J. Comp. & Math. Sci. Vol.2 (3), 573-580 (2011)
R-multiples of Fractionary Fuzzy Ideals of Integral Domains *
SOURIAR SEBASTIAN and GEORGE MATHEW
*
Department of Mathematics, St. Albert’s College Ernakulam, Kochi- 682018, Kerala India Department of Mathematics, B.C.M College, Kottayam-686001, Kerala, India ABSTRACT
The concept of fuzzy modules was introduced by Negotia and Ralescue9 and has been further developed by Golan4 and Bambri and Kumar1-2. In this paper we give a definition for valuation rings in terms of fuzzy modules of the quotient field. Certain fuzzy modules of the quotient field, called fractionary fuzzy ideals, was introduced by Malik and Mordeson in their paper8 and another definition for valuation rings in terms of fractionary fuzzy ideals was given by Kim and Park5. We prove the equivalence of these definitions. We also study the properties of multiples of fractionary fuzzy ideals of an integral domain especially that of a Dedekind domain. Keywords: Valuation rings, Dedekind domains, localization, fuzzy ideals, fuzzy submodules, fractionary fuzzy ideals.
1. INTRODUCTION Zadeh18 defined a fuzzy subset µ of a given universal set X as a function µ: X ⟶[0,1]. Rosenfeld10 used this idea to introduce the notion of fuzzy groupoids and fuzzy groups and derived their elementary properties. Liu6-7 defined and studied fuzzy ideals in groups and rings. The concept of fuzzy modules was introduced by Negotia and Ralescue9 and has been further developed by Golan4 and Bambri and Kumar1-2. Some works done in this area by the authors are reported in11-17.
In this paper we give a definition for valuation rings in terms of fuzzy modules of the quotient field. Certain fuzzy modules of the quotient field, called fractionary fuzzy ideals, was introduced by Malik and Mordeson in their paper8 and another definition for valuation rings in terms of fractionary fuzzy ideals was given by Kim and Park5. We prove the equivalence of these definitions. We also study the properties of multiples of fractionary fuzzy ideals of an integral domain especially that of a Dedekind domain.
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2. BASIC CONCEPTS
3. R-MULTIPLES OF FUZZY MODULES
Throughout this paper, R denotes a commutative ring with unit element, unless otherwise stated. Algebraic terms and notations used here are either standard or as in3 and19. Given a universal set X, a function μ :X ⟶ [0, 1] is called a fuzzy subset of X. For a fuzzy subset µ of R and t ∈ [0,1] the set µt = {x ∈ R : µ(x) ≥ t} is called the level set of µ corresponding to t. If µ and ν are fuzzy subsets of X, we write µ ⊆ ν if µ(x) ≤ ν(x) ∀ x ∈ X. Recall that if R is a ring and M is an R- module, then a fuzzy subset µ : M ⟶ [0, 1] is called a fuzzy R- submodule of M if (i) µ(x + y) ≥ µ(x) ∧ µ(y) ∀ x, y ∈ M (ii) µ(rx) ≥ µ(x) ∀ x ∈ M and r ∈ R and (iii) µ(0) = 1. We denote the set { x ∈ R : µ(x) = µ(0) } by µ∗. For fuzzy subsets µ and ν of R, the fuzzy subsets µ ∪ ν, µ ∩ ν are defined as follows:
3.1. Definition. Let M be an R- module and μ be a fuzzy R- sub module of M. If r ∈ R, then the function rμ : M ⟶ [0, 1] defined by (rμ)(x) = μ(rx) is called an R-multiple of µ.
(µ ∪ ν) (x) = µ(x) ∨ ν(x) and (µ ∩ ν) (x) = µ(x) ∧ ν(x) ∀ x∈ R. Here ∨ and ∧ denote the max (or sup) and min (or inf) operators respectively. An integral domain R is said to be a valuation ring if for any two ideals A and B of R, we have either A ⊆ B or B ⊆ A. That is, if ‘⊆’ is a total order on the set of all ideals of R. 2.1. Proposition[19]. For an integral domain R the following are equivalent (i) R is a valuation ring (ii) If a, b ∈ R, then either (a) ⊆ (b) or (b) ⊆ (a) (iii) If K is the quotient field of R, then for every x (≠0) ∈ K, either x ∈ R or x -1 ∈ R∎
3.2. Proposition. Let M be an R- module and μ and ν be fuzzy R- submodules of M. Then (i)
rμ is a fuzzy R- submodule for every r∈R
(ii)
If r ∈ R, then μ ⊆ ν ⇔ rμ ⊆ rν.
(iii) If r, s ∈ R, then r(sμ ) = (rs) μ. In particular r(rμ ) = r2μ . (iv)
If r ∈ R, then (rμ) ∩ (rν) = r (μ ∩ ν) and r( ∩ μi) = ∩ (rμi) i
i
Proof. (i) Let x, y ∈ M and r ∈ R. Since μ is a fuzzy R- sub module of M, using the definition of rμ, (rμ)(x + y) = μ(r(x + y)) = μ(rx + ry) ≥ μ(rx) ∧ μ(ry) = (rμ)(x) ∧ (rμ)(y) (rμ)(sx) = μ(rsx) ≥ μ(rx) = (rμ)(x) (rμ)(0) = μ (0) = 1 Therefore rμ is a fuzzy R- sub module of M for every r ∈ R. (ii) Suppose µ ⊆ ν. Let r ∈ R. Then for x ∈ M, (rµ)(x) = µ(rx) ≤ ν(rx) = (rν)(x). ∴ rµ ⊆ rν. (iii) Let r, s ∈ R. Then (r(sµ))(x) = (sµ)(rx) = µ(srx)
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Souriar Sebastian, et al., J. Comp. & Math. Sci. Vol.2 (3), 573-580 (2011)
= µ(rsx) =((rs)µ)(x). ∴ r(sµ) = (rs)(µ). In particular, r(rµ) = r2µ. (iv) Let r ∈ R. Then ((rμ) ∩ (rν))(x)=(rμ)(x) ∧ (rν)(x)=μ (rx) ∧ ν(rx) = (μ ∩ ν)(rx) = (r(μ∩ν))(x) ∴ (rμ) ∩ (rν) = (r(μ ∩ ν)). In general, ∩ i
rμi = r( ∩ μi). ∎ i
4. VALUATION RINGS AND R-MULTIPLES OF FUZZY SUBMODULES OF THE QUOTIENT FIELD
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4.3. Proposition. An integral domain R is a valuation ring iff for any given fuzzy R- submodule µ of the quotient field K, the set of fuzzy submodules { rµ : r ∈ R } is linearly ordered. Proof. Let R be a valuation ring with quotient field K and µ be a fuzzy Rsubmodule of K. Let r1, r2 ∈ R. Since R is a valuation ring either (r1) ⊆ (r2) or (r2) ⊆ (r1). Suppose (r1) ⊆ (r2). Then r1 = rr2 for some r ∈ R. Now (r1µ)(x) = µ(r1x) = µ(rr2x) ≥ µ(r2x) = (r2µ)(x) ∀ x ∈ K. ∴ r2µ ⊆ r1µ. On the other hand, if (r2) ⊆ (r1), then r1µ ⊆ r2µ. ∴ { rµ : r ∈ R } is linearly ordered.
Proof. We have x ∈ (rµ)t ⟺ (rµ)(x) ≥ t ⟺ µ(rx) ≥ t ⟺ rx ∈ µt. Hence r(rµ)t ⊆ µt. On the other hand, if x ∈ µt, then then x/r ∈ (rµ)t. Hence x ∈ r(rµ)t. ∴ µt ⊆ r(rµ)t. Thus r(rµ)t = µt for all t ∈ [0, 1] ∎
Now to prove the converse, assume that given any fuzzy R- submodule µ of K, the set of fuzzy submodules { rµ : r ∈ R } is linearly ordered. To prove that R is a valuation ring, let r1, r2 ∈ R. By assumption, either r1µ ⊆ r2µ or r2µ ⊆ r1µ. If r1µ ⊆ r2µ, then by proposition 4.2, r2µt ⊆ r1µt ∀ t ∈ [0, 1]. If µ is chosen to be the characteristic function of R, then µ∗ = µ1 = R. It follows that r2R ⊆ r1R. i.e., (r2) ⊆ (r1). If r2µ ⊆ r1µ, then in a similar way we get (r1) ⊆ (r2). Hence R is a valuation ring ∎
4.2. Proposition. Let R be an integral domain and K its quotient field. If µ is any fuzzy R- submodule of K, then for r1, r2 ∈ R, r1µ ⊆ r2µ ⟹ r2µt ⊆ r1µt ∀ t ∈ [0, 1].
If R is an integral domain and K its quotient field, let χ W denote the characteristic function of a subset W of K and for t ∈ [0, 1), χ R(t) denote the fuzzy
4.1. Proposition. Let R be an integral domain and K its quotient field. If µ is any fuzzy R-submodule of K and r ∈ R, then r(rµ)t = µt for all t ∈ [0, 1].
Proof. By proposition 4.1, r1(r1µ)t = µt and r2(r2µ)t = µt. Therefore r2r1(r1µ)t = r2µt and r1r2(r2µ)t = r1µt. i.e., (r1r2)(r1µ)t = r2µt and (r1r2)(r2µ)t = r1µt. Now suppose r1µ ⊆ r2µ. Then (r1µ)t ⊆ (r2µ)t ∀ t ∈ [0, 1]. It follows that (r1r2)(r1µ)t ⊆ (r1r2)(r2µ)t ∀ t ∈ [0, 1]. Hence r2µt ⊆ r1µt ∀ t ∈ [0, 1] ∎
subset of K defined by χ R(t) (x) = 1 if x ∈ R and χ R(t) (x) = t if x ∈ K – R . For d ∈ K, and t ∈ [0, 1], we let dt denote the fuzzy subset of K defined by dt(x) = t if x = d and dt (x) = 0, otherwise. If µ and ν are fuzzy subsets of K, then the fuzzy subsets µ ∘ ν and µν of K are defined by
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(µ ∘ ν) (x) = ∨ { µ(y) ∧ ν(z) : x = yz, y, z ∈ K }∀ x∈ K. n
(µν)(x) = ∨ { ∧ (µ(yi) ∧ ν(zi)) : yi, zi ∈ K, 1
= χ R (x). ∴ a χ (a) = χ R
.
(1)
i=1
n
≤ i ≤ n, n ∈ N, x = Σ yi zi, } ∀ x ∈ K. i=1
4.4. Definition[8]. Let R be an integral domain K its quotient field. A fuzzy Rsubmodule μ of K is said to be a fractionary fuzzy ideal of R if there exists d ∈ R and t ∈ [0,1] such that d1∘ μ ⊆ χ (Rt) . 4.5. Theorem. Let R be an integral domain and K be its quotient field. Then the following conditions are equivalent. (i) R is a valuation ring (ii) Given any fuzzy R- submodule µ of K, the set of fuzzy submodules { rµ : r ∈ R} is linearly ordered. (iii) The { 0, 1 } valued fractionary fuzzy ideals of R are linearly ordered. Proof. The equivalence of conditions (i) and (ii) is proved in proposition 4.3. We now prove that conditions (ii) and (iii) are equivalent. Assume (ii). Let μ and ν be fractionary fuzzy ideals of R. Case I : μ = χ (a) and ν = χ (b) where χ (a) and χ (b) are the characteristic functions of the principal ideals (a) and (b) in K. We have (a χ (a) ) (x) = χ (a) (ax) = 1 if ax ∈ (a)
0 if ax ∉ (a)
= 1 if x ∈ R
0 if x ∉ R
Similarly, b χ (b) = χ R . Again, (b χ (a) )(x) =( χ (a) )(bx) = 1 if bx ∈ (a)
0 if bx ∉ (a)
But bx ∈ (a) ⟺ bx = ra, for some r ∈ R ⟺ x = ra/b ∈ (a)/b. ∴ b χ (a) = χ (a)/b (2) Now condition (ii) implies that { r χ (a) : r ∈ R } is linearly ordered. Therefore either a χ (a) ⊆ b χ (a) or b χ (a) ⊆ a χ (a) . Hence using (1) and (2), either
χR ⊆
χ (a)/b or χ (a)/b ⊆ χ R . Therefore either R ⊆ (a)/b or (a)/b ⊆ R which means either (b) ⊆ (a) or (a) ⊆ (b). Hence either χ (b) ⊆
χ (a) or χ (a) ⊆ χ (b) . i.e., μ ⊆ ν or ν ⊆ μ . Case II : Let μ = χ A and ν = χ B where A and B be integral ideals of R. Suppose χ A ⊈ χ B . Then A ⊈ B. Therefore there exists a ∈ A but a ∉ B. Take b ∈ B and b ≠ 0. By case I, χ (a) ⊆ χ (b) or χ (b) ⊆ χ (a) . If χ (a) ⊆ χ (b) , Then (a) ⊆ (b), hence a = rb ∈ B which is not possible. ∴ χ (b) ⊆
χ (a) . Hence (b) ⊆ (a). ∴ b = ra ∈ A. Since this is true for all b ∈ B, we get B ⊆ A. Hence χ B ⊆ χ A . Thus either χ A ⊆ χ B or
χ B ⊆ χ A i.e., μ ⊆ ν or ν ⊆ μ .
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Souriar Sebastian, et al., J. Comp. & Math. Sci. Vol.2 (3), 573-580 (2011)
Case III : Let μ = χ A and ν = χ B where A and B be fractionary ideals of R. Since A and B are fractionary ideals, we can take A = I1/d1 and B = I2/d2 where I1 and I2 are integral ideals of R and d1, d2 ∈ R. By case I, either χ (d1 ) ⊆ χ (d2 ) or χ (d2 ) ⊆
χ (d1 ) , hence (d1) ⊆ (d2) or (d2) ⊆ (d1). ∴ d1/d2 ∈ R or d2/d1 ∈ R. If d1/d2 ∈ R, then (d1/d2)I2 and I1 are integral ideals of R. Applying case II to these ideals we get either I1 ⊆ (d1/d2)I2 or (d1/d2)I2 ⊆ I1 . i.e., either I1/d1 ⊆ I2/d2 or I2/d2 ⊆ I1/d1. Hence χ A ⊆ χ B or χ B ⊆ χ A . Thus, in all cases, the {0, 1} valued fractionary fuzzy ideals of R are linearly ordered. This proves (iii) We now assume condition (iii). In order to prove (ii), let μ be any fuzzy R- submodule of K. Let r1, r2 ∈ R. By assumption, either χ (r1 ) ⊆ χ (r2 ) or χ (r2 ) ⊆
χ (r1 ) , hence (r1) ⊆ (r2) or (r2) ⊆ (r1). If (r1) ⊆ (r2) , then by the proof of proposition 4.3, r2μ ⊆ r1μ. Similarly if (r2) ⊆ (r1), then r1μ ⊆ r2μ . Therefore the set { rμ : r ∈ R } is linearly ordered. This proves (ii). Hence the theorem. ∎ 4.6. Proposition. If V is a valuation ring, and P is a prime ideal of V, then the localization Vp of V at P is also a valuation ring. Proof. Let K be the quotient field of V. Then K is the quotient field of VP also. Let μ be any fuzzy VP- submodule of K. Consider the set of submodules { xμ : x ∈ VP }. We prove that this set is linearly ordered. Note that μ is a fuzzy V-
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submodule of K also, for, if r ∈ V and z ∈ K, then μ(rz) = μ((r/1)z) ≥ μ(z), since r/1 ∈ VP. Since V is a valuation ring, by proposition 4.3, { rμ : r ∈ R } is linearly ordered. Let x = r1/s1, y = r2/s2 ∈ VP. Then r1s2, r2s1 ∈ V. Therefore either (r1s2)μ ⊆ (r2s1)μ or (r2s1)μ ⊆ (r1s2)μ. But then since 1/s1s2 ∈ VP, by proposition 3.2, we have (1/s1s2) (r1s2)μ ⊆(1/s1s2 )(r2s1)μ or (1/s1s2) (r2s1)μ ⊆ (1/s1s2 )(r1s2)μ. i.e., (r1/s1)μ ⊆ (r2/s2)μ or (r2/s2)μ ⊆ (r1/s1)μ. i.e., xμ ⊆ yμ or yμ ⊆ xμ. Therefore the set {xμ : x ∈ VP } is linearly ordered. By proposition 4.3, VP is therefore a valuation ring. ∎ 5. R- MULTIPLES OF FRACTIONARY FUZZY IDEALS 5.1. Definition[8]. Let R be an integral domain. Then a fractionary fuzzy ideal μ of R is said to be fuzzy invertible if there exist a fractionary fuzzy ideal μ' of R such that μμ' = χ (Rt) for some t ∈ [0,1]. The following theorem gives a necessary and sufficient condition for fuzzy invertability. 5.2. Theorem[8]. Let R be an integral domain and μ be a fractionary fuzzy ideal of R. If Sup { μ(x) : x ∈ K∖μ∗ } exists, then μ is fuzzy invertible if and only if μ∗ is an invertible fractionary ideal of R ∎ 5.3. Proposition. Let R be an integral domain and K be its quotient field. Let μ be a fuzzy R- submodule of K. Then, (i) μ is a fractionary fuzzy ideal of R ⇔ rμ is a fractionary fuzzy ideal of R for any r ∈ R.
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(ii) If Sup { μ(x) : x ∈ K∖μ∗ } exists and μ is a fuzzy invertible, then Sup { (rμ)(y) : y ∈ K∖μ∗ } exists and rμ is fuzzy invertible. Conversely if Sup { (rμ)(y) : y ∈ K∖(rμ)∗} exists and rμ is fuzzy invertible, then Sup { μ(x) : x ∈ K∖μ∗ } exists and μ is fuzzy invertible. Proof. (i) Suppose µ is a fractionary fuzzy ideal of R. Then there exist d ∈ R and t ∈ [0, 1] such that d1∘ μ ⊆ χ (Rt) . ∴ μ(x/d) ≤ t ∀ x ∈ K – R. Since (rμ)(x/rd) = μ(x/d), we get (rμ)(x/rd) ≤ t ∀ x ∈ K – R. Hence (rd)1 ∘ (rμ) ⊆ χ (Rt) . ∴ rμ is a fractionary fuzzy ideal. Conversely suppose rμ is a fractionary fuzzy ideal of R for r ∈ R. Then there exist c ∈ R and t ∈ [0, 1] such that c1∘ (rμ) ⊆ χ (Rt) . ∴ (rμ)(x/c) ≤ t ∀ x ∈ K – R. Hence μ(rx/c) ≤ t ∀ x ∈ K – R. But since μ is a fuzzy R- submodule, μ(rx/c) ≥ μ(x/c). ∴ μ(x/c) ≤ t ∀ x ∈ K – R. Hence c1 ( t) ∘ μ ⊆ χ R . ∴ μ is a fractionary fuzzy ideal of R. (ii) By proposition 4.1, we have r(rμ)t = μt for all t ∈ [0, 1]. In particular r(rμ)∗ = μ∗. ∴ (rμ)∗ = (1/r)μ∗. Assume that Sup { μ(x) : x ∈ K - μ∗ } exists. Let Sup { μ(x) : x ∈ K∖μ∗ } = t. If y ∈ K ∖(rμ)∗, then y ∉ (rμ)∗ ∴ (rμ)(y) = μ(ry) < 1, hence ry ∉ μ∗ or ry ∈ K∖μ∗. ∴ (rμ)(y) = μ(ry) ≤ t. Hence Sup { (rμ)(y) : y ∈ K∖(rμ)∗ } exists. Moreover if μ is fuzzy invertible, then by theorem 5.2, μ∗ is an invertible fractionary ideal of R. It follows that (rμ)∗ = (1/r)μ∗ is invertible. Applying theorem 5.2 again, we can conclude that rμ is fuzzy invertible. By similar steps we can prove that if
Sup { (rμ)(y) : y ∈ K∖(rμ)∗ } exists, then Sup { μ(x) : x ∈ K∖μ∗ } exists and if in addition, rμ is fuzzy invertible, then μ is also fuzzy invertible ∎ 5.4. Proposition. Let R be an integral domain and K its quotient field. Let µ be a fuzzy R- submodule of K. If μ ≠ 1K and μ∗ ≠ (0), then (r1) ⊊ (r2) ⟹ r2μ ⊊ r1μ. Proof. In course of proving proposition 4.3, we have proved that (r1) ⊆ (r2) ⟹ r2μ ⊆ r1μ. Let us assume that μ ≠ 1K and μ∗ ≠ (0), and that (r1) ⊊ (r2). Then r1 = r2b where b is a non unit of R. Choose 0 ≠ x ∈ μ∗ but x/b ∉ μ∗. This is possible since μ ≠ 1K and μ∗ ≠ 0. For, μ∗≠ 0 implies that there exists 0 ≠ x ∈ μ∗. If x/b ∈ μ∗ for all 0 ≠ x ∈ μ∗, then μ∗/b ⊆ μ∗. At the same time μ∗ ⊆ μ∗/b, since r/s ∈ μ∗⟹ μ(r/s) = 1 ⟹ μ(rb/s) ≥ μ(r/s) = 1 so that μ(rb/s) = 1 ⟹ r/s = rb/sb = (rb/s)/b ∈ μ∗/b. Either of them imply that μ∗/b = μ∗ or μ∗ = bμ∗. But then b becomes a unit and that is not possible. Now take y = x/(r2b). Then (r1μ)(y) = μ(r1y)= μ(r1x/r2b) = μ(x) = 1, since x ∈ μ∗. But (r2μ)(y) = μ(r2y) = μ(r2x/r2b) = μ(x/b) < 1, since x/b ∉ μ∗. Thus (r2μ)(y) < (r1μ)(y). Hence (r2μ) ⊊ (r1μ) ∎ 5.5. Remark. The converse of the above proposition is not generally true as is shown by the following example. 5.6.Example. Consider the ring of integers Z. Let ∞
J = ∪ (3)/2k = (3) ∪ (3)/2 ∪ (3)/22 ∪ ............... k=1
Then J is an Z- submodule of the quotient field Q. Also J is not a fractional ideal. Let µ
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be the characteristic function of J in Q. Then µ is a fuzzy submodule of Q and µ1 = µ∗ = J. We have 2J = J, hence 2µ = µ. At the same time 3J ⊊ J, hence µ ⊊ 3µ. Thus 2µ ⊊ 3µ. But (3) ⊈ (2) ∎ 5.7. Definition. Let R be an integral domain and K its quotient field. We say that a fuzzy R- submodule µ ≠ 1K and µ∗ ≠ (0) of K is minimal with respect to r ∈ R, if rµ ⊋ µ and that there exists no r1 ∈ R such that rµ ⊋ r1µ⊋ µ. In other words, rµ ⊇ r1µ ⊇ µ for r1 ∈ R must imply that either (r) = (r1) or (r1) = R. 5.8. Proposition. Let R be a Dedekind domain and K its quotient field. Let µ ≠ 1K and µ∗ ≠ (0) be a fractional fuzzy ideal of R. Then µ is minimal with respect to r ∈ R ⇔ (r) is maximal in the set of all proper principal ideals of R. Proof. If (r) is not maximal in the set of all proper principal ideals of R, then there exists r1 ∈ r such that (r) ⊊ (r1) ⊊ (1). But then by proposition 5.4, µ ⊊ r1µ ⊊ rµ. Hence µ is not minimal with respect to r. Conversely if µ is not minimal with respect to r, then there exists r1 ∈ R such that µ ⊊ r1µ ⊊ rµ. Hence by proposition 4.2, rµt ⊆ r1µt ⊆ µt ∀ t ∈ [0, 1]. Since µ is a fractionary fuzzy ideal of R, µt is a fractionary ideal of R for some t. Since fractionary ideals are invertible in a Dedekind domain, multiplying with its inverse, we get (r) ⊆ (r1) ⊆ (1). Now if (r) = (r1), then by proposition 4.3, r1µ = rµ and if (r1) = (1), then µ = µ1. Since µ ⊊ r1µ ⊊ rµ, this leads to a contradiction. Therefore (r) ⊊ (r1) ⊊ (1) so that (r) is not maximal ∎
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5.9. Remark. We know that (r) is maximal iff r is irreducible. Hence if R is a Dedekind domain and µ is a fractionary fuzzy ideal of R, then the set of all minimal elements of µ is equal to the set of all irreducible elements of R. REFERENCES 1. Bhambri S. K., Kumar R. and Kumar P., On fuzzy primary submodules, Bull. Cal. Math. Soc. 86, 445-452 (1994). 2. Bhambri S. K., Kumar R. and Kumar P., Fuzzy prime submodules and radical of a fuzzy submodule, Bull. Cal. Math. Soc. 87, 163-168 (1995). 3. Gopalakrishnan N. S., Commutative Algebra, Vikas Publishing House, New Delhi, (1984). 4. Golan J. S., Making modules fuzzy, FSS 32, 91- 94 (1989). 5. Kim H. and Park Y. S., Fuzzy Multiplicative Ideal Theory, J. Fuzzy Math.14, 245-254 (2006). 6. Liu W. J., Fuzzy Invariant subgroups and Fuzzy Ideals, Fuzzy sets and systems 8, 133- 139 (1982). 7. Liu W. J., Operations on Fuzzy Ideals, Fuzzy Sets and Systems 11, 31-41 (1983). 8. Malik D. S. and Mordeson J. N., Fractionary fuzzy ideals and fuzzy invertible fractionary fuzzy ideals, J. Fuzzy Math.5, 875-883 (1997). 9. Nagoita C. V. and Ralescu D. A., Appllications of Fuzzy sets to Systems Analysis, Wiley, New York,54-59 (1975). 10. Rosenfeld A., Fuzzy Groups, J. Math. Anal. & Appl. 35, 512-517 (1971). 11. Souriar Sebastian and S. Babusundar, On the chains of level subgroups of
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