On generalizations of classical primary submodules over commutative rings

Abstract

Let $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ be a function where $\mathcal{S}(M)$ is the set of all submodules of R-module M. A proper submodule N of M is called a ϕ-classical primary submodule, if for each m ∊ M and a, b ∊ R with abm ∊ N-ϕ(N), then am ∊ N or bⁿm ∊ N for some positive integer n. Some characterizations of classical primary and ϕ-classical primary submodules are obtained. It is shown that N is a ϕ-classical primary submodule of M if and only if for every m ∊ M−N, (N:m) is a φ-primary ideal of R where (ϕ(N):m) = φ(N:m). Moreover, we investigate relationships between classical primary, ϕ-classical primary and ϕ-primary submodules of modules over commutative rings. Finally, we obtain necessary and sufficient conditions of a ϕ-classical primary submodule in order to be a ϕ-primary submodule.

Keywords:

• classical primary submodule
• ϕ-classical primary submodule
• ϕ-primary submodule
• φ-primary ideal

Public Interest Statement

In this paper, we extend the concept of classical primary submodules to the context of ϕ-classical primary submodules. Some characterizations of classical primary and ϕ-classical primary submodules are obtained. Moreover, we investigate relationships between classical primary, ϕ-classical primary and ϕ-primary submodules of modules over commutative rings.

Competing interests

The authors declare no competing interest.

1. Introduction

Throughout this paper, we assume that all rings are commutative with 1 ≠ 0. Let R be a commutative ring and M be an R-module. We will denote by (N:M) the residual of N by M, that is, the set of all r ∊ R such that rM ⊆ N. Let I be a proper ideal of R. Then $\sqrt{I}=\{r\in R:r^n\in I$, for some positive integer n} denotes the radical ideal of R. A proper ideal I of R is called a weakly primary ideal if whenever 0 ≠ ab ∊ I for a, b ∊ R, then a ∊ I or $b\in \sqrt{I}$. The notion of weakly primary ideals has been introduced and studied by Atani and Farzalipour (2005). Anderson and Badawi (2011) generalized the concept of 2-absorbing ideals to n-absorbing ideals. According to their definition, a proper ideal I of R is said to be an n-absorbing ideal of R if whenever a₁a₂ … a_n+1 ∊ I for a₁, a₂, …, a_n+1 ∊ R, then there are n of the a_i’s whose product is in I. Later, Badawi, Tekir, and Yetkin (2015) generalized the concept of weakly primary ideals to weakly 2-absorbing primary ideals. According to their definition, a proper ideal I of R is said to be a weakly 2-absorbing primary ideal of R if whenever 0 ≠ abc ∊ I for a, b, c ∊ R, then ab ∊ I or $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$ Clearly, every weakly primary ideal is a weakly 2-absorbing primary ideal. Also, Tekir, Koc, and Oral (2016) generalized the concept of quasi-primary ideals to 2-absorbing quasi-primary ideals. According to their definition, a proper ideal I of R is said to be a 2-absorbing quasi-primary ideal of R if $\sqrt{I}$ is a 2-absorbing ideal of R. Thus, a 2-absorbing quasi-primary ideal is quasi-primary.

Let $\phi :\mathcal{I}(R)\to \mathcal{I}(R)\cup \{\mathrm{\varnothing }\}$ be a function where $\mathcal{I}(R)$ is a set of ideals of R. A proper ideal I of R is called a φ-prime ideal of R as in Anderson and Bataineh (2008) if whenever ab ∊ I − φ(I) for a, b ∊ R, then a ∊ I or b ∊ I. Darani (2012) generalized the concept of primary and weakly primary ideals to φ-primary ideals. A proper ideal I of R is said to be a φ-primary ideal of R if whenever ab ∊ I − φ(I) for a, b ∊ R, then a ∊ I or $b\in \sqrt{I}$. Clearly, every φ-prime ideal is a φ-primary ideal. Later, Badawi, Tekir, Ugurlu, Ulucak, and Celikel (2016) generalized the concept of 2-absorbing primary ideals to φ-2-absorbing primary ideals. According to their definition, a proper ideal I of R is said to be a φ-2-absorbing primary ideal of R if whenever abc ∊ I − φ(I) for a, b, c ∊ R, then ab ∊ I or $ac\in \sqrt{I}$ or $bc\in \sqrt{I}$. Thus, a φ-primary ideal is φ-2-absorbing primary.

In 2004, Behboodi introduced the concepts of a classical prime submodule. A proper submodule N of an R-module M is said to be a classical prime submodule of M if whenever abm ∊ N for a, b ∊ R, m ∊ M, then am ∊ N or bm ∊ N. (see also Azizi, 2006; Azizi, 2008; Behboodi, 2006, in which, the notion of classical prime submodules is named “weakly prime submodules”). For more information on classical prime submodules, the reader is referred to (Arabi-Kakavand & Behboodi, 2014; Behboodi, 2007; Behboodi & Shojaee, 2010; Yılmaz & Cansu, 2014). Later, Baziar and Behboodi (2009) introduced the concepts of a classical primary submodule. According to their definition, a proper submodule N of M is said to be a classical primary submodule of M if whenever abm ∊ N for a, b ∊ R, m ∊ M, then am ∊ N or bⁿm ∊ M for some positive integer n. Clearly, every classical prime submodule is a classical primary. Also, Behboodi, Jahani-Nezhad, and Naderi (2011) introduced the concepts of a classical quasi-primary submodule. According to their definition, a proper submodule N of M is said to be a classical quasi-primary submodule of M if whenever abm ∊ N for a, b ∊ R, m ∊ M, then aⁿm ∊ N or bⁿm ∊ N for some positive integer n. Thus, a classical primary submodule is classical quasi-primary. The notion of weakly classical primary submodules has been introduced and studied by Mostafanasab (2015). A proper submodule N of an R-module M is said to be a weakly classical primary submodule of M if whenever 0 ≠ abm ∊ N for a, b ∊ R, m ∊ M, then am ∊ N or bⁿm ∊ N for some positive integer n. Mostafanasab, Tekir, and Oral (2016) introduced the concepts of a weakly classical prime submodule. According to their definition, a proper submodule N of M is said to be a weakly classical prime submodule of M if whenever 0 ≠ abm ∊ N for a, b ∊ R, m ∊ M, then am ∊ N or bm ∊ N.

Zamani (2010), generalized the concept of prime and weakly prime submodules to ϕ-prime submodules. Let $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ be a function where $\mathcal{S}(M)$ is the set of all submodules of M. Recall that a proper submodule N of M is called a ϕ-prime submodule of M as in Zamani (2010) if whenever am ∊ N − ϕ(N) for a ∊ R, m ∊ M, then m ∊ N or a ∊ (N:M). Also, Ebrahimpour and Mirzaee in (2017), generalized the concept of semiprime and weakly semiprime submodules to ϕ-semiprime submodules. According to their definition, a proper submodule N of M is said to be a ϕ-semiprime submodule of M if whenever a² m ∊ N − ϕ(N) for a ∊ R, m ∊ M, then am ∊ N.

Motivated and inspired by the above works, the purposes of this paper are to introduce generalizations of classical primary submodule to the context of ϕ-classical primary submodule. A proper submodule N of M is said to be a ϕ-classical primary submodule of M if whenever abm ∊ N − ϕ(N) for a, b ∊ R, m ∊ M, then am ∊ N or bⁿm ∊ N for some positive integer n. Some characterizations of classical primary and ϕ-classical primary submodules are obtained. We show that N is a ϕ-classical primary submodule of M if and only if for every m ∊ M − N, (N:m) is a φ-primary ideal of R with (ϕ(N):m) = φ(N:m). Moreover, we investigate relationships between classical primary, ϕ-classical primary and ϕ-primary submodules of modules over commutative rings. Finally, we obtain necessary and sufficient conditions of a ϕ-classical primary submodule in order to be a ϕ-primary submodule.

2. Some basic properties of ϕ-classical primary submodules

The results of the following theorems seem to play an important role to study ϕ-classical primary submodules of modules over commutative rings; these facts will be used frequently and normally, we shall make no reference to this definition.

Definition 2.1.

Let M be an R-module and let $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ be a function where $\mathcal{S}(M)$ be a set of all submodules of M. A proper submodule N of M is called a ϕ-classical primary submodule, if for each m ∊ M, a, b ∊ R with abm ∊ N − ϕ(N), then am ∊ N or bⁿm ∊ N for some positive integer n.

Remark 2.2.

It is easy to see that every classical primary submodule is ϕ-classical primary.

The following example shows that the converse of Remark 2.2 is not true.

Example 2.3.

Let $R=\mathbf{Z}$ and $M={\mathbf{Z}}_2\times {\mathbf{Z}}_3\times \mathbf{Z}$. Clearly, M is an R-module. Consider the submodule $N=\{([0],[0],0)\}$ of an R-module M. Define $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ by $\varphi (K)=\{([0],[0],0)\}$ for every submodule K of M. It is easy to see that N is a ϕ-classical primary submodule of M. Notice that $2·3([1],[1],0)\in \{([0],[0],0)\},$ but $2([1],[1],0)\notin \{([0],[0],0)\}$ and $3^n([1],[1],0)\notin \{([0],[0],0)\}$ for all positive integer n. Therefore, N is not a classical primary submodule of M.

Throughout the rest of this paper, M is an R-module and $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \left\{\mathrm{\varnothing }\right\},$ $\phi :\mathcal{I}(R)\to \mathcal{I}(R)\cup \{\mathrm{\varnothing }\}$ are functions. Since N − ϕ(N) = N − (N ∩ ϕ(N)) and I − φ(I) = I − (I ∩ φ(I)) for $N\in \mathcal{S}(M),I\in \mathcal{I}(R),$ without loss of generality, we will assume that ϕ(N) ⊆ N and φ(I) ⊆ I.

Theorem 2.4.

Let M be an R-module. Then, the following statements hold:

(1)	If N is a ϕ-classical primary submodule of M, then (N:m) is a φ-primary ideal of R for every m ∊ M − N with (ϕ(N):m) ⊆ φ(N:m).
(2)	If φ(N:m) ⊆ (ϕ(N):m) and (N:m) is a φ-primary ideal of R for every m ∊ M − N, then N is a ϕ-classical primary submodule of M.

Proof

1. Let a, b ∊ R such that ab ∊ (N:m) − φ(N:m). Then abm ∊ N and $ab\notin \phi (N:m).$ Since (ϕ(N):m) ⊆ φ(N:m), we have $ab\notin (\varphi (N):m).$ Clearly, abm ∊ N − ϕ(N). By assumption, am ∊ N or bⁿm ∊ N for some positive integer n. Therefore, a ∊ (N:m) or bⁿ ∊ (N:m) for some positive integer n. Hence, (N:m) is a φ-primary ideal of R.

2. Let a, b ∊ R and m ∊ M − N such that abm ∊ N − ϕ(N). Then ab ∊ (N:m) and $ab\notin (\varphi (N):m).$ Since φ(N:m) ⊆ (ϕ(N):m), we have $ab\notin \phi (N:m).$ It is clear that ab ∊ (N:m) − φ(N:m). By hypothesis, a ∊ (N:m) or bⁿ ∊ (N:m) for some positive integer n. Clearly, am ∊ N or bⁿm ∊ N some positive integer n. Hence, N is a ϕ-classical primary submodule of M.        □

The following example shows that the converse of Theorem 2.4 is not true.

Example 2.5

1. Let $M=\mathbf{Z}\times \mathbf{Z}\times \mathbf{Z}$ be a $\mathbf{Z}$-module. Define $\phi :\mathcal{I}(R)\to \mathcal{I}(R)\cup \{\mathrm{\varnothing }\}$ by $\phi (I)=\left\{\begin{array}{c}\left\{0\right\};I=\left\{0\right\}\\ 2\mathbf{Z};I=2\mathbf{Z}\\ 3\mathbf{Z};I=3\mathbf{Z}\\ \mathrm{\varnothing };\text{otherwise}\\ \end{array}\right.$ for every ideal I of R. Consider the submodule $N=\left\{0\right\}\times 2\mathbf{Z}\times 3\mathbf{Z}$ of M. Clearly, (N:(m₁, m₂, m₃)) is a φ-primary ideal of R, where (m₁, m₂, m₃) ∊ M − N. Define $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ by $\varphi (K)=\left\{(0,0,0)\right\}$ for every submodule K. Notice that 2⋅3(0, 1, 1) ∊ N − ϕ(N), but $2(0,1,1)\notin N$ and $3^n(0,1,1)\notin N$ for all positive integer n. Therefore, N is not a ϕ-classical primary submodule of M.

2. Let $M={\mathbf{Z}}_{10}$ be an ${\mathbf{Z}}_{10}$-module. Define $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ by $\varphi (K)=\left\{[0]\right\}$ for every submodule K. Consider the submodule $N=\left\{[0]\right\}$ of an R-module M. Clearly, N is a ϕ-classical primary submodule of M. Define $\phi :\mathcal{I}(R)\to \mathcal{I}(R)\cup \{\mathrm{\varnothing }\}$ by φ(I) = ∅ for every ideal I of R. Notice that $[2][5]\in \left\{[0]\right\}=(N:[1])-\phi (N:[1]),$ but [2] ∉ (N:[1]) and [5]ⁿ ∉ (N:[1]) for all positive integer n. Therefore, (N:[1]) is not a φ-primary ideal of R.

Theorem 2.6

Let (ϕ(N):m) = φ(N:m) for all m ∊ M − N. Then N is a ϕ-classical primary submodule of M if and only if (N:m) is a φ-primary ideal of R for all m ∊ M − N.

Proof

It is clear from Theorem 2.4.        □

Theorem 2.7

If N is a ϕ-classical primary submodule of an R-module M, then $(N:r)=\left\{m\in M:rm\in N\right\}$ is a ϕ-classical primary submodule of M for every r ∊ R − (N:M) with (ϕ(N):r) ⊆ ϕ(N:r).

Proof

Let a, b ∊ R and m ∊ M such that abm ∊ (N:r) − ϕ(N:r). Then, rabm ∊ N and $abm\notin \varphi (N:r).$ Since (ϕ(N):r) ⊆ ϕ(N:r), we have rabm ∊ N − ϕ(N). By assumption, arm ∊ N or bⁿm ∊ N for some positive integer n. Therefore, am ∊ (N:r) or bⁿm ∊ N ⊆ (N:r) for some positive integer n. Hence, (N:r) is a ϕ-classical primary submodule of M.        □

Remark 2.8

Let N be a ϕ-classical primary submodule of M, and $r_1\in R-\left(N:M\right),$$r_2\in R-\left(\left(N:r_1\right):M\right),\dots$ with $(\varphi (N):r_1)\subseteq \varphi (N:r_1),\left((\varphi (N):r_1):r_2\right)\subseteq \varphi \left((N:r_1):r_2\right),\dots .$ Then (N:r₁), (((N:r₁):r₂)…), … are ϕ-classical primary submodules of M andN ⊆ (N:r₁) ⊆ (((N:r₁):r₂)…) ⊆ …..

A submodule N of an R-module M is said to be irreducible if N is not the intersection of two submodules of M which properly contain it.

Proposition 2.9

Let N be an irreducible submodule of an R-module M. For every r ∊ R if (N:r) = (N:r²), then N is a ϕ-classical primary submodule of M.

Proof

Let a, b ∊ R and m ∊ M such that abm ∊ N − ϕ(N). Suppose that $am\notin N$ and $b^{n}m\notin N$ for all positive integer n. Clearly, N ⊆ (N + Ram) ∩ (N + Rbⁿm) for all positive integer n. Let m₀ ∊ (N + Ram) ∩ (N + Rbⁿm). This implies that m₀ ∊ N + Ram and m₀ ∊ N + Rbⁿm. Then, there exist r₁, r₂ ∊ R and n₁, n₂ ∊ N such that n₁ + r₁am = m₀ = n₂ + r₂bⁿm. Since an₁ + r₁a²m = am₀ = an₂ + ar₂bⁿm, we have a²r₁ m ∊ N.It follows that r₁m ∊ (N:a²). By the assumption, r₁m ∊ (N:a). So, r₁am ∊ N. Thus, N = (N + Ram) ∩ (N + Rbⁿm). Now since N is an irreducible of M, we have am ∊ N + Ram ⊆ N or bⁿm ∊ N + Rbⁿm ⊆ N, a contradiction. Hence, N is a ϕ-classical primary submodule of M.        □

Corollary 2.10

Let R be a Boolean ring. If N is an irreducible submodule of M, then N is a ϕ-classical primary submodule of M.

Proof

It is clear from Proposition 2.9.        □

Theorem 2.11

Let M, M^′ be two R-modules and let f:M → M^′ be a homomorphism. Suppose that ${\varphi }^{\prime }:S(M^{\prime })\to S(M^{\prime })\cup \{\mathrm{\varnothing }\}$ is a function. Then, the following statements hold:

(1)	If N′ is a ϕ′-classical primary submodule of M′ and $f(\varphi (N))\subseteq {\varphi }^{\prime }(f(N))$, then f−1(N′) is a ϕ-classical primary submodule of M.
(2)	Let f be If N is a ϕ-classical primary submodule of M and f(ϕ(N)) ⊆ ϕ′(f(N)), then f(N) is a ϕ′-classical primary submodule of M′.

Proof

1. Let a, b ∊ R and m ∊ M such that $abm\in f^{-1}(N^{\prime })-\varphi \left(f^{-1}(N^{\prime })\right).$ Since f is homomorphism, $abf(m)=f(abm)\in N^{\prime }$. Clearly, $abf(m)=f(abm)\notin {\varphi }^{\prime }(N^{\prime })$ so abf(m) ∊ N^′ − ϕ^′(N^′). By assumption, f(am) = af(m) ∊ N^′ or f(bⁿm) = bⁿf(m) ∊ N^′ for some positive integer n. Thus am ∊ f⁻¹(N^′) or bⁿm ∊ f⁻¹(N^′). Therefore, f⁻¹(N^′) is a ϕ-classical primary submodule of M.

2. Let a, b ∊ R and m^′ ∊ M^′ such that abm^′ ∊ f(N) − ϕ^′(f(N)). Since f is surjectivity, there exists m ∊ M such that m^′ = f(m). Therefore, f(abm) = abf(m) ∊ f(N). So abm ∊ N. Clearly, $abm\notin \varphi (N).$ It implies that abm ∊ N − ϕ(N). By assumption, am ∊ N or bⁿm ∊ N for some positive integer n. Thus, am^′ ∊ f(N) or bⁿm^′ ∊ f(N). Hence, f(N) is a ϕ^′-classical primary submodule of M^′.        □

Let N be a submodule of an R-module M and let $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ be a function. Define ${\varphi }_N:S(M/N)\to S(M/N)\cup \{\mathrm{\varnothing }\}$ by$${\varphi }_N(K/N)=\left\{\begin{array}{cc}(\varphi (K)+N)/N\hfill & ;\varphi (\text{K})\ne \mathrm{\varnothing }\\ \mathrm{\varnothing }\hfill & ;\varphi (\text{K})\text{=}\mathrm{\varnothing },\end{array}\right.$$

for every submodule K of M with N ⊆ K (Zamani, 2010). Zamani (2010) gives relations between ϕ-prime submodules of M and ϕ_N-prime submodules of M/N. This leads us to give relations between ϕ-classical primary submodules of M and ϕ_N-classical primary submodules of M/N.

Theorem 2.12

Let N, K be two submodules of M. If K is a ϕ-classical primary submodule of M, then K/N is a ϕ_N-classical primary submodule of M/N.

Proof

Let a, b ∊ R and m ∊ M such that ab(m + N) ∊ (K/N) − ϕ_N(K/N) = (K/N) − (ϕ(K) + N)/N = (K − ϕ(K))/N. Clearly, abm ∊ K − ϕ(K). By assumption, am ∊ K or bⁿm ∊ K for some positive integer n. Therefore, a(m + N) ∊ K/N or bⁿ(m + K) ∊ K/N for some positive integer n. Hence, K/N is a ϕ_N-classical primary submodule of M/N.        □

Theorem 2.13

Let N, K be two submodules of M. If K/N is a ϕ_N-classical primary submodule of M/N, then K is a ϕ-classical primary submodule of M.

Proof

Let a, b ∊ R and m ∊ M such that abm ∊ K − ϕ(K). Then, ab(m + N) = abm + N ∊ (K − ϕ(K))/N = K/N − (ϕ(K) + N)/K = (K/N) − ϕ_N(K/N). By the given hypothesis, a(m + N) ∊ K/N or bⁿ(m + N) ∊ K/N for some positive integer n. Thus, am ∊ K or bⁿm ∊ K for some positive integer n. Hence, K is a ϕ-classical primary submodule of M.        □

Now, by Theorem 2.12 and Theorem 2.13, we have the following corollary.

Corollary 2.14

Let N, K be two submodules of M. Then K is a ϕ-classical primary submodule of M if and only if K/N is a ϕ_N-classical primary submodule of M/N.

Proof

The proof is similar to Theorems 2.12, 2.13 and so the details are left to the reader.        □

3. Properties of ϕ-classical primary submodules

Let S be a multiplicatively closed set in R and let T be a set of all pairs $(x,s),$ where x ∊ M, s ∊ S. Define a relation on T by (x, s) ∼ (x^′, s^′) if and only if there exists t ∊ S such that t(sx^′ − s^′x) = 0. Then ∼ is an equivalence relation on T (Larsen & McCarthy, 1971). For (a, s) ∊ M × S, denote the equivalence class of ∼ which contains (a, s) by $\frac{a}{s}$ and denote a set of all equivalence classes of ∼ by S⁻¹M. Then S⁻¹M can be given the structure of an S⁻¹R-module under operations for which $\frac{m}{s}+\frac{n}{t}=\frac{tm+sn}{\mathit{st}},\frac{a}{s}\frac{m}{t}=\frac{\mathit{am}}{\mathit{st}}$ for all m, n ∊ M and s, t ∊ S, a ∊ R. The S⁻¹R-module S⁻¹M is called the module of fractions of M with respect to S, its zero element is $\frac{0}{1}$ and this is equal to $\frac{0}{s}$ for all s ∊ S (Larsen & McCarthy, 1971). We know that every submodule of S⁻¹M is of the form S⁻¹N for some submodule N of M (Sharp, 2000).

Let S be a multiplicatively closed set in R and let $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ be a function. Define ${\varphi }_S:\mathcal{S}(S^{-1}M)\to \mathcal{S}(S^{-1}M)\cup \{\mathrm{\varnothing }\}$ by$${\varphi }_S(S^{-1}N)=\left\{\begin{array}{cc}{S}^{-1}\left(\varphi (N)\right)\hfill & ;\varphi (N)\ne \mathrm{\varnothing }\\ \mathrm{\varnothing }\hfill & ;\varphi (N)=\mathrm{\varnothing }.\end{array}\right.$$

Zamani (2010) gives relations between ϕ-prime submodules of M and ϕ_S-prime submodules of S⁻¹M. This leads us to give relations between ϕ-classical primary submodules of M and ϕ_S-classical primary submodules of S⁻¹M.

Theorem 3.1

Let S be a multiplicatively closed subset of R. If N is a ϕ-classical primary submodule of M with S ∩ (N:M) = ∅ then S⁻¹N is a ϕ_S-classical primary submodule of S⁻¹M.

Proof

Since S ∩ (N:M) = ∅, we have S⁻¹N is a proper submodule of S⁻¹M. Let a₁, a₂ ∊ R, s₁, s₂, s₃ ∊ S and m ∊ M such that $\frac{a_1}{s_1}\frac{a_2}{s_2}\frac{m}{s_3}\in S^{-1}N-{\varphi }_S(S^{-1}N).$ Then there exists s ∊ S such that sa₁a₂m ∊ N. If sa₁a₂m ∊ ϕ(N), then $\frac{a_1}{s_1}\frac{a_2}{s_2}\frac{m}{s_3}=\frac{s{a}_1{a}_{2}m}{s{s}_1{s}_2{s}_3}$∊ S⁻¹ϕ(N) = ϕ_S(S⁻¹N), a contradiction. Now if $s{a}_1{a}_{2}m\notin \varphi (N),$ then sa₁a₂m ∊ N − ϕ(N). By assumption, a₁sm ∊ N or $a_2^{n}sm\in N$ for some positive integer n. Thus, $\frac{a_1}{s_1}\frac{m}{s_3}=\frac{s{a}_{1}m}{s{s}_1{s}_3}\in S^{-1}N$ or ${\left(\frac{a_2}{s_2}\right)}^n\frac{m}{s_3}=\frac{s{a}_2^{n}m}{s{s}_2^n{s}_3}\in S^{-1}N$ for some positive integer n. Hence, S⁻¹N is a ϕ_S-classical primary submodule of S⁻¹M.        □

Theorem 3.2

Let S be a multiplicatively closed subset of R. If S⁻¹N is a ϕ_S-classical primary submodule of S⁻¹M such that S ∩ Zd(N/ϕ(N)) = ∅, S ∩ Zd(M/N) = ∅, then N is a ϕ-classical primary submodule of M.

Proof

Since S ∩ Zd(N/ϕ(N)), we have N as a proper submodule of M. Let a, b ∊ R and m ∊ M such that abm ∊ N − ϕ(N). Then, $\frac{a}{1}\frac{b}{1}\frac{m}{1}=\frac{\mathit{abm}}{1}\in S^{-1}N.$ If $\frac{a}{1}\frac{b}{1}\frac{m}{1}\in {\varphi }_S(S^{-1}N)=S^{-1}\varphi (N),$ then there exists s ∊ S such that sabm ∊ ϕ(N) which is a contradiction. If $\frac{a}{1}\frac{b}{1}\frac{m}{1}\notin {\varphi }_S(S^{-1}N),$ then $\frac{a}{1}\frac{b}{1}\frac{m}{1}\in S^{-1}N-{\varphi }_S(S^{-1}N).$ By assumption, $\frac{a}{1}\frac{m}{1}\in S^{-1}N$ or ${\left(\frac{b}{1}\right)}^n\frac{m}{1}\in S^{-1}N$ for some positive integer n. If $\frac{a}{1}\frac{m}{1}\in S^{-1}N,$ there exists s ∊ S such that sam ∊ N. Thus, $s\left(am+N\right)=sam+N=N.$ Since S ∩ Zd(M/N) = ∅, we have am ∊ N. Now if ${\left(\frac{b}{1}\right)}^n\frac{m}{1}\in S^{-1}N,$ then it is clear that bⁿm ∊ N. Hence, N is a ϕ-classical primary submodule of M.        □

In view of Theorem 3.1 and Theorem 3.2, we have the following result.

Corollary 3.3

Let S be a multiplicatively closed subset of a commutative ring R and let S ∩ (N:M) = ∅, S ∩ Zd(N/ϕ(N)) = ∅, S ∩ Zd(M/N) = ∅ for $N\in \mathcal{S}(M).$ Then, N is a ϕ-classical primary submodule of M if and only if S⁻¹N is a ϕ_S-classical primary submodule of S⁻¹M.

Proof

The proof is similar to Theorem 3.1 and Theorem 3.2.        □

In the following result, we give an equivalent definition of ϕ-classical primary submodules.

Theorem 3.4

Let N be a proper submodule of M. The following conditions are equivalent:

(1)	N is a ϕ-classical primary submodule of M.
(2)	For every a, b ∊ R, (N:ab) ⊆ (ϕ(N):ab) ∪ (N:a) ∪ (N:bn) for some positive integer n.

Proof

(1 ⇒ 2) Let m ∊ (N:ab). Then abm ∊ N. If abm ∊ ϕ(N), then m ∊ (ϕ(N):ab) ⊆ (ϕ(N):ab) ∪ (N:a) ∪ (N:bⁿ). If $abm\notin \varphi (N),$ then abm ∊ N − ϕ(N). By assumption, am ∊ N or bⁿm ∊ N for some positive integer n. Therefore, m ∊ (N:a) or m ∊ (N:bⁿ) for some positive integer n. Hence, (N:ab) ⊆ (ϕ(N):ab) ∪ (N:a) ∪ (N:bⁿ) for some positive integer n.

(2 ⇒ 1) Let a, b ∊ R and m ∊ M such that abm ∊ N − ϕ(N). Then, m ∊ (N:ab) and $m\notin (\varphi (N):ab).$ Since (N:ab) ⊆ (ϕ(N):ab) ∪ (N:a) ∪ (N:bⁿ) for some positive integer n, we have m ∊ (N:a) ∪ (N:bⁿ) for some positive integer n. Clearly, am ∊ N or bⁿm ∊ N for some positive integer n. Hence, N is a ϕ-classical primary submodule of M.        □

Theorem 3.5.

Let N be a proper submodule of M. The following conditions are equivalent:

(1)	N is a ϕ-classical primary submodule of M.
(2)	For every a ∊ R and m ∊ M if $a^{n}m\notin N$ for all positive integer n, then (N:am) = (ϕ(N):am) ∪ (N:m).
(3)	For every a ∊ R and m ∊ M if $a^{n}m\notin N$ for all positive integer n, then (N:am) = (ϕ(N):am) or $\left(N:am\right)=\left(N:m\right).$

Proof

(1 ⇒ 2) Clearly, (ϕ(N):am) ∪ (N:m) ⊆ (N:am). On the other hand, let r ∊ (N:am). Then m ∊ (N:ar). Thus by Theorem 3.4, m ∊ (ϕ(N):ar) or m ∊ (N:r) or $m\in (N:a^{n_1})$ for some positive integer n₁. This implies that r ∊ (ϕ(N):am) or $a^{n_1}m\in N$ or r ∊ (N:m). Since $a^{n}m\notin N,$ we have r ∊ (ϕ(N):am) ∪ (N:m). Therefore, (N:am) ⊆ (ϕ(N):am) ∪ (N:m) and hence (N:am) = (ϕ(N):am) ∪ (N:m).

(2 ⇒ 3) By the fact that if an ideal (a subgroup) is the union of two ideals (two subgroups), then it is equal to one of them.

(3 ⇒ 1) It is obvious.        □

Theorem 3.6

Let N be a proper submodule of M. The following conditions are equivalent:

(1)	N is a ϕ-classical primary submodule of M.
(2)	For every a ∊ R, m ∊ M and every ideal I of R if aIm ⊆ N − ϕ(N), then Im ⊆ N or anm ∊ N for some positive integer n.
(3)	For every m ∊ M and every ideal I of R if $Im⊈N,$ then (N:Im) = (ϕ(N):Im) or $\sqrt{(N:Im)}=\sqrt{(N:m)}.$
(4)	For every ideals I, J of R if IJm ⊆ N − ϕ(N), then Im ⊆ N or $J\subseteq \sqrt{(N:m)}.$

Proof

(1 ⇒ 2) Let a ∊ R, m ∊ M and let I be an ideal of R such that aIm ⊆ N − ϕ(N). Then I ⊆ (N:am) and $I⊈(\varphi (N):am).$ If aⁿm ∊ N for some positive integer n, then we are done. Let $a^{n}m\notin N$ for all positive integer n. Therefore by Theorem 3.5, we have I ⊆ (N:am) = (N:m), i.e. Im ⊆ N.

(2 ⇒ 3) Let I be an ideal of R and m ∊ M such that $Im⊈N.$ It is easy to see that (ϕ(N):Im) ⊆ (N:Im). Assume that r ∊ (N:Im). Then rIm ⊆ N. If rIm ⊆ ϕ(N), then r ∊ (ϕ(N):Im) so (N:Im) = (ϕ(N):Im). Now if $rIm⊈\varphi (N),$ then rIm ⊆ N − ϕ(N). By assumption, Im ⊆ N or rⁿm ∊ N for some positive integer n. Since $Im⊈N,$ we have rⁿm ∊ N. Therefore, $r\in \sqrt{(N:m)}.$ Clearly, $(N:Im)\subseteq \sqrt{(N:m)}.$ Hence $\sqrt{(N:Im)}=\sqrt{(N:m)}.$

(3 ⇒ 4) Suppose that IJm ⊆ N − ϕ(N), where I, J are ideals of R. Let $Im⊈N.$ By assumption, i.e. (N:Im) = (ϕ(N):Im) or $\sqrt{(N:Im)}=\sqrt{(N:m)}.$ Since IJm ⊆ N − ϕ(N), we have J ⊆ (N:Im) and $J⊈(\varphi (N):Im).$ This implies that $J\subseteq (N:Im)\subseteq \sqrt{(N:Im)}=\sqrt{(N:m)}.$

(4 ⇒ 1)It is obvious.        □

Let M be an R-module. The M is called a multiplication module if every submodule N of M has the form IM for some ideal I of R (El-Bast & Smith, 1988). Note that, since I ⊆ (N:M) then N = IM ⊆ (N:M)M ⊆ N. Thus $N=\left(N:M\right)M.$ Let N₁ and N₂ be two submodules of M with N = I₁M and N₂ = I₂M for some ideals I₁ and I₂ of R. A product of N₁ and N₂ denoted by N₁N₂ is defined by N₁N₂ = I₁I₂M. The following theorem offers a characterization of ϕ-classical primary submodules.

Theorem 3.7

Let R be a noetherian ring and let N be a proper submodule of a multiplication R-module M. Then the following conditions are equivalent:

(1)	N is a ϕ-classical primary submodule of M.
(2)	If K1K2 ⊆ N − ϕ(N) for some submodules K1, K2 of M, then K1 ⊆ N or $K_2^n\subseteq N$ for some positive integer n.

Proof

(1 ⇒ 2) Suppose that K₁, K₂ are submodules of M. Since M is multiplication, there are ideals I₁, I₂ of R such that K₁ = I₁M and K₂ = I₂M. Let m ∊ M. Then I₁I₂m ⊆ I₁I₂ M = K₁K₂ ⊆ N − ϕ(N). By Theorem 3.6, i.e. I₁m ⊆ N or $I_2\subseteq \sqrt{(N:m)}.$ Therefore I₁M ⊆ N or $I_2^{n}M\subseteq N$ for some positive integer n. Hence, K₁ ⊆ N or $K_2^n\subseteq N$ for some positive integer n.

(2 ⇒ 1) Let m ∊ M and I₁mI₂m = I₁I₂m ⊆ N − ϕ(N) for some ideals I₁, I₂ of R. Thus by part 2, i.e. I₁m ⊆ N or $I_2^{n}m\subseteq N$ for some positive integer n. Thus I₁m ⊆ N or $I_2\subseteq \sqrt{(N:m)}.$ By Theorem 3.6, N is ϕ-classical primary submodule of M.        □

We are finding additional condition to show that a classical primary submodule is a ϕ-classical primary submodule of an R-module M.

Theorem 3.8

Let ϕ(N) be a classical primary submodule of M. Then N is a ϕ-classical primary submodule of M if and only if N is a classical primary submodule of M.

Proof

Suppose that N is a classical primary submodule of M. Clearly, N is a ϕ-classical primary submodule of M. Conversely, assume that N is a ϕ-classical primary submodule of M. Let a, b ∊ R and m ∊ M such that abm ∊ N. If $abm\notin \varphi (N),$ then abm ∊ N − ϕ(N). By assumption, am ∊ N or bⁿm ∊ N for some positive integer n. Now if abm ∊ ϕ(N), then am ∊ ϕ(N) ⊆ N or bⁿm ∊ ϕ(N) ⊆ N for some positive integer n. Hence, N is a classical primary submodule of M.        □

Definition 3.9

(Bataineh & Khuhail, 2011) Let M be an R-module and let $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ be a function where $\mathcal{S}(M)$ be a set of all submodules of M. A proper submodule N of M is called a ϕ-primary submodule, if for each m ∊ M, a ∊ R with am ∊ N − ϕ(N), then m ∊ N or $a\in \sqrt{(N:M)}.$

Remark 3.10

It is easy to see that every ϕ-primary submodule is ϕ-classical primary.

The following example shows that the converse of Remark 3.10 is not true.

Example 3.11

Let $R=\mathbf{Z}$ and $M=\mathbf{Z}\times \mathbf{Z}.$ Consider the submodule $N=\left\{0\right\}\times 4\mathbf{Z}$ of M. Define $\varphi :\mathcal{S}(M)\to \mathcal{S}(M)\cup \{\mathrm{\varnothing }\}$ by $\varphi (K)=\left\{(0,0)\right\}$ for every submodule K of M. It is easy to see that N is a ϕ-classical primary submodule of M. Notice that $4(0,1)\in \left\{0\right\}\times 4\mathbf{Z},$ but $(0,1)\notin \left\{0\right\}\times 4\mathbf{Z}$ and $4\notin \sqrt{(\left\{0\right\}\times 4\mathbf{Z}:\mathbf{Z}\times \mathbf{Z})}.$ Therefore N is not a ϕ-primary submodule of M.

We provide some relationships between ϕ-classical primary submodules of an R-module M and ϕ-primary submodule of M. However, these results require that M be a cyclic R-module.

Theorem 3.12

Let M be a cyclic R-module. If N is a ϕ-classical primary submodule of M, then N is a ϕ-primary submodule of M.

Proof

Let r ∊ R and m₀ ∊ M = Rm for some m ∊ M such that rm₀ ∊ N − ϕ(N). Then there exists s ∊ R such that m₀ = sm. Therefore, rsm = rm₀ ∊ N − ϕ(N). By assumption, sm ∊ N or rⁿm ∊ N for some positive integer n. Thus, m₀ ∊ N or rⁿ ∊ (N:M). Hence, N is a ϕ-primary submodule of M.        □

Now, the following result follows immediately from Theorem 3.12.

Corollary 3.13

Let M be a cyclic R-module. Then, N is a ϕ-primary submodule of M if and only if N is a ϕ-classical primary submodule of M.

Proof

This makes the same assertion as Theorem 3.12.        □

Now, we are finding additional condition to show that a ϕ-primary submodule is a ϕ-classical primary submodule of an R-module M.

Lemma 3.14

Let N be a ϕ-classical primary submodule of M and let m₁ ∊ M and m₂ ∊ M − N such that φ(N:m₂) is a primary ideal of R and (ϕ(N):m₂) ⊆ φ(N:m₂). Then, (N:m₂) = (N:rm₂) for all $r\in (N:m_1)-\sqrt{(N:m_2)}.$

Proof

Let a ∊ (N:m₂). Then, a(rm₂) = r(am₂) ∊ N for all r ∊ R. This implies that (N:m₂) ⊆ (N:rm₂). On the other hand, let a ∊ (N:rm₂). Then (ar)m₂ = a(rm₂) ∊ N. Clearly, ar ∊ (N:m₂). If $ar\notin \phi (N:m_2),$ then ar ∊ (N:m₂) − φ(N:m₂). By assumption, a ∊ (N:m₂) or rⁿ ∊ (N:m₂) for some positive integer n. Now by our hypothesis, a ∊ (N:m₂). Thus, (N:rm₂) ⊆ (N:m₂). Now if ar ∊ φ(N:m₂), then a ∊ (N:m₂) or rⁿ ∊ (N:m₂) for some positive integer n. Again, by the assumption, a ∊ (N:m₂). Therefore (N:rm₂) ⊆ (N:m₂) and hence (N:m₂) = (N:rm₂).        □

Theorem 3.15

Let N be a ϕ-classical primary submodule of an R-module M. For any m₁ ∊ M and m₂ ∊ M − N such that φ(N:m₂) is a primary ideal of R and $(N:m_1)-\sqrt{(N:m_2)}\ne \mathrm{\varnothing },$ we have N = (N + Rm₁) ∩ (N + Rm₂).

Proof

Let m₁ ∊ M and m₂ ∊ M − N. Clearly, N ⊆ (N + Rm₁) ∩ (N + Rm₂). For another direction, we show that (N + Rm₁) ∩ (N + Rm₂) ⊆ N. Let m ∊ (N + Rm₁) ∩ (N + Rm₂). This implies that m ∊ N + Rm₁ and m ∊ N + Rm₂. Then there exist r₁, r₂ ∊ R and n₁, n₂ ∊ N such that n₁ + r₁m₁ = m = n₂ + r₂m₂. Since $(N:m_1)-\sqrt{(N:m_2)}\ne \mathrm{\varnothing },$ there exists $r\in (N:m_1)-\sqrt{(N:m_2)}.$ Thus, rm₁ ∊ N. Clearly, rr₁m₁ ∊ r₁N ⊆ N. Now since rm = rn₁ + rr₁m₁ = rn₂ + rr₂m₂, we have rr₂m₂ ∊ N. It follows that r₂ ∊ (N:rm₂). By Lemma 3.14, r₂ ∊ (N:m₂) so that r₂m₂ ∊ N. Therefore, m = n₂ + r₂m₂ ∊ N and hence N = (N + Rm₁) ∩ (N + Rm₂).        □

Theorem 3.16

Let N be a submodule of M such that φ(N:m) ⊆ (ϕ(N):m) for all m ∊ M − N. For any m₁ ∊ M and m₂ ∊ M − N such that $(N:m_1)-\sqrt{(N:m_2)}\ne \mathrm{\varnothing },$ we have N = (N + Rm₁) ∩ (N + Rm₂). Then, N is a ϕ-classical primary submodule of M.

Proof

Let m ∊ M − N. To show that (N:m) is a φ-primary ideal of R. Let a, b ∊ R such that ab ∊ (N:m) − φ(N:m). Assume that bⁿ ∊ R − (N:m) for all positive integer n. Since ab ∊ (N:m), we have b ∊ (N:am).Clearly, $b\in (N:am)-\sqrt{(N:m)}.$ This implies that $(N:am)-\sqrt{(N:m)}\ne \mathrm{\varnothing }.$ By the assumption, N = (N + Ram) ∩ (N + Rm). Now since am ∊ (N + Ram) ∩ (N + Rm), we have am ∊ N. Therefore a ∊ (N:m). Thus, (N:a) is a φ-primary ideal of R. By Theorem 2.4(2), N is a ϕ-classical primary submodule of M.        □

Lemma 3.17

Let N be a ϕ-classical primary submodule of M and m₂ ∊ M − N such that φ(N:m₂) is a primary ideal of R. For each r ∊ R, m ∊ M and m₀ ∊ M − N if rm ∊ N − ϕ(N), then N = (N + Rm) ∩ (N + Rrⁿm₀) for some positive integer n.

Proof

Let r ∊ R, m ∊ M and m₀ ∊ M − N such that rm ∊ N − ϕ(N). It is clear that N ⊆ (N + Rm) ∩ (N + Rrⁿm₀) for any positive integer n. On the other hand, we show that (N + Rm) ∩ (N + Rrⁿm₀) ⊆ N for some positive integer n. We divide our proof into two cases.

Case 1. If rⁿm₀ ∊ N for some positive integer n, then Rrⁿm₀ ⊆ N. Therefore, (N + Rm) ∩ (N + Rrⁿm₀) ⊆ N + Rrⁿm₀ ⊆ N.

Case 2. If $r^n{m}_0\notin N$ for all positive integer n, then $r\notin \sqrt{(N:m_0)}.$ Since rm ∊ N, we have r ∊ (N:m). Clearly, $r\in (N:m)-\sqrt{(N:m_0)}\ne \mathrm{\varnothing }.$ By Theorem 3.15, N = (N + Rm) ∩ (N + Rm₀). Now since (N + Rrⁿm₀) ⊆ (N + Rm₀), we have (N + Rm) ∩ (N + Rrⁿm₀) ⊆ (N + Rm) ∩ (N + Rm₀) = N. Therefore, $(N+Rm)\cap (N+R{r}^n{m}_0)\subseteq N.$ Hence, N = (N + Rm) ∩ (N + Rrⁿm₀) for some positive integer n.        □

Proposition 3.18

Let N be a ϕ-classical primary submodule of M such that (ϕ(N):m) is a primary ideal of R for all m ∊ M − N. Let m ∊ M − N be such that $(N:m)-\bigcup _{m_i\in M-N}\sqrt{(N:m_i)}\ne \mathrm{\varnothing },$ then $N=(N+rm)\cap \bigcup _{m_i\in M-N}(N+R{m}_i).$

Proof

The inclusion $N\subseteq (N+rm)\cap \bigcup _{m_i\in M-N}(N+R{m}_i)$ is clear. For the other inclusion, let $x\in (N+rm)\cap \bigcup _{m_i\in M-N}(N+R{m}_i).$ Then there exist m^′ ∊ M − N, r₁, r₂ ∊ R and n₁, n₂ ∊ N such that n₁ + r₁m = x = n₂ + r₂m^′. By hypothesis, there exists a ∊ (N:m) such that $a\notin \sqrt{(N:m^{\prime })}.$ Since an₁ + ar₁m = ax = an₂ + ar₂m^′, we have ar₂m^′ ∊ N. If $a{r}_2{m}^{\prime }\notin \varphi (N),$ then ar₂m^′ ∊ N − ϕ(N) and as N is a ϕ-classical primary submodule of M, we have r₂m^′ ∊ N or aⁿm^′ ∊ N for some positive integer n. Since $a\notin \sqrt{(I:m^{\prime })},$ it follows that r₂m_i ∊ N. On the other hand, if ar₂m_i ∊ ϕ(N), then ar₂ ∊ (ϕ(N):m_i). Since (ϕ(N):m) is a primary ideal of R and $a\notin \sqrt{(N:m^{\prime })},$ it follows that r₂ ∊ (ϕ(N):m^′) so

r₂m_i ∊ ϕ(N) ⊆ N. In either case, we have x = n₂ + r₂m^′ ∊ N. This shows that$N=(N+rm)\cap \bigcup _{m_i\in M-N}(N+R{m}_i).$        □

Now, the following result follows immediately from Proposition 3.18

Corollary 3.19

Let N is a ϕ-classical primary submodule of M such that (ϕ(N):m) is a φ-primary ideal of R where m ∊ M − N. If $(N:m)-\bigcup _{m_i\in M-N}\sqrt{(N:m_i)}\ne \mathrm{\varnothing },$ then N is not irreducible.

Proof

By Proposition 3.17, $N=(N+rm)\cap \bigcup _{m_i\in M-N}(N+R{m}_i).$ Since m ∊ M − N, we have N ⊆ N + rm and $N\subseteq \bigcap _{m_i\in M-N}(N+R{m}_i).$ Hence N is not irreducible.        □

We are finding additional condition to show that a ϕ-primary submodule is a ϕ-classical primary submodule of an R-module.

Theorem 3.20

Let N be an irreducible submodule of an R-module M. The followings are equivalent.

(1)	For m2 ∊ M − N if φ(N:m2) is φ(N:m2) is a primary ideal of R, then N is a ϕ-primary submodule of M.
(2)	N is a ϕ-classical primary submodule of M.

Proof

(1 ⇒ 2) It is obvious.

(2 ⇒ 1) Let r ∊ R and m ∊ M such that rm ∊ N − ϕ(N). Assume that $r^n\notin (N:M)$ for all positive integer n. By Lemma 3.16, i.e. N = (N + Rm) ∩ (N + Rrⁿm₀) for some positive integer n. Since N is an irreducible submodule of M, we have N = N + Rm or N = N + Rrⁿm₀. Now since $r^n\notin (N:M),$ we have $N+R{r}^n{m}_0⊈N.$ Therefore, m ∊ Rm ⊆ N + Rm ⊆ N and hence N is a ϕ-primary submodule of M.        □

Funding

The authors received no direct funding for this research.

Additional information

Notes on contributors

P. Yiarayong

P. Yiarayong is an assistant professor of Department of Mathematics, Faculty of Science and Technology, Pibulsongkram Rajabhat University, Phitsanuloke, Thailand. One of his research orientations deals with abstract algebra, semigroup, LA-semigroup, LA-ring and fuzzy subset.

M. 

M. Siripitukdet is an associated professor of Department of Mathematics, Faculty of Science, Naresuan University, Phitsanuloke, Thailand. He research interests include operation research, abstract algebra, semigroup and fuzzy subset.