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/p^{2} + Z〉 and 〈1/p + Z〉 of the Z-module Z_{p}_{∞} 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 N_{1}, N_{2}be two submodules of M with (N_{1} :_{R} M) and (N_{2} :_{R} M) prime ideals of R. Then N_{1} ∩ N_{2}is a quasi 2-absorbing submodule of M.
ProofSince (N_{1} ∩ N_{2} :_{R} M) = (N_{1} :_{R} M) ∩ (N_{2} :_{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} I^{n}M) = (N :_{R} I^{n}^{+1}M), 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} I^{2}M) = (N :_{R} I^{3}M). It is clear that (N :_{R} I^{2}M) ⊆(N :_{R} I^{3}M). Since N is quasi 2-absorbing submodule, (N :_{R} I^{3}M)I^{3}M ⊆N implies that (N :_{R} I^{3}M)I^{2}M ⊆N or I^{2}M ⊆N. If (N :_{R} I^{3}M)I^{2}M ⊆N, then (N :_{R} I^{3}M) ⊆(N :_{R} I^{2}M). If I^{2}M ⊆N, then (N :_{R} I^{2}M) = R = (N :_{R} I^{3}M).
(c) Let M be a finitely generated Laskerian R-module. Then $\sqrt{(N{:}_{R}M)}=(rad(N){:}_{R}M)$ by [9, Theorem 5]. Now the result follows from the fact that $\sqrt{(N{:}_{R}M)}$ 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} Ann_{R}(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 {K_{i}}_{i}_{∈}_{I} be a chain of quasi 2- absorbing submodules of M. Then ∩_{i}_{∈}_{I}K_{i} is a quasi 2-absorbing submodule of M.
ProofClearly, (∩_{i}_{∈}_{I}K_{i} :_{R} M) ≠ R. Let a, b, c ∈ R and abc ∈ (∩_{i}_{∈}_{I}K_{i} :_{R} M) = ∩_{i}_{∈}_{I} (K_{i} :_{R} M). Assume to the contrary that ab ∉ ∩_{i}_{∈}_{I} (K_{i} :_{R} M), bc ∉ ∩_{i}_{∈}_{I} (K_{i} :_{R} M), and ac ∉ ∩_{i}_{∈}_{I} (K_{i} :_{R} M). Then exist m, n, t ∈ I such that ab ∉ (K_{n} :_{R} M), bc ∉ (K_{m} :_{R} M), and ac ∉ (K_{t} :_{R} M). Since {K_{i}}_{i}_{∈}_{I} is a chain, we can assume without loss of generality that K_{m} ⊆K_{n} ⊆K_{t}. Then
$$({K}_{m}{:}_{R}M)\subseteq ({K}_{n}{:}_{R}M)\subseteq ({K}_{t}{:}_{R}M).$$As abc ∈ (K_{m} :_{R} M), we have either ab ∈ (K_{m} :_{R} M) or ac ∈ (K_{m} :_{R} M) or bc ∈ (K_{m} :_{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^{−1}N is a quasi 2-absorbing S^{−1}R-submodule of S^{−1}M. Furthermore, if S^{−1}N is a quasi 2-absorbing S^{−1}R-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,
$$({S}^{-1}N{:}_{{S}^{-1}R}{S}^{-1}M)={S}^{-1}((N{:}_{R}M))$$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 R_{i} be a commutative ring with identity and M_{i} be an R_{i}-module for i = 1, 2. Let R = R_{1} × R_{2}. Then M = M_{1} ×M_{2} is an R-module and each submodule of M is in the form of N = N_{1} × N_{2} for some submodules N_{1} of M_{1} and N_{2} of M_{2}.
Theorem 2.14
Let R = R_{1}×R_{2}be a decomposable ring and let M = M_{1} × M_{2}be an R-module, where M_{1}is a multiplication R_{1}-module and M_{2}is a multiplication R_{2}-module. Suppose that N = N_{1} × N_{2}is a proper submodule of M. Then the following conditions are equivalent:
(a) N is a quasi 2-absorbing submodule of M;
(b) Either N_{1} = M_{1}and N_{2}is a quasi 2-absorbing submodule of M_{2}or N_{2} = M_{2}and N_{1}is a quasi 2-absorbing submodule of M_{1}or N_{1}, N_{2}are prime submodules of M_{1}, M_{2}, respectively.
ProofSince (N_{1} × N_{2} :_{R}_{1×}_{R}_{2}M_{1} × M_{2}) = (N_{1} :_{R}_{1}M_{1}) × (N_{2} :_{R}_{2}M_{2}), the result follows from [11, Theorem 1.2] and Lemma 2.13.
Theorem 2.15
Let R = R_{1} × R_{2} × ··· ×R_{n} (2 ≤ n < ∞) be a decomposable ring and M = M_{1} × M_{2} ··· × M_{n} be an R-module, where for every 1 ≤ i ≤ n, M_{i} is a multiplication R_{i}-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) Either$N={\times}_{i=1}^{n}{N}_{i}$such that for some k ∈ {1, 2, …, n}, N_{k} is a quasi 2- absorbing submodule of M_{k} and N_{i} = M_{i} for every i ∈ {1, 2, …, n} {k} or$N={\times}_{i=1}^{n}{N}_{i}$such that for some k,m ∈ {1, 2, …, n}, N_{k} is a prime submodule of M_{k}, N_{m} is a prime submodule of M_{m}, and N_{i} = M_{i} 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 = M_{1} × ···× M_{n}_{−1}. We show that the result holds when M = K × M_{n}. By Theorem ??, N is a quasi 2-absorbing submodule of M if and only if either N = L × M_{n} for some quasi 2-absorbing submodule L of K or N = K × L_{n} for some quasi 2-absorbing submodule L_{n} of M_{n} or N = L × L_{n} for some prime submodule L of K and some prime submodule L_{n} of M_{n}. Notice that a proper submodule L of K is a prime submodule of K if and only if $L={\times}_{i=1}^{n-1}{N}_{i}$ such that for some k ∈ {1, 2, …, n − 1}, N_{k} is a prime submodule of M_{k}, and N_{i} = M_{i} for every i ∈ {1, 2, …, n − 1} {k}. Consequently we reach the claim.