New approaches on some fuzzy algebraic structures

Abstract

Starting from the study of the papers [1 , 14], we have seen that we can change some concepts and give some new definitions on fuzzy algebraic structures. Our goal by introducing these definitions is to get the classical and well known ones when the fuzzy set is just the characteristic function. As a second purpose we were tempted to generalize as long as it is possible some results known in the classical set theory. We get many interesting results which concern fuzzy relations, fuzzy subgroups of a given group, fuzzy ideals of a ring. We also define ideal generated by an element in the commutative case, prime and maximal ideals. Some relationships have been made and many other still open and which will be of our interest in next studies.

Keywords

Fuzzy sets fuzzy subgroup fuzzy operation fuzzy equivalence relation fuzzy ideal fuzzy prime ideal fuzzy maximal ideal

1 Preliminaries

This section contains some definitions and properties related to fuzzy subgroups, perfect T-vague groups and fuzzy equivalence relations that will be needed later. Some properties of fuzzy maps and T-vague operations are also stated so that perfect T-vague groups could be set in this context.

Definition 1. Let I = [0, 1], a mapping T : I × I ⟶ I is a t-norm if,

for all a ∈ I, T (1, a) = a,

for all a, b ∈ I, T (a, b) = T (b, a),

for all a, b, c ∈ I, T (T (a, b) , c) = T (a, T (b, c)),

for all a, b, c, d ∈ I, T (a, b) ≤ T (c, d) if a ≤ c and b ≤ d.

Throughout the paper T will denote a given t-norm.

Definition 2. Let (G, ★) be a group, e its identity. A fuzzy subset μ of G is a fuzzy subgroup of G if and only if

μ (e) =1,

μ (a ★ b) ≥ min {μ (a) , μ (b)} , ∀ a, b ∈ G,

∀a ∈ G, μ (a) = μ (a^-1) .

It is said normal if in addition μ (a ★ b ★ a^-1) = μ (b) .

Definition 3. Let (G, ★) be a group. A fuzzy subset μ of G is a T-fuzzy subgroup of G if and only if $T (μ (a), μ (b)) \leq μ (a ★ b^{- 1}), \forall a, b \in G .$

Definition 4. A fuzzy binary relation or just a fuzzy relation on a set X is a mapping $R : X \times X ⟶ [0, 1] .$

Definition 5. A fuzzy relation $R$ on a set X is a fuzzy equivalence relation if and only if for all a, b, c of X the following properties are satisfied,

$R (a, a) = 1$ (Reflexivity)

$R (a, b) = R (b, a)$ (Symmetry)

$R (a, c) \geq \min {R (a, b)), R (b, c)}$ (Transitivity)

Example 6. Let μ be a normal fuzzy subgroup of a group (G, ∗). The relation R defined on G by, R (a, b) >0 if and only if μ (a ∗ b^-1) >0 is a fuzzy equivalence relation on G.

Definition 7. Let (R, + , ×) be a ring and 0_R its identity for +. A fuzzy subset μ of R is a fuzzy ideal of R if

μ (0_R) =1,

μ (a) = μ (- a) ,

μ (a + b) ≥ min {μ (a) , μ (b)} , ∀ a, b ∈ R,

μ (a × b) ≥ max {μ (a) , μ (b)} , ∀ a, b ∈ R .

It is said prime, if in addition ∀a, b ∈ R $\begin{matrix} (μ \neg = \underline{1} and μ (a \times b) > 0) \\ \Rightarrow μ (a) > 0 or μ (b) > 0 . \end{matrix}$ In the case where μ is prime with μ (1) =0, μ is called a proper prime ideal.

Example 8. It is clear that any non null constant fuzzy subset $\underline{α} \neg = \underline{1}$ is a prime fuzzy ideal of the ring (R, + , ×) .

Definition 9. Let (R, + , ×) be a ring and 0_R its identity for +. A fuzzy ideal of R is said maximal if for all fuzzy ideal ν of R the following implication holds, $μ \leq ν \Rightarrow ν = μ or ν = \underline{1},$ where $\underline{1}$ is the constant fuzzy subset defined by $\underline{1} (x) = 1, \forall x \in R .$

2 Main results

2.1 Fuzzy operations

Definition 10. A fuzzy operation on a set E is a mapping f : E × E × E ⟶ [0, 1] such that, for x, y ∈ E there exists a unique element z ∈ E and f (x, y, z) >0.

A fuzzy operation on a set E is said,

Commutative if ∀x, y, z ∈ E,

f (x, y, z) = f (y, x, z).

Associative if ∀x, y, z, a, b ∈ E, f (x, y, a) >0, f (y, z, b) >0 and f (a, z, α) = f (x, b, β)

imply α = β .

having an identity or a neutral element e ∈ E if

f (x, e, x) = f (e, x, x) , ∀ x ∈ E.

Proposition 11. If E has an identity e for f and f (x, e, x) >0, ∀ x ∈ E, then this identity is unique.

Proof. Suppose that e and e′ are two distinct neutral elements of a set E for a fuzzy operation f.

Since e is an identity then f (e′, e, e′) = f (e, e′, e′) >0.

Again since e′ is an identity then f (e, e′, e) = f (e′, e, e) >0.

We then get f (e, e′, e′) >0 and f (e′, e, e) >0 and by the definition of the fuzzy operation, necessary e = e′. □

Definition 12. Let f : E × E × E ⟶ [0, 1] be a fuzzy operation on a set E and e be an identity of f. An element x ∈ E admits x′ ∈ E as a symmetric element if f (x, x′, e) = f (x′, x, e).

Definition 13. Let f : E × E × E ⟶ [0, 1] be a fuzzy operation on a set E. An element a ∈ E is left (resp. right) regular or cancelable if for any elements x, y, z ∈ E, the equality f (a, x, z) = f (a, y, z) (resp. f (x, a, z) = f (y, a, z)) implies x = y. It is regular or cancelable if it is left and right regular.

Proposition 14. If f a fuzzy operation on E has an identity e, if for all x, y, z ∈ E we have f (x, y, e) = f (x, z, e) >0 and x is left regular then x has at most one symmetric element.

Proof. Suppose that an element x has two symmetric elements x′ and x″. Since f (x, x′, e) = f (x, x″, e) >0 and since x is left regular we get x′ = x″. □

2.2 Fuzzy subgroups

Proposition 15. Let (G, ★) be a group and μ a fuzzy subgroup of G. If there exists in G an element a such that μ (a) = θ ∈ I then μ (e) ≥ θ.

Proof. We have μ (a) = μ (a^-1), so μ (e) = μ (a ★ a^-1) ≥ min {μ (a) , μ (a^-1)} = μ (a) ≥ θ. □

Corollary 16. μ (e) =1 if and only if $\exists a \in G | μ (a) = 1 .$

Proof. The proof is trivial. □

Remark 1. It is easy to see that if {μ_i} _i≥1 is an infinite family of fuzzy subgroups and a is a member of each of them, a can be an element which is not a member of their intersection. It suffices to take an element a such that $μ_{i} (a) = \frac{1}{i}$ in which case we have inf {μ_i (a)} =0 and then a is not a member of the intersection.

Proposition 17. The intersection of a finite or infinite family of fuzzy subgroups is a fuzzy subgroup.

Proof. Let {μ_i} be a family (finite or infinite) of subgroups of a group G. and let μ = ⋀ _iμ_i

$μ (0) = inf_{i} {μ_{i} (0)} = inf_{i} {1} = 1$ ,

$\forall a \in G, μ (a^{- 1}) = inf_{i} {μ_{i} (a^{- 1})} = inf_{i} {μ_{i} (a)} = μ (a)$ since for all i one has μ_i (a^-1) = μ_i (a),

∀a, b ∈ G, $μ (a . b) = inf_{i} {μ_{i} (a . b)} \geq inf_{i} {inf {μ_{i} (a), μ_{i} (b)}} .$ Indeed in the case where the family is infinite we can permute ⋀_i and the function inf(. ,.) on two elements. The last expression is then equal to $inf {inf_{i} {μ_{i} (a), μ_{i} (b)}} = inf {μ (a), μ (b)} .$ Moreexactly if we set $x = inf_{i} {inf {μ_{i} (a), μ_{i} (b)}}$ and $y = inf {inf_{i} (μ_{i} (a)), inf_{i} (μ_{i} (b))}$ , it is easy to prove that x ≤ y and y ≤ x. □

Definition 18. Let G be a group and μ be a fuzzy subgroup of G. μ is called a fuzzy subgroup generated by an element a ∈ G if we have,

μ (a) > 0,

∀x ∈ G, μ (x) > 0 implies x = aⁿ for some $n \in ℕ^{*} .$

Example 19. Let G be the group $ℤ / 8 ℤ$ and a = 2 then the mapping μ given by, is a fuzzy subgroup generated by 2.

Example 20. Let G be a group then the mapping μ given by,

μ (e) =1 and μ (x) =0 is is the fuzzy subgroup generated by e.

Although the T-fuzzy subgroup structures are not the object of our study, however we are tempted to introduce the following example.

Example 21. As an example of T- fuzzy subgroup let us consider the subgroup of example 19 and let us set T (a, b) = a . b. T is a t-norm. It is easy to see that the chosen fuzzy subgroup μ verifies $T (a, b) \leq μ (a - b), \forall a, b \in ℤ / 8 ℤ$ and then it is a T-fuzzy subgroup of $ℤ / 8 ℤ$ .

Definition 22. Let G be a group and μ a fuzzy subgroup of G. μ is called a cyclic fuzzy subgroup of G if it is generated by an element a of the group G.

Lemma 23. Let G be a finite cyclic group generated by a and μ be a fuzzy subgroup of G. If μ (a) = θ ¬ =0 then μ (x) ≥ θ , ∀ x ∈ G \ {e}.

Proof. G is cyclic of order n then G = {e, a, a², ⋯ , a^n-1}. By induction on k we will prove that μ (a^k) ≥ μ (a) , ∀ k ≤ n - 1.

For k = 0, μ (e) =1 ≥ μ (a).

For k = 2, μ (a²) = μ (a . a) ≥ min {μ (a) , μ (a)} = μ (a).

Suppose that for k ≤ k - 2 we have μ (a^k) ≥ μ (a), so μ (a^k+1) = μ (a^k . a) ≥ min {μ (a^k) , μ (a)} = μ (a) and the result follows. □

Example 24. Consider the group cyclic group of order 8, C₈ = {e, a, a², ⋯ , a⁷} and let μ be a fuzzy subgroup of C₈. If μ (a) = θ ¬ =0 then μ has one of the following structures.

$1 \geq α \geq θ \geq 0 .$

$1 \geq α \geq β \geq θ \geq 0 .$

Proof. By the lemma 23, we have μ (a^k) ≥ μ (a) , ∀ k ≥ 0. From the condition μ (x) = μ (x^-1), we have μ (a⁷) = μ (a) , μ (a⁶) = μ (a²) , μ (a⁵) = μ (a³).

On the other hand μ (a) = μ (a⁷) = μ (a²a⁵) ≥ min {μ (a²) , μ (a⁵)} and μ (a) = μ (a⁷) = μ (a³a⁴) ≥ min {μ (a³) , μ (a⁴)}.

If μ (a²) ≤ μ (a⁵) = μ (a³), by the first inequality we have μ (a) = μ (a⁷) = μ (a²) = μ (a⁶).

If μ (a³) ≤ μ (a⁴) from the second equality we get μ (a) ≥ μ (a³) and then μ (a) = μ (a⁷) = μ (a²) = μ (a⁶) = μ (a³) = μ (a⁵) .

If μ (a³) ≥ μ (a⁴) then by a similarly manner we get μ (a) = μ (a⁷) = μ (a²) = μ (a⁶) = μ (a⁴) . But μ (a) = μ (a⁶) ≥ μ (a³) then μ (a) = μ (a⁷) = μ (a²) = μ (a⁶) = μ (a⁴) = μ (a³) = μ (a⁵) .

The first table follows.

If μ (a⁶) = μ (a²) ≥ μ (a⁵) = μ (a³), by the first inequality we have μ (a) = μ (a⁷) = μ (a⁵) = μ (a³).

If μ (a³) ≤ μ (a⁴) the only information we get is that μ (a⁴) ≥ μ (a²) ≥ μ (a). The second table follows.

If μ (a³) ≥ μ (a⁴) then by a similarly manner we get μ (a) = μ (a⁷) = μ (a³) = μ (a⁴). But μ (a⁴) ≥ μ (a²) = μ (a⁶) thenμ (a) = μ (a⁷) = μ (a²) = μ (a⁶) = μ (a⁴) = μ (a³) = μ (a⁵) . We get again the first table. □

Example 25. In the case of C₉ we have only one possible table of a fuzzy subgroup of C₉, which has the form, where 1 ≥ α ≥ θ ≥ 0.

Proof. By a same considerations as in the above example it is easy to remark as 8 = 2 +6 = 3 +5 = 4 +4 to deduce that $\begin{matrix} μ (a) & = & μ (a^{8}) \geq (μ (a^{2}) \land μ (a^{6})), \\ μ (a) & = & μ (a^{8}) \geq (μ (a^{3}) \land μ (a^{5})), \\ μ (a) & = & μ (a^{8}) \geq μ (a^{4}) . \end{matrix}$

But μ (a⁴) ≥ μ (a) then μ (a) = μ (a⁴). In the other hand μ (a) = μ (a⁴) ≥ μ (a²) and μ (a⁷) = μ (a²) ≥ μ (a) then we finally get $μ (a) = μ (a^{2}) = μ (a^{4}) = μ (a^{5}) = μ (a^{7}) = μ (a^{8})$ and $μ (a^{3}) = μ (a^{6}) \geq μ (a^{2}) .$

The result follows. □

At this stage we can make the following.

Conjecture 26. We conjecture that if G is a finite cyclic group generated by a there are a few number of different structures of fuzzy subgroups of G. □

Proposition 27. Let (G, ★) be a group, e its neutral element and μ a fuzzy subset of G such that μ (e) =1. μ is a T-fuzzy subgroup of G if and only if ∀a, b ∈ G the following properties hold

μ (a) = μ (a^-1),

T (μ (a) , μ (b)) ≤ μ (a ★ b).

Proof. The proof is trivial. □

Proposition 28. Let (G, ★) be a group, e its neutral element and μ a fuzzy subgroup of G such that μ (e) =1. Then the core H of μ (i.e. the set of elements a of G such that μ (a) =1) is a subgroup of G in the usual meaning.

Proof.

Since μ (e) =1 then e ∈ H.

Let a, b ∈ H. Since μ (a ★ b) ≥ min {μ (a) , μ (b)} = min {1, 1} =1 and μ (a ★ b) ≤1, it follows that μ (a ★ b) =1 and then a ★ b) ∈ H.

For any a ∈ H, we have μ (a) =1. But μ (a) = μ (a^-1) and then a^-1 ∈ H . □

Proposition 29. Let G be an abelian finite group of order n, for any a ∈ G if μ : G ⟶ [0, 1] is defined by $μ (a) = \frac{ord (a)}{n}$ , then μ is a fuzzy subgroup of G.

Indeed if an element a ∈ G is such μ (a) =1, then G is cyclic and a is a generator for G.

G is of prime order if and only if ∀a ¬ = e, μ (a) =1 .

Proof. Since G is abelian then ord (ab) = l . c . m (ord (a) , ord (b)). From the inequality $\begin{matrix} l . c . m (ord (a), ord (b)) & \geq & \max {ord (a), ord (b)} \\ \geq & \min {ord (a), ord (b)} \end{matrix}$ we deduce that μ is a fuzzy subgroup of G.

The rest of the proposition is trivial. □

Remark 2. For the above mapping μ (a) measures the Thickness of the subgroup generated by a with respect to the group G. But we know that this can be deduced just from the orders, the proposition can then appears superfluous. In this case we can take it as an example of fuzzy subgroup.

Example 30. As a second example of fuzzy subgroup, let us denote by r_a ∈ {0, 1, 2} the remainder of the euclidian division (long division) of a by 3 in $ℤ$ . For any element $a \in ℤ,$ we set, $μ (a) = {\begin{matrix} 1 & if r_{a} = 0 \\ 1 / 3 & if r_{a} \neg = 0 \end{matrix}$

It is easy to verify that μ is a fuzzy subgroup of $ℤ$ .

Proposition 31. Let G be an abelian group of order p^k, k ≥ 3 and p prime. If any element of G is of order less than p^k-2, T is defined by $\begin{matrix} T : & [0, 1] \times [0, 1] & ⟶ & [0, 1] \\ (x, y) & ⟼ & x . y \end{matrix}$ and μ is a fuzzy subgroup defined on G as in the Proposition 29, then μ is a T-fuzzy subgroup of G.

Proof. Let a, b ∈ G such that ♯a = p^m, ♯ b = pⁿ then $T (μ (a), μ (b)) = μ (a) . μ (b) = \frac{♯ a}{p^{k}} . \frac{♯ b}{p^{k}} = \frac{p^{m}}{p^{k}} . \frac{p^{n}}{p^{k}}$ . □

2.3 Fuzzy ideals

Proposition 32. If μ is a fuzzy ideal of R then for all θ ∈ [0, 1] the set I = {x ∈ R | μ (x) ≥ θ} is an ideal of the ring (R, + , ×).

If in addition ∀a, b ∈ R, a × b ∈ I implies $μ (a \times b) = μ (a) . μ (b)$ then I is prime.

Proof.

μ (0_R) =1 ≥ θ then 0_R ∈ I.

Let a, b ∈ I then μ (a + b) ≥ min {μ (a) , μ (b)} ≥ θ so a + b ∈ I.

By definition ∀a ∈ I one has μ (a) = μ (- a) so if a ∈ I so is -a.

Let a ∈ I and b ∈ R then μ (a × b) ≥ max {μ (a) , μ (b)) ≥ μ (a) ≥ θ and then (a × b) ∈ I .

For the second part of the proposition, suppose that a × b ∈ I and b ∉ I then from θ ≤ μ (a × b) = μ (a) . μ (b). At this step there are two cases. If θ = 0 then since μ (a) ≥0, a is an element of I. If θ ¬ =0, μ (b) ¬ =0 otherwise μ (a × b) =0 and a × b ∉ I. From θ ≤ μ (a × b) = μ (a) . μ (b) we can deduce that $θ = \frac{θ}{1} \leq \frac{θ}{μ (b)} \leq μ (a) and a \in I . □$

Let us give an example of fuzzy prime ideal which is not maximal.

Example 33. The function μ defined on $ℤ / 8 ℤ$ in the Example 19 is a fuzzy prime ideal. To see that μ (a . b) ≥ max {μ (a) , μ (b)} and μ (a . b)> implies μ (a) >0 or μ (b) > 0, it suffices to draw the multiplication table and affect each cell with the value of element obtained in the corresponding cell by the product.

If we change the value of μ (4) =0.6 in the example by μ (4) =0.7, the obtained function is still a fuzzy ideal great than the precedent and different from the constant fuzzy ideal $\underline{1}$ . It follows that the fuzzy ideal is not maximal.

Proposition 34. Let (R, + , ×) be a ring and 0_R its identity for +, the intersection α ∧ μ of two fuzzy ideals α and μ of R is also a fuzzy ideal of R.

Proof. Let α, μ be two fuzzy ideals of R and let ν = α ∧ μ.

ν (0) = α (0) ∧ μ (0) = min {α (0) , μ (0)} = min {1, 1} =1 .

Let a ∈ R, then ν (- a) = α (- a) ∧ μ (- a) = min {α (- a) , μ (- a)} = min {α (a) , μ (a)} = ν (a).

Let a, b ∈ R, then $\begin{matrix} ν (a + b) & = & α (a + b) \land μ (a + b) \\ = & \min {α (a + b), μ (a + b)} \\ \geq & \min {\min {α (a), α (b)}, \min {μ (a), μ (b)}} \\ = & \min {\min {α (a), μ (a)}, \min {α (b), μ (b)}} \\ = & \min {α (a) \land μ (a), α (b) \land μ (b)} \\ = & \min {ν (a), ν (b)} . \end{matrix}$

Let a, b ∈ R, then $\begin{matrix} ν (a \times b) & = & α (a \times b) \land μ (a \times b) \\ = & \min {α (a \times b), μ (a \times b)} \\ \geq & \min {\max {α (a), α (b)}, \max {μ (a), μ (b)}} \\ \geq & \max {\min {α (a), μ (a)}, \min {α (b), μ (b)}} \\ = & \max {ν (a), ν (b)} . \end{matrix}$ It follows that ν is an ideal of R. □

Proposition 35. Let μ be a fuzzy ideal of a ring R and let $\begin{matrix} β \in \sqrt{μ} & = & {α : R ⟶ [0, 1] | \exists n \in ℕ \\ and α^{n} = μ}, then, \end{matrix}$

β is an ideal,

if μ is prime then β is also prime,

if μ is maximal so is β.

Proof. Since $μ \in \sqrt{μ}$ , the set $\sqrt{μ}$ is not empty.

Let $β \in \sqrt{μ}$ then there is a natural n such that βⁿ = μ and

βⁿ (0) = μ (0) =1 and the only root of the equation xⁿ = 1 in [0, 1] is 1, soβ (0) =1.

For a ∈ R one has βⁿ (- a) = μ (- a) = μ (a) = βⁿ (a) so $\begin{matrix} β^{n} (- a) - β^{n} (a) = 0 \Leftrightarrow \\ (β (- a) - β (a)) (β^{n - 1} (a) + β^{n - 2} (a) β^{1} (- a) \\ + \dots + β (a) β^{n - 2} (- a) + β^{n - 1} (- a)) = 0 . \end{matrix}$

If $\begin{matrix} (β^{n - 1} (a) + β^{n - 2} (a) β^{1} (- a) + \dots \\ + β (a) β^{n - 2} (- a) + β^{n - 1} (- a)) = 0, \end{matrix}$ then all the terms are null and then β^n-1 (a) = β^n-1 (- a)) =0 and therefore β (a) = β (- a)) =0.

If $\begin{matrix} (β^{n - 1} (a) + β^{n - 2} (a) β^{1} (- a) + \dots \\ + β (a) β^{n - 2} (- a) + β^{n - 1} (- a)) \neg = 0, \end{matrix}$ then (β (- a) - β (a)) =0 and consequently β (- a) = β (a)).

βⁿ (a + b) = μ (a + b) ≥ min {μ (a) , μ (b)} = min {βⁿ (a) , βⁿ (b)}, from which we can deduce that βⁿ (a + b) ≥ min {βⁿ (a) , βⁿ (b)} and since x ⟼ xⁿ is an increasing function from [0, 1] to [0, 1]. Necessary β (a + b) ≥ min {β (a) , β (b)} .

βⁿ (a × b) = μ (a × b) ≥ max {μ (a) , μ (b)} = max {βⁿ (a) , βⁿ (b)} .

By the same arguments as in c), we deduce that β (a × b) ≥ max {β (a) , β (b)} .

It follows that β is an ideal of R.

Suppose that μ is a prime ideal and suppose that β (a × b) >0 then βⁿ (a × b) >0 and therefore μ (a × b) >0. Since μ is prime we get μ (a) >0 or μ (b) >0 and then βⁿ (a) >0 or βⁿ (b) >0. As remarked above, we have necessary β (a) >0 or β (b) >0.

Now suppose μ maximal and let ν be an ideal such that β ≤ ν then for all $n \in ℕ, (β)^{n} \leq (ν)^{n} \Leftrightarrow μ \leq (ν)^{n}$ and since (ν) ⁿ is also an ideal then by the maximality we have either μ = (ν) ⁿ or $(ν)^{n} = \underline{1}$ . The consequences are then, either β = ν or ν = 1. □

Proposition 36. Let μ be a fuzzy ideal on a ring R and a be an element of the center of R. The fuzzy set, μ_a : R ⟶ [0, 1] defined by μ_a (x) = μ (a . x) is a fuzzy ideal of R. Indeed,

if μ is prime then μ_a is prime.

if μ is maximal and the mapping b ⟼ ab, from R to R is surjective the ideal μ_a is also maximal.

Proof.

Let us prove first that μ_a is a fuzzy ideal.

μ_a (0) = μ (a . 0) = μ (0) =1 .

μ_a (- b) = μ (a (- b)) = μ (- (ab)) = μ (ab) = μ_a (b) .

μ_a (b + c) = μ (a (b + c)) = μ ((ab) + (ac)) ≥ min {μ (ab) , μ (ac)} = min {μ_a (b) , μ_a (c)}.

μ_a (bc) = μ (a (bc)) = μ ((ab) c) = μ (b (ac)since a is in the center of R. If μ (ab) ≥ μ (ac) then μ_a (bc) = μ (a (bc)) = μ ((ab) c) ≥ max {μ (ab) , μ (ac)} since μ ((ab) c) ≥ max {μ (ab) , μ (c)} and then it is always great then μ (ab).

Let b, c ∈ R such that μ_a (bc) >0 then μ (a (bc)) >0 but μ (a (bc)) = μ ((ab) c) >0 and since μ is prime then either μ (ab) >0 or μ (c) >0. μ (ac) ≥ max {μ (a) , μ (c)} >0 so we get either μ_a (b) = μ (ab) >0 or μ_a (c) = μ (ac) >0 and μ_a is a prime ideal.

Suppose now that α is an ideal such that μ_a ≤ α. Since μ_a (b) = μ (ab) ≥ μ (b). From the condition μ_a ≤ α, we deduce that μ (b) ≤ μ (ab) = μ_a (b) ≤ α (b) , ∀ b ∈ R. But μ is maximal so μ = α or $α = \underline{1}$ . And as μ (b) ≤ μ_a (b) ≤ α (b) , ∀ b ∈ R it follows that μ_a = α or $α = \underline{1}$ and μ_a is maximal. □

Example 37. As an example of fuzzy ideal and as in Example 30, let us denote by r_a ∈ {0, 1, 2} the remainder of the euclidian division (long division) of a by 3 in $ℤ$ . For any element $a \in ℤ,$ we set, $μ (a) = {\begin{matrix} 1 & if r_{a} = 0 \\ 1 / 3 & if r_{a} \neg = 0 \end{matrix}$

We have seen that μ is a fuzzy subgroup of $ℤ$ . Indeed for all $a, b \in ℤ$ , from the properties of the remainder in the Euclidian division and by performing all the possible values of r_a and r_b, we have μ (ab) = r_ar_b ≥ max {r_a, r_b} and then μ is a fuzzy ideal of the ring $(ℤ, +, \times)$ .

2.4 Fuzzy subgroups and fuzzy relations

Proposition 38. Let G be a group and μ be a fuzzy subgroup of G. The fuzzy relation $R$ defined on G by, $R (a, b) = μ ({ab}^{- 1}) \forall a, b \in G$ is an equivalence fuzzy relation on G.

Proof.

∀a ∈ G, we have $R (a, a) = μ ({aa}^{- 1}) = μ (e) = 1$ then $R (a, a) = 1$ and so $R$ is reflexive.

∀a, b ∈ G, if $R (a, b) = μ ({ab}^{- 1}) = μ (({ab}^{- 1})^{- 1}) = μ ({ba}^{- 1}) = R (b, a)$ . It follows that $R$ issymmetric.

Let a, b, c ∈ G we have ac^-1 = (ab^-1) (bc^-1) so $R (a, c) = μ ({ac}^{- 1}) = μ (({ab}^{- 1}) ({bc}^{- 1})) \geq \min {μ ({ab}^{- 1}), μ ({bc}^{- 1})} = \min {R (a, b), R (b, c)}$ and the relation is then transitive.

□

Definition 39. The relation defined in the Proposition 38, will be called the fuzzy equivalence relation on the group G modulo the fuzzy subgroup μ and will be denoted by $R_{μ}$ .

Proposition 40. Conversely suppose that $R$ is a fuzzy equivalence relation on a group G. The mapping μ : X ⟶ [0, 1] defined implicitly by $μ ({ab}^{- 1}) = R (a, b)$ is a fuzzy subgroup of G.

Proof. Let a, b ∈ G we have $μ (ab) = μ (a (b^{- 1})^{- 1}) = R (a, b^{- 1}) \geq \min {R (a, e), R (e, b^{- 1})} = \min {μ (a), μ (b^{- 1})} .$

On the other hand, since $μ (b^{- 1}) = μ ({eb}^{- 1}) = R (e, b) = R (b, e) = μ ({be}^{- 1}) = μ (b),$ then $μ (ab) \geq \min {R (a, e), R (e, b^{- 1})} = \min {μ (a), μ (b)} .$ The result follows. □

Let us return to the general notion of fuzzy relations not necessary related to fuzzy subgroups.

Remark 3. If $R$ is fuzzy relation on a set X then the number $R (x, y) \in [0, 1]$ can be interpreted as the degree of relationship between x and y. So we will say that x and y are in real relationship if $R (x, y) > 0$ .

Definition 41. Let a be an element of a set X and $R$ be a fuzzy equivalence relation on X. We denote by $\bar{a}$ the set of all elements of X which are in real relation with a and call it the real equivalence class of a.

Example 42. Let X = {a, b, c} and $R$ be the relation defined by the following table,

It is easy to see that $R$ is a fuzzy equivalence relation and for example $\bar{a} = {a, b, c}, \bar{d} = a = {d} .$

Proposition 43. If a, b are two elements of a set X endowed with a fuzzy equivalence relation $R$ such that $R (a, b) \neg = 0$ then any element x ∈ X such that $R (a, x) \neg = 0$ verifies also $R (x, b) \neg = 0$ .

Proof. Suppose that there exists an element x ∈ X such that $R (a, x) \neg = 0$ and $R (x, b) = 0$ . Since $R$ istransitive then $R (b, x) \geq \min {R (a, b), R (a, x)} \neg = 0$ ,contradiction. □

Proposition 44. If a, b, c are distinct elements of a set X endowed with a fuzzy equivalence relation $R$ . If $R (a, b) \geq R (a, c) \geq R (b, c)$ then necessary $R (a, c) = R (b, c)$ .

Proof. Suppose that a, b, c are such $R (a, c) > R (b, c)$ . By the transitivity one has $\begin{matrix} R (b, c) & \geq & \min {R (a, b), R (a, c)} \\ = & R (a, c) > R (b, c), \end{matrix}$ contradiction. □

Remark 4. From the above proposition if a, b and c are such that $R (a, b) \geq R (a, c) \geq R (b, c)$ , then either the three values are equal or $R (a, c) = R (b, c)$ .

2.5 Fuzzy ideals and fuzzy relations

Proposition 45. Let (R, + , ×) be a ring and μ be a fuzzy ideal of R. The fuzzy relation $R$ defined on R by, $R (a, b) = μ (a - b) \forall a, b \in R$ is an equivalence fuzzy relation on R.

Proof.

∀a ∈ R, we have $R (a, a) = μ (a - a) = μ (0) = 1$ then $R (a, a) = 1$ and $R$ is reflexive.

∀a, b ∈ R,, we always have $R (a, b) = μ (a - b) = μ (- (a - b)) = μ (b - a) = R (b, a)$ and then $R$ is symmetric.

Let a, b, c ∈ R. From a - c = (a - b) + (b - c), we have $R (a, c) = μ (a - c) = μ ((a - b) + (b - c)) \geq \min {μ (a - b), μ (b - c)} = \min {R (a, b), R (b, c)}$ and the relation is transitive. □

Proposition 46. Conversely suppose that $R$ is a fuzzy equivalence relation on a ring (R, + , ×) satisfying $R (ab, 0) \geq \max {R (a, 0), R (b, 0)}$ . The mapping μ : R ⟶ [0, 1] defined implicitly by $μ (a - b) = R (a, b)$ is a fuzzy ideal of R.

Proof.

$μ (0) = μ (a - a) = R (a, a) = 1 .$

$μ (- a) = μ (0 - a) = R (0, a) = R (a, 0) = μ (a - 0) = μ (a) .$

$μ (a + b) = μ (a - (- b)) = R (a, (- b)) \geq \min {R (a, 0), R (0, (- b))} = \min {μ (a), μ (b)} .$

$μ (ab) = μ (ab - 0) = R (ab, 0) \geq \max {R (a, 0), R (0, (- b))} = \max {μ (a), μ (b)} .$ □

Proposition 47. Let a be an element of a commutative ring (R, + ,.) and μ : R ⟶ [0, 1] be a mapping satisfying the following conditions.

$θ \geq μ (a) \geq \frac{θ}{2}$ ,

μ (0) =1,

$μ (x) = {\begin{matrix} μ (a) & If x = a . y for some y \in R \\ θ - μ (a) & otherwise . \end{matrix}$ Then μ is an ideal.

Proof.

If x = a . y then -x = a . (- y) and so μ (- x) = μ (a).

If x ¬ = a . y, ∀ y ∈ R then -x ¬ = a . z, ∀ z ∈ R and consequently $μ (- x) = 1 - μ (a) = μ (x) .$

First, from the first condition we remark that μ (a) ≥ θ - μ (a). So for x, y ∈ R one has $μ (x + y) = {\begin{matrix} μ (a) & If x + y = a . t \\ for some t \in R \\ θ - μ (a) & otherwise . \end{matrix}$ When x + y = a . t for some t ∈ R, we havethree cases, (that is (x = at, y ¬ = az) , (x ¬ = at, y = az) and (x = at and y = az). But in all these cases, we always have μ (x + y) = μ (a) ≥ min {μ (x) , μ (y)}.

If x + y = ¬ = a . t it follows that at least one of the elements x and y is not a multiple of a and then min {μ (x) , μ (y)} = θ - μ (a) ≤ θ - μ (a) = μ (x + y) .

Let x, y ∈ R, then $μ (x . y) = {\begin{matrix} μ (a) & If x . y = a . t for some t \in R \\ θ - μ (a) & otherwise . \end{matrix}$ In the case where x . y = a . t, it is trivial that for all the values of μ (x) and μ (y), we always have μ (x . y) = μ (a) ≥ max {μ (x) , μ (y)}.

If x . y ¬ = a . t neither x nor y is not a multiple of a and then $\begin{matrix} \max {μ (x), μ (y)} \\ = θ - μ (a) \leq θ - μ (a) = μ (x . y) . □ \end{matrix}$

Definition 48. Let a be an element of a ring R. The ideal generated by a is the least ideal μ : R ⟶ [0, 1] such that $\begin{matrix} μ (a) > 0, and μ (x) = μ (a) \\ if and only if \\ x = ay for some y \in R . \end{matrix}$

Remark 5. For any element a in a commutative ring R, we can always define an ideal in relationship with a it suffices to take the ideal μ such that $μ (x) = {\begin{matrix} 1 & If & x = 0 \\ μ (a) & If & x = a . t for some t \in R \ {0} \\ 0 & otherwise . \end{matrix}$

Indeed we have

either x = at and then -x = a (- t) or x ¬ = at and -x ¬ = at′ and all the two cases one has μ (x) = μ (- x).

For x, y ∈ R, $μ (x + y) = {\begin{matrix} 1 & if x + y = 0 \\ μ (a) & if x + y = at \\ 0 & otherwise \end{matrix}$

More exactly in the case where x + y = 0 we always have μ (x + y) =1 ≥ min {μ (x) , μ (y)}.

If x + y = at then μ (x + y) = μ (a) ≥ min {μ (x) , μ (y)} since the elements x and y can not be simultaneously null in which case μ (x + y can be less than 1 = min {μ (x) , μ (y)}.

If x + y ¬ = at, t ∈ R then μ (x + y) =0. Also in this case at least one of the elements x or y can not be a multiple of a so then min {μ (x) , μ (y)} =0 and the result follows.

Let x, y ∈ R then $μ (x . y) = {\begin{matrix} 1 & if x . y = 0 \\ μ (a) & if x . y = at \\ 0 & otherwise \end{matrix}$

If the first case it is clear that if x . y = 0 then μ (x . y) =1 ≥ max {μ (x) , μ (y)} (It does not depend on R is an integral domain or not).

If x . y = at then neither x nor y is null thenμ (x . y) = μ (a) ≥ max {0, μ (a)} ≥ max {μ (x) , μ (y)} .

If x . y ¬ = at, ∀ t ∈ R then neither x = at′ nor y = at″ so μ (x) =0 = μ (y) and μ (x . y) =0 ≥ max {μ (x) , μ (y)} =0.

Definition 49. Let μ, β be two ideals of a ring R. The least ideal α of R such that μ ∨ β ≤ α is called the sum of the ideals μ and β and will be denoted by μ + β.

Example 50. Let μ, ν be two fuzzy ideals of a ring R generated by a and b respectively. the sum μ + ν is the least ideal α satisfying max {μ (x) , ν (x)} ≤ α (x) , ∀ x ∈ R . It is easy to see that α verifies,

α (0) ≥ max {μ (0) , ν (0)} = max {1, 1} =1 so α (0) =1,

α (- x) ≥ max {μ (- x) , ν (- x)} = max {μ (x) , ν (x)} = α (x) and α (x) ≥ max {μ (x) , ν (x)} = max {μ (- x) , ν (- x)} = α (- x), so α (- x) = α (x),

α (x . y) ≥ max {μ (x . y) , ν (x . y)} ≥ max {max {μ (x) , μ (y)} , max {ν (x) , ν (y)}} = max {μ (x) , μ (y) , ν (x) , ν (y)} = max {max {μ (x) , ν (x)} , max {μ (y) , ν (y)}},

α (x + y) ≥ max {μ (x + y) , ν (x + y)} ≥ max {min {μ (x) , μ (y)} , min {ν (x) , ν (y)}},

Question 51. Is any maximal fuzzy ideal μ of a ring (R, + , ×) a prime fuzzy ideal?

Footnotes

Acknowledgments

The author acknowledges with thanks to the Deanship of Scientific Research, Majmaah University, KSA, the providing supports in the completion of this research work under the grant N°38/1/2.

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