On Radical of Intuitionistic Fuzzy Primary Submodule

Abstract

In this paper, we further study the theory of Intuitionistic fuzzy submodules and we will define intuitionistic fuzzy primary submodule with the help of the definition of a radical submodule, and we also study the properties of these submodules. Furthermore, homomorphic image and preimage of intuitionistic fuzzy primary submodule are investigated. This manuscript is online on arXiv by the link https://arxiv.org/abs/2207.09772.

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Throughout this paper, $R$ is a commutative ring with unity and $M$ is an unitary R-module. A proper submodule $P$ of $M$ is called a prime submodule provided that for any $r\in R$ and $m\in M$, $rm\in P$ implies that $m\in P$ or $r\in P:M=r\in R|rM\subseteq P$. The set of all prime submodules of M is called the prime spectrum of M and is denoted by Spec(M). The Weakly prime submodules have been introduced by Behboodi and Koohy in [1]. Also, the spectrum of weakly prime submodule has been investigated in [2]. Following this, the concepts of primal submodules over the noncommutative ring and nearly prime submodules have been presented in [3, 4].

One of the structures that play an important role in mathematics is the structure of fuzzy algebra, which has been used extensively in many fields such as computer science, information technology, theoretical physics, and control engineering. Since the introduction of fuzzy sets in 1965 [5], researchers have done extensive research on various concepts of abstract algebra in the space of fuzzy sets, such as Rosenfeld [6], who in 1971 was the first to define the concept of fuzzy subgroups of a group. Many extensions of this concept have been proposed since then (especially in recent decades). In 1975, Relescu and Naegoita [7] applied the concept of fuzzy sets to the theory of modules. After presenting this definition, different types of fuzzy submodules were examined in the last two decades. In 1986, Atanassov [8] introduced intuitionistic fuzzy sets based on the degree of membership and degree of nonmembership, provided that the total membership and nonmembership should not exceed one another. In 1989, Biswas [9] applied the concept of intuitionistic fuzzy sets to group theory and studied the intuitionistic fuzzy subsets of a group. In the last few years a considerable number of papers have been done on fuzzy and intuitionistic fuzzy submodules in general, and fuzzy and intuitionistic fuzzy prime and primary submodules in particular. Hur et al. [10] introduced the notion of intuitionistic fuzzy prime ideals and intuitionistic fuzzy weakly completely prime ideals in a ring. The concept of the fuzzy prime submodule and fuzzy primary submodule was studied by Mashinchi and Zahedi in [11]. Also, intuitionistic fuzzy submodules and their properties were studied by many mathematicians [12–14]. In this paper, and in the first step we define the intuitionistic fuzzy radical of an intuitionistic fuzzy submodule of an R-module M and so we present some of its properties similar to the studies in the module theory. Also and related to this definition, we will define intuitionistic fuzzy primary submodule and investigate their properties. We show that for any intuitionistic fuzzy primary submodule of an R-module M, with sup property its radical will be an intuitionistic fuzzy prime ideal of R.

2. Definitions and Preliminary Results

In this paper, all rings are commutative with unity $1\ne 0$ and all R-modules are unitary. We use the symbol $\theta$ for the zero element of a R-module. Let $X$ be a nonempty set. An intuitionistic fuzzy subset is a function $A={\mu }_A,{\nu }_A$, which ${\mu }_{A}x$ denotes the degree of membership and ${\nu }_{A}x$ denotes the degree of nonmembership of $x\in X$ to subset $A$, such that $0\le {\mu }_{A}x+{\nu }_{A}x\le 1$ for all $x\in X$, can be denoted by $A=x,{\mu }_{A}x,{\nu }_{A}x$. It is clear that when ${\mu }_{A}x+{\nu }_{A}x=1$, then $A$ will be a fuzzy subset of $X$. The class of all intuitionistic fuzzy subsets of $X$ is denoted by $\text{IFS}X$ [8, 14].

In this section, we give some basic concepts of intuitionistic fuzzy sets and intuitionistic fuzzy modules. For simplicity, we sometimes denote each intuitionistic fuzzy set $A=x,{\mu }_{A}x,{\nu }_{A}x$ by $A={\mu }_A,{\nu }_A$.

Definition 1 (see [8]).

Let $X$ be a nonempty set and $A={\mu }_A,{\nu }_A,B={\mu }_B,{\nu }_B$ be intuitionistic fuzzy sets of $X$. Then,$$\begin{array}{c}\text{(1)}& A\subseteq B\iff {\mu }_{A}x\le {\mu }_{B}x,{\nu }_{A}x\ge {\nu }_{B}x,\hspace{1em}\text{for\hspace{0.17em}all}x\in X,\\ A\cup B=x,{\mu }_{A}x\vee {\mu }_{B}x,{\nu }_{A}x\wedge {\nu }_{B}x|x\in X,\\ A\cap B=x,{\mu }_{A}x\wedge {\mu }_{B}x,{\nu }_{A}x\vee {\nu }_{B}x|x\in X.\end{array}$$

Definition 2 (see [15]).

Let R be a ring and $A\in \text{IFS}R$. Then, $A$ is called an intuitionistic fuzzy ideal of R if for all $x,y\in R$, the followings hold:

(1) ${\mu }_{A}x-y\ge {\mu }_{A}x\wedge {\mu }_{A}y,{\nu }_{A}x-y\le {\nu }_x\vee {\nu }_{A}y$

(2) ${\mu }_{A}xy\ge {\mu }_{A}x\vee {\mu }_{A}y,{\nu }_{A}xy\le {\nu }_{A}x\wedge {\nu }_{A}y$

The class of intuitionistic fuzzy ideals of R is denoted by IFI(R).

Definition 3 (see [12]).

Let $M$ be an R-module and $A\in \text{IFS}M$. Then $A$ is called an intuitionistic fuzzy submodule if

(1) ${\mu }_A\theta =1,{\nu }_A\theta =0$

(2) ${\mu }_{A}x+y\ge {\mu }_{A}x\wedge {\mu }_{A}y,{\nu }_{A}x+y\le {\nu }_{A}x\vee {\nu }_{A}y$ for all $x,y\in M$

(3) ${\mu }_{A}rx\ge {\mu }_{A}x,{\nu }_{A}rx\le {\nu }_{A}x$ for all $x\in M$ and $r\in R$

We denote the class of all intuitionistic fuzzy submodule of an R-module $M$, by $\text{IFM}M$.

Notice that, when $R=M$, then $A\in \text{IFM}M$ if and only if ${\mu }_A\theta =1,{\nu }_A\theta =0$ and $A\in \text{IFI}R$.

Now, for an intuitionistic fuzzy submodule, we define an intuitionistic fuzzy radical.

Definition 4.

Let $M$ be an R-module and $A\in \text{IFS}M$. The intuitionistic fuzzy subset $\sqrt{A}=\sqrt{{\mu }_A},\sqrt{{\nu }_A}$ of $R$ is called the intuitionistic fuzzy radical of $A$ and is defined as follows:$$\begin{array}{c}\text{(2)}& \sqrt{{\mu }_A}r=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m,\sqrt{{\nu }_A}r=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^n.m,\end{array}$$for all $r\in R$, $m\in M$, $n\in \mathbb{N}$.

It is clear that when $M=R$, then the intuitionistic fuzzy radical of an intuitionistic fuzzy ideal of $R$ will be defined by: $\sqrt{A}=\sqrt{{\mu }_A},\sqrt{{\nu }_A}$ where $\sqrt{{\mu }_A}r={\vee }_{n\in \mathbb{N}}{\mu }_A{r}^n$ and $\sqrt{{\nu }_A}r={\wedge }_{n\in \mathbb{N}}{\nu }_A{r}^n$. Now we show that for any intuitionistic fuzzy submodule $A$ of $M$, $\sqrt{A}$ is an intuitionistic fuzzy ideal of R.

Proposition 1.

Let $A={\mu }_A,{\nu }_A$ be an intuitionistic fuzzy submodule of $M$ then $\sqrt{A}=\sqrt{{\mu }_A},\sqrt{{\nu }_A}$ is an intuitionistic fuzzy ideal of R.

Proof.

Let $r_1,r_2\in R$ then$$\begin{array}{c}\text{(3)}& {\sqrt{\mu }}_A{r}_1+r_2=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^{2n}.m\vee \underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^{2n+1}.m.\end{array}$$

Now, we have$$\begin{array}{c}\text{(4)}& \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^{2n}.m=\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{\displaystyle \sum _{i=0}^{2n}{C}_i^{2n}{r}_1^{2n-i}{r}_2^i}.m\ge \underset{m\in M}{{\displaystyle \wedge }}\begin{array}{c}{\mu }_A{r}_1^{2n}.m\wedge {\mu }_A{r}_1^{2n-1}.r_2.m\wedge \cdots \wedge {\mu }_A{r}_1^n{r}_2^n.m\wedge \\ {\mu }_A{r}_1^{n-1}{r}_2^{n+1}.m\wedge \cdots \wedge {\mu }_A{r}_2^{2n}.m\end{array}\\ \ge \underset{m\in M}{{\displaystyle \wedge }}\begin{array}{c}{\mu }_A{r}_1^{2n}.m\wedge {\mu }_A{r}_1^{2n-1}.m\wedge \cdots \wedge {\mu }_A{r}_1^n.m\dots \\ \wedge {\mu }_A{r}_2^{n+1}.m\wedge \cdots \wedge {\mu }_A{r}_2^{2n}.m\end{array}\\ =\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge {\mu }_A{r}_2^{n+1}.m\ge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge {\mu }_A{r}_2^n.m\\ =\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2^n.m.\end{array}$$

Also,$$\begin{array}{c}\text{(5)}& \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^{2n+1}.m=\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{\displaystyle \sum _{i=0}^{2n+1}{C}_i^{2n+1}{r}_1^{2n+1-i}{r}_2^i}.m\ge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^{2n+1}.m\wedge {\mu }_A{r}_1^{2n}.r_2.m\wedge \cdots \wedge {\mu }_A{r}_1^n{r}_2^{n+1}.m\wedge \cdots \wedge {\mu }_A{r}_2^{2n+1}.m\\ \ge \underset{m\in M}{{\displaystyle \wedge }}\begin{array}{c}{\mu }_A{r}_1^{2n+1}.m\wedge {\mu }_A{r}_1^{2n}.m\wedge \cdots \wedge {\mu }_A{r}_1^{n+1}.m\\ \cdots \wedge {\mu }_A{r}_2^{n+1}.m\wedge \cdots \wedge {\mu }_A{r}_2^{2n+1}.m\end{array}\\ =\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^{n+1}.m\wedge {\mu }_A{r}_2^{n+1}.m\ge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge {\mu }_A{r}_2^n.m\\ =\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2^n.m.\end{array}$$

Therefore,$$\begin{array}{c}\text{(6)}& \sqrt{{\mu }_A}{r}_1+r_2=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^{2n}.m\vee \underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r_1+r_2}^{2n+1}.m\ge \underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2^n.m\vee \underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2^n.m\\ =\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2^n.m\\ =\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1^n.m\wedge \underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2^n.m=\sqrt{{\mu }_A}{r}_1\wedge \sqrt{{\mu }_A}{r}_2.\end{array}$$

Similarly, we show that $\sqrt{{\nu }_A}{r}_1+r_2\le \sqrt{{\nu }_A}{r}_1\vee \sqrt{{\nu }_A}{r}_2$.$$\begin{array}{c}\text{(7)}& {\sqrt{\nu }}_A{r}_1+r_2=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r_1+r_2}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{\displaystyle \sum _{i=0}^n{C}_i^n{r}_1^{n-i}{r}_2^i}.m\le \underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_1^n.m\vee {\nu }_A{r}_1^{n-1}{r}_2.m\vee \cdots \vee {\nu }_A{r}_1{r}_2^{n-1}.m\vee {\nu }_A{r}_2^n.m\\ \le \underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_1^n.m\vee {\nu }_A{r}_1^{n-1}m\vee \cdots \vee {\nu }_A{r}_2^{n-1}.m\vee {\nu }_A{r}_2^n.m\\ \text{since\hspace{0.17em}}A\text{\hspace{0.17em}is\hspace{0.17em}an\hspace{0.17em}intuitionistic\hspace{0.17em}fuzzy\hspace{0.17em}submodule\hspace{0.17em}of\hspace{0.17em}}M\\ =\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_1^n.m\vee {\nu }_A{r}_2^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_1^n.m\vee \underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_2^n.m\\ =\sqrt{{\nu }_A}{r}_1\vee \sqrt{{\nu }_A}{r}_2.\end{array}$$

And, $\sqrt{{\mu }_A}{r}_1{r}_2={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r_1{r}_2}^n.m={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r}_1^n{r}_2^n.m$$\ge {\vee }_{n\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r}_1^n.m=\sqrt{{\mu }_A}{r}_1$. Similarly, $\sqrt{{\mu }_A}{r}_1{r}_2\ge \sqrt{{\mu }_A}{r}_2$. Hence $\sqrt{{\mu }_A}{r}_1{r}_2\ge \sqrt{{\mu }_A}{r}_1\vee \sqrt{{\mu }_A}{r}_2$.

Also, $\sqrt{{\nu }_A}{r}_1{r}_2=\sqrt{{\nu }_A}{r}_1$ and $\sqrt{{\nu }_A}{r}_1{r}_2\le \sqrt{{\nu }_A}{r}_2$, hence $\sqrt{{\nu }_A}{r}_1{r}_2\le \sqrt{{\nu }_A}{r}_1\wedge \sqrt{{\nu }_A}{r}_2$. Next for $r\in R\sqrt{{\mu }_A}-r={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in M}\sqrt{{\mu }_A}{r}_1{\mu }_A{-r}^n.m={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r}^n.m=\sqrt{{\mu }_A}r$;$$\begin{array}{}\text{(8)}& \sqrt{{\nu }_A}-r=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{-r}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^{n}m=\sqrt{{\nu }_A}r.\end{array}$$

Therefore, $\sqrt{A}$ is an intuitionistic fuzzy ideal of $R$.

In the following proposition, we will give some properties of intuitionistic fuzzy radicals.

Proposition 2.

Let $A$, $B\in \text{IFM}M$ then

(1) $\sqrt{\sqrt{A}}=\sqrt{A}$

(2) $A\subseteq B⟹\sqrt{A}\subseteq \sqrt{B}$

(3) $\sqrt{A\cap B}=\sqrt{A}\cap \sqrt{B}$

(4) $\sqrt{\sqrt{A}+\sqrt{B}}\subseteq \sqrt{A+B}$

Proof.

(1) Since $\sqrt{\sqrt{A}}=\sqrt{{\sqrt{\mu }}_A},\sqrt{{\sqrt{\nu }}_A}$, so by Proposition 1$\sqrt{\sqrt{A}}$ is an intuitionistic fuzzy ideal of $R$ then $\sqrt{\sqrt{{\mu }_A}}r={\vee }_{n\in \mathbb{N}}\sqrt{{\mu }_A}{r}^n={\vee }_{n\in \mathbb{N}}{\vee }_{n^{\prime }\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r^n}^{n^{\prime }}.m$$={\vee }_{n\in \mathbb{N}}{\vee }_{n^{\prime }\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r}^{n{n}^{\prime }}.m={\vee }_{n\prime \prime \in \mathbb{N}}{\wedge }_{m\in M}{\mu }_A{r}^{n\prime \prime }.m=\sqrt{{\mu }_A}r$. It is clear that $\sqrt{\sqrt{{\nu }_A}}r=\sqrt{{\nu }_A}r$, since the proof is the dual of the proof of the first part. Hence $\sqrt{\sqrt{A}}=\sqrt{{\mu }_A},\sqrt{{\nu }_A}=\sqrt{A}$.

(2) By 3, $A\cap B=m,{\mu }_{A}m\wedge {\mu }_{B}m,{\nu }_{A}m\vee {\nu }_{B}m:m\in M$, hence$$\begin{array}{c}\text{(9)}& \sqrt{{\mu }_A\wedge {\mu }_B}r=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A\wedge {\mu }_B{r}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m\wedge {\mu }_B{r}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m\wedge \underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_B{r}^n.m=\sqrt{{\mu }_A}r\wedge \sqrt{{\mu }_A}r.\end{array}$$

  Similarly,$$\begin{array}{c}\text{(10)}& \sqrt{{\nu }_A\vee {\nu }_B}r=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A\vee {\nu }_B{r}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^n.m\vee {\nu }_B{r}^n.m=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^n.m\vee \underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_B{r}^n.m=\sqrt{{\nu }_A}r\vee \sqrt{{\nu }_A}r.\end{array}$$

  Therefore, $\sqrt{A\cap B}=r,\sqrt{{\mu }_A}r\wedge \sqrt{{\mu }_B}r,\sqrt{{\nu }_A}r\vee \sqrt{{\nu }_B}r=\sqrt{A}\cap \sqrt{B}$

(3) ${\mu }_A\subseteq {\mu }_A+{\mu }_B$ because ${\mu }_A+{\mu }_{B}x=\vee {\mu }_{A}y\wedge {\mu }_{B}z|y+z=x$$\ge {\mu }_{A}x\wedge {\mu }_B\theta ;{\mu }_B\theta =1={\mu }_{A}x$. And similarly, ${\mu }_B\subseteq {\mu }_A+{\mu }_B.$

In the same way, we have ${\nu }_A\supseteq {\nu }_A+{\nu }_B$ because ${\nu }_A+{\nu }_{B}x=\wedge {\nu }_{A}y\vee {\nu }_{B}z|y+z=x\le {\nu }_{A}x\vee {\nu }_B\theta ;{\nu }_B\theta =0={\nu }_{A}x$. Also ${\nu }_B\supseteq {\nu }_A+{\nu }_B$.

Now, we have $\sqrt{{\mu }_A}\subseteq \sqrt{{\mu }_A+{\mu }_B}$, $\sqrt{{\mu }_B}\subseteq \sqrt{{\mu }_A+{\mu }_B}$ , hence. $\sqrt{{\mu }_A}+\sqrt{{\mu }_B}\subseteq \sqrt{{\mu }_A+{\mu }_B}⟹\sqrt{\sqrt{{\mu }_A}+\sqrt{{\mu }_B}}\subseteq \sqrt{\sqrt{{\mu }_A+{\mu }_B}}=\sqrt{{\mu }_A+{\mu }_B}.$

And, also $\sqrt{{\nu }_A}\supseteq \sqrt{{\nu }_A+{\nu }_B}$ , $\sqrt{{\nu }_B}\supseteq \sqrt{{\nu }_A+{\nu }_B}$ hence $\sqrt{{\nu }_A}+\sqrt{{\nu }_B}\supseteq \sqrt{{\nu }_A+{\nu }_B}⟹\sqrt{\sqrt{{\nu }_A}+\sqrt{{\nu }_B}}\supseteq \sqrt{\sqrt{{\nu }_A+{\nu }_B}}=\sqrt{{\nu }_A+{\nu }_B}$. Therefore $\sqrt{\sqrt{A}+\sqrt{B}}\subseteq \sqrt{A+B}$.

Definition 5 (see [16]).

Let $A$ be a submodule of $M$. Then intuitionistic fuzzy characteristic function ${\chi }_A={\mu }_{{\chi }_A},{\nu }_{{\chi }_A}$ is defined as follows:$$\begin{array}{c}\text{(11)}& {\mu }_{{\chi }_A}x=\begin{array}{cc}1,\hfill & \text{if\hspace{0.17em}}x\in A,\hfill \\ 0,\hfill & \text{if\hspace{0.17em}}x\notin A,\hfill \end{array}{\nu }_{{\chi }_A}x=\begin{array}{cc}0,\hfill & \text{if\hspace{0.17em}}x\in A,\hfill \\ 1,\hfill & \text{if\hspace{0.17em}}x\notin A.\hfill \end{array}\end{array}$$

Proposition 3.

Let $A$ be a submodule of $M$, then $\sqrt{{\chi }_A}={\chi }_{\sqrt{A}}$ where ${\chi }_A$ is intuitionistic fuzzy characteristic function of $A$.

Proof.

Let $r\in R$, then $\sqrt{{\chi }_A}r={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in M}{\mu }_{{\chi }_A}{r}^n.m,{\wedge }_{n\in \mathbb{N}}{\vee }_{m\in M}{\nu }_{{\chi }_A}{r}^n.m$. Now we consider two cases:

  Case 1: $r^n.m\in A$ then ${\mu }_{{\chi }_A}{r}^n.m=1,{\nu }_{{\chi }_A}{r}^n.m=0$ and hence $\sqrt{{\chi }_A}r=\mathrm{1,0}$. By $r^n.m\in A$ it follows that $r\in \sqrt{A}$ and then ${\chi }_{\sqrt{A}}r=\mathrm{1,0}$ this mean that $\sqrt{{\chi }_A}={\chi }_{\sqrt{A}}$.

  Case 2: $r^n.m\notin A$ then ${\mu }_{{\chi }_A}{r}^n.m=0,{\nu }_{{\chi }_A}{r}^n.m=1$ and $\sqrt{{\chi }_A}r=\mathrm{0,1}$. Now since $r^n.m\notin A$ for all $n\in \mathbb{N}$ then $r\notin \sqrt{A}$ hence ${\chi }_{\sqrt{A}}r=\mathrm{0,1}$ and therefore for this case $\sqrt{{\chi }_A}={\chi }_{\sqrt{A}}$.

Definition 6 (see [17]).

If $A\in \text{IFS}R$, then $A$ is said to have the sup property if for every $\varnothing \ne Y\subseteq R$, there is a $y_0\in Y$ such that ${\text{sup}}_{y\in Y}{\mu }_{A}y={\mu }_A{y}_0$ , ${\text{inf}}_{y\in Y}{\nu }_{A}y={\nu }_A{y}_0$.

Definition 7.

Suppose that $A\in \text{IFS}X$. Then the set $X_A^{\alpha ,\beta }=x|x\in X$ such that ${\mu }_{A}x\ge \alpha \text{\hspace{0.17em}and\hspace{0.17em}}{\nu }_{A}x\le \beta$, where $\alpha ,\beta \in \mathrm{0,1}$ with $\alpha +\beta \le 1$, is called $\alpha ,\beta$-cut set (crisp set).

Proposition 4.

Let $A\in \text{IFM}M$ and suppose that $A$ has sup property, then ${\sqrt{M_A}}^{\alpha ,\beta }=\sqrt{M_A^{\alpha ,\beta }}$, where $M_A^{\alpha ,\beta }=x\in M|{\mu }_{A}x\ge \alpha ,{\nu }_{A}x\le \beta ;\alpha +\beta \le 1;\alpha ,\beta \in \text{Im}{\mu }_A\times \text{Im}{\nu }_A$.

Proof.

Let $A$ be an intuitionistic fuzzy submodule of $M$ with sup property, then$$\begin{array}{c}\text{(12)}& {\sqrt{M_A}}^{\alpha ,\beta }=r\in R|\sqrt{{\mu }_A}r\ge \alpha ,\sqrt{{\nu }_A}r\le \beta =r\in R|\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m\ge \alpha ,\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^n.m\le \beta \\ =r\in R|\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^{n_1}.m\ge \alpha ,\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^{n_2}.m\le \beta \text{\hspace{0.17em}for\hspace{0.17em}some\hspace{0.17em}}{n}_1,n_2\in \mathbb{N}\\ \text{Since\hspace{0.17em}}A\text{\hspace{0.17em}has\hspace{0.17em}sup\hspace{0.17em}property}\\ =r\in R|{\mu }_A{r}^{n_1}{r}^{n_2}.m\ge \alpha ,{\nu }_A{r}^{n_2}{r}^{n_1}.m\le \beta \text{\hspace{0.17em}for\hspace{0.17em}all\hspace{0.17em}}m\in M\\ \text{Since\hspace{0.17em}}{\mu }_A{r}^{n_1}{r}^{n_2}.m\ge {\mu }_A{r}^{n_1}.m\ge \alpha ,{\nu }_A{r}^{n_1}{r}^{n_2}.m\le {\nu }_A{r}^{n_2}.m\le \beta \\ =r\in R|n^{″}=n_1+n_2;{\mu }_A{r}^{n^{″}}.m\ge \alpha ,{\nu }_A{r}^{n^{″}}.m\le \beta ,\text{for\hspace{0.17em}all\hspace{0.17em}}m\in M\\ =r\in R|r^{n\prime \prime }.m\in M_A^{\alpha ,\beta },\text{for\hspace{0.17em}all\hspace{0.17em}}m\in M,\text{for\hspace{0.17em}some\hspace{0.17em}}{n}^{″}\in \mathbb{N}\\ =r\in R|r^{n\prime \prime }.M\subseteq M_A^{\alpha ,\beta }\text{for\hspace{0.17em}some\hspace{0.17em}}{n}^{″}\in \mathbb{N}=\sqrt{M_A^{\alpha ,\beta }}.\end{array}$$

The above-given proposition is also true for strict level cuts, where the sup property is not required.

3. Intuitionistic Fuzzy Primary Submodules

In this section, we will give some characterization of an intuitionistic fuzzy primary submodule. We begin with a definition.

Definition 8.

Let $A={\mu }_A,{\nu }_A$ be an intuitionistic fuzzy submodule of $M$. Then, the intuitionistic fuzzy subset $\overline{A}={\mu }_{\overline{A}},{\nu }_{\overline{A}}$, is defined.$$\begin{array}{}\text{(13)}& {\mu }_{\overline{A}}r=\underset{m\in M}{{\displaystyle \wedge }}{\mu }_{A}r.m,{\nu }_{\overline{A}}r=\underset{m\in M}{{\displaystyle \vee }}{\nu }_{A}rm,\end{array}$$where $\overline{A}=A:M$.

Proposition 5.

Let $A$ be an intuitionistic fuzzy submodule of $M$. Then, $\overline{A}$ is an intuitionistic fuzzy ideal of $R$.

Proof.

Let $r_1,r_2\in R$ then$$\begin{array}{c}\text{(14)}& {\mu }_{\overline{A}}{r}_1-r_2=\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1-r_2.m=\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1.m-r_2.m\ge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1.m\wedge {\mu }_A{r}_2.m=\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_1.m\wedge \underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}_2.m={\mu }_{\overline{A}}{r}_1\wedge {\mu }_{\overline{A}}{r}_2.\end{array}$$

It is clear that ${\nu }_{\overline{A}}{r}_1-r_2\le {\nu }_{\overline{A}}{r}_1\vee {\mu }_{\overline{A}}{r}_2$. Since the proof is the dual of the proof of the first part, hence we got that ${\mu }_{\overline{A}}{r}_1-r_2\ge {\mu }_{\overline{A}}{r}_1\wedge {\mu }_{\overline{A}}{r}_2$, ${\nu }_{\overline{A}}{r}_1-r_2\le {\nu }_{\overline{A}}{r}_1\vee {\mu }_{\overline{A}}{r}_2$. Now, suppose that $r_1,r_2\in R$. Then, we have ${\mu }_{\overline{A}}{r}_1{r}_2={\wedge }_{m\in M}{\mu }_A{r}_1{r}_2.m\ge {\wedge }_{m\in M}{\mu }_A{r}_1.m={\mu }_{\overline{A}}{r}_1$, similarly ${\nu }_{\overline{A}}{r}_1{r}_2\ge {\nu }_{\overline{A}}{r}_2$. Now ${\mu }_{\overline{A}}{r}_1{r}_2\ge {\mu }_{\overline{A}}{r}_1\vee {\mu }_{\overline{A}}{r}_2$ and also ${\nu }_{\overline{A}}{r}_1{r}_2\le {\nu }_{\overline{A}}{r}_1\wedge {\nu }_{\overline{A}}{r}_2$. Therefore, $\overline{A}$ is an intuitionistic fuzzy ideal of $R$.

Remark 1.

If $A\in \text{IFM}M$ then its radical is equal to the radical of $\overline{A}$, because$$\begin{array}{c}\text{(15)}& \sqrt{\overline{A}}=r,\sqrt{{\mu }_{\overline{A}}}r,\sqrt{{\nu }_{\overline{A}}}r=r,\underset{n\in \mathbb{N}}{{\displaystyle \vee }}{\mu }_{\overline{A}}{r}^n,\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}{\nu }_{\overline{A}}{r}^n\text{Since\hspace{0.17em}}\overline{A}\text{\hspace{0.17em}is\hspace{0.17em}an\hspace{0.17em}intuitionistic\hspace{0.17em}fuzzy\hspace{0.17em}ideal\hspace{0.17em}of\hspace{0.17em}}R\\ =r,\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m,\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\nu }_A{r}^n.m=r,\sqrt{{\mu }_A}r,\sqrt{{\nu }_A}r=\sqrt{A}.\end{array}$$

Proposition 6.

Let $M$ be an R-module and $A\in \text{IFM}M$. Then $M_{\overline{A}}^{\alpha ,\beta }=\overline{M_A^{\alpha ,\beta }}$, where $M_A^{\alpha ,\beta }=x\in M:{\mu }_{A}x\ge \alpha \text{\hspace{0.17em}and\hspace{0.17em}}{\nu }_{A}x\le \beta$.

Proof.

$$\begin{array}{c}\text{(16)}& \overline{M_A^{\alpha ,\beta }}=M_A^{\alpha ,\beta }:M=\text{Ann}\frac{M}{M_A^{\alpha ,\beta }}=r\in R|r.m+M_A^{\alpha ,\beta }=M_A^{\alpha ,\beta },\hspace{1em}\text{for\hspace{0.17em}all\hspace{0.17em}}m\in M\\ =r\in R|r.m+M_A^{\alpha ,\beta }=M_A^{\alpha ,\beta }\text{,}\hspace{1em}\text{for\hspace{0.17em}all\hspace{0.17em}}m\in M\\ =r\in R|r.m\in M_A^{\alpha ,\beta },\hspace{1em}\text{for\hspace{0.17em}all\hspace{0.17em}}m\in M\\ =r\in R|{\mu }_{A}r.m\ge \alpha ,{\nu }_{A}r.m\le \beta ,\hspace{1em}\text{for\hspace{0.17em}all\hspace{0.17em}}m\in M\\ =r\in R|\underset{m\in M}{{\displaystyle \wedge }}{\mu }_{A}r.m\ge \alpha ,\underset{m\in M}{{\displaystyle \vee }}{\nu }_{A}r.m\le \beta \\ =r\in R|{\mu }_{\overline{A}}r\ge \alpha ,{\nu }_{\overline{A}}r\le \beta =M_{\overline{A}}^{\alpha ,\beta }.\end{array}$$

Definition 9 (see [14]).

Let M and N be modules over the same ring R and let $f$ be a mapping from M to N. Let $A={\mu }_A,{\nu }_A$ be an intuitionistic fuzzy submodule of $M$ and $B={\mu }_B,{\nu }_B$ be an intuitionistic fuzzy submodule of $N$. Then, the image of the intuitionistic fuzzy submodule $A$ of $M$, $fA=f{\mu }_A,{\nu }_{A}f\in N$ can be defined for all $y\in N$ as follows:$$\begin{array}{c}\text{(17)}& f{\mu }_{A}y=\begin{array}{cc}\vee {\mu }_{A}x|x\in R,fx=y,\hfill & \text{if}{f}^{-1}y\ne \varnothing ,\hfill \\ 0,\hfill & \text{otherwise,}\hfill \end{array}{\nu }_{A}f=\begin{array}{cc}\wedge {\nu }_{A}x|x\in R,fx=y,\hfill & \text{if}{f}^{-1}y\ne \varnothing ,\hfill \\ 1,\hfill & \text{otherwise,}\hfill \end{array}\end{array}$$and inverse image of an intuitionistic fuzzy submodule $B$ of $N$ can be defined as $x\in M,f^{-1}B=Bfx$.

Theorem 1.

Let $f:M⟶M^{\prime }$ be an epimorphism of R-modules. If $A\in \text{IFM}M$ then $\overline{A}\subseteq \overline{fA}$.

Proof.

First, we show that $fA$ be an intuitionistic fuzzy submodule of $M^{\prime }$. Let $u,v\in R$, then there exists $x,y\in R$ such that $fx=u,fy=v$ hence $fx-y=u-v$. Now, we have the following equations:

(1) $\begin{array}{c}f{\mu }_{A}u-v={\vee }_{z\in f^{-1}u-v}{\mu }_{A}z\ge {\vee }_{x-y\in f^{-1}u-v}{\mu }_{A}x-y\ge \\ {\vee }_{\begin{array}{c}x\in f^{-1}u\\ y\in f^{-1}v\end{array}}{\mu }_{A}x\wedge {\mu }_{A}y={\vee }_{x\in f^{-1}u}{\mu }_{A}x\wedge {\vee }_{y\in f^{-1}u}{\mu }_{A}y\\ =f{\mu }_{A}u\wedge f{\mu }_{A}v\end{array}$.

(2) $\begin{array}{c}f{\mu }_{A}r.m={\vee }_{m^{\prime }\in f^{-1}r.m}{\mu }_A{m}^{\prime }\ge {\vee }_{m^{\prime }\in f^{-1}m}{\mu }_{A}r.m^{\prime }\ge \\ {\vee }_{m^{\prime }\in f^{-1}m}{\mu }_A{m}^{\prime }=f{\mu }_{A}m\end{array}$.

  Similarly, ${\nu }_{A}fu-v={\nu }_{A}fu\wedge {\nu }_{A}fv\text{\hspace{0.17em}and\hspace{0.17em}}{\nu }_{A}fr.m={\nu }_{A}fm$.

(3) $f{\mu }_A{0}_{M^{\prime }}={\vee }_{m\in f^{-1}{0}_{M^{\prime }}}{\mu }_{A}m={\mu }_A{0}_M$,${\nu }_{A}f{0}_{M^{\prime }}={\wedge }_{m\in f^{-1}{0}_{M^{\prime }}}{\nu }_{A}m={\nu }_A{0}_M$.

Thus, we prove that $fA\in \text{IFM}{M}^{\prime }$. Let $r\in R$, then$$\begin{array}{c}\text{(18)}& \overline{f{\mu }_A}r=f{\mu }_A:fMr=\underset{m^{\prime }\in fM}{{\displaystyle \wedge }}f{\mu }_{A}r.m^{\prime }=\underset{m^{\prime }\in fM}{{\displaystyle \wedge }}\underset{m\in f^{-1}r.m^{\prime }}{{\displaystyle \vee }}{\mu }_{A}m\ge \underset{m^{\prime }\in fM}{{\displaystyle \wedge }}\underset{m\in f^{-1}r.m^{\prime }}{{\displaystyle \vee }}{\mu }_{A}r.m\text{Since\hspace{0.17em}}m\in f^{-1}{m}^{\prime }\text{then\hspace{0.17em}}r.m\in f^{-1}r.m^{\prime }\ge \underset{\begin{array}{c}{m}^{\prime }\in fM\\ m\in f^{-1}{m}^{\prime }\end{array}}{{\displaystyle \wedge }}{\mu }_{A}r.m=\underset{\begin{array}{c}{m}^{\prime }\in fM\\ m\in \cup f^{-1}{m}^{\prime }\end{array}}{{\displaystyle \wedge }}{\mu }_{A}r.m=\underset{\begin{array}{c}{m}^{\prime }\in fM\\ m\in \cup f^{-1}{m}^{\prime }\end{array}}{{\displaystyle \wedge }}{\mu }_{A}r.m\\ =\underset{m\in M}{{\displaystyle \wedge }}{\mu }_{A}r.m={\mu }_{\overline{A}}r.\end{array}$$

It is clear that $\overline{{\nu }_{A}f}r={\nu }_{\overline{A}}r$. Therefore $\overline{A}\subseteq \overline{fA}=\overline{f{\mu }_A},\overline{{\nu }_{A}f}$.

Theorem 2.

Let $f:M⟶M^{\prime }$ be a homomorphism of R-modules. If $B\in \text{IFM}{M}^{\prime }$, then $\overline{f^{-1}B}\supseteq \overline{B}$. Equality holds if $f$ is an epimorphism.

Proof.

Let $r\in R$, then$$\begin{array}{c}\text{(19)}& \overline{f^{-1}{\mu }_B}r=f^{-1}{\mu }_B:f^{-1}{M}^{\prime }r=\underset{m\in f^{-1}{M}^{\prime }}{{\displaystyle \wedge }}{f}^{-1}{\mu }_{B}r.m=\underset{m\in f^{-1}{M}^{\prime }}{{\displaystyle \wedge }}{\mu }_{B}fr.m=\underset{m\in f^{-1}{M}^{\prime }}{{\displaystyle \wedge }}{\mu }_{B}rfm=\underset{fm\in M^{\prime }}{{\displaystyle \wedge }}{\mu }_{B}r.fm\\ \ge \underset{m^{\prime }\in M^{\prime }}{{\displaystyle \wedge }}{\mu }_{B}r.m^{\prime }={\mu }_{\overline{B}}r.\end{array}$$

Similar to the previous part, we can show that $\overline{f^{-1}{\nu }_B}r\le {\nu }_{\overline{B}}r$. Therefore, $\overline{f^{-1}B}\supseteq \overline{B}$. Furthermore, if $f$ is an epimorphism then $f^{-1}\overline{B}=\overline{B}$.

Proposition 7.

Let $A,B\in \text{IFM}M$. Then, the following hold:

(1) $\overline{\overline{A}}=\overline{A}$

(2) $A\subseteq B⟹\overline{A}\subseteq \overline{B}$

(3) $\overline{A\cap B}=\overline{A}\cap \overline{B}$

(4) $\overline{A+B}\supseteq \overline{A}+\overline{B}$

Proof.

Let $r\in R$, then

(1) Since we can consider $\overline{A}$ as R-module, then ${\mu }_{\overline{\overline{A}}}r={\wedge }_{r^{\prime }\in R}{\mu }_{\overline{A}}r.r^{\prime }={\mu }_{\overline{A}}r$, and also ${\nu }_{\overline{\overline{A}}}r={\vee }_{r^{\prime }\in R}{\nu }_{\overline{A}}r.r^{\prime }={\nu }_{\overline{A}}r$ (Since $1\in R$). Hence $\overline{\overline{A}}={\mu }_{\overline{\overline{A}}},{\nu }_{\overline{\overline{A}}}={\mu }_{\overline{A}},{\mu }_{\overline{A}}=\overline{A}$

(2) Since $A\subseteq B$ then ${\mu }_A\subseteq {\mu }_B$ and ${\nu }_A\supseteq {\nu }_B$, hence ${\mu }_{\overline{A}}r={\wedge }_{m\in M}{\mu }_{A}r.m\le {\wedge }_{m\in M}{\mu }_{B}r.m={\mu }_{\overline{B}}r⟹{\mu }_{\overline{A}}\subseteq {\mu }_{\overline{B}}$, and ${\nu }_{\overline{A}}r={\vee }_{m\in M}{\nu }_{A}r.m\ge {\vee }_{m\in M}{\nu }_{B}r.m={\nu }_{\overline{B}}⟹{\nu }_{\overline{A}}\supseteq {\nu }_{\overline{B}}r$. Therefore, $\overline{A}\subseteq \overline{B}$.

(3) We know that $A\cap B={\mu }_{A\cap B},{\nu }_{A\cup B}$, so ${\mu }_{\overline{A\cap B}}r={\wedge }_{m\in M}{\mu }_{A\cap B}rm={\wedge }_{m\in M}{\mu }_{A}rm\wedge {\mu }_{B}rm$$={\wedge }_{m\in M}{\mu }_{A}rm\wedge {\wedge }_{m\in M}{\mu }_{B}rm={\mu }_{\overline{A}}r\wedge {\mu }_{\overline{B}}r={\mu }_{\overline{A}}\cap {\mu }_{\overline{B}}r$. Similarly, ${\nu }_{\overline{A\cup B}}r={\nu }_{\overline{A}}\cup {\nu }_{B}r$ and so we get $\overline{A\cap B}=\overline{A}\cap \overline{B}$.

(4) It is clear that ${\mu }_A\subseteq {\mu }_A+{\mu }_B$, ${\mu }_B\subseteq {\mu }_A+{\mu }_B$ and ${\nu }_A\supseteq {\nu }_A+{\nu }_B$, ${\nu }_B\supseteq {\nu }_A+{\nu }_B$. So ${\mu }_{\overline{A}}\subseteq {\mu }_{\overline{A+B}},{\mu }_{\overline{B}}\subseteq {\mu }_{\overline{A+B}}$. Hence ${\mu }_{\overline{A}}+{\mu }_{\overline{B}}\subseteq {\mu }_{\overline{A+B}}$ and ${\nu }_{\overline{A}}\supseteq {\nu }_{\overline{A+B}},{\nu }_{\overline{B}}\supseteq {\nu }_{\overline{A+B}}$ and so ${\nu }_{\overline{A}}+{\nu }_{\overline{B}}\supseteq {\nu }_{\overline{A+B}}$. Therefore $\overline{A+B}\supseteq \overline{A}+\overline{B}$.

In this part of the paper, we define the intuitionistic fuzzy primary submodule.

Definition 10.

Let $A$ be an intuitionistic fuzzy submodule of R-module $M$, then $A$ is called intuitionistic fuzzy primary submodule of $M$, if for $r\in R,m\in M$$$\begin{array}{c}\text{(20)}& {\mu }_{A}r.m={\mu }_{A}m,{\nu }_{A}r.m={\nu }_{A}m\text{or}{\mu }_{A}r.m\le \sqrt{{\mu }_A}r,{\nu }_{A}r.m\ge \sqrt{{\nu }_A}r.\end{array}$$

Example 1.

Consider ${\mathbb{Z}}_4$ as a $\mathbb{Z}-\text{module}$. Define $A={\mu }_A,{\nu }_A$ by$$\begin{array}{c}\text{(21)}& {\mu }_{A}z=\begin{array}{cc}1,\hfill & \text{if\hspace{0.17em}}z=\overline{0},\hfill \\ 0,\hfill & \text{if\hspace{0.17em}}z=\overline{1},\hfill \\ \frac{1,}{2,}\hfill & \text{if\hspace{0.17em}}z=\overline{2},\hfill \\ 0,\hfill & \text{if\hspace{0.17em}}z=\overline{3},\hfill \end{array}{\nu }_{A}z=\begin{array}{cc}0,\hfill & \text{if\hspace{0.17em}}z=\overline{0},\hfill \\ 1,\hfill & \text{if\hspace{0.17em}}z=\overline{1},\hfill \\ \frac{1}{2},\hfill & \text{if\hspace{0.17em}}z=\overline{2},\hfill \\ 1,\hfill & \text{if\hspace{0.17em}}z=\overline{3}.\hfill \end{array}\end{array}$$

It is clear that $A$ is an intuitionistic fuzzy submodule of ${\mathbb{Z}}_4$. Since $\sqrt{{\mu }_A}z={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in {\mathbb{Z}}_4}{\mu }_A{z}^{n}m$ and $\sqrt{{\nu }_A}z={\wedge }_{n\in \mathbb{N}}{\vee }_{m\in {\mathbb{Z}}_4}{\nu }_A{z}^{n}m$, then$$\begin{array}{c}\text{(22)}& \sqrt{{\mu }_A}z=\begin{array}{cc}1,\hfill & \text{if\hspace{0.17em}}z=0,\hfill \\ 1,\hfill & \text{if\hspace{0.17em}}z\in 2,\hfill \\ 0,\hfill & \text{if\hspace{0.17em}}z\notin 2,\hfill \end{array}\sqrt{{\nu }_A}z=\begin{array}{cc}0,\hfill & \text{if\hspace{0.17em}}z=0,\hfill \\ 0,\hfill & \text{if\hspace{0.17em}}z\in 2,\hfill \\ 1,\hfill & \text{if\hspace{0.17em}}z\notin 2,\hfill \end{array}\end{array}$$for all $z\in \mathbb{Z}$. Thus, $A={\mu }_A,{\nu }_A$ is an intuitionistic fuzzy primary submodule of $\mathbb{Z}$.

Now, we would like to investigate the relationship between intuitionistic fuzzy primary and primary submodules of a module.

Proposition 8.

Let $A$ be a submodule of $M$. Then, its intuitionistic fuzzy characteristic function ${\chi }_A$ is an intuitionistic fuzzy primary submodule if and only if $A$ is a primary submodule of $M$.

Proof.

Let $A$ be a primary submodule of $M$ and $r\in R,m\in M$. By 5 we know that ${\chi }_A={\mu }_{{\chi }_A},{\nu }_{{\chi }_A}$ where$$\begin{array}{c}\text{(23)}& {\mu }_{{\chi }_A}x=\begin{array}{cc}1,\hfill & \text{if}x\in A,\hfill \\ 0,\hfill & \text{if}x\notin A,\hfill \end{array}{\nu }_{{\chi }_A}x=\begin{array}{cc}0,\hfill & \text{if}x\in A,\hfill \\ 1,\hfill & \text{if}x\notin A.\hfill \end{array}\end{array}$$

If ${\chi }_{A}rm=\mathrm{1,0}$ then $rm\in A$. Since $A$ is a primary submodule, then $m\in A$ or $r\in \sqrt{A}$. Now, we consider the following two cases:

  Case 1: $m\in A$ then ${\chi }_{A}m=\mathrm{1,0}$ and hence ${\chi }_{A}rm={\chi }_{A}m$

  Case 2: $r\in \sqrt{A}$ then ${\chi }_{\sqrt{A}}=\mathrm{1,0}$, but by Proposition 3, $\sqrt{{\chi }_A}={\chi }_{\sqrt{A}}$ hence ${\chi }_{\sqrt{A}}=\mathrm{1,0}$, therefore ${\chi }_{A}rm={\chi }_{A}r$

If ${\chi }_{A}rm=\mathrm{0,1}$ then $rm\notin A$ and so $m\notin A$, hence ${\chi }_{A}m=\mathrm{0,1}$. Thus, ${\chi }_{A}rm={\chi }_{A}m$ or ${\chi }_{A}rm\le \sqrt{{\chi }_A}r$.

Conversely, let ${\chi }_A$ be an intuitionistic fuzzy primary submodule of $M$ and suppose that $rm\in A$, $r\in R,m\in M$. Then, ${\chi }_{A}rm=\mathrm{1,0}$. Now, since ${\chi }_A$ is an intuitionistic fuzzy primary submodule then$$\begin{array}{}\text{(24)}& {\chi }_{A}rm={\chi }_{A}m\text{or}{\chi }_{A}rm\le \sqrt{{\chi }_A}r={\chi }_{\sqrt{A}}r.\end{array}$$

If ${\chi }_{A}rm={\chi }_{A}m$ then ${\chi }_{A}m=\mathrm{1,0}$ and so $m\in A$. But If ${\chi }_{A}rm\le {\chi }_{\sqrt{A}}r$, then$$\begin{array}{c}\text{(25)}& {\mu }_{{\chi }_A}rm\le {\mu }_{{\chi }_{\sqrt{A}}}r,{\nu }_{{\chi }_A}rm\ge {\nu }_{{\chi }_{\sqrt{A}}}r⟹{\mu }_{{\chi }_{\sqrt{A}}}r\ge 1,{\nu }_{{\chi }_{\sqrt{A}}}r\le 0⟹{\mu }_{{\chi }_{\sqrt{A}}}r=1,{\nu }_{{\chi }_{\sqrt{A}}}r=0⟹{\chi }_{\sqrt{A}}=\mathrm{1,0}⟹r\in \sqrt{A}.\end{array}$$

Proposition 9.

Let $M$ be an R-module and $A$ be an intuitionistic fuzzy primary submodule of $M$ with sup property. Then, $M_A^{\alpha ,\beta }$ is a primary submodule of $M$.

Proof.

Since $A\in \text{IFM}M$, so $M_A^{\alpha ,\beta }$ is a submodule of $M$. Now let $r\in R,m\in M$ and $rm\in M_A^{\alpha ,\beta }$, then ${\mu }_{A}rm\ge \alpha ,{\nu }_{A}rm\le \beta$. Since $A$ is an intuitionistic fuzzy primary submodule of R-module $M$ then $\alpha \le {\mu }_{A}rm={\mu }_{A}m,\beta \ge {\nu }_{A}rm={\nu }_{A}m$ or $\alpha \le {\mu }_{A}rm\le \sqrt{{\mu }_A}r,\beta \ge {\nu }_{A}rm\ge \sqrt{{\nu }_A}r$. Hence ${\mu }_{A}m\ge \alpha ,{\nu }_{A}m\le \beta$ or $\sqrt{{\mu }_A}r\ge \alpha ,\sqrt{{\nu }_A}r\le \beta$. This mean that $m\in M_A^{\alpha ,\beta }$ or $r\in {\sqrt{M_A}}^{\alpha ,\beta }$. But $A$ has $\text{sup\hspace{0.17em}property}$ and by Proposition 4 we have ${\sqrt{M_A}}^{\alpha ,\beta }=\sqrt{M_A^{\alpha ,\beta }}$, therefore $m\in M_A^{\alpha ,\beta }$ or $r\in {\sqrt{M_A}}^{\alpha ,\beta }$, and it follows that $M_A^{\alpha ,\beta }$ is a primary submodule of $M$.

If $M_A^{\alpha ,\beta }$ is strict level cut or $A$ is finite valued intuitionistic fuzzy submodule, then the above-given proposition holds without assumption of sup property. We know that $A$ is finite valued intuitionistic fuzzy submodule of $M$, if $\text{Im}\mu =\mu x:x\in M$, $\text{Im}\nu =\nu x:x\in M$ be finite.

Proposition 10.

Let $A$ be an intuitionistic fuzzy submodule of $M$, such that every level cut of $A$ is a primary submodule of $M$, then $A$ is an intuitionistic fuzzy primary submodule of $M$.

Proof.

Let $r\in R,m\in M$, and let ${\mu }_{A}r.m=\alpha ,{\nu }_{A}r.m=\beta$, then$$\begin{array}{c}\text{(26)}& r.m\in M_A^{\alpha ,\beta }⟹m\in M_A^{\alpha ,\beta }\text{or\hspace{0.17em}}r\in \sqrt{M_A^{\alpha ,\beta }}⟹m\in M_A^{\alpha ,\beta }\text{or\hspace{0.17em}}{r}^n,M\subseteq M_A^{\alpha ,\beta }\text{for\hspace{0.17em}some\hspace{0.17em}}n\in \mathbb{N}⟹{\mu }_{A}m\ge \alpha ,{\nu }_{A}m\le \beta \text{\hspace{0.17em}or\hspace{0.17em}}{\mu }_A{r}^n.m\\ \ge \alpha ,{\nu }_A{r}^n.m\le \beta \text{\hspace{0.17em}for\hspace{0.17em}all\hspace{0.17em}}m\in M⟹\alpha ={\mu }_{A}r.m\ge {\mu }_{A}m\ge \alpha ,\beta \\ ={\nu }_{A}r.m\le {\nu }_{A}m\le \beta \text{\hspace{0.17em}or}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m\ge \alpha ,\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^n.m\le \beta \\ ⟹{\mu }_{A}r.m={\mu }_{A}m,{\nu }_{A}r.m={\nu }_{A}m\text{or}\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\mu }_A{r}^n.m\ge \alpha ,\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}^n.m\le \beta \\ ⟹{\mu }_{A}r.m={\mu }_{A}m,{\nu }_{A}r.m={\nu }_{A}m\text{or}\sqrt{{\mu }_A}r\ge \alpha ,\sqrt{{\nu }_A}r\le \beta \\ ⟹{\mu }_{A}r.m={\mu }_{A}m,{\nu }_{A}r.m={\nu }_{A}m\text{or\hspace{0.17em}}{\mu }_{A}r.m\le \sqrt{{\mu }_A}r,{\nu }_{A}r.m\ge \sqrt{{\nu }_A}r.\end{array}$$

This means that $A$ is an intuitionistic fuzzy primary submodule of $M$.

Definition 11.

Let $A$ be an intuitionistic fuzzy submodule of the R-module $M$. The support of $A$ is defined as $A^{\ast }=x\in M|{\mu }_{A}x>0,{\nu }_{A}x<1$.

In the following, we investigated related between intuitionistic fuzzy submodule $A$ and support of $A$.

Proposition 11.

Let $A$ be an intuitionistic fuzzy primary submodule of the R-module $M$. Then $A^{\ast }$ is a primary submodule of $M$.

Proof.

Suppose that $rm\in A^{\ast }$ for $r\in R,m\in M$. Then, ${\mu }_{A}x>0,{\nu }_{A}x<1$. Since $A$ is an intuitionistic fuzzy primary submodule, then

${\mu }_{A}rm={\mu }_{A}m$ or ${\mu }_{A}rm\le \sqrt{{\mu }_A}r$, ${\nu }_{A}rm={\nu }_{A}m$ or ${\nu }_{A}rm\ge \sqrt{{\nu }_A}r$, thus ${\mu }_{A}m>0$ or $\sqrt{{\mu }_A}r>0,{\nu }_{A}m<1$or $\sqrt{{\nu }_A}r<1$. Then $m\in A^{\ast }$ or $r\in \sqrt{A^{\ast }}$. Therefore we prove that $A^{\ast }$ is a primary submodule of $M$.

Remark 2.

The support of an intuitionistic fuzzy submodule may be a primary submodule, but it need not be an intuitionistic fuzzy primary submodule. So the converse of the above-given proposition need not be true. Note the following example.

Example 2.

Let $R=M=\mathbb{Z}$. Define an intuitionistic fuzzy subset $A={\mu }_A,{\nu }_A$ in $\mathbb{Z}$ by$$\begin{array}{c}\text{(27)}& {\mu }_{A}x=\begin{array}{cc}1,\hfill & \text{if\hspace{0.17em}}x\in 6\mathbb{Z},\hfill \\ 0.3,\hfill & \text{if\hspace{0.17em}}x\in 2\mathbb{Z}-6\mathbb{Z},\hfill \\ 0,\hfill & \text{if\hspace{0.17em}}x\in \mathbb{Z}-2\mathbb{Z},\hfill \end{array}{\nu }_{A}x=\begin{array}{cc}0,\hfill & \text{if\hspace{0.17em}}x\in 6\mathbb{Z},\hfill \\ 0.5,\hfill & \text{if\hspace{0.17em}}x\in 2\mathbb{Z}-6\mathbb{Z},\hfill \\ 1,\hfill & \text{if\hspace{0.17em}}x\in \mathbb{Z}-2\mathbb{Z},\hfill \end{array}\end{array}$$

It is clear that $A$ is an intuitionistic fuzzy submodule of $\mathbb{Z}$ and $A^{\ast }=2\mathbb{Z}$, which is a primary submodule of $\mathbb{Z}$. We show that $A$ is not an intuitionistic fuzzy primary submodule of $\mathbb{Z}$. Consider $r=2,m=3$. Then, ${\mu }_{A}rm={\mu }_{A}2.3={\mu }_{A}6=1\ne {\mu }_{A}3=0,{\nu }_{A}rm={\nu }_{A}2.3={\nu }_{A}6=0\ne {\nu }_{A}3=1$ and ${\mu }_{A}rm={\mu }_{A}2.3>\sqrt{{\mu }_A}2={\vee }_{n\in \mathbb{N}}{\wedge }_{m\in \mathbb{Z}}{\mu }_A{2}^n.m=0.3,$${\nu }_{A}rm={\nu }_{A}2.3<\sqrt{{\nu }_A}2={\wedge }_{n\in \mathbb{N}}{\vee }_{m\in \mathbb{Z}}{\nu }_A{2}^n.m=0.5$. Therefore, $A$ is not an intuitionistic fuzzy primary submodule of $\mathbb{Z}$.

Definition 12 (see [18]).

An intuitionistic fuzzy ideal $A$ is said to be an intuitionistic fuzzy weakly primary ideal if for any $x,y\in R$$$\begin{array}{c}\text{(28)}& {\mu }_{A}x.y={\mu }_{A}x,{\nu }_{A}x.y={\nu }_{A}x\text{or}{\mu }_{A}x.y\le {\mu }_A{y}^n,{\nu }_{A}x.y\ge {\nu }_A{y}^n,\end{array}$$for some $n\in \mathbb{N}$

Proposition 12.

Let $A$ be an intuitionistic fuzzy primary submodule of $M$ with sup property, then $\overline{A}$ is an intuitionistic fuzzy weakly primary ideal of $R$ and $\sqrt{A}$ is the intuitionistic fuzzy prime ideal of $R$.

Proof.

Let $r_1,r_2\in R$, then

${\mu }_{\overline{A}}{r}_1{r}_2={\wedge }_{m\in M}{\mu }_A{r}_1{r}_2.m={\wedge }_{m\in M}{\mu }_A{r}_1{r}_2.m$, ${\nu }_{\overline{A}}{r}_1{r}_2={\vee }_{m\in M}{\nu }_A{r}_1{r}_2.m={\vee }_{m\in M}{\nu }_A{r}_1{r}_2.m$. Now since $A$ is an intuitionistic fuzzy primary submodule, then$$\begin{array}{c}\text{(29)}& {\nu }_A{r}_1{r}_2.m={\nu }_A{r}_1.m\text{or}{\nu }_A{r}_1{r}_2.m\ge \sqrt{{\nu }_A}{r}_2⟹⟹{\nu }_{\overline{A}}{r}_1{r}_2=\underset{m\in M}{{\displaystyle \wedge }}{\nu }_A{r}_1.m\text{or}{\nu }_{\overline{A}}{r}_1{r}_2\ge \underset{m\in M}{{\displaystyle \wedge }}\sqrt{{\nu }_A}{r}_2=\sqrt{{\nu }_A}{r}_2\\ ⟹{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or}{\nu }_{\overline{A}}{r}_1{r}_2\ge \sqrt{{\nu }_A}{r}_2=\underset{n\in \mathbb{N}}{{\displaystyle \vee }}\underset{m\in M}{{\displaystyle \wedge }}{\nu }_A{r}_2^n.m\\ ⟹{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or}{\nu }_{\overline{A}}{r}_1{r}_2\ge \underset{m\in M}{{\displaystyle \wedge }}{\nu }_A{r}_2^{n_1}.m\text{for\hspace{0.17em}some}{n}_1\in \mathbb{N}\\ \text{Because}A\text{has\hspace{0.17em}sup\hspace{0.17em}property}\\ ⟹{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or}{\nu }_{\overline{A}}{r}_1{r}_2\ge \underset{m\in M}{{\displaystyle \wedge }}{\nu }_A{r}_2^{n_1}.m\ge \underset{m\in M}{{\displaystyle \wedge }}{\nu }_A{r}_2^{n_1}{r}_2^{n_2}.m,\end{array}$$and also$$\begin{array}{c}\text{(30)}& {\nu }_A{r}_1{r}_2.m={\nu }_A{r}_1.m\text{or\hspace{0.17em}}{\nu }_A{r}_1{r}_2.m\ge \sqrt{{\nu }_A}{r}_2⟹⟹{\nu }_{\overline{A}}{r}_1{r}_2=\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_1.m\text{or}{\nu }_{\overline{A}}{r}_1{r}_2\ge \underset{m\in M}{{\displaystyle \vee }}\sqrt{{\nu }_A}{r}_2=\sqrt{{\nu }_A}{r}_2\\ ⟹{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or\hspace{0.17em}}{\nu }_{\overline{A}}{r}_1{r}_2\ge \sqrt{{\nu }_A}{r}_2=\underset{n\in \mathbb{N}}{{\displaystyle \wedge }}\underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_2^n.m\\ ⟹{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or\hspace{0.17em}}{\nu }_{\overline{A}}{r}_1{r}_2\ge \underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_2^{n_1}.m\text{for\hspace{0.17em}some\hspace{0.17em}}{n}_1\in \mathbb{N}\\ \text{Because}A\text{has\hspace{0.17em}sup\hspace{0.17em}property}\\ ⟹{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or\hspace{0.17em}}{\nu }_{\overline{A}}{r}_1{r}_2\ge \underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_2^{n_1}.m\ge \underset{m\in M}{{\displaystyle \vee }}{\nu }_A{r}_2^{n_1}{r}_2^{n_2}.m,\end{array}$$for some $n_1,n_2\in \mathbb{N}$. If we consider $n^{\prime }=n_1+n_2$ then we have$$\begin{array}{}\text{(31)}& {\mu }_{\overline{A}}{r}_1{r}_2={\mu }_{\overline{A}}{r}_1,{\nu }_{\overline{A}}{r}_1{r}_2={\nu }_{\overline{A}}{r}_1\text{or\hspace{0.17em}}{\mu }_{\overline{A}}{r}_1{r}_2\le {\mu }_{\overline{A}}{r}_1^{n^{\prime }},{\nu }_{\overline{A}}{r}_1{r}_2\ge {\nu }_{\overline{A}}{r}_1^{n^{\prime }}.\end{array}$$

This proves that $\overline{A}$ is an intuitionistic fuzzy weakly primary ideal of $R$. Next, by Remark 1 we have $\sqrt{\overline{A}}=\sqrt{A}$ and since $\sqrt{\overline{A}}$ is intuitionistic fuzzy prime ideal then $\sqrt{A}$ is an intuitionistic fuzzy prime ideal of $R$.

The opposite of the above proposition is true if $A$ is an ideal.

Theorem 3.

Let $A\in \text{IFI}R$ with sup property. Then, $A$ is an intuitionistic fuzzy primary submodule if and only if $A$ is an intuitionistic fuzzy weakly primary ideal of $R$.

Proof.

Suppose that $A$ is an intuitionistic fuzzy primary submodule of $R$, then clearly $A$ is an intuitionistic fuzzy primary ideal of $R$. Now since $A$ is an intuitionistic fuzzy primary submodule, then for $r_1,r_2\in R$$$\begin{array}{c}\text{(32)}& {\mu }_A{r}_1{r}_2={\mu }_A{r}_1,{\nu }_A{r}_1{r}_2={\nu }_A{r}_1\text{or}{\mu }_A{r}_1{r}_2\le \sqrt{{\mu }_A}{r}_2,{\nu }_A{r}_1{r}_2\ge \sqrt{{\nu }_A}{r}_2⟹\\ ⟹{\mu }_A{r}_1{r}_2\le \underset{n\in \mathbb{N}}{{\displaystyle \vee }}{\mu }_A{r}_2^n,{\nu }_A{r}_1{r}_2\ge \underset{n\in \mathbb{N}}{{\displaystyle \wedge }}{\nu }_A{r}_2^n\\ ⟹{\mu }_A{r}_1{r}_2\le {\mu }_A{r}_2^{n_1},{\nu }_A{r}_1{r}_2\ge {\nu }_A{r}_2^{n_2}\mathrm{for}\text{\hspace{0.17em}some}{n}_1,n_2\in \mathbb{N}\\ \text{Since\hspace{0.17em}}A\text{has\hspace{0.17em}Sup\hspace{0.17em}property}\\ ⟹{\mu }_A{r}_1{r}_2\le {\mu }_A{r}_2^{n_1}\le {\mu }_A{r}_2^{n_1}{r}_2^{n_2},{\nu }_A{r}_1{r}_2\ge {\nu }_A{r}_2^{n_2}\ge {\nu }_A{r}_2^{n_1}{r}_2^{n_2}\\ ⟹{\mu }_A{r}_1{r}_2\le {\mu }_A{r}_2^{n^{\prime }},{\nu }_A{r}_1{r}_2\ge {\nu }_A{r}_2^{n^{\prime }}\mathrm{consider}{n}_1+n_2=n^{\prime }.\end{array}$$

Therefore, we obtain that$$\begin{array}{}\text{(33)}& {\mu }_A{r}_1{r}_2={\mu }_A{r}_1,{\nu }_A{r}_1{r}_2={\nu }_A{r}_1\text{or}{\mu }_A{r}_1{r}_2\le {\mu }_A{r}_2^{n^{\prime }},{\nu }_A{r}_1{r}_2\ge {\nu }_A{r}_2^{n^{\prime }}.\end{array}$$

Hence $A$ is an intuitionistic fuzzy weakly primary ideal of $R$.

Conversely, let $A$ be an intuitionistic fuzzy weakly primary ideal of $R$, then

${\mu }_A{r}_1{r}_2={\mu }_A{r}_1,{\nu }_A{r}_1{r}_2={\nu }_A{r}_1$ or ${\mu }_A{r}_1{r}_2\le {\mu }_A{r}_2^n,{\nu }_A{r}_1{r}_2\ge {\nu }_A{r}_2^n$ for some $n\in \mathbb{N}$. Therefore, ${\mu }_A{r}_1{r}_2={\mu }_A{r}_1,{\nu }_A{r}_1{r}_2={\nu }_A{r}_1$ or ${\mu }_A{r}_1{r}_2\le {\vee }_{n\in \mathbb{N}}{\mu }_A{r}_2^n,{\nu }_A{r}_1{r}_2\ge {\wedge }_{n\in \mathbb{N}}{\nu }_A{r}_2^n$. Thus, ${\mu }_A{r}_1{r}_2={\mu }_A{r}_1,{\nu }_A{r}_1{r}_2={\nu }_A{r}_1$ or ${\mu }_A{r}_1{r}_2\le \sqrt{{\mu }_A}{r}_2,{\nu }_A{r}_1{r}_2\ge \sqrt{{\nu }_A}{r}_2$ for all $r_1,r_2\in R$. Hence $A$ is an intuitionistic fuzzy primary submodule of $R$.

The above-given theorem shows that the concept of the intuitionistic fuzzy primary submodule is the generalization of the intuitionistic fuzzy primary ideal.

Example 3.

Let ${\mathbb{Z}}_{12}$ as $\mathbb{Z}-\text{module}$. Define $A=m,{\mu }_{A}m,{\nu }_{A}m$ such that $m\in {\mathbb{Z}}_{12}$. By$$\begin{array}{c}\text{(34)}& \begin{array}{c}{\mu }_{A}m=1,\\ {\nu }_{A}m=0,\end{array}\hspace{1em}\text{if\hspace{0.17em}}m\in \overline{0},\overline{4},\overline{8},\begin{array}{c}{\mu }_{A}m=\frac{1}{2},\\ {\nu }_{A}m=\frac{1}{2},\end{array}\hspace{1em}\text{if\hspace{0.17em}}m\in \overline{2},\overline{6},\overline{10},\\ \begin{array}{c}{\mu }_{A}m=0,\\ {\nu }_{A}m=1,\end{array}\hspace{1em}\text{if\hspace{0.17em}}m\in \overline{1},\overline{3},\overline{5},\overline{7},\overline{9},\overline{11}.\end{array}$$

We get that $A$ is an intuitionistic fuzzy submodule of ${\mathbb{Z}}_{12}$. Now $\overline{A}=x,{\mu }_{\overline{A}}x,{\nu }_{\overline{A}}x$ for $x\in \mathbb{Z}$ defined by the following equation:$$\begin{array}{c}\text{(35)}& \begin{array}{c}{\mu }_{\overline{A}}z=\underset{m\in {\mathbb{Z}}_{12}}{{\displaystyle \wedge }}{\mu }_{A}z.m=1,\\ {\nu }_{\overline{A}}z=\underset{m\in {\mathbb{Z}}_{12}}{{\displaystyle \vee }}{\nu }_{A}z.m=0\end{array}\hspace{1em}\text{if\hspace{0.17em}}z\in \overline{4},\begin{array}{c}{\mu }_{\overline{A}}z=\underset{m\in {\mathbb{Z}}_{12}}{{\displaystyle \wedge }}{\mu }_{A}z.m=\frac{1}{2},\\ {\nu }_{\overline{A}}z=\underset{m\in {\mathbb{Z}}_{12}}{{\displaystyle \vee }}{\nu }_{A}z.m=\frac{1}{2},\end{array}\hspace{1em}\text{if\hspace{0.17em}}z\in \overline{2}-\overline{4},\\ \begin{array}{c}{\mu }_{\overline{A}}z=\underset{m\in {\mathbb{Z}}_{12}}{{\displaystyle \wedge }}{\mu }_{A}z.m=0,\\ {\nu }_{\overline{A}}z=\underset{m\in {\mathbb{Z}}_{12}}{{\displaystyle \vee }}{\nu }_{A}z.m=1,\end{array}\hspace{1em}\text{otherwise}.\end{array}$$

Clearly $\overline{A}$ is an intuitionistic fuzzy ideal of $\mathbb{Z}$. Now we know that $\sqrt{\overline{A}}=\sqrt{A}$ defined by $\sqrt{A}=\sqrt{{\mu }_A},\sqrt{{\nu }_A}$ and$$\begin{array}{c}\text{(36)}& \begin{array}{c}\sqrt{{\mu }_A}z=1,\\ \sqrt{{\nu }_A}z=0,\end{array}\hspace{1em}\text{if\hspace{0.17em}}z\in \overline{2},\begin{array}{c}\sqrt{{\mu }_A}z=0,\\ \sqrt{{\nu }_A}z=1,\end{array}\hspace{1em}\text{otherwise}.\end{array}$$

It is easy to show that $Az.m=Am\text{or}Az.m\le \sqrt{A}z$ for all $m\in {\mathbb{Z}}_{12},z\in \mathbb{Z}$. Hence $A$ is an intuitionistic fuzzy primary submodule of ${\mathbb{Z}}_{12}$.

Proposition 13.

Let $f:M⟹M^{\prime }$ be an epimorphism of R-module $M$ to R-module $M^{\prime }$. If $A\in \text{IFM}M$, then $\sqrt{A}\subseteq \sqrt{fA}$. Equality holds if $A$ is constant on kernel $f$.

Proof.

By Theorem 1 we have $\overline{A}\subseteq \overline{fA}$, therefore $\sqrt{\overline{A}}\subseteq \sqrt{\overline{fA}}$ and hence $\sqrt{A}\subseteq \sqrt{fA}$. Also $\overline{A}\subseteq \overline{fA}$ if $A$ is constant on kernel $f$, hence $\sqrt{\overline{A}}\subseteq \sqrt{\overline{fA}}$ and therefore $\sqrt{A}=\sqrt{fA}$.

Proposition 14.

Let $f:M⟶M^{\prime }$ be a homomorphism of R-module $M$ to R-module $M^{\prime }$. If $B\in \text{IFM}{M}^{\prime }$, then $\sqrt{f^{-1}B}\supseteq \sqrt{B}$. Equality holds if $f$ is an epimorphism.

Proof.

By Theorem 2, $\overline{f^{-1}B}\supseteq \overline{B}$, hence $\sqrt{\overline{f^{-1}{A}^{\prime }}}\supseteq \sqrt{\overline{B}}$ and therefore $\sqrt{f^{-1}B}\supseteq \sqrt{B}$. If $f$ is an epimorphism, then $\overline{f^{-1}B}=\overline{B}$ and hence $\sqrt{f^{-1}B}=\sqrt{B}$.

Theorem 4.

Let $f:M⟶M^{\prime }$ be an epimorphism of R-module $M$ to R-module $M^{\prime }$, and let $A$ be an intuitionistic fuzzy primary submodule of $M$, which is constant on kernel $f$, then the image $f$, $fA$ is an intuitionistic fuzzy primary submodule of $M^{\prime }$.

Proof.

By Theorem 1$fA$ is an intuitionistic fuzzy submodule of $M^{\prime }$. Let $r\in R,m^{\prime }\in M^{\prime }$, then$$\begin{array}{c}\text{(37)}& f{\mu }_{A}r{m}^{\prime }=\underset{m\in f^{-1}r{m}^{\prime }}{{\displaystyle \vee }}{\mu }_{A}m,f{\nu }_{A}r{m}^{\prime }=\underset{m\in f^{-1}r{m}^{\prime }}{{\displaystyle \wedge }}{\mu }_{A}m.\end{array}$$

Since $A$ constant on kernel $f$ then$$\begin{array}{c}\text{(38)}& \mathrm{if}m\in f^{-1}r{m}^{\prime }⟹fm=r{m}^{\prime }=rf{m}_1\mathrm{There}\mathrm{is}{m}_1\in M\mathrm{such}\mathrm{that}f{m}_1=m^{\prime }⟹fm-fr{m}_1=0⟹m-r{m}_1\in \text{ker}f⟹{\mu }_{A}m={\mu }_{A}r{m}_1,{\nu }_{A}m={\nu }_{A}r{m}_1.\end{array}$$

Therefore, $f{\mu }_{A}r{m}^{\prime }={\vee }_{m_1\in f^{-1}{m}^{\prime }}{\mu }_{A}r{m}_1,f{\nu }_{A}r{m}^{\prime }={\wedge }_{m_1\in f^{-1}{m}^{\prime }}{\mu }_{A}r{m}_1$, but $A$ is an intuitionistic fuzzy primary submodule of $M$ and hence$$\begin{array}{}\text{(39)}& {\mu }_{A}r{m}_1={\mu }_A{m}_1,{\nu }_{A}r{m}_1={\nu }_A{m}_1\text{or}{\mu }_{A}r{m}_1\le \sqrt{{\mu }_A}r,{\nu }_{A}r{m}_1\ge \sqrt{{\nu }_A}r.\end{array}$$

Now if ${\mu }_{A}r{m}_1={\mu }_A{m}_1,{\nu }_{A}r{m}_1={\nu }_A{m}_1$ then$$\begin{array}{c}\text{(40)}& f{\mu }_{A}r{m}^{\prime }=\underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \vee }}{\mu }_A{m}_1=f{\mu }_A{m}^{\prime };f{\nu }_{A}r{m}^{\prime }=\underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \wedge }}{\mu }_A{m}_1=f{\nu }_A{m}^{\prime }.\end{array}$$

And, if ${\mu }_{A}r{m}_1\le \sqrt{{\mu }_A}r,{\nu }_{A}r{m}_1\ge \sqrt{{\nu }_A}r$ then$$\begin{array}{c}\text{(41)}& \underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \vee }}{\mu }_{A}r{m}_1\le \underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \vee }}\sqrt{{\mu }_A}r⟹f{\mu }_{A}r{m}^{\prime }\le \underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \vee }}\sqrt{{\mu }_A}r=\sqrt{{\mu }_A}r\le \sqrt{f{\mu }_A}r,\\ \underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \wedge }}{\nu }_{A}r{m}_1\ge \underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \wedge }}\sqrt{{\nu }_A}r⟹f{\nu }_{A}r{m}^{\prime }\\ \ge \underset{m_1\in f^{-1}{m}^{\prime }}{{\displaystyle \wedge }}\sqrt{{\nu }_A}r=\sqrt{{\nu }_A}r\ge \sqrt{f{\nu }_A}r.\end{array}$$

Therefore, we got that $f{\mu }_{A}r{m}^{\prime }=f{\mu }_A{m}^{\prime },f{\nu }_{A}r{m}^{\prime }=f{\nu }_A{m}^{\prime }$ or $f{\mu }_{A}r{m}^{\prime }\le \sqrt{f{\mu }_A}r,f{\nu }_{A}r{m}^{\prime }\ge \sqrt{f{\nu }_A}r$. This show that $fA$ is an intuitionistic fuzzy primary submodule of $M^{\prime }$.

Theorem 5.

Let $f:M⟶M^{\prime }$ be a homomorphism of R-module $M$ to R-module $M^{\prime }$. If $B$ is an intuitionistic fuzzy primary submodule of $M^{\prime }$, then $f^{-1}B$ is an intuitionistic fuzzy primary submodule of $M$.

Proof.

Clearly $f^{-1}B$ is an intuitionistic fuzzy submodule of $M$. Let $r\in R,m\in M$, then$$\begin{array}{c}\text{(42)}& f^{-1}{\mu }_{B}rm={\mu }_{B}frm={\mu }_{B}rfm,f^{-1}{\nu }_{B}rm={\nu }_{B}frm={\nu }_{B}rfm.\end{array}$$

But $B$ is an intuitionistic fuzzy primary submodule, hence$$\begin{array}{c}\text{(43)}& {\mu }_{B}rfm={\mu }_{B}fm=f^{-1}{\mu }_{B}m,{\nu }_{B}rfm={\nu }_{B}fm=f^{-1}{\nu }_{B}m\text{\hspace{0.17em}or}\\ {\mu }_{B}rfm\le \sqrt{{\mu }_B}r\le \sqrt{f^{-1}{\mu }_B}r,\\ {\nu }_{B}rfm\ge \sqrt{{\nu }_B}r\ge \sqrt{f^{-1}{\nu }_B}r.\end{array}$$

Thus, we show that$$\begin{array}{c}\text{(44)}& f^{-1}{\mu }_{B}rm=f^{-1}{\mu }_{B}m,f^{-1}{\nu }_{B}rm=f^{-1}{\nu }_{B}m\text{\hspace{0.17em}or}\\ f^{-1}{\mu }_{B}rm\le \sqrt{f^{-1}{\mu }_B}r,\\ f^{-1}{\nu }_{B}rm\ge \sqrt{f^{-1}{\nu }_B}r.\end{array}$$

This proves that $f^{-1}B$ is an intuitionistic fuzzy primary submodule of $M$.

4. Concluding Remarks

In this paper, we defined the radical of an intuitionistic fuzzy submodule, and then we studied the intuitionistic fuzzy primary submodules with the help of this concept. We also presented some properties of homomorphic image and preimage of these submodules.

References

[1] M. Behboodi, H. Koohy, "Weakly prime modules," Vietnam Journal of Mathematics, vol. 32 no. 2, pp. 185-195, 2004.

[2] J. Goswami, H. K. Saikia, "On the spectrum of weakly prime submodule," Thai Journal of Mathematics, vol. 19 no. 1, pp. 51-58, 2019.

[3] J. M. Abulebda Lamis, "Uniformly primal submodule over noncommutative ring," Journal of Mathematics, vol. 2020,DOI: 10.1155/2020/1593253, 2020.

[4] N. D. H. Nghiem, S. Baupradist, R. Chinram, "On nearly prime submodules of unitary modules," Journal of Mathematics, vol. 2018,DOI: 10.1155/2018/7202590, 2018.

[5] L. A. Zadeh, "Fuzzy sets," Information and Control, vol. 8 no. 3, pp. 338-353, DOI: 10.1016/s0019-9958(65)90241-x, 1965.

[6] A. Rosenfeld, "Fuzzy groups," Journal of Mathematical Analysis and Applications, vol. 35, pp. 512-517, 1971.

[7] C. V. Negoiţă, D. A. Ralescu, Applications of Fuzzy Sets to Systems Analysis, 1975.

[8] K. Atanassov, "Intuitionistic fuzzy sets," Fuzzy Sets and Systems, vol. 197, 1986.

[9] R. Biswas, Intuitionistic fuzzy subgroups, vol. 10, pp. 37-46, 1989.

[10] K. Hur, S.-Y. Jang, H.-W. Kang, "Intuitionistic fuzzy ideals of a ring," The Pure and Applied Mathematics, vol. 12 no. 3, pp. 193-209, 2005.

[11] M. Mashinchi, M. M. Zahedi, "On L-fuzzy primary submodules," Fuzzy Sets and Systems, vol. 49 no. 2, pp. 231-236, DOI: 10.1016/0165-0114(92)90328-2, 1992.

[12] B. Davvaz, W. A. Dudek, Y. B. Jun, "Intuitionistic fuzzy -submodules," Information Sciences, vol. 176 no. 3, pp. 285-300, DOI: 10.1016/j.ins.2004.10.009, 2006.

[13] P. K. Sharma, G. Kaur, "Intuitionistic fuzzy prime spectrum of a ring," International Journal of Fuzzy Systems, vol. 9, pp. 167-175, 2017.

[14] S. Rahman, H. K. Saikia, "Some aspects of Atanassov’s intuitionistic fuzzy submodule," International Journal of Pure and Applied Mathematics, vol. 77, pp. 369-383, 2012.

[15] P. Sharma, "Translates of intuitionistic fuzzy subring," International Review of Fuzzy Mathematics, vol. 6, pp. 77-84, 2011.

[16] P. K. Sharma, K. Gagandeep, "Residual quotient and annihilator of intuitionistic fuzzy sets of ring and module," International Journal of Computer Science and Information Technology, vol. 9, 2017.

[17] K. Hur, S.-Y. Jang, H.-W. Kang, "Intuitionistic fuzzy subgroupoids," International Journal of Fuzzy Logic and Intelligent Systems, vol. 3 no. 1, pp. 72-77, DOI: 10.5391/ijfis.2003.3.1.072, 2003.

[18] D.-S. Lee, C.-H. Park, "On intuitionistic fuzzy w-primary submodules," Honam Mathematical Journal, vol. 29 no. 4, pp. 631-640, DOI: 10.5831/hmj.2007.29.4.631, 2007.