bcM ⊆P. Thus either P/(T + abM) is a minimal quasi 2-absorbing submodule of M/(T + abM) or P/(T + bcM) is a minimal quasi 2-absorbing submodule of M/(T + bcM) or P/(T + acM) is a minimal quasi 2- absorbing submodule of M/(T + acM). Therefore, there are only a finite number of possibilities for the submodule P. This is a contradiction.
Recall that Z(R) denotes the set of zero divisors of R.
Proposition 2.11
Let N be a submodule of a finitely generated R-module M and S be a multiplicatively closed subset of R. If N is a quasi 2-absorbing submodule and (N :R M) ∩ S = ∅︀, then S−1N is a quasi 2-absorbing S−1R-submodule of S−1M. Furthermore, if S−1N is a quasi 2-absorbing S−1R-submodule and S ∩ Z(R/(N :R M)) = ∅︀, then N is a quasi 2-absorbing submodule of M.
ProofAs M is a finitely generated R-module,
by [13, Lemma 9.12]. Now the result follows from [11, Theorem 1.3].
Lemma 2.12
Let f : M → Ḿ be a monomorphism of R-modules. Then N is a quasi 2-absorbing submodule of M if and only if f(N) is a quasi 2-absorbing submodule of f(M).
ProofThis follows from the fact that (N :R M) = (f(N) :R f(M)).
Lemma 2.13. ([7, Corollary 2.11])
Let N be a submodule of a multiplication Rmodule M. Then N is a prime submodule of M if and only if (N :R M) is a prime ideal of R.
Let Ri be a commutative ring with identity and Mi be an Ri-module for i = 1, 2. Let R = R1 × R2. Then M = M1 ×M2 is an R-module and each submodule of M is in the form of N = N1 × N2 for some submodules N1 of M1 and N2 of M2.
Theorem 2.14
Let R = R1×R2be a decomposable ring and let M = M1 × M2be an R-module, where M1is a multiplication R1-module and M2is a multiplication R2-module. Suppose that N = N1 × N2is a proper submodule of M. Then the following conditions are equivalent:
(a) N is a quasi 2-absorbing submodule of M;
(b) Either N1 = M1and N2is a quasi 2-absorbing submodule of M2or N2 = M2and N1is a quasi 2-absorbing submodule of M1or N1, N2are prime submodules of M1, M2, respectively.
ProofSince (N1 × N2 :R1×R2M1 × M2) = (N1 :R1M1) × (N2 :R2M2), the result follows from [11, Theorem 1.2] and Lemma 2.13.
Theorem 2.15
Let R = R1 × R2 × ··· ×Rn (2 ≤ n < ∞) be a decomposable ring and M = M1 × M2 ··· × Mn be an R-module, where for every 1 ≤ i ≤ n, Mi is a multiplication Ri-module, respectively. Then for a proper submodule N of M the following conditions are equivalent:
(a) N is a quasi 2-absorbing submodule of M;
(b) Eithersuch that for some k ∈ {1, 2, …, n}, Nk is a quasi 2- absorbing submodule of Mk and Ni = Mi for every i ∈ {1, 2, …, n} {k} orsuch that for some k,m ∈ {1, 2, …, n}, Nk is a prime submodule of Mk, Nm is a prime submodule of Mm, and Ni = Mi for every i ∈ {1, 2, …, n} {k,m}.
ProofWe use induction on n. For n = 2 the result holds by Theorem 2.14. Now let 3 ≤ n < ∞ and suppose that the result is valid when K = M1 × ···× Mn−1. We show that the result holds when M = K × Mn. By Theorem ??, N is a quasi 2-absorbing submodule of M if and only if either N = L × Mn for some quasi 2-absorbing submodule L of K or N = K × Ln for some quasi 2-absorbing submodule Ln of Mn or N = L × Ln for some prime submodule L of K and some prime submodule Ln of Mn. Notice that a proper submodule L of K is a prime submodule of K if and only if such that for some k ∈ {1, 2, …, n − 1}, Nk is a prime submodule of Mk, and Ni = Mi for every i ∈ {1, 2, …, n − 1} {k}. Consequently we reach the claim.