- Research
- Open Access

# A fuzzy characterization of QF rings

- Wenxi Li
^{1}, - Jianlong Chen
^{2}and - Liang Shen
^{2}Email author

**2014**:184

https://doi.org/10.1186/1029-242X-2014-184

© Li et al.; licensee Springer. 2014

**Received:**22 February 2014**Accepted:**28 April 2014**Published:**13 May 2014

## Abstract

Let *R* be a ring. *R* is called a quasi-Frobenius (QF) ring if *R* is right artinian and ${R}_{R}$ is an injective right *R*-module. In this article, we introduce (weak) fuzzy homomorphisms of modules to obtain a fuzzy characterization of QF rings. We also obtain some fuzzy characterizations of right artinian rings and right CF rings. These results throw new light on the research of QF rings and the related CF conjecture.

**MSC:**03E72, 16L60.

## Keywords

- fuzzy homomorphisms
- weak fuzzy homomorphisms
- artinian rings
- QF rings

## 1 Introduction

Recall that a *fuzzy subset* of a nonempty set *X* is a map *f* from *X* into the closed interval $[0,1]$. The notion of fuzzy subset of a set was firstly introduced by Zadeh [1]. Then this important ideal has been applied to various algebraic structures such as groups and rings and so on (see [2–9] etc.). In this article, we introduce some special fuzzy subsets of modules to characterize quasi-Frobenius (QF) rings.

QF rings were introduced by Nakayama [10] as generalizations of group algebras of a finite group over a field. A ring *R* is called *quasi-Frobenius* (*QF*) if the right *R*-module ${R}_{R}$ is both artinian and injective. QF rings became an important algebraic structure because of their beautiful characterizations and nice applications (see [11–16] etc.). For example, a ring *R* is QF if and only if every right *R*-module can be embedded into a free right *R*-module. Many results of QF rings have been applied into coding theory. During the progress of research on QF rings, many important conjectures arose. One of them is the CF conjecture (see [17, 18] etc.). It says that every right CF ring is right artinian. Recall that a ring *R* is called *right CF* if every cyclic right *R*-module can be embedded into a free right *R*-module.

Firstly, we introduce the fuzzy homomorphism and weak fuzzy homomorphism of *R*-modules in Section 2. Then in Section 3, we use weak fuzzy homomorphisms to give a characterization of injective right *R*-modules. We also obtain some new fuzzy characterizations of right artinian rings. In Section 4, we give a fuzzy characterization of right CF rings. We also give an approach to the CF conjecture through fuzzy viewpoints. Then based on the results we have obtained, we finally get a fuzzy characterization of QF rings.

## 2 Definitions and examples

Throughout the paper, *R* is an associative ring with identity and all modules are unitary. For a subset *X* of a ring *R*, the right annihilator of *X* in *R* is $\mathbf{r}(X)=\{r\in R:xr=0\text{for all}x\in X\}$. We write ${M}_{R}$ to indicate that *M* is a right *R*-module. Let ${M}_{R}$ and ${N}_{R}$ be two right *R*-modules. ${Hom}_{R}(M,N)$ denotes the set of all right *R*-module homomorphisms from ${M}_{R}$ to ${N}_{R}$. $A\times B$ means the Cartesian cross product of two sets *A* and *B*. We use $Im(f)$ to denote the image of a map *f*. For much more notations one is referred to [19].

*f*of ${M}_{R}\times {N}_{R}$.

- (1)
$\mathrm{\forall}x\in M$, $\mathrm{\exists}y\in N$ such that $f(x,y)>0$;

- (2)
$\mathrm{\forall}x\in M$, $\mathrm{\forall}{y}_{1},{y}_{2}\in N$, $f(x,{y}_{1})>0$ and $f(x,{y}_{2})>0$ implies ${y}_{1}={y}_{2}$;

- (3)$\mathrm{\forall}{x}_{1},{x}_{2}\in M$, $\mathrm{\forall}y\in N$,$f({x}_{1}+{x}_{2},y)\ge sup\{min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}:{y}_{1}+{y}_{2}=y,{y}_{1},{y}_{2}\in N\};$
- (4)
$\mathrm{\forall}x\in M$, $\mathrm{\forall}y\in N$, $\mathrm{\forall}r\in R$, $f(xr,y)\ge sup\{f(x,{y}_{1}):y={y}_{1}r,{y}_{1}\in N\}$.

**Definition 2.1** If *f* satisfies (1), (2), (3), and (4) of the above conditions, *f* is called a *fuzzy homomorphism* from ${M}_{R}$ to ${N}_{R}$. If *f* satisfies (1′), (2), (3), and (4) of the above conditions, *f* is called a *weak fuzzy homomorphism* from ${M}_{R}$ to ${N}_{R}$. We will use ${FHom}_{R}(M,N)$ (resp., ${WFHom}_{R}(M,N)$) to denote the set of all fuzzy homomorphisms (resp., weak fuzzy homomorphisms) from ${M}_{R}$ to ${N}_{R}$. It is clear that ${FHom}_{R}(M,N)\subseteq {WFHom}_{R}(M,N)$.

**Example 2.2**Let $u\in {Hom}_{R}(M,N)$. ${f}_{u}$ is a fuzzy subset of $M\times N$ constructed by

Then ${f}_{u}\in {FHom}_{R}(M,N)$.

*Proof*The conditions (1) and (2) of ${f}_{u}$ are satisfied obviously. Let ${x}_{1},{x}_{2}\in M$ and $y\in N$. If $y=u({x}_{1}+{x}_{2})$, it is clear that

Hence ${f}_{u}$ satisfies the condition (3). For the condition (4), let $x\in M$, $y\in N$, $r\in R$. If $y=u(xr)$, it is clear that $1={f}_{u}(xr,y)\ge sup\{{f}_{u}(x,{y}_{1}):y={y}_{1}r,{y}_{1}\in N\}$. If $y\ne u(xr)$ and $y={y}_{1}r$, then ${y}_{1}\ne u(x)$. So ${f}_{u}(x,{y}_{1})=0$. Hence $0={f}_{u}(xr,y)\ge 0=sup\{{f}_{u}(x,{y}_{1}):y={y}_{1}r,{y}_{1}\in N\}$. Thus, ${f}_{u}\in {FHom}_{R}(M,N)$. □

**Remark 2.3**Let $f\in {WFHom}_{R}(M,N)$.

- (i)
According to the condition (3), $\mathrm{\forall}{x}_{1},{x}_{2}\in M$, $\mathrm{\forall}{y}_{1},{y}_{2}\in N$, $f({x}_{1}+{x}_{2},{y}_{1}+{y}_{2})\ge min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}$. According to the conditions (3) and (4), $\mathrm{\forall}x\in M$, $\mathrm{\forall}y\in N$, $f(0,0)\ge min\{f(x,y),f(-x,-y)\}=f(x,y)=f(-x,-y)$.

- (ii)
Set $t\in (0,1]$ and ${M}_{t}=\{x\in {M}_{R}:\mathrm{\exists}y\in N,f(x,y)\ge t\}$. If ${M}_{t}$ is not empty, according to the conditions (3) and (4), ${M}_{t}$ is a right

*R*-submodule of ${M}_{R}$. - (iii)
Let ${K}_{R}$ be a submodule of ${M}_{R}$ such that $f{|}_{K\times N}\in {FHom}_{R}(K,N)$. Then by the conditions (1) and (2), for each $x\in K$, there exists a unique ${y}_{x}\in N$ such that $f(x,{y}_{x})>0$. Now define a map $u:K\to N$ with $u(x)={y}_{x}$. Again by the conditions (2), (3), and (4), it is not difficult to see that $u\in {Hom}_{R}(K,N)$.

**Definition 2.4** Let ${f}_{1},{f}_{2}\in {WFHom}_{R}(M,N)$, we say ${f}_{1}\le {f}_{2}$ if ${f}_{1}(x,y)\le {f}_{2}(x,y)$ for all $x\in M$, $y\in N$. ${f}_{1}\wedge {f}_{2}$ is defined by $({f}_{1}\wedge {f}_{2})(x,y)=min\{{f}_{1}(x,y),{f}_{2}(x,y)\}$, $\mathrm{\forall}x\in M$, $y\in N$. It is easy to prove that if ${f}_{1},{f}_{2}\in {WFHom}_{R}(M,N)$, then ${f}_{1}\wedge {f}_{2}\in {WFHom}_{R}(M,N)$. It is also clear that ${f}_{1}\wedge {f}_{2}\le {f}_{1}$ and ${f}_{1}\wedge {f}_{2}\le {f}_{2}$.

**Definition 2.5** A weak fuzzy homomorphism $f\in {WFHom}_{R}(M,N)$ is said to be *extendable* if there exists $g\in {FHom}_{R}(M,N)$ such that $f\le g$.

**Example 2.6**Let ${M}_{R}$ be a non-artinian right

*R*-module. Then

*M*has an infinite descending chain of submodules $M={M}_{1}\u228b{M}_{2}\u228b{M}_{3}\cdots $ . Set $N={\bigcap}_{i}{M}_{i}$. Now we define a fuzzy subset

*f*of $M\times M$ by

Then $f\in {FHom}_{R}(M,M)$.

*Proof* It is obvious that *f* satisfies the conditions (1) and (2).

For the condition (3), let ${x}_{1},{x}_{2},y\in R$ with $y={y}_{1}+{y}_{2}$, where ${y}_{1},{y}_{2}\in R$. We only need to consider the following three cases.

Case 1: $min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}=0$. It is clear that $f({x}_{1}+{x}_{2},y)\ge min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}$.

Case 2: $0<min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}<1$. We can suppose that $f({x}_{1},{y}_{1})=1-\frac{1}{j+1}$ or 1, $f({x}_{2},{y}_{2})=1-\frac{1}{k+1}$ and $j\ge k$. Then $I\subseteq {I}_{j}\subseteq {I}_{k}$ and ${x}_{1}={y}_{1}\in {M}_{j}$ or *N*, ${x}_{2}={y}_{2}\in {I}_{k}$. So ${x}_{1}+{x}_{2}={y}_{1}+{y}_{2}\in {I}_{k}$. Hence $f({x}_{1}+{x}_{2},y)\ge 1-\frac{1}{k+1}=min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}$.

Case 3: $min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}=1$. Then ${x}_{1}={y}_{1}\in N$ and ${x}_{2}={y}_{2}\in N$. So $f({x}_{1}+{x}_{2},y)=1\ge min\{f({x}_{1},{y}_{1}),f({x}_{2},{y}_{2})\}$.

From the above three cases, it is clear that *f* satisfies the condition (3).

Finally, let $x,r,y\in R$ with $y={y}_{1}r$, where ${y}_{1}\in R$. If $f(x,{y}_{1})=0$, then $f(xr,y)\ge f(x,{y}_{1})$. If $f(x,{y}_{1})=1-\frac{1}{k+1}$ for a positive integer *k*, then $x={y}_{1}\in {M}_{k}$. So $xr={y}_{1}r\in {M}_{k}$. Hence $f(xr,y)\ge 1-\frac{1}{k+1}=f(x,{y}_{1})$. If $f(x,{y}_{1})=1$, then $x={y}_{1}\in N$. So $xr=y\in N$. Thus $f(xr,y)\ge f(x,{y}_{1})$. Therefore, $f(xr,y)\ge sup\{f(x,{y}_{1}):y={y}_{1}r,{y}_{1}\in R\}$ for all $x\in R$, $y\in R$ and $r\in R$. Then *f* satisfies the condition (4). □

**Definition 2.7** A weak fuzzy homomorphism $f\in {WFHom}_{R}(M,N)$ is said to be *bounded* if there exists $t\in (0,1]$ such that, $\mathrm{\forall}m\in M$ and $\mathrm{\forall}n\in N$, $f(m,n)\ge t$ or $f(m,n)=0$.

**Example 2.8**Let ${M}_{R}$ be a non-noetherian right

*R*-module. Then

*M*has an infinite ascending chain of submodules ${M}_{1}\subseteq \u0337{M}_{2}\subseteq \u0337{M}_{3}\cdots $ . Now we define a fuzzy subset

*f*of $M\times M$ by

Then $f\in {WFHom}_{R}(M,M)$. In particular, *f* is extendable and not bounded.

*Proof*By a similar discussion as that in Example 2.6, we have $f\in {WFHom}_{R}(M,M)$. It is obvious that

*f*is not bounded. Now set

It is clear that $g\in {FHom}_{R}(M,M)$ and $f\le g$. So *f* is extendable. □

## 3 Fuzzy characterizations of injective modules and artinian rings

According to Baer’s Criterion, a right *R*-module ${M}_{R}$ is said to be *injective* if every homomorphism from a right ideal *I* of *R* to ${M}_{R}$ can be extended to a homomorphism from ${R}_{R}$ to ${M}_{R}$.

**Theorem 3.1** *Let* *R* *be a ring and* ${M}_{R}$ *a right* *R*-*module*. *Then* *M* *is injective if and only if every* $f\in {WFHom}_{R}(R,M)$ *is extendable*.

*Proof*(⇐) Let

*I*be a right ideal of

*R*. Suppose $u\in {Hom}_{R}(I,M)$, we will show that

*u*can be extended to a homomorphism $v\in {Hom}_{R}(R,M)$. Firstly we construct a fuzzy subset of $R\times M$ by

By a similar proof of Example 2.2, $f\in {WFHom}_{R}(R,M)$. Since *f* is extendable, there exists some $g\in {FHom}_{R}(R,M)$ such that $f\le g$. Now define $v:R\to M$ via $v(x)=y$, where $g(x,y)>0$. According to Remark 2.3(iii), $v\in {Hom}_{R}(R,M)$. It is easy to see that $v{|}_{I}=u$. This shows that *M* is an injective right *R*-module.

(⇒) Assume that *M* is an injective right *R*-module and $f\in {WFHom}_{R}(R,M)$. Set $I={\sum}_{0\ne t\in Imf}{R}_{t}$. By Remark 2.3(ii), *I* is a right ideal of *R*. Now define $u:I\to M$ via $u(x)=y$, where $f(x,y)>0$. According to Remark 2.3(iii), $u\in {Hom}_{R}(I,M)$. Since *M* is injective as a right *R*-module, *u* can be extended to a homomorphism *v* from ${R}_{R}$ to ${M}_{R}$. By Example 2.2, we have ${f}_{v}\in {FHom}_{R}(R,M)$. It is clear that $f\le {f}_{v}$. So *f* is extendable. □

Next we will give some new fuzzy characterizations of right artinian rings. Recall that a fuzzy subset *μ* of a ring *R* is called a *fuzzy left* (*right*) *ideal* of *R* if *μ* satisfies: (i) $\mu (x-y)\ge min\{\mu (x),\mu (y)\}$; (ii) $\mu (xy)\ge \mu (y)(\mu (xy)\ge \mu (x))$ for all $x,y\in R$. A fuzzy subset *f* is called *finite valued* if Im*f* is a finite set. If Im*f* is an infinite set, *f* is called *infinite valued*.

**Theorem 3.2**

*The following conditions are equivalent for a ring*

*R*.

- (a)
*R**is right artinian*. - (b)
*Every fuzzy right ideal of**R**is finite valued*. - (c)
*For every*$f\in {WFHom}_{R}({R}_{R},{R}_{R})$,*f**is finite valued*. - (d)
*For every*$f\in {FHom}_{R}({R}_{R},{R}_{R})$,*f**is finite valued*.

*Proof*(a) ⇔ (b) See [[6], Theorem 3.2].

- (b)⇒ (c) Suppose $f\in {WFHom}_{R}({R}_{R},{R}_{R})$. We can define a fuzzy subset
*μ*of*R*by$\mu (x)=\{\begin{array}{ll}t,& f(x,y)=t>0\text{for some}y\in R,\\ 0,& \text{others}.\end{array}$

*μ*is a fuzzy right ideal of

*R*. Since $Imf\subseteq Im\mu \cup \{0\}$ and

*μ*is finite valued,

*f*is finite valued.

- (c)
⇒ (d) is obvious.

- (d)⇒ (a) Assume that
*R*is not right artinian. Let $R={I}_{1}\u228b{I}_{2}\u228b{I}_{3}\cdots $ be a descending chain of right ideals of*R*and $I={\bigcap}_{i}{I}_{i}$. Define a fuzzy subset*f*of $R\times R$ by$f(x,y)=\{\begin{array}{ll}1,& x=y\in I,\\ 1-\frac{1}{n+1},& x=y\in {I}_{n},x\notin {I}_{n+1},\\ 0,& \text{others}.\end{array}$

By Example 2.6, $f\in {FHom}_{R}({R}_{R},{R}_{R})$. But *f* is infinite valued. This is a contradiction. So *R* is right artinian. □

## 4 Fuzzy characterizations of right CF rings and QF rings

In this section, we will firstly give a fuzzy characterization of right CF ring. It is well known that a ring *R* is right CF if and only if for every right ideal *I* of *R*, there exist ${x}_{1},\dots ,{x}_{n}\in R$ such that $I=\mathbf{r}({x}_{1},\dots ,{x}_{n})$ (see [[14], Lemma 7.2]).

**Theorem 4.1** *A ring* *R* *is a right CF ring if and only if for every bounded and extendable* $f\in WFHom({R}_{R},{R}_{R})$, *there exist* ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$ *such that* $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$.

*Proof*(⇒) Suppose

*R*is a right CF ring and $f\in WFHom(R,R)$ is bounded and extendable. Then there exist $F\in {FHom}_{R}({R}_{R},{R}_{R})$ and $s\in (0,1]$ such that $f\le F$ and $f(x,y)\ge s$ for all $x,y\in R$ with $f(x,y)>0$. Set $I={\sum}_{0\ne t\in Imf}{R}_{t}$. By Remark 2.3(ii),

*I*is a right ideal of

*R*. Since

*R*is a right CF ring, there exist ${c}_{1},\dots ,{c}_{n}\in R$ such that $I=\mathbf{r}({c}_{1},\dots ,{c}_{n})$. To be convenient, we assume that ${c}_{1}=0$. Since $F\in {FHom}_{R}({R}_{R},{R}_{R})$, for the identity $1\in R$, there exists $c\in R$ such that $F(1,c)>0$. We now can construct fuzzy subsets ${F}_{1},{F}_{2},\dots ,{F}_{n}$ of $R\times R$ by

- (a)
If $x\in I$, according to Remark 2.3(i), there exists $y\in R$, such that $f(x,y)>0$. As $F\in {FHom}_{R}({R}_{R},{R}_{R})$, by the condition (4), $F(x,cx)\ge F(1,c)>0$. Then by the condition (2), we have $F(x,y)=0$ for all $y\ne cx$. Since $f\le F$, $f(x,y)=0$ for all $y\ne cx$. This implies that $f(x,cx)=f(x,y)>0$. Therefore, ${F}_{i}(x,cx)=f(x,cx)>0$. If $x\notin I$, let $y=(c-{c}_{i})x$, then ${F}_{i}(x,y)=s>0$. So ${F}_{i}$ satisfies the condition (1).

- (b)
If $x\in I$ and ${y}_{1},{y}_{2}\in R$ with ${F}_{i}(x,{y}_{1})>0$ and ${F}_{i}(x,{y}_{2})>0$, then ${y}_{1}={y}_{2}=cx$. If $x\notin I$ and ${y}_{1},{y}_{2}\in R$ with ${F}_{i}(x,{y}_{1})>0$ and ${F}_{i}(x,{y}_{2})>0$, then ${y}_{1}={y}_{2}=(c-{c}_{i})x$. So ${F}_{i}$ satisfies the condition (2).

- (c)
Let ${x}_{1},{x}_{2},y\in R$ with $y={y}_{1}+{y}_{2}$, where ${y}_{1},{y}_{2}\in R$.

If $min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}=0$, then ${F}_{i}({x}_{1}+{x}_{2},y)\ge min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}$. If $min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}>0$, we only need to discuss the following three cases.

Case 1: ${x}_{1},{x}_{2}\in I$. Then ${y}_{1}=c{x}_{1}$ and ${y}_{2}=c{x}_{2}$.

Case 2: ${x}_{1}\in I$ and ${x}_{2}\notin I$. Then ${y}_{1}=c{x}_{1}$, ${y}_{2}=(c-{c}_{i}){x}_{2}$ and ${x}_{1}+{x}_{2}\notin I$.

So $min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}=s$. Since $I=\mathbf{r}({c}_{1},\dots ,{c}_{n})$, ${c}_{i}{x}_{1}=0$. Then ${y}_{1}+{y}_{2}=c{x}_{1}+(c-{c}_{i}){x}_{2}=(c-{c}_{i})({x}_{1}+{x}_{2})$. Thus, ${F}_{i}({x}_{1}+{x}_{2},y)=s=min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}$.

Case 3: ${x}_{1}\notin I$ and ${x}_{2}\notin I$. Then ${y}_{1}=(c-{c}_{i}){x}_{1}$ and ${y}_{2}=(c-{c}_{i}){x}_{2}$.

So $min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}=s$ and ${y}_{1}+{y}_{2}=(c-{c}_{i})({x}_{1}+{x}_{2})$. If $({x}_{1}+{x}_{2})\in I$, then ${y}_{1}+{y}_{2}=(c-{c}_{i})({x}_{1}+{x}_{2})=c({x}_{1}+{x}_{2})$ because ${c}_{i}{x}_{1}={c}_{i}{x}_{2}=0$. Thus, ${F}_{i}({x}_{1}+{x}_{2},y)=f({x}_{1}+{x}_{2},y)\ge s=min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}$. If $({x}_{1}+{x}_{2})\notin I$, then ${y}_{1}+{y}_{2}=(c-{c}_{i})({x}_{1}+{x}_{2})$. So ${F}_{i}({x}_{1}+{x}_{2},y)=s\ge min\{{F}_{i}({x}_{1},{y}_{1}),{F}_{i}({x}_{2},{y}_{2})\}$.

- (d)
For all $x,{y}_{1}r\in R$ with $y={y}_{1}r$, where ${y}_{1}\in R$. If ${F}_{i}(x,{y}_{1})=0$, ${F}_{i}(xr,y)\ge {F}_{i}(x,{y}_{1})$. If ${F}_{i}(x,{y}_{1})>0$, we consider the following two cases.

Case 1: $x\in I$. Then ${y}_{1}=cx$ and ${F}_{i}(x,{y}_{1})=f(x,{y}_{1})$. Thus, $xr\in I$ and $y={y}_{1}r=cxr$. So ${F}_{i}(xr,y)=f(xr,y)\ge f(x,{y}_{1})={F}_{i}(x,{y}_{1})$.

Case 2: $x\notin I$. Then ${y}_{1}=(c-{c}_{i})x$ and ${F}_{i}(x,{y}_{1})=s$. Since $xr\in I$, ${c}_{i}xr=0$. Then $y={y}_{1}r=(c-{c}_{i})xr=cxr$. Thus, ${F}_{i}(xr,y)=f(xr,y)\ge s={F}_{i}(x,{y}_{1})$. If $xr\notin I$ then $y={y}_{1}r=(c-{c}_{i})xr$. So ${F}_{i}(xr,y)=s\ge {F}_{i}(x,{y}_{1})$.

Hence ${F}_{i}$ satisfies the condition (4).

From the above, ${F}_{i}\in {FHom}_{R}({R}_{R},{R}_{R})$ for $i=1,2,\dots ,n$.

Next we show that $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$.

Case 1: $x\in I$. If $y=cx$, then ${F}_{i}(x,y)=f(x,y)$, $i=1,2,\dots ,n$. So $f(x,y)=min\{{F}_{1}(x,y),{F}_{2}(x,y),\dots ,{F}_{n}(x,y)\}$. By (a) in the above proof, $f(x,cx)>0$. If $y\ne cx$, then $f(x,y)=0$. So ${F}_{i}(x,y)=0$, $i=1,\dots ,n$. Hence $f(x,y)=min\{{F}_{1}(x,y),{F}_{2}(x,y),\dots ,{F}_{n}(x,y)\}$ for every $x\in I$.

Case 2: $x\notin I$. By the definition of *I*, $\mathrm{\forall}y\in R$, $f(x,y)=0$. If $y\ne (c-{c}_{j})x$ for some $j\in \{1,2,\dots ,n\}$, then ${F}_{j}(x,y)=0$. Thus, $f(x,y)=min\{{F}_{1}(x,y),{F}_{2}(x,y),\dots ,{F}_{n}(x,y)\}$. If $y=(c-{c}_{i})x$ for any $i\in \{1,2,\dots ,n\}$, since ${c}_{1}=0$, $cx=(c-{c}_{1})x=(c-{c}_{2})x=\cdots =(c-{c}_{n})x$. So ${c}_{i}x=0$, $i=1,2,\dots ,n$. Hence $x\in \mathbf{r}({c}_{1},\dots ,{c}_{n})=I$. This is a contradiction.

So $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$.

*I*be a right ideal of

*R*. We can construct a fuzzy subset

*f*of $R\times R$ by

It is easy to see that $f\in {WFHom}_{R}({R}_{R},{R}_{R})$ and it is bounded and extendable. So there exist ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$ such that $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$. Then there exists ${c}_{i}\in R$ such that ${F}_{i}(1,{c}_{i})>0$, $i=1,\dots ,n$. By the condition (4), for each $x\in R$, ${F}_{i}(x,{c}_{i}x)\ge {F}_{i}(1,{c}_{i})>0$, $i=1,2,\dots ,n$. But for every $x\in I$, ${F}_{i}(x,0)\ge f(x,0)=1$. By the condition (2), ${c}_{i}x=0$, $i=1,2,\dots ,n$. Hence $x\in \mathbf{r}({c}_{1},\dots ,{c}_{n})$. This shows that $I\subseteq \mathbf{r}({c}_{1},\dots ,{c}_{n})$. On the contrary, if $x\in \mathbf{r}({c}_{1},\dots ,{c}_{n})$, then ${F}_{i}(x,0)={F}_{i}(x,{c}_{i}x)>0$. So $f(x,0)=min\{{F}_{1}(x,0),\dots ,{F}_{n}(x,0)\}>0$. Hence $x\in I$. This shows that $\mathbf{r}({c}_{1},\dots ,{c}_{n})\subseteq I$. Therefore, $I={\mathbf{r}}_{R}({c}_{1},\dots ,{c}_{n})$. So *R* is a right CF ring. □

The following proposition can be looked on as an approach to the CF conjecture.

**Proposition 4.2** *Let* *R* *be a ring*. *If for every extendable* $f\in {WFHom}_{R}({R}_{R},{R}_{R})$, *there exist* ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$ *such that* $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$. *Then* *R* *is right artinian*.

*Proof* By Theorem 4.1, *R* is a right CF ring. It is well known that a right CF and right noetherian ring is right artinian. So we only need to prove that *R* is right noetherian.

*R*is not right noetherian, then there is a strictly ascending chain ${I}_{1}\subseteq \u0337{I}_{2}\subseteq \u0337{I}_{3}\cdots $ of right ideals of

*R*. Let

*f*be a fuzzy subset of $R\times R$ constructed by

By Example 2.8, $f\in {WFHom}_{R}({R}_{R},{R}_{R})$. *f* is extendable and not bounded. So there exist some ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$ such that $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$. According to the condition (1), there exist ${c}_{i}\in R$ such that ${F}_{i}(1,{c}_{i})>0$, $i=1,2,\dots $ . Then by the condition (4), ${F}_{i}(x,{c}_{i}x)\ge {F}_{i}(1,{c}_{i})>0$. So ${F}_{i}(x,y)=0$ or ${F}_{i}(x,y)\ge {F}_{i}(1,{c}_{i})$ for all $i=1,2,\dots ,n$. Thus, $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$ is bounded. This is a contradiction. So *R* is right noetherian. Thus, *R* is right artinian. □

**Remark 4.3** According to Theorem 3.2, Theorem 4.1 and Proposition 4.2, the CF conjecture is equivalent to saying that every extendable $f\in {WFHom}_{R}({R}_{R},{R}_{R})$ of a right CF ring *R* is bounded.

At last, we obtain a fuzzy characterization of QF rings.

**Theorem 4.4** *R* *is a QF ring if and only if*, *for every* $f\in {WFHom}_{R}({R}_{R},{R}_{R})$, *there exist* ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$ *such that* $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$.

*Proof* Suppose *R* is a QF ring. Then *R* is right artinian and the right *R*-module ${R}_{R}$ is injective. By Theorem 3.2, every $f\in {WFHom}_{R}({R}_{R},{R}_{R})$ is finite valued. So *f* is bounded. Since ${R}_{R}$ is injective, by Theorem 3.1, *f* is extendable. As QF rings are right CF rings, by Theorem 4.1, there exist ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$ such that $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$.

Conversely, if for every $f\in {WFHom}_{R}({R}_{R},{R}_{R})$, there exist ${F}_{1},\dots ,{F}_{n}\in {FHom}_{R}({R}_{R},{R}_{R})$, such that $f={F}_{1}\wedge {F}_{2}\wedge \cdots \wedge {F}_{n}$. By Proposition 4.2, *R* is right artinian. According to Definition 2.4, $f\le {F}_{1}$. Then by Theorem 3.1, ${R}_{R}$ is injective. So *R* is a QF ring. □

## Declarations

### Acknowledgements

The authors would like to thank the referees for their nice suggestions and comments. It is supported by NSFC (No. 11371089), NSF of Jiangsu Province (No. BK20130599), NSF of Anhui Province (No. 1408085MA04), the Project-sponsored by SRF for ROCS, SEM and Specialized Research Fund for the Doctoral Program of Higher Education (No. 20120092110020).

## Authors’ Affiliations

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